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Full text of "A system of practical mathematics; being no.xvi. of a new series of school-books"

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Piniiiiiiii 



6000364648 



I'i-Jlf e. Vl 



A SYSTEM 



OF 



PRACTICAL MATHEMATICS, 



PJLIftT I. 



CONTAINING 



ALGEBRA AND GEOMETRY. 



BEING 



N^ XVI. 



OF 



A NEW SERIES OF SCHOOL-BOOKS, 



BT THE 



SCOTTISH SCHOOL-BOOK ASSOCTATION. 



9ttllfe|ely tot t|e 9iMoti&tUm, ts 
WILLIAM WHYTE AND CO., 

BOOftSJEELLJiBB TO THB QUKXN DOWAOXR, 
13, GEORGE STREET, EDINBURGH. 

HOUXSTOH AND STONEMAN, LONDON ; W. GRAPEL, AND G. H. AND 
J. SMITH, UTERPOOL; ABEL HETWOOD, MANCHESTER; 

J. ROBERTSON, DUBLIN. 



MDCCCXhY. 



••-I \ -'\K\ 




\ 






PREFACE. 



The following is the First Pari of a Treatise on 
Practical Mathematics, and comprehends that por- 
tion which does not require the use of Tables. In 
adding another to the many existing Treatises on 
this subject, it may be proper to state the objects 
that have been kept in view in- its composition. 
These have heen^ Jirst^ To exclude all useless matter, 
and thereby to keep the work within a small com- 
pass; secondly, To make it as entirely demonstra- 
tive as possible, without reference to any other work 
on Mathematics. IFor this purpose, as well as for 
its own intrinsic usefulness, a Treatise on Geometry 
is introduced, in which, by adopting a different order 
of the propositions from that used in Euclid's Ele- 
ments, and using symbols for certain expressions of 
frequent occurrence, an unprecedentedly large quan- 
tity of geometrical truths is presented, without in 
any instance detracting from the fulness of the de- 
monstrations, which are always given at length. 

The article on Algebra, it is hoped, will be found to 
be sufficiently extensive for most practical purposes; 
and the pupil that has thoroughly studied it, will 
find himself well prepared for entering on the study 
of larger works. In many instances exercises have 
been introduced of such a nature, as not only to 
iUnstrate the rules, but to assist in reducing cq^iAaaxl 




DEFINITIONS. 

Art. 1. Algbbba is a biancL of mathcmatica in nhich 
ealcnlations are performed by means of letters wkich de- 
note Dumbera or quantities, and signs which, indicate ope- 
lationB to be performed on them. 

2, The first Jetters of the alphabet, as a, b, e, &c., are 
lUed to denote known quantities, and the latter letters, as 
t,f, t, &c., to denote unknown ones. 
_ 3. The sign + (named phis), indicates that the quanti- 
lia between which it stands are to be added together ; thus 
3-f £ denotes the sum of the quantities a and b. 

4. The aign — (named minun), indicates that the num- 
ber or quantity placed after it is to be subtracted from that 
pited before it; thus a — h denotes the remainder left by 
fsloag the quantity b from a. 

5. Theaign x (^aB,Taed multiplied 177(0), indicates that the 
^UntitieB between which it stands are to be multiplied the 
one by the other; thus aXcdenotesthata is to be taken as 
rflen as there are units in c, or that c is to be taken as often 
u there are imita in a^ This symbol is however seldom 
)<%d, as a.e, or simply ac written as the letters of a word, 
iidicales the same thing. 

6. The sign -^ (named divided by), indicates that the 
'IMntJty before it is to be divided by that placed after it ; 
tliua u-^c denotes that a la to be divided by c. This sym- 
^1 is also seldom used, as division is more commonly de- 
iDicd by placing the dividend above a line as the numera- 
toi of a fraction, and the divisor below it as its denomina- 
tor; thus ~ is the same as a-i-i. 

7. The sign ^ (read equal, or i» equalto), indicates that 
i^ quantities before it are er[ual in value to those after it; 
thus 4x3+7=9x2+1, for each is equal to 19. 

8. The quantities before and after the eigti = are to- 
Ktilet called an equation; that portion which stands before 

~ ' 1 ^ being called the first side of the efYaatitm, a.iA 
* aafierit the second. 




9. The symbol J denotes that tlie nmnber oyer which 
it is placed is to have its square root extracted; thus 
Vl6 "'•^''^'^^ '^^ ^1'''"^^ rootoflfi, which is 4, and the 
t/^ denotes the square root of a, that is a number that, be- 
ing multiplied into itself, would produce a. 

10. In the same manner, the cube root of a number as 
a is denoted by ^a, the fourth root, by Jl/o, and so on. 

11. A number placed before a letter or combination of 
letters is called a coejicieiii; thus 3tt denotes three times a, 
and 3 is called the coefficient of a. The first letters of the 
alphabet are frequently called the coefficients of the latter 
letters; thus, in the expression 3cx, 3c is called the coeffi' 
cient of jr. 

12. When the same letter enters several times as a mul- 
tiplier into an expression, instead of repeating the letter it 
is only written once, and a figure written after it to indi- 
cate the number of times it enters as a multiplier; thus 
a", «', «■*, &c., denote respectively the second, third, and 
fourth powers of ti, and tte smaU figures, 2, 3, 4, &e., 
placed after the letters, are called the exponents or indices 
of the letters. 

13. Fractional exponents are also osed to indicate roots; 
thus, instead of ^j; x^ , is written, for ^ x^ 3.3, for ^ ^^ 
sc*, and so on to any extent; fractional exponents, where 
the numerator is not nnr', are also used; thus ,r', x^, &c,, 
the former of which denotes that x is to be raised to the 
second power, and then the third root of this power ex- 
tracted, and the latter denotes that j: is to be raised to the 
fifth power, and then the square root of this power extract- 
ed; or generally the numerator of the fractional exponent 
denotes a power to which the quantity is to be raised, and 
the denominator indicates the root of this power which is 
to be eitracted. 

14. When ^ = J, it is frequently written thus, a:h:i 

e : d, and read, a is to t as c is to d, and the four qnantitiea 
are said to constitute a prop'irlion or analog!/; the terms a 
and d are called exlreme>i, and and c j/ieans. 

15. The symbol -■. ia used instead of the words Cherefore 
ortOTtatvptrti^iy, which occur very ftequently in mathemalioal 
leasoning; and the symbol ■,■ instead of became. 

Ifi. Litte quantities are such as are expressed by means 
of tie game ietters, and llie same powers of these letters, 



ALGEBRA. 11 

and unlike quantities are expressions which contain dif- 
ferent letters or different powers of the same letters; thus 
3a^c^ and Ja^c^ are like quantities, whilst Sbx'yand Wxy* 
are unlike. 

17* A simple quantity consists of one term, as Acx; a 
compound quantity consists of two or more terms connected 
hj the signs -f* o' — > ^^s 16a^(;-|-a5 and ISc'o?' — icd 
are compound quantities. 

18. A vinculum^ bar , or parenthesis ( ), is used to 

collect seyeral quantities into one; thus a + ad or (a+x)d 
denotes that the sum of a and ^ is to he multiplied into d; 

also ^4ac — h^ or {4ac — P)^ indicates the square root of 
the remainder left by subtracting the square of b from four 
times the product of a multiplied into c. 

19. The reciprocal of a quantity is the quotient arising 

from dividing unity by that quantity; thus - is the recipro- 
cal of o^ and can also be written a"^ ; in the same man- 
ner the reciprocals of a^, x^, 2", are -g- or a"^,~5 or x" , 

~ or 2"^, where n may represent any number either whole 

or fractional, and is used as a general symbol for any expo- 
nent. 

20. Find the numerical values of the following expres- 
sions, when a=8, 6=4, c=3, fl?=2, c^l5,/^0. 

1. ac-\'(b — d)e — bed . . . = .30 

2. a(6c-f c)— «?(56— c) . . =102 

3. (a— l)(6--l)(c— l)(c;+c) . . = 210 

4. anJce-\-b — CtJce-^-b . . =: 35 

5. a\be-^e)'^f^2^d+b . . = 960 

6. a^(2b+e+4)^+bi(e+c^d)^ . =14 

In the following equations, the first and second sides will 
always give the same numerical value, if the same value be 
given to the letters on each side : verify this. 

'J.f^^x^ + xy+y^. 

8. (a-^aj)(a— a;)=a^ — x^. 

9. (a + b—c)(a^b+c) =a2— 6'-*— c'* + 2bc. 
10. x^^y*=(x^y){x^+x^y+xy^+y^). 



ALOKBBA. 



ADDITION 



21. Is commonlj dirided into three cases; — \st. When 
the quantities arc like, and have like signs; 2rf, When the 
<]uaiititie9 are like, but have unlike signs; 3d, When the 
quantities are not nil like, and have unlike signs. 

" ' Rui,E. — Add the coefficients together, and to 
anex the literal part. 



tiQa^'bcxi,' 



I tbe 

^^H 3a — oc 5V^^+? 

^^M a — 7«c (ir"+/)^ 

^^1 Sa — 9ac 19 (x'+y)^ 

^^M 8a — 4ViH? 

^^^ _5a — 3^ fl 3 (a'^+y')^ 

Sum, 26a — 29ac 35 (a:'+y')^ 

Ei. I. Find the sum of ax^+^aa:^+3ax^+4^(u:s^ 

|a«*. Ans. Q^ax^. 

2, FindtheBttmof2a»w:'+4awu:'+7^anij;'+4Jaj7t:c'. 
Ans. ij^amx'. 

3. Find th e sum crf^V«^+F+V^N^ + V^Hy* 

Ans. 44^x--\-i/*. 
Find the sum of 12(j:«_r)3^(3;»_:^3^7(,^4 j^i 

Ans. 4H(j'^— 3)i 
22. Cabe II. When the quantities are like, hut hare 
unlike signs. 

RiTLE. — Add the coefGoients of the plus quantities inti> 

one sum, and those of the miiiua quantities into another, 

thrar difference is the coefBcient of the sum, and is plus if 

tarn of the plus coefficients be the greater, and minus 



ALGEBRA. 13 

EXAMPLES. 

If^ 2d. Zd. 

7acx 4a^T+P («+ft)(a?'+y*)^ 

4acx —4aJT+p 10(a+6)(a?*+y»)^ 

-^ — 7aV r+^ 8(a+bXx^ + y«)i 

i. Find the sum oia,Jmx — ^afjmx-^ la^mx — xkijmx 
-^-iajmx —jjmx. Ans. 4^a^nix. 

2. Find the sum of ^mpx^y^ — Ampx^y^ — Qmpx^i/^ + 
^ah/px^y^ — Qmpx^y^ + "^mpx^y^ . Ans. — Smpx^y^ , 

3. Find the sum of 3(a;«— y«)^+.7(a;«— 3^«)^— 4(a;«— 

Ans. (x^—y^)i 

4. Find the sum of 5(6— c)a?^— 7(6— c)a?^+4(J--c)a;^ 

-4(6-.c)a;^+7(6--^>^--3(6— c)A Ans. 2(6— c>^* 

23. Case III. When both the signs and the quantities 
are unlike, or some like and others imlike. 

Rule. Find the sum of each parcel of like quantities 
bj the last rule, and write the seyeral results after each 
other, with their proper signs. 

Note. The above rules will be obvious from the following consi- 
derations: — The first rule is simply this, that any number of quanti- 
ties, as 4, and 5, and 7, of the same kind, will make 4+5+7, or 16 
quantities of the same kind. In the second, it must be remembered, 
that minus quantities are svbtractive, while plus ones are additive, and 
that to add and then subtract the same quantities, is the same as to 
perform no operation at all; therefore to add a greater quantity, and 
then subtract a less, is the same as to add their difference; and to 
subtract a greater, and then add a less, is the same as to subtract 
their difference, which is the rule. Again, it is evident that the third 
role just enables us to combine several accounts in the second into 
one sum. 



w 



GXAiiPLE. Here we begin ivlth tlie fin 

3x1/ +4a: — Sac terni,whithcontainBay. Wi 

— 4<ij;^-i-4ry — 3az therefore collect all the (iByV: 

"J ex — Soa" — 2xy into one sum; then find th( 

— 5ni +4(a; ^-aa^ sum of the {azYs, and so oi 

3ji; 4-5cj +3aj:' till wo hare collected all thi 

8i:i/—4az+ ]3fj;— 5aj-' terms. 

1. Find the snm of 4,t'3+8jy+4/, 3j-^+4j;y+y', 
2.1:' — 4j-(/+2/, andj:'^— 7-. Ans. 10i" + 8j-y+6j^^ 

2. Find the sum of 3a6+4u«— c,/, — 66v'+8crf+4a^ 
-|.6&e — 4ut+5e(i, and 46i.--|-2cii. 

Ans. a(iJ+4ae+12cii+4fia 

3. Find the sum of ia'+lOub+ll'', 3a-+4dl+i', 
2a'+Gul,+4b',ania'—6ab—7IA Ans. 10(i'+14a/i+56'. 

4. Find the sum of a—3ah^+2/,, Z<t—iah^—b, 
■7a+ 14ah^—3l<, and a~4ah^+'^. 

Ans. — 2<(+3A^+26. 



SUBTRACTION. 

24. RtTLE. Change the aigns of the quantitiea to he sub- 
tracted, or suppose them changed, and then collect the 
quantities, as in addition. 

Note. This rule will appear evident from the foUi 
rationa : — If we are roiiiiired ta Bubtracl 3a — b Cram So, this meaitt 
that we are to take away (3a— A) from oa; if liien wa take AWayS^ 
the remninder will ba io— Su ; but it ia evident we have hero l**— 
avay too muub by & ; we must then add b, and we will have for 
true remninder So — 3a+b=2a+b, where it is evident that the K_ 

Iaf the Bnbtrahend have besn changed, and tlien the quantitieB ■»!• 
kfited by addition. The sune proccBS of rcaaooing will apply to 
[ 



From 7a 4- 5nc^— Sic— 4ic', 

Take Ha^Sac^ + iljc —QUK 

Rem. 5a+8ac'— 7te+aic*. 

1, From 'i'' + 2iib+b'', take n* — 2<il,+b'. An«. 

S. From x'^+Sj^j/ -j- ary* + ifi, takea-'— Sa'^y + 3j?y'- 

Ans. feV+SJ'*. 
3. From x^—a^ take a^—3ax+x\ 
*' Am. *:i-^»4.2ojr— a*— «*, 

'4. Prom 3bx'—4cx+5A take Sjt'— 46j»+3m. 

Ana. 7^*— 7< 



ALGEBRA. 15 



MULTIPLICATION. 

25. In algebraic multiplication three things are to be 
attended to; first, the sign; second, the coefficient; and 
third, the literal part of the product. 

Rule. When the signs of the factors are like, the sign 
of the product is plus, and when the signs are unlike, the 
sign of the product is minus. The product of the coeffi- 
cients of the factors is the coefficient of the product. And 
the letters of both factors, written after each other as the 
letters of a word, form the literal part of the product, the 
letters being commonly arranged in the order of the alpha* 
bet 

KoiB 1. That like signs give plus, and unlike give minus, can be 
shown in the following manner : — When +6 is to be multiplied into 
+a, the meaning is, t£at +6 is to be added to itself as often as there 
are units in a, and therefore the product is 4-a6 ; if now (6 — 6), 
which is evidently =0, be multiplied by a, the product must be s=0; 
bat +&x+a has been shown to be +a6, and that the whole product 
may be =o, the other part must be — ab; therefore — 6x+a= — ab. 
Again, since the product of two factors is the same, whichever be 
considered as the multiplier, +ax — b= — ab, and if (a — a), which 
equals 0, be multiplied into b, the product must be ; but it has been 
shown, that +ax — 6= — ab, therefore that the product may be 0, 
-—ax -—6 must be equal to -{-ab. Hence like signs give plus, and unlike 
tigns give minus. 

Note 2. When the same letter appears in the multiplicand and 
multiplier, it will appear in the product with a power equal to the 
sum of its powers in each factor ; for a^xa^ denotes (aaa)x(aa)=:. 

(uma=a^=a^^'^'^\ that is, its power in the product is the sum of its 
powers in the multiplier and multiplicand. 

Multiplication naturally divides itself into three cases. 

26. Case I. When the multiplicand and multiplier are 
both simple quantities. 

Rule. Multiply the coefficients together for the coeffi- 
cient of the product, and the letters together for the literal 
part, and prefix the proper sign. 

Example. 5(ic x 3a6^= 15a'6^c. 

1. Multiply da^c, by 7ah''(?. Ans. 2laW. 

2. Multiply — 5(icd, by 3bcd, Ans. — I5al)c^d'^. 

3. Multiply — 4aV, by —7bcd. Ans. 2Sa^(^d. 

4. Multiply 7 OCX, by — Socy. Ans. — 2la^c^xj/. 

27. Case II. "When the multiplicand is a compound 
and the multiplier a simple quantity. 

Rule. Multiply each term of the multiplicand by the 
multiplier, and write the several products after e?Le\i o\5cL'ei, 
mib their proper signs. 



16 AU»BItA. 

ExAMPLB. Multiply 3ac+2bd+5c'', by 4,ib. 
Multiplicand, 3ae+2l>d+Sa''. 
Multiplier, 4alt. 

Product, 12d'bc+8ub''d+m^. 

1. Multiply 7a''4.4o6*^-6^ by Sac. 

Ana. 2]d^c+12a^b\-i.3ab*e. 

2. Multiply 3a=+4a=6—7<»c, by 5a'6, 

Ans. \5a^b+20a*b^~-3Wbc 

3. Multiply iafl+§a'6—|ac«, by 4ai.". 

Ans. 2a-b-c+^a^b^—3d'b^e'. 

4. Multiply fa'+lafi+SJ", by 3«6. 

Ans. 2(i^i+4a'6*+|ai'. 

5. Multiply 7a'c' — 56c^ — 4ni, by faic. 

Ans. 5ia36c3— 3ja6V— 3a'6V 
28. Case III. When both multiplicand and multipLiei 
arc compound quantities. 

RuLK. Multiply al! the terms of the multiplicand by 
each term of the multiplier, and collect tbe several pro- 
ducts into one sum by addition. 
Example. Multiply 3(t'— fti6+36', by a~b. 
Multiplicand, 3a' — 6iib-\-3b''. 

Multiplier, a — 6, 

Product by a, 3a'—6d'b+3ab\ 
Product by b. —3a %+6 <ib^— W. 

Total product, 3a»— 9a^6+9a6'— 3&». 

1. Multiply o+ 6, by a+6. Ana. a^ + Sab + S*. 

2. Multiply a—b, by o— 6. Ana. a''—2ab+b* 

3. Multiply a+i, by 0—6. Ans. a-— 6' 

4. Multiply a*+a6+6*. fcy «— *■ Ana. a'— fc>. 

5. Multiply a' — ax+x^, hy a+r. Ans. e^^t^. 

6. Multiply 7ab + 3ae+4d, by 3a— 2c. 

I Ans. 2]o''6+9o=c+]2rtrf— I4a6e— 6ac^— 8«/. 

L 7. Multiply 5fl+3r+3/,ty3a + 2r. 

^ Ans. i5a''+l9ar+3ai/+6x*+23y. 

8. Multiply «+iy— 2, by |:ot.3-/. 
« ., ,., Ans. :ij.=+3ir^-4^+liv'- _,. 

9. Multiply js+3^j/+xi/^+,iA by K— J/. Ans. ar* — y*. 

10. Multiply^*— a<i+a;"— / + !, bya" + s-— 1. 
Ans. i«— 3:4+xS— j:'+2jr — 1. 

. Multiply ax-\-bx^ + ax», hy l + j-+j;* + j/', 

g4-cr». 

PB. Tu KonBM I. The square of the mun of tn'o quanti- 

^ eqitat to tbe sum uf the B^uuies o{ «ttcli of the qoan- 



ALOEBBA. 17 

tities, together with twice their product. (See Art 28^ 
Example 1.) 

30. Thbobbm II. The square of the difference of two 
quantities is less than the sum of their squares hj twice 
^eir product. (See Art. 28, Example 2.) 

31. Theobem III. The product of the sum and diffe- 
rence of two quantities is equal to the difference of their 
squares. (See Art 28, Example 3.) 

Note. The above theorems are veiy important, and should be 
committed to memory. 

12. Write by Theorem I. the squares of the following 
quantities, and verify the results by multiplication, (o;-}-^) , 




13. Write by Theorem II. the squares of the foUowinj^ 
quantities, and Terify the results by multiplication, (a — d?) , 
(2a— ^)^ (3a;— 2y)^ (a^— 60^ {aG—yy, (3a*— 4c)^ 

Ans. {a^—2ax^x% (Aa^^Aah+h'')X^x'—\2xy+Ay*), 
(a4_2a262+6*), (a«c*— 2a<jy+y«), (9a*— 24a*c+16c0. 

14. Write by Theorem III. the product of the follow- 
ing quantities, and yerify the results by multiplication, 
(a+a;)(a— a;), (2a+x){^a—x\ (3a:+2y)(3a:— 2y), (a*— 
h^W + h% (4a2— 9/)X(4a^+9c2). 

Ans. {a^ — x% (4a2 — a;*), {9x'' — Ay% (a* — b^), 
(16a*— 81c*). 

15. Find the continued product of (a — x) (a-^-x) (a^+o?*) 
(a*+a?*). Ans. a® — a^. 

16. Find the continued product of (2a + a;) (2a — x) (4a* 
+a:*) (16a*+a?*). ' Ans. 256a8— a;8. 

17. Find the continued product of (a?* +xy+y^) {x — y) 
{a^+y^). Ans. x^ — /. 

18. Find the continued product of (a^ — x^y+xy'^ — y^) 
ix+y) (x^+t^). Ans. x^^ — x^y^+x^y^ — y^^. 



DIVISION. 

32. Division being the converse of multiplication, natu- 
rally gives the following Bides, 

1st, If the signs of the dividend and divisor are like, the 
sign of the quotient is -{- ; and if unlike, the sign of the 
quotient will be — . 

2d, Divide the coefficient of the dividend by that of the 
divisor, for the coefficient of the quotient. 

3d, 8i9ce anf quantity divided by itself giYea 1 fox cjvxo- 



18 

tient, and a quantity multiplied by 1 gives that same quan- 
tity for product, it follows, tliat the letters which are com- 
mou to the di-risor and dividend, with the same exponent 
in each, will not appear in the quotient. 

4th, The letters which aie in the dividend, and not ii 
the diyiBor, will appear as multipliers in the quotient 
while letters which are in the divisor, and not in the divi- 
dend, will appear as denominatora of a fraction u 
quotient, 

5th, "WTjcu the same letter is in both dividend and di^i-J 
sor, with different exponents, it will appear in the quotient 
with an exponent equal to the difference of its exponents, 
and in the denominator of a fraction, when the exponent of 
the divisor is the greater; thus, 

a-^a =:a >or—,a ~a =a 

33. Cask I. When the dividend and divisor are both 
simple quantities. 

Uui.K. Place the dividend as the numerator of a frac- 
tion, and the divisor as its denominator, then divide by the 
I above general rules. 

I ExAJiFLE. Divide lia-l/c, by 

I Proof. 2a-hox7l'-=lia^i'c. 

I 1. Divide Sa'x^, bj — 2oa''. Ans, — 4ax. 

I 2. Divide — ZJp^nu-*, by Opvix. Ans, —3ja^. 

^^^ 3. Divide —84j;yi^ by— 12^3/j. Ana. 7^^ 

^^^^ 4. Divide 120aV/, by 30<i.e^^. Ans. 

^^^f 5. Divide —iOx^^/h, bysJff'^z^. Ans. — 5«^ 

I 6. Divide 36^Vj. by 20j-«//Jii Ans. IJj^M^^* 

34. Casb n. When the dividend is a compound, and the 
divisor a simple quantity. 

EiTLB. Divide each term of the dividend by the divisot, 
as in the last case, and the sum of the separate quotients 
will be the answer. 

Example. Divide 3a'b+4u%-:, by 4ab. 



Sa'b+ia'bc 



4a& 



Jfiride J^a'^'f— 24a6*L-3— 10<i^6=c^,hy 3aie. 
;^2a^<r— SiaiV— loA^ _ ^^^^ _^|,t*— '^3^ , 



ALGEBRA. 19 

1. Divide 3a4— 6a3+4a«— 8a, by 2a«. 

3 4 

Ans. -a* — 3a-f2 . 

2. Divide 12a3—36a26+36a62— 12^.3, by 4a6. 

Ans. -7 9a -4- 96 . 

3. Divide ^Sa^h^X'-^6a^l»jc^+14al?x*, by 2ahx. 

Ans. 2ia3b—lSabx+7b^Ji^. 

4. Divide 4()p2^y2—3ap*a:V—60i?2:i:»^, by —10j5^^ 

Ans. — ixif+3j3»3i^if+6a^i/^. 

5. Divide 7a36^c^— 12a*63o-.14a62c*, by 4aMA 

Ans. -j^ aM— SftM— S^aM. 

6. Divide 4aV— 8a2ca:+4a2^, by 4a2. 

Ans. (^—.2cjc+a^. 

35. Oasb III. When both dividend and divisor are com- 
pound quantities. 

RuLB. Place the quantities as in division in arithmetic, 
arranging both dividend and divisor according to the 
powers of the same letter. Divide the first term of the di- 
vidend by the first term of the divisor, and put the result 
vfith its proper sign for the first term of the quotient. 
Multiply the terms of the divisor by this quotient, and sub- 
tract the product from the dividend; to the remainder 
bring down as many terms of the dividend as may be ne- 
cessary, then divide as before, and so on till the work is 
finished. 

bxamples. 

Divisor. Dividend. Quotient 

a+x)a^ + Sa^x+3ax^+x^(a^+2ax+x^ 
a^+a^x 

2a^^x+3ax^ 
2a^x+2ax^ 



ax^^x^ 
ax^+x^ 

Divisor. Dividend. Quotient 

a -'— 2ax + a;« )a4— 4a5ar+6a V— 4aa?5 + a;4 (Gt2__2aic + a;^ 

a^—2a^x+a^x^ 

— 2a^a;+ oa^x^ — Acix^ 

a'^x^ — 2ax'^ ^x^ 




Here it ia evident the -quotient ■would 
but go on to Infinity, the power of .r incn 
each step. 

This quotient givea several interesting Beries, by subsl 
tuling fractional values for ir ; for csample, if we make 
Buccessirely equal to ^, ^, ^, ^, |, we niU have the folloi 
ing results : — 



I 



ri=srri=2=l + i+i+J+TV- &c. to infinity. 

, - ._, =j^j=4=l + J+A+H+A'. + &o->oi«l 

t) /^(i— 2nd Igiyes |— ^+^+4TTV\\.-V&ft.toial 



^^ ALGEBBA. 21 

1. Ditide a* + Sa^x+iOa^x" + lOa'x^ +5ax* +x> by 
fl'+aar+o;''. Ans. a^ +3a''x+3ax' + x^. 

2. Dmde a^— 5a* +\Oa'—lOa'+ 5a— I by a'— 3a« 
+3a— 1. Ana. «"— 2«+ 1. 

3. Dmdear* — a* hy x'-\-x^a-^xa' + a'. Ana. x — a. 

4. Divide j:^+j)* by x*— j:'y+arV"— ^^i'^+i'''- 

Ans. x+i/. 
G. Dmde 9j;6_46x=+95*« + 150j! by ^2—4^— 5. 

Ans. 9.e*— 10.r' +oj-»— 30x. 

6. Diride 25a;9— «»— 2j;'— &f" by 5:c=— 4j:". 

Ans. Sx^+'ir'+Sx+Q. 

7. Dindejc*— ^x'+iT'+la;— 2 by |j^— 2. 

Ana. Jj:' — Jar^ + l. 
37< Veri^ by diTision the truth of the follofring ex- 

preuJDiu : — 






»*+fl 



=x*—x'a+x^a''—xa'' + a*——^. 

The eight expreaaions given above are particular illua- 
tatioiiB of the four follovruig TIteorema: — 

38. Thborbm I. The difference of the same powers of 
1*0 quantities is alwaja divisible by the difference of the 
quantitiea themselvea, whether the Mponents be even or 
odd. See Ist and 3d. 

39. Theorem U. The difference of the same powera 
irf two quantities is divisible by the sura of the quantities 
"ithoat a remiuader, if the exponent be even, but not if 
Ihe exponent be odd. See 2d and 4th. 

40. Tbrorku HE. The Bum ofthesamcpovfet^oH^o 
quantities is nerer dmsihh by the difference of tW <\aB.u< 




^ 



titles without remainiler, whether the exponents be odd or 
even. See 5th and 7th. 

41. Theorhm IV. The sura of the same powers of two. 
quantities is divisible without remainder, by the sum ' ' 
quantities, if the exponent be odd, but not if it be 



r ALGEBRAIC FRACTIONS. 

42, The Rules for the management of Algebraic Frac- 
tions are the same with those in common aritlimetic. 

Case I. To reduce a mixed quantity to the fonn of » 
fraction. 

Rule. Multiply the integral part by the denominator 
of the fraction, and to the product annex the numeratoi 
with the same sign as the fraction; under this sum place 
the former denominator, and the result is the fraction re- 
quired. 

Example, ReduceSa + j; to the form of a fraction, 

2nxi+j inb+x 
I ~ 6 ■ 

Here 2(/, the integral part, is multiplied hy i, and .v, the 
numerator, is added to the product, the sum of which forma 
the numerator; under which we write the former denomi- 
nator b, which gives -^ — -j the fraction required. 

43. If the numerator be a compound quantity, and ibe 
fraction be preceded by the sign minm, the signs of ( " '" 
terms must be changed by the rule for subtraction, (Art. 

24); thus, (i+.r — ~ — = 

I. Reduce3a-{ — to the form of a fraction. 3a»+ja 

" ^^ Ana, 

, Reduce « — ''+"j' to the form of a fraction. 

d — 

, Reduce I5a*+j: — to the form of a fraction. 

Ans. _-7 



ALGEBRA. 23 

5. Reduce x—a to the form of a fraction. 

Ans. . 

X 

6. Reduce a-}- a:- to the form of a fraction. 

a — X 

44. Case II. To reduce improper fractions to whole or 
mixed quantities. 

Rule. Divide the numerator hj the denominator, as 
far as an integral quotient can be obtained ; then write the 
remainder for the numerator, and the former denominator 
for tlie denominator of the fraction. 

Example. Reduce -. . 

4a 

=3a+l_j-. 

Here 12a* divides by 4a, and gives an integral quotient, 
so also does 4a divided by 4a ; but 3c will not give an in- 
tegral quotient when divided by 4a, therefore it is written 
as a fraction at the end with its proper sign. 

1. Reduce to a mixed quantity. Ans. 3j: . 

2. Reduce to a mixed quantity. Ans. 4x+ — . 

^ ^ . 10a«6+2a8— 36« . . , 

3. Reduce ^ to a mixed quantity. gra 

Ans. 56-4-1 — r-g. 

. ^ . 12d*— 6a3+9a» ^ • . , r 

4. Reduce ^ to an mtegral form. 

. ^ , *»+3A+3«»+a» ' . , Ans. 4a«-2a+3. 

5. Reduce — ^— ^ — -rri *o an mtegral form. 

Ans. a; 4- a. 

45. Lemma. To find the greatest common measure of 
two algebraic quantities ; that is, the greatest quantity that 
will divide them both without leaving a remainder. 

Rule. Arrange the quantities according to the powers 
of the same letter, as in division^ then divide that which 
contains the highest power of that letter by the other> and 
the last divisor by the remainder continually, till there be 
BO remainder, the last divisor used is the greatest common 
measure. 

Note 1. If all the terms of one of the quantities be divisible by 
any simple quantity which does not divide every term of the oMher, 
nnce it cannot form m part of the common measure, it may \>q «\x\lc^i^ 



r 



94 

out of all the tfrmH ly dividing by that aimpte qiULQtity before (ha 
ganeral division ia perfomted. 

Note 2, If tho coefficient of the leading term of the dividend bs 
not divisible by the coefficient of the leaJiag term of the divisor, 
every tBrm of the dividend may be multiplied liy Buuli a nnmbOT 
will make its coefficient diviaible by lliat of the iliviBor without i 
mainder; by this meoiiB fmctions are avoided. 

Note 3. Place the i^uantitics in two coIudhib, bo thai the a. 
which ia to be the divisor may aland in the left hand column, k 
the other in the right; then divide the right hand colnma by the le 
and place the quolieut ailer the right hand column ; then divide t 
left band column by the Tenmlnder, (reduced, if necessary, bf 
Note 1 ), and place Uie quotient before the left hand column ; pro- 
ceed in this manner till there be no remainder. 

Example. Required the greatest 
the qoantities x" +2ax-\-a^, and a;'— 



;c+a 






r+a 



x'+2ax'+ t 



by dividing 

by — 2««, 



x+a 

46. The process of finding the greatest commoD measaro 
can often be much more elegantly performed, by reducing 
the quantities into factors by the theorems, at the end m 
multiplication and division ; then the product of all the 
factors that are common to both quantities is the greatest 
common measure sought. The above example wrought ii 
this way tvonld be as under. 

Thence x+a being the only factor common to both, ii 
the greatest common measure. 

1. Find the greatest commou measure of a' — x*, and' 
o>— 2a«Ji+oj;«— 2a:=. Ans. a»+«*. 

2. Find the greatest common measure of x^+i/^, and 
3!*+x^y — xy' — >/^. Abb. x+y. 

3. Fiad the greatest common measure of 4Rr^ + IBj- — 15, 
and 24j;^— 22^^ + 173^5. Ans. 12i— S. 

4. Find the greatest common measure of x" — a% and 
iI!*-ftr'o+6j;'a''+4a-n=+n*. Ans. j; + . 

47. Case HI. To reduce fractions to their lowest terms. 
Edle. Divide bofh numeralor and denominator by 

their greatest common meosuTe, and the quotients will be 
the fraction in its lowest terms. 



Example. Reduce - 



-3i'a+3lo* 



-1 to its lowest terms. 



ALGEBBA. 25 



Ans. 



Here it will be found, that x — a is the greatest common 
measure ; therefore we hare the following : — 

jr3— ai;*a+3«o"— aS-S-Cj:— a) ~ x'^2ax+t^' 

1 . Reduce ^r to its lowest terms. Ans. -^-^ . 

oox—cx 3^ — f 

2. Reduce -52^2^^:es to its lowest terms. Ans. ^i?. 

3. Reduce — -z — . » to its lowest terms. 

"^-^^ Ans. Jtt?^. 

4. Reduce -j — — -=-= — . to its lowest terms. * m"+ii" 

5. Reduce — » - ^>, - — to its lowest terms. 

Ans -2*^^ 

4x"+6ar+9a^ 

48. Case IY. To reduce fractions with different deno- 
minators, to equiyalent ones having a conmion denomi- 
nator. 

Rule 1. Multiply both numerator and denominator of 
each fraction by the product of the denominators of all the 
others ; the resulting fractions will haye a conmion deno- 
minator. 

Rule 2. Multiply the terms of each fraction by the 
quotient arising from dividing the least common multiple 
of all the denominators by its denominator ; the resulting 
fractions will have the least common denominator. 

49. Note. The least common multiple of several quantities is the 
least quantity that is divisible by each of them, without leaving a 
remainder. It may be found by resolving each of the quantities 
into its simplest factors, and then multiplying together all the sepa- 
rate factors that appear in the whole, using the highest power of 
each that appears in any one quantity. For example, let us find the 
leastcommonmiiUtipleof (or' — a^W(ar+a:a+a^)(jr — a),(^ — 2ar+a*), 
ss.{x — a)*, 3^ — €?=z{x-\-a){x — a), it will be the product of (a:^+a:a+a") 
(f — a)'(j:+a), because these are all the different factors which ap* 
pear in the whole, and {x — a) is used in the second power, because 
it appears in that power in one of them. « n 

Sample. Reduce to a common denominator i— , ztt* 

4a iJtX 



A ^ 




^3^- 




By Rule 1, 


Zx \2x 3jf 
Aa ^ \2x ^ Zx 




2a ^ 4a ^^ 3x 
12x ^ 4a ^ 3x 



108:e» 

Uiaa^ 

24(^x 



w 



B7 Rule S. The least common multiple of all the de- 
nominators is Hirjj; therefore the fractions redaced to 
their least common denominator aaa 

3x 3j _ Sj* 

<■ ^ 3* ~ 13ot' 

2a a _ 2fl" 

fi ^ a ~ ISol' 

a» 4a 4a' 

^ X 7- = 1^- 

, . , 3* 3a , Se 

denoremator ^, — , and — . 

a'3x' X 

. 9j« 2b* lii 

, . , 0-* a+i , . 

denominator —-7, — -.. and - 



d ' a+b' a—i' 



3. Bedace ton 

4. Reduce to their least common denominator 

, 4(Kr . 7a'— To J 3aj+3i* 

fmd-— -_ Ans. --3--j-> -,— ^. 



_ 7« 3« 



&. Reduce to their least connuon denominator - 
6. Reduce to a common denominator 






i ADDITION OF ALGEBRAIC FRACTIONS. 

I £0. RuLK. Reduce (he fractions to a common denomiaa- 

I tor; then add their numerators, and under 

I their common denominator; the fraction so formed will b* 

I tlie fraction required. 



■~» 



J^bcAM pi«E. Add together the fbllcMiuifir fiaciioM e ^. , 
-V and t;^^ Here the least common denominator of 
9U the fractiQns is a«— $« ; henqe the fr^tionp If^qme 

WM?' ^SIS*^ ^ i^^ ' ^ - "^"^ 'V™ ^ tfiewfore 

1. Add iogether ^ , 7 , and i-. Ans. ^^^^-^. 

2. Add together ---^, and ^--^. Ans. -^— rj • 

3. Add together -,j^, ;^^, and ^. 

Am. ' fl*__^ • 

i Add together ^^ and ^-?-- Anfl. j^__\q ' 

5. Add together ^, g, j, and ^. Ans. ^, or Ji«» 



SUBTEACTION OP ALGEBRAIC FRACTIONS. 

51. Rule. Reduce the fractions to a common denomi- 
nator, then take the difference of the numerators, under 
which write the common denominator, and the result will 
be the fraction required. 

ExAMPiiB. From -, take -r-. Here the fractions re- 
duced to a common denominator are rr-7 and 7iri.> there- 
fore their difference is ,^ . . 

1. From -jr-, take -— . Ans. x, 

2. From -7- take — . Ans. ^q- 

I 1 2c 

3. From — take — r- -A:ns. g 4 . 

II * 

4. From , — take ~. — -». Ans. -: . «■. 

1 — a- 1 — x" 1 — X 

. ^ jn+n . , m — » ^ Aimu 

5. From take —rr- Alia. -^^ — ^*« 






LTIPLICATION OF ALGEBEAIC FRACTIONS. 
RuLB. Multiply the numerators together for the 
numerator of the product, and the denominalots together 
for the dtnominator of the product. 

ExAuPLE. Multiply -^ by -— -. Here we hare 3a x 6 
=3a6 for the numerator, and (a-{-h) X (_a — 6)=a' — 6' foi' 
the denominator j hence the product required is — ,- r,. 

Note. Cancel faotore whioli Eppcur iu bqth termB. 

1. Multiply -£ by 3^. 






fi. Multiply ^^ by ^^. 

3. Multiply^ by ^. 

4. Multiply ^:^ by ^j. 

a b , 

6. Multiply ^, —^< and 
6. Multiply |, -, 2- e' °°' 



2? 
19oJ 

ns, ^,_^. 

Ans. — ;. 
2a+6 
■'^■{^/- 
ah 

An^. 



DITISION OP ALGEBRAIC FRACTIONS. 

53, Rule. Multiply the dividend bj the divisor 
verted, and the product ^ill be the quotient required. 

Note. The reason of Uie above rule will appear from the follov- 
jng ejiample: — Lot it be proposed to divide -j by j- : for -^ write i^ 

then it will be V, wbere llio valae will not be altered by iaalti{i1p[ig 
omnefator and denominator by d, which gives -^=nx- ; bat n i 
eijna] to ?-, therefore nx -=-^ x - , which 19 tlie rule. 

1. Divide j^ by ^. Ans. ^ 

Divide ~ by^. Ani-J 






lDmd.i=ikj^. i,..i!f=i. 



5- DiTide - 



Am- — -. 



6. Dmde -^- by y. Ans. — 

I WTiat is the sum and difference of "-^ and "— ? 

Ans. Sum, a; difference, 
i!. What is the aam and difference of - g-' and -"""^I 
Ana. Sum, a^+it'; and difference, 2(Ke. 
9- What is the sum and difference of — and ~» 

J+M, i— » 

Ans. Sum, J^ ■ difference, -^^zx- 
111. ffliat is the product and quotient of — and ^^? 

Ans. Product, ; quotient, . 

II. Iteduce to their simplest forms the following ^fc 

prmonB:— 1st. — j±^ 2d, 2. |"^+^; 3d, -^-^ 
Ads. 1st, =?.; 2d, =:x; 3d, ='-^; and 4th, : 



INVOLUTION AND EVOLUTION. 
54. Involution is the raising of powera, and Evoi 
noN the extraction of roots. 

loTolution is performed by the continued maiti plication 
cf the quantity into itself, till the number of factors amount 
Id the number of units in the index of the power to which 
He quantity is to be iuTolved; thus, the square of 
'iXo=a'; the Sth power of x is xxxxx-=.3:,^ ; ^iie 
power of I2a) is 2x2x2x2x2 X aaaaa=32a5 . 



H 





I 


3 


16 


25 36 


7 






27 




125' 216 


343 






HI 


256 


625 I2fl6 


2101 




3'2 


•2i3 


1024 


3125 1 7776 


16807 



Ais' 7» 
4096 1 6561 



Iloots, . 
Squorea, . 
CnbeS, ■ 
4th pDwora, 
Stfi powers, 

55. In raising simple algebraic quantities to any power, 
obsetrc the following rules: — Ist, Raise the nnmerieaJ oo- 
eflicient to the given power for the coeffieient, 2d, Multi- 
ply tiie exponents of each of the letters bjthe power to whiolv' 
the quantity is to be raised. 3'^, If the sign of the givdii' 
quantity be plus, the signs of all the powers will be plmj 
but if the sign of the given quantity he minus, all the even 
powers will be plus, and the odd powers will be 

These rules are exemplified in the following Table, 'which 
the puinl should fill up for himself; 



BOOTS AND 

Boots . . . 

Squu'es . . 
Cubes . . . 
4 th powers 
£th powers 



MPLR ALGEBRAIC QUANTITIES. 



. 




+'• 


+ 4^ 


2o 
"35 
4fl» 


1 

4^' 




a' 


— «^ 




8a' 

27r> 


ii^ 


.. 


It' 


+ X' 


+ Jb? 


816* 


ii 


2«'„ 




he four! 


~32i» 

h power. 


2435' 

Ans. 


1 
"33^ 



Ans. 343a>»i»i». 

Ans. „^a*6*c'V: 



9. Eaise Ja^px^ to the third power. 

3. Kaise ^abc^x to the fifth power. 

4. Raise ^a'b*e to the eighth power. Ans. oTgO*^'^ 

2401o'fcV 



i-to (he fourth 






Ans. - 



S. Raise 7 

56. When it is required to raise a compound qnantl^ 
tb any power, it can always be effeeted by raulliplying the 
quantity successively by itself; but (here is another meAtit 
of finding the powers of a binomial, commonly called thV 
binomial theorem, first given in all its generality by Sr' 
Isaac Newton, by which the power required can be writtM 
at CBce, m'thout going through aU tbc mtenoedUte 



This ^eorem will now be gircn, and the atndeDt can xerify 
its results by actual multiplication in the mean lime, as Iha 
^eral proof would oat be undeiBtood *.t thi» Btage of the 
piqiU'a advanoement. 

SINOIIIAL TDEDREM. 

07- Let it be required to raise {it-\-x) to tbo nth power, 
ithere n may be any number, and a and x any quantities, 
eithK aimple or cnnpound. The first term nil! be a'. And 
die Kcond will be obuined frcim it, by multiplying by 

— , and will therefore be na"~^x. The third will be ob- 

labed ftom the second, by multiplying hy -^ — , and in 

the nme manner, the 4tb, 5th, 6tb, 7th, will be obtained 
by multiplying the 3d, 4tb, 5th, and 6di sueceuJTely by 

-r -, -T- -, -T- -. and — -, and go on, so that 

the teiulting series will he (a-t-^)'— 

*+ 1-3-3 ^ ^ 1-2 



!lH )(.-3)(-») (.-4) 



it!^^3?-\-Scr:., where, by substituting 



cr of 1 



isUeod of n, 2, 3, 4 5, sucaessively it becomes 
(fl+ j;) 3 =a» + 3«'« 4-3ar' + a;' . 

From the above, it is evident, (1st), That the power 
n in the first term is the same as the power to which the 
hinomial is raised. (2rf), That the powers of a decrease 
hy unity in each successive term, whereas those of x in- 
ctBise by unity, till the last term, where it is equal to the 
power to whi<Ji the hinomial is raised. (Stf), That the 
number of terms in the expansion is always one more than 
the power of the binomiah (4'/i), That ihe co-eiBcieat of 
tie second term h always equal to the power to which the 
liinomial is raised, and that the successive co-efficieats cai; 
be ohtained by multiplying the co-efficient of the previous 
leriB into the power of a in that term, and dividing by the 
nnmhcr of terms from the beginning of the cxpajiaion; 
thus 10, the co-effieient of the third term in the expansion 
pf {o-4-«)'' cftn be obtained br mnltiplying 5, the eo-effioi- 
Mit of the second term, into 4, the power of a mlWt tetro, 
§tAiinSagbyS, the Bumb^ ot' one teimfioi%t]b.c\»^'&^ 



ning of the cspansion. In the same manner, may all the 
other CO- efficients he oh tain ed, 

K the Bign of the second term of the biaoiiiial were 
minus, since the odd powers of a minus quantity are minus, 
and the even powers are pins, the terms which contain an 
odd power of the second term will he minus, (Art. 55), and 
those which contain even powers of that term will he pins. 
Thus, 

If it were recjuired to cspand (a+b — e)", it might be 
effected by first considering (6 — c) as one quantity, anA 
then raising it to the power denoted by its index in eacli 
term, and separating into single terms. Thus, 



3a»(i— c)= 



+3a^b—3a^c 

63_3i'c+36c=— c 



36'c+36c'— 6aJc. 
J . What is the 4tL power of (u— 6) ? 

Ans. a*— 4o%+6a=6'— 4n6'+e*i 

2. What is the 3d power of (4a— 2j-) ? 

Ans. 64a'— g6a''ic+48aa:s— &(». 

3. "What is the 9th power of ^/x+v t 

Ans.3:a+3^V+3^'+y'- I 

4. What is the Sth power of (2« — x) ? 

Ans. 32^^— BOtt''j;+80«^^_40aV'+10tw:* 

5. What is the 3d power of {a—{x+!/)] f 
Ans. a' — a:' — i/^ — Sa^x—da'^Jz+Sax" +30^" — 3*"| 



EVOLUTION. 

58. Cask I. When the given quantity is simple. 

Rule. Estract the given root of the numerical c 
cieot for the coefficient of the root, then divide the eip<^ 
uents of eacli of the literal factors by the name of the root, 
nod the several results connected by the sign of mxiltiplicft- 
tion will be the root sought. 

E xample. ■ Extract the 4th root of Sla^a*. 

"e 4th root of 81 is 3, the 4th root of a' is a'=a, end 
biaotof x' is x^=3fl, .•.3x<J.V,i^-=3aj:* is tli* 



ALGEBRA. 33 

loot sought. In the same maimer may the roets of the 
following quimtities be foond. 

1. What is the 2d or square root (^ Ida^'b^ifi ? 

Ans. 4a6'A 

2. What is the 3d or cube root of ?^? Ans. ^. 

a What is the 4th root of ^'? Ans. ^. 

• ar z 

4. What is the 5th root of ^^? Ans. ^. 

5. What is the 4th root of -j-r^ ? Ans. -j — , 

59. Casb II. When the given quantity is compound, 
and it is required to extract its square root. 

HuLE. Arrange the terms according to the powers of 
the same letter, so that the highest power may be first, and 
the next highest in the second term^ and so on. Extract 
the square root of the first term by Case I., and place its 
root for the first term of the root. Subtract its square from 
the first term, and there will be no - remainder. Bring 
down the next two terms for a diyidend, and for a divisor 
take twice the part of the root already found. Divide the 
first term of the dividend by the divisor, and place the quo- 
tient both in the root and in the divisor. Multiply the 
divisor thus completed by the term last placed in the root, 
and subtract the product from the dividend; to the re- 
iBainder, if any, bring down the next two terms, and pro- 
ceed thus till the root is found. 

The above rule will be obvious, by observing that the 
square of a+x is a^+2ax+3o^, and that consequently the 
square root of a^+2(zx+x^ will be a+x; but after we 
hare subtracted the square of a, the remainder is 2ax+x^ 
^{2a-}-x)Xf the first term of which remainder, if divided 
bjr 2a, will give the quotient x ; and x being added to 2a, 
and then the sum multiplied by x, will leave no remainder. 

£xAMPLE. Extract the square root ofa^+2ah+b^ + 
2ac+2bc+c\ 

a^-+2ab+b^+2ac,+2bc+c^\a+b+c root. 

Divisor 2a+ 6 I 2ab'\'b^ 

I 2ab + b'' 

2a+2b + c 2ac4-26c-fc^ 

2ac+2bc-^c^ 



I 



Jlere, after (a+b) lias been obtained in tLe root, it u 
evident that the remainder can be written 2(i»+6Jc+<^=s 
{2(a+i) + c]c; whetB a + i takes the place of ain ttQ 
remainder, and is of the same form as {2a-\-x)x. 
1. Extract the scjuare root of Qa'+Gai+b'. 

Ans. 3a4-fi< 
S. Extract the square roct of a^-f 8aj-+lfiE^ 

Abb. a-|>4a!. 

3. Extract the equare root of a''c^-\-4a^c.r-\.4x-^. 
Ans. a'c+S* 

4. Extract the Bquare root of 4a*+12a6+96^ + 16fM!+ 
24fe+16<:^ Ans. 2a+36+4o. 

5. Extractthe square root of a-^+4j-'+2,H+9.r' — 4x+^ 

AnB. a^+2j;=— a:-J-2. 

60. Case III. When the piven quantity is compound 
and it is reqoirei to extract its cube root. 

Bui.B. Arrange the terms as in Case II. Find tht 
cnbe root of the first term, '\Thtch will be the first term of 
the root. Subtract its cnbe from the first term, ivbicli 
leave no remainder. Bring down the next term, and divide 
it by three times tJie square of the root already found ; ttw 
quotient v«ill bo the eecond term of the root. Raise tlww 
two terras to the third power, and subtract the result from 
the given quantity; if there be a remainder, divide its fiat 
term by three times the square of the first part of the ntt 
as before, and thus proceed till the work is finished. 

The third power of {a+x) is a^+3a^i(;+3a*'+j^J 
hence it is evident, that the cube root of tr'-f Sa'x^-So^ 
+ie' IS a+jr ; taking away the cube of a, the first term <f 
the root, the remainder is3a'ie+3(u:''+j?3; the first term 
of which divided by 3a', gives the quotient 2. 

ExAMPLK. Extract the cube root of a^+6Ei^.^l2aV 

««+fla*j:+12aV+8r-\ | a''+2x. Root. 

The first term of the remainder is 6«^j-, which diri^ej 
by 3t^, which is three times the square of «", gives 2Kf«f 
quotient; and (i'+2j; raised to the third power, gives flit 
"idnantity whose root was to be extracted, and no remainderir 
^ that a''-\-2x is the root eought. 
li. Extract the cube root of 27a'—^'J<i^x+9a!c' — a». 

Ans. 3a — f> 
. Extract the cube root of a'^+6(('6= + 12a'i*+86'. 

Am, a'+3i». 



3B 

3L BBdmct the iX^iotit^m^^ik^yJ^VJxf^^f. 

Am. « — df. 

61. Any toalt lAoAem may be eztnMttd by the follow- 
in? foimula : let n be the nmiae ef the root, which will be 
i Imp the eqoare root> 3 for the cube root, and so om ; then 
haying arranged the ttnni at in the Sfnarc and cube rooti^ 
extiact ihe toot of tbe first term by Case L> and let a re- 
present that roOft^ then the second term dirided by n(f^\ 
w31 give A0 aecond term of the root ; the first and second 
terms of the root being raised to the nth power^ and sob* 
tnstsd from the given quantity, the remainder, if there be 
any, will be such, that its first term divided by the same 
divisor wiB givo tfao third term. The exercises in the 
square and cube root may be wsoaght by this role. 



BQtTATlOKS. 

h Am EqvaUon. is a pr<^siti(m which declares the 
o^iiaiity of two quantities expressed algebraically. 

This is done by writing the two quantities, one before 
md the other after the fiigB(=): thus 44^- je=3jc — 4 is an 
tquadon, which asserts the equality of 4+«; and Sa? — 4. 

8. A jSim^ EquatUm iS one which, being reduced to its 
simpiest fetrn^ contains otaly the first power of the unknown 
fwott^. 

3i A Qtimdrailic E^fucstkn is one which, being reduced 
ts its simplest form, contains the squajre of the unknown 
foantity. 

4* When ah Equation, after being reduced to its simplest 
bmii oontains the third power oi the unknown quantity, it 
isoslled a Ctibio Equation, 

5. A Pure QtutdraUc is one into which only ihe square 
of the unknown quantity enters. 

6L An Actfected Qttadratic is one which contains both the 
first and second powers of the Unknown quantity. 

7* Thk ^esoiuHoh of EquaHone is the determining from 
•ome giten quantities the value or values of those that are 
unknown* 

The resolution of equations is effected by the application 
of one ^r more of ihe following axioms :— « 

AXIOHB. 

(1.) If equal quantities be added to equal quantities^ thc( 
sums will be equa). 



(2.) If equal quantities be taken from equal quaatideE, 
the remaindeTS will be equal. 

(3.) If equal quantities be multiplied by the same or 
equal quantities, the products will be equal. 

(4 ) If equal quantities be divided by the same or equal 
quantities, the quotients will be equal. 

(5.) If the same qaantiljbe added to and snbtracted 
from another, the value of the lattei will not be altered. 

(6.) If a quantity be both multiplied and divided by tha 
same quantity, its value will not be altered. 

(7-) The same powers and roots of equal quantities are 

From the above axioms the following rules for the re- 
solution of equations can be derived: — 

63. Rdlb 1. Any quantity can be taken from one ade 
of an equation to the other, by changing its sign. 

This rule is derived from asioms 1st and 2d, as will ap- 
pear evident from the following example; 

Let 3ie — 4=2a!+6; if 4 be added to both sides the 
equality will still exist by axiom first, but the equation 
will become 3j;^2j;+6-f-4; where the — 4 has been 
taken to the other side and its sign changed; so that tak- 
ing a minus quantity from one side to the other, and 
changing its sign, is equivalent to adding that quanti^ to 
both sides; if now 2x be taken from both sides, the equa- 
tion will become 3x — 2x^10, where by taking 2a; ftom 
both sides of the equation, it has disappeared from the 
second side, hut has reappeared on the first; hence, taking 
a plus quantity from one side, and placing it on the other, 
with its sign changed, is equivalent to subtracting equals 
from equals, and it has just been shown, that taking a 
minus quantity from one side to the other, and making it 
plus, is equivalent to adding equals to equals. The solu- 
tion of the above equation will now stand as under : 

3x — 4=2x+6, the given equation. 

Sr:=z2x-^10, by transposing — 4 and adding 4 and 6- 

3x — 2ij;^IO, by transposing 2x. 

.■. x^lO, by performing the subtraction on the first side. 

64. Rdlr 2. If, after all the unknown quantities are 
msposed to the first side, and the known ones to the 

1, the unknown quantity have a coefficient, it may be 
away by dividing both sides of the equation by it. 

is rule is evidently the same as axiom 4th. 
MPLB. Given ir + 27=48— 3j-, to find the value 

^x-f-Q^—iS — 3x, given ecjiial^on. 



ALOEBBA. 37 

4^-f.3jr=48 — 27» by transposing — 3x and 27- 
Tx=2\i by collecting the tenns. 
•*. a;=:3, by applying the role. 
65. Bulb 3. If there are fractions in any of the terms, 
they may be taken away by multiplying all the terms by 
each of the denominators in succession ; or by multiplying 
ail the terms at once, by the least common multiple of aU 
the denominators. 

It is eyident that this rule is merely an application of 
axiom 3d, and points out when that axiom may be applied. 

ExAMPi«B. ^+2 "^7 +^=2^> ^ ^^^ ^® value of a;. 
;v-f-^ -f- J +4=2x, the given equation. 

4x'{'2x+x+ l6=zSx, by multiplying both sides by 4. 
Jx-^-I^^lSx, by collecting the like terms. 
.-. 16=0?, by subtracting 7x from both sides. 
66. BuLE 4. If the value of any root of the unknown 
quantity can be found from the equation, raise both sides 
to the power denoted by the root, and the value of the 
\mbiown quantity will be found. This is evident from 
aiiom 7th. 

I Example. Given a:* + ix=z^ + ^ > ^^ fi^d *be value of x, 

I x*+ia:=:| +4, the given equation. 

j 2a;^-f. arzrar+S, by multiplying by 2. Rule 3d. 

2a;^=8, by taking x from both sides. 

ap*=4, by dividing by 2. Rule 2d. 
.'. x=:l6, by the Rule. 
67. Rule 5. If^ after the equation has been reduced to 
its simplest form by the preceding rules, the value of some 
power of the unknown quantity is obtained, its value may 
oe found by extracting the corresponding root of both sides. 
This is also evident from axiom 7th. 

Example. Given — -— =a?+12; to find the value o£x. 

— — =a?+12, the given equation. 

o 

a^+3a;=3ar+36, by multiplying by 3. Rule 3d. 
x^=:36, by subtracting Sx from both sides. 
.*. ;r=6, by extracting the square root. 
The previous rules will be found sufficient for the solu- 
tion of equations containing only one unknown quantity; 
the following solutions are added as examples of t\ieii ^i^^- 
plication. 



w 



1. GiTen 5j^^+34=4r+36 ; to find the vnlof of j-. 
By transposition, 5x — 4a;=36 — 34. Ilule 1st. 

.*. by collecting the terms, x=2. 

2. Given 4aj! — 5i=3(ic+2(;; to find the vahitatjr. 
By transposition, 4a<e — 3djr=2c+&b. Rule Ut 

By collecting, (iaSd)x=2c+5b. 



9. Given -^+2=12 — -; to find (he Tslne rfar. 

By multiplying all tlie terms by 6, the leait comma 
multiple of the denomiuittors, it becomes 2^ — lO-i-3x^7^ 
-aB+20. Bale 3d. 

And by transpoMtion, 2j: + ir + 2.r=72 + 20 + JO, 
Sale Iftt. 

Hence, by collecting the tenus, it becomes Jx^lOQ, 

.: by division, by Rale 2d, j;=I4|. 

68- Scholium 1. If the felotion between ,b ,a»d tin- 
known quantities be not giren in the form of an eguatioiif 
but of a proportion, it may be changed into the form of an 
equation, by making tbe first term divided by the second. 
^ibe third divided by the fourth, — see detinitiooa ; — M 
by making the product of the eicCremcs^that of the means. 

For when a : b :: c: d,hy definition, t ^=^-}, and multiply- 
ing both sides by irf, it becomes ad^bc ; or the product of 
the extremes is equal to that of the means. 

Given -h— : —7 — :: 7 : 4 ; to find the ralne of x. 



Sj multiplying extremes and means, 10j:-{-ii^~ 



Multiplying by 4 it becomes 40j-|-32=126— Jj. 
by transposition, 47j'=94. Kule 1st. 
nd bence, a-=2. Rule 2d. 

69. 80HOLIDM 2. When nn equation appears under the 
form of the equality of two fractions, it may frequently be 
solved with much elegance, by making the sum of the nu- 
merator and denominator on the first side, divided by tlieir 
ditfercncc, equal to the sum of the numerator and denomi- 
nator on the second, divided by their difi'erenue. If only 
one side be a fraction, it may be reduced to the abon fotiD ' 
&r RntJn^ 1 for a denominator on tbe integral side. Th« 
tthore principle may be denwnatiated l\wa ■. 



j— 1=^—1. Ax. 2. 
.-. -y-=:-j^. (3.) And by diTidiB^ 
i?) we obtain ^zJr^^lld' "^^ "^'^ S"'^"* aboTe. 
ExufPiB. Given 4-— S^' = - ; tD find 

•Ja — ^a—x " 

'WBy Bqwmng, ^— ^,^„ -, 

BylnTerting, — — ="iT^^+°i' 

Brfncing the first side, 1 — -:=. .' j. 
Imspasing, '—1^3"+^-;- 

Stdnoing tbe first iide, . , ^-. 

1. Given 3!— 6+25— 36; to fintl a. 
a. ©Ten 3«— 5=23— *; to tind x. 
3. GiTcn 7*— 3=5j+ 13 ; to find a-. 
* Given 3jr+5=10j.— 10; to find a;. 
5. Given 2r+ll:^7j-— 14; to find 3:. 
8. Given 15a'— 24=20 +^:c; to find a 

7. GiTen i+~ — ^rz4j-— 17; to find ^,. 



Ans!; 



tafe 



1 



^^H 8. Given*— 2 + 3 — ^=7; to find a:. Ans. x=ia 
^^m 9. G:ren|+{+| + |+~=46;fofind«. Ans.a;=^ 
^^H 10. Given^+I — |=J;tofindj-. Ans. x=^, 

I: 






11. Given -g- +'^-=14+^; to find r. Ans.iF=13. 

12. Given-— H — — =16' ■■ ; tofindg. Ans.3;=7. 

13. Given « ^^5| '^+1' toGaix. Ans.a-=S. 

; to find .r. Ana. . 

15. Given - + j— <^; *" fi"*^ ^' -Ana. ,r=- 

16. Given (tr+6^=E3--j-«'; to find x. Ans. a'=(i+fi. 

17. Given bx-^-2x—a=3i:+Se; tofind.r. Ans.a:=:f±?", 

18. Given 3j;— | +c:c=-^_— ; to find x. 

PK0BLKM9 IN SIMPLE EQUATIONS. 

ExAUPLB 1. What number is that to wliicli, if its faaff 

id its fifth part be added, the sura will he 34 1 

Let a- represent the number sought; then its half will Iw 

^, and ita fiftli part vrill be r Therefore we will have by 

tbe question, i+-^ + |= 34, 

and multipl.vinp! by 10; Jac+&r+2^=:340, Kule 3d; 
tence by collecting the terms, 17if=340, 

.-. x= 20, by Rule 2d. 

2. What number is that whose third part exceeds ita I 
seventh part by 4 ? j 

Let 3; represent the number sougbt; then its tbiid part ] 

will be 3 , and its seventh pajt ■= . Therefore by the q 



oiaJiiptj^'ng both sides by 21, ^J^-Sa=84, Eule 3d; 

ACTce coUectine the terms, \x=SA, 



ALGRBBA. 41 

3. A cistern can be filled by two pipes.in 12 bours, and 
bytbe first of tbem alone in 20 bours: in wbat time would 
it be filled by tbe second alone? 

Put X for tbe time requiredl 

12 

Tben — would be tbe quantity supplied by tbe second 

* 12 

in 12 bours; and — would be tbe quantity supplied by 

the first in 12 bours. 
But in twel?e bours tbe two running together filled tbe 

12 12 , 
astern, .-.—+—= 1, 

and multiplied by 20a?, 240+12a:=20a:, Rule 3d; 
by transposition, &c. 240= Sx, Rule 1st; 

.-. 30= a?, Rule 2d. 
4. How mucb rye, at four shillings and sixpence a bushels 
must be mixed witb 50 bushels of wheat, at six shillings 
a bushel, that tbe mixture may be wortb five shillings 
a bushel? 
Let X be tbe number of bushels of rye. 
Then 9a?=: tbe price of the rye in sixpences. 

600=: the price of the wheat in sixpences. 
(50-|-it?)10= the price of tbe mixture. 

.-. 9x+ 600=500 + 1 Oo: by the question. 
Hence 100=a; by transposition. 

70. Scholium. The transferring of problems into algebraic 
equations will be facilitated by studying carefully the fol- 
lowing remarks: — 1st, If the sum of two numbers be given, 
and one of the numbers be represented by x, then the other 
will be tbe sum minus x. 2dj If the difference of two num- 
bers be given, and the less be represented by x, the other 
^ be a? plus the given difference; and if the greater be 
represented by x, the other will be x minus the difference. 
^, If the product of two numbers be given, and one of 
them be represented by x, the other will be the product di- 
vided by X. 4th, If the quotient of two numbers be given, 
and one of them be represented by x, the other will be 
the quotient multiplied by x. 5tk, If the sum of two 
numbers be represented by «, and the less number by x, 
then the difference of the numbers will be s — 2x ; and if 
the greater be represented by x, their difference will be 
(a»— 5). 

19. Wbat number 13 that which being incxeaa^3L\i^ \V% 
half and itB third part, the sum will be 154? Alia. ?k^. 



^k£ 



20. What numbei U that to TCliich, if its third and foartl 

pBitB be added, the sum will exceed its sislh part by 17 1 

Amis 

31. At a. certain election/ 311 persons voted, and tbi 
successful candidate had a majoritj of 79: liow many Tof< 
for each? Ars. 195 and llfl 

22. ^That number is that from which, if 50 be sal 
tracted, the remainder tviU be equal to ils half, logethi 
with its fourth and sixth parts ? Ans. 60(1 

23. A hnsband, on the day of his marriage, irtn Ari 
as old as his wife, but after they had lived together 1 
years, he was only twice as old: what were their ages on tl 
marriage day? Ans. Husband's, 45; wife's, ll 

24. It is required to divide L.3(M> between A, B, and < 
BO that A may have twice as much as B, and G as man 
as A and B together: what was the share of each? 

Ans. A's, L.lOOr Bs. L.50,- and C's, L-lS 

25. A cistern can be filled with water by one pipe in 1 
hours, and by another in 8: in what time would it he fillfl 
bj both running together? Ans. 4^ honn 

26. Two pieces of cloth, which together measared 4t 
yards, were of equal value, and the one sold at 3s., and tb 
other at 78. a-yard: how many yards were of each? 

Ans. 12 yards at 78., and 38 at 3 

27. A has three times as much money as B, and if 
give him L.50, A will have four times as much as B: fin 
the money of each. Ana. A'a, L.750; Bs, LSM 

38. After 34 gallons had been drawn from one of tn 
equal casks, and 80 from the other, there remained thiii 
as much in the first as in the second: what did each eo) 
tain when full? Ans. 103 galloa 

29. A person paid a bill of L.lOO with half-guineas --^ 
crowns, using in all 202 pieces: how many pieces y 
there of each sort? Ans. IBO half-guineas, 22 

30. There is a cistern which can be supplied wit 
from three different pipes; from the lirst it can he filled i 
8 hours, from the second in 16 hours, and from the this 
in 10 hours: in what time will it be tilled if the three pi] 
bo all set running at the same time? 

Ans. 3 hours 28^f mjnntol 

31. Solve the above question generally on the suppoi^ 
— that the first pipe can fill the cistern in a hours, tl^ 



•ndja b, and the third in 



Ana,- 



AtOEBRA. 48 

SIMULTANEOUS EQUATIONS. 

71. When two or more lyiknown quantities are to be 
determined, there must always be as many independent 
equations as there are unknown quantities ; and since the 
values of these unknown quantities must be the same in 
all the equations, the values are said to subsist simulta* 
neouslj, and the equations are called simultaneous equa- 
tions. 

Case I. 

To determine two unknown quantities from two inde- 
pendent equations. 

72. Rule I. Make the coefficients of one of the un- 
known quantities the same in both equations, then bj add- 
ing or subtracting these equations, there will result an 
equation containing only the other unknown, whose value 
may be found by the previous rules. 

Note 1. The equations are to be subtracted when the quantity 
whose coefficient is rendered the same in both equations, has the 
same sign in each, and added when it has opposite signs. 

Note 2. The coefficients of either of the unknown quantities 
may always be rendered the same in both equations, by multiplying 
the first equation by the coefficient of the unknown quantity, which 
is to be made to disappear in the second equation, and the second 
equation by the coefficient of the same letter in the first. By this 
means the coefficients of that quantity will evidently be the same in 
both, for it will be the product of its two coefficients in the original 
equations. 

73. Rule II. Find a value of one of the unknown 
quantities in terms of the other from each of the equations, 
and make these values equal to each other, which will give 
an equation containing only the other unknown, from which 
its value can be found by the previous rules. 

74. EuLE III. Find a value of one of the unknown 
quantities in terms of the other from one of the equations, 
and substitute this value instead of it in the other, from 
which there will again result an equation containing only 
one unknown quantity, which may be solved as before. 

75. Rule IV. Multiply one of the equations by a con- * 
ditional quantity n, then add this product and the other 
equation together : let n fulfil the condition of making the 
coefficient of one of the unknown quantities 0, tVi^ii VJckfc 
equation will onljr contain the other unknown ; del^imvck^ 
i£f value of n that fuMls the above condition, and ^tjJa^^Lv- 

^ tbiB ralue instead of it in the resulting eq\xal\OTi, ^kA. 
the ralue of the other will be determined. 



or on. 
tated 

I WlUB 

^^ (3)- 







leUtion to one 


of llie unknoHii ouantiliea, and then in relation to 


other, whidi will give 




each J or the tt, 


of one ol the unknown qnantitiea being found, its 


vulnecanboBuli. 


tated instead of it in either of the given equations, from which 






Example 


Given 3j;+5tf=35, and 


7.^-4y=19: 


find the values of ;c 


ind y. 








By Hule 1st. 






1 


3xJf fii/= 35. 


-^^ 




2 


7^- 4y= 19. 




<I)X 7 


3 


2U+35y=245. 


^^^1 


(2)X 3 


4 


21^— I2y— 57. 


^^^1 


(3)-C4) 


5 


47^=188. 


^^^1 


(5J-47 


6 


■■■ v= 4. 


^^H 


C1)X 4 


7 


]2x+20^=I40. 


^^H 


(2)x 5 


8 


35j:— 20(/= 95. 




(7) + (8) 


9 


Alx =235. 


^^H 


■(9)^47 


10 


.-. « = 5. 
By Eule 2d. 


M 


From (1.) 


by tiaD 


sposttion aud division 


' ^ 3 * 


And from 


(8) 




X———. 



-by _ 19+45 



Ax. iBt. 



_35— Sj_ 



245— 35y=.->7+J2;/ by mult, by 21. 
188=^47y by irnnsposition. 
.■- A=.y by dividing by 47. 
And substituting Ibis value of y ia (1.) tve oblaio 
3ir+20=35, 
Hence 3j-=15, 

And jr=5, as before. 
By Eule 3d. 
It bas already been found from (I), that a:=: 
ftfltitdting this value, instead of jr in (2) it becomes 
7(??=i')-4,=.9. 
245— 35^— 12i/=57, by mult, by 3. 

]88=47i', by transposition. 
.-. 4=y. 
In the same manner may x be found from either eqi 
fion, hy dubsCiluting a \aiue of y found from (he otl 
equation. 



ALGEBRA. 45 

By Rule 4th. 

Multiplying (1) by n, and adding (2), we obtain 

(3»+7)a:+(5«— 4)y=357i+19, which, if the coefficient 

of y be 0, becomes (Sn'\'7)xz=z35n+19 ; and since the 

coefficient of y is =0, 5n — 4=z0; hence n=J ; substitnt- 

ing this Talae of ti in the equation (37i+7)a:=35w4-19, it 

becomes 9|a?=47. 

Hence 47a;=235. 

And .*. ir=5. 

Next, let n be such as to render the coefficient of ar=0, 

then the equation will become (5n — 4)y=357i+19; and 

7 
since 3n+7=0 .•. niz — « , substituting this ralue instead 

of 71 in the equation^ and changing the sign8> it becomes 

15|y=62f. 
Hence 47^=188. 
And .*. y=4. 

1 p. f 3a:-}-2y=56 ) to find the values of x and y. 
1. ijiven S2a:+3y=54j Ans. a;=12, y=10. 

2 r*vp i^^ — 7y= 8l to find the ralues of a? and y. 

\^4x — y=34j Ans. ii?=10, y=:6. 

3 0* J 3^+ 2^=^^ ) to find the values of x and y. 

J Jip-}- .y=9 J Ans. a?=9, j^^H. 

4 p. J 3a;4- iy=38 \ to find the values of x and 15^. 
^iven ^ i^^2.y=12 f Ans. x=\2, y=4. 

e p. J 0?- — ^y*^=20\ to find the values of a? and y, 

^ ^^ \x +y =10 J Ans. a:=6, y=4. 

g p. j x-\-y=8 \ to find the values of a? and y. 

(a; — j/=d f Ans. xz=z^(s'\'d), yz=^{s — d). 

1 -^-r-^^4-2j?=16 / to find the values of x 




7. Given < ^_, ^^^^„ > and 3^. 

Ans. ic=6, y=3. 

( , , 1 to find the values of x and y. 

The above equations may all be solved, by substituting 
in the answer to the (8) the proper values of a, h, c, a\ b', 
and c', with their proper signs; only (7) would require to 
he reduced to the proper form before the substitution can 
be made. 

Case II^ 

76, To determine the values of three or more unknown 
quantities from as many independent equationa ?l^ VScietfe 
are unknown quantities. 



^^ 



Rule I, Multiply eauh of the equations by sucli multi- 
pliers as will make the resulting coefficient of one of the 
unknown quantities the same in all thg equations] then b^ 
adding or subtracting these equations, a new set of equa- 
tions, one less in number, and containing one unknown 
quantity leas than before, will be obtained; with which 
proceed as before. 

77. Rule II. Find a value of one of the unknown quan- 
tities, from each of the equations, in terms of the other 
ujiknowiiBj then equa.te these values, and a new aet of 
equations will be obtained, containing one unknown quaii> 
tity less than formerly; with which proceed as before. 

78. KuLE HI. Find a value of one of the unknown 
quantities from one equation, and substitute this tbIvs, 
instead of it, in the others, which will evidently give one 
equation less than formerly, containing all the unknown 
quantities, except that whose value was found, with which 
equations proceed as before. 

79. Rule IV. Multiply all the equations except one 
bv some conditional multipliers, as w, ji, p, &c., then add 
all the multiplied equations and the unmultiplied one to- 
gether, and a new equation will be obtained, in which if 
we make all the cocflicients of the unknown quantities ex- 
cept one equal to U, a set of equations among the condi- 
tional multipliers will he obtained, which will determtns 
their values, and these values substituted in the resulting 
equation will give the vuluc of the other unknown quantity 
whose coefficient was not considered to be 0. 

The above rules, as well as tliose given in Case I. are all' 
evidently true from the axioms. 



EXAMPLE. 



> 



find the values of x, y» 



Given J2x+7ff--8i= 

(4^+ y+ 2= — J 
Multiplying the first equation by 4, the seoond by 6 
&e third by 3, they become by Rule 1, 
1 1 I".'— 8i<+24j=88 



, ^2x+7i/~8r=24[- " 



(2)-Cl) 

(2)_(3) 
(4).^ 2 

(5)-;- 3 
(6)xl3 
(7JX25 



12j.'.f42y— 48. 

50'/—72z=56. 
39tf— 51.'=.';4. 
25v— 36j=28. 
13>— 17^=18. 
325;/— 4«Hit=364. 



144. 



AliGSBSA. 

43z=86. 

25y=100. 

12a:-l32+48=88. 
12jp=:72. 



47 



(»M8) 10 

(10)^43 11 

Bf subeC. in (6) 1^ 
Transposition 13 

(13) -5-25 14 

By subst. in (1) 15 
Transposition 16 

(16)-5-12 1 17 i.-.ar=6. 

The ts<rfation bj Rules 2d and 3d are left as exercises 
for the pnpils. The solution bj Rule 4th is as follows : 
sraltipljing the 'first of the given equations by m, the 
Beooiid bj n, and adding these products to the 3d of the 
given equations^ we obtain the following: (3in-|-2n-f-4)a; 
-(2iii^7n— 1> + («m— 8n+ l)=22wt + 24*+30. 

Let now m and n be such as to make the coefficient of 
y and r each equal to 0, and the following equations are 

obtained^ in which m and n are the unknown quantities. 

1 1 2?/i— 7n— 1=0. 

2 6w^— 8n+l=0. 

3 6w— 21/1— 3=0. 

4 13/^+4=0. 

5 13w=— 4. 

6 n=—fs.^ 
In the same manner^ or by substitution^ we obtain m= 

15 
— rr; and since the coefficients of y and z are each=0, we 

We (3wi+2n+4)a:=22m+247i+30, 
in wiiich^ bj substituting the values of m and n, there results, 
f ^^ ^ ,A\ _ 380 96 ^ 
^"^-"18" + ^r -"""26 ""13 +^' 
hence 14 J2r=r9||. 

43ar=258, by mult, by 26. 

In the same manner, if the coefficients of x and z be 
equated to 0, we can find .the value of y ; or if the coeffici- 
ents ofx and y be equated to 0, we can find the value of 2. 
(^+y — z=10^ to find the values o£ x, y, 
1. Given < x — y+2= 6 J- and e. 

yy+z — a?= 2J Ans. a;=8, 2^n6, 2=4. 

*+ y+ 2=35'! to find the values of a?, 
ar4-2y-|-32!=66 >- y, and z. 

i^+§^y+i2;=13j Ans. a?=:12, y=15, z=8. 



(1)X3 
(2)-(3) 
Transp. 
(5)-13 



2. Given 



3. Given 



'^^_2g I to find the values of ar, y, and z. 
I'll^QQ I ^^^' ^=16, l/=\^> «\^- 



^y+z=22) 



a -J 2s 






■+y4-«=9(> \ to find the valaes of ST, 
■«— 20=3./— 40 J- jj, and z. 

+20=2;4.5 } Aas.x=aa,i,=30,i=55. 

ISOBLEMS PRODUCING GQDATIONS WITH TWO OR MORE 
UNKNOWN QUANTITIBB. 

1. Find two numbers fluch that the first with twice the 
second shall be 2] , and twice the first, with half the second, 
shall be equal to 14. Ans. 5 and 8. 

2. Find two numbers such that one half the first, with i 
one third the second, may be 7, B.ni one third of the first, i 
with one fourth of the second, may he 5. Ans. 6 and 12. I 

3. Find two numbers such, that if 5 he added to the 1 
first, the sum will be twice the second, and if four times 
the second be increased by 3, the sum will be three times 
the first. Ans. 13 and S. 

4. Find a fraction such, that if its numerator be h 
ed by 1, and its denominator diminished hy 2, its tbIuq will I 
be ^, and if its numerator be increased by 3, and its de- fl 
nominator diminished hy 2, the raluc will GtiU be |. I 

Ans.A.| 

5. Find a number, consisting of two digits, the sum of 
whose digits is equal to ^ of the number, and the product 
of whose digits is equal to ^ of the number, Ans. 36. 

6. There is a certain number consisting of two figures, 
which is equal to 6 times the figure in the unit's plaee, and 
if 27 be added to the number, the digits will be inTerted: 
what is that number? Ana. 36. 

7. There is a certain number consisting of three figures, 
the sum of the digits is 7i twice the sum of the extrcnu 
digits is equal to 5 times the meau, and if 297 he subtract- 
ed from the number, the digits wiU be inverted: what 18 
the number? Ans. 421. 

8. Find three numbers, so that tjie first, with lialf the 
other two, the second with one-third of tho other two, and 
the third with one-tburth of the. other two, mnv each be 
equal to 34. Ans. 10, 22, 36. 

9. Find a number consisting of three figures, whose 
digits are in arithmetical progression, such that if this 
number be divided hy (he sum of its digits, the quotient . 
will be 41^; and if from the number 193 he subtracted, the J 
digits will be inserted. Ans. 432. I 

10. If A and B together can perform a piece of work in I 
8 days, A and C together in 9 days, and B and C together ' 
iaIOdajB, in what time will e.ich of them perform it alone? 

Ans. A in 14s J, B in \7U' a»i ^ '^^^5\^^- 



QUADRATIC EQUATIONS. 

80. Qdadratic Equations may he divided into two 
kinds, namely, snch as contain only llie square of the un- 
known qaanlity, and those which contain both the square 
and first power of the unknown quantity; the first are 
called Pwff Qvadralke, and their solution may be effected 
hy Rule V. of Simple Equations, with one unknown 
qnantity; the second are caWei Ailfidfd Quadratics, and 
ihtysie of one of the four following Ibrms: — 

a^' + 6x=+c, I 

Wf -~^=~- s 

JUl Hiese are included in the general fonnula. ax^:±:lix ■ 
^:tc; and we proceed now to solve this equation. If the 
first terra were multiplied by 4a, it would become the 
square of 2iw; let both sides be multiplied hy this quan- 
tity, viz, 4ni and the general equation becomes ia'x'dtz4abx 
^:t4ac; the first side of this equation evidently wants 
something of beiof; the square of a binomial, of which the 
first term is 2(w^; let this qunnlity be p^, then adding this 
^oantity to hoth sides, the equation becomes 4a''x'^=4a/ix 
+p"=p'^:4ae : now if the first side he a complete square, 
its square root can he no other than 2'iie=t=p, (Art. 29 ;) 
hence squaring this, its square must be identical with the 
first ride of the hist equation, but its square is 4a'x':±: 
iapx^p'^ .-. to 4aV-'^4aJ,E+/)^; hence taking away 
the common terms from both sides, 4apx^4abJ.; and divid- 
ug both sides hy 4ax, we have p=b, and therefore the 
^ quantity which must be added to both sides of 4a''x'' 
I ^i(thx^':^4ac, so as to makeit a complete square, is b^, 
I Hat is, the square of the coefficient of it. If now the root of 
liotli sides be extracted, weobtain 2ax^=b=2*^ ^ b'^4ac, 

2a " 

fwtoola in which, if we insert the proper values of a, b, 
Ind e, with their proper signs, (the sign of a being always 
+), we will have two values of a:, both of which fulfil the 
wnditions of the algebraic equation. 

From the above inves/ig'alion the following iu\e\& ia- 
tired.- — 



w. 



wiU 
Hen 

I 



Rule. 

1. Transpose all tlie terms containing the 
{[uantity to one Gitle of the ei^uation, and so tliut tlie tei 
containing the square of tlie unknown q^uantity may be po 
sitire, and the known tenns to the other. 

2. Multiply both sides of this equation by four times tk 
coefficient of x'. 

3. Add the square of the coefficient of x in the fin 
equation to both sides, then will the first side be a complt 
Kquare. 

4. Extract the root of botb sides, and the result wifl I 
a. simple equation, which may be solved by the previw 
rules. 

82. ScHOLimu. When the equation, after being trsii 
posed aa in the Rule, can he divided by the coefficient 
j:'', without introducing fractions, it may be solved convi 
niently aa follows : — Perform the division, then add tl 
gq^uare of half the new coefficient of a: to both sides, wild 
wdl make the first side a complete square; then 
the square loot of both sides, and the equation will be a 
duced to a simple one, which may be solved as before. 

EsAUPLB 1. Required (he values of j^ in thi 

363:*— 4fe=]it20, by multlpljlng by 12=(4x3), 

36*'' — ifiLC+16=1936, by adding (4/ to both sides, 
Gx — 4^^44, by extracting the square root, 

63;=r4it44, by transposition, 

r^8, or — (>§, by dividing by (i. 

Either of these values substituted in the given equatii 
will make the sides equal, and they are therefore both nxi 
of the equation. 

Example 2. Find two numbers whose sum is 100, i 
whose product is 2059. 

Let j:= the one ; then since their sum is 100, the ol 
may be represented by 100 — x; and hence their prodn 
will be x(IOO— ^), which by the question is equal to 20S 
Hence IOOj: — a:^=2059. 

x'' — 100j-= — 2059, by changing the signs. 

ie»_100j^+25O0=441, by scholium. 

X ^-50=^^1, by extracting the root. 

.-. «=50±21=7I, or 29, which are the ti( 

required, and therefore x has come out either tb 

r or the less part. The solution by the rule is lei 

Hie eiercise of the pupil. 



AIiGBBBA. 51 



1. OiTen x'^zix+iS ; to find x. Abb. x=9, or 

2. Giyen 5x^+7 +4x=^5; to find o^ 

Ana. x=:2, or -— 2|« 

3. Oiren ^'»^=:5-^-dr ; to find x. 

j^_^ Ans. a?=r5, or — 11. 

4. GiTen 4x — 14= — - ; to find x. Ans. ar=:4, or — J. 

5. GiT€iL 3« 4-^=2-1 ; to find x. 

6. GiTen 4— — r- = — r-^; to find x, Ans. a:=6, or X, 

7. Oiven i^'=?i^4-^ ; to find x. Ans. <e=l^, or — |. 
& Giyen Sst;^ +^+6^=301 ; to find ^. 

J2Q Ans. a:=:3, or — 4. 

9» Giren a?— 1=5 — •; to find x. Ans. orizll, or — 10. 

10. Giyen - z=24jt; — TOO; to find a?. Ans. a:=70, or 50. 
U» Giyen --7= -7o;tofinda^ Ans.;r=:4f0r— *4. 

12. Giyen — - + "ZI^^ I * *^ ^^ ^• 

13. Giyen r + -tt- =9i to find a?. Ans. xzsl2, or 6. 



QUESTIONS PBODUCING QUADRATIC EQUATIONS. 

1. What two numbers are those whose difference is 15, 
and half of whose product is equal to the cube of the less I 

Ans. 3 and 18. 

2. What two numbers are those whose sum is 100, and 
vluMe product is 2059 ? Ans. 71 and 29. 

3. Find two numbers, so that their difference may be 8, 
ind their product 240. Ans. 20 and 12. 

4 Haying sold a piece of cloth for L.24, I gained as 
mach per cent, as the cloth cost me ; what was its prime 
C08t? Ans. L.20. 

5. A grazier bought as manj oxen as cost him L.480, 
and retaining 6 to himself, sold the remainder for the same 
sum, bj which he gained L.4 a head on those sold. How 
many oxen did he buy, and what did he pay for each ? 

Ans. 30 oxen, at L.16 each. 

6. A labourer dug two trenches, one of which was 4 
yards longer than the other, for L.20, and each trench cost 
as many shillings a-yard as there were yards in its length. 
fiow many yards were in each ? Ans. 12 yards and \6 ^^i&!&. 

7* Tb»pUte of a Jooking-glaaa i» 24 inches by \6 \ it \fr 



to be framed by a fraine of uniform width throaghoat 
whose surface aball be equal to the surface of the glaaSi 
Required the breadth of the frame. Ans. 4 inches. 

8. There are three numbers in geometrical progression. 
The sum of the first and second is 10, and the difference 
of the second and IJiiid is 24. AVhat are the numbers? 
Ana. 2, 8, E 

y. A and B set off af the same time to a place at t 
distance of 300 miles. A travels at the rate of one roilci 
Lour faster than B, and arrives at his joumey'a end 
hours before him. At what rate did each travel per honrt 
Ana. A travelled 6 miles per Lour, and B travelled 3. 

]0. A and B distribute L.1200 each among a certioii 
number of persons. A relieves 40 persons more than B, 
and B gives L..5 a-piece to each more than A. Howmanj 
persons were relieved by A and B respectively? 

Ans. 120by A. andSObyE 

U. A person bought cloth for L.33, )5s., which he soli 
again at L.2, 8s, per piece, and (gained as much by the 
bargain as one piece cost him. liequired the number of 
pieces. Ans. 15i 

12. A company dine together at an inn for L.3, ISi 
One of them was not allowed to pay, and the &haTe of cad 
of the rest was, in consequence, Lalf-a-crown more than i 
all had paid. How many were in the company. Ans. t 

13. A draper bought two pieces of cloth for L.3i ft 
The one was fi yards longer than the other, and eachli 
them cost as many sbilllDgs a-ynrd as there were yaidili 
the piece. What was the length of each? 

Ans. 2 yards an 

14. Two girls carry 100 eggs to market. One of then 
had more than the othei-, but the sum which each re- 
ceived was the same. The first says to the second, if 
had had as many eggs as you, I should have received li 
pence. The other answers, if I had had your number, I 
should have received 6^ pence. How many eggs hu 
each, and what did each ivceive ? 

Ans. The first girl had 40, and the second 60 
received 10 pence. 

15. Find three numbers having equal differences, to 
that their sum may be 9, and the sum of their fourtti 
powers 707- Ans. 1, 3, and ' 



OVJDBATIC EQUATIOVS, WITH TWO WM^KNO^IA 1^ ».MTITIB8. 

S3. In solving quadratic equalioTis -wM^i Wo ^tJmvwww 
gaaadties, it is necessary, frequents, to &Da.a.-^5i'*e«'t«* 



ALGEBRA. 53 

»f the unknown quantities in terms of the other unknown, 
ind then suhstitute this valae, instead of it, in the other 
equation, and then solve this equation for the other un- 
mown quantity ; then the remaining unknown quantity 
irill easily be found, either by substituting this value, or 
otherwise, as may be found most conyenient. If the sum, 
or the sum of the squares of the two quantities be given, 
and their product be either given or can be found, the sum 
and difiEerence of the quantities can be found, and then 
their values determined by the solution of a simple equa- 
tion. ^ 

Example 1. Given a?+y=21, and — =2; to find x 
and y, ^ 

From the second equation we find xzz2y^ ; substituting 
this value instead of x in the first it becomes 

2y^+y=2l, 

Hence 16y^+8y+l=169» ^7 comp. sq, 

4y4-l==*=13, extract the root 
. \ y =3, or —3^. 
These values substituted instead of y in the first equa- 
tion give a;=18, or 24 J. 

DxAMPiiE 2. Given a;+y=10, and ii?y=24. 
' Squaring the first gives x^ + 2xy + y ^ = 1 00. 
Pour times the second 4iy=96. 

Hence by subtraction, x^ — 2xy^y^z=.4. 
By extracting the root, x — ^y=2 (a). 

By the first equation, a?+y=10 (6). 

.-. 2x=\2{a) + {h). 
and a?=:6. 

/. 2y=8 (6)-C«) 
and y=4. 
Example 3. There are two numbers, the diflference of 
whose cubes is 117) ^nd the difference of the numbers 
themselves is 3, what are those numbers ? 

Let «=the greater, and y=:the less ; then by the ques- 
tion, x^ — y'=117. 

And X — y=3. 
Divide the first equation by the second, and there results 

x^^xyJ^y'^z=!3>d (a). 
Square of the 2d x^^2xy^y^=id (h). 

,\ 3a;y=30, by subtraction, 
and xy=ilO (c). 

a?+y=z7, by extract. lOOt. 

But x—y=:'^^ 

Hence x=5, and yz=:2. 



194 ALGEBRA. 

Example 4. Wliat two nnmbers are tbo§e whose 
multiplied hy the greater is 77i and whose difference 
tijJiedby thele8siBl2? 
Lei a;:=the greater, and //^tbe less. 
Then {.T+,^yT=j^''+^i/=77. 
And {x—;/}y=xi/—;/'=12. 
Assume x-=.vy, then by substituting (his value insti 
t£xia each of the equations, they become 
And w/'J— y«=12. 
The first of these being divided by the second, gii 
■Co 



Hence ]2eif + ]2c=77t^77- 

.-. ]2««— 65,.=r— 77- 
I 'Completing the square, 5761'"— 31 20f + 422n =529. 
\ Extracting the root, 2\v — 65:=^::23. 



Either of these values will fulfil the conditions of i 
question. Thefirst giveaa'=— ^2, and i/=-=J2, 
the second gives ir=7. and y=4. 




BXBRCI8GS. 

a'+y^IS \ to find x and ?/ 



21 t 
V 



Ans. ir=:lf, ws 
7>'to find .rand;/. ^ 

( ar^=18f Ans. 3-=9, y: 

'3;=+j/^=169 J to find j: and »/. 

X—y='j\ Ans. x=12, y; 

jr' — s(^=;72\ to find x and y. 
k''+V=108( Ans. «;=9,y: 

fl:4-2y=I5l to find jt and -/. 
a^^+4/=113 ]■ Ans. :r=8 or 7, ,/=3i 01 
*+Jf=^5^) tofindrand,/, 

^=-^\ Ans, x=16,y=] 



:[• 



Ana. fl;=8, t/=i 
of two numbeis ina\V\B^i\i3 iVc pwi 



AIX>£BBA. 55 

240, and their sum multiplied by the less is 160; what are 
hese numbers ? Ans. 12 and 8. 

10. What two numbers are those, of which the sum 
nultiplied by the ^eater is equal to 220, and the difiference 
nultiplied by the less is equal to 18 ? 

Ans. 11 and 9, or 10^2 and ^"2. 

11. Find three numbers such that their sum multiplied 
bj the first may be 48, their sum multiplied by the second 
may be 96^ and their sum multiplied by the third may be 
112. Ans. 3, 6, 7. 

12. Fmd two numbers such that five times their differ- 
ence may be equal to four times the less^ and the square of 
Ihe greater, together with four times the square of the less, 
may be 181. Ans. 9 and 5. 

13. What two numbers are those, the sum of whose 
squares is 34, and the sum of whose fourth powers is 706 ? 

Ans. 5 and 3. 

14. What fraction is that which is double of its square, 
aad whose numerator increased by one, and the sum mul- 
tiplied by its numerator, is equal to the denominator dimi* 
Hfihed by one, and the remainder multiplied by the deno- 
ninator ? Ans. \, 



SURDS. 



84. Those roots which cannot be expressed in finite 
terms, are called surds or irrational quantities, 

'Rius the V2, X/a^y s/xT or 2*, a% a;", are surds, for 
their values cannot be expressed in finite terms. 
. All surds may be expressed by means of fractional expo- 
lents, in which it is eyident that the value of the surd will 
Aot be altered by multiplying or dividing both numerator 
and denominator of the fractional exponent by the same 

-umuci , vttuo M =8^=(8^)^, or d^zza~^ for a "raised to 



mr *nr 



the rth power, is a"* , and the rth root of this is a "'^ ; but 
a quantity raised to the rth power, and then the rth root 
extracted, is the quantity itself; hence a surd is not altered 
in value by multiplying both numerator and denominator 

of its exponent hy the same number, and since a**^ \ia% ^il- 

fi» 

vadf been shown to be equal to a"*, the surd is not ?\l«feA. 
ralue by diriding both numerator and denomiualoi o? W^ 



lexponent by Ihe same number. The following operations 
npon surds depend upon this principle. 

85. 1st, To reduce a quantity to thefonn of a given intrd. 

BuLG, Ruise the quantity to the power denoted by th< 
exponent of the surd, and indicate the extraction of 
same root. 

Example. Reduce 2(i to the form of the cube toot. 

Here 2a raised to the tbii'd power becomes 8a^, and 

dicating the extraction of tlie cube root, (8a')^, the form 

required. 2o lo* fii 

1. Express each of the quantities, ax, Zay, — , ^, tj 

and ^y^ separately in the form of the square root. An& 

3. Express each of the quantities, 2a*c, -^, 4ay^ 3aVi 
and — ■ separately in the form of the cube root- Anfc 

3. Express each of the quantities, — 2(i,— , ^, — -j, 
and —J- separately in the form of the fifth root. Am 

Note. IF a surd hsTe a coefficient, the whule may be expRB 
ui Uie foriu of a sud, by rulHuig the cuefficicut lo the p«w<ff d 
noted by the sui'd, aud multiplying this power intu the sari, tb 
placing the eymbol af the root over tbo whole product. IMi 

4. Express 4ja, 3^^, ^{ax^)^, and 3(3:»y')i,inft* 
form of simple surds. Ana. ^'iGa, (27<jj;»)3, (625<w«)*i 
and (Sli'y^)''. 

8ti. 2d, To reduce a stird to its eimplmt forra. 

Rule. Resolve, if possible, the quantity into two ^ 
tors, one of whicb shall be a complete power, the root of 
which is denoted by the surd; place the root of this fatUlt 
before the symbol, and it will be the form required. If th» 
Burd have a denominator, multiply both numerator and de* 
BotBiBator of Ihe fraction by such a quantity as will malu 
the denominator tlie power deuotci'b^ V\\e ^xud, then ex- 
tmct its root, and place it wu\iout Xte s'jmWV. 
■KxiMp/.E, Reduce ^ 27a? to i^ Min^XeA S.-ina. 



ALGEBRA. 57 

Here 27a'=9a'X3a, the first factor is a square, ex- 
tractiiig its root^ and placing it without the symbol, we ha?e 

SatjSof the form required. 

1. Reduce V32a^, l/sTa^, ^12^, and (180a'a;«^)i to 
their simplest forms. Ans. 4a^ V2a, Sa^^Sa, 5,Jb, and 

2. Reduce 4/1250^, (Oda^x^y^, and (72a?*y')*, to 
their simplest forms. Ans. Bxy^lOxy^, 2af6aa;*)'*,and 

3. Reduce ^-y* v ""V^' ^^ (~27"J ' *^ *^®"^ ^^°^'' 
plest forms. Ans. -y-Ay7a> ^4/3^, -—a/So. 

flieir simplest forms. Ans. i(x^z)^, y^^^^)^' X^21ay)*, 
and |(10a^«z)*. 

87. 3d> ^<> reduce surds having different indices to other 
eguivalent ones having a common index. 

Rule. Reduce the fractional indices to a common de* 
nominator, then inyolve each quantity to the power denoted 
\fj the numerator of its fractional exponent, and over the 
rkiilts place for exponent one for a numerator, and the 
common denominator for a denominator. 

Example. Reduce (2a)^ and (3a^)3 to equivalent 
suds haying a common index. 
Since i=B> ^^^ 3=1 » *^® quantities are equivalent to 

(2a)^ and (3a®)®, raising each of these quantities to the 
power denoted by the numerator of its fractional! exponent, 

they become (8a' )^ and (9a*) 6, which is the form re- 
quired. 1 1 

1. Reduce (ac)^ and 5^ to equivalent surds having a 

common index. Ans. (a^c^y and (125)6. 

2. Reduce 4^ and 3* to equivalent surds having a com- 
mon index. Ans. (256)^^^ and (243^^. 

3. JReduce 4/4a^a; and ^/So^^ to equi\a\eT\V. swcSi^ 
faring a common index. Ans. ^^25Qa^x^ and^^^Ta'^x^ 



w 



4. Keduce (-3-) and (—Y '" •'q'^'^'alent surds haying 
index. Ana. ("^ 

2" I and f --J- 1* to equivalent Hurda Lst^ 

ing a common index. Ana. ("i^) ^^^ (3^) • 

I 8fi. To ai.ld or sahtrad, surd*. 

I Edlh. Reduce the surds to their simplest form. 

j if they have the same radical quantity in each, the sum of, 

\ the coefficients pre6Ked to this radical will he their stUn; 

and the diiFerence of the coefficients prefixed to the la^' 

eal will be their difference. But if they have different ibA' 

cal quantities, their sum can only be indicated by placing 

the sign 'phiA between them, and their difference by ph 

the sign miiiug between them. 

( The reason of this rule is obvious, for the radical qasn- 

tity may he represented by a letter, and then the rule will 

I be identical with that of addition and subtraction in algt' 

I Example. What is the sum and difference of -JM 

I :md VIM Here 'v'2a8= VTi4x2=12v'2,_aBd V^ 

I ='/Qix2=W2; hence their sum is 20^2, and QA' 

(difference is A-J2. 
I _ 1. Find the sum and difference of 3v'32 and ^Vm 
(3€ 



Ana. s«m30V2. diff. eja' 
j 2. Find the sum and difference of 3^/54 and ,^250. 

Ans. mmU^2,i\^.iiJi 

\ 3. Find the sum and difference of ^^Aa^ and^'H'So. 

Ans. sum (2a+4),^3a, diff. (2a— 4)4&. 

. Find the sum and difference of Vao and -J^ 

Ans. sum I'/o, diff, ^] 
6. Find the sum and difference of (Sfi.i^)^ and (98a)i 

Ans. sum {Ga-Ja^l ^^a), diff {Ga-J'^'jJ^)'' 
6. Find the sum and difieteucc -at (^\0W1«,«'^^ uia| 
(300as)i. 



AliSKBEA. 59 

Abs. ram (lOaWjM+lOa'/^), ftnd diff. lOa'^/TO^ 
10a VSa. 

TO MtTLTIPLY 8UBDS. 

39. Rule. Reduce the surds, if necessary, touoommon 
iex^ then moltiplj the coefficients together for a coeffici- 
;, and the surd quantities together for the eurd^ oyer 
ich place the common index. 

Example. Midtiply S^/lO hy 2 4/l§. 

:2=:6, the coefficient, and ^10x^/12=^/120. 

6v I20is the result; which, however, can he simplified; 

! 4^120=4^8 X 15z=2,yi5; hence the quantity in its 

^k8tformisl24/l5. 

Examlk2. Multiply Vo'by 4^6. 

ere Ja^a^=J=ia^)K ojidj^b=(b)^=h^=:ih^y^ .-. 

Ty ^b=z ^a^)^ X (b^y^ia^V'yk 

\. Multiply 5 V5 hy 3V8. Ans. 30\/lO. 

2. Multiply (18)* hy S^/i Ans. lO^^. 

8. milii^lj VTO by ^i5. , ^ 

Ans. (233^65)* or jy225000- 

i Multiply 5a^ by |ai. Aas-^a^^. 

t Multiply (a +6)^ by (a— 5>i Ans. (a»— 5')*. 

m r , _^ 

0. Multiply a by a . Ans. a ^ * 



TO DIVIDE SURDS. 



90. Rule. Reduce the quantities, if necessary, to a 
tinmon index, then divide the coefficients and the surd 
lantities- separately as in rational quantities. 

•Example. Divide ahs/<^<^ hy b^/be. 
are ab-^b=ia, the coefficient, and ac-i^bc^z-r the surd 

3/a 

the quotient is a^ 6 . 

EbiUMPLE 2. Divide ^^ac by 2sjhc. 

re the quantities reduced to a common index \)ecoTCL^ 
9^^, and 2 (3V')s ... the coefficient of tlxe cjao^xcuX 



60 xiOBWtii ■ 

is - J and the aurd {i^]^=-{j3yi which reduced to its 



_3(A0i 






aimpleet form is - (a^i^c)*, and hence the quotient is 



II. Divide 10 v'27 by Sv^S. Am. 15. 

Ls. Divide ^4/6 bj |4^a Am. J. 

|3. Divide ii/iShj 2^Q. Ans. 2:^/2. 

m4. Divide ^^^a by f^^nt. Ana, — ^^_ 

|5. Divide I V'^ by J^o*- Am. {J. 
quir 



. Divide (a'— 6*)^ by (_a—l)k Ans. {a+6)i 



INVOLUTION OF SURDS. 

^i- 91. Bulb. Raise the coefficient of the surd to the re- 
quired power, find then multiply the exponent of tbe sori 
by the exponent of the power. 
Example. Find the third power of Sv'ac. 
Here we raise 2 to the tliird power, which gives 8, sinl 
then nmltipiy ^, the exponent of the surd, by 3, the expo- 
Bent of the power, which gives- .'. the third power of 
I/^is8(<ic)^=8(«Vx'^)^=8((c(„c)^ or 8ac-/^ 
1. Rjiinn ^Ini-^s to iTip BPcfind Dowpr. Ans. 4f(wl'' 

I 



Raise 2{ac)3 to the second power. Am. 4(iK)' 

Raise 4(60^)"^ to tlie third power. 

Ans. 646cj^*'M 
8. Raise ^v6 to the fourth power. Am. jV 

4. B^se a*b* to the sixth power, Ans. 0*6'. 

5, R^se 1 4- ^■'' to the third power. 
Ana. l+3•/i+ar^-a■^'*. 

RaiKP /.l^Sv'a) to tlie BccDnd jowct. 



ALOEBSA. 61 



EVOLUTION OF SURDS. 



92. Rule. Extract the required root of the coefficient, 
md then multiply the fractional exponent of the surd by 
;he fractional exponent of the root 

Example^ Extract the square root of Qv^oT. 

Here the square root of 9 is 3, and the fractional expo- 
Qent of the surd is i, which we are to multiply by ^ the 
exponent of the root, which gives ^ ; hence the quantity 

sought is 3(ab)*' 

1. Extract the square root of 9^3* Ans. dJ^/s 

2. Extract the square root of 36 v^2. Ans. &i^2 

3. Extract the cube root of 84/5. Ans. 2^* 

4. Extract the cube root of 27>s/7« Ans. 3x7^. 

5. Extract the fourth root of 644/4. Ans. 2 x (256)^^ 

EQUATIONS CONTAINING 8UBDS, ETC. 

93. In equations containing surds, before the solution 
can be effected, the surds must be cleared away; to effect 
this, transpose all the terms which do not contain surds to 
(Hie side of the equation, and the surds to the other, then 
laise both sides to a power denoted by the index of the 
BQid, and if there was only one term containing a surd, the 
nird will be cleared away, if there be more than one, the 
operation must be repeated. 

If an equation appear under the form xz±za^ x±:h, 
or 3^''z±zaaf*^Ci it can be solved as an adfected quadratic, 
by solving first for.the power in the second term, and then 
for the quantity itself. 

Example. Given Va;+9=v^a?+]. 

Squaring both we have a; + 9=ir + 2 v^ -^ L 

Bj transposition 2V^a;=ft 

.-. V^=4. 

And squaring a;=16. 

Example 2. Given !x^•\'0^:sl^2, to find the value of a?. 
Here the equation comes under the form ic^'+ar^rzc, since 
the exponent of the first term is double its "poY^et \tv \\i<^ 

second; hence we mv^t solve for x^. Tlie opeiraXVoTV X«*^ 
teas follows: ' . f ^^ 



im. 



a*+3!*-}- J=— , by completing Bquare. 
a'^+^;=:t— , by extracting tlie toot. 
.'. x^^8 or — 9, by franspositioij. 
and a:'=G4 or 81, by squaring, 
hence a-— 4 or 34/3, by extracting the cube root. 

KXEBCISES. 

1. Gi¥en ■/3jH-'1-5, to find j-. Ans. jtsjl 

2. Given v'4+5.t=2+. JSJ; to find *. Ans. «=;12, 
a GiTen ^2.c+10+4=8i to find J. Ans. »-^27. 

4. Given ^— +5=7; to find x. Ana. ar=l 

5. Given J4j:+17+G^x+S=8^^+3; to find «. 

Anfcic=l& 

6. Given. •J'ic-^t/j: 7 =—7= : tafind^ Ans.ir=l6. 



I. Given*/? -1-3;+-:^ =-7^} to find k 



9. Given ^x:;:^+|=Xl±i> ; to find x. Ans. j=8 

,_ , 9a a 

10. Given V«+ Vir+ar; 'J-^, tofind a^. Ana. ar= -. 

11. Given^ie — V^=\'aj;, tofindj;- Ans. 3:= /j "^jy 

12. Given V'x+a+A/a— x=&, to find x. , • 

____^ Ana. :.= I C4a-ft')*. 

13. Given vT+W*^+^=l+a',tofind3r. Ans.iPsS. 



Oiven — - — = — ^_, to find x. Ana, !e=.^ 

Given ~7^~/t — 5- *•* fi"*^ J^- Ana. ^=tf ^^) 



ALGEBRA. 63 

18. CBven m;+2,J»zz24, to find x. Ana. ;rszl6, or dd. 

19. Given a:* — 2x^zzx, to find ar. Ans. x:=4, or 1. 

20. Given a^+x^zzS, to find a?. Ans. a?=32, or —243. 

21. Givep a»--2aii=rl33, to find a;. Ang. a:z=49, or '-|-\ 

22. Given (a;+ 12)*+ (a? +12)^=6, to find a?. 

Ans. a7=:4, or 69. 



ARITHMETICAL PROGRESSION. 

94. Definition. If a series of quantities increase or de- 
crease, by the constant addition or subtraction of the same 
q[iiantitj, then the quantities are said to be in arithmetical 
proffresdon ; and the quantity thus added or subtracted is 
called the common difference. Thus, 2, 5, 8, 11, &c., is an 
increasing arithmetical progression, where the common 
diflbence is 3; and 19, 17> 15, 13, &c., is a decreasing 
arithmetical progression, in which the common difi'erence 
U2. 

In general, if a represent the first term, I the last term, 
i the common difference, n the number of terms, and s the 
mm of all the terms^ the progression may be thus ex- 
pressed : — 

1st term, 2d, 3d, 4th, last. 

a, a+c?, a+2d, a^-^d, L 

or a, (3^— d?, a — 2c?, a — 3fl?, h 

Where the first represents sm increasing, and the second a 
decreasing series. 

In each of the series it may be observed, that the coeffi- 
cient of d is always one less than the number of that term 
in the series; hence the Tith or last term is equal to 

a-f n — \d, that is, Z=a+.w — \d> 

95. To find s ; observe that the series may be written 
hj beginning with the last term and subtracting d; thus I, 

l^ lr^2d% I — 3dt I — 4c?, I — 7j-r-l(Z, where it is obvious, 
tka^ Ir-.^^fi.^ld^a; writing the series thea in both forms, 
and then adding ; thus, 

«=Z+^— c?+^— 2rf+; — M,.. + l-^n — Ic?, 

where the number of terms on the second side ia eVAeii^'^ 
f/ .-. 2s=(a+0n, and s=z(a+l) |. In which a\i\i!iVA.tvA^ 



ing instead of I its talue a4-(n — 1)(/, we obtain 

Fiom the cqunlions found above, namely, 
l=a-^n^d, and g=[2a+ii— Irf]|, 
by substitution and reduction the following tbeorenn 
may be deduced, from which it will appear, that any thretf 
of the five quantities, a, d, I, n, s, being giren, the remain- 
ing two may be found. 



r 



].t. 


P+.X'-.) 

28 


2d, 
4th, a= 
6th, != 
8th, 1-. 
10th, rf 
12th, i 

Hth, 71 


«=,'-V''- 


3d, 
5lh, 


=Jv'(2i+<i)--&i.+l* 

_ 2s 


Jlh, 

llih, 
131h, 


25— Gun 
-.(.-I)' 


161b, 
IBll, 
17lb, 


«=i{i'(2»-j)'+ai.+<i-2.i. 



19th, s='_^+J(;+a). 20lh, s=[2f~«— ]rfl|. 

The above 20 theorems are suf&cient for the solution of 
any question that ean be proposed in arithmetical progres- 
sion; the pupil should deduce the theorems from tbetwi 
given equaiions, it being one of the best exercises in liUnt' 
analysis that can be given. 

KXBBCIBEB. 

1. The first term of an arithmetical series is 5, the o 

iDoa difference 4, and the number of terms 12; find lilt 

Jast term, and the sum of the sericB. 

-Afipljr Theorem 5th to fiud t, attAT:\\eOTem\?tVVVoSwA< 



ALGEBRA. 65 

2. Giyen the fint tenn 3, the last term 51, and the 
common difference 2, to find the number of terms and the 
som of the series. 

Substitute in Theorem 14th to find n, and in Theorem 
19th to find 8. Ans. n=25^ and 8=675. 

3. Given the sum of an arithmetical series 12^100^ the 
first term I, and the common difference 2. Find the last 
term, and the number of terms. 

Substitute in Theorem 8th to find If and in Theorem 
15th to find n. Ans. Z=:219, and n=110. 

4. A person was forty years in business, and spent the 
&nt year L.80, and increased his expenditure annually by 
L4. What did he spend the last year^ and how much 
during the whole forty ? 

Aiu. He spent the last year L.236^ and during the whole 
forty jears L.6320. 

5. The first term of a decreasing arithmetical series is 9, 
the common difference ^, and the number of terms 21; find 
&e sum of the series. Ans. 119. 

6. A man is to receive L.300 at twelve several pay- 
ments, each payment to exceed the former by L.4. He is 
willing to bestow the first payment on any one that can 
tell him what it is. What must the arithmetician have 
tor his pains ? Ans. L.3. 



GEOMETRICAL PROGRESSION. 

%. Definition. If a series of terms be such that each 
can he produced from the immediately preceding one, by 
multiplying by the same number, the series is called a 
geometrical progression ; and the series is an increasing or 
decreasing one^ according as the multiplier is greater or 
jess than unity. Thus, I, 2, 4, 8, 16, &c., is an increas- 
ing geometrical series, where the common multiplier is 2 ; 
a&d 243, 81, 27> 9^ 3, &c., is a decreasing geometric^ 
Knes, in which the common multiplier is ^. 

The common multiplier is called the ratioy and is com- 
monly represented by r; and if a be put for the first term, 
the general representation of a geometrical series will be 
the following: a, ar^ ar^^ ar^, ar\ &c. ; and if n be put 
for the number of terms, and s for the sum of the series, 
we will have 

Mult^lxhotb aides of(l) by r, and it becomes 



c 



B AZMSBRJi. 

Subtract (1) from (8), and thi 

IT — g^ar' — a. The other terms destroj one another. 
Hence ()^l>=a(r"— 1). 

... ^^'^^jfi, by dividirg by ;^. (3.J 

Thia IB the formula for s, when r is greater than unity 

but if r is less than unity, the first term of the series wil 

he the greatest, and the proper espression for s is obtained 

by subtracting (2) from ( 1 ), which gives 

I — w^« — ar". 

Hence (]—)■>= 3(1—1-). 

('--). . . , .:*" 

If now we represent the last tenn by ^ itia(I) 
that l=-ii-^K 

From these two equations, namely, l=ai*~*, and |3 
■ ■! or= -r-- ', the following theorems may be ik 

rived, 




The above theorems are given as exercises in Htvtl 
analysis, and should bM he deduced from the fith and fllll| 
which were previously proved. When r is less than'I] 
tlie term r" may be rendered less than any (juantit^ th^ 
can he named, however small, by taking » sufficiently 
great; so that (4) in the caso of n=infinity, will becoiu 

le expression for the sura ( 

BB continued to infinity. 

EXERCISES. 

7. Gtren the first teim of a geomeV.n'sA »ftTie%\, tlw 
'toinoa ratio 2, and the -nttm^w* ol tenaa W, ^ fesA. "'^ 
ftena and the mm of the 8ou«s- 




GbbtUvte in !]nieoreiii9 9tli and 6&, and we hare 

Ana. lzs&l% «=1023. 

2f. Tbe sum of a geometrical progression, whose first 
tefm 18 1, and hoM term 128> i9 255. What is the com* 
von. ratio? Ans. Theorem 4th girea r=2. 

3. Find the sum of the geometrical series, I9 i> i> j» &c., 
QOntSmied to infinity. Ana. 2. 

4U Find the sum of the geometrical series, whose first 
tan 13 If and common ratio §, when oontinned to infinity. 

Ans. 5. 

5u Find the sum of the geometrical series whose first 

term is m and commooi ratio ^ when continued to in- 

Ibltj. * Ans. mn. 

(k A servant agreed with bis master to serre him for 
tvelre months, upon this condition, that for his first 
month's serrioe he should receive a farthing, for the 
Kcond a penny, for the third fourpence, and so on. What 
lid his wages amount to at the expiration of his seryice ? 

Ans. L.5825, 8s. Sjd. 
7* There are three numbers in geometrical progression 
vLose sum is 53, and the sum of the first and second is to 
the sum of the first and third as 2 is to 5. Required the 
imnbers. Ans. 4, 12, 36. 

8. There are three numbers in geometrical progression. 
Che sum of the first and second is 15, and the sum of the 
list and last is 25. What are the numbers ? 

Ans. 5, 10, 20. 



GEOMETRICAL RATIO. 

97. The geometricsd ratio between two numbers is de- 
ennined by dividing the one number by the other. The 
pu^tient is &e value of the ratio. The number divided is 
oiled the arUecedenU and the divisor the conseqtcent of the 
itio. Thus the ratio of 9 to 6 is §=1^, in which 9 is the 
itttecedent and 6 the consequent, and the value of the 
atip is 1^, 

Ratio may therefore be considered as a fraction, the nu- 
nerator of which is the cmUcedentt and the denominator 
he consequent of the ratio. 

When the antecedent is greater than the consequent, it 
is called a ratio of greater ineqiuiliiy^ and when the aiA.^- 
iedent Is less than tbe xonaeqneiA, it is called a ial\Q ol 

96. SSnce ratios can be expressed by fractions, Wve^ ^«ii 



be compared with each other by reducing the fractions tv 
a common denoininator ; then that nill be the greater 
ratio which haa the greater numerator. Ratios are com- 
monly written by placing two points between the antece- 
dent and consequent; thus a: h expresses the ratio of a to if 
and is read a is to h. 

99. Proposii'mn ls(. A ratio of greater Inequality 
minished by adding the same quantity to both its terms; 
whereas a ratio of leeser inequality is increased by adding 
the same quantity to each of its terms. 

For — is a ratio of greater inequality, and if c he addd 
to each of its terras, it becomes — -— . Eeducingtheseratiol 

to a common denominator, the first becomes r 

and the second — — , which is evidently lesa tlmn lis 

first by Y — :■ 

Again, let be a ratio of lesser inequality; addetO 

each of its terms, and it becomes ; reducing t) 

to a common deoomlnator, they become . — r — , 

— p- -TT — , where the second is evidently greater than the ; 

100- Fropoiition 2d. A ratio of greater inequality il 
increased, and a ratio of lesser inequality diminished, li} 
subtracting the same quantity from each of its terms. 

Let — be a ratio of greater ineqniJity, take c from eaCl 
of its terms, and it becomes ; reducing these to i 

common denominator, the first becomes — ^ — . — , anJ 
the second — ^ — , which is greater than the former b] 

Again, let — be a ratio of lesser inequality; tabs i 
from each of its terms, and it becomes ; redodl) 



.(a-c) 



these to a common denominator, they become - 
and — ' — — — , wLicli U efideaWj \e«a V\iM:i the (ami 



iOJ. Frop. ScL A ratio is not altered bj multiplying 
or diyiding its terms by the same quantity. 
Let a : 5 be any given ratio, then it is identical with 



r=^ = r- Q- E. D. 

' MO O 



PROPORTION. 

102. The equality of two ratios constitutes a proportion; 
lius if a : & be equal to c: d, the two constitute a pro{>or- 
ion^ and are written thus; a:h: : c: d, or a: hz=c : d, 
md read^ aisto&ascistoc?; consequently^ since the 

ratio of a to & is 7 ^ and the ratio of c to c^ is -^^ we have 

r =: J, in which a and c are called antecedents, and 5 and d 

sonseqnenCs : also a and d are called extremes, and c and 
h means. Art. 14. 

103. Prop, 1. In every proportion the product of the 
Extremes is equal to the product of the means. 

For if a : 6=c : (f, then -=-, multiplying both sides 
by hd we have ad=zhc. Q. E. D. 

NoTB. If a : bsJ) : c, then b is called a mean proportional be- 
tween a and c, and c is called a third proportional to a and b; and 

[by Prop. 1) it is evident that l^=ac; hence b=^ac, or a mean 
proportional between two quantities, is the square root of their 
piodnct. 

Prop. 2. Two equal products can be converted into a 
proportion by making the factors of the one product the 
extremes, and the other the means. 

For if a€?=6c, dividing both sides by hd, we have -=- ; 
kence aihzicid. Q. E. D. * "^ 

104. Prop, 3. K four quantities be proportional, they 
are also proportional when taken inversely; that is, the 
second is to the first, as the fourth to the third. Since 

7=-:, 1-5-7 =:l-i--; •*•-=-> and hence hiaszdic, Q. E. D. 
a o a a c 

105. Prop. 4. If four quantities be proportional they 
are also proportional when taken alternately; that is, the 
first is to the third as the second is to the fourth. Eot 

mce-=^, if both Bides he maltipiied by - and t\ie co\sv- 

moD factors cancelled from the numerator aixd denomVn^- 



y 



70 

tor on both sides, we hare -=- ,: a: e:^h:d. Q. K. D. 

106. Frtip. 5. Wlieii four quan^ties are proportional, 
they are also proportional Iry c07/iposilioji; that is, the EQU 
of the first and second ia to the second, as the sum of ths 
third and fourth is to the fonrth. 

„. c a c , , , a+b c+d 

Since r=: J I r+^~Z + ''> nance -^=^— -. 
b d b ' d ' b d 

Therefore (i+t:6=c+(;:d. Q. R Dj 

107- Pvop. 6. When four quantities are proportional, 

they are also proportional b>/ division; that is, the difier- 

ecce of the first and second is to the second as the differ*- 

ence of the third aod fourth is to the fourth. 

Sincere-;, 7 — 1=-: — 1; hence -7—=— 3-. 

b d b d b a 

Therefore a—b : b=c—d .d. Q. E. D, 

108. Prop. 7- When four quantities are proportional,' 
fhey are also proportional by mixing; that is, the Gam of^ 
the first and second is to their difference as the sum of th 
third and fourth is to their difference. 

For Prop. 5th ^-=^. and Prop. 6th ^-^. 

...2±-^X — =— X— ■ 
b a — b d e — d' 

and therefore a+t:((—5=c+d:i>—rf. Q. E.1 

109. Frop. 8. Quantities ore proportional to tiu 
equimultiples. 

Let a and h represent any quantities and ma and < 
an^ equimultiples of them., then a : b=ma : ini/. 

For-=-^; therefore a: h=ma: m6. (%.KI 

Where m is any quantity, whole or fractional. 

1 10. Prop. 9. The like powers and roots of propoi 
quantities are proportional. 

Since ~ =2 . ^'= J .-■ a' •■ 6"=c- : if . Q. E. 

Where n may be either whole or fractional, and 
quently represent either a power or a root 

111. Prop, 10. If two proportions have the sames 
cedents, another proportion may beformcd, having the 
seqnents of the one for its antecedents, and the consequoil 
of the other for its consequents. 

For if a : 6 :: c : t£, and tt : e : ; c :/, 
lien ^ = J 1 and by inversion. 



AI^OBBSA. 71 

l»aice|x^:i=axf .'.5 = {; 

wherofore e : 6=/: di where e and /, the conse* 
pients of the one, are the antecedents, and h and d, the 
xmseqnents of the other, are consequents. Q. E. D. 

112. Prop. 11. If the consequents of one proportion 
be the antecedents in another, a third proportion will arise, 
liaTing the same antecedents as the former, and the tome 
Donsequents as the latter. 

Let a:&=c:d^ and l:<=:(i:/; then a:e=zc:f; for 

from the first ^ = j> and from the second - = ^ ; henoe 

mtdtiplying these e^aals tdgel^eir, -gyc -^^^^ f' *** " 
=: p ; hence a : e=ze :/. Q. E. D. 

113. Prop. 12. I^ there be any number of propor- 
fionals, as one antecedent is to its consequent, so is the 
mm of all the antecedents to the sum of all the conse- 
{bents. 

Let a : h=zc : d=e :fi=g : h ; then 

For ah=:bay and Prop. 1st ad=bc, afzzhe, and ahzzhg; 
therefore by adding equals to equals, we have ab-{'ad+af 
'{'ah^ba-{'hc+be-{'hg ; hence a(J>-^d-\-f+h)=zb(a+c 
^e-\'g) ; and therefore by Prop. 2d we obtain 

aib=ia+o+e+g:b+d+f+h. Q. E. D. 

114. Definition. When any number of quantities is 
n continued proportion, the first is said to have to the 
Uid the duplicate ratio of the first to the second, and the 
list is said to have to the fourth the triplicate ratio that 
lie first has to the second. 

115. Prop. 13. The duplicate ratio is the same as the 
"atio of the squares of the terms expressing the simple ratio ; 
md the triplicate ratio is the same as the ratio of the cubes 
)f the t^rms expressing the simple ratio. 

Let a: h^b : €=c : d, then a : c=a^ : h^. 
And a : cP=a : &*. 

A h t a h a a a c^ 

lence a : c::=a^ : b^* 

abtaaaac? 

ft ^c ^5 ""6 ^6' ^b •*• " "6»* 

»d hence a : d=€^ : ^. Q^ TL T>. 



n 



116. Prop. 14. The product oF the like terms ofanfl 
mmetical proportions are themselves proportional. 

ta:b=e: d, thea^=^. 

J n 

: lc=l: m, then 7=-. 
and consequently ~ ^ X 7 =4 xf X -■ 

o y^J h a h m 

hence aei : hfk=cgl : dJtm. Q. E. D. 

117- i'njp. 15. If there he threemagnitudes, n, 6j e.anJ 
other three, rf, e,/, Buch that a : 6=rf : e, and 6:c=e:/, 
then a : c=d : f. 

For 7 =~ and - =- , hence -- x - =- X -,. 
be c / bee/ 

and therefore 7=-;,. 

consequenlly a : c=d : / Q. R D. 



1 



the 

turn 
amc 



INTEEEST. 

118. Interest is the allowance given for the loan or 
forhearance of a sum of money, which is lent for, or bf- 
comes due at a certain time ; this allowance heing gens- 
rally estimated at so much for the use of L.lOO for a year. 

The money lent is called the principal, the sum pud foe 
its use is called the interest, the sum of the principal aiA 
interest is called the amownt, and the interest of L.100 fw 
one year is called the rate per cent. 

Interest is either Simple or Compound. 

Simple interest is that which is allowed upon the ongt' 
Dal principal only, for the whole time or forbearance. 

119. PfionLBM I. To find the simple interest of aD/ 
sum for any period, and at any given rate per cent. 

Let r=the interest of one pound for a year, jd= 
principal or sum lent, (=the time of the loan in yean, 
the interest of the given principal for the given time, 
the amount of the givenprincipaland ils interest Ibr^ 
le ( ; then we will obviously have the following relatioO' 
among the quantities ; 1 : pt=;r : i .", i=pii. (I.) 
and hence a=p+prt=p{l+rt). {2.) 



ALGEBRA. 7^ 

' ' ' ' 

and r=^=^ = - (3.) 

Bj means of the above five formul2e all the circumstances 
connected with the simple interest of monej are readily 
determined. But as the rules for the calculation of simple 
interest are generally given in reference to the rate per cent. , 
instead of the rate per pound, as above, the formulae may 
lie all changed into those relating to rate per cent., by mak- 

ing r represent the rate per cent, and substituting -— - in- 
stead of r throughout all the formulae; and the student is 
requested to write in words the rules which the five for- 
ibulse contain, by which he will be led to see the advan- 
tage of algebraic formulae. 

1. What is the interest of L.560 for 3 years, at 4^ per 
cent.? . Ans. L.75, 128. 

2. What is the amount of L.420 for 6 years, at 3 per 
cent.? Ans. L.495, J 2s. 

3. What principal laid out at interest for 5 years at 4 
per cent, will gain L.60 ? Ans. L.300. 

4. What principal laid at interest for 10 years at 3^ per 
cent, will amount to L.607, 10s. ? Ans. L.450. 

5. In what time will L.500 amount to L.800 at 4 per 
cen,t. ? Ans. 15 years* 

6. At what rate per cent, will L.200 amount to L.344iii 
18 years ? Ans. 4 per cent. 



COMPOUND INTEREST. 

120. In compoimd interest, the interest is added to the 
principal at stated intervals or periods, and this amount is 
made the principal for the next period. Hence if H repre- 
sent the amount of one pound at the end of one period, 
nnce this is the sum laid at interest during the next period, 
we will evidently have the following proportions to find the 
4Uxiount of L.1 at the end of any number of periods. 

1 : R-=iR : R'^, amount at the 2d period. 

1 : R=:R^ : R^, amount at the 3d period. 

1 : R:=zR^ : R^, amount at the 4th period. 

1 : jB=jff"~* : R^, amount at the nth period. 
Prom which it appears that the amount of one pound at 
the end of any number of periods is R raised to the power 
denoted by the number of periods, and it is plam l\idX. \\i^ 
amount of /? pounds will be p times the amoxwvt o? otiei 
pound; bence^ representing the amount of p pouixis >a'S A., 
and the number of periods by t, we will have 



I 



-y^ 



irlog. jl=lfig.;> + nog. ^. (1.) 

log.p=l«e.A-l\<,g.J!. (2.) 

1 log. i;=log. j(_lBg. p. (3.) 
log. J— lo g. 



(« 



Iog.ii=^5---j3J. (5,, 



121. The interest is generally conrerted into prindpa 

yearly, but sometimes half-yearly, and aomelimea even 
quarterly. If r represent the simple interest of L.l for* 
yeai', and » the numW of years for which the calculating 
is to be miide, then whea the inlerCBt is conTertiMe inb^ 
principal half-yearly, Ji and ( will have the following viJues: 

Jl-=l-\- ■=, (=;2re; and when it is convertible ([Qartedff 

^=1+ ^, t=4n; also J?=l+ ^, and t=mn. when tl 

interest is convertible into principal m. times per amran. 

nxEncises. 

1. tVhat will be tlie amount of L.1000 in ten years, kt 
H per cent, compuutid interest? Ans. L.1628, 17s. 9^4 

2. What principal laid at compound interest will amousl 
to I<.700 in eleven years, at 4 per cent. 1 

Ans. L.454, Us. I|A 

3. In what time will L.365 amount to L.400, at 4 pM 
cent, compound interest? Ans. 2 years 122 da» 

4. At what rate per cent, compound interest will LJO 
amount loL.63,lGs.3jd.injive years? Ans. 5 per ceni 

f). In what time will a sum of money double itself, 4 
a per cent, compound intetest? Ans. 14'3yeHl 

6. In what time will n sum of money double ilsetfi B 
4 per cent, compound interest. Ana, 17"67yeaM 

7- In wlmt time nt compound interest, reckoning 5 pc 
cent, per annum, will L.IO amount to l/.lOO? 

Ans. 4719 yeai*. 

a. What will be the compound interest of L.lOO for 
f weKe years at 4 per cent., if the interest be payable year- 
ly? what if payable half-yearly? and what if payablt 
quarterly ? 

Ans. L.60, 2a. OJd., L.60, 16a. lOid., and L,61, 48. 51ill 



ALGK&HA. 7d ' 



ANNUITIES. 

122. Ammuitxes signify anj interest of money^ rents, 
or pensions, payable from time to time, at particular pe- 
riods* The most general diTision of annuities, is into an- 
nuities certain^ and annuities contingent ; the payment of 
the latter depending upon some contingency, such, in 
particular, as the continuance of a life. 

Annuities have also been divided into annuities in ^jo«- 
Bodon, and annuities in reversion, the former meaning 
Mch as have commenced, or are to commence immediately, 
and the latter such as will not commence till some particu- 
lar future event has happened, or till some given period of 
time has expired. 

Annuities may be farther considered as being payable 
yearly^ half-yearly, or qtiarterly. 

The present value of an annuity is that sum, which 
being improved at compound interest, will be sufficient to 
piy the annuity. 

The present value of an annuity certain, payable yearly, 
md the first payment of which is to be made at the end of 
I year, is computed as follows : — 

Let the annuity be supposed L.1; the present value of 
^ first payment is that sum in hand, which being put to 
nterest, will amount to L.l in a year; in like manner, the 
iRsent value of the second payment, or of L.l to be re- 
vived two years hence, is that sum, which being put to 
nterest immediately, will amount to L.l in two years, and 
io on for any number of years or payments ; and the sum 
)f the values of all the payments will be the present value 
if the annuity. 

123. Let the interest of L.1 for one year be represented 
>7 r, then L.l will amount to 1+r in a year, and the sum 
kat will amount to one in one year, which call x, will evi- 
leatly bear the same proportion to L.l, that L.l bears to 
i+r; hence we have the following proportion: — 

x: 1=1 : l + r; 

.*. ar= — , value of 1st payment. 

In the same manner, that sum which in two years will 
imoont to L.1, is evidently that sum which in one year 

wU amount ^o -—-. Stating the proportion so lliat Wi^ 

oaDtitf sought majr stand last, we have 



:1 : — , present value of lat pajmentl 

= 7-7 'cij:"-.!' preaent value of 2 

= T- — ^ : 7T~~ra. present value of Zd payment. 



J24. Tlie present vnlue of an annuity of L.l for n yea 
y. IS therefore the sum of the series. 

This is evidently a geometrical series, in fvhicli the fii 

term is :^ — . and the common ratio is also — - ; heiicefini 

ing its sum as in geometrical progression, and putting ji_ 
the sum (hat is the present value of the annuity, vfb hart 

^= f^r + O+ry + (l+if + ■'■ (f+0-- (i 



(I.)-(2.) 
C30x(l+M 



(3 



125. If the annuity is to continue for ever, then »fc 

leomes infinite, and also (1-|- ?■)■"; hence t— — loaybecffl 

I Bidered as 0, and therefore we have for the present toIi 
t of an annuity of L.] , payable for ever, 

p= -, value of a perpetuity of L.l. 

It is plain, that if the annuity be a pounds instead ofM 
'■ will just bi! a times aS great as before; and therefore l! 
esent value of an annuity of a pounds, payable for » je* 
jiriilbe 



ALGEBRA. 77 

iuid that of a perpetuity of a poonds will be 

a 

I 

126. When an annuity is only to commence n yean 
hence, and then continue for t years, it is called a deferred 
annuity, and it is plain that its present value will be the 
difference between the present value of an annuity to con- 
tinue for ^ 4- n years, and another to continue for n years ; 
but we have seen (5) that the present value of an annuity 

of L.1 to <;ontinue t+n years, is ,, ^ , and that 

• '' r(l+r)i+n 

the present value of an annuity to continue n years is 
rfi-L v> * *^® difference of these expressions is therefore 

^e present value of an annuity commencing n years hence 
sad continuing afterwards for t years; reducing these ex- 
pressions to a common denominator and subtracting. We 

™^® V^ — ^: — > and therefore the present value of an an- 
unity to commence n years hence, and afterwards to con- 
tinue for t years, is p=z j- — t^t~* ^md if the annuity be a 

pounds per annum, instead of one, it is plain that the whole 
result will be a times as great ; therefore the present value 
"of an annuity of a pounds per annum deferred for n years, 

and then payable t years, is p— °*-V ^^ i 

r(l+ry+» 

If the annuity be payable for ever after n years, then t, 
and consequently (l+r)' become infinite, dividing both 
numerator and denominator of the above expression by 

(1+r)', and observing that — — r- becomes 0, we have for 

tke present value of a perpetuity of a pounds deferred for 

n years, »= -j- -.-. 
^ ^ ^ r(l+r)» 

127. To find the amount of an annuity left unpaid any 
number of years, at compound interest. Let A be the an- 
nuity, then the amount of the first payment which is fore- 
borne for n — 1 years will be ^(l + r)**"^; of the second for 
«— 2 years will be -4 ( 1 + r)"-'* ; &c. 
.•..the whole amount=^(l + (l+r)+(l+r)« + , &c.) to 
n terms; ^/ \ 

or the amount = 7 { ( 1 + r^—l h Art 96. {^?>.) 



k 



Ex. 1. What is ttio present Tfllue of apentlon of L.IOI^ 
payable yearly, for 20 years, at 5 per cent, compound in- 
terest ? Ana. L.1246, 4s. SJd, 

3, What is the present value of a perpetual annuity a 
L. 100, payable yearly, interest al 5 per tent. ? Ana. L.2OO0; 

3. What is the present value of an annuity of L-ICK^ 
payable half-yearly for 20 years, interest at S per cent pa 
annum, also payable half-yearly ? Ans. L.125.5, 2a. 1(^4 

4. What is the present value of a perpetuity of L.10 
per annum, payable half-yearly, interest at 5 per cent, peg 
annum, being also payable half-yearly? Ans. L,20u(t 

5. What is the present value of an annuity of I^lOOtfl 
commence JO years hence, and then continue for 30 yeai^ 
interest at 4 per cent. ? Ans. L.l 1 68, 3s. 7^0. 

6. What is the present value of an annuity of L.50, 1 
commence 8 years hence, and then to continue foi A 
years, interest at 5 per cent. ! Ans. L.589, 12s. 89 

7. In what time will a pension of L.50 amount to 
l,.100O, interest at 5 per cent. ? Ans. 14-2 yearf. 

8. To what sum will an annuity of L.24 amount in f* 
Tears, tvhea improved at 5 p?r cent.? 

Ana. L.793, llg.8 



L 



1. It is required to divide each of the numbers 11 a 

17 into tvta parts, go that the product of the first parted 

each may be 45, and of the second 48. Ans. 5, 6, and^ft 

2. Divide each of the number321 and BO into twopaiU 
so that the first part of 21 may be three limes as great c 
the first part of 30, and that the sum of liie squares of fV 
remaining parts may be 585. Ans. 18, 'd, and 6, ft ^^ 

3. A gentleman left L.210 to three servants, to bedividid 

in continued proportion, so that the first shall have L4f^^ 
more than the last : find their legacies. ' 

Ans. L.120, I..60, and h» 

4. Tliere are two numbers, whose product is 45, ani & 
difiercnce of their squares is to the square of their diSiE! 
euue HB 7 is to 2: what are the numbers! Ans. and &■ 

5. A and B engage in partnership with a cnpitai ct 
L.lOO: A leaves his money in the partnership for3 monlbh 
and B for 2 months, and each takes out L.99 of cspitaiasf 
profit; determine the original conlribution of each, 
ui Aaa.Ah.4&,i • ' 



PLANE GEOMETRY. 



GsoMETBY is that branch of Mathematics which treats of 
the properties of measurable magnitudes. 

Magnitudes are of three kinds, viz. lines having length 
onljy surfaces having length and breadth^ and solids having 
length, breadth, and thickness. 

Thai branch of Geometry which treats of lines And sur- 
£ieet is called Plane Geometry, and that which treats of 
the properties of solid bodies is called Solid Geometry. 

DEFINmONa 

1. A point is that which has position but no magnitude. 

2. A line is length without breadth. 

3. The extremities of a line are points. 

4. A straight line is that which lies evenly between its 
extreme points. 

5. A w,perficies is that which hath only length and 
beadth. 

6. The extremities of a superficies are lines. 

7- A plane superficies is that in which any two points 
Mng taken, the straight line between them lies wholly in 
(hat superficies. 

8. A plane rectilineal angle is the inclination of two 
ttraight lines which meet in a point, but are not in the 
tame straight line. 

Note. When there A 

in levaral angles at one 

punt, as at B, each of 

^ angles must be 

ouned by three letters, 

and the letter at the 

angular point must be 

fbeed between the 

other two ; thus, the 

sngla formed by the lines AB and BD meeting in the point B, is 

called the angle ABD or DBA; also the angle foi*med by the straight 

Jmes DB and BC^is called the angle DBC or CBD; but when there 

is only one angle at the point, as at "E, it may be caWed ^Irn^V^ >^« 

angle at £1 




9. When a straight line Etanding on ati- 
>ther straight line makes the adjacent 
angles equal to one andihcr, each of the 
angles is called a riffkl angle; and the 
Rlriiifiht line which stands on the other is - 
calk'd a pei-pendiculnr to it. 



10. Aa dbluse angl 
greater than a right ai 



5 that which i 




11. An acute angle is that which is less 
than a right angle. 

12. A term or hovnd'iry is the extremity of any thing. 
I.S. A_;^nisthat -which is enclosed by one or inei 

boundaries. 



14. A circle iit a plain figure 
hounded hy one line, which is 
c-illed the clrcurafeience, and is 
such that all straight lines drawn 
from a certain point within it to 
the circunifweuce are equal to 



15. And ihis point is called the r,enirt. 

16. The tfinmettr of a circle is a straight line dni 
through the centre, and terminated both ways by the tj 
cumference. 

17. A nemicirale is the figure contained by a H 
meter and the part of the circuiu fere ace cut off by ti 
diameter. 

IS. A straight line drawn from the centre to the drcmi* 
ference of a circle ia'callrd a iWi'ms. 1 

19. A straight line which is terminated both wajtU 
Ihe circumference, hut docs not pass through the centce^,^ 
called a chord. , 

20. The part of the circumference cut off by the choid 14 
lied an «rc. 
31. The figui 




y21. The figure hoi 
mail. 
ga. Jteefiliiieatji-^i 
^ht tiaes. 



hounded by the chord and arc is called A 
those which are contained hj 



N.AJU OEOMXTBT* 81 

23. Trilateral figures, or triangles, are contained bj 
three straight lines. 

24. Quadrilateral figures are contained by four straight 
lines. 

. 25. MtUiilateral figures, or polygons^ are contained bj 
more than four straight lines. 

26. An equilateral triangle has all its sides 
equal. 



27^ An isosceles triangle has two equal 
ades. 

28. A scalene triangle has three un- 
cqualsides. 



29. A right angled triangle is that which 
has one right angle. 






30. An obtuse angled triangle is 
that which has one obtuse angle. 

31. An acute angled triangle has 
ftU its angles acute. 



32. Of four- sided figures, a square is that which 
has all its sides equal, and all its angles right 
angles. 





33. A rectangle is that which has all its 
angles right angles^ but all its sides are 
Bot equal. 




34. A rhomhus is that which has all its 
sides equals but its angles are not right 
sngles. 

35. A rhomboid is that which has its 
opposite sides equal to one another^ hut 
^ its sides are not equal, nor are its 
angles right angles. 
V 36. All other four-sided figures besides these are called 
trapeziums. 




n PliARK GBOMETRr. 

37- Faral/rl etraifffil lines are sach 

ae, being in the same plane, and being 

produced ctct bo far both ways, do 
not meet. 

3B. A pnralfflnffjvim is a four-sided figure whose oppo- 
site sides are parallel. 

^!l. A ]>o«l?tIate requires us to admit tlie possibiliiy of 
doing Bomefhing, witliout feeing sboirn how to do it. 

40. A pfoposition is a distinct portion of science, and ik 
eilher a prMeni or a llieoyem. 

41. A prdtlrm, is an operation proposed to be perfoimcA 

42. A theorem is a truth which il is proposed to prore. 

43. A lemnia is a preparatory proposition to render whit 
follows more easy. 

44. A corollary is an obvious consequence resulting fno 
a preceding proposition. 

45. A scholium is an observation, or remark upon some- 
Ibing preceding it. 

46. An ariom is a self-evident trutL 

47- The side opposite to the right angle of a right-angM 
triangle is called the h>/pQtemi*e ; one of the sides alniutths 
right angle is called the basf; and the remaining side ii 
called the perpendicular, 

48. In a triangle vrhich is not right-angled, any aAe 
may be called the ba«f; the intersection of the other ivo 
sides is called the vertex; and the angle at that point tbi 
vfTtk'ol angle. 

49. The space contained vrithin a figure is called ill 
mrfaee; and in reference to that of another figure vriA 
which it is compared, is called its arta. 

50. A polj/gon is a figure contained by more than fiiur 
itraight lines; when its sides are all equal, and also ill 
angles, it is called a ifffidar polygon. 

51 . A polj-gon of five sides is called a petdagotx; thai df 
six sides, a /ie.ca30M; that of seven sides, fi AiTifajon; ihst 
of eight sides, on octagon; that of nine sides, a noivagott; 
■hat of ten sides, a decagon; that of eleven aides, an unitr 
cagon; (hat of twelve sides, a dodecagon; (hat of fifiew 
sides, a qnindfcaijon. 

POSTULATKB. 

1. I*t it he granted tha( a alraight Hoe way be Jraini 
from any point to any other poinl. 

2. Let it be granted thnt a terminated straight line naf 

'e produced to any length in a straight line. 



PLANS QEOMMTBY. 83 

3. Let it be granted that a circle maj be described from 
Dj centre^ and with anjr radius. 

AXI01I8. 

1. Things that are equal to the same thing are equal to 
ach other. 

2. If equals be added to equals, the sums are equals. 

3. If equals be taken from equals, the remainders are 
qwla, 

4. If equals be added to unequals^ the sums are un- 
guals. 

5. If equals be taken from unequals, the remainders are 
nequals. 

6 Things which are double of the same, or equal things. 
He equal to one another. 

7* Things which are halves of the same, or equal things, 
le equal to one another. 

8. Magnitudes which coincide with one another^ that 
I, which exactly fill the same space, are equal to one another. 

9. The whole is greater than its part, and equal to all its 
arts taken together. 

10. Two straight lines cannot inclose a space. 

11. All right angles are equal to one another. 

12. If two magnitudes be equal, and one of them be 
reater than a third, the other is also greater than the third. 

13. If two quantities be equal, and one of them be less 
lan a third, the other is also less than the third. 

14. If there are three magnitudes, such that the first is 
reater than the second, and the second greater than the 
lird, jnuch more is the first greater than the third. 

15. If there are three magnitudes, such that the first is 
!ss than the second, and the second less than the third, 
rach more is the first less than the third. 

16. Through the same point there cannot be drawn two 
laight lines parallel to the same straight line without co- 
iciding. 

EXPLANATION OF SYMBOLS. 

. means angle. 

s „ angles. 

I „ right angle. 
'/.« „ right angles. 
= „ equal to. 

II „ parallel to. 
^ „ triangle. ■ . 

„ inaDgles, j 



means parallelogram. 

I „ straight line. 

::^ „ greater than. 

^ „ less than. 

4. „ perpendicular. 

„ because. 

,, therefore. 



0^8 



w 



Phoposition I.— TnKoHBM. 



Tlie angles ACD and DCB, 
vrhicli one straight line, DO, makes 
with another. AB, on one side of it, 
are c;ithcr two right angles, or are 
together equal to two right angles. 

If the U ACD and DCB be equal, 
each of them is a j'i, (Def. 9.) ; but if they are not equali 
coDceive CE to be drawn ->- to AB, then the La ACE and 
ECB are two t'Ls; but tbe three is ACE. ECD, and 
DCB. are together = the two La ACE and ECB, and qIm 
to the two Is ACD and DCB; .-. the two is ACD and 
DOB are together = the two La ACE and ECB, but the 
two Ls ACE and ECB are two I'Ls; .: the two La ACD 
and DCB are together equal to two r'/.s. Q. E. D. 

Cor. 1. AH tbe angles that can be formed at the point 
C, in the straight line AB, on one side of it, are together 
equal to two right angles. 

For the L ACD ts = the two Ls ACE and ECD, K 
that tbe Eum of the Ls is not increased hj drawing the 
{ EC, and in the same manner it ma;^ he shown the sun 
of the Ls would not be increased by drawing anj numbei 
of \s through the point C> 

Cor. 2. All the angles formed at the point 0, on tbe 
other side of tbe line, by any number of lines meeting ii 
C, will also be equal to two right angles. 

Cor. 3. Kence all the angles formed round a point ij 
any number of lines meeting in it, are together equal '' 
two right angles. 

SoBoL-iun. For tbe purposes of calculation, tbe ea- 
cumference of every circle is supposed to be dirided 
3t)0 equal parts, called degrees, and each degree is 
posed to he divided into 60 equal parts, called min 
and each minute into 60 equal parts, called seconds. Tit 
grees, minutes, and seconds, are dislingulsbed by the fff" 
lowing marks : — 7" 3' 24", which is read 7 degrees, 3 Wl" 
DUles, aud 24 seconds. 

In the same manner all the 
are divided into the same numlier of di 
and seconds. Since then the circle entirely 
surrounds its centre, and is similarly sitn- 
Bted to it in every direction, tbe portion 
of the circumference intercepted between 
*ifo Jines druivn from tbe centre to the 
m/rura/erence, is the meaaute oS \\\e a."n^< 



s round about a, pMil 
, minul* 



©• 




PliAKX OKQUBTRY, 85. 

t the centre; thus the angle AOB is measured hj the 
itercepted arc AB^ and the angle COB is measured hj 
le arc CB. 

Since all the angles round a point are (Cor. 3.) equal to- 
e^er to four right angles^ and also to 300*, the numerical 
teasure of a rignt angle is 90^. 

Proposition II. — ^Thborem. 

B; at a point B^ in a 
laight line AB^ two 
ther straight lines^ CB 
ad BD^ on opposite sides 
f AB^ make the adjacent 
Bgles ABC and ABD 

pgether equal to two 

\^t Angles^ these two 
traig^t mies are in one and the same straight line. 
For if BD he not in the same | with CB, let BE he in 
lie same | with it ; then since CBE is a |» and AB makeii 
J with it, the two Ls ABC and ABE are together = two 
*U; hut the two Ls ABC and ABD are also together = 
iro r^Ls hj supposition ; .*. the two Ls ABC and ABE 
re = the two Ls CBA and ABD; take away the common 
ngle ABC, and there remains the Z.ABE = the L ABD, 
be less = the greater, which is impossiMe ; .'. BE is not 
a the same | with CB, and in the same manner it can he 
hown that no | can be in the same | with CB, except BD, 
rhieh therefore is in the same | with it. Q. E. D. 

Proposition III. — ^Theoresi. 

If two straight lines, AB and CD, ^ 
!Ut each other in the point E, the ver- 
ical or opposite angles, AEC, DEB, ^^ -e 

iie equal. 

For the two Ls AEC and AED, 
fihich the | AE makes with the | CD, are together equal to 
:wo r^Ls; and the two Ls AED and DEB, which the 

DE makes with the | AB, are also together equal to two 
^Ls; .*. the two Ls AEC and AED, are together equal 
to the two Ls AED and DEB ; take from each the com- 
mon Z.AED, and there remains the LAEC=DEB; in 
the same manner it may he demonstrated that the two 
U AED and CEB are equal. Q.. E. D. 





Proposition 

To make a triangle AEB, 
ivhose three sides shall be 
equal to the three given 
straight lines, AB, C, and 
D. 

From the centre A, with 
n radius equal to G, describe 
the circle EFII, (Post. 3), 
and from the centre B, with 
a radius equnl to D, describe 

a circle EGII; and from the — ^ ■ 

point E, where the circles 

cat each other, draw the |« AE and EB. (Post. 1), 

AEB will be the triangle required. 

Because AE is the radius of the circle EFH, and it wat 
described with a radius ^C, •'. AE is =C, and because 
BE is the radius of the circle EGII, and it was described 
with a radius =D, .-. EB is =D. Hence the a AEB 
has its three sides = to the three {s AB, 0, and D. 

Cor, I. If the \s C and D were equal, the triangle woall 
be isosceles, 

Cor. 2. If the given lines, AH, C, and D, were att' 
equalj the triangle would be equilateral. 

Cor. 3. If C- and D were together less than AB, tba 
circles would not intersect, and the construction would b« 
impossible ; hence any two sides of a triangle are together 
greater than the third, which is established in a di&cent 
manner in (Prop. 13.) 

PflOPOsiTioN V. — Theohku. 

If two triangles, ABC and DEF. have two sides, AC 
and CB, and the contained angle ACB in one, respectireljr 
equal to two sides, DF and FE, and the contained ai^ 
DFE in the other, the triangles are equal in every i^iped. 

For conceive the point C to 
be laid on the point F, and 
the |CA on the [FD, then, 
■.■ these lines are =, the point 
A will coincide with the point .^ 
D. And since CA coincides 
with FD, and the iC is = the LF, the line CB wiO 
fall on the line FE ; and ■.• CB and FE are =, tdr 
point D will coincide with the point E. Htiice sinM 





(be pointf A and B coincide with the points D and E, 
the line AB will coincide with the line DE^ and the a 
iBC will coincide with the aDEF; .*• the two a« are 
squal, and have all the parts of the one equal to the cor- 
responding part^ of the other, namely the side AB = the 
tide DE, the LA = the LD, and the LB = the Z.E. 

Q. E. D. 
Pbo?osition VL — ^Theorbm. 

In any two triangles, ABC, DEF^ if two angles, A and 
B in the one, he respectively equal to two angles, D and 
E in the other, and the sides AB and DE, which lie be- 
tween these equal angles be also equal, the triangles are 
^oai in all respects. 

For conceiye the point A to be 

lud on the point D, and the side 

AB on the side DE, then, *.* these 

Knes are equal, the point B will ^ / \; Br 

coincide vrith the point E. And 

» AB and DE coincide, and the AA is = the LD, the 

ride AC will coincide with DF, and for a like reason BC 

will coincide with EF. .•. since AC fails on DF, and BC 

im EF, the point C must coincide with the point F ; and 

•'.the two A^are in all respects equal, having the other 

ides, AC, BC, = the two, DF, EF, and the remaining 

LO s= the remaining Z.F. Q. E. D. 

Pboposition VII — Theorem. 

In an isosceles triangle, ABC, the 
iDgles A and B, opposite the equal 
does BC and AC, are equal. 

For, conceive the Z.C to be bisected 
)y the line CD, then the two a« ACD 
Bid BCD have ACzzBC, and CD 
iommon to both, and the LACD = ^' 
ie ABCD ; .*. they are equal in every respect, (Prop. 5), 
uid have the Z. A = the LB, *.• they are opposite to the 
»mmon side CD. Q. E. D. 

Cor. 1. If the equal sides AC and BC be produced to 
B and F, the angles E AB and ABF on the other side of 
^e base will also be equal. 

For, the two Ls CAB, BAE, are together = two /Ls, 
iProp. 1), and the two Ls CBA and ABF are also toge- 
ler = to two t^Ls; .•. the two Ls CAB, BAE, are =: the 
^0 Ls CBA, ABF, (Ax. 1) ; and it was proved i\i^\ l\v<^ 




nf 



pt iN^' «ft9uiiMr. 



» 







id CAB and CBA are equal ; .-. the angles EAB a 
ABF are aha equal, (As. 3.)- 

Cor. 2. The line tliat liisecls the vertical angle of I 
iaoscelps triangle also bisects the baae. and cuts it at * 
anftles. 

For, the as ACD and BCD w^re shown to be e 
in every respect ; .-. AD is = BD, and iADC is = BD( 
and tbey are adjacent Ls ; hence (Def. 9) each of them ; 
a right angle. 

Cor. 3. Every equilateral triangle is also equiangular. 

pKOPOSiTioN VIII, — Theorem. 

If two angles, CAB and CBA, of a triangle be equal, t 
sides, CB and CA, opposite them nill also he equal. 

For, if AC be not = CB, let AC ^ 

be ::^CB, and let AD be the part of 
AC that is = BC, join BD, (Post. 1 ) ; 
then the ah ADB. CBA, have AD 
=CB, and AB common; and the 
£DAB contained by the two sides of 
the one is = the iCBA contained by 
the two sides of iheothec; .-.(Prop.C) Ai__ 
the aABD is = the aABC, the 
less := the greater, which is impossible ; .-. AC is not :?^I 
and in the same manner it may be shown it ib not k 
Lence AC is = BC. Q. K ? 

Cor. Every equiangular triangle is also equilaleraL 

Proposition IX. — Theoeksi. 

If two triangles. ABC and DEF, have Ibe three ndca' 
i respectively equal to the three sides of the otT 
ingles shall be equal in all respects, and haTe d 
equal, that are opposite to equal sides. 
Let ABC and DEF be two 
having AC=Dr, CB=FE, 
and AB=DE, and let AB and 
DE be the sides which aj-e not &-■ 
less than any of the olhcis. Con- 
ceive the side DE to be applied 
to the side AB, so that the point 
D roay coincide with A, and Ihc 
Uae DE with AB, then the point E will coincide with] 
jDE is = AB ; but \et t\ie levte^^ ^a\\ va vlw Q^poii 
rtion, from C as at G, jtim GC -, \Ui:Ti-.- CV,vfcT;-1! 



.A 



PLANS OEOMBTRY. 89 

and FE is = "BG, being the same line in a different posi- 
tion, CB is =BG, (Ax. 1); .-. the Z.BCG is = LBGC, 
(Prop. 7); again, ••• AC is=DF and DF is =AG, .-. AC 
is =AG-, and hence the L ACG is = the Z. AGC ; .-. also 
the whole AACB is = the LAGB, (Ax. 2); but LAGB 
is the LDFE in a different position; /. the LACB is =: 
the LDFE ; and since AC, and CB, and the LC, are re- 
spectively = DF and FE and the LF, (Prop. 5), the LA. 
is = the LD, and the LB is =: the Z,E. Hence the angles 
are equal that are opposite to the equal sides. Q. E. D. 
Cor. 1. The areas of the triangles are also equal. 
Cor. 2. To make at the point F, in the straight line 
DF, an angle equal to ACB. Construct a A^FE, 
(Prop. 4.) having its three sides equal to the three straight 
lines AC, CB, BA, namely FD=AC, FE=CB, and 
ED=AB; the LDFE will be = the LACB by the pro- 
position. 

Proposition X. — ^Theorem. 

If a side BC of a triangle ABC be produced to D, the 
exterior angle A CD" will be greater than either of the in- 
terior opposite angles CAB or ABC. 

Conceive AC to be bisec- 
ted in E, and BE joined, 
and the line BE produced 
to F, so that EF may be 
=BE and join FC, and 
prodace AC to G ; then the 
A« AEB and CEF have AE 
and EB, and the contained 
LAEB in the one = CE 
and EF, and the contained 
iCEF in the other.; .-. (Prop. 5) the Z.EAB is = the 
iECF; but LACD is greater than LECF ; .-. (Ax. 13) 
the Z,ACD is greater than the ABAC. 

If the side BC were bisected, and a similar construction 
made below the base, it might be shown in the same man- 
ner, that the iLBCG, which is = the LACD, (Prop. 3), is 
7^ the LABC; /. the LACD is ;:^ either of the LsCAB 
or ABC. Q. E. D. 

Cor. Any two angles of a triangle are together less than 
tiFo right angles. 

For the LACD is -z^ the LBAC, and if the LXCB \i^ 
added to each, the two IsACD and ACB (thatia, Ivjo 
^« Prop. l.)are:^ the LsBAC and ACB ; and l\ie sam^ 
^Jght be shown of any other two angles. 




PnoPosiTtoN XI. 

The grealer side of e?erj triangle has the greater angle 
opposite lo it. 

Iftheside ACof the AABC 
be :^ the side AB, tlie i.ABC 
willbe^tiie i.ACB. 

For make ADrrAB, anil join 
BD. then tlie /.ADB is = the 
Z,ABD(Prop.7); but the /.ADB 
is z:^ the iACB (Prop. 10) ; .-. 
the i.ABD is z^ tbe IaCB (As. 13), still 
- whole /.ABC z:^ the Z.AC'B. 




Q. EB 
the graatei 



Cor. The greatest aide of any triangle hai 
angle opposite lo it. 

Proposition XII Theo«em. 

Ifthe angle ABC of the triangle 
ABO be greufer llian the angle 
ACB, the side AC opposite the 
grealer angle will be greater than 
the side AB oppnsile llie Ices. ■_ 
Or the greater angle of every triangle has the greater ai4f 
opposite lo it. 

For if AC he not z^ AR, it most either be =: it 
less; AC is not = AB, for then the i.B would be = i 
LQ (Prop. 7). which it is not; neither is AC -:iAB,J 
then the IV, would be ^i: the Z.C, nhieli it is not {PB 
II)..-. ACisi^AB. 

Cor. Tlie greatest angle of eyery triiingle has tbe gresh 
Bide opposite to it. 

pKOPosiTioN XIII. — TnaoaBM, 

In any triangle ABO, the snmofany 
two of its aides, as AB and AC, is greater 
than the Temainiog side BO. 

Produce AB to D, so that AD may be 
= AC, and join DC; then ■.■ AD is = 
AC (by Const.), the L ACI> is=the i. ADC 
(Prop. 7), .-. in ihe AOBC tbe i-BCD is 
z:^ Ihe /.BDC. hence the side BD is::^ 
tbe side BC (Prop. 12); but BD is = AB and AC, •■■ i. 
is = AC, .-. BA and AC are together zp- BC (As. 12.) 
Q. E 

Cor. Tbe differenee of two sides ofa triangle is leMtl 
the third side. 

For,sinceAB and ACare:^BC,ifACbc taken from« 
there remains AB z^ the ditfctence of BC and AC, (Al. 




FLAKE GEOMETBY. 



91 





1-E 



Proposition XIV. — Theorem. 

If two triangles, ABC, 

DEF, have two sides, AB, 

BC, of the one respectively 

equal to, DE and EF, two 

sides of the other, hut the 

angle ABC, included hy the 

two sides of the one, greater 

than the angle DEF, includ- 
ed by the corresponding sides 

of the other; then the side 

AC is greater than the side DF. 

Let ABG be the part of the Z. ABC, which is z= DEF, 
and let BG be = EF or BC. Then the A^ ABG, DEF, 
are equal in all respects, (Prop.5), and have the side AGr= 
DF. And as BC and BG are =r, the L BGCis = the L BCG 
(Prop.7); but the Z.BCG is^^-theZACG, .-. also the LBGC 
i8:7-the LACG, (Ax. 12); much more then is the LAGC 
T^ the L ACG, and hence (Prop. 12), the side AC is 7:^ 
AG, and .-. also z:^ its equal DF (Ax. 13). Q. E. D. 

Proposition XY. — Theorem. 

If two triangles, ABC, DEF, 
iave the two sides AC, CB of the 
one respectively equal to two sides 
DF, FE of the other, but the re- 
maining side AB of the one greater 
than the remaining side DE of the A.^ 
other; the angle ACB will be 
greater than the angle DFE. 

For, if the L ACB be not ^^ the ZDFE, it must either 
he equal to it or less; the ZACB is not ■= DFE, for then 
(Prop. 5), the base AB would be = DE, which it is not ; 
neither is the ZACB .^ the ZDFE, for then (Prop. 14), 
the base AB would be ..^ the base DE, which it is not, 
.-. the ZACB is z;^ the ZDFE. Q. E. D. 

Proposition XVI. — Theorem. 

If a straight line, 
HF, fall upon two 
Other straight lines, 
AB^ CD, and make 
the alternate angles 
AEF, EFD, equal 
to one another, the 
straight lines AB and CD are parallel. 




B D 





92 PLANE GBOMETBY. 

For, If AB and CD lie not || , they will meet ^hea p 
duced, either towards B, D, or A, C; suppose that the 
meet towards B, D, in the point G, then EGF is a ^, an 
itseilerior /.AEF is ^:^ the iEFGyrop. JO); but Z.AK 
is = /.EFG, .■. AB and CD do not meet towards B, T 
and in the Siime inauuer it mny be shown that they do n 
meet towards A, O, .-. (Def.-37), AB is parallel to CD. 
Q. E. I 

Cor. 1. If the esterior anffle HEB be equfd to the ii 
rior angle EFD, AB is jiarallei to CD. ^_ 

For, since iHEB it = i-EFD, and also to Z.AEi 
(Prop. 3), .-. /.AEF is = Z,EFD, and they are alteiaat 
angles; therefore (Prop. 16), AB is parallel ti 

Cor. 2. If the two angles BEF and El'D be togethi 
equal to two right angles, AB and CD are jiarallel. 

For, since the two U BEF and EFD are = two i*.U 
by (SopO. an* Ilie t™" ^* BEF and AEF are tngelher = 
two r'L» (Prop. I), the two U BEF and EFD are =: ' 
the two Ls BEF and AEF, and takinc; away the comBU 
angle BEF, there remaina the iAEF = Ibe iEFD.JU^ 
they are alternate angl^; hence AB is parallel to "CD 
tProp. 16.) 

Proposition XVIT.— Thkorem. 

If two parallel straight lines, AB ir. 
and CD, be cut by another line EF, in 
the points G and 11, the alternate 
angles AGH and GHD wilt be tqual, 
lh(^ exterior angle EOB vrili be equal 
to the interior opposite angle GHD; 
and the two interior angles BGH and GHD on the Mil 
aide of the line will be together equal to two right angl« 

If the ZAGII he not = the ZGHD, let LG be d«fH 
making the /.LOU = the Z.GHD, and produce LG to St. 

■.■ the LUiU. is = the ^GUD, and they are altenuW 
Id, .-. LM is II to C D, (Prop. 1 6). But AB is given || toCD- 

.'. through the same point G there are drawn tw 
straight lines LM and AB, {| to the same straight line CDi 
whicli is imposBihIe. (Ai. 16.) Hence the Z.AGH. " 
unequal to the ZGIID, tbnt is, it is equal to it. 

Then, sinc-e the /.AGH is = the lGIID, and also t» 
the iEGB (Prop. 3), the /EGB is = the lGUD. .: t 
eilerior Is equal to the interior opposite angle on the sal 
jade of the line, 

UAgitiD, since the ^EGB is = the Z.GIID, to eacb 
e equals add the i.lJOH, Xben ■will the two It 



^ t\vo i, BGII and GIID ; bat the i 

are t-.^-ether = two t'/b (Prop. 1) 

»d GIID are togetiier equal to two right 

h,(v.^. , Q. E. J>. 

-""■■"gilt lines. LM and CD. being cut by a 

■* "le t«,o interior angles LGII and GHC, 

« «ae of iiig (,^nj^g |jj^^_ logelher less than two 

™?e two lines will meet if produced far' 

" "(le where the angles formed are less than. 

y do not meet on tha* side, they must eitbi 
f will meet being produced on the other side , 
\, ftr then the two La LGII and GHC would 
= two rL't, whith they are not: neither do 
g produced towards M and I>, for then the 
E and GHD would be two /,« of a A. !"»d .-. 
" > (Prop. 10); but the font Lf LGH. GHC, 
1), are together = fonr I'Ls (Prop. 1). of 
, I.GH and GHC, are together leas than 
i two, MGH aud GIID, are ^^ two I'Ls, 
t CD do not meet towards M aud D, aud 
» been shown that tlwy are not 1|, they must 
(rartia L aud C. Q. E. D., 

pBOPOsiTiON XVIII, — Theori 
A a poi"* A to draw 
t line parallel to a given 
ine BC. 

lake any po'"' L>, jmn 

at llie iioiiit A make ] 

P^o 9). tIi«^D'\'i = 

C ^d prodLiue KA to F, ■.- the iEAD = thj 

bid they are alternate angles, .•■ (Prop. ]6J,^ 



I 



bence through the p 



een dravvn 



II to BC. 



e BC of o 
BC be P"""" 
). the e»te- 
ACD "i" 

** ana tl.e 



ighC Un^^ 



■ angles oi ^ 

Je are togethei c(iu,a\ lo t^io u^ 




^\\\. ■Mi'^fi. 



Throngh C draw (Prop. 18), CE || to AB. Then V 
Vis II to CE, and AC mecls tliem, Ihe alleraale /.sBACa 
ACE arc= (Pror- 17); and .' AB is 1| to CE, and I 
fells upon lliem, the exterior /.ECD is = the interioi 
ABCbuttheilACEis^ the iUAC,.. the whole iAC 
is = the two li CAB and ABC. 

To each of these equals add the ZACB, ,■. (be two 
ACD, ACB are = the three la CAB, ABC, and BC 
but the two Z» ACD, ACB are together = 
three angles CAB, ATl^;, and BCA arc togetlier equal 
two right angles. 

Cor. 1. If two angles in one triangle be equal to t 
angles in another triimgle, the remaining angles ofthi 
triangles are equal. 

Cor. 2, If one angle in a triangle he equal to an aiq 
in another triangle, the sum of ibc remaining angleS' 
each triangle are'equal. 
"■■ Cor. 3. If one angle in a triangle be a riglit angle, 1 
l^her tTvo angles are together cqnal to a right angle; il 
\ence each of them is an acute angle. 
I Cor. 4, Every triangle has at least ttvo acute angles. 
> Cor. 5. Hence from this proposilion, and (Prop. S), 
'angles hare two angles iu Ihe one equal to tf 
ingles in the other, and it corresponding side equal id eac 
e equal in al! respects. 



Prop( 



>; XX.— Thi 



If two lines AB 
and BC meeting in a 
point B, be respcc- 
ti»rly parallel to DE 
and EP meeting in a 
point E, the included 
angles ABC and DEF are equal. 

For join B, E, and produce BE to G. Since AB i* 

>E, and GB falls nn them, the Z,<'IiA is = the /.OB 

~ op. 17): foralikereasonthe ZCBG is :r the ZFH 

aking equals from equals, there remains the iABC 

LVEF. Q. E. 

pBOPosiTioN XXI. — ^Theorem. 

Pill figure he produced, I 
so forjued will be togelli 



buaJ to four right angles. 



FLANE GXOMBTJIT. 



9i 




Take any point o, 
and draw oh \\ to 
EK, oM II to AF, 
oN II to BG, oP II J 
to CH, and oQ || to 
DJ. Then since 
KA and AB are re- 
spectively II to ho 
and oM^ the La is 
= the La' (Prop. 
20) ; for a like rea- 
fton the Lb is =z the 
Lh\ Lc =z L<f, Ld 

= Ld\ and Le = 

U\ .'. the sum of 

an Ae Z« a, h, c, </, 

^; are equal to the 

wm of all the Z# a', 6', c', d\ and «' ; but the Ls a\ h\ c\d\ 

aad t' are together equal to four 7^ is (Prop. J, Cor. 3), .-. 

all the exterior Ls a, h, c, d, and e, are together equal to 

four right angles. Q. E. D. 

Proposition XXII. — Theorem. 

All the interior angles of any rectilineal figure are to- 
gether equal to twice as many right angles as the figure 
has sides, wanting four right angles. 

For (Figure to Prop. 21) erery interior Z.EAB, together 
with its adjacent exterior LBAK, are together equal to two 
^Ls; ,\ all the interior, together with all the exterior, are 
€qaal to twice as many /Ls as the figure has sides ; but 
ail the exterior Ls are = four r^Ls, (Prop. 21) ; .-. all the 
interior are = twice as many right angles as the figure has 
sides, wanting four right angles. Q. E. D. 

Cor. 1, All the interior angles of any quadrilateral 
figure are together equal to four right angles. 

Cor 2. If the sum of two angles of a quadrilateral figure 
be equal to two right angles, tlie sum of the remaining 
angles is also equal to two right angles. 

Proposition XXIII. — Theorem. 

Of all straight lines, drawn from the point A to the 
straight line BC, the perpendicular AD is the least; AE, 
which is nearer to the perpendicular, is less than AF, 
,which is more lemotej and there can only be Atclyjiv \.nso 




L 



9S VEotMB GSOMKVKT. 

equal strtiiglit lines, as 
AE and AD, one on 
each side of the per- 
pendicular. 

Since AD is -u Ic 
CU, the LADE is a 
y'L; .-.the /.AED is 
^ a 7'L rProp. 19 
Cor. 3); and .-.the side 

AD is ^: A E. Again, since tl.e Z AED is ^::r a I'l. the 
AEFia^a^'i, (Prop. 1), but tlie iAEU is ^ the L] 
AFE, (Prop. 10); .-.the ZAEF is p' the ZAFE, anil 
.'. the side AF is ^' the side AE. In the same 
it mny be shown, that AC is z^ AF. Again, if DB b» 
= EIJ, ABwillbe=AE; forinthe two A« ADB ' 
ADE, the two sides BD and DA are = the two sides 
and DA, and they contain equal is, for each of tbem 
r-Z; therefore AE=AB, (Prop.5): and besides AB,lhB« 
cannot be drawn any other line from the point A^AE|. 
for if it were nearer to the -I- it would be less, and if mo» 
remote it would be greater. Q. E. B 

Proposition XXIT. — Theorem. 

The opposite sides and i], 
angles of any parullelo- 
gram are equal to each 
other, and the diagonal 
divides it info two equal 
triangles. That is, 
DC=AB, BC = AD, 
ZDAB=ZDCB, ZADC=ZABC, and AADB=ADCR 

■■■ DC is II to AB, and DB meets them, the ZCDBii 
the ZABD, and :• AD is || to CB, and DB meets thelii 
the ZADB is = the ZCBD, .-. in the two A« ADB ani 
DCB, there are two Z» CDB and CBD, respeclively = 
two Is ABD and ADB in the other, and the uAe Dft 
Ijinp between the equal angles, is common to bodl. 
.-. (Prop, (i) DC is = AB, AD is = CB, ZDCB w=i 
DAB, and the ADCB is = the A^AD : also 
equal Z^ ADB and CBD, there be added the equal it 
CDB and ABD, the whole ZADC will he = the whoh 
ZABC. Q. E. D. 

Cor. I. The lines which join the extremities of two 
equal and parallel straight lines towards the same pEii1% 
are ibemsvives equal and parallel. 
For if CD and AB be equal aai ^aro\W, the A» CDB 




and ABD have two sides, CD and DB of (he one = fwo 
AB and DB in the olher, and Ihe contained Z.CDB is 
= the contained /ABD. (Prop, 17), -■, the base CB is 
= the base AD, and the iCBD is = the iADB, (Prop, 5), 
.-. AD and CB are ||, (Prop. 16,) 

Cor. 2. If one angle of a parallelogram be a right angle, 
all the other angles will he right angles, and the figure niU 
be a rectangle. 

Cor. 3. Any two adjacent angles of a parallelogram are 
together equal to two right angles. 



PaoPoaiTioN XXV.— TnEOiiK: 

Every quadrilatpral ABCD ivhich n 
has its opposite sides AB and DC 
equal, and also AD and CB equal, is 
a parallelogram, or has its opposite , 
sides parallel. 

For join DB. and then the two As DCB and BAD have 
fte side DC=AB and CB=AD, and the siiilt- DB com- 
mon ; .-. (Prop, 9) the A" '"'^ equal in all rt^pecls, and 

, have the Ls equal that are opposite to equal sides; .'. the 
LABD is = the Z.CDB. and hence DC ia || to AB; also 

I the /.CBD is = the Z.ADB, and hence AD is || to OB, 
(Prop, le), and ABCD is therefore a parallelcgram (Def. 
38). Q, E. D. 



P 



pROPOsJTroN XXVI. — Theorem. 



A B 



Parallelograms. DABC and FABE, 
Mid also triangles, CAB and FAB, ' 
itonding on the same base AB, and be- 
tween the same parallels, AB and DE, 
ue cqnal to each other. 

For, since DA and AF are re5pectively || to CB and 
BE, the i. DAF is = the Z.CBE (Prop. 20), also since ED 
Mb on the two \\b AD and CB, the Z.FDA is = the 
iECB, (Prop. 17); .■- the A« FDA and EOB hare two 
^in the one = two Ls in the other, and the aide AD = 
lie corresponding side BC, (Prop. 24); .-. the A^'DA is 
= tlie AECB, (Prop. 6); if now each of these = Ashe 
liiien separately from the whole figure DABE, there will 
ttmaia in the one case the iZZ?ABCD, and in the other 
4b £:Z7FABE, these i — 7 x are .-. equal, (Ax. 3); and since 
ttie ACAB is half of the one. (Prop. 2-4), and l\ie C^fiiiS 
lulf of the other, theaa ^4 are also equal, (,Ait. T)- 



i 



w 



Proposition XXVII. — Theorem. 






PamHeloRranis.AnCDandEFGII, 
and also triangles, CAB and EFG. 
standing upon equal basps AB and 
EF, and lying between ihe same pa- 
rallels AF and DG^, are equal to each 
other. 

For, since AB is = EF, (Hyp.), and EF is = HQ 
(Prop. 24), .-. AB is = HG, and Bince AB and HQ at 
joined towards the same parts by AH and BG, ■'■ ABGl 
is a CZ7. (Prop. 24, Cor. 1), and it is equal to (b 
ZZZ7ABCD, -.■ Iliey are on tlie same base AB, and b 
tween the same ||s AB and DG, (Prop. 26); it is si 
equal to the CZJEFGH. ■-■ tliey are on the same ba«eH{li 
and between the same |ls HG and AF, (Prop. 2ti), .-.th* 
djABCD is = the C^fEFGH ; again, since the A* AW 
and EFG are halves of these equal [=js. (Prop. 24), tiu 
AABO is = the AEFG, (Ax. 7-) Q. E. D. 

Proposition XXVIII. — Theorem 
If a parallelogram ABCD, and a triangle 
EBC, be upon file same Lase BC, and be- 
tween (be same parallels, the parallelograo) 
is double of the triangle. u C 

For, join AC, and tbe AABC is = the 
AEBC, (Prop. 2r>), but tlie c^ABCDis double of tl 
AABC (Prop. 24), .-. tlie cuABCD is also double 
Ihe AEBC, (Ax. 7). Q- E- ^■ 

Cor. If ihu base of a parallplogram be half thai. of' 
triangle, and they lie between the same parallels, tlietl 
nJlelugram will be e<^ual to the triangle. 



Proposition XSIX. — Theorem. 
Equal triftngles, ABO and DEF, up- a 



the same side of equ-.il bases. BC and 
EF, that are in tbe some straight line, / \ 1 '^• 

between the samo paralkU, UF Z \ , ,1 .l. .A 

and AD. ^ C e I 

For, if AD be not || to BF, let AG be drawn throoithj 
II to BF, and join GF. then the A« ABC and GEF « 
upon = bases BC and EF, and between t]ie same ||*Bi 
and AG, .-. the AABC is = the AGEF, (Prop. 27J. bi 
the AABC is = tbe ADEF,(Hyp.); .-.tbe AGEFis: 
ihe ADl^. the less = the greater, whieh is imporaibl 
'■ AG is not II to BG, and ihe sainft 'm&'ij be sbown vf u| 



PLANE OROAIETBY. 99 

other line imssiiig through A^ except AD, Ivhich therefore 
18 II to BF. • Q. E. D. 

Gor. 1. Eqnal triangles on the same base and on the 
same side of it, are between the same parallels. 

Cor. 2. In the same manner it might be shown that 
equal triangles between the same parallels are upon equal 
bases. 

Cor. 3. Since parallelograms are the doubles of triangles 
on the same base, and between the same parallels, (Prop. 
28). this proposition and its corollaries are also true of 
parallelograms. 

Scholium. The preyious four propositions, with the 
corollaries of the last two, are also true, if instead of the 
words *' between the same parallels," we substitute " hav- 
ing the same altitude," for the altitude is the perpendicu- 
lar distance between the parallels, which is everywhere the 
ftune, since two perpendiculars would be opposite sides of 
a parallelogram, and therefore equal, bj (Prop. 24). 

Proposition XXX. — Theorem. 

If two triangles, as 
ABC and DEF, have 
two sides AB, BC of 





the one, equal to two 

ttdes DE, EF of the 

other, and the con- ^ E 

tained angles ABC, DEF together equal to two right angles, 

the triangles are equal. 

For, conceive the point C to be applied to F, and the line 
CB to FE, the point B will coincide with E, and lot A fall 
as at G, •/ the La ABC, (that is, GEF), and DEF are to- 
gether = two /Is, GE and ED are in the same straight line, 
(Prop, 2). Again, *.• GE and ED are equal, the A« CrEF 
and DEF' are equal, (Prop. 27). Q. E. D. 

Proposition XXXI. — Theorem. 

Parallelograms ABCD, EFGH, 
are equal, which have two sides 
AB, BC of the one, equal to two 
aides EF, FG of the other, and 
the contained angles ABC, EFG 
also equal. 

For, draw the dijigonals AC, EG, 
the A« ABC, EFG have the sides AB, BC, and the 
xmtained L ABC in the one = the sides EF, FG and con- 
ained Z.EFG in the other, ,\ the As ABC, TEEG ^\fe 




' an 



equal in al! respects, (Prop. H); consequently the 
ABCD, Et'GH, which ate doubles of these A«. (Prop. 24 
are also equal to each other, (Ax. (i ) Q. E. 

Cor. 1. KectanglcB contained by equal Btraigtit Iji 
" e equal to each other. 

Cor. 2. The squares JeGcribed on equal lines are eqi 

each other. 



PaopoBiTioN XXXII. — Theorem. 



n 



If ABCD he a parallelogram, and FH 
and GE parallelograius about its diago- 
nal DB; and CO and OA the remain- ^ 
ing parts which make up the whole 
figure, and are therefore called comple- 
ments; the complements CO, OA are ■^ 
equal to each other. ^^ 

For, since a i::r7 is bisected by its diagonal, the ADCI 
is = the ADAB, the A^GO is = the ADKO, andtl 
AOFB is = the AOHB, .-. the two As HGO and OF 
are together = the two A« I>EO and OHU; taking thei 
pquals from the equal A" DCB and DAB, there remi' 
the complement CO = the complement OA. Q. £, 

Qef. A quadrilateral £gure which has two of its H 
parallel, but the remaining sides not parallel, is called 
trapezoid. 

Phoposition SXXIII.^ — Theobem. 

A trapezoid ABCD is equal 
tallelogram of the same 
base is equal to half the 
lallel sides AB and DC. 

Join DB, then the AADB is = a i^^ haTingthesM 
altitude, and its base = one half of AB. (Prop. 28, Ca 
also the AEDC is = a i — 7 having the same ijlitude,* 
its base = one-half of DC, (Prop. 28, Cor.) .'. the wbt 
figure ABCD is = a iz:d n-liose base is half the 8un«. 
AB and CD. and its altitude =. the distance bctweeof 
parallels AB, DC. Q. B. ; 

Def. A rectangle is said to be contained by two of i 
adjacent sides. 

Def. The rectangle contained by two lines, is a 
tangle nhich has these lines, or lines equal to them, fi 
two adjacent sides. 

Def. The rectangle contained by two lines AB and C 

written for brevity thus: the rectangle AB-CD, 
e Jeeci'ihed on a line AB, is ^suVVea AB*. 



is equal to a pn- n 

i ahiiude, whose /\ I 

le sum of its pa- / N.I 

• Ai J( 



PLANE GEOMETRY. 



101 



Proposition XXXIV. — Theorem. 

The rectangle contained by two lines, 
A6 and AD, is equal to the several rec- 
tangles contained by AB, and the parts 



ISLE 







D ] 


? E 



AlF, FE, ED, into which the other AD B & u c 
is divided. 

Make the rectangle AC, having AB and AD for its 
i^acent sides, and through F and E draw FG, EH || to 
A.B, then each of the figures, AG, FH, EC, as also AC, 
is a rectangle, and the three, AG, FH, and EC, are toge- 
aier =AC. But AG is the rectangle AB-AF, FH is the 
rectangle AB-FE, for it is contained by FG and FE, and 
FGis=AB; and EC is the rectangle AB*ED, for it is 
contained by EH and ED, of which EH is =FG= AB, 
(Prop. 24) ; and it is evident that the whole is the rec- 
tangle AB-AD ; .-. the rectangle AB-AD is =AB-AF4- 
iBFE+ABED. Q. E. D. 

Proposition XXXV. — ^Theorem. 

If a straight line, AB, be divided into two a c B 
arts in the point C, the rectangles AB*AC ~ 
f AB'BC, i^hall be equal to the square of 
IB. 

For on AB, describe the square ADEB, 
ffld through C draw CF || to AD or BE, 
ben each of the figures, AF, CE, is a rectangle, and they 
le together =AE. But AF is the rectangle AB'AC, for 
t is contained by AD and AC, and AD is =AB, being 
ides of a square ; also CE is the rectangle AB-BC, for it is 
ontained by EB and BC, and EB is =rAB, being sides of 
isquare; .*. the rectangles AB'AC+AB'BC, is =AB^. 

Q. E. D. 

Proposition XXXVI. — ^Theorem. 

If a straight line, AB, be divided into 

wo parts in the point C, the rectangle 

)ontained by the whole, AB, and one of 

he parts, BC, is equal to the square of 

hat part, BC, together with the rectangle 

iC'CB, contained by the two parts. 

Let CE be a square, described on CB, produce EF to 
[), and through A draw AD || to CF, then AF and AE 
ire rectangles, and AE is =CE+AF. But AE is the 
rectangle AB'BC, for it is contained by AB, BE, and BE 
a zrCB, hewg sides of a square ; CE is bj coTV&\.i\xc.\AaTL 




the squiire on CB, anil AF is the rectangle AC-CB. for 
is contained by AC-Cl", and CF ia =CB, being sides of 
square; .-. the rectangle AB-BC ia =CB*+AC-CB. 

Q. E. 
Cor. Uence also the rectangle AB-ACz= AC + AC-( 

Phoposition XXXVII.— Thbobbk. 

The square of tlie sum of two lines. AC, _ 

CB, ia equal to tlie sum of the squares of 

the linos, together with tivice the rectangle 

contained hy the lines. That is, AB''= 

AC-+BC+2ACCB. 

For AB3=AD'AC+AT?-BC. (Prop. 35.) 

L and AB-AC=AC'' + ACCR.l ,p „„ . 

I also AB-BC=BC^ + AC-CB. f U"p.Jt..J 

P .-. AB«=;AC^ + BC«+2.\C-CB. (Ax. 2.) 
' The diaKRim shows how the Bqnivvo described upon i.^— 

sum of the lines may he divided into the several sqiun 

and rectangles. 

Cor. 1. If AB be considered as a line divided into ti 

parts in the point C, the proposition becomes the followia 

If a straight line he diviJed into two parts, the squinvi 

the whole line is equal to the sum of the squares of W 

parts, together nith twice the rectangle contained by a 

Cor. 2. If AB and CB be equal, it is plain thai til. 
rectangle AC'CB will be a squ.irc. IIi?nce tlie squareH 
a line is equal to four times the square on half tiic line. 

■ Proposition XXXVIII. — Theorem. 

P The square of the difference AC, of 

two lines AB, BC, is less thnn the sum A ? i 

of their squares, (AB' + BC'^), by twice 

the rectangle contained by tiie lines 

AB, BC. 

Let AB he one line, and BC another, then AC ia thiil 

difference, and .'. from (Prop. 37, cor.) 
AB«=:AC''+CB''+2AC-CB. 
To eacU of these equals add CB". 
.-. AB'' + CB'=AC« + 2CB' + 2ACCE. (Ax.a) 
But 2ABBC=2CB« + 2AC-CB. (Prop. 36.) 
Take these equals from the former, and ne have 
AB"+CB»— 2ABBC— ACS. (At. 3.) 
Hence the squnrc of the difference is less than theiun 
of the squares by twice the lecUn^Ve. Q. E. tt 



PLAKX OBOMETRV. lOS 

Dor. 1. The square of the sum, te- 
ther with the square of the difference - — ? 1 ^ 

two lines, is equal to twice the sum 
the squares of iSie lines. 

Let AB be one line, and BC another, then AC is their 
m; make BD=BC, then AD is their difference. 
AC«=AB« + BC«+2AB-BC. (Prop. 37.) 
id AD«=AB« + (BC«=BD2)— 2ABBC. (Prop. 38.) 
iding these equals together, we obtain 
iC« + AD«=2AB« + 2BC«. 
Cor. 2. If a line AB be bisected 

1 C, and divided unequally in D, the '^ 1 — ? — ? 

am of the squares of the two unequal 
arts, AD, DB, are together equal to twice the square of 
lalf the line AC, and twice the square of CD, the line be- 
ween the points of section. 

For AC may be considered as one line, and CD as ano- 
ber, then AD will be their sum ; and since CB is = AC, 
)B will be their difference ; .-. AD^ + DB^ is =2AC2 + 
CD«. (Cor. 1.) 
Cor. 3. If a straight line AB, be 

iiected in C, and produced to D, the "^ 5- 

|oare of the whole line AD, thus pro- 

nced, and the square of BD, the part 

loduced, will be together equal to twice the square of 

alf the line AC, and twice the square of CD, the line 

Mide vLp of the half, and the part produced. 

For AC can be considered as one line, and CD as ano- 
her, then AD will be their sum; and since AC=CB, 
\D will be their difference; .-. AD^+DB^ is =2AC« 
|.2CD«. (Cor. 1.) 

Cor. 4. The square on the sum of two lines is greater 
ban the square on their difference, by four times the rec- 
tngle contained by the lines ; for the square on the sum 
f the lines is greater than the sum of their squares, by 
mce their rectangle, (Prop. 37), and the square on their 
lifierence is less than the sum of their squares by twice 
keir rectangle. (Prop. 38.) 

Cor. 5. The sum of the squares of two lines, AB, BC, 
B eqtbl to twice the rectangle contained by the lines, to- 
[ether with the square of their difference. 

For (Prop. 38) AB^ + BC^— 2ABBC=:AC2; add to 
ach of these equals 2AB'BC, and we hare AB2 + BC^= 
JABBC + AC«. 

Cor. 6. The square on the sum of two lin^B, K^^^C, 






I Fot 



mis equal to four times the rectangle contained by<tlie Un 
, together with the square of the difference of the lines. 

Far adding 2AB-HC to both sides of the result in Cor. 
gives AB«+BC' + 2AB-BC=4ABBC + AC«. 

The first of these equnls is (Prop, 37); the square 

I scribed on a line AD, (Cor. .5), which is the sum of 
snd BC, and the second is four times the rectangles 
tained by the lines, together with the square on , 
(Cor. 3), which is the difference of the hnea- 
Cor. 7. If a straight line AB be bi- 
■ected in C, and divided unequally in D, ■ " 
the rectangle AD-DB, together with the 



N 

N 



of section is equal to the square on (AC or CB) half the JIi 
For AB"=AD3 + DB^+2ADDB. (Prop. 37). 
but AB«=r4AC'', (Prop. 37, Cor. 2). 
and AD^ + DB^— 2ACJ+2CD".{Prop.38,Cot. 
Substitute these Talues of AB" and AD^+DB' in 
first, it becomes 4AC«=2AC''+2CD»+2AD-DB. 

Talte 2AC" from each of these equals, and take 
halves of both sides, and we have 

AC"=CDi' + ADDB. 
If a straight line AB be W- , ^ 

Bected in C, and produced to D, the 1— 

rectangle AD-DB contained by the whole 
line thus produced, and the part produced, together " 
: on CB half the line, will be equal to the sau 
on tbe line CD, made up of half the line and 
part produced. 

For ADDB=ACBD + CB-BD+BD=, (Prop. 34,) 
but since AC=CB, substitute CBBD for .\C-Bft- 
and add CB' to both sides, and it becomes 
AD•DB+CB^=2CB■BD+BD^+CB■'=CD^(P^op.37 
Scholium. The last four propositions, with their co» 
lories, are also true arithnielically or algebraically, if pi 
dnct be substituted for rectangle, number or quantllj 
line, and increased for produced, and the student BflO 
prove iheir truth arithmetically by taking a number * 
subjecting it to ihe processes described in each of the M 
positions and corollaries, and proving the idenlilyDt t 
results. The following are the algebraic processes tlui 
equivaleot to these propositions: — 

Let AC=a, and CB=6, then AB=a+t. 
(Prop. 35.^ a{a+h) + b(a+h)={a+b)''. 
.(Prop. 36.) a{a+b)=a''+ab. 
miProp. 37.) (a+b)3=;a'-V4abJrl>''. 



rLAm oxomTBT. 10r> 

Cot. t. The ume aJgebnucallj as tbe proposition. 
Cor. 2. (2a)''=4a''. 
(?^o^88'.) {a— S)»=o'— 2o6+6» |aB=o, CB=6, AC-Co— i)}. 
Cor.I. {fl+ft)«+(i.— 6)»=2i^+2i' {AD^(o— 6).AC=(o+i)}. 



Cot.*. («+fr)»-{ci-4)«=*n6. 

Cor. S. «»+i»=-2D6+(fl— *)*. 

Cor. 6. («-i-ft)'-4o6+{a— iV. 

Cor. 7. (a-t-iXa— t)+i'=o* AC or CB=a, CD_b. 

Cor. S. (2a+i)t-l-a*=i(a+&)* AC=CB=a, BD-b. 

Cw. 9. The rectangle under the sum and difference of 
two ILnes is eqoal to the difference of their squares. 

For ADDB+CB'=CD', (Cor. 8), take CB' from both 
Biei,wd we hare AD-DB=CD'— CB'; now AD is th« 
nnof CD and CB, for AC is = CB, and BD is their 
diflcrence. 

Jn the aboTe it will be remarked that several of the co- 
nflaries assnme the same algebraical forms; this arises 
fana the different wa^a in which a line made np of two 
DBrtsmaybe considered: either the parts of the line may 
te considered as gepamte lines, and the whole line as their 
nun, or the whole maj be considered as one line, and one 
if the parts as another, then the remaining part will be the 
Itflerence of the two lines ; or when a line is divided into 
IWD equal and also into two unequal parts, half the line 
nay be considered as one line, and the distance between 
lie middle of the line and the point of unequal section as 
mother, then tbe greater of the two unequal segments wiU 
w tbe sum, and the lets will be tbe difference of the two 

PnoPoaiTioN XXXIX. — Theorrh. 

The square BQ described on t 
^potennse BC of a right-angled t 
ogle is eqnal to the sum of the ^v 
iqoares LB and KC described on AB 
IM AC, the sides that contain the 
ight angle. 

Draw AE ]| toBF or CG, join AF, 
HO, AG, and BI, '.• the Li LAB and 
BAC are t'U LA and AC are in the 
nme straight line, (Prop. 2); for a like reason KA and AB 
tre in the same straight line. Again, each of the Is ABH, 
DBF are r*/*, being Zsinasqnate; to each add the ^ABC, 
dien the whole ^HBC is s: (he Z ABF; also A&-^%,«iu^ 




BC=BF. .-. HB, BCare = AB, BF.andiHBCliasIie 
proTed = the ZABF, ■. the AHBO is = the AAB 
(Prop. S). But the square LB is double of the AHB 
and the rZjBF. is double of tiie AABF, (Prop. 28) 
the square LB is = the cz;BE, (Ai. 6). In the s 
mauncr it caa bo shown that the square KC is = 
r — j CK, .'. the whole square BG is ^ the two squares I 
and KC, that is, the square on the hjpotemiee BC is : 
the sum of the squares on AB and AC, the sides that ci 
tain the right angle. 

Cor. 1 . Hence the square of one of the sides of a rig 
angled triangle is equal to the square of the hjpotonnse 
minished by the square of the other side, or equal to ( 
rectangle under the sum and difference of the hypotenn 
and the other aide, (Prpp. 3S, Cor. 9), which is thns e 
pressed, AB'=BC'— AC"-{BC+AC)(BCU-AC). 

Cor. 2. If the hypotenuse and one side of a, rightaoj^ 
triangle be respectively equal to the hypotenuse and a liii 
in another, the remiuning side of the one is equal to thel 
maining side of the other, and the triangles are equd i 
eTery respect. 

Pbopobition XL. — Thborem. 

In any obtuse angled triangle, ABC, the 
equate of the side AC subtending the ob- 
tuse angle ABC, is greater than the sum of y" 
the squares on the other sides, .AB, BC, y^ 

which contain the obtuse angle, by twice ^ ^ 

the rectangle contained by the base AB and 

the distance BD of the perpendicular from the I 

angle; or AC'=AB'+BC'+2AB-BD. 

For AC*=AD'+CD^ since ADC is a t'L. (Prop. 38) 

But AD'=AB'+2ABBD+BD', (Prop. 37). 

.-. AC'=AB'+2AB-BD-|-BI>'+CD^. 

And BD'+CD'=CB', since CDB is a t'L (Prop.S 

.-. AC=AB'-|-BC'+2ABBD. 
Otherwise, 

AP' -AB'+D B'+SAB-BD; add CD' to both, 

and AD'+CIJ-=AB=-|-BD'-|-CD'+2AB-BD. 

.-. AC'=AB'-1-BC''+2ABBD. 

Proposition XLI. — Tbeobbk. 

In any triangle ABC, the square on a side CB tabteoi 
iag an acute angki is leu t\iim XW «\3l«l of th« 



PIiAMB GBOMBTRT. 



107 





n the sides AC^ AB^ that 
ODtain the acute angle, by 
wice the rectangle contained 
)j the base AB and distance 
(D of the perpendicular from 
he acute angle CAB. 

Or CB«=AB"+AC"— 2AB-AD. 
Since BD is the difference of AB and AD, (Prop. 38), 

BD>=AB"+ AD'— 2AD-AB; add CD* to both, 
and BD'+CD*=AB>+CD*+AD«-.2ADAB. 
/. CB«=AB«+ AC*— 2ADAB. 

Scholium. If in the last two propositions the sides op- 
lonte to the angles at A, B, C, be denoted bj a, 6, c, 
?ropodtion 40 will be b^^a^+c^+2cBD, which, by 
nuuposing a^ and c^, and dividing by 2cy gives BD= 

— , from which^ if the three sides be given, the dis- 

nce of the perpendicular from the obtuse angle may be 
nmd. 
Similarly from Proposition 41, by the same substitution, 

"BD— ^i^I_" 
" 2c 

Proposition XLII. — ^Theorem. 

The difference of the squares of the sides AC, CD, of a 
iangle, is equal to the difference of the squares of the 
stances, AD, DB, of the perpendicular, from the extre- 
ities A, B, of the third side AB. 

Or, AC«— CB«=AD«— DB«. 
For AC2=AD2 + CD«, (Figure to last Prop.) 
andBC«=BD2 + CD2. 

.-. AC«— BC"=AD2— BD2; since CD^ from CD^ 
ivesO. 

Cor. The rectangle contained by the sum and difference 
' two sides of a triangle, is equal to the rectangle con- 
bed by the base and the sum or difference of its seg- 
mts, according as the perpendicular falls without or 
ifhin the triangle. 

For AC«— BC«=AD«-^BD«, by the proposition. 
.-. (AC + BC) (AC~BC) :=: (AD+BD>(AD— BD.) 
hop. 38, cor. 9). 

Or, (AC+BC)(AC— BC)=sAB(AD+DB.) Figure 1. 
and(AC+BC)(AC— BC)=AB(AD^DB.) ¥\gQi«i^, 




I 

» 



The sum of the squares of two sides, C 

AC, CB, of a triangle ACB, is equal to y^ 

twice the square of hulf the base AD, and / / iX 

twice the square of the line CD, joining / j j \ 
the vertex and the middle of the base. ^ BB 

Or AC'+CB2=2AD»+2CD2. 

Draw CE perpendicular to AB, and consider ADC 
ooe triangle, and CDB as another; then 

AC2=AD2+DC2 4.2AD-DE. (Prop. 40;) 

CBa^AD'+DC^— 2AD-DE, (Prop. 41), AD W 
=DB. 

.-. ACi!+CB2=2AD«+2DC», (As. 2.) 

Pbopobition XXiIV. — Throrbh. 

In any isosceles triangle ABC, th( 
■j^uare of a line CD, drawn ftom the ver- 
tex C lo any point D in the base, togethei 
with the rectangle contained by the seg- 
ments AD, DB, of the base, is equal to — - 
the square of one of the equal sides of the 
triangle. Or CDa+AD-DB^CB*. 

For CBS— CDs=BES— EDa. (Prop. 42.) 

= (BE + ED) (BE— ED) = BD-Di 
(Prop. 38, cor. 9.) 

.-. CB==CD»+BDDA, by adding CD= to both sidw 



Proposition XLV. — Thborkw. 



^ 



In any parallelogram the diagonals 
bisect each other, and the sum of the 
squares of the four sides is equal to the 
iquares of the two diagonals, 

V AD is II to BC, the Z.ADE is = ^_ 

the /.CBE. and the Z.DAE is = the iECB ; .-. since A! 
is =BC, (Prop. 6). AE is =EC, and BE is -ED i 
wnce AB is =DC, and AD is =BC, (Prop. 24), I 
squares of AB and AD are together = the squares oFI 
and BC, but because BD is bisected in E, BA'+AD 
2BE»+2AE», (Prop. 43); .-. DC+BC are also sSf^ 
+2AE»; and hence BA»+AD*+DCHBC« are =41 
+4AE=BI>i+AC». (Prop. 37, cor. 2.) 



109 





PLANS OEOMBTBT. 

Any portion of the circnm- 
of a circle^ as ACB, is called 

tiaigbt line AB^ which joins the 

des of an arc^ is called the clwrd 

irc. 

pace contained bj the arc ACB 

chord AB is called a segrnenL 

figure CAD, bounded by two 
C, AD, and the intercepted arc 
ailed a sector. 

3 radii be perpendicular to each 
A AC, AB, the sector is called 



ngle in a segment is fiontsaned by 
dght lines drawn from any point 
re of the segment to the extre- 
f the same arc. -^ 



Dgle on a segment or on on arc is contained by two< 
lines drawn from the extremities of the segment or 
my point in the remaining part of the circumfe- 
thus the angle ABC is said to be in the segment 
r on the segment or arc AC. 
ngle at the circumference of a circle, is one whose 
point is in the circumference ; and an angle at the 
8 one whose angular point is at the centre. 
'ur segments are those that contain equal angles, 
ctilineal figure is said to be inscribed in a circle 
ts angular points are in the circumference of the 

rcle is said to be inscribed in a rectilineal figure 
le circle touches all the sides of the figure. 

Pboposition XL VI. — ^Theoiiem. 

(traight line CD, drawn through 
tre C of a circle, bisect a chord 
kich does not pass through the 
it will cut it at right angles ; and 
it it at right angles it shall hi- 

, let AB be bisected in D, and 
L, CB. 




I 




110 PLANE CEOMETHT. 

Th«i in the As ADC, BDC, AD is =DB, and DC ci 
man to both, also AC ia =;Cfi, being radii of the a 
circle ; -■. the /ADC ia =z the £BDC, (Prop. 9), and th( 

:e adjacent Is, hence each of them is a right angle. 
Nest let CD he at right angles to A B, AB is hiaectt 
iD, 

For since AC ia =CB, and CD common, and the 
ADC and BDC j' Ls, .: AD is =DB, (Prop. 39, cor, 2. 

A straight line which bisects a chord at 
angles passes through the centre. 

Proposition XLVII Theorkm. 

ITie angle ACB, at 
the centre of a cirule, 
is double of the angle 
ADB at the circumfe- 
rence, standing on the ^ 
Kame arc AB. 

Join DC, and pro- 
duce it to F, then -.- 
AC is =CD, the /CAD is = the ZCDA, -■- the two i 
CAD and CDA are together double of the /.ADC, butti 
ZACF is also = the two /.a CAD and CDA, (TWf, V 
.: the /ACF is double of the /.ADC ; in the same maa 
it uan be shown, that the /FCB is double of the a 
CDB; .-. the whole, or the remaining /ACB, is douM» 
the whole, or the remaining /.ADB. (The ■ 
applies to the first figure, and rfmaining to the seeond.] 

Cor. 1. The angles is the same segment of a (urcle, I 
the angles standing on equal arcs, are equal to each oths 

Cor, 2. An angle at the circumference of a circle 
measured by half the arc on which it stands. For by tl 
proposition it is half the angle at the centre, and the an| 
at the centre is measured bj the arc on which it ataa 
(Prop. 1. cor. 3.) 

Pbopobition XLVIIl. — Tbgorrn 
The opposite angles of a quadrilateral 
I figure inscribed in a circle are together 
equal to two right angles. 

For the three /s of the A ABC, name- 
ly ABC, BCA, and CAB, are together 
lal to two right angles, (Prop. 19), 
J the /BCA is = the /BDA, (Prop. 
; cor. 1), also the /.BAC w = the 




ThAXim OEOBfSTBT. 



Ill 




BDC ; .*. the ZADC is equal to the two angles B AC and 
€A; and adding the ZABC to each, the two is ADC 
Bd ABC are together = the three Is ABC, BAC, ACB, 
lat is, to two / is. 

Otherwise, 

The angle ABC is measured bj half the arc ADC, 
Prop. 47« cor. 2), and the Z. ADC is measured by half the 
t9 ABC; therefore the two Is ADC and ABC are 
leasured by half of the whole circumference, and are . 
Iierefore together equal to two r* Ls. 

Cor. If a side CB of a quadrilateral figure, inscribed 
BL a circle, be produced, the exterior angle ABE is equal 
the interior angle ADC ; for each of them, together 
nth the angle ABC, makes up two right angles. 

Pboposition XLIX. — Theorem. 

Equal chords, AB, CD, in a circle, are 
ignaUj distant from the centre; and if 
M distances, EF, EG, from the centre. 
It equal, the chords, AB, CD, are equal. 

For draw the radii, EB, ED, and ••• 
IF and EG are -^ to AB and CD, these 
iMords are bisected in the points F and 
X (Prop. 46.) Since then the chords 
ItB and CD are equal, their halves FB and GD are also 
nal, and the radii EB, ED, are equal, and the angles at 
r and G are r* Z«; .*. the A< EFB, EGD, are equal in 
fwy respect, (Prop. 39, cor. 2); hence EF is =EG. 

In the same manner, if £F be given zzEG, in the 
'I' A« FEB, GED, we have EB and EF in the one 
=ED and EG in the other; .-. FB is =GD, (Prop. 39, 
ar. 2), but AB is double of FB, and CD is double of GD ; 
'. AB is =CD, (Ax. 6.) 

Proposition L. — Theorem. 

The diameter AB is the greatest chord 
n a circle, and that chord ED, which is 
learer to the centre, is greater than the 
iiord GH, which is more remote; and 
ionverselj the greater chord is nearer to 
lie centre than the less. 

For since C is the centre of the circle, 
EC is =CA, and CD is =CB, .-. EC and CD are =AB, 
Imt EC and CD are greater than ED, (Prop. 13), .-. AB 

Again, since the chord ED is nearer to the cenlx^ Wv'dXL 




118 

GH, CF^CL, lut CF^+FD= are =CL'+LH». for ei 
is equal to the square of the radius ; .■. since GF'.c^CL*', 
FD^r^LH*, and -■. FD is t^JM, and hence ED::p-G^ 
(Prop. 46, and As. 6,) 

Nest let the chord ED he ^GH. FC shall he ^:CI^ 
for since DF'^4-FC»^HI;!-t-CL=, of which FD^^LH* 
.-. FC'^CL^ : hence FC^CL. 

Cor. 1. The angle at the centre of the circle, aubtendcj 
hy the greuCer chord ED, is greater than the angle ivb 
tended by the less, GH. 

For in the two A* ECD, GCH, the two sides EC, Cft 
in the one, are equal to the two sides GC, CH, in tbi 
other, hnc the hase ED is ^^ the base GH ; .-. the ^EOS 
ia^theZGCH. (Prop. 15.) 

Cor. 2. The angle EUD is tneaaured hy the are Eft 
and the angle GCH by the arc GU. Therefore the greater 
chord cuts off the greater arc. 

Cor. 3. Equal chords cut off equal arcs. 

For if the chords were equal, the angles at the C 
would he equal, and consequently the arcs which meaauil 
I them. 

^^^L Proposition LI. — Theor; 

^^^H A straight line AB, drawn perpendi- P 

^^H colar to the extremity of a radius CA. ia j' ^(T 

a tangent to the circle at that point; that / /^N 

is, it touches the circle without cutting it. { C^ U 

For take any point E in AB, and join V / 

EC. thensince ACis -L- to AB.itisless N. 7 

than CE. (Prop. 23), but CF is =CA, -^ 

.-. CEp»-CF, and hence the point E is 

without the circle, and E beiug any point on the line ABi 

.■- AB is without the circle. 

Cor. A Etniight line drawn perpendicular to the U 
gent AB, from the point of contact A, will pass through ll* 
centre. For it must coincide wilh CA, which is J- to AR 

Proposition LU. — Tseobem. 

An angle ADB in a semicircli 
right angle, and an angle DBA in ; 
ment greater than a semicircle i 
than a right angle, aad an angle AED M 
in a segment less than n semicircle is 
greater than a right angle. 

Join DC, and produce AD to F, then 
■-■ AC=CD, the iCAD = ihe LCDi, 




rJLAHX QXOMBTRT. 113 

and -.- DCzuCB, the ZCBD = the ZBDC, .% the whole 
LADB is = the two Ls DAB and DBA, but the ZFDB 
k also = the two Ls DAB and DBA, (Prop. 19); /. the 
£ADB is = the ZBDF, and they are adjacent is, .*. each 
of them 18 a /Z, (Def. 9); hence the ZADB is a r*l, .•. 
&e ZABD (which was shown to be =: the ZCDB), is less 
than a /Z. And again, since AEDB is a quadrilateral in- 
Kribed in a circle, the two Ls AED and DBA are together 
= twoVZ*, (Prop. 48); .*. since DBA has been prored 
leiB than a /Z» AED must be greater than a /Z* 

SoHOLiUM. The proposition might have been demon- 
rtiated thus: — Since the ZADB is measured, (Prop. 47» 
Oor. 2), bj half the semicircle AGB, it is a r'Z; since the 
[DBA is measured bj half the arc AED, which is .^ a 
MDudicle, itia^^^Af^L; and since the Z AED is measured by 
half the arc ABD, which ib^p^ sl semicircle, it is :::^ a /Z. 

PiKOPOSITION LIII. — ^ThBOREM. 

Parallel chords AB, CD in a circle, 
btercept equal arcs AC, BD. 

For join CB, then ••• AB is || CD, 
md CB meets them, the Ls ABC, BCD 
ue equal, but the Z ABC is measured bj 
Wf the arc AC, (Prop. 47, Cor. 2), and 
the ZBCD is measured by half the arc 
BD, .•. the arc AC is = the arc BD. 

CJor. If the adjacent extremities A, B, and C, D, of two 
iqual arcs AC, BD, be joined, the chords AB and CD will 
K parallel. 

Proposition LIV. — Theorem. 

If a tangent ABC be parallel to a 4. 
^ord DE, the intercepted arc DBE is 
Msected in the point of contact B. 

Draw FB from the centre to the point 
sf contact B, then FB is -J- to AB, and 
•'•it is also -L to DE; hence, since FG 
i« -L to DE, DE is bisected in G, 
(Prop. 46); .-. in the A« DGF, EGF, 
DG is = GE, DF is z= FE, and GF is common, .-. the 
ZDFG is = the ZEFG, (Prop. 9); hence the arc DB, 
which is the measure of the one, is =: the arc BE, which 
1*8 the measure of the other, .•. the arc DBE is bisected in 
B, the point of contact. 





w 



E OBOllETRT. 



Proposition LV. — Theokem. 



igle ABD, formed by a tangent AB, and a chotA 
DB drawn from the point of contact, is equal to the angle 
in the alternate aegment BED, and is measured by half Sie 
intercepted arc BU. 

For (figure laat PropOj let DE be |{ to AB, then the an 
BD is = the arc BE, (Prop. 54), .-. the /.BDE, meaawed 
by half the one, is =; the Z.BED, measured by half the 
other; but the iBDE is = the iDBA, (Prop. 16),) .-. abo 
the Z.ABDis = the /.BED; hut the Z.BED is measured bj 
half the arc BD, .'. the Z.ABD formed by the tangent B^ 
and the chord DB is also measured by half tbe arc BD. 

Cor. Tbe angle DBC fonned by the tangent CB, and 
the chord DB drawn from the point of contact, is measured 
by half the intercepted arc BEHD. 

For the part DBE is measured by half the arc DHE, 
(Prop. 47, Cor. 2), and the part CBE is measured by half 
the arc BE, .-. the whole /DBC is measured by half itw 
arc BEHD. 

Proposition LVI. — Tbeobem. 

If two chords AB and CD intersect 
within a circle, the angle AEC formed at 
their point of intersection is measured 
by half the sum of the intercepted 
AC, DB. 

Join DA, then the /AEC being the 
exterior angle of the AAED is = to the _ 
two L» DAE and EDA, but the /DAE 
is measured by half the arc DB, and the /EDA is IB*«" 
Bured by half the arc AC, (Prop. 47, Cor. 2); .-. the /AEC 
is measured by half the sum of the arcs AC and DB. 

Proposition LVIL — Theorem. 
If two chords, AB, CD, meet when 
produced in a point E, the angle AEC 
formed at their intersection is measured 
by half the difference of the intercepted 
arcs AC and BD. 
_ -For, join BC, then the /AEC isequalto 
I ttie diflerence of the ie ABC and BCD. 
I fProp. 19); but the /ABC is measi^ed by 
f half tbe arc AC, and the /BCD by half the arc BD. .% A 
J /AEC is measured by half the difference of the arcs A( 
fjuid BD. 





PLAVE OEOMBTRT. 115 

Cor. If a chord CD produced, meet a tangent £F in 
&e point E^ the angle C£F is measured hj half the differ- 
ence of the arcs CF and FD. 

For the LGEC is = the difference of the Z« GFC and 
FCE, (Prop. 19), and the ZGFC is measured by half the 
arc CF, (Prop. 65), and the ZFCE by half the arc DF, 
(Prop. 47» Cor. 2); .'. the LCEF is measured by half the 
diftrenee of the arcs CF, FD. 

ScHouusr. In the application of proportion to geo- 
metrical quantities, since there is no geometrical principle 
bj which it can be ascertained how often one line is con- 
tained in another, the algebraical definition of proportion 
win not apply to geometrical magnitudes, and therefore it 
is necessary to substitute another, which is as follows: — 

De£ The first of four magnitudes is said to have the 
nme ratio to the second, that the third has to the fourth, 
wlien any equimultiples whatever of the first and third be- 
ilif taken, and any equimultiples whatever of the second 
and fourth — ^if the multiple of the first be less than that of 
^ second, the multiple of the third is also less than that 
cf the fourth; or if the multiple of the first be equal to that 
of the second, the multiple of the third is also equal to that 
of the fourth; or if the multiple of the first be greater than 
tfaatof the second, the multiple of the third is also greater 
Oaii that of the fourth. 

Note. A multiple of a quantity is the result of repeating that 
quantity a certain number of times ; and equimultiples are the re- 
xiliB of repeating several quantities the same number of times. 

In order Uiat the propositions on proportion that were demon- 
rtnted in the treatise on Algebra, may be made available for the 
derdopment of geometrical properties, it will be necessary to prove, 
fliat if quantities are proportional by the algebraical definition, they 
tte also proportional by the geometrical; and conversely. 

Let now a, 6, c, d, be four quantities that are proportional by the 

tlgebraieal definition, that is, such that j- = -5, they are also propor- 
tbnal by the geometrical definition, that is, if ma ^^ nfe, mc Z^ nd; 
if aa =s aft, iwc = nd, md ii ma^^nb, mc^^nd; since ^=-^, -^ = 
jjj, for the second is obtained from the first by multiplying both sides 
rfth« equation by - ; now it is evident that if wia be greater than 
*> -^ is greater than unity, but "^ ^ = ^^> therefore -^ is greater 

Aan unity, which can on]|r be true by mc being greater than mf ; 
Iirace, when the multiple of the first is greater than that of th« se- 
omd, the multiple of the third is also greater than the multiple of 
the fourth. 



w 



FLAn QSOHETBT. 



In the Bamo manner, if ma be = ni, ~ = unity, but then ^ 

liich can only be by mr being =m nd; henco, when tho, 
multiple of [Jie firtit in equal to tlie luultipio of the second, the 
tiple of the third in also equal to that of tJiu fourth. 

Again, if ma be lees than nb, —^ ia less than unity, therefore —r 
IB aiso less than unity, which can only be by tnc being lesa than aJ} 
hence, when the muIUple of the first is lees than that of the 
(he multiple of the third is also lees than that of the fourth. 

Therefore, if four quantities be proportional by the algebrwcal 
fiaition, thuy are also praportional by the geometricnL 

GtDversely, let a, b,c,d,hB four magnitudes, such that whitlevtf 
nuiubeis n oud n may be, ma is greater than, equal to, or lea ' 
nfr, according as mc ia greater than, equal to, or less than ad, 
j-= J, for if not, let j-^-j therefore . — — , and hence na k 
greater than, equal to, or leas than tib, according aa wc 
than, equal to, or less than ne; that is, nd aud ne poaseHi 
properties, or i/ = e, and consequently ^= ^. 

Therefore, if four quantities he proportional by the 
definition, they are also proportional by the algcbi^ical. 

Coutioqueutly, whatever was ilentonatrated to be true of quantitw 
that are propDrtional by ooe dcfinitioti. is also true of those that Ut 
proportional by the other, for each defiuidon has been showa to odd- 
tain the other. 

Lemma. Quantities that liare the same ratio to the ttat 
quantitjnre equal to one another; and equal qoanti^- 
have the same ratio to the same qunntitj. 

Let a and h have the same ratio to c, thea a=6; Ex 
since a has the same ratio to c that h has to c, -^ -. Mul- 
tiplying hy c, we have a=6. 

Nest, let 0=6, and c be a third quantity, \)itaa-.e=h:t\ 

for since a^xh, if each he divided by e, we hare ' — {' 

Def, Similar rectilineal figures are those which haw 
their several angles equal each to each, and the sides abod 
these equal angles proportionals. 

Def. Reciprocal figures, viz, triangles and paralleln- 
grains, are such as have their sides about tvro of ihtir 
angles proportional in such a manner, that a side of tht 
first is to a side of the second as the remaining side of ''' 
■econd is to the remaining side of the first. 

Def. A strjight line is said to be cut in extreme and 
n ratio, when the whole is to the gteati 
^ greater segment to the less. 




PLAHK OEOMETRT. ll? 

Def. The altitude of any figure is the straight line 
drawn from its yertex perpendicular to the hase. 

Pboposition LVm. — Theobem. 

Triangles ABC, ACD, and paral- ^ ^ 

Wograms EC, FC, haying the same 
altitude, are to one another as their 
bases BC, CD. 

Produce the base both "ways, and ^ „ ^ t, ^ 
take BG, GH, HI, any number of "^ ^ ^ -^ ^ 
lines, eac^ equal to BC, then IC is a multiple of BC, take 
DK, KL, any number of lines each equal to CD, then CL 
18 a multiple of CD ; join AG, AH, Al, AK, AL, then *.* 
BC, BG, GH, HI, are aU equal, the A« ABC, ABG, 
AGH, AHI, are all equals (Prop. 27); .'. what multiple 
Merer the base IC is of the base CB, the same multiple is 
the AAIC of the A ABC; again, since CD, DK, KL, are 
afl equal, the A« ACD, ADK, AKL, are all equal, (Prop. 
27); .*. what multiple soeyer the base CL is of the base 
CD, the same multiple is the AACL of the A ACD. 

Now, if the base IC be greater than the base CL, the 
AAIC is greater than the AACL; if equal, equal; and if 
less, less; .*. there are four magnitudes, namely, the two 
bases BC, CD, and the two A« ABC, ACD, and of the 
Vase BC, and the A ABC, the ist and 3d, there haye been 
taken any equimultiples whateyer, yiz. the base IC and the 
AAIC, and of the base CD and the A ACD, the 2d and 
4th, there haye been taken any equimultiples whateyer, yiz. 
the base CL and the AACL, and it has been shown that 
as the base IC "p^, =, or -*i:r the base CL, 
so is the AAIC z^^ =, or ^^^ the AACL, 
.-. A ABC : A ACD = base BC : base CD, 
and •.• ^=I7CE = 2AABC, and £Z=7CF z= 2AACD, 
.-. AABC : AACD = £Z=7CE : £Z=7CF, (Alg. 109.) 
and •.• AABC : AACD = base BC : base CD, 
.-. ZZZ7CE : £=7CF = base BC : base CD. 
Cor. Triangles and parallelograms haying equal alti- 
tudes are to one another as their bases. 

Proposition LIX. — Theorem. 

If a straight line DE be drawn parallel to one of the 
ndes BC ef a triangle ABC, it will cut the sides AB, 
AC, or these sides produced proportionally; and if th^ 
tides AB, AC, or these sides produced, be cut ]^ro]^ortioiL- 
ally, in the points D, E, the straight line Y7\i\c\i ^o\ii*& ^^ 




¥ 



318 FLAJTB flBOJIETBT. 

points of section will be 
parallel to the third side 
BC of the triangle ABC. 

First let DE be drawn 
II to BC, a side of the 
AABC; thenBD:DA b'= 
= CE:EA;fc ■ * "" 
DC; 

Then ADBEzzADCE, ■.- thej are on tbe aane htm 
DE, and between the same ||s DE and BC, (Prop. 26.) 

And ■." A^l^li is another mi^nitude, 

.-. ABDEr AADE=ACDE: AADE. (Lem.,p.ll6. 

But ABDE: AADE=BD:DA,1 p^^^ „ 

and ACDE : AADE^CE ; EA. P"P- ***■ 

.-. BD:DA=CE: EA. (Alg. 111.) 

Secondly, let the sides AB, AC, of the AABC. or tlua 
flidesproduced, be cut in the points D, E, bo that BD;DA3 
CE:EA; then DE is II to CB. 

The same construction being made, 

V BD.-DA=CE:EA, 

and BD : DA=AJ3DE : AADE,\ „^ -„ 

and CE : EA= ACDE : AADE, f ' P" ** 



ABDE ; AADE=ACDE t AADE ; that ia. 
A« BDE, CDE, have the aarae ratio to the AADE, 
and .-. ABDE=ACDE. (Lem., p. 116.) 
and they are on the same base DE, and oa the un 
ide of it; .-. DE is || to BC. (Prop. 29.) 



^^^H PRorosiTioN LX. — Theorem. 

F If the vertical angle BAC of a 

triangle be bisected by a straight line ^ 

\ AD, which also cuts the base, it will ^}Cl 

divide the base into segmenls BD, ^■"''W'^V / 

' DC, which have the same ratio aa the ^ I V 

othersides, BA, AC; andif the seg- B d C 

ments of the base have the same ratio 

which the other Ejdea of the triangle have to each other, ihi 
' right line drawn from the vertex to the poiot of 

1 divides the vertical angle into two equal parts. 

First, let the Z.BAC be bisected by the straight liJ« 
I AD; then BD:DC=BA:AC. For draw CE || DA 

and let BA produced meet CE in E. 

Then ■-• A D is II to EC, the exterior iB AD = the il 

tenor /.AEC. and the alternate b DAG, ACE, « 

equaX; but the ^BAD is = the ZDAC ; .-. the ZAEC 
also = th e Z A C E ; hence the s\4e k^ w ■=. t\a uda AC 




TI.A.NE 6E0METBT. 119 

md •.• AD is || to EC, a side of the ABCE, BD : DC= 
U : AE. (Prop. 59.) 

but AE=AC ; .-. BD : DC=BA : AC. 

Secondly, let BD : DC=BA : AC ; join AD, then the 
BAG is bisected bj AD, that is, the ZBAD = the 
DAC. 

The same construction being made as before, 

vBD:DC=BA:AC, 

and that BD : DC=BA : AE, (AD being || EC.) 

.*. BA : AC=BA : AE; hence AC= AE. (Lem., p. 116.) 

md .-. the LAEC=: the Z ACE, 

But the ZAEC= the ZBAD, and the ZACE=ZCAD. 

Prop. 17.) 

•.theZBAD=the LCAD. 

Proposition LXI. — ^Tifeorem. 

Ilendes about tiie e^al angles of 

[niaogidar triangles are propor- 

mk^ the sides opposite to the 

[oal angles being the antecedents 

consequents of the ratios. 

Letthe A«AEB,BDC, having the 

UB= the ZDBC, the LABE= 

eZBCD, and the LAEB=: the Z.BDC, be so placed that 

B and BC maj be one continuous straight line. Then *.* the 

HAB, together with the ^BCD, (which is =the ZABE), 

less than two r^is, AE and CD, will meet, if produced, 

Vop. 17, cor.); let them meet in F, then since the 

iBA=the LBCB, and the ZEABzz the ^iDBC, (Prop. 

, cor. 1.), FD is II EB, and EF is || BD, •. also 

)=EB, and EF=BD, (Prop. 24.) Now since EB is || 

the side FC of the AAFC, AE : EF=: AB : BC, (Prop. 

); but EF=BD, 

.-. AE : BD=AB : BC, and by alternating, (Alg. 105.) 

AE: AB=BD:BC. 
^^in, since BD is || AF a side of the A-^^C, 
AB : BC=FD : DC, (Prop. 59), but FD=EB, 
•. AB : BC=EB : DC, and hy alternating, 
AB:EB=BC:DC. 

•. also AE : EB=BD : DC. (Alg. 112.) 
2lor. Equiangular triangles are similar. 

Pkoposition LXII. — Theorem. 

tf the ffldes of two triangles about each of ilieit ^ii^<^ 
jwgot^xauik, the triangles shall be eqiuan^xilox, ^tA 



..^.™ M 



^ 



M 



lIQO FIiAHE OKOSOtTBT. 

hare those angles equal that are oppo- 
site to the sides that are the antecedents 
or consequents of the ratio. 

Let the A« ABC, DEF, have theii 
Bides proportionals, 

so that AB : BC=DE : EF; BC : CA 
=EF:FD: 
and consequently (Alg. 112,) BA: AC=ED:DF, 
then /,A=/.D, /.B=Ze, and LC^IV. 
Let the A^FG on the other side of the base EF 
equiangular to the AABC, so that the Z.ABC may be 
the /.GEF. the iBCA= the ZEFG, and consequently d 
ZBAC= the LEGF ; 

then since the A^ ABC, GEF, are equiangalax 

EC : CA=EF ; FG. but BC : CA-EF : FD, (by hjp, 

.-. EF ; FG=EF : FD ; hence FG=FD, (Lem„ p. liB 

again, BC : B A=EF : EG, and BC : BA=EF : ED ; 

.-. EF : EG=EF : ED ; hence EG=ED. (Leni., p. lli 

Wherefore in the two triangles EDF and EOF, I 

three sides of the one are Tespectively equal to the &I 

sides of the other; -■- the angles are = that are oppowtc 

the equal sides, (Prop. 9); that is, the /.DEF =( 

Z,GEI-, but the £GEF= the ZABC; .-. the IDEVizi 

/ABC. 

In the same manner it can he proved, that the iBAC 
the /.EDF, and the iACB to the /DFE. 

Proposition LXll I. — Tdk o hem. 
■ Triangles which have an equal 
angle included between propor- ^ 

tional sides are similar. 

Let the /.a A and D be equal ; 
if we have AB : AC=DE : DF, 
the A« ABC, DEF, are similar. ■_ 

Take AG=DE. and draw GE 
\[ to BC, then (Prop. 17) the /AGE= the /ABC, a 
the /.AEG= the /ACB; .-. the A» ABC. AGE, J 
equiangular, hence (Prop. 61), AB : AC= AG : AE ; 
hutAB: AC=DE:DF; 

.-. AG : AE=DE : DF, (Ax. 1); and since AG H 
DE, AE is =DF. .-. the two A* AGE and imFlii 
two sides, AG, AE, of the one, = two sides, DE, DP, 
ither, and the LA is = the Z.D. .-. the AAGE is eqi 
every respect to the ADEF, (Prop. 5) ; .■. the IM 
= (he ZDEF, and the /AEG is = the /DFE. Hot 
sAown, that the /.AGE \a = tiic LKBC. and that 



A 




PJUANS GBOaiETRy. 121 

:AEG is = the ZACB; .-. also (Ax. 1) the ZABC is = 
he ZDEF, and the ZACB is = the ZDFE. Hence the 
t^s are equiangular^ and .*. similar. (Prop. 61, cor.) 

Pboposition LXIV. — Thborsm. 

Equal parallelograms^ AB, BC^ 
rMch hare one angle FBD of 
lie one, equal to one angle £BG of 
be otherj hare the sides ahout the 
qnal angles reciprocally propor- 
bnalj and parallelograms which 
ATe. one angle of the one, equal 
one angle of the other, and the 

ides about the equal angles reciprocally proportional, (viz. 
OB : BE=GB : BF), are equal. 

Let the sides DB, BE, he placed in the same straight 
Joe; then since the LDBF = the LEBG, add the ZFBE 
)cach; .-. the two L$ DBF+FBE, are = the La GBE+ 
BE; but DBF+FBE, are = two /Za, since DBE is a 
Bught line, (Prop. 1); .-. the Z« GBE + FBE, are to- 
ither equal to two t^Is, and hence (Prop. 2) GBF is a 
nnght fine. 

Now since ZZI7AB=ZIZ7BC, and FE is another ZZZ7, 
dzAB r zz=7FE=zz=7BC : ZZI7FE. (Lem. p. 1 16.) 
But ZII7AB : zz=7FE=DB : BE, and ZZZ7BC : ZZZ7FE 
:GB : BF, (Prop. 58), and .-. DB : BE=GB : BF. 
Next, let the sides about the = Ls be reciprocally pro- 
rtional, that is, let DB : BE=:GB : BF, the n=7AB will 

= the ZZI7BC. 

For ••• DB : BE=GB : BF, and since by (Prop. 58), 
B : BE=ZZI7AB : £=7FE, and GB : BF=ZZ=7BC :/=7 
E; .-. /ZZ7AB : £iZ7FEz= zz=7BC : ZZZ7FE. (Alg. 102.) 
lliat is, the ZZZ7 a AB and BC have the same ratio to 
eci7FE; .-. zz=7AB=z=z7BC. (Lem. p. 116.) 
Cor. 1. Since triangles are the halves of parallelograms, 
<m the same base, and having the same altitude,, equal 
angles, which have one angle of the one equal to one 
l^e of the other, have the sides about these angles reci- 
xally proportional ; and if two triangles have one angle 
the one, equal to one angle of the other, and the sides 
out these angles reciprocally proportionsJ, the triangles 
> equal. 

Cor. 2. If the angle DBF were a right angle, the pa- 
Idogiams would be rectangles; hence the «\de% <)i Vn^ 




I ISS 

equal rectangles can be conTerted into a propDrtian, bj 
makiiig the sides of the one the extremes, and the sides is 
the other the means; and if four straight lines lie propor- 
tional, the rectangle contained by the estremes is equal to 
the rectangle contained by the means. 

Cor. 3. If the means of the proportion be equal, the 
rectangle formed by them nill be a square. Henro, 
" when three straight lines are proportional, the rectangle 

L contained by the extremes is equal to the square on ibe 

^^K mean;" and conrersely. 

^^1 PsOFOaiTioN LXT — Theorem. 

If two chords, AB, CD, cut one ano- n 

ther in a point £, vithin a circle ACD, 
the rectangle contained by the segments 
of the one shall be equal to the rectangle 
contained by the segments of the other. 
That is, the rectangle AE-EB— CE-ED. 
Join CA and BD, then the LCAE is 
= the Z.BDE ; for they stand on the 
same arc CB, and the ZCEA is = the iBED, (Prop.3); 
hence the As AEC, BED, ate equiangular, (Prop. 19, 
cor. 1); hence AE:EC=DE:EB, (Prop. 61.) 
_ And .-. AE-EB=CE-ED. (Prop. 64, cor. 2.) 

^^H Proposition LXVI.^Theorem. 

^^^ If from a point E, without a circle 
ABC, two straight lines be drawn, cut- 
ting the circle, the rectangle contained 
by the whole AE, and the external seg- 
ment, EB of the one, will be equal to the 
rectangle contained by the whole, EC, 
and the estemal segment, ED, of the 
other. That is, AE-EB— CEED. 
Join AC and BD, then the i^EDB is 

= the /.EAC. (Prop. 48, cor.), and the LE Is . 

.-. the As'EBD, ECA, are equiangular, (Prop. 19. cor. Ilj 
I hence AE:EC=DE;EB, {Prop. 61.) 

^^^ And .-. AEEB=CEED. (Prop. 64, cor. 2.) 

I 




PnoPoaiTiON LX V II, — Theorem. 
' If feom a point A witliottt a. circle, there be drawn 




PIiANS GBOMETBT. 123 

secant AB, and a tangent AD, Hie 
ctei^le contained hj the \rh6ie secant 
B, and its external segment AC, will 
i equal to the square on the tangent 
D. That is, ABAC=AD^ 

Join DB and DC, then the ZADC 

ontained hj the tangent AD and the 

Old DC) is = the LDBA, (Prop. 55), 

;d the ^A is convmon to the two A< 

DC, ABD. 

.'.these A» are equiangular, (Prop. 19, cor. 1); and 

nee BA : AD=AD : AC. (Prop. 61.) 

.-. AD*= AC-AB. (Prop. 64, cor. 3.) 

P^iOFOsiTioN LXVin.— ^Th^orbm. 

Ji a point' £ he taken in the circnm- 
ence of a circle, and straight lines A£, 
{ be drawn to the extremities of the ^/ 
meter AB, and also EC perpendicular 
the diameter, then A£ is a mean pro- 
lional between the diameter and the 
scent segment AC ; EB is a mean 
portional between the diameter and 
adjacent segment BC ; and EC is a mean proper tional 
ween the segments of the diameter AC, CB. Or AE^ 
^B-AC, EC'rrAC'CB, and BE^=ABBC. 
Por the A« AEB, ACE, are equiangular, since the Ls 
B, ACE, are r^Ls, and the ZA common, (Prop. 19, 
1); also the ^s AEB, ECB, are equiangular, since 
rhaye the La AEB, ECB, t^Ls, and the LB common, 
op. 19, cor. 1); hence also the ^s ACE, ECB, are 
iaugular, since each of them has been shown to be 
iangular to A^EB. Now, since the A« BAE, EAC, 
equiangular, 

A : AE= AE : AC ; .-. AE^iziABAC. 
Jid since the A« BCE, EC A, are equiangular, 
C : CE= CE : C A ; .-. CE^zzBCCA. 
jnd since the A« ABE, EBC, are equiangular, 
B : BE=BE : BC ; .-. BE2=AB-Be. 
or. 1. Since AEB is a right-angled triangle, and EC 
:awn from the right angle perpendicular to the hypo- 
ise; " the triangles on each side of a perpendicular cm 
hypotenuse, drawn from the right angle of a right 
ed triangle, are similar to the whole, and Ic^ on^ ^AiQ- 




I 



Qter; each aide is a mean proportional between the hj| 
tenuse and its adjacent segment, and the pcrpendicnlai 
proportional between the segments of the hyp 
tenuae." 

Cor- 2. This furnishes another proof of Proposit 
for since AE^=AB-AC, and BE==ABBC ; 

AE'+BE^=AB-AC+ABBC=AB=. (Prop. 35- 



Proposition LXIX. — Theorem. 



^<-^ 



^ 



Triangles ABC. DBE, which have 
one angle, ABO, of the one, equal to 
one angle, DBE, of the other, are to 
one another in the ratio compounded \ / 

of the ratios of the sides about the \ / 
equal angles. That is, AABC : V 

ADBE=ABBC : DBDE. « 

Let the A» ABC. DBE be so placed that AB and I 
may he in one Straight line, then DB and BC will also 
in one straight line; and join AD. then (Prop. 58). 

AABC .- AABD=CB : BD and 

AABD : ADBE=AB : BE ; .-. compounding theM' 
tios, and omitting froro the first and second terms A^l 
we have AABC : AI>BE=ABBC : DBBE. (Alg. 11 

Cor. Parallelograms which haye one angle of the 4 
equal to one angle of the other, are to one another 
the ratio which is compounded of the ratios of the t 
about the equal angles. For they are double of the 
of which the property has just been demonstrated. 

Proposition LXX. — Trkohbm, 

Similar triangles ABC, DEF, b 

are to one another as the 
squares of their corresponding 

For by last Theorem ' 

■-■ the A* fe equiangular, the Z.A being ^ Ou i} 
AABC : ADEF=ABAC: DEDF, 
but AB ; AC=DE : DF, (Prop. 61). 

.-, AB : DE=AC : DF, (Alg. 105), 
and AB : DE=AB : DE, from equality. 

.-. AB' ! DE»=AB-AC: DEDF, (Alg. 1]6). 
I .-.AABC: ADEF=AB";DE", (Alg. 102). 




PLAIIS OBOMETRT* 125 

Oor* EquJangglar parallelograms are to one another as 
e Benares of uieir corresponding sides, since thej are the 
rabies of equiangular triangles. 

Pboposition LXXI. — Theorkm. 

milar polygons ABCDEI, 
urHIKj maj he divided 
to the same number of 
milar triangles, haying 
le same ratio that the 
)l7gons have; and the 
djgons are to one an- 
ther as the squares of their corresponding sides. 
Let E and K be corresponding angles, AB and FG cor- 
ssponding sides, draw the diagonals EB, EC in the one, 
ad KG, KH in the other; and let A and F, B and G, 
and H, and D and I be the other corresponding angles. 
Ince the polygons are similar, and A and F equal Ls, 
AB : AE=FG : FK, and hence (Prop. 63), the A« 
BE, FGK are equiangular; .-. the Z.ABE = the ZFGK ; 
ke away these equal Ls &om the equal U ABC, FGH, 
kd there remain the equal U CB£, HGIC Now since the 
1 ABE is similar to the AFGH, .-. AB : BE=FG ; GK, and 
le polygons being similar, AB : BC=FG : GH; .•• (Alg. 
11), EB : BC=KG : GH; but the Ls CBE, HGK were 
toved equal, .•. the ^BCE is similar to the A^HK; 
mce the Ls BCE, GHK are equal: but the Ls BCD, 
HI are also equal ; .*. the remaining Ls ECD, KHI are 
[oaL Again, since the A^ BCE, GHK are similar, .-. 
C : CE=GH : HK; and since the polygons are similar, 
C:CD=GH:HI; .-. (Alg. Ill) EC:CD=KH:HI; 
It the Ls ECD, KHI are equal, .-. the ACDE is simi- 
X to the AHIK, (Prop. 63.) 
Since the A« ABE, FGK are similar, we have 
AABE : AFGK=BE2 : GK^ (Prop. 70); and since 
the A« BCE, GHK are similar, we have 

ABCE : AGHK=BE2 : GK^ (Prop. 70). 
/. AABE : AFGK= ABCE : AGHK, (Alg. 102;. 
By the same mode of reasoning we should find A^CE : 
iGHK=ACDE: AHIK. 

And from this series of equal ratios, (Alg. 113), we con- 
^ude that the sum of the antecedents, that is the A^ 
^BE, BCE, CDE, or the whole polygon ABCDE is to 
je sum of all the consequents, that is the A* ^^^^ 
'HK, HJK, or the whole polygon FGHIK a& onei «ftXft- 



^IF 




cedent, ABE is to one consequent FGK, or (Prop, 70) 
.AB'':FG^; -■. tLe polygons are to one anoiber as tl 
squares of their corresponding sides. 

Cor. In like manner it may be proved that similu 
figures of any number of aides are to one another as thff 
squares of their corresponding sides ; therefore, universally 
similar rectilineal figures are to one another as the square* 
of their corresponding sides. 

Proposition LXXII. — ^TaEonKM. 

Regular polygons, in circles, having the same diu 
of sides, are similar figures. 

Let ABCDEF, 
KLMNOP, be two 
regular polygons, in- 
Bcrihed in circles hav- , 
ing the same n 
ber of sides, they 
similar to one 

Since the chords AB, BC, i^c., are all equal, the audi 
at the centre are all equal, (Prop. 50, cor. 3); .-, ea(£c 
ihem is a sixth part of 3^0°, for the same reason eachll 
the Ls at Q are ^^ a sixth of 360°, .-. the remaiaiag {j ( 
the As AGB, KQL are togelher equal, (Prop. 19) ; an 
being isosceles, they are equal to one anoiher ; and nu 
each of the angles of the polygons is double of the ^ at ik 
base of the A' ^^ polygons are equiangular ; and since BJ 
the sides of the polygons are equal, it is evident that th( 
sides about the equal angles are proportional, .-. the pol)' 
gons are similar. 

Cor. 1. The polygons are to each other as the squares 
the radii of the circles in ivbicb they are inscribed. 

For they are to each other (Prop. 71) as AB^ : KL 
und the As AGB, KQL being similar, AB' : KU: 
AG' : KQ^ 

Cor. 2. The polygons arc to each other as the Bqnarw< 
the perpcndicuhirs GR, QS from their centres up^ tl 
BJdea. 

For the As AGE, KQS are evidently similar, .-. AG': 
KQ'=GR'' : QS-'; but it was shown (Cor. J) that " 
polygons are to one another as AG* : KQ' ; .-, tiey u 
one anoiher na GR' : QS-. 

Cor. .?. The perimeters, or sums of all the sides, are tfl 
oae another as the perpendicuWa horn \.\te centres. 



ThAKlR OEOHETRT. 12? 

For since the /^s AGR, KQS are similfir, GR : RA=: 
»8: SK, and al^ematelj, (Alg. 105), GR : QS=RA : SK; 
at RA is the half of AB, and SK the half of KL ; /. if n 
i the number of sides of the polygons, 2n X AR will be 
\e perimeter of the first, and 2n x KS the perimeter of 
le second ; but (Alg. 109) AR : KS=2n x AR : 2n X KS, 
(Alg. 102) GR : QS=2n x AR : 2/i x KS. 

Pkoposition LXXIII Leuma. 

If from a quantity there be taken its half, and horn the 
ifiainder ks half, and so on continually, there will at last 
smsin a quantity less than any given quantity, however 
nail that quantity may be. 
Let A be the quantity on which the operation is to be 

fidTonned, the seTeral remainders will be -—, — =-rj» 

'^-^9 and generally the remainder after n subtractions 

31 be "-jp. Let also B be the quantity than which the 
It remiunder must be less, and mB a multiple of B 

aiUer than A; then A ..^i. niB, .*. — .^^B. 

m 

A A A 

ike noyif n such that 2** zp^ m, then—- .^ — , but — -c:! 

2" TO m 

A 
I •-. also much more is— .^ B ; hence a remainder may 

obtained less than B, however small B may be. 
Cor. If the quantity taken away at each subtraction be 
nre than a half, for a stronger reason, the quantity ulti- 
itely left will be less than any given quantity. 

Paoposition LXKIY. — ^Thsorbm. 

If a series of polygons be inscribed in 

nrcle, each having double the number 

sides of the preceding, the difference 

tween the radius and the perpendi- 

lar from the centre upon any side, is 

s than half the difference between 

e radius and the perpendicular from 

e centre on a side of that polygon 

lich has half the number of sides of the former. 

Let AB be a side of a polygon inscribed in a c\ie\e> ^oA 

C a ade of that which has double the numbei ot ^\j^> 







iSB FLAKK GEOMCTRT. 

then GI, the iJifference between the radius and lie per- 
pendicular DI, is .^ the half of EC, the difference betnecti, 
the radius and the perpendicular from the centre, nn tf 
side of lha.t vhich has half the number of sides. 

For the A» AEC, AIF are equiangular, Kince the LA 
common to both, and the icat E and I are j'Z", ■•■ AE;EC 
=AI : IF; hut since AEC is a »■■/., AC ^ AE, .•. AI tb» 
half of AC (Prop. 46) is ^^ the half of AE ; /. since the 
third term is :^ half the first, the fourth is also ^=- half 
the second, that is, IF p^ the half of EC. 

Ag^n, since DEF is a /i, DF ^ DE, but DC=DG, 
.-. the remainder EC ^ FG ; hence, from a quantjtj FG 
.^ EC there has been takep FI z^ the half of EC, .-. the 
remainder IG ^^ half EC. 

In the same mannerj if AG were joined, and a perptn- 
dicular drawn from the centre I> upon it, the diSetenci 
between this perpendicular and the radius would be kil 
than the half of IG, and still more less than the fourth of 
EC. 

Cor. By continually doubling the number of sides of tlw 
polygons, the difference between the radius and the perpen- 
dicular ftom the centre, on a side of the ultimate polygoDi 
may be made less than any given line- 

For it has been shown in the proposition that this il 
equiyalent to taking away from EC more than its half, ani 
from the remainder more than its half, and so on contunir 
ally; .'. (Prop. 73, Cor.) the quantity ultimately leflUlOi 
than any given quantity. 



Proposition LXXV Theoi 



If there be two similar poly- 
^ ins, the one inscribed in a circle, 
and the other described about it, 
the perimeter of the inscribed is 
to that of the described as the' 
perpendicular from the centre 
on a side of the inscribed is to 
the radius of the circle; and 
the polygons are to each other 
as the squares of these lines. u p a 

Let ABCDEF be a regular polygon iiiii:riLei in » 
circle, and GHIKLM a polygon descdbud about the 
circle, hafing its sides parallel to the sides of the in- 
scribed, then (Prop. 20) these polj'gons will he et|iri- 
tngular, and since their sidea mc ei^viaV, ihej aie uoulit 




PLANE 6EOMETST. 129 

bo, let ON be drawn from the centre perpendicular to 
lB« and produced to P^ OP will be perpendiculfur to GU, 
nee AB 18 II GH. 

.'• (Prop. 7% Cor. 3) the perimeter of the inscribed is 
> the perimeter of the described as ON is to OP; and 
Prop. 7^9 Cor. 2) (the inscribed polygon) : (the described 
dygon)=ON^ : OP^ 

Cor. 1. If the number of sides of the polygons be con- 
naally doubled, their perimeters will become more nearly 
^uid dian by any giyen difference. 
For thej are to one another as ON : OP, which is ulti- 
mtdy a ratio of equality. (Prop. 7^* Cor.) 
Cor. 2. The polygons diemselyes, upon the same supposi- 
ion, will become more nearly equal than by any given 



Oor. 3. Since the circle is always greater than the in- 
cribed, and less than the described, polygon, each of them 
nO ultimittely become more nearly equal to the circle than 
J any giyen difference. 

Cor. 4. The perimeters of each of the polygons will 
Itimately coincide with, and consequently be equal to, 
iie circumference of the circle. 

Cor. 5. Circles are to one another as the squares of 
kir radii. 

For (Prop. 72, Cor. 1) the polygons inscribed in them 
le to one another as the squares of the radii of the circles 
1 which they are inscribed, and these polygons are ulti- 
late equal to the circles ; .*. the circles are one to another 
8 the squares of their radii. 

Pbofosition LXXVI. — Problem. 

Haying given the perpendicular from 

lie centre of a circle upon a side of a 

gore inscribed in it, to find the per- 

endicular upon the side of a figure 

aving double the number of sides. 

Let CD be the given perpendicular, 

nd OF the required one. 

Join AE ; then because AB is a dia- 

oeter, AEB is a semicircle ; .-. the ZAEB is a t^L (Prop. 

>'2); but since CF is perpendicular to EB, AE and CF are 

I; .-. the A» AEB, CFB, are similar; and since AB is 

buble of CB, .-. AE is double of CF; and since AD is 

imendicular to EG, ADE is a /L, and .-. = the LAEB, 

M the angle at A is common to each of the ^s> "BklL^ 




^ 



iSD Kiuaoaomnnr. 

EAD, hence they are equiangular; .'. AB : AE=AE:, 
AD, (Prop. 61), and AE-=ABAD; but AE was proTcd 

to be double of CF, and (Prop. 3?, cor. 2), AE'^=4CF'A 
nibBtituting this value, we haTe 4GF^=AB-AD; .-. CF* 
=ziAB'AD, and conaeiuently CV-^^AliAD. 

Scholium. If a numerical result be required, let EG Iw 
the Bide of an equilateral triangle, and let the radias of the. 
circle be 1 , then CD is equal to ^ ; for if EC were joined, 
EC6 fTOuld evidently be an equilateral triangle, and the 
perpendicular on CB would bisect it, (Prop. 7. cor. 2.) 
.'. AD would be (l+i)::zS, and AB would be two; pu^ 
ting these valueS in the expression above, it becomes CF= 

Generally, if p he the perpendicular from the centre, 
upon a side of a figure inscribed in a circle, p, the perpen- 
dicular upon a aide of one having double the number "i 
jhj the perpendicular on a side of one having four 
the number of aides, and ;i„, the perpendicular on " 
ide of one having 2" number of aides as the first; then 

Pi= J^-f. P,= y ^ . and p.= J'—--- 

Also, {Prop. 68), EB=V-AB\BD=Va{I— p), la^iii 
being one ; in the same manner, if s be a side of the o^ ' 
nal polygon g,, aside of that of double the Dumber of sidi . 
and generally a„ a side of one having 2*" the number rf 
rides as the first, we will have «„+i— */2(l— p„), and if 
n be the number of sides of the original polygon in& 
which we commence, the whole perimeler of the polygon 
will be =n2-+>j2il—p„). Hence, {Prop. 72, cor. 3), 
if this operation be repeated often enough, the result ffitl 
be the circumference of the circle to radius one. 



Proposition LXSVII. — Peoblkm. 



To divide a given line AB 
into two parts, such that the 
greater part, Ah', shall be a 
mean proportional between the 
whole line, AB, and the re- 
maining part, BF. 

Let BC be drawn J- to AB, ~ ' » 

from the extremity B. of the 

aa-aigbc Une AB, and =: the half of AB; join AC, wJ 
produce it to E, and from l\\e ccnue C, Vv'CB.\lifi wdirf 



OXOMXTBY OF PLANXS. J 31 

^ describe the semicircle DBE« and from AB cut off 

JF^ AD ; then since AB is perpendicular to CB, it is a 

ine;ent to the circle at B ; .*. (Prop. 67)t EA : AB= AB : 

D ; but since DE and AB are each double of CB, they 

re equal, and AFs AD ; .-. (Alg. 107), EA— AB : AB= 

B— AD : AD, which, from the aboTe mentioned equali- 

es, giyes AF : AB=BF : AF; hence, (Alg. 105), AB : AF 

:AF : BF. 

Cor. Since AB=DE, the first proportion gives AE : ED 
:ED : AD ; •*• the line A£ is also divided in the same 
amner. 

Scholium. A line divided as in the proposition, is said 
) 1)6 divided in extreme and mean ratio, and also to be 
It in medial section, or to be divided mediidl j. 



GEOMETEY OF PLANEa 

Dt^FiNiTiovs. — ^I. A solid is that which hath length, 
tsdth, and thickness. 

II. The boundaries of a solid are superficies. 
IIL A straight line is perpendicular, or at right angles 
• a plane, when it makes right angles with every line 
hich meets it in that plane. 

IV. A plane is perpendicular to a plane, when the 
might lines drawn in one of the planes perpendicular to 
« common section of the two planes, are perpendicular 
the other plane. 

Y. The inclitiaiton of a straight line to a plane is the 
fate angle contained by that straight line, and another 
awn from the point in which the first line meets the 
ane, to the point in which a perpendicular to the plane 
Bwn from any point in the first line, meets the same 
ane. 

VJ. The inclination of a plane to a plane is the acute 
igle contained by two straight lines drawn from any the 
me point of their common section, at right angles to it, 
le upon one plane, and the other upon the other plane. 
VII. Two planes are said to have the same, or like 
clinations to one another, which two other planes have, 
hen their angles of inclination are equal to one another. 
YIII. Parallel planes are such as do not meet qiv^ 
(Other though produced. 



IX. A Htraiglit line and a plane are parallel, if they do I 

not meet when produced. ] 

■". The angle formed liy two interaecting platiea « 



called a dihedral angle, and is meaaurei 

XI. Any two angles are said ( ' 
when they are either both greatt 
a right angle. The 
same or equal circles, when they i 
both not greater than a quadr; 



Def. 

of the same affeeliim, 

both not greater than 

! term is applied to arcs of the 

e either both greater or 




I PRorosiTiON LXXVIII.^ — Theorem. 

Any three atrmght lines which 
meet one another, not in the same 
point, are in one plane. 

r^t the straight lines AB, BC. 
CD meet one another in the points, 
B, C, and E, AB, BC, CD are in 
the same plane. 

Let any plane pass through the 
Straight line AB, and let the plane, produced if necessary, 
be turned about A B, till it pass through the point C. Theni 
because the points B and C are in the plane, the straigbl 
line BC is in the plane, (Def. 7) ; and because the pointt 
C and E are in the plane, the | CE or CD is in the 
plane, and by hypothesis AB is in the plane ; -■. the thr» 
straight lines AB, UC, CD are all in one plane. 

Cor. 1. Any two straight lines that cut one anothtr 
are in one plane. 

Cor. 2. Only one plane can pass through three point^- 
or through a straight line and a point. 

Cor. 3. Any three points are in one plane. 



PflOPDBITION LXSIX.— 

' If two planes cut o 



^^^B> If two planes cut one another, th( 

P common section is a straight line. 

Let two planes AB, BC cut 
another, and let B and D be 

' points in the line of their 

section. From B to D draw the 
straight line BD, then since B and D are points JD tilt 
plane AB, the line BD is in that plane, (Det^ 7) : for th* 
same reason it is in the plane CB ; .-. being in each of lli> 

I planes, it is their common section ; hence the comuiiis 

cecfioit of the two planes ia a sWa\^t\me. 



OSOM KTRT OF PLANB8. 



133 




Proposition LXXX. — ^Thbobbm. 

If a straight line stand at right angles to each of two 
tnight lines in the point of their intersection, it will also 
leat right angles to the plane in which these lines are. 
Let PO he -i- to the lines AB, 
DD, at their point of intersection 
), it is -i- to their plane. 
For take OB=OD, and join 
PB, PD, and BD, and draw any 
ine FE, meeting BD in "E, and 
join P£. 

Tben ".- BO=:DO^ and PO common to the two A« 

POB^ POD, and the contained angles r^ls, the base PB= 

PD, .*. the A^BD is isosceles, and the AOBD is isosceles 

hj construction. Now •/ POD is a r'Z, PO^ + OD' = PD*, 

bat OD«=OE'»+DEEB, (Prop. 44), and PD«=PE*+ 

DEEB, .-. PO^+OE«+DEEB=PE'+DEEB; take 

the rectangle DE'EB from both, and there remains P0^4- 

OE*=PES .-. the ZPOE is a right angle, and PO is at r^Ls 

to EO ; in the same manner, if the line EO had been in any 

if the other angles BOG, CO A, or AOD, it could be demon- 

teted to be at /Za to PO ; .•. when PO is at r^is to two 

taight lines at the point of their intersection, it is at 

0|^t angles to ererj Ime in that plane. 

Cor. 1. If a plane be horizontal in any two directions, it 
b so in eyery direction. 

Cor. 2. The perpendicular PO is less than any oblique 
Kne, as PB, (Prop. 23), and therefore the perpendicular 
measures the shortest distance from the point P to the 
plane. 

Proposition LXXXT ^Theorem. 

If three straight lines meet all in one point, and a 
straight line stand at right angles to each of them in that 
point, these three straight lines are in one and the same 
plane. 

Let the | AB stand at r^ U to each 
of the \s BC, BD, BE, in B, the 
point where they meet ; BC, BD, and 
BE are in one and the same plane. 

If not, let, if possible, two of them, 
as BD, BE, be in the same plane, and 
BC aboTe it, and let a plane pass 
dnongh AB^ BC, and cut the plane in w\i\c\i'&T> vcA^SS^ 




are in the [ BF, (Prop. 79) ; then since AB is at r* 
each of the \a BD. BE, at the point of their intersection, il 
is also at / Li to BF, (Prop. 80), which ' " 
plane ; .-. the ^ABF is a r-/ ; but by hypoiheBis, the lABO 
is also a t'L; hence the ZABF=: the iABC, and thej 
are hoth in the same plane, nhieh is impossible; .'> th« 
I BC is not above the plane in which are BD and BE, and' 
in the same manner it may be shown (hat it is not below 
it. Wherefore the three [s BC, BD, BE, are in one mi 
the same plane. 

PnoPosiTioN I, XXXII, — Theorem. 

If two straight lines be at right angles to the same plani 
they eball be parallel to one another. 

Let the [g AB, CD be at i-' Ls lo the 
tame plane; AB is || CD. 

Let them meet the plane in the 
points B. D, and draw the | BD, to 
which draw DE at r" Zs tn the same 
plane; and make DE=AB, and join 
BE, AE, AD. Then ■.■ AB is -J- to 
the plane, it shall make r* is with 
erery | which meets it and is in that plane, (Def. 3) j '■ 
BD, BE, which are in that plane, do each of them n 
AB ; .-, each of the b ABD, ABE is a r'Z- For i 
same reason each of the Ls CDB, CDE is a /Z ; aw 
AB=DE, and BD common, the two sides AB, BD 
■= the two ED, DD, and they contain r* L/. .: the l 
AD=BE. (Prop. 5). Again, :■ AB=DE, and BE=AD, 
and the base AE common to the Aa ABE, EDA, the 
ZABE = the ZEDA, (Prop. 9); but ABE is a /i, 
EDA is also a i-'Z, and ED is J. DA ; but it is also 
to each of the two BD, DC : wherefore ED is at r* Z* t* 
each of the three |« BD, DA, DC, in the point in whidi 
they meet, .-. these three \s are nil in the same planer 
(Prop. SI ) ; but AB is in the plane in which are BD, DA, 
(Prop. 78). since any three |s which meet one another «t 
in one plane. -■. AB, BD. DC are in one plane; toA 
each of the U ABD. BDC is a ^'Z ; hence (Prop. Ifl, 
cor. 2) AB is II CD. 

pBOTOsmos LXXXIII Theobem. 

If two straight lines be parallel, and one of them ia rf 
right angles Co a plane, the other shall also be at i^l 
angles to (he same plane. ' 




OSOMBTBT OF PLANES. 



135 





Let AB» CD be two || straight 
lies, and let one of tfaem AB be at 
^« to a plane ; the o^ier CD is at 
Ls to the same plane. 
For, if CD /be not -L. to the plane 
• which AB is -l., let DG be J- to 
. Then (Prop. 82) DG is || AB; 
DG and DC are both || AB, and are drawn through 
le same point D^ which is impossible. 

Pboposition LXXXIV. — Theorem. 

If two straight lines be each of them parallel to the 

one straight liney though not both in the same plane with 

, ihey are parallel to one another. 

Let AB; CD be each ^ j^ ^ 

wm II EF, and not in the ^ r -b 

me plane with it ; AB shall k^ -\o .y 

HI CD. . • / " 

In EF take any point G, ^ ^ » 

m which draw, in the plane. passing through EF, AB, 
e I GH at r^ £« to EF ; and in the plane passing through 
F, CD, draw GK at r*. Ls to the same EF. And be- 
Qse EF is -L- b6th to GH and GK, EF is -i- to the 
me HGK, passing through them ; and EF is || AB ; .-. 
B is at 7^ ^« to the plane HGK, (Prop. 83). For the 
BMi reason, CD is likewise at t* Le to the plane HGK. 
Bnce AB, CD are each of them at / Z« to the plane 
SK. .% (Prop. 82) AB || CD. 

PnoposiTroN LXXXT. — ^Theorem. 

If two straight lines AB, BC, 

ieting -one another, be pa- 

lei to two others DE, EF, i 

it meet one another, though 

t in the same plane with the 

it two, the first two and the 

ler two shall contain equal angles. 

Take BA, BC, ED, EF, all equal to one another ; and 

n AD, CF, BE, AC, DF ; •.• B A is = and || ED, 

AD is both = and || (Prop. 24, cor. 1) to BE; for 

i same reason CF is = and || BE ; .*. AD aod CF 

i each of them = and || BE. But |« that are || the 

ne I aire || one another, (Prop. 84); .*. AD is ^ C¥ *, 

1 it ia eguai to it, and AC, DF, Join tkem tAHvax^ 




tbi 



le parts, and .-. AC is = and || DF ; and ••■ AB, 
= DE, EF, and the base AC to tiie base DF, .-. 
le ZABC = the ZDEF, (Prop. 9.) 



Propobition LXXXVI.- 



■Problbm. 

plane, fion 



k 



To draw a straight line perpendicular to ; 
a given point above it. 

Let A be the given point above the plane BH ; it is re- 
quired to draw from the point A a I -^ to the plane BH. 

In the plane draw any | BC, and 
from the point A draw AD J- to 
BC. If AD be also -i. fo the plane 
BH, the thing required is done; 
but if it be not, from the point D 
draw, in the plane BH, DE at i' /.s " 

to BC ; and from the point A draw AF -^ to DE, sod 
through F draw HG |] BC; and •.- BC is at t' ti lo 
ED and DA, BC is at r' Is (Prop. 80) to the plane passise 
through AD, DE ; and GH ia || BC, .-. (Prop. 83) Gl 
is at r* is to the plane through AD, DE ; and is .*. J 
to every 1 meeting it in that plane. But AF, which ti 
in the plane through AD, DE, meets it; wherefore GBii 
-L to AF, and DF is also -i- to AF ; hence AF is -»- te 
Mch of the Is GF. FD, and ■.■ AF ia -i- to the plane ia 
which GH and ED are, (Prop. 80), that is to theplue 
BH. Hence from the given point A, above the plaoe 
BH, the I AF is drawn J- to that plane. 

Oor. !• If it be required from a point C in a plau 
to erect a perpendiculaT to that plane, take a point A 
above the plane, and draw AF perpendicular to the plane; 
if from C a line be drawn parallel to AF, it will 1« 
the perpendicular required. 

For, being parallel to AF, it will be perpendicular to tbe 
same plane to which AF is perpendicular. 

Cor. 2. From the same point, whether without or in ■ 
plane, there can only be drawn one perpendicular to tin' 

Proposition LXXXVII. — Tbeorbm. 

Planes CD, EF, to which the same straight line AB !■ 
perpendicular, are parallel to one another. 

If not, they shall meet one another when produced; W 
them meet their common section, will be a \ GH, i" 
which take any point K, and join AK, BK ; then, v AB 
w ^ to the plane EF, il is ^ (Oet V), Ka the | BK, whidl 






GKOIIBTKT OP rLAKES. 

a thatpIoBe; .-. ABK ia a t'L; tor 
■ame reason, BAK ia a r'Z ; wher»< 
I the two /j ABK, BAK. of the 
\BK, are together equal to two r'lt, 
ich la impomble, (Prop. 10, cor.) ; .'. 
{danea CD, EF, thoneh ^rodoced, do 
meet one another ; UuU ia, thej aie 
illel, (Def. &> 



PsoroeiTioN I/XXXYIII.— Thborbh. 

f two atroight lines AB. BC, which meet one another, 
MfwctiTel; parallel to other twoDE, £F, which meet one 
diet, but are not in the same plane with the first two; 
phM which passes thtDi^;h the first two is parallel to 
t wUdi passes through the latter. ' ^ 

Ora^h the pcrint B draw BG -«- to b_— — ffc^r^'" 




-Jj^--~^ 



I^MW which, passes through EF, 

^lIVopi 86), meeting it in G; and 

«gh 6 dnw GK if EF, and GH 

D. And -.■ BG is -I- to the plane 
it is a. (I>e£ 3) to GK and ^ 
which meet it in that plane ; ■'• 
of the Ls BGH, BGK ia a r'L. And since AB and 
are each || ED, they are || one another, (Prop, 
and BGH ia a r'L, ■■■ GBA ia also a r'L Agwn 
is II BC, for each of them is || EFi .-. since KGB 
r'L GBC is also a I'Z, (Prop. I?)- Since .-. BG is 

'Z« to each of the linea BC and BA, it ia ~i--to the 

le AC, and it was drawn -J- to the plane DF. Since 

1 BG is -<- to each of the planes AC, DF, ihese planes 

parallel, (Prop. 87). 

PnopoaiTioiT LXXXIX, — Tbeobem. 

r two parallel planes AB, CD, be out by a third 
GH, their common sectiass EF, GH, with it are 
dlel. 

'or the \s EF and GH are i: 
e plane, namely EFGH, which 
I the planes AB and CD ; and 
f do not meet though produced : 
the planes ia which they are do 
meet : .*• EF and GH are parallel 



.Ml" 



SSaKETBT OK PUHKB. 




"^ 


&. 




" 




^^ 




" 



^ 



PflopOBixroN XC— Tbkobem. 

If two parallel planes AB, CD be cut by a third plan* 
EH, they have the same inclination to that plai 

Let the 1* EF and Gil 
bo the common BecHons of 
the plane EH, with the two 
planes AB and CD ; and 
from K, any point in EF, 
draw in the plane EH ' 
the I KM at / Le to 
EF, and let it meet Gil in I. ; draw also KN at / l» tff' 
EF in the plane AB; and through the |s KM, KN, lets 
plane be made to pass, cutting the plane CU in the line LO. 

And ■■■ EF and GH are the common sections of th» 
plane EH, with the two parallel planes A B and CD, EF i» 
tl GH, (Prop. 89). But EF is at r-^s to the plai 
passes through KM and KN, (Prop. 80) ; ■.■ it ia at r-Z* 
to the lines KJt and KN ; .-. GH is abo at r>la to ths 
same plane, (Prop. 83) ; and it ia .-. at r'Zir to LM, LO, 
which it meets in that plane, .*. since LM and LO are it 
T^Ls to LG, the common section of the two planes CD aod. 
EH, the Z.OLM is the inclination of the plane CD to ll 
plane EH, (Def. 6). For the same reason the Z.NK1 
is the inclination of the plane AB to the plane EH. Bi 
V KN and LO are parallel, being the common sectioDsof&a 
parallel planes AB and CD, with a third plane ; the interiw 
iNKM is = the exterior /.OLM ; that is, the inclinatia« 
of the plane AB to the plane EH is =: the inclination d 
^ plane CD to the same plane EH. 

PnoFosiTiON XCI. — Theorem. 



AB, CD be cut by parallel planH 
points A. E, B ; C, F, D ; the; 
liime ratio : that ia, AE : EBa 



If two straight 
GH. KL, MN, in 
shall be cut in 
CF : FD. 

Join AC, BD, AD, and let AD meet the 
plane KL in the point X; and join EX, XF. 7] 
-.' the two parallel planes KL, MN, are cut 
by the plane EBDX, the common sections 
EX,BDareparallel,(Prop.89). Forthesame 
reason, -.■ the two parallel planes GH, KL, ^f 
are cut by the plane AXFC, the common ii 
sections AC, XF ate parallel. And ■■• EX 
w // BD, a Bide of the A^^Ti. KE-.EB=AX: 




op.59). Again, v XFis [| AC, aside of the AADC, 
:!XD = CF : FD. .-. AE : EB=CF : FD, (Alg. ] 

tpROPOSITIOM XCIl, — Theohem. 
ight line AB be perpendicular to a plane Gti 
plane CE toucliing it will be perpendicolai to (" 
« plane CD. 

'or let CBG be their line of com- 
1 section, and from any point G, 
CG, let EG be drawn -i- to it in 
plane OE. Also, let BF. Gil be < 
t« CQ in the plane CD. Then 
LB 18 J- to the plane CD, ABF 
i /L ; and Bince EG is (| AB, 
I GH is II BF, the ZEGH ia also a v'l, {Prop, fl 
t the Z.EGH is the inclination of the plane CE to the 
le CD, (Def. 6) ; .-. the plane CE is at lY* to tlwj 
le CD. "^ 

PaoposiTioN XCIII. — Theokkm. 

' two planes AB, BC, cutting one 

her, be each of them perpendicular 

k third plane ADC, dieir common 

ton BD will also be perpendicular 

be Bame plane. 

h>m D in the plane ADC, draw 

-L to AD, and DF -l to DC. v 

is X to AD, the common section of ^ 

ilanea AB and ADC,and -.- the plane 

is at I'La to ADC, DE is at I'ls 

be plane AB, and .-. also to the | BD in that plane- 

the same reason DF is at r'ls to DB. And since 

is at t'Is to both the lines DE and DF, it is at i^Li 

e plane in which DE and DF are,' that is, to the plane 

-JProp. 80). 




ijProi 



SOLID GEOMETEY, 



.^golid angle ia that which is formed by more tliid 
plane angleB at the Bame point, but not in iVe aax 



k 



BOt-tD eSUUETBT. 

I solida bounded by planes are Eimtlar, when 
theii solid angles are equal, and tbeir plane figures similai^ 
each to each. 

III. A parallelopiped is a solid bounded by six planei, 
of which the opposite ones are parallel. If the aojaceat 
sides be perpeni^culai to one another, it is a reclangnlar 
parallelopiped. 

IV. A ciile is a rectangular parallelopiped, of wIucIl 
six sides are squares. 

Y. A prism is a solid of which the sides are paislldo 
grams, and the ends are plane rectilineal figures. 

VI. A pyramid is a solid of which the sides are triangle^ 
having a common vertex, and the base any plane re»' 
tilineal figure. If the base be a triangle, it is a triangoltt 
pyramid. If a square, it is a square pyramid. 

VII. A cyUndei- ii a solid described by the rerolntioK 
of a rectangle about one of its sides remaining fixed ; whiA 
side is named the axis ; ajid either of the circles describe^ 
by its adjacent sides the base of the cyhnder. 

VIII. A cone is a solid described by the revolution of i 
right angled triangle about one of its sides remaining fixe^ 
which is called the axis ; and the circle described by Ihtf 
other side is the base of the cone- • 

IX. Cones or cylinders are similar when they are d« 
scribed by similar figures. 

Pboposition XCIV — Theohem. 

If a solid angle A be contained by three plane a ^ 
BAG, CAD, DAB, any two of these are together greatlt 
than the third. ^ 

If all the Ls be =, or if the two 
greater be =, the proposition is evi- 
dent. In any other cose, let BAC be 
the greatest /., and' let ISAE be cut 
off frota it, = DAB. Through any i 
point E in AE, let the | BEC be 
drawn in the plane of the Z.BAC to meet its sdes il 
B and C. Make AD=EA, and join BD, DC. •.- ih 
A* BAD, BAE have AD=AE and AB common, an 
the included Ls BAE, BAU=, the base BE is = BE 
(Prop. 5). But the two sides BD, DC are IP' BE, EC, . 
DC is :^ EC ; and since the side AE is = AD, and A( 
common to the two At ACD, ACE, but the base DCj" 
EC, .: (he ^CAD is ^^ EAC, (Prop. 14). Conseqiwnll 
the sum of the is BAD, CM) \f.^ '^'i LBAC. ^ 



St*n» BWHTBTBI'.' 




Pbopo9ition XCV. — Teieohem. 

II tlie plane angles which form any soUd 
e, are together leas than four right angles, 
et there be a solid L at the point A con- 
ed by the plane U BAC, CAD, DAE, 
3, these L», taken together, ale ,^ four 
For, through any point B in AB, let a 
.e be extended to meet the aides of the , 
[ ^ in the lines of common section BC 
, DE, EB, therehy forming the pyramid BCDE — A^^ 
from any point O within the rectilineal figure BCDI^^ 
ch is the base of the pyramid, let the \» OB, OC, OD, 
, be drawn to the angular points of the figure, nhich 
I thereby be divided into as many ^$ as the pyramid 
aides. Then '■■ each of the solid Ls at the base of the 
imid is contained by three plane i«, any two of them 
together ^ the third, (Prop. 94.) Thus, ABE, ABC, 
together z^' EBC. .■. the Z« at the bases of the A* 
ch have their vertex at A, are together ^^^ the L> 
le bases of the As which have their vertex at O. But 
he Li of the former are together = all the Zs of ihe 
rr. .'. the remaining L» at A are together -^ (he re- 
nins L» at 0, that is, -^ four right angles, (Prop. 1^ _ 

i 

Proposition XCVI. — Theobem. ~ 

' two solid angles he each of them contained by three 
e angles, and have these angles equal, each to each, 
alike fiiluated, the two solid angles are equal, 
et there be a solid angle at A contained by the three 
e U BAC, CAD, DAB, and a solid angle at E, con- 
ed by the three plane U PEG, GEH, HEF, = 
ler, each to each, and alike situated, these solid a 

^or let the [» AB, AC. 
. EF, EG, EH, b ■■ 
tl, and let their e 
es he joined by the 
i BC. CD, DB, EG, 
, HE; and thus there 
formed two isoscele 
unids, BCD-A, and 
a-E. Vpon the bases BCD. FGH, let tte - 




143 



souiD oaoHZmT. 



EL, fall from the Tertices A, E. Then ■■■ the I'Ld A' 
AKB, AKC, AKD, have equal hj^oteniues, and the Hde 
AK common, their other aides,KB,KC,KD, (Prop. 39, cor, 
2). are also equal ; .-. K is the centre of the circle that cir- 
cumecribes the ^ BDC> In like maaner, L is the centre rf 
the circle that circumBcribea the A FHQ, But these A* 
are equal in every reapect. For the sides BC, FG, are- 
equal, '.■ they are the bases of equal and similar A* BAQ 
EFG; and for the same rnnsou, CD=:GH, andBD=HF. 
If. .■. the pyramid BCD-A be applied to the pyramid 
FGH-E, their bases BCD and FGH will coincide. Al» 
the point K will fait on L, ■.■ in the plane of the circh 
about FGH, no oilier point than the centre ia equaUjr 
distant from the three points in the circumfereDMi 
the perpend ionlai KA will coincide with LE, (Prop. 8% 
cor. 2), and the point A will coincide with E, ■■• the t'Li 
A» AKB, ELF, have equal hypotenuses, (Prop. 39, cor. 2), 
andBK=FL. But the points B, D, 0, coincide wi* 
F, H, G, each with each ; .-. the |s AB. AC. AD, coineife 
with EF. EG, EH, and the plane U BAG, CAD, DAft 
with FEO, OEH, HEF, each with each. Consequeallf 
the solid i» themselves coincide and are equal. 

Cor. 1. If two solid angles, each contained by thrrt 
plane angles, have their linear sides, or the planes that 
hound them, parallel eaeli to each, the solid angles an 
equal, (Props. 84 and 89.) 

Cor. 2. If two solid angles, each contained by the samfl 
number of plane angles, have their linear or plane ai» 
parallel, each to each, the solid angles are equal. For eftcK 
of the solid angles may be divided into solid is, each oon- 
tained by three plane Li, and the parts being eqTuJ, and 
alike situated, the wholes are equal. 

I PaoposiTioN XCVII. — Theorem. 

If two triangular prisms. ABC-DEP, and GHK-LMH, 
have the plane angles BAG, CAD, DAB, and ^GKj, 



L 



KGL. IXJH, and ale 
GH, GK, OL, about two of 
their solid angles at A and G 
equal, each to each, and alike 
situated, the prisms are equal 
and similar. 

For, since the Bolid Ls A, 
O, are each contained by 
three plane Za, wbicb ait 



T Bides AB, AG, AD, »U' 



*Liit*L 



MUD OEOMETBT. 1 43 

lal eacli to each, these solid ig beinj; applied to each 
er, coincide, (Prop. 96.) .-. the U AB, AC, AD, coiii- 
e with the \g GH, GK, GL, each with each. But AB 
nciding nith GH. and the point D with L, the |DK 
at feifupon LM, (AS.16J, and BE on HM; .-. the 
nt £ coincides with M, and the point t' with X. Con- 
[oently the rectilineal figures which bound tlie one 
sm, coincide with, and are equal and similar to the 
tiliDcai figares which bound the other, euch to each ; 
! solid Zs of the one coincide with and nre equal to the 
id l» of the other, each to each ; and the sulida them- 
ves wte equal and aimilar. 

Cor. 1. If two triangular pyianiids have the plane 
glea and linear sides about two of their sulid angles 
ual. esefa to each, and alike situated, the pyramids aie 
nal and Gimiinr. 

Cor. 2. If two triangular prisms have ihe linear sides 
init two of their solid angles both equal and parallel, 
;h la e»ch, the prisms are equal and simitar. 
Cor. 3. If two pyramids have the linear sides about 
lir vertical angles both equal and parallel, each to 
I jvjraniids are equal and similar. 



Proposition XCVIII. — Theokkm. 



""^^ 



the opposite sides, as ABCD, EFGIl, of a parallelopi- 
I ABCD-EFGH, are similar and equal parallelograms, 
1 the diagonal plane divides it into two equal and simi- 
pnsinB. 

Por, since the opposite planes, AF, 
ir.areparallel.Andcutbya third plane 
)) the common sections AD, UC, are ^ 
'allel, (Prop. 89) ; for a similar reason 
> and BC are parallel ; hence the 
lire ABCD is a i — ? . 
[n like manner, all the figures which 
and the solid are i — i s. Hence AD 
EH, and DC=HG, and the iADC is =EHG, (Prop. 
); .-. the As ADC, EHG, (Prop. 5), and conse- 
ently the c=?s ABCD, EFGH, are equal and similar. 
Again, -.- AE, CG, are each — and || DH; they arc 
and II one another; .*. the figure ACGE is a c=7,(Prop. 
, cor. I), and divides the whole parallelopipeil AG into 
o triangular prisms ABC-EFG. and ACD-EGH, and 
ese prisms are equal and similar, (Prop. 9T)-, •.- ftva 
ine If and iiaear sides about the aoliil i.s a^. V n.i\i.\> 




I4-I aClLID OEOMETBT. 

are equal eacli to each. For the plane Ls EFG, ADC, aw. 
:= one another, each being =^ ihe ZEHG, and the linear 
sides FG, DA, ate := one anotber, each being =EH, and. 
so of the others. 

Cor. I. The opposite solid angles of a parallelepiped 

Cor. 2. Every rectangular parallel opiped is bounded bf 
tec tangles. 

Cor. 3. The ends of a prism are similar and equal 
figures, and their planes parallel. 

Cor. 4. Eyery parallelopiped is a quadrangular prisnii 
of nbich the ends are parallelograms, and conversely. 

Cor. 5. If tno parallelepipeds have the plane anglei 
and linear sides about two of their solid angles equal, eacli 
to each, and alike situated, or the linear sides, about two 
of their solid angles, both equal and parallel, each to eaeb, 
the solids are equal and similar. 

Cor. 6. If from the angular points of any rectilineal 
figure, there be dravrn straight lines above its plane, all 
equal and parallel, and their extremities be joined, Ihs 
figure BO formed is a prism. If the rectilinear figure bes 
parallelogram, the prism is a parallelopiped. If the recti- 
lineal figure he a rectangle, and the straight lines be per- 
pendicular to ila plane, the prism isarectangular parallelo- 
piped. 

Cor. 7- Every triangular prism is equal to another 
triangular prism, having its base equal to the half of ona 
of the sides of the prism, and its altitude the perpendicnlu 
distance of the opposite edge. The prism EEG-BACs 
ABF-DCG, (formed by the diagonal plane AFGD), 
which has for its bitse half the side ABFE, and for its alti- 
tude the perpendicular distance of the edge CG ; for thej 
'. are each half of the parallelopiped FD. 

I Proposition XCIX. 

Parallelepipeds ABCD-EFGH, and ABCD-KLMN 
upon the same base ABCD, and be- __ 

tween the same parallel planes AC, /]_ 

EM, are equal. //y 

Because the \» EF. GH, KL, MN, 'fr^—/Mf 
are || AB, CD, they are i| one A i\///n,/ 
another, and being all in the same \\y/ /// 
plane, NK, ML, being produced, will \^Em/-''' 
.-. meet both EF and HG ; let them WW' 
meet the former in 0, P, an4 Xbe\aV- * 



SfWitB OEOSfliTBT. 



149 



in R. Q, and let AO. BP, CQ, DR, be joined. Tlie 
re ABCD-OPQR, is a parallelopiped. For, by hypo- 
lis. the plane Ea !s || AC, the plane of the paralleh 
:-IIQ is II the plane of lUe parallels AB-EP, and 
plane AD-NO parallel to BC-MP. Hence AEO- 
IR and BFP-CGQ are two triangular prisma, whicL 
e (he linear sides AE, AD. AO. about the solid /A, 
0. equal and || the Hnear sides BF, BC\ BP, about 
solid Lli, (Prop. 98), each to each. Consequently 
se prisms are equal, (Prop. 97), and each nf them being 
en away from the whole aoHd ABCD-EPQH, the re- 
iaders, the parallelopipeds AG, AQ, are equal. In the 
le manner, the parallelopiped s AM and AQ may be 
)Ted equal, .'.the parallelepipeds AG, AM, are equal 
one another. 

Cor. Triangular prisma, upon the same base, and be- 
een the same parallel planes, are equal. 
For, if two planes be made to pass, the one through 
!, EG, and the other through AC, KM, they will bisect 
paratlelopipeds AG, AM, (Prop. 98) ; .-. the prisms 
>C-EHG, ADC-KNM, will be equal. 

Pro POSITION C, — Theorem. 



»araIIelopipeds ABCD-EFGH, and ABKL-OPQR. 



i parallel pb 



L equal bases, and between the s 
:, EO, are equal. 
I^ASB I- When the bases have 
) side AB common, and lie be- 
!eo the same parallel lines AB, 
Z, these paraUelopipeda AG) 
I, are equal. 

?0T let the planes EAL, FBK, 
Bt the planeHC of the former, 
the lines of common section j 
[, KN, and the plane HF in 

lines EM, FN. Then since EA, AI„ are || FB, BK 
irpianesareparallel,and the plane EB ia || the planeHK; 
he figure AN is a parallelopiped. But ADL-EHM, 
1 BCK-FGN, are two triangular prisms, which have 

linear sides, AD, AE, AL, about the solid Z.A, both = 
1 II the linear sides, BC, BF, BK, about the solid /.B, 
h to each ; .-. these prisms are equal, (Prop. 97, cor. 2) ; 
I each of them being taken away from the whole solid, 
{KD-EFNH, the remainders, the paraUelopipeda AG, 
*, are equal. But AN is =AQ, (Prop. 99) ; .•• ^G^a 
4Q. 




^ 



SOLID OBOMETBT. 

Cabs II. When the bases ABCD, 
CEFG, are equiangular, haying the 
/DCB = the IGCE, place DC, CE, 
iu one |, then GC, C13, are also in one 
]. Also let their sides be produced 
to meet in the points II, K. Since ^ 
the c^s AC, CF, are equal, they are 
uompleraentsof thei^Z? AF, and the |HCK is itsdiagonuL 
Upon the base AF let a parallelepiped be erected, of ihs 
same altitude with those upon AC, CF, and let it be cat 
hy planes || its sides, and touching the lines DCE, GCB| 
these planes divide the whole parallelepiped into four otha 
paralleiopipeds, upon the bases DG. AC, BE, CF. But the 
diagonal plane touching HK divides each of the poruUelo- 
pipeds upon AF, DG, BE, into two equal prisms, (Piop> 
1)8). .-. the prisms upon HGC, CEK, are together =: the 
priBins upon HDC, CBK; and these heiag taken &«s; 
from the ^ prisms upon HFK, HAK, the reniainderh tlw 
parallelepipeds upon AC, CF, are equal. Gonsequentlj 
any two paralleiopipeds upon these bases, and betneeu (ht 
same parallel planes, are equal, (Prop. 99). 

Cabb III. When the bases are neither equiangular, nor 
have one side common, a parallelogram van be described 
on the base of the one equal to it, (Prop. 26), and eqi^ 
angular to the other, by which it is reduced to the secood 
case. Therefore, nniversally, paralleiopipeds upon equl 
bases, and between the same parallel planes, are equaL 

Cor. i. Paralleiopipeds of equal bases and equal alti- 
tudes are equal. 

Cor. 2. Prisma, Upon equal bases, which are either bolk 
triangles, or both parallelograms, and of equal altttodeii 

Cor, 3. Two prisma, upon equal bases, the one a 
angle, and the other a parallelogram, and having the S 
altitude, are equal. 

Cor. 4. A prism upon any rectilineal base ts equal 
parallelepiped having an equal base and the same aldluds.' 

Proposition CI. — Ti 
Paralleiopipeds EK, AI„ of equal altitudes, are to oai 
another aa their bases EFGH, ABCD. 

Produce EF' both ways to N and R. and make the liM 
FQ such that the altitude of the £r=7 EG is to the allitntto 
of the c=7AC as AB is to FQ, and complete the / — i¥ft 
aitd the paralleiopiped FS, Hence the cjFW is = tin 
'iZZ'AC, (Prop. ti4) ; and \\ie YM&\\e\o'g\^i» VS aad Al 



aoioD omoiMSfTKr. 



W A' / 


-^r- 


• /t , 


— t 




1 












/ / V I 


— 


lT ^ 


* ' 



ting equal bases and altitudes, are equal. Take FA aaj 
iltiple of FQ, and FN any multiple of FE, and compleU 
t l=ts Wa, EM, and MN, and the solids QT, EP, and 
<■ ; it is manifest, tbat since FtJ, QB, are equal, the unit 
JV, WR, are equal, (Prop. 2?) ; and that, since FW, WR, 
i eqoal, (he solids FS, QT, are also equal, (Prop. 100); 
what multiple soerer the base Gli is of FVV, the mum 
ultiple is the solid FT of the solid FS. In the same maif 
iT it may be sbovrn, that wh.-kt multiple soever the bma 
N is of the base EO, the same multiple is the solid KN 
the solid EK. Now, if the base NQ be :::^ the base GE, 
e solid NK will be ;:^ the solid FT; if equal, equal; aad 
less, less. -■. EK : FS = the biise EG : the base FW; 
d since the base FWz=AC, and the solid FS=:AL; the 
id EK : the solid ALz: the base EG : the base AC. 
Cor. 1. Prisms standing on their ends, and of equal alti- 
des, are to one another as their bases. 
Cor. 2. Parallelepipeds, npon the same or equal bases, 
I to one another as their altitudes. 
For the parollelopipeds KN, KR, on the same base KF, 
•■to OSM another as NG ; GB, that is, as FN : Fit. 

^P^fBi^Ilslopipeds DE, KL, wbicb bave a solid angle 

«f the one equal to a solid angle G of the other, are to 

e another in (he ratio compounded of the ratios of the 

ear sides BA. UC, BE of the one, to the linear sides 

.•', GH, GL of the other, each to each, about these solid 

gles. 

For, let AR, CB, EB be 

jduced to M, N, O, 80 that 

a=GF. BN=GH and BO 

GL. With the lines EB, 

M, and BN, let the parallel- 

iped EP be completed; and 

th the lines OB, BM, and 

lV, the parallelopiped OP. 

ten -.- the ratio DE : OP is 



Pboposition CII. — Theoheii. 




i 



148 90i.it> a^oMenrri 

, compounded of the ratios DE : EP, and EP : PO, of whicH 
the former ia = AC : MN, and the latter ia = EM : MO. 
(Prop. ]0I), or EB : BO, (Prop. 101, cor.2)j the ratio of 
DE : OP is the aanie nitli that nliich is compounded of 
AC : MN, and EB : BO, or of AB : EM, CB : BN, and EB : 
BO. But OP is = KL, (Prop, 98, cor. 5) ; -.■ these paral- 
lelopipedB have the plane is and linear aides about their 
solid is B and G respectively equal and alike situated. 
Consequently, the ratio DE : KL ia the same with that 
which is compounded of AB : BM, CB : BN, and EB : 
BO ; or of AB : FG. CB : HG, and EB : LG. 

Cor. I. Two rectangular parallelopipeds are to one ' 
another in the ratio compounded of the ratios of the linear 
aides of the one to the linear sides of the other, each to 
each. And anj ratio compounded of three ratios, (whose 
terms are straight lines), is the same with the ratio of the 
rectangular parallelopipeds under their homologous terms. 
Cor. 2. Two cubes, or, in general, two similar parallel- 
--■ -' 'o one another in the triplicate ratio of their 



Cor. 3. Similar parallelopipeds are to one another at 
the cuhes of their homologous linear sides. 

Cor. 4. If four straight lines be in continued proportion, 
the first is to the fourth as. the cube of the first to the cube 
of the second. 

Cor. 5. The rectangular parallelopipeds, under the cor- 
responding terms of three analogies, are proportional. 

Cor. 6. If four straight lines be proportional, their cubes 
are also proportional ; and conversely. 

Cor. 7. Rectangular parallelopipeds, and cOQSequentljr 
any other parallelopipeds, are to one another in the ratio 
compounded of the ratios of their bases and altitudes. 

Cor. 8. Parallelopipeds whose bases and altitudes an 
reciprocally proportional are equal, and conversely. 

Cor. 9. Prisms are to one another in the ratio com- 
pounded of the ratios of their bases and altitudes. There- 
fore the 2d cor., prop. 101, and 8th Cor. of this, may be 
applied to prisms. 

_ PROPOBITION cm. — TheoBEM. 

Every triangular pyramid may be divided into two 
equal prisms, which are together greater than half the 
irhole pyr,iaid, and two eqaal pyramids, which are nmilu 
to tho whole and to one auothet, 




T tOLID GXOHBTBr. 4^ 

Let BCD-A be any triangular pyra- 
mid. I*t its lioear sidea, AB, AC, 
AD. be bisected in E, F, G, and the 
points of section joined ; and let the 
sides of the base CB, BD, DC, be bi- 
sected in 11, K, L, and the points of 
section joined. Also let EH, EK, and ^^ 
LG, be drawn, v the two sides, AB, 
AC, of the AABC, are bisected in EF, the |EF is = and 
II CU, or HB, (Prop. 59), half the remaining side BCj 
and so of the other \s that join the points of section. Hence 
EC, CG, and conseqaently EL, are i m , and the planes 
EFG, BCD, are parallel. .-. the solid EFG-HCL. upon 
the triangular base HCL, is a prism. And the solid EHK- 
OLD is a triangular prism = (Prop. 98, cor. 7) a prism on 
the base, (IIKL=HCL). and having the same altitude as 
the former prism ; .■• the two prisms EFG-HCL, and 
EHK-GLD, are equal to one another, and together = a 
prism on the base HCL, and having the same altitude as 
the pyramid BCD^A. .-. since tbe base HCL is the fourth 
part (Prop. 70J of the base BCD, the whole solid HCDK- 
EFG is the fourth part of a prism on the base BCD, and 
having the same altitude as tbe pyramid. The two re- 
maining solids EFG-A, and B!1K-E, are triangular 
pyramids. They are equal and simitar to one another, 
and also similar to the whole, because the ^s which 
bound the one, are equal and aimiliir to the A^ which 
bound the other, and also similar to those which bound 
the whole, each to each, and alike situated. But either of 
these pyramids is, fur the same reason, = the pyramid 
KLD-jG, and .■'. .^^ either of the two prisma. Conse- 
quently, the solid HCLK-EFG is ::^^ the two pyramids 
EFG-A, BHK-E, and :^ half the whole pyramid. 

Cor. 1. By taking from the whole pyramid the two 
equal prisms, and from each of the remaining pyramids 
two equal prisms, formed in like manner, there will re- 
main at length a magnitude less than any proposed mag- 
nitude. (Prop. 73). 

Cor. 2. Since the magnitude taken away from each of 
the remaining pyramids is equal to a prism on a fourth 
part of its base, and having the same altitude as tbe pyra- 
mid ; the solids thus taken from both will be equal to a 
prism on the fourth part of their base, or the sixteenth 
' part of the original base, and its altitude equal to that of 
the original pyramid. 

Cor. 3. For ibe same reason the soUda laVcu Uokv ^% 



150 SOUD OKOU8TKT. 1 

four remaining pyramids will be equal to a prism taTlng 
its base a, aixty^fourth part of the original base, and its 
altitude equal to the altitude of the original pyramid. 

Cor. 4. Therefore all the solids thus token away, that 
is, the whole pyramid, is = (^+^'g + Bij+-^j+&c.) of 
a prism, having the same base and altitude as the py- 
ramid. But C+ + t\t + A + t4¥+ "'C'. to infin.) = *s 
(Alg. 96). 

Cor. 5. A triangular pyramid is equal to the third part 
of a prism on the same base, and haviDg the same alti- 
tude. 

Cor. 6. Since (Alg. 109) quantities are proportional to i 
their equimuKiples, whatever has been proved of prisms, 
in regard to their ratios, will also be true of the pyraraidi 
ou the same base, and having the same altitude. 

Cor. 7. A polygonal pyramid is equal to the third part 
of a prism on the same base, and having the same alti- 
tude. For it can be divided into as many triangular pyra- 
mids as there are sides in the polygon, and earii of these 
pyramids will be the third part of a prism ou the sane 
base, and having the same altitude ; .'- the sum of all the 
pyramids, that is, the polygonal pyramid, vrill be the third 
part of a prism upon the sum of all the triangles, that ii< 
on the whole polygon. 



I • Pboposjtion civ. — Theorku. 

A cone is the third part of a cylinder on the same base, 
and having the same altitude. 

For, if a series of polygons be inscribed in the circular 
base of the cone or cylinder, each having its number of 
sides double of the former, and on each of these a pyramid 
and prism, having the same altitude as the cylinder, lie 
erected ; the polygon will ultimately be ^ the circle, 
(Prop. 75, cor. 3}, while at the same time the prism will 
be equal to the cylinder, and the pyramid to the cone ; but 
the pyramid is the third part of the ptism ; .■. also the cone 
is the third part of the cylinder. 



H Proposition CV. — Tbeoreu. 

A sphere is two-thirds of a cylinder, having its altitude 
and the diameter of its base each equal to the diameter of 
the Bpbere. 



' SOLID GBOMBTRT. 

Let CDLB be a quadrant of a „ 
circle, DCBE a square described 
about it, and DCE a right angled 
triangle, having the side DE="" 
and oonsequeotlj anj line, as I 
DE, will be =K0, and GN win 
be=:GC. 

If the irhole figure thus formed revolve about DC, as a 
fixed axis, the figure DCBE will generate a cylinder. 
(Def. 7), the ACDE will generate a cone, (Def. 8), and 
the quadrant will generate a hemisphere ; now these 
figures may be conceived to be made up of an infinite num- 
ber of lamina, represented in tbickncBS by the line GI or 
the line KM, the solids generated by the several parts of 
the line GI will be as the squares of their generating lines; 
bat the generating lines are in the cone GN, in the circle 
GH, and in the cylinder GI. Now the square of GI is = 
the sum of the squares of Gil and GN, for GN is equal 
to GC, and the squares of CG and GH are equal to the 
square of the radius ^ the square of GI. .-. the lamina 
thus added to the cone and sphere are together equal to the 
lamina added lo the cylinder, and the same is evidently 
true at any other point. Hence the cone and hemisjihere 
together are equal to (he cylinder ; but the cone was shown 
(ftop. 104) to be one-third of the cylinder, therefore the 
hemisphere is two-thirds of the cylinder ; .-. the whole 
sphere will be equal to two-thirds of a cylinder, having its 
altitude double of the line DC, that is, equal to the diame- 
ter of the sphere; and it is evident, that the diameter of 
the base being the double of CD, is also equal to the dia- 
meter of the sphere. 

Cor. 1. The portion of the sphere, together with the 
portion of the cone lying between the lines CB and Gr,are 
together equal to the portion of the cylinder lying between 
' the same lines. 

Cor. 3. Any portion of the sphere, together with the 
corresponding portion of the cone, is equal to the corrc- 
tponding portion of the cylinder. 



L-A^ 



J 




GEOMETRICAL EXERCISES. 

L 1. If a straight line bisect unnther at right aogles, every point of 

lie first line wiU be equally diataat from the two extremities of the 

secDDd line. 

2. If Etraight lines be drawn, biaecting two sides of a triaogtc at 
rigbt angles, and from the point of their intersection a. pErpendicukr 
be drawn to the thicd side, it will bisect the tliird side. 

3. If two angles of a triangle bo bisected, and from the point 
where the bisecting lines cut one another, a str^glit liue he draon 
to the third angle, it will bisect the thii'd angle. j 

I. The difieroQce of aay two aides of a triangle is less thui Ili« I 
third side. I 

5. The aum of two aides of a triangle la greater than twice the 
straight line drawn from tho vertex to the middle of the base. 

6. If the opposite aides of a quadrilateral ligore be equal, Iha 
figure ia a parallBlogrQjn. 

7. If tho opposite angles of a (quadrilateral figure be equal, tlu 
figure is a paralldugrain. 

8. If a. straight line bisect the diagonal of a, parallelogram, It will i 
bisect the parallelogram. 

9. The diagonals of a rhombus bisect one another at right angles. 

10. If a straight line biaect two sides of a triangle it will be p»- 
rallel ta the third aide, and equal to the half of it, and will cut off * 
triangle equal to one-fourth part of the original triangle. 

I I . The diagonals of a right angled paraUElogram. are equal 

IQ. From a given point between two iudefiuite straight lines gtvca 
iu position, but not parallel, to draw a line which aboil be ter 
Dated by the given Imes, and bisected in the given point. 

1 3. If the sides of a quadrilateral figure be bisected, and the 
jaoent points of blsectioii joined, the figure so formed will be a 
rallelogram, equal to half of the quadrilateral figure. 

14. If a point be takeo either within or without a rectangle, uri 
etmight Hues drawn from it to the angular points, the som a! Ibt 
equares of those drawn to the extremities of one dii^^iuU wilt beeqiul 
to the sum of the squares of those drawn to (he extremities of tU 

15. In any quadrilateral figure, tlie sum of the squares of llx 
diagonala, together with fonr times the square of the line joEni 
tlieir middle points, is equal to the aura of the squares of the «da 

16. If the vertical angle of a ti'iangle be two-thirds of two ri^ 
angles, the sqnare of the base will be equal to the sum of thesqnxM 
of the aide, iucreased bj the rectangle contained by the aides; in* 

if the vertical angle bo two-thirds of one right angle, the squire <i I 
the base will be equal to the sum of the squares of the ude^ dim- I 
niahed by tho rectangle contained by the sides. I 

!7. To bisect a triangle hyaline drawn from B given point in OM I 
of Its aides. I 

1 3. A perpendicular drawn from an angle of an eqnilaMnl I 
triangle hi the opposite ^de, is equal to three times the radius at (tir 
inacribed cii'clei. 



GCOMETRICAIi EXEBCISE8. 153 

1 9. If perpendiculan be drawn from the extremities of a diame- 
' to any chord in the circle, they will cut ofif equal segments. 

20. If, from the extremities of any chord in a circle, straight lines 
drawn to any point in the diameter to which it is parallel, the 

n of their squares will be equal to the sum of the squares of the 
;ments of the diameter. 

21. If, from two angular points of any triangle, straight lines be 
kwn, to bisect the opposite sides, they will divide each other into 
pients, having the ratio of two to one, and, if a line be drawn 
ongh the third angle and their point of intersection, it will bisect 

third side, and divide the triangle into six equal triangles. 
!2. If two triangles have two angles equal, and other two angles 
ether equal to two right angles, the sides about the remaining 
rles will be proportional 

S3. If a circle be described on the radius of another circle, any 
Bight line drawn from the point where they meet to the outer 
comference is bisected by the interior one. 
24. The rectangle contained by two sides of a triangle, is equal to 
i rectangle contained by the perpendicular from the contained 
^le apon the third side, and Ihe diameter of the circumscribing 
sle. 

f5. If a straight line be drawn bisecting the vertical angle of a 
ingle, the rectangle contained by the two sides will be equal to the 
are of the bisecting line, together with the rectangle contained by 
segments of the base. 

:6. A straight line drawn from the vertex of a triangle to meet 
base, divides a parallel to the base in the same ratio as the 

J. 

7. If the three sides of one triangle be perpendicular to the three 
8 of another, the two triangles are similar. 

8. If from any point in the diameter of a circle produced, a tan- 
t be drawn, a perpendicular from the point of contact to the dia- 
er will divide it into segments, which have the same ratio which 
distances of the point without the circle from each extremity of 
diameter have to each other. 

9. A straight line drawn from the vertex of an equilateral 
ogle, inscribed in a cirde to any point in the opposite circumfe- 
9e, is equal to the two lines together which are drawn from the 
emities of the base to the same point. 

0. If the interior and exterior vertical angles of a triangle be 
cted by straight lines, which cut the base, and the base produced, 
' will divide it into tluree segments, such that the rectangle con- 
9d by the whole \)ase thus produced, and the middle segment, 
I be equal to the rectangle contained by the two extreme seg- 
itSb (A line divided in this manner is said to be divided barmo- 

Jly.) 

1. The side of a regular decagon inscribed in a circle, is equal to 
greater segment of the radius divided medially; the side of a reg- 

hexagon is equal to the radius ; and the side square of a regu- 
pentagon is equal to the side square of a regulai* hexagon, tu- 
ter with the side square of a regular decagon. 






PRACTICAL GEOMETRY. 
Problem I. 



I 

I J 

1 



To diTide a given straight line AB 
into tvro equal parts. 

From the centres A and B, with 
the Bame radius ^- half of AB, de- j 
Bcribe arcs intersecting in D and E, 
and draw the pCE, it will bisect 
AB in the point C. 



I PsoaLEM II. 

To divide a given angle ABC into two 
'equ:il parts. 

From the centre B with any radius, de- 
scribe the arc AC, and from the centres A 
and C with the same radius, describe a 
intersecting in D, and join BD; the anglti 
ABC will be bisected by the |BD. 

PHOBI-EM III. 

To trisect a right angle ABC) that is, to divide it inM 
fliree equal parts. 

From the centre B ^ith any radius, 
describe the arc AC; and from the centre 
C with the same radius, cut the arc in D, 
and from the centre A with the same ra- 
dius, cut the arc in E, and join BD and 
BE, and they will trisect the angle as 
required, 

Pkoblem IV. 

To erect a perpendicular from a given point A, in* 
given line AC. 

CiSB I. When the point is 
near the middle of the line. 

On each side of the point A 
lake any two equal distances. Am, 
An. Fromthecentreswi, «, with 



any radius greater thai 

describe two arcs intersect- 



in D. Through A and D draw the itraight line AO, 

t it will be the perpendicular required. 

^ASK II, When the point is near the end of the line. 

•"roni the centre A with any ra- 

9, descrihe an arc mnr. From | 

centre m with the same radius, 

1 the compasses twice o*er on ,-'^ "'''? 

arc at n and r. Again, from 

centres n and r with the same B ^g ^O 

.us, describe arcs intersecdng in 
Then draw AD, and it will be the perpendicular re- 



Anoth^ Method. 



rrom any point n ae a centre, with 
idins ■=n\, describe an arc. not less 
n a semicircle, cutting the line in m 

A. Through m and n draw the 
aeter mnr, cutting the arc in r, and »■ 

Ar, and it will be the pcrpendicu- 
Kqoired. 

'or. If the point from which the perpendicular is to be 
TO were (be eitremity of the line C, it would only be 
^ssary to take iiC as a radius instead of nh., and the 
I parts of the construction would be the some. 



Anotfier Method. 
rom any scale of equal parts take 
stance equal to 3 divisions ; then 
L the centre A, with a radius ^ 
risions, describe an arc, and from 
centre tr, with a ladiua =: 5 di- 
ins, describe another arc, cutting b— 
Former in D, and join DA, and 
Jl be the perpendicular required, 



Another Method, 
ith a marqnois square, which is a right-angled triangli 
)ut of ivory or wood, apply the right angle to the point 
tnd make one side coincide nith AB, a line drai 
g the other side will be the perpendiculai required. 



B» given 
^npeipeni 



Pbobleh V, 
given point C| without a 



straight line AB. 



I 



] 



w 



156 ntACTCOAL GIDUBTKr. 

Case I. When the point is nearly opposite to the 
middle of the line. 

From the centre C describe an 
arc, cutting the straight line AB 
in m and n. From the centtes 
m and n describe arcs, with any 

radius greater than ^mn, inter- A V^.^_J^_^_^ b 

secting in s. A straight line 

drawn through the points C and X'' 

a will be perpendicular to AB. 

Or, apply one of the sides containing the right angle of 
a. marquois equare to AB, and slip it alonj; the line, till the 
point coincide with the other side ; then a line drawn 
along this side wilt be perpendicular to AB. 

Ca3b II. When the point is nearly opposite to the end 
of the line. 

From C draw any h'ne Cm, and 
bisect it in w. From the centre n 
with the radius Cn, describe thi 
semicircle CDm, cutting the line AI 

P ; join CD, and it will be the "" »»■ 

trpendicular required. 



Aiw/ker Melliod. 



From A, or any point in AB as 

centre, describe an arc through 
the point C. From any other point 
m, describe another arc through C, 
cutting the former in n. Through 
C and n draw the line CD«; and 
CD will be perpendicular to AB. 

N.B. — This can also be done as described in Case L, by 
the marquois square. 

Problem VI. 

At a given point E in the line ED, to make an ao^ 
equal to a given angle ABC. 

From the point j, 

B with any radius, ^c _ ^^ 

describe the arc ^^C^ 

mn, cutting AB ^■^ \ 

and CB in m and 8-=^=- is — : 

n. Prom ihe centre 
E with (iae radius Bm, describe die 



PRACTICAL GEOBfETRT. 157 

\e mn^ and lay it from r to «, and through E« draw the 
[ght line Est; then the angle FED will be = the 
BC. 

r the angle be gi^en in degrees. Draw a straight line 
ID, and using a scale of chords from the centre E, with 
dius == the chord of 60^^ describe an arc rs; then take 
the chord of the number of degrees required from the 
i, and from the centre r, with the chord of the giren 
e as a radius, cut the arc in s, and through the points 
nd 8 draw the |EF, and the IFED will contain the 
n number of degrees. If the angle be greater than 90^, 
m be divided into two parts, and one part laid off, and 
I the other part, from its extremity ; it is generally 
t conyenient to lay off the chord of 60^, and then the 
rd of the excess aboye 60^. 

Problem YII. 

draw a line parallel to a given line AB. 

kSK I. When the parallel Hne is to be at a given dis- 
e C, firom the given line. 

rom any two points m and n ^ g 

le line AB, with a radius ^ '^^ ^"^ 

1 to C, describe the arcs r . 

8. DrawDE to touch these ^"^ ^""^^ 



without cutting them, and c 

11 be the parallel required. 

kSE II. When the parallel line is to pass through a 

1 point C. 

om any point m in the line 

with the radius mC, describe D 7 r"B 

arc Cn, From the centre C / 

the radius Cm, describe the A. — 1 -1 — b 

Tw, Take the arc Qn in the 

>asses, and apply it from m to 0. Through C and 

DE, and it will be the parallel required. 

B, — ^This problem is more easily effected with a pa- 
. ruler. 

Problem VIII. 

\ divide a given line AB into . 5 ^» 

)ropo8ed number of equal parts. 

trough A and B draw h/m and 

tarallel to each other. In each 

le lines Km, Bn, beginning at 

d B, set off as many equal parts 

.e Mne AB is to be divided into. 






in>s 



TK ACTIO AL 0£OHBTBT. 



Join A, 5 ; 1, 4; 
ed as required. 



,3; 3,2; &c; and AB will be dirid* 



Another Method. 




% 



Through the point B draw 
tte indefinite line CD, making 
anyanglenithAB. Take any 
point D in that line, and 
through it draw DE || to AB, 
and set o£f from D as many 
equal parts DK, KH, HG, &c., 
DS the line AB is to be divided w ir a. ^ 

info. Through the points E, A, 

draw the hnc EA, and produce it to meet CD In C ; ll* 
lines drawn from C, through the points F, G, H. &C., will 
divide the line AB into the required number of parts. 



Phoblbm IX. 




^_ lin 



To find the centre of a given circle, o 
described. 

Let ABCD be the given circle ; take 
any three points, A, fi, C, and from the 
centres A and B, with a radin* p^ half 
the distance AB, describe arcs intersect- 
ing in j- and 3, and from the centres B 
snd C, with a radius :;^ half the dis- . 
tance BC, describe arcs Intersecdng in 
m and n. Then through the points ar 
and mn, dratv straight lines intersecting 
be the centre required. 

Pkoblem X, 



To describe the circumference of a circle through anj 
three given points A, B, C, provided they are not in lli^ 
same straight line. 

Use the three points the same as in the last problt 
then a circle described from the centre <>, with a radial 
equal to the distance oA, will pass through the other tnv 
and will be the circle required. 

Cor. Since the three points are not in the same stra 
line, they may be the three angular points of a triaii 
"snce this problem also shows how a cucle may bB 

{bed about a given triangVe. 





fbactecaij geomethy. 

Phoalbm XI. 

To draw a tangent to a given circle, 
ihat shall pass through a given ]ioint A.*' 

Case I. When tie point A is in the 
circumference of the circle. 

From the point A, to the centre of 
the circle, draw the radius oA. Then 
through the point A, draw BC -^ to o.4, 
and it will be the tangent rec[uired. 

Cask II. When the point A is without tli 

To the point A, from the centre 
draw the straight line oA, and bise 
it in m. From the centre m, with the 
radius mA or mO, describe the semi- 
circle ABO, cutting the given circle 
in h. Then through the points A and 
B draw the line AB, and it will be 
the tangent required. 

Frobledi Xn. 

To two given straight lines A, B, to find a third ] 
poitional. 

From any point C draw \«, making 
any iFCG, and make CE= A, CD 
and CG each =B. Join ED, andc 
draw GF || to ED, then CF will be 
the third proportional required. For 

(CE=A):(CG = B) = (CD=;m: e 

CF. 

y.S.—Ii EG had been made =B, then DP, by the 
same construction, would have been the third propor- 

Pboblem XIII. 

To three given straight lines. A, H, C, to find a fourth 
proportional. 

From any point 
D, draw two straight 
lines, making any_ 
ZFDH, and make 
DE=A, EF = B, - 
DG=:C. Join EG, - 
and through F draw 
FH||Er ■• 




PKACTICAI. eKOHKTBT. 




Pboblbh XIV. 

Between two giren straight lines. A, B, to find a mean 
proportional. 

Draw any straight line g 

CD, in which take CE= /^T^X 

A, and ED=B. Bisect ^ / \ 

CD in F, and from the B J ' V 

cenlre F, with the radius *^ 

FD or FC, describe a semicirek. From E draw EG J- 
to DC, DieelJDg the semicircle in G, then EG is the mean 
proportional required. 

2^.£. — Another method may readily be deduced from 
(Prop. 68) Geometry. 

PnODLEM XV". 

To describe an equilateral triangle 
on a given line AB. 

From the centres A and B, with the 
radius AB, describe arcs intersecting in 
C. Draw AC and BC, and ACB will 
be an equilateral triangle on the re- j 
quired line. 

NoiE. An isoBcelea iriangla may lie deacribod on AB, by takiat 
as mdtua the length of one of the equiil sides. Far the method et 
diacribing a. triangle, having its sides equal to three given lines, Wt 
Geometry, (Prop, i.) 

Pbohi-em XYI. 

given 



To describe a square up( 
line AB. 

Draw BC -i- to and =iAB, and from 
the centres C and A, witha radius =AB, 
describe area intersecting in D, and join 
DC, DA, and the figure ABCD wiU be 
a aquare. 

Another Method. 

From the centres A and B, with the 
radius AB, describe arcs crossing at E. " 
Bisect EA in wi; then from the cenlre 
E, with the radius Em, cross the arcs 
BD, AC, in C and D. Join AD, DC, 
md BC ; the figure so formed will be 




PRACTICAL GEOMETRY. 



161 




Problbm XVII. 

t describe a rectangle^ whose length and breadth shall 
espectivelj, equal to two given lines, AB and C. 
t the point B, in the given line 
erect BD=C. From the point 
rith a radius =AB, describe an 
and from the centre A, with a 
LS :=C, describe another arc, 
Dg the former in £ ; join ED 

A.E : the figure so formed will 

le rectangle required. 

TB. Any parallelogram may be described in nearly the same 
ler, by drawing BD so as to make the proper oblique angle, 
may be effected, by taking the chord of SO"" from a scale of 
Is, and from the centre B, with the chord of 60°, describe an 
nd with the chord of the number of degrees that the angle is to 
In, cut off the portion required, and draw the line BD through 
tremity. 

Problem XVllI. 

inscribe a circle in a given triangle ABC. 
sect any two of the Lst as A 
C, by the lines Ao and Co. 

from the point of intersec- 
9, let fall the -t- on upon 
r of the sides, and it will be 
idius of the required circle. 




Problem XIX. 

a given circle, to inscribe an equilateral triangle, a 

;on, or a dodecagon. 

)m any point A in the circumfe- 

, as a centre, with a distance equal 

! radius Ao, describe the arc BoF. 

AF, AB, and BF, also bisect AF 




en BF laid three times round the £' 

, will form an equilateral triangle. 

)r AF laid six times round will 

a regular hexagon ; and AG laid 

e rimes round will form a regular dodecagon. 

R. The side of a regular hexagon is equal to th^ t%i- 

)f the circle in which it is inscribed. 



FB A OTTO At. OXOMBTST. 





I 



PROBLEU XX. 

To inEcribe a sqaaie in a given circle. 

Draw two diameters, AC and 13 B, 
4- to each other, and join AB, BC, 
CD, and DA; then ABCD will be 
the square required. 

Cor. A circle may be dessribed 
about a given square ABCD, by draw- 
ing the diagonals, AB, CD, and their 
point of intersection, O, will be tlie *- 

centre ; from which, with the radius 
OA, describe a circle, and it will be the circle required. 

Problem XXI, 

To describe a square about a given circle. 

Draw any two diameters, AB, CD, _ „ 

at ■>'Ls to one another; then through 
the points C and D draw FG, EH || to 
AB, or -L to CD, and through tlie 
points A and B draw FE and GH |t to 
CD, or -L- to AB; and the figure FEHG 
will he a square described about the »l 

Problem XXII. 

To inscribe a regular polygon of any proposed nomlin 
of sides in a given circle. 

Divide 360° by the number of sides of 
the polygon, and make the /.AoB at 
the centre, containing the number of de- 
grees in the quotient. Then if the points 
A, B, be joined, and the chord AB ap- 
plied to the circumference the proposed 
□umber of times, it will form the polygon 
required. 

Pros L KM XXIII. 

On a given line AB, to form a reguli 
proposed number of sides. 

Divide 180" by the number of sides 
of the figure, and subtract the quolient 
from 90°. Make the Z,sOAB and DBA 
each equal to the difference fio found ; 
and from the point of intersection O, 
vith the radius OA or OB. describe a 
circle, and the chord AB \»ei.\i^ a^^\wi 



□13 

m; 




if anj 



to the circumference the proposed number of times, will 
form the polygon required. 

Pboblbm XXIV. 

To make a triangle et^ual to a given trapezium ABCD. 

Draw the diagonal DB, and „ 

thiough C draw CE || to DB, 
meeting AB, produced in E, 
and join DE; the AADE will \ 
leequaltothetrapeziumABCD. 
For the ADEB is = the ADCB, (Gee 

Problem XSV. 

To make a- triangte equal to any right lined figur 
ABCDEFG. 

Produce the side AB both _ 

irajs at pleasure, and draw the 
diagonals EA, £0, and by the 
last problem make the AE1A= 
tie trapezium AGFE, and the 
^EHB=: the trapezium BCDE. 
Draw IK || to AE, and HL ij _ 
to EB ; then join KE and EL, K 
aad the ^KE]. will bo equal ti 
the figure ABCDEFG. 



PnoBLEM xxvr. 

To describe an oral or figure resembling 
giren straight line AB. 

Take any points, C, D, at equal 
distances from A, B, and on CD 
describe two isoaceles A*i CED, 
CFD, and produce the sides to 
llie points p, o, n, in ; then from 
tbe centres C and D, with the 
ndiusAC or DB, describe area 
bounded by the sides of the is- 
oweles A* produced ; and from 
the centre F, willi radius Fm or 
Fn, describe the arc mii, and fro 
radios Ep or Eo, describe the ar 
formed will be on oval. 




1 ellipse o 




le centre E. with the 
1, and the figure thus 



J 



r^ 



GENEBAL EXERCISES IN ALGEBEA. 



I 



1. Find the sum of 3a* + 4alj + b\ 4a'— 406+*=, 
'+4ai+4i', 7a"— 36', and 5o- + 10a6 + 56^ 

Ans. 18a" + 14a6+ffi'. 

2. Find the sum of 16<ic+]6c' + 4ai'', 12c'+7ac+a\ 
6a'+4c% l3ac+lUa'+c^ Ua^'+^ac+lc^ and3a''+ 
12ac+lV. Ans. ie9a''+70ete+i9<:'. i 

3. Findjhe sum of I4a+jac—5c, lSa^5J^+8e, | 
7a— 14V^, 3c+5a, 7c— 6Va^+5a, 13n+14Vac+3c, 
and4^ac— 6a+4c. Ans. Slo— 6Vac+20r, ' 

4. Find the sum of 4a + 76— 3c, a + 4o6— c, 5a— 3aS 
4-4c, 3a—alj+c+3d, ia+7c—id, and 5d— c+4a— 2i. I 

Ans. 21a+56+7c+4d. 

5. Find the difference of 1^+ i2ab+13b^, and 7a'+ 
116". Ans. 8a=+12a6+2i'. 

6. From 17((c+5a'+6c°, take 17(ic— 5q"+3c' — id. 
Ana. I0(t'^+3tfH*i ' 

7. From the sum of 3a'-f2(wr + 2a:^, and 5n" + 7(ij:— r. 
take the sum of 7a=— 7aj;+a;', and 12aj;— 3a»— 4**. 

Ans. 4a"+4(W+lr- 

8. From a^ + b^ +e' + 2al>+2ac+2bc, take 2a6+2ao- 
26c— a'— 6"— c*. Ans. 2a'+26' + 2c"+4ic. 

9. Multiply a+6 by c — d. Ans. ac+be — ad— M. 

10. Multiply a— &+C— <; by ffl+6—c—i;. 
Ans. a'—b'^ — c'+rf^ — 2ad+2he. 

11. Find the product of (a:— 10)(«+l)(j: + 4). 
Ans. it"- 5j-^— 46x— 40. 

12. Find the product of a'+oc+e'^, and a" — ac+e'- 
Alls. a'+oV+f'. 

13. Find the continued product of a-f-l, a-f-S, a+S, 
and« + 4. Ans. (x'+]0o=+35a=+S0a+2t 

H. Find the continued product of a — x, a-\-x, fl'+M 
+ a:", anda'- «j; + a:^. Ana. a^—if. 

15. Multiply j;'^+10j-y + 7. bya;''- aEi/+4. 
Ans. a:*+4A— tiOA''-i-nx-— ary+aa 

16. Divide fi^- 3it:=+12*:'+14:c, by 3x. 
Ans. 2^^- a:*+4iB+4|. 

J7. Divide ^'— 81, by r~3. Ans. r' +-ix-- +9x+^. 
i8. Divide a'+6^ by a+d. Ans. a''^al-\-h'. 



19. Divide a^+y^ by 3-+?/. 

Ans. x*—3^i/+^'V^ — ^y^+i/*- 

20. Divide 3:=+;EV+a'/+/. by *'+/■ Ans.a.^+y^ 

21. Divide j^ — f>x*+qa^ — qii'-^px — l,\iyx — I. 

Ana. 3;'-(p-lK+(5-p+J>^_(p_l>+l. 

22. Divide^_4.'+'i^-lf _^i^+27.b7;-- 
1+3. Aw. -^'^-Sji'-f j+9. 

Projuscuous Exehcises. 

23. Find the sum of (he products (a+!i+c)(a+b~c), 
(„+6+c){(.^+^), and (a+b+c)(^i+h + c). 

Ana. a'-\-b-^-i~c'+2(a+Sac+2bc. 

24. Find the sum of the products {ci+x){a+x), {a — x) 
(a^r}.and (a+:e){a^-x). Ans. 3a^+A 

25. Find the sum and difference of (a+aXa-f «), and 
(a_x) X (a^-x). Ana. Sura 2a'+2:c", diff. 4ax. 

26. Find the sum and difference of (a+b+c)la+l + c). 
aBd(a+5-<-)(«+6-c). 

Ans. Bum 2a''' + 2i' + 2c''+4a6, and diff. 4ac+4bc. 

27. Find tiie sum of the products of {a+b+c+d) 
(a+b+c—d), and (a+J— c-J-rf) (a — i-fc+rf), and alau 
Ibeit difference. Ans. S\if» 2a''+2ab+2ac+4bc+ 
2ad, and their diff. 2ab+2ae—2ad+2b^+2c^—2d^. 

28. Uo«v much doea the product of (x-f fi-f 2c)(a4-£ 
+2c), exceed the product of (a — 6 — 2c)(a — b—2c). 

Ans. 4ab+ai 

29. Divide a'^— «*, by a^—r'. Ans. a+a^^i'+j-. 



— & by a* — 1/ 



Ans. a^+aH^+ah'^+b^l 



31. Find the greatest common measure of a* +ai — 12i-, 
and a* — 5a6+fK-^ Ans. a— 36. 

32. Find the greatest common measure of 6a'.\.'Jai — 
3i',aiid6u- + H,fi-f-3?A Ana. 2a+3J. 

33. Find the leaat common multiple of sf* — a\ and 
/'—a*. Ans. x^ + ji^a—xa^—a^. 

34. Find the least common multiple of a — I, a^—i, 
a— 2, and a^ — 4. Ana. a^ — 5(t^ + 4. 

35. Find the least common multiple of 2^—1, 4a'— 1, 
and4a'+l. Ana. 16a*— 

36. Reduce to its lowcfit terms .77^ — :. Xws. — . 
jy — d) Tib — t 



EXEaCISES IN ALGEBRA. 



37- Reduce to its lowest teTms 

i8. Reduce to its lowest terms ■— ~ *" . 



!9. Reduce to its lowest terms 



ac+ax+bx+bc 
n'+P+c'+2a6+5ge+at(r 



f 

^^H 40. Reduce to its lowest terms 

■ 41. Prove Iha. 1+ "itg=' = (wiwK.tt -l) 
H 42.P,.„thatl_ia^ = &±J=9fc£:!0. 
^B-'«.Pri)T6tli.lbjArt,.(29-3I),AJg.J/j'_pt!f]'|, 
" o^bemi«c.dloth.fom&i^i*±^]ft£:«f±t?l 

44. Prove that if. In the list example, s= ■■, Qieex- 
■pression may be reduced to the form sfs — (t)(s — 6)(j_o). 

45. Show that a -i- ^ 1^^ ~ + t ' ^^f^^pt a^6. 

46. Show that a?+l:^a'+ffl, except a=\. 

147. If fl.^i:a:, show that {x.iraf—^.^i<i3?. 
48. Multiply a;— ^ by ^ + ^. Ans. f — ^. 

49. Mdt,ply;j+ ^r-2-,-T^. ^y?-^- +2^- 
, a* 4c»d« , He'd* *W 
.n IT 1^ 1 0*— 91+20 , o*— 13n+42 , a^— ll.+SS 
SO. M,Jt,ply-jc5^, I7 ^^j— . A»..-^j— 



' EXEBCISES IN ALGEBBA. IB? 

5.T Find the sum and difference of |- — ^](^ + j^ )■ 

, [I a\/m ii\ , Sim 2an , Sac 2flm 
^-1 (a + ^}(n- Sj- ^'- -^ - ST- ^^-^ ^ - li^- 

54. Show that — -—— - - ib equal to ^. 

Involution and Evolution. 

55. Bequired the square of a^^ — j:+l. 

Ans. a;'— 2jr'+3j:'— 2x+l. 

56. Required the cube of \+2x+33:'', 

57. Show that the cube of a — -is a' 5— 3f« ). 

58. Extract the square root of 4jc* — '12j^+25x'' — 24a: 
f 16. Ans. 2x^~-3x+4. 

59. Extract the square root of a' — 2a^4- V — o +^'!1' 

Ans. n'— ^+f 

60. Extract the cube root of -^ — ^ +3abc^~bV. 

'' ' , "=' ^ 

Aub- -t — he. 

61. Extract the cube root of «*— fb^+lSj;*— 203^+15a:^ 
— 6a;+l. Ans. a:^ — 2r-J-l. 

62. If two numbers differ by 1, prove that the difference 
of their squares is the sum of the two numbers. 

63. If two numbers differ by any number (a), prove 
that the difference of their squares is the sum of the num- 
bers multiplied by their difference (a). 

64. If the sum of two fractions =1, prove that their, 
difference is equal to the difference of their squares. 

B5. Prove that the square of any odd number diminished 
by 1 is divisible by 8. 

66. Prove that the product of two odd numbers la odd ; 
Md that the product of two even numbers, or of an even 
and an odd number is even. 

tl EXEBCISGS IN SlUFLE EHUATIONS. 

^. GiTen 8a;— 4=13— |;r, to find x. Ans. x=2. 

te.Givcn2K+7+|-=6j:— 23,tofinaiE. Xua.ai^V^, 



EXERCISES IN ALGEBRA. 



Ang. x=U 



71. GiTeii| + |+ j=7j:— 734+1, to find*. 

o„4.T K»j.o Ana. x=:i» 

72. Giren -^ — ^ =36, to find a:. Am. x^ 

73. Giren —— 7""= -— , tofinda:. Ans. ;i:=| 

74. Given 1- - — ■ =19 , to find x. 

^ ^ * Ana. J^=J 

76. Given aj:-\-e=bx-i-d, to find x. Ans. ^= — j 

76. Given tM;+6'=o''+6*, to find x. Ans. j;=a+J 

77- Given(«+^)(6+.r)— u(J + p)=^'+'^'. tofiudj 

AnB. x=^ 

70. If f _1_ '^ =Sab. x=: ^}^ 



79. Il^=ic+d+-. 




^ t«+J 


™-"»+t-7:-='- 




^_ .(.+a.->| 


'"■"^+lr+:^ + 


^— /■■ 


'- U/„, 


^■«{^:jg:lJ?} 




..=16 J 

J-=35. t 


«-"{i:TgsS} 




.= 1« 

J=!l& 


84. If-;x+2=38j- 
(j,+,=46j 




.= 6 
,,=14 1 
;=32. 


85. IfJ*||==20il 
Ij+J«=34 J 




,=32 
■=10. 


(2x+5,-70=- 
86. If-! 6i_ y+3.= 

I7I+6J+ ! = 


288) 
227!- 
297) 


.=13 . 



PBOBLEMS PRODUCING SiMPLE EqI 



87- How much money have I in my pocket, when tfce 

I fomrlt and fifth of the same together amount to 1 Is. 3d. 

Ana. L.1, as, 

K. In a company of 266 persons, consialing of officers. 

merchantB, and students, there were four times us many 

I merchants, and twice as many officers, as students. How 

many were of each class ? 

Aus. 38 students, 152 merchants, and 76 officers. 

89. Divide the number a into three such parts, that the 
second may be m times, and tho third u times as great as 

the first. . _ a ma na 

l+m+n' l4ni+n' l+m+n' 

90. A father gives to his five sons L.IOIW, which is to 
be divided among them according to their ages, so that 
«ach of the four elder may receive L.20 more than his 
next younger brother. How much will the youngest re- 
ceive? Ans. L.KiO. 

91. A courier who goes a miles a-day, started n days 

Itefore another, who goes h miles a-day. In what time will 

the second overtake the first? , "" , 

Ans. days. 

92. A person paid the sum of L.2, ISs. in crowns and 
shillings, using ]? pieces in all. How many of each sort 
"Was used ? Ans. 9 crowns and 8 shillings. 

93. One of my acquaintances is now 40 years old, and 
lis son 9 years. In how many years will ihis man, who 
is now more than four times as old as his son, be only 
twice as old ? Ans. In 22 years. 

94 A transcriber was asked how many sheets he wrote 
treekly: he answered, " I only work four hours a-day, and 
cuiuot finish 70 sheets, which I could wish to do ; but if 
1 could wort ten hours a-day, then I should write weekly 
exactly as many above 70 sheets as I now write less ttiau 
tkat number." How many sheets did he write weekly? 

Ans. 40. 

95. Find two numbers, such that if the first be multi- 
Jlied by 2, and the second by 5, the sum of the products 
■"ill be 31; and if the first be multiplied by 7, and the 
wlher by 4, the sum of the products will be 68. 

Ans. First 8, and second 3. 

96. It is required to find a fraction, such that if 3 be 
'kubtracted from the numerator and denominator, tlie valut 
ef the remaining' fiaction will be ^, and if ^ be aA4ei \o 




r. What is the ftactionl 

Am. ^. 

97- The Bum of two numbers is =a, the difference of 

! Aeir squares =6, "What are the numbers ? 

, il»+& , a— • 

Ans. -5—, and -^. 



. A, B, and C, one L^lf)0 amongst them, and u 
one of them can paj this sura alone. But when tliej 
unite, it can be done in the following ways: by B putliar 
^ of his property to all A's; or by C putting § of bn 
property to that of B ; or by A putting | of his piopeitf 
to tbat of C. How much did each possess ? 

Aas. AL.1530, BL.15^0, CLllJft 
100. Required to find three numbers, which possess ih 
following properties: — If fi be added to the first ani 
second, the sums are to one another as 2 to 3 ; if 2 h 
added to the first and third, then the sums are to 
another as 7 to 11; but if 36 be subtracted fiom the « 
and third, the remainders are to one another aa GkT 
What are the numbers? Ans. 30, 48, aodSO 



Quadratic Equations. 

101. Ifa:"+6«=27. ^-S, or— 9. 

103. I{x'—7x+3i=0. x=6^oi^. 

103. If a;*— SJic^lB, x=Q, or — 2j 

104. USx''— 3^=65. 3;=5,or— 4|! 

105. If -^ = ~. :c=14, or — 1 

1+60 3:1^5 



106. If «+ -~ =3a^4. r=5, o 

107. l(a\l+bV)=b(2a^:i+b). X-. 



x=: — , er —• 



109. If(a— SV— (a+6>+26=0. x=i,ia^. 
Hi. If»"_7i'=8. x=^,i>t— 1. 



•m A%aKBKk. 



m 









114. If|"' 



^^tul i"-2ij-/= 31 I 

^K" ti»"+2«-j-y"=ioi I 

^^^Tboblems producing Quai 



"boblems producing Quadratic Eqitations. 

U7« What number is h, whose half multiplied by its 
Ihird part, gires 864 ? Ans. "^2. 

118. It is required to find a nmnlier, such thut if we 
Ent add it to 94, then subtract it from 94, and afterivards 
moltiplj this remainder by the former sum, the prcduci: 
may be 8512. What number is it? Ang. 18. 

119. What two numbers are those whose product is 144, 
and whose quotient is 9 ? Ana. 3ti and 4. 

120. What two numbers are those whose product is 
=1, and whose quotient is =?/^ . <— r ■ /7i 

^ Ans. V ah and ^ - . 

121. A gentleman left L.2I0 to three servants, to be 
divided in geometrical progresaion, so that the first shall 
We L.90 more th^ the last. Find their legacies. 

Ans, L.120. L.60, L.30. 

122. A certain capital yields. 4 per cent. ; if we multi- 
ply the number of pounds in the capital by the number of 
pounds in the interest for fite months, we obtain 1 1 7041 ^. 
What is the capital ? Ans, 2650. 

123. There are two numbers, one of which is greater 
than the other by G, and whose product is 240. What 
uamberB are they? Ans. 12 and 20. 

124. It is required to find a number, whose sqiiiue ex- 
ceeds its simple potrer b/ 306? Xtvs, Vft 



■3' ' 



F^ 



EXBRCrsSH IN ALGEBBA. 



125. It ie required to find a number, Bxich that iTwe 
multiply its third part by its fourth part, and to the pro- 
duct add five times the number required, this sum eicesdi 
the number 200 by as many as the number sought is !e« 
than 280? Ans, &. 

126. A person buys some pieces of cloth, at equal prices, 
for L.60. Had lie got 3 more pieces for the same buNi 
each piece would have cost him a pound less. ITow mai^ 
pieces did he buy? Ans. 12, 

127. A person dies, leaving cliiidren, and a fortune of 
L.46,800, which, by the will, is to be divided equBliy 
among them. It happens, however, that immediately aftfi 
the death of the father, two of his children also die. Ili 
consequently, each child receives L.1950 more than he or 
she was formerly entitled to by the will, bow many chil- 
dren were there ? Ans. Eight. 

128. Two retailers jointly invest L.500 in business, M 
which each contributes a. certain sum ; the one let hii 
money remain five months, the other only two, and eaeh 
received L.4.'iO capital and profit. How much did each 
advance ? Ans. One L.200, the other L30B. 

i2Q. What two numbers are those whose sum is 4\. 
and the sum of whose squares !^01 ? Ans. 2f> and I^- 

130. A capital of L. 5000 stands at 4 percent, compoiud 
interest. "VThat will it amount to in forty years? 

Ans. L.24005, 2s. l^d. 

131 . How long must L.3600 remain at 5 per cent, cob- 
pound interest, so that it may become as much as L.5(KI0< 
at 4 per cent, for twelve years? Ans. Ifiyears, ]36daji 

132. What capital, at 4 per cent., will fifteen yean 
hence be equal in value to L.4500, at 6 per cent, for nine 
years? Ans. L.422I-483. 

133. A town contains 20,000 inhabitants, and we kno" 
that the population has regularly increased -jjn yearh. 
What was its population ten years ago ? Ans. 14,882, 

134. In how many years will the population of a place 
become ten times as great as it is at present, if the yeailj 
increase amount to three persons in a hundred? 

Ans. 78 years nearlj. 

135. What is the present value of an annuity of L.20, 
to continue for forty years, reckoning interest at the rale 
of 6 per cent, per annum. Ans, L,300, 18s. 66, 

136. What annuity, m^rovcd o,t the rate of 4 per cent. 
per annum, compound inteieBt, "nVi lA &e coi w iwelTt 
j-ears amount to L.500, l^s. 'i\&.'< Kto,- \.3a,"w.-Sfe. 



A SYSTEM 



or 



PRACTICAL MATHEMATICS, 



II. 



OONTAININO 

LOOABITHMIC ASITHllEnC^ TRIOONOMETBT, MENSURATION OP 

AND DISTANCES^ NAYIOATION, MENSURATION OF SURFACES AMD 
aOLIDSy LAND-SURTETING, SPECIFIC ORAYITT, AND GAUGING. 

WITH A COPIOUS AND HIGHLY ACCURATE SET OF 

STEREOTYPED LOGARITHMIC TABLES. 
BEING PAET SECOND OP 

N*>. XVI. 

or 

A NEW SEEIES OF SCHOOL-BOOKS. 

BTTHB 

800TT1SH SCHOOL-BOOK ASSOCIATION. 



WILLIAM WHYTE AND CO. 

B0OK8BLLBR8 TO THB QUBBN DOWAOKR, 

13^ GEORGE STREET, EDINBURGH. 

BOVLBION AND STONEMAN, LONDON ; W. GRAFEL, AND G. H. AND 

J. SMTTHy LIVERPOOL; ABEL HETWOOD, HASCTaili&TlS.'B.^ 

FINLAT AND CHARLTON, NEWCASTLE. 

HDOOCXIiVI. 



bdinbuboh: 

ARDBSW JACK. PBWTBB* 



PREFACE 



Iv publishing ** Part II. of a System of Practical Mathe- 
inatics,'' the Committee beg to state, that they have used 
e^eij efibrt in their power to secure perfect accuracy in the 
Answers to the questions, which are all given €u they can be 
obtained from the Tables at the end of the volume. This is a 
very important point to both Teacher and Student ; and yet 
it is one that has very generally been overlooked in Treatises 
on Practical Mathematics. 

The article on Looartthiis contains all that is necessary* 
to be known in regard to their practical applications to com- 

: i&on numerical calculations. 

r The article on Tbioonombtbt embraces the modem im- 
provements in the science. By adopting the definitions of the 

. trigonometrical ratios, now used in all theoretical treatises on 
the subject, it became necessary to demonstrate all the Rules 
in a difiFerent manner from that followed in other practical 
treatises on the subject ; and the consequence is, that the de- 
monstrations have thereby been greatly simplified. The 
method of surveying by Rectangular Co-ordinatesy here given 
for the first time, will be found very useful in extensive sur- 
veys^ and also in Marine Surveying. The rule for finding the 
angles of a triangle when the three sides are given, is new. 

The articles on Mensuration of Surfaces and Solids, 
Und Surveying, and Specific Gravities, contain clear and 
perspicuous Rules, illustrated by suitable Examples, and a 
eollection of Exercises, sufficiently numerous to render the 
Papil expert in performing the various calculations, and the 
Practical Measurer acquainted with the moftt tt^ij^T^'^^^ 
methods of takings dimensiona» 



V FBEFACK. 

In order to procure a practically useful treatise on Gauoinc 
the article on this subject has been entirely written by a prac 
tical man, an intelligent Supervisor of Excise. 

All the other articles, both in Parts I. and 11., have beei 
written by an eminent Mathematician in the Scottish metro 
polls, who is also distinguished as a public Lecturer on tli< 
science. 

For the conyenience of Teachers, and the use of such Stu- 
dents as do not enjoy the benefit of an instructor, a Key has 
been published simultaneously. with, the work. Besides the 
solutions at leogth of all the Exercises in the book, there is 
appended to it a valuable Supplement, containing demonstra- 
tions not given in the work, and much other useful matter. 



CONTENTS OF PART IT. 



inns, . • . . • .1 

I of Tables of Logarithms, . . 2 

ic ArithmiBtic, . . • .6 

Q of Tables of Log. Sines, Tangents, &c., . 1 1 

of Latitude and Departure, . . .15 

nterest and Annuities, . . . 16 

Trigonometrt, . . 19 

f ' • • « • •J" 

Right-angled Triangles, . . 22 

Oblique-angled Triangles, , . .24 

tion of useful Theorems, . . 29 

I Angles when the sides are given, . . 32 
8 -for the Sine and Cosine of the sum and difference 

jigles, - . ' . . . 33 

Trigonometry, Art. 51-85, . . 34 

Mensuration of Heights and Distances, . 37 

yy Rectangular Co-ordinates, , .42 
aduced from the angular bearings of three distant 

rhose distance is known, • . 46 

JUS Exercises, . . . .48 



Nayigation, . . - 49 



ng, 

ailing, 
Wng, . 
tude Sailing, 





53 


• 


^A. 


k 


^^ 


• 


\\i. 


■ s 


b^ 



w. 



I 



SURPACKS, 

To find the Area of a Parallel agram, . 

To find tlie Area of a Triangle, 

To Hod tho Area of a Trapezoid, 

To find the Area of a Trapeziara, 

To find tho Area of any Irregular Figure, 

To find the Area of a Regular Polygon, 

To find the Dioiueter aud CircumfereDoe of a Circle, the o 

from the other. 
To find Che length of an Arc of a Circle, 
To find the Area of a Circle, 
To find the Area of a Sector of a Circle, 
To find the Area of a Segment of a Circle, 
To find the Area of a Zone, 
To find the Aroa of a Circular Ring, . 
To find the Area included between two Arcs of different Circle^ 

having a common Chord, . . . i 

^B* find the Area of a Figure bounded by a curve and atraight 



To 

To 

i 



or Solids, , ) 

Ta find the Solidity of a Priam, . . . ] 

To flad the Surface and Solidity of a Cylinder, . , I 

To find the Surface and Solidity of a Pyramid, . < 

To find tlie Surface and Solidity of a Cone, , . 1 

To find the Surface and Solidit]' of the FruBtutu of a Pyramid 

or Cone, . . . . . ( 

To find the Solidity of a Wedge, . , . i 

To find the Solidity of a Priamoid, . . .1 

To find the Curve Surface of a Spliere, or any Segment or Zone 

of it, . . . , ,1 

To find the Solidity of a Sphere, ... 1 

To find the Solidity of a Segment of a Sphere, . . I 

To End the Solidity of a Zone of a Sphere, . 
To find the Soliditj- o( a Circular Spindle, 
To find the Solid Content of the Middle Frustum or Zone of ■ 

Circular Spindle, .... 
To find the Solidity and Surface of a Circular Ring, . 

To find the Surface and Solidity of each of the Regolar Bodw% M 



a Field ot (hree aiea, 
ueasure a Field ot four aiiw. 



C01ITB1IT8. TU 

P«ge 
toiiieasiize»Fiflid«f oMratib^nfoiirsidMy 1<M 

.To survey a pieee of Land in the form of an irreg^ular beU^ 107 

To measure a Lake or Wood without entering it, .106 

To find the breadth of a River without croaBing it, . IM 

BcBcription of the Plane Table, . . • 110 

To survey with the Pkne Table, . . Ill 

To survey with a Theodolite^ • . . 114 

To find the SurfiMse, and draw a Phoiy of hiUy or sloping 

ground, . • . . 115 

To deduce the true angle at a station from angles measured 

near it, . . . . .116 

To find lite leogiSityf a line ^oh ^fte level of the sea ^onesponding 

to a line measured at an elevated level, . . 117 

On flie Division of Land, . . . . 1 1)B 

Misedluieoiis Ezer^bes in Ltod Bltfteyilig, l!3i 

Snworic Gmajoi^ 138 

^tMe of Specific Gravities of Bodies, . . ib. 

Oiren the Specific Gxftvity and !ISkAidity <^ a Body, to find its 
Weight, ..... 126 

CKrcn the Specific Gravity and Woigitt Of a Body, to find its 
Solidity, . . • . .127 

To find the Specific Gravity of a Body, its Weight and Solid 
Content being given, . . . . ib. 

Td find the Specific Gravity of a Body without knowing its So- 
li^, . . . . . . 128 

To find the Specific Gravity of a Fluid, . ' 130 

To find the quantity of each ingredient in a given compound of 
two ingredients, • . . ib. 

Gaugihg, . • 131 

Deseription of the Sliding Rule, . • . 132 

To find the content of Solids of greater ^epth than one inch, 137 
Ganging open Vessels, . . 139 

To Gauge a Malt Cisterii, &c., ... ib. 

To Gauge a Vessel in the form of a FHistuin of a Pyramid or 

Gone, ..... 140 

To Gauge, Hx, and Tabulate a Vessel, which is nearly circular, 141 
To Gauge a Vessel nearly rectangular, • . \V^ 

To Gmx^ » Copper with a riaing crown, • , \^4 

To Qat^ a Copper with a &Umg crown, • . . \i^^ 

. T0 Oai^ a Veaa^ with a £aU or dnp, . \K^ 




'o Gauge and Taliulaie a,\ 
OrdinateE, 
To Gange a. SlM, 
Matt Gauging, 
Cask Gaugiug, . 
Ullaging of Casks, 
Miacellaiieous Questions is 
General ExHrciaea, . 






Loganthmic Sines, Tangeata, and Secanls, for every point ani 

quarter point of the Compaaa, 
IjOgarithniB nf all niunbors, iraui 1 1o 100, , 
LogBrilhmBofallDQmbere, from 100 to 1,000,000, 
Logarithmic Sines, Tangents, and Secants, for every degree 

and minute of the Quadrant, 
Natural Sines and Coainoa for every degree and Minute of the 

Quadrant, . . . • . 

Difference of Latitude and Departure, 

Compound InterSBt and Aiinuitiee, , . . 

Length of Circular Arcs, &a, . . . it. 




OF LOGARITHMS. 

1. Logarithms are a kind of artificial numbers, invented 
bj Lord Napier, of Merchiston Castle, near Edinburgh, to 
racilitate certain calculations, such that while the natural 
numbers increase in geometrical progression, their loga- 
rithms increase in arithmetical progression. By this arti- 
fice the operation of multiplying numbers is reduced to that 
of adding their logarithms ; that of dividing numbers to 
that of subtracting their logarithms ; the raising of powers 
to that of multiplying the logarithms ; and the extracting 
of roots to that of dividing the logarithms of the numbers. 

2. If n:=a*, x is called the logarithm of the number n to 
the base a ; or the logarithm of a number to a given base 
is that power of the base which is equal to the given num- 
ber. 

3. The base may be any number whatever except 1, but 
file most convenient base is 10, which is the base of the 
common logarithms given in this work ; but a simple illus- 
tration of the nature of logarithms may be given, by taking 
ike base 2, or 3, or 10, and only using those numbers of 
the natur^ series whose logarithms are whole numbers. 

Thus to the base 2. 

0. L 2. 3. 4. 5. 6. 7. 8. log. 

1. 2. 4. 8. 16. 32. 64. 128. 256. nat. num. 

Or to the base 3. 

0. 1. 2. 3. 4. 5. 6. 7. 8. log. 

1. 3. 9. 27. 81. 243. 729. 2187- 6561. nat. num. 

Or to the base 10. 

0. 1. 2. 3. 4. 5. 6. 7. 8. log. 

1.10. 100. 1000. 10,000. 100,000. 1,000,000.10,000,000.100,000,000 

nat. num. 

In either of the above series it will be found, that if two 
logarithms be added together, such as 3 and 5, their sum 
S points oat the number which is the product of the num- 
kefs of which 3 and 5 are the logarithms ; or if the diiFe- 
mnce of two logarithms, as of 7 and 5, be taken, the re^ 
aiainder, 2, will point out the quotient arising from divid- 
iiK the number of which 7 is the logarithm, b]f tha^t q^ 
iUck 5 is tfad h^nthm; hence. 




_ I of the lor/arithna of two itumbera it the foy-- 

«W(Am of their product, and the difference of the logarithm 
of two nv/mheri ie the logar Uhm of their quotient ; wfiicli mt^' 
be demonstrated generally as follona, bj the notation 
in(Art.2); for let ii=a', and n'=a'', then their produrt. 
^»ea }in'=a'xa''=a"'+ ^>, where a:-f-*', or the snm of 
the logarithms of the numhers, is evidently the ]ogarilhin({ 

their product: also their quotient gives -^ — .^a'**^, 

where x — x', or the difference of ihe logarithms of the nuj 
hers, is the logarithm of ibeir quotient. 

5. The logarithm of a. potver or root of any number 
the logarithm of the number multiplied by the espoaent 
which indicates that power or root; for let «=a'repreB«it 
any number having ils logarithm x to the base o, and 1*1 
m represent the exponent of any power, (m^ an inleget)i 
or root, (m^ a fraction) ; then we have 7(™=;(a') "=: 
therefore if x be the logari tbm of n, »w is the logarithi 
its With power or root ; hence by giving proper values t 
we have the following niles for raising numbers to 
power, or extracting any root. 

6. To square any number by logarilbms, multiply Oil 
logarithm of the number by two, and the product wiUbt 
the logarithm of its square. 

7. To find the cube of any number by logarithms, 
tiply (he logarithm of the number by three, and the prodncl 
will be the logarithm of its cube. 

8. To find the square root of any number by logarilhiDS, 
divide the logarithm of the number by two, and the quo- 
tient will be the logarithm of its square root. 

9. To extract the cube root of a number by logarilbmii 
divide the logarithm of the number by three, and the quO' 
tient will be the logarithm of its cube root. 



k 



Description of the- Tablks of Looabithnj 

10. Since the base of the logarithms here used U 10. 
any number will be equal to IC, where x is the logarilia 
of the number; if j = 0, then 10"=!, therefore the loj- 
ariibmof lis 0; if ^=1, (hen ]0'=10, the logarithm of 10 
18 1, bence the logarithms of all numbers between J UmI 
W are greater than ani \eM \Wti\, V\«rt is, they tn 






LOGABITHMS. O 

decimal fractions. In the same manner, by making a;=2, 
it is evident that the logarithm of 100 is 2, and therefore 
that all numbers between 10 and 100 will have their loga- 
rithms 1, with a decimal fraction annexed ; but all num- 
bers between 10 and 100 are written with two figures, 
bence the int^pral part of the logarithm is one less than 
the integral places in the number, and this rule may in the 
same manner be proved universal. For this reason the 
decimal part (or mantissa) only of the logarithms is in- 
serted in the tables^ and the integral part (or index) is pre- 
fixed by the following rule : 

11. Find how many places from the unit's figure the first 
significant figure of the number is^ that prefixed to the 
dedmal part found from the tables will be the true loga- 
liUun. If the first significant figure be to the left of the 
unit's place, the index is plus, but if to the right, the index 
is, n^nus, and the sign ( — ) must be placed over it; thus 
the logarithm of 150 is 2176091, while that of 0015 is 

I 2*176091, where the ( — ) applies only to the index, not to 

- the mantissa, it being always ( + )• 

' 12. To find the logarithm of any whole number under 
100. 

Look for the number under N, in the first page of the 
logarithms, then immediately on the right of it is the loga- 
rithm sought, with its proper index. Thus the log. of 56 
18 17481&, and the log. of 91 is 1-959041. 

13. To find the logarithm of any number between 100 
' and 1000. 

iind the given number in some of the following pages 
' of the table, in the first column, marked N, and imme- 
diately on the right of the number stands the decimal part 
ti the logarithm, in the column marked 0, at the top and 
bottom, to which decimal prefix the proper index, (Art. 11). 
Thus the log. of 457 is 2-659916, and the log. of 814 is 
i^l0624. 

14. To find the logarithm of any number consisting of 
four places. 

Find the first three figures in some of the left-hand co- 
himns of the pages, as in the last case, and the fourth 
figure at the top or bottom of the table ; then directly un- 
do the fourth figure, and in the straight line across the 
S, from the first three figures of the number will be 
d the decimal part of the logarithm sought, and the 
index must be supplied by (Art. 11). Thus the log. of 
7^384 is 0*868292, the log. of 793*5 is 2-899547, au^Xk^X 
efHMa38iff«»3E39A 



_ ' 1&. To find tbe logarithm of any number consistiog ol 
fire or six placee. 

Find the logarithm of the first four left-hand 6gures, as 
in the last article, to which prefii the proper index bj 
(Art. 11); then from the right-hand column, marked D, 
(meaning tabular difference), take the number oppoaite to 
that logarithm, and multiply it by the remaining f^rea ol 
the natural number ; point off from the right-hand side d 
the product as many figurea as there are in the multiplier; 
then add the rest of the product to the logarithm befoie 
found, and the sum is the logarithm required. 
Ex. 1. Required the logarithm of 348G3. 

Log. of 34860 by the table is 4.54232? 

Diff. 125 X 3=37-5, therefore ad.i 37 

I Gives log. of 34863 =4-542364. 

Note. It maj be remarked here, that wheu the firat flgm 
pointed off in the product nf the difference is S, we may tUber tab 
the remnining iigureE, or tiie reiaaJnin); Rgtires increased by one, M 
in either caao the crrer is |, in the last fi^re or the logarithm; Iwl 
if the figure or figures pointed ofl" exceed "5, the remaning Sgora 
must slwaya be iucreaaed by unity, as the error by this meara vU 
be leBB than j, ju the last figure of Clje logarithm, whereas by onJt- 
tiag then, it wuuld be greater than a half. 

Ex. 2. Required the logarithm of 46-8375. 

Log. of 4683 by the table is 1-670524 

Diff. 93x75=69-73. ■■• add 70 

Hence log. of 46 8375 =1-670594 

Ex. 3. Required the logarithm of •0076452.^. 
Log. of -007645 by the table (Art. 1 1) is 3-883377 

Diff. 57x25=14-25, .-.add 14 

Hence log. of '00764525 =3-883391. 



EXERCI8K3. 




The log. of 4fi is 


=1-662758 


The log. of 97 is 


=1-986772 


The log. of 286 is 


=2-456366 


. The log. of 901 is 


=2954725 


. The log. of 7569 is 


=3-879039 


. The log. of 2344 is 


=3-369958 


. The log. of 46874 is 


=4-670932 


. The log. of 954625 is 


=3-979832 


. The log. of 3-56773 is 


=0-552394 


To find the logarithm of a 


nilgar fraction, or 


number. 





r iiDOAHiTaHB. G 

Reduce the volgar fraction to a decimal ; then £nd the 
decimal part of its logarithm hy the preceding rules, and 
preiis the proper index, as found by (Art. II.) Or, from 
ibe logarithm of the numerator subtract the logarithm of 
the denominator, and the remainder will be the logarithm 
of the fractioa eought, (Art. 4.) A mixed number may be 
reduced to aa improper fracdon, and its logarithm found 
in the 81 



Es. 1. Required the logarithm of ^■^, =-2916. 
From log. of 7 by the table ' =0845098 
Take log. of 24 =1-380211 

Hence log. of a\, or -29 1 (J, is =1 ■464887. 

NoTB. If (he logarithm of thiR example be taken out &vm ths 
dednul froetion, since it repeats G, which is the di-citnsl of |, n-e 
must add | of the t&bulBT diifereDce to the logarithio of the first 
Imit figuTSB, in order to obtmn the true logarithm ; thus log. of 
'3916=T-4647HK, and the tabular dttTorence is 14!), two-thirds of 
■hich ia 98, which beiug added to the log. foniierlj found, ^vea 
H6itta7, the some as baforc 

El. 2. Required the logarithm of 231= 'V''=23'25. 

From log. of 93 by the table =1-968483 

Take log. of 4 

Hence log. of =i% or 23-25 
17. To £ud the natural numher answering to any given 



Look for the decimal part of the given logarithm in the 
different columns, until yon find either it exactly, or the 
next lesE. Then in a line with the logarithm found, in the 
left-hand column marked N, you nil! have three figures of 
the number sought, and on the top of the column in which 
ibe logarithm found stands, you have one figure more, 
nhich annex to the other three ; place the decimal point 
to that the number of integral figures may be one more 
llian the units in the index ot the logarithm, (Art. 11), and 
if the logarithm was found exactly, you have the number 
(cquired. 

If the logarithm be not found exactly in the table, sub- 
tract the logarithm next less than it found in the table from 
tlie given one, and divide the remainder, with two ciphers 
annexed, hy the tabular difference, which will give other 
Iwo figures, (if the quotient give only one figure, the first 
is a cipher), which annex to the four figures found by in- 
apection, and it will be the number sought, eice^t \Ve \q.- 
dex show (hat fienumfaer consists of more ihaivavifeaAMts, 



* 



I 



in which case annex ciphers to the number formerly Found, 
till it contain the required number of integral places. ThU 
will be the number sought wilb as great accuracj aa it can 
be obtained by logaritbma carried to six decimal places, and 
will be always nearer to the true number than by a mil- 
lionth of itself, however many integral places the numbtf 
nay contain. 

Required the number answering to the log. 
3785624. 

The given logarithiu, 3-785fi24 

The next less tab. log. is the log. of ol 04= 3-785615 
Remainder, ^ 9. 

Tab. diff. =71)900(13 to the nearest unit. 

Hence the number sought is 6I04']3, since the index 3 
shows that it consists of 4 integral places. 

In like manner (be natural number corresponding to 
Ibe fulloiving logarithms may be found. 

1. Tlie number corresponding to the log. 5-314782, ii 
2UfJ-)34. 

2. The number corresponding to the log. 3-290035, ii 
■00195. 

3. The number corresponding to the log. 2-531907, » 
34()'335. 

4. The number corresponding to the log. 0-357912, il 
2-27988. 

5. The number corresponding to the log. 6-486S05, it 
3063410. 

6. The number corresponding to the log. 1-282169, l> 
1^1915. 



LoaAniTUMic ARirrtiuBTic. 



18. To PERFORM Multiplication by Looarithhs. 
EuLK. Add together the logarithms of all the factors. 
and the sum will be the logarithm of the product. If 
flome of the indices be negative and some positive, add the 
positive indices and the carriage from the decimal part Into 
one sum, and the negative indices into another, and their 
difference will bf the index, which will have the same sign 
as the greater suia. 
Ex, 1. Required the fioduct of 371 an<i 84-5. 
LoE. ot S1\=.0-^of£ftlV 
Log. 0^84.5 =\-^ie6Vl^ 
Log. of product, -a^^ AS>5='i-AamV- 



Logarithms. 7 

Ex. 2. Required tlie product of 56*125, and 743*75. 

Log. of56125=l-749156 
Log. of 74375=2-871427 
Log. of product, 41 743- =4-620583: 

Note. In finding the number corresponding to the logarithm of 
the prodact in thiB example, we obtain 41742*9, with a remainder of 
84, which is greater than half the divisor 104; we therefore increase 
the last figure by unity, which giyes the result in the text. The same 
should be done in eyery case where the remainder is greater than 
half the divisor. 

Ex. 3. Required the continued product of 356*225, 
•6385, -07425, and 8*42725. 

Log. of 356*225=2-551724 

Log. of -6385 =1*805161 

Log. of -07425 =2*870696 
Log. of 8-42725 =0*925686 

Log. of product, 142*32 =2*15326?. 

NoTB. In the above example we add the decimal part of the log- 
arithms, which are all positive, and having 3 to carry, we add it to* 
the positive index 2, which makes 5, then adding the negative in- 
dices, we find their sum 3, which being taken from the positive in- 
dex 5, leaves 2, which is the index required, and is positive, because 
the greater sum is positive : if the sum of the negative indices had 
bem greater than the sum of the positive, we would have subtracted 
fte sum of the positive from that of the negative, and marked the 
lemainder negative: if the sum of the positive and negative indices 
vere equal, of course the index would be 0. 

BXBRCISBS. 

1. Required the product of 75*825 and 84 75. 

Ans. 642616. 

2. Required the product of -75, -0625, and 846*25. 

Ans. 39-6678. 
a Required the product of 4*825, -0225, and 0145. 

Ans. 00157415. 

4. Required the product of 7{, 2}, 31^ and 20 A. 

Ans. 1012*45. 

5. Required the product of '7345, 734*5, and 73*45. 

Ans. 39625-6. 

6. Required the product of -25, -7325, and 01725. 

Ans. -00315891. 

JP. To Pemfobm Division by Looabitbl^^. 

BuLE. From the logmthm of tte di^denSi «vi\i\x«^cX. 



the logarithm of the diviaor, and the reminder will be 
the logarithm of the quotient. Or, take the arithmetical 
complement of the divisor, and to it add the logarithm of 
the dividend, the sum, rtjictiny ten frmn Ike index, will be 
the logarithm of the qnoiient. 

'llie arithmetical complement of a logariCbm is the diffe- 
rence between it and 10 of an index, and is contracted 
into (nr. co.). It is most conveniently found hj beginning 
at the left-hand side of the logarithm, and salitracting the 
index and the succeeding figures from 9, except the lait 
significant figure, which must be subtracted from JO. 

Ex. Divide 74^-625 by 42 6125. 

Log. of 743-625=2-87l353 2-871353 

Log. of 42-6 125 = 1-62953 6 ar. co. = 8 370464 

Quotient, I7-4'>09=1'241817 i2418J7. 

Ex. 2. Divide 398^ by 7439. 
Log, of 398-125=2-600019 2-600019 

Log. of 7439 =3-871515 ar. co. = 6-128485 

Log. of quotient, *06351 85 =2728504 2^728504. 

Ex. 3. Divide I by -78.>4. 

Lc^. of 1=0-000000 0-000000 

Log. of ■78.'j4=l 895091 ar. co. =10-104909 

Quotient, 1-2 7324=0- 104909 



Note. When we have to subtract a mlnaa ini 
by the rules of Algebra, Jjeeausa Bubtraetiog u n 
adding a plus quantity; and if the minuend ba> 
must Le eubtracled. 

EXBRCISES. 

1. Divide 385 by 29-25. 

2. Divide 463-28 by 474. 

3. Divide 364 x 54-75, by 7854. 

4. Divide 463x25-25, by 3-1416. 

5. Divide 1 by 3-1416. 
fi. Divide 3-1416 by 180. 



To Solve a Pkopoetion bv Logaritbus. 

Hulk. From the sum of the logarithms of the second 
md third terms subtract the logarithm of the first, and (he 
■emainder will be the logariihra of the answer, in the same 
'enomination as tbc third term. Or, take the arithmetkil 
implement of the first leim, Kadi \o \t &dd the loganthnu 



0- J 04909. 

ex, we most add il 
inus is the samcu 



Ans. 13-1634, 
Ans. 9-77386. 
Ans. 25374-3, 
Ans. 3721-27. 
Ans. -318309, 
Ans. -017453, 



of the second and third terms, and the sum, rejecting 10 
from the index, wilt be the logarithm of the answer. 

If the proportion be compound, add the logarillims of 
all the second terms and the third together, and from the 
sum aulitract the sam of tbe logorilhms of the first leims, 
and tbe remainder will be the logarithm of the answer. 
Or, take the arithmetical complement of each of the first 
terms, and the logarithms of the second terms and the 
third, then add them all together, and the sum will be the 
logarithm of the answer, after rejecting as many 10s from 
the index as there were arithmetical compleraeuta taken. 

Ex. 1. If 3125 yards cost L.l, 5b., what will 73«62!> 
jud» eost 1 

B As 3-125 log. 0-494850 ar. co. 9-505150 
I : 730625 log. 1-863695 log. I-B65695 
^:L.l-25 log 096910 log. 009f)9IO 
:L.29-225 =l-4fi57a5 1-465755. 

The liTst method ia wrought in one line, by kilding the 
■emd and third togurithma together, &nd froin the suiu Bubtnivtiii); 
Ibg first, and doing bo in eacli column spparately as we prucend ; 
nd the second is wrought by ulding tliu three liaes together, nnd 
abtracting 10 from the index. The answer obtained is L.3!<'-2-25 = 
1.S9, 4s. 6d., which may be verified by conunoD Britlimetii!. 

Ex. 2. If 30 masons build a wall 60 feet long, 20 feet 
Ugh, and 2 feet thick, in 12 days, working 8 hours per day, 
law many days should 20 masons take to build a wall 100 
ket long, 15 feet high, and 3 feet thick, when they work 
10 hours per day? 

; 12 days. 



I 



60 ft, 


100 ft. 




20 ft. 


15 ft. 




2 ft. 




3 ft 




10 ho. 




8 ho 




20=1-301030 




Log. 30=1-477121 


60=1-778151 




., 100-2000000 


20=1-301030 




„ 15=M760!>1 


2=0 301030 




3=0477'2I 


10=1-000000 




8=0-903090 


3-68]i 


41 




12=1-079181 



7-11 2604 

Subtract sum of logs, of the first terms =5-681241 
The answer. Log. 27=r43r:fli3 

Rbiurk. In this eiMinpIe tbe logarithms ot tbe ftrtl, Uivra* tttfc 
patdovo by tbiuaaelveB, and added inJo one aum.and llic\<igB,ivftwv« 
Bfti/eteeondandiliird by theniseJves, and added inW, aauVXiM ', *>"■>■ 
U Bra sum IB auboscted from the second, and tUc remamiet \^ 



^ 



10 LosAWTSItS. 

tliB lognritbro or the anawpr, »-hich, bowBver, is less than tlie loga- 
ritlim of 37, hy 1 in the Ust plaice. Thifi urkeB from the loniiuii- 
aUutee, that tUo lagai-ithms of 30, S, and IS, are loo small in lb» 
last figure bj ,%, and tliat of 12 also too Email by ^'g, the i * 

j'l,; bat the logarithm of GO in tbc first terms is al. 

L, whieh being taken from the former, leaves ,'g toi 
lit, whieh in the tables is made lo increaw the last tigiu* 
)>y unit}', liocnuec it is greater than a lialf. 



1. If 25 labourers can dig a trench 220 yards long, 
3 feet 4 inches wide, and 2 feet 6 inches deep, in 32 dajs, 
of nine hours each; how many would it require to digs 
trench half a mile long, 2 feet 4 inches deep, and 3 Ktt 
<i inches wide, in 26 days, of eight honrg each ? 

Ans. 98 labourem 

2. If 3 men, working ten honrs a-day, can reap a field, 
nieasurinjt 150 yards by 240 yards, in five days, how manj 
men, working twelve hours a-day, can reap a £eld meaaur- 
ing J92 yards by 300 yards, in four days? Ans. 5 ntta. 

3. If 27 men can do a piece of work in 14 days, wott- 
ing 10 hours a-day, how many hours a-day must 24 boy* 
work, in order to complete the same in 45 days, the work 
of a. hoy being half that of a man ? Ans. 7 boon. 

4. If 120 men, in 3 days of 12 hours each, can " 
trench 30 yards long, 2 feet broad, and 4 feet deep ; 
muny men would be required to dig a trench 50 yatdf 
long, 6 feet deep, and 4^ feet broad, in 9 days of 15 boiu* 
each? Ans. 180 

21. Involution and Evolution bv Logabithms. 
RuLB. Multiply the logarithm of the number by lie 
I exponent of the power or fractional esponent of the tool 
to be eitracted, and the result will be the logarithm of the 
power or root required. (See Art. 5.) 
I Ei. 1. What is the third power of 72'85; or wbl « 
I (72-85)'? - 

Log. of 7285 =1-862430 



Log. of 38fl625-=5-5872y0 
Ei. 2. What is the fourth power of -05326 T 
Log. of -05326 =2726320 



Log. of ■oooooaowv\=?.'2!ssa'«a 

«E«r*H.K. In this emmv>6, *<: ^ni«^^™^,^'«^,'^>-Zii3 
«ddwg the 2, ,hieh w« cans irotu <.\« is=v«>^\ ,«vv.>i»^.*» 



lOGABITHMS. 1 1 

of the index, we subtract it, because it is the index only that is nega- 
tive, the dednuJ part being always positive. See Art. 10. 

Ex. 3. Find the square root of 7854 ? 
Log. of 7854 =1-89509 1 

Log. of -886227=7947545 
Ex. 4. What is the fifth root of -34625? 

Log. of -34625 =1-539390 

I 
J 



Log. of -808869=1-907878 

Remabk. In the last two examples, where the index is negative, 
it most be remarked, that in dividing by the denominator of ^e in- 
dex of tiie root, the negative index is increased by as many as will 
make it divisible without a remainder, and then the same number 
18 considered as a positive remainder, and taken in with the decimal 
part of the logarithm; thus the same quantity is first subtracted by 
considering it negative, and then added by considering it positive; 
and hence the value of the quotient is not altered. This must al- 
ways be done when a logarithm with a negative index is divided. 

EXERCI3ES. 

1. What is the sixth power of 1055 ? Ans. 1 -37883. 

2. What is the tenth power of 1-125 ? Ans. 3-24736. 

3. What is the seventh power of 5-125? Ans. 928662. 

4. What is the square root of 2625 ? Ans. 51-2347. 

5. What is the cube root of 17? Ans. 2-57128. 

6. What is the tenth root of 5386-25 ? Ans. 2-361 18. 

7. What is the cuhe root of f ? Ans. -658633. 

8. What is the fifth root of J? Ans. 72478. 

9. What is the fourth root of i^^^^^^? Ans. 1 05967. 

180 

10. Find the value of /5^^H1M?\I? Ans. 25-4327- 

Y 19x52 ) 

11. Find the value of (l-05)i'^ x705 x |? Ans. 361-733. 

12. Find the value of ( !f l^^'lf o^ ^*? Ans. -849433. 



EXPLANATION OF TABLES. 

Of ILoGABiTHMic S1NB8, Tangbnts, and Skck^t^. 

J. To £nd the logarithmic sine, tangent, ox aec^aX.^ ^l 
aajr number of degrees and minutes. 



19 

B.DLE. If the Dumber of degrees be less than 45", wek 
them at the top of the page ; then in a line with the given 
number of minutes in the left hand marginal column, 
marked M at the top, aud under the word sine, tangent, 
or secant] you have the logarithmic bidc, tangent, or 
secant, of the proposed number of degrees and minutes. 

If the number of degrees he above 45°, and under 90°, 
seek them at the bottom of the page; then in a line wilh 
the minutes in the right kand taatgvaaX column, marked M 
at the bottom, and above the ivord sine, tangent, or secant, 
you have the logarithm sought. 

Exactly in the same manner is the logarithmic cosine, 
cotangent, or cosecant of any number of degrees and mi- 
nutes less than 90° found, in the columns marked cosine, 
cotangent, or cosecant, at the top or bottom; the name 
being alwai/B found at the top, when the degrees are found ^ 
the top, and at the bottom, tvketi the degrees are found at iht 
bottom. 

When the degrees exceed 90°, subtract the angle from 
180°, and take out the sine, tangent, secant, &c., of the 
remainder ; or subtract 90° from the angle, and instead of 
the sine, tangent, secant, &c., of the angle, take out the 
cosine, colangent. cosecant, &c^ of the remainder, awl 
conversely. 



Angle. 


Siue. 


Tangont. 


Secwil. 


15= 12' 


9-418615 


9-434080 


1 0-0 154(15 


70= 16' 


9-973716 


10-445259 


10-471543 


21 => 54' 


9-571695 


9-604223 


10-03252y 


36° 15' 


9771815 


9-865240 


10-093425 


108» 18' 


9-977461 


10-480642 


10-503081 


164= 39' 


9-631593 


9-675564 


10-043971 



2. To find the sine, tangent, or secant, of any number 

of degrees, minutes, and seconds. 

Rule. Find the sine, tangent, or secant, corresponiliiig 
to the given number of degrees and minutes, as befolti 
and from the adjacent column, marked D, take out itis 
tabular difference, corresponding to the same angle, vrhicli 
multiply by the seconds in the given angle ; cut off t*" 
decimal places from the right of tbe product, and add lh« 
Temaiuiug figures to the logarithm furmerly found, and i< 
will be the logarithm reijuiied. 



KEFJCAITATIOII OF TABLES. 




^niied the logarithmic tangent of 24° 3' 16"? 
Tangent of 24° 3' =9-649602 DifF. 566 



^^Correcti 


n for 16" = 


91 


9056 


^gangent 


of24°3'16"= 


9tJ4Hti»3 




w 


EXERCISES. 




Angles. 


Sine. 


Tangent. 


Secant. 


36° 12' 51" 


9-771444 


9-864670 


lU'093-227 


430 18' 43* 


9-836305 


9-974394 


10-138089 


69° 56' 13" 


9'972813 


10-437449 


10-464637 


98= 17' 40" 


9-995434 


10-836287 


10'840853 


52> 26' 19" 


9-899109 


10-114057 


10'2 14947 


1340 45' 6' 


9'851359 


10'003765 


10-152405 



The cosine, cotangent, and cosecant of any number of 
degrees, minutes, and seconds, are to be found in the same 
manner; except that the proportional part for the seconds 
most be subtracted. 



EXERCISES. 



Angles. 
330 14' 24" 
21° 46' 52" 
34° 17' 32" 
32" 5' 35" 
73° 43' 8" 
129° 16' 11" 



9-922405 
9-967832 
9-917072 
9-927979 
9-447701 
9-801384 



10-398387 
10-166244 
i 0-202642 
9465476 
9'912545 



Cosecant. 
10-261103 
10-430554 
10-249172 
10'274663 
10-017775 
10-111161 



3. To find the angle corresponding to any giTen loga- 
lithmic sine, tangent, 01 eecant; cosine, cotangent, or co- 
eecant. 

R111.E. Find in its rcspectiTe column the sine, tangent, 
or secant nearest to that given, but less, if it be not found 
exactly in the table ; take out the degrees and minutes 
which correspond to this logarithm ; then subtract it from 
the given logarithm, and to the remainder annex livo 
ciphers, nnd divide it by the tabular difference adjacent to 
the nearest logarithm found in the table ; the quotient will 
be (he seconds of the angle, which annex to the degrees 
and minutes formerly foiind, and it will be the angle 
■ought 

But if it he a cosine, cotangent, or coB&can^, ^n.&^n\^ 
respectire column the nearest to it, tut gie-A^eT,\^ \*.« ■n.'A, 



I 



14 KXPLAJIATION or TABLES. ^ 

found cxnctly, and taking out the degrees and minntea 
correspontJing lo it, subtract the given logarithm from it; 
to the remainder annex two ciphera, and divide bj the 
tabular diffotence opposite the nest greater minute ; the 
quotient will be the aeconds, which annex to the degrees 
and minutes formerly found, and it will be the angle 
sought' 

Example. It is required to find the angle correspond- 
ing lo the logarithmic tangent 9943765, and to the logs- | 
rithmic cosine 9-496724. 

Firsl, By inspection of the columns of tangents, we find 
that the nest less than it is 9-943732, which correspondi 
to 41" ]8', and is less than the given one by 13. To this | 
remainder we annex two ciphers, which mnkca 1300; this I 
we diTide by the tabular dilference 427, and the quotient I 
is 3, which is therefore the seconds; and hence the angle ] 
sought is 41° ] 8' 3". 

Second, By inspection of the columns of cosines, we find 
that the next greater cosine is 9-496919, which corre- 
sponds to 71° 43', and is greater than the given one by 19a; 
to this we annex two ciphers, which makes 19500; ihii 
we divide by the tabular difference, corresponding to 43', 
viz. 637. which gives 31" nearly; hence the angle sought 
is 71° 42' 31". 



Sin Re. Angles. Tutigcnts. Ai]gli?B. 

9-346373=] 2° 49' 59" 9-764825=30= 1 1' 36" 

9782599=37= 18' 49" 9-978546=43=35' 7" 

9-952864=630 47' 8" 10- 6 3-5 887 =760 58' 42" 

Secants. Coaiaes. 

10'468357=70° 6' 55" 9'367285=i76° 31' 43" 

10'004763= 8= 28' 1 1" 9-846349=45° 24' 3*/' 

10-I47364=44=34'52" 9 167249=81= 32' 54" 

Cotangents. Angles. Cosecants. Angles, 

9-874639=53= 9' 24" 10-438674=21= 21' 27" 

10-104738=38= 9' 25" 10-017643=73= 46' 43" 

9-964286=47= 21' U" 10-114763=50= 9' 18" 

Note. If the erne, taDgeiit, &c, be knsmt to be tliat'ofatiehsN 

angle, the correapoDdmg ajigte, as round above, must be subbwltd 
from IBO% to gel tbo true angle. 



^^'1 



To find the natural sine, tangent, secant, &c.j of wf 
.fcer of degrees, minatea, and seconds. 



^p 



Kt«>AITjLTI01I or TABLES. 15 

RuLB. Find the logarithmic sine, tangent, secant, &c„ 
of the given angle, subtract 10 from the index, and take 
out the numher corresponding to the remainder from the 
tables of logarithms of numbers, and it will be the sine, 
tangent, &c., required. 

The natural sine and cosine of any angle may be more 
readily found from the tables of natural sines and cosines, 
the arrangement of ivhich is the same as that of the loga- 
rithmic sines, &c., except that the differences are oat given. 
These must be found by subtraction, and the corrections 
made by the following proportions: 60' is to the given 
number of seconds, as the diiFerence for 60'' is to the cor- 
rection required ; and the difference for 60" is to the given 
difference, as 60" is to the seconds required. 



Example 


. What is 


the natural sine of 12" 34' 12"? 


Natural 


ineofl2''34' 


217575 


Difference, -2178.59- 


-217575=284 




60:12: 


284 : 57, o 


284 XA^ 


-f-57 


Natural 


ineofl2°34'ia" 


217632 


Ex. 2. Find the angle 


corresponding to the natural co- 


tiae -843647? 










-843704=c 


sine 31- 28'. 




The differe 


ce between this and the sive 


n cosine is 57, 


Md the difference belwee 


n COS. 3P 28', an 


d COS. 31= 29'! 


iil5fl; hence 156:57: 


60:23 nearly; 


therefore the 


mgle sought is 31° 28' 22 








BXERCISES. 




HakSnes. 


Angles. 


Nat. Coainea. 


1 Angi™. 


■354638= 


=20° 46' 17' 


-576438=54 


47' 58" 


•459637= 


=27° 21' 49' 


■274869=74 


2' 45" 


■M763g= 


=57° 57' 21' 


■542873=57 


7' 14" 


■646739= 


=33= 8' 37' 


■984733=10 


V 23" 



■>. DiFPERBNCE Ol' LATITUDE AND DePABTUHE. 

In this table are inserted the hypotenuse, sides, and 
angjea of ail right-angled triangles, whoso acute angles 
are either whole degrees, points, or quarter points of the 
compass. The hypotenuse is represented by the distance, 
placed at the top and bottom of the page, the side opposite 
the angle by the column marked Dep., (Departure), and 
the side adjacent the angle by the column mavked. ^£1.^.1^ 
(Difference of Latitude^. It is principuWy \isei ^ot ^ivi- 



i 



^ 



\ ing the sides of triangles when the hypotenuse and an 
angle are given, but may be used for other purposes. 
Wlien the angle ia found on the left side of the page, the 
name of the sides must be read at the top, and when the 
angle is found on the right hand side of the page, the 
name of the side is found at the bottom. The numbers, 
1, 2, 3, &c., at the top or bottom of the page, may be ac- 
counted 10, 20, 3(1, &c., if the decimal point in the co- 
lumns under them he removed one place to the right; oi 
they may be accounted 100, 200, 300. &(;., by remoring 
the decimal point two phices to the right. 

Example. Find the sides of a triangle, the bypotenuae 



being 346, and one 


of the acute angle 


28°? 


Hypgteliuiie. Angle 
300 28° 

40 28- 
6 38° 


Side opposite. 
140-84 
18-779 
2-8168 


Side adjacent 
264-88 
35-318 
5-2977 


346 28° 


162.4358 


305-4957 



6. Another important use may he made of this table, m 
follows: — In that column marked 6, at the fop and botton, 
if the decimal point be removed one place to the right, the 
difference of latitude will express the length of a degree of 
longitude in nautical* miles at that place whose latitude u 
the same as the course. 

Example. Find the length of a degree of longitude on 
the parallel of 57°! and also on the parallel of 34°f 

Since 57° is found at the right-hand side of the page, 
the name must be found at the bottom ; hence the lengdi 
required is 32678 miles. 

Again, 34° being found at the left hand side of the pag^i 
the name is found at the top ; and opposite to 34°i anil 
under 6, in the proper column, is found 49-742 miles, the 
length required. 

I Tablbb of Interest and Annuities. 

7. Table VII, gives the sum to which L.l will amoaBt, 
if improved at compoimd interest, for any number of yeaw 
not esceeding 60, at 2i, 3, 3|, 4, 4^, 5, and 6 percent 
The amount of any other sum may therefore be found, bj 
multiplying the amount of L.l for the given time and ret* 
Lf the principal, espressed in pounds, ( Alg. 120,) 

Thus the amount ot T.,30, \0b., M\J3fl-5, for 17 jvoh 



at 6 per cent., is Ia82'1295765 ; for under 6 per cent, and 
opposite to 17 years, is found 2()92773. wbich being mul- 
tiplied by SO.'J. gives 82-1295760. 

8. Table Vlll, pves the sum, irhich being improved at 
compnond interest, will amount to L.l in sinv (fiven mini' 
ber of years not exceeding fiO, or tbe sum which should be 
paid down immediately, as an equivalent for L.l, to be 
paid at the eipiration of tbe given number of years. The 
present value of any given sum is found by multiplying 
the present value of L. 1 by that num expreiised in pounds. 

Thus the present value of L.2.'). due 14 years hence, nt 
3^ per cent., is tbe sum found in the table under H^ per 
wnt., and opposite to 14 years, via. ■6177f!2 multiplied by 
25, orL.l.'i-4445.i. 

9. Table IX. gives the amount of an annuity of L.I. 

*ben deferred for any number of years not exceeding &}. 

nnd improved earh year as it becomes due, at compound 
intereat, at 2i, 3, 3^, 4, 4^-, fi. and C per cent. Tbe 
amount of any annuity is found by multiplying the tabu- 
lar amount, corresponding to the given rate and time, by 
the annuity cEpressed in pounds. 

Thui the amount of an annuity of L.RO per annum, 
deferred for 24 years, and improved at :H per cent., is 
L.36-666i32a x 80 =L, 2933 32234. 

10. Table X. gives the present value of an annuity if 
1*1, payable tor any number ot years not exceeding 60, 
when money can be improved at tbe rate of 2^, 3, 3^, 4. 
44. 5, or 6 per cent., the first payment being due one 
year hence. The present value of any annuity is found 
by multiplying the tabular value, corresponding to the 
given time and rate per cent,, by the annuity expressed in 

Thus the present value of an annuity of L.3fl, to be paid 
at the end of each year, for 40 years, when money can be 
improTcd at compound interest, at the rate of 2^ per cent, 
i» the tabidar value, viz. L.25102775x30=L.753-08a2S. 

11. Table XL contains the length of circular arcs to ra- 
diuv-l, for any number of degrees from 1 to 30, and for 
every JO degrees from 40 to 180; also the fourth and fifth 
ulunins contain the lengths of arcs for any number of 
BiinateB or seconds, the figures given being tbe last of 7 
tic^inial places, so that tbe necessary number of ciphers 
moat be prefixed to put them in their proper -pVaces, vit^B. 
liBed aloDe. The Jength of an arc to any ot\ieT raA\\i6 \* 

fituad hf Srst Sniiing the ienplh to radiua \ , ani ftiftvi. 

la/fipfyiag bj- tbe given radius. 



I 



18 EXPLANATION OF TABtES. 

Thus, let it be required to find the length of an arc of 
74" 26' 43" to radius 12. This is eridentty equal to tli< 
sum of the lengths of the ares of 7p°. i", 20', &, 40", 
and 3"; hence length of 7O''=:l-2217305 
4'= -0698132 
20'= -0058178 
6'= -0017453 
40"= -0001939 
3"= -OOOOUS 



Xength of arc 74'" 26' 43"=1-2993I52 to radius 1. 

12 

Length of arc74°26'43"=15'3917824 to radius 12. 

12. Table XII. contains the logarithms of the amount 
of L.l at the end of one year, to ten decimal places, for 
eTeiy fourth per cent., from \ to 6, it is a necessary addi* 
tion to the ordinary logarithmic tables, for obtaining cor- 
rect results in c^uestions on compound interest and annui- 
ties, -when the number of years for whicli the calcutatioa 
is made is great. Thus, for example, to £nd the Brntmnt 
of one penny, improved at compound interest, at the ntt 
of 5 per cent., since the beginning of the Christian era, or 
for a period of 1845 years. We take the logarithm of thi 
amount of L.l at .5 per cent, that is, the logarithm «f 
1-05, and multiply it by 1845, then subtract the logantbrn 
of 240, the number of pence in a pound, to find tlu 
amount in pounds; hence L.105=-02lia92991 x 1846= 
39'094256a'195, from irhich subtract 2-3fi021I, the loga- 
rithm of 240, and the remainder is 36'7I404(>, which ii 
the logarithm of the .tmount in pounds, n'hich therefbs 
consists of 37 places of figures, the first six of which on^ 
we can find correctly from logarithms carried to six deoi- 
mal places, the other mast therefore be filled up vHh 
ciphers, as we have no means of knowing what &tj 
are; L.5 1 76620,000000,000000,000000,000000,000000. il 
therefore the amount required. 

Rehibji . The valoe of a cubic inch of pure gold is about L.4a'434i 
mod if wo consider the earth as a, globe, whose diameter is TSll 
miles, we will find that the above sura would be equal in ralouM 
about 194135001) globes of pure gold, each as large bb our (Mtkl 
wbilo tha simple interest for the same time, and at tlxi aamewlik 
would only amount to /s. Old. 

13. Table XIII. is insttted to facilitate the talcing oH 
of the logarithms of Be^eraV TivHft\«i«, <A b«Qji«x,v«coiP- 

r*nce in calculation. lU uw Sa saSiciKAXi *«%««», ■«*<■ 
out further cxplaaatioD. 



E TRIUUNOMCTBI. 



PLANE TRIGONOMETRY. 

Artiglb 1. Plane trigonometry Lns for ita object the 
ilotion of the following problem : — Of the three sides and 
:ree angles of any plane triangl-e, any three (except tha 
aee angles) being given to determine ihe other three. 

2. This is effected by means of trifjonoraetvical tables, 
liich contain the ratios of the sides of a right angled tri- 
)gle to every minute of the <^uadmnt. 

3. In trigonometry all the angles round a point are 
Tided into 360 equal parts, called degrees, each decree 
to 60 equal parts, called minute*, and each minute into 
I equal parts, called stcondx. Degrees, minutes, and se- 
nds, are respectively designated by these characters — 
', "; thus the eipression, 36° 1 4' 32", represents an arc 
angle of thirty-six degrees, fourteen minutes, and thirty- 
o seconds. 

4. In the trigonometrical tables, each side, in succession, 
oaed as the measuring unite, (called radius), and tha 
otients arising from dividing the other sides by it. are 

t in tlie tables under the following names, in referenca _ 
one of the angles ; viz. sine, cosine, tangent, ci 
»at, and cosecant, which are contracted thus : e 

5. Let ABC be a right angled tri- 
gle, right angled at C, and let the 
les opposite to the angles A, B, 0, 

designated by the letters a, b, and c, 
rpectively; then the ratios put into - 
i tables for every minute of the angle 
, are the following: — 




90 Tr-A!»B Ti 



11. Cosec.A= 



. 'i ^T"*' 



m 



siile uppoeile' 

13. Badias contracted R= - = 

13. (e)+(7)l!i.»'^ = = x;- = ^ = tan.(8), ... £ 
= tan. I h 

J4. (7)+(6)Biw."^^ = 'x;; = ^=col. (9); .-. j; 
= <:ot. 1 A ). 1 

J5. 1 H-{S) gira ^ =1 X ; = ; = oot. (8); ■■• j^ = 

V l+(9)Bir.. J^=1X J = J=1«,.(S); .-. i = , 

17. i+(iO)g;™__i^=ix|=j=<»..(7)i.-.-i= , 

_I8.^1+(11) Bive. ;ji.^ =1 X J= = = ■m.(6); .-.^ ; 
39. l-^(6) gives r — - =1 X -= -^eosec. (11); .-. — I 

= COaec. "'" on aa. I 

^SO. l+(7)Bi...^-^=lx; = f=.,c.(10); 

Erop.39.) ^'' -^ ■^ 

Hence cos- A=; J 1 — »in.'' A, and 

22. The relative mafpiitmies of 
the trigonometrical ratios may lie 
rpprespntpd as in the annexed dia- 

I pram, ivhere rudius = AO, OU. 
I OC, or OE. and in reference to the 

r angle AOB, or the arc ^^. which is 

its measure ; ]iD= the sine, OB or 

0D= the cosine, AF= the tangent, 

CI= the cotangent, 0F= the secant, and 01 = 

23. AD or R— cosine, is called the verted tine. 

24. CG or R— sine, is called the covened sine. 
2i>. ED or R+ cosine, is called the iuversed «n«. 

26. The difference between an angle and 90* is calW 
its coraplemenC; thus in the ilii^ram above, COA being 
a right angle, the iOOB 19 the complement «f A* 
£AOfi. The names, cosine, cotangent, and cosecant, we 
contractions I'or sine of the complement, tangent of the 
•wrapleinent, and eecanl o? l\\e coTo^VmeTv*.. 





SI 

27- The difference between an angle and 180° » 
its supplement. In the above diagrani, LHOJi ii 
plement of the IBOA. 

28. If XX' and YY' be two 
lines at right angles to each other 
is the point of their intersec- 
tion O; then itll lines measured 
along XX', or on lines parallel to ^ 
ihem, Ijing towards the right of 
YY', are called +. whilst those 
lying towards the left are called 
— ; thus OC and OF are +, whilst 
OE and OD are — ; also those measured n 
YY', above XS', as BC. B'D, are called - 
below XX'. as B"E, B"T, are called — . 

29. If a line, as OB, revolve Ihroueh all the four quad- 
rants, XOY, YOX', X'OY', and Y'OX, and in every 
position have a -J- drawn from its extremity B, upon 
the line XX', the sine of the angle through which it has 
passed, beginning in the position OX, is that perpendicular 
with its proper sign, divided hy OB, and the cosine is the 
distance from O to the foot of the perpendicular, with its 
proper sign, divided by OB, Ilence the sine will have the 
same sign as the peTpeudicular, and the cosine will have 
the some sign as the line intercepted between O and the 
foot of the perpendicular. The sine will therefore he + 
in the first and second quadrants, since BC and B'D are -|- , 
and — in the third and fourth, since B"E and B"'F are 
— ; whilst the cosine will he + in the first and fourth 
quadrants, since 00 and OF are -f, and — in the second 
and third, since OD and OE are — . The signs of the 
other trigonometrical ratios can be determined from these, 
since each of them can be expressed in terms of the sine 
and cosine, (Art. 13-20). 

30. The Bine of an angle w = the sine of Us supplement, 
and Ifie cosine uf an angle m = — tlie cosine o/ its supplt- 

For if the ZBOX be = the iBOX', then the Z.B'OX 
being the supplement of B'OX', will also bp the supple- 
ment of BOX ; but if the Ls BOX and B'OX', be equal, 
the triangles BOG and BOD will be similar; .-. ^ will 
be = 1^, or sin. BOX = sin. BOX = sin.. box:. 

-Alao ^ will be = — — _, or the cosine at a.ii aa'^t "»a 
gua/to — the cosiae of its, au-iplement. 



S2 PLANE TRIGONOMETBT. 

31. In the tables of logRritlimic sines, tan^nts, and 
KcantB, &c., it is the Ingaritlims of these ratios multi- 
plied by 10.000.000,000 ihat is inserted, and R is the pre- 
ceding multiplier, it is plain that the ratios are all increas- 
ed in the same proporlion ; and if we divide each of them 
by R. vie will have the same values as before ; hence ne 
obtain ain A cos A 6 



a 



11 



These being all pairs 
;en in the form of proporti 






, thej may be writ- 



,B follows :- 
a(=hyp.:perp.), 
B : COS. A=e : 6(= hyp. : base), 
^- R; tan. A=£:a(^ base : perp.), 

^b B : cot. A:=a : £(= perp. ; base), 

^" B : sec. A^l/ : i-(^ base : hyp-), 

'■ E: cosec. A^a; t'(^ perp. : hyp,). 

Note. From the above six proportions olher six may be obluii»ir 
by making tha third term of eacli tho first, the fuurth the second, 
tho first tho third, and the second the foartb. 

32. If the side used as the denominator in the Taloeof 
each of these ratios be called Badins, and the name of the 



ratio arising from dividin] 
■written on that side, we 
tions included in the two 
cieut for solving all the ca 

33. Cask I. Given ii 

and one of the acute angli 
third angle. 

Bulb. Make the given side radius, 
responding names on the other sides, thi 
ing propott 



I 



The 



n the 



c given side 



riven side, (Badius), 
the required side, 



each of ihe other sides by it k 
1 have all the above propor- 
owtng rules, which are suffi- 
of right angled triangles that 

right angled triangle a siJ« 
o find the other sides and the 



quired side. 
34. Case 11. Given in a right angled triangle, two 

to find the ancles and the third side, 
\uLti. Make either of ttie ^\\en eidea radius, and -niit' 

-— ■ J 



K.AMX SUOOVOmTKT. S3 

the con-espondiag names on the other aides ; then state the 
folloniDg prnportion : 

The side made ritdius. 

Is to the other given side. 

Is to the name on that side. 
This result being found in the trignnometricat (ahlea, 
will determine the vnlue of the angle to which the names 
were referred ; and since all the angles of every triangle 
are equal to two right angles, and one of the angles is a 
right angle, the other will be the complement of that just 
found. The angles being thus found, the third side can he 
calculated hj (33). 

EXAMPLE. 

The base AG of a right angled triangle is 240, and the 
perpeDdicuIar BO is 264. Required the angles and the 

Construction. From a scale of equal 
parts, lay off AC=240, and -J- to it draw 
BC=264, join AB, and ACB iviU he the 
triangle required. 

Rule (34) gives either ^ 

AC:CB:3lt:tan. A; or, 

BC:CA = R:cot. A, ortan. B. 

35. In calculating a proportion by logarithms, it is 

most convenient to take the avit/imeticaC complement of 

the logarithm of the first term; that is, subtract it from 10 

or 20, according as it is the logarithm of a number or of a 

trigoDonietricBl ratio; to this add the logarithms of like 

aecond and third terms, and reject 10 or 20 from the index 

for the logarithm of the answer. The following examples 

will aU be wrought in this way: 

Ar. CO. Log. AC 240= 7'6197a9 
Log. CB 264= 242Ifi04 

I Log. R =20000000 

Log. tan. A=47° 43' 35"=I004l3y3 
Or ar. co. Log. CB 264= 7 57fi396 
I^g.AC240= 2 380211 
Log. R = 10000000 

>>g.tan.B=42' I6'25"1 _ (..o^o^m 
angles being thus obtained, the \i5^oVeia\i&e 




I 



thu 

I 



PLANE TRIOONOHKTBT. 

e found, (33), by making either AB or CB radiuB; 

L : sec. A=AC : AB. or E : cobm. A=:BC : AB. 

Ar. CO. Log. R =10000000 

I^g.scc.A =l0\72i97 

I^y. AB 240 = 2;^380211 

Log. AB 356-7)16= 2^532408 

Ar. CO, Log. R =10-000000 

Log. coaec. A =]ll 130803 

Log. CB2fi4 = 2'4^i604 

Log. AB 356-7^6= 25.'J2407 

Note. The hjpotenuae might alao be found without findi^ 

angles, for (Geo., prop. :13). AB*=A(?+BC; .-. AB =jA^+BC', 
Similarly AC=^AB°— BS, imd ItC=.^ A.B'~/iC. 



^^ 49= 



1 . In the right angled triungle ABC, right angled at C 
given AC 300, and BC 221, to find the acute Hnglfannd 
the hypotenuse. Ans. Z A 36° 22' 40", i.B 53° 37'20". 
and the hypotenuse, 372-6J3- 

2. Given the base 560 feet, and the angle at the biw 
required the perpendicular and hypotpnuse. 

Ans. Perp. 648-7^6' feet. Hyp. 857027 feeL 
ren the hypotenuse 641, and the angle at the base 
to find the Lase and perpendicular, and the le- 
maining acute angle. Ads, Base 542-21. Feip. 341-889. 
Other acute I. 57" 46'. 

4. The perpendicular of a right angled triangle is aOOi 
and the angle at the hase 49"; required the reinwiiin|> 
parts. Ans. Hyp. 662 306. Base, 434643. Third. LH'. 

5. The hypotenuse is 100, and the perpendicular fiO; 
requited the base and the acute angles. Ans. Base. 60. 
Angle at the base, 36=52' 11-4". Vertical angle. 53» 7* 
486". 

6. Given the hypotenuse 580, and the base 361, to. 
find the other pints, Ans. Angle at the base. al'Sff 
27". Vertical Hngle, 38° 29' 3:¥'. Perp. 453939. 

36. Thkohkm. In any plane tri- 
nngle. the sides are to one another . 
the sines of the opposite angles. 

Let ABC heanv phine triangle, dra 
BD -t- to AC. Call the sides oppo- 
w(o the la A, B, C, tea^i:i;U.v:\i, a, t, ■^' 



PI.ANB TRIGONOIIETRT. 25 

and c, (a notadon which will be frequently adopted), and 
the perpendicular p. Then froio the right angled A< ABD 
and CBD^ we have 

(6.) Sin. A= -, Qiud sin. C= ^ ; .•. sin. A -fc- sin. C= 

c • a' 

Or . ' =^ X - = -,or8in.A:;8in.C=:a:c,(AIg.l^). 

In the same manner it can be proved^ that 

sin. A : sin.'Bzza : 6, 
and that sin. B : sin. 0=6 : c; hence 

g^„ sin. A , sin. A . . • r>i<f^ • x« ^ 

37. a-z^c - — 77 =6 - — - ; sin. A=: sm. C - = sin. B r . 

am. G (prn. B c o 

6=c -: — - =a - — r ; sm. B=: sm. C -=: sm. A -. 
8121.C sin. A c a 

sin. C ,8in.C .^ . .<? .t><? 
c:=za - — 7 =6 -r— :=; ; sm. C= sm. A •«- = sm. B 7 . 
sin. A sin. B a u 

38. Rbmark. The above theorem enables us to find 
the sides and angles of any triangle^ when there are given 
either two sides and an angle opposite to one of them, or 
two angles and a side opposite to one of them. If, how- 
ever^ the data be two sides, and the angle opposite to. the 
less, there are two. solutions^ and hence this case is called 
umbigtuyus. 

Let AB be the greater side, 
and make the angle BAD = the 
given angle, which must neces- 
sarily be acute, since it is oppo- 
site to the less side ; then from 
the centre B, with a distance == j^^ 
the less side, describe an arc^ ^ ^ 

and it will either cut the line AD in two points, or meet 
it in one ; in the last case the triangle would be right 
angled, and there would be only one solution. In the tirst 
case» let the arc cut the line AD in the points C and D, 
and join BC, BD; then each of the triangles ABC and 
ABD have the given data ; therefore the third side may 
either be AC or AD, and the angle opposite to the greater 
side may either be ACB or ADB. But since BC=BD, 
the ZBCD= the ZBDC, and the ZACB is the supplement 
of the Z.BCD; .•• the Z.ACB is the supplement of the 
ZADB. But (30), the sine of an angle is = the sine of 
its supplement ; .*. both the angle corresponding to the 
tine in the tables and its supplement mutit be laVeu^oxX^ii^ 




I 



26 FLANR TRIGonOMETBX. 

angle opposite the greater side; then there will also )>« 
found two values of ihe third angle, viz. ABC and ABD. 
one oftvhich gives the tliird side AC, and the other AD. 
Hence the aides and angles of both triangles are deter- 
mined. 

39. The following rules are evidently deducihle from the 
theorem (3fj), the hrst being that by which we find an 
ogle when we know two sides and an angle opposite lo 
ae of them ; and the second that by which we find a side, 
'ticn there are given two angles and a side. 
Rule I. The side opposite (he given angle. 

Is to the side opposite the required angle. 
As the sine of the given angle 
Is lo the sine of the reijuired angle. 
BuLE II. The sine of the angle opposite the given aide, 
Is to (he sine of the angle opposite the required 

side, I 

As the given aide i 

Is to the required side. 1 

Example. In the triangle ABC, given AB=450, 1 

BC=4I1, and the /,C=60% to find the remaining anglei 

and ihe third side. 

Eule 1 . Gives 450:411=: sin. 60" ; sin. A j 

Ar. CO. Log. AB 450=7-3*6787 

Log. BC 41 1=2(513842 

Log. sin. 60°=z!l'937531 

Log. sin. A=52= 16' 34"=9^8y8160 

The third angle Bz;67o 43' 26", 

(Geo. prop. lit). 

Again, by Rule 2, sin. 60° : sin. 67" 43' 26' 




.-AC. 

Ar. CO. Log. G 



I. 60": 



i;(AB=450) 

: Log. cosec. 60° =10-062469 
Log. sin. 67= 43' 26"= 9-966317 
Log. 450 — 2-633213 

Log. AC=480-838 =a-bB1993 



EXfiRcrsEs. 
1. Givenoneside215, another248, and iheangleonpo- 
^te the latter 74°, to find the remaining angles and tb« 
ide, Ans. I. opposite, 215=50' 26' 40', Other, 
1^49° 33' 20". Third side, 196343-. 

iven one side 215, another 169, and the angle oppo- 
e to the former 72°; to find the remaining angles and 



the Uiirf side. Ans. Z. opposite. lefc^iS* 22' 51". Other, 
i=5&- 37' 59". Third Bide. 159(»22. 

3. GiTcn one angle 64° 13', another 45" 27', and the 
side opposite the latter 1046 links; required the third 
angle, and the other two sides. Ans. Third ^70° 20'. 
Other sides, 1321-66 and ]3J)2'17 links. 

4. Gi»eii one angle 49° 15', another 70° 18', and the 
aide lying between them 5230 feet ; required the sides o[>- 
pOGite to the giren angles, and the third angle. Ans. bides, 
566014 and 45545. Third, LGty 2?'. 

5. Oiven one side 800 feet, another 605, and the angle 
opposite ta the latter 37° 1^'; to find the other paria. 
Ana, / opposite the greater side, 53° 8' 17". or 12b° 51' 
43". OdM-r Z.89° 37' 43", or 15° 54' 17". Third side, 
999-88, or 274-01 feet. 

6. GiTen one angle 90° 33' 26', another 39° 39^ 20". 
Skod the side opposite to the former :H002i to find the 
third angle and the other sides. Ans. Third Z49° 47' 14 ". 
Other sides, 22926 and J9l-58^. 

40, When two sides and the contained angle are given, 
use the folloiving JitUe : 

Xhe anm of the tno given sides, 

l9 to their difference. 

As the tangent of half the sum of the other two angles. 

Is to the tangent of half their difference. 

Dbvonst RATION. Let ABC he 
a triangle, of which the side BC(a) 
i8:^AC(6): produce HC to D, so 
that CD may =AC, and AC to E, 
M that CE may be =BC ; join BE 
and AD, and produce AU to F. 
BD ia=(a+6),and AE is=(a— 6). 
Because the two /.s CBE and CEB 
aretogether=CAB and CBA, eachs^ 
of them beiag^lhe iACD, (Geo. 
prop. 19); .'. since CBE and CEB are equal to one ano- 
ther.eachofthemishalf ihesumof the^aAand B. Again, 
theiCABis^=-theiCEB,oritsequalCBE,bytheiABE, 
while the ^CBA ia .^ the LCBE, by the same iABE; 
.'. the /.ABF = half the difference of the Li A and B. 
AU9 the A« BpF and EAFare equiangular, for CD=CA, 
.'.the /.CDA = theiCAD = the iEAF, (Geo. prop. 3); 
also the iDBF = the LAEF, consequently the /DFB = 
the lAFE, .: each of ihem is a right angle; and since 
the ^DBF and AEF are eqiiiaBgiOu, BD : DP = ¥.k 




k 



28 PLANB TRICON OMF.TRT. 

: AF, (Geo. prop. 61), and alternately, (Ale. 105). 
BD:EA=DF:AF 
_DF AP 

P*' ~bf'bf 

I' =tan.DFB:tan. ABF; 
that ii, {a+b) : (,a—b)= tan. i(A+B) : tan, K*— B). 

NoTK 1 . Since the angle C ia given, 4( A + B ) can be found, 
for it is ^{180°— C)=(90'— iC). The first three terms are 
therefore known, consequenily the fourth can be found ,' 
then l(A+B) + J(A~B)=A, the greater angle, which is 
always opposite to the greater aide, (Geo. prop. 11); and 
i(A + B)— i(A— B)=B, the less angle. The angle* 
being thas Found, the third side can he calculated fay (39.) 
Note 2. The third side can be calculated without find- 
ing the angles A and B; for !n the /v,ABE, we hare, 
fay (39), sin. ^(A— B) : sin. ^A + B) ^(a— 6): c = AB. 
And in the ABDA, since sin. BAD = cos. ABF = cos. 
^(A— B), and sin. BDA= cos. DBF= cob. i(A+ B). 
Cos. i(A— B) : COS. ^(A + U)^(a + 6) : c=AB. 
NoTB 3. The third side may also fae found without 
calculating the angles hy the following formula: c= 
Vt'+6" — ^"b COS. 0, which is easily deducihte from 
(Geo. prop. 41) and (7); but the form is not suited to 



Example. Let the 
side AC of the triangle 
ACB be 210-3, the side 
BC i6Q% and the ZC 
1I0°1'20"; requiredthe 
ii A and B and the side jg-g 2 

AB. 

4(80— /.CllO" 1' 20")=34° 59' 20"= half the Bom of the 
2* A and B. 

Side AC —210-3 
Side CB = 160-2 
Sum of Bides=370-5 Ar. co. Log. =7-431212 
Dif of 8ide8= 50-1 Log. =1-699838 

^ sum of the is 34° 59' 20" Log. tan.=9-84504S 

5°_24'_24'^ Log. tan.=8^976aB 

.-. iB 40"° 23' 44" I 

aniiA 29° 34' 56" I 




TLAHS TRIOONOBfBTRT. 29 

And by (39), sin. A : sin. C=CB : A B. 
Log. cosec. A=29» 34' 56"= 10-306561 
Log. sin. C=110*» V 20" = 9-972925 
Log. BC= 160-2 = 2-204663 

Log. AB=304-9 = 2-484149 



XZBRCI8R8. 

L Ohen in the AABC, the i\de AC 241, the side BC 
73, and the inclnded angle C 103^, to find the remaining 
ngles, and the third side AB. 

Ans. LA 31 o 3' 23", Z B 45° 5& 37", and AB 326-753. 

2. Given in the triangle ABC, AC=79, BC=67, and 
he indnded LC 85* 16', to find the remaining it, and the 
hiid side AB. 

Ans. IB 52* 28' 6", and LA 42* 15' 54", and AB 99-2795. 

3. In the AABC, given the side AC 3450 links, the 
ide BC 4025 links, and the LC 91o 30'; to find the re- 
nainingi^^ and the third side A B. Ans. LA 48° 3*2' 77", 
IB 39^ 67 52-3". and AB 5369-37. 

4. In 4e AABC, given the side AB 800, the side AC 
749, and the Z.A 80^ 10'; to find the remaining Is and the 
third side BC. 

Ans. LC 52* 9' 25-4", ZB 47° 40' 34-6", and BC 998-161. 

41. Theorem. Twice the cosine square of half an 
mgle is equal to one plus the cosine of the whole angle ; 
ind twice the sine square of half an angle is equal to one 
ninns the cosine of the whole angle ; or 2 cos. ^ ^A=rl-|- 
sos. A, and 2sin.^^A=l — cos. A. 

Let BAC be the ZA, draw BC ^ ^B 

JL to AC, and from centre A, 

irith the radius AB, describe the 

lemicircle EBD; produce AC 

[K>th ways to £ and D ; then EC ]£' 

=AB + AC,andCD=AB— AC, 

Hso (Geo. prop. 68) EB^z=EDxEC=2AB (AB+AC), 

and BC2=ECxCD=:(AB+ACXAB— AC). Also the 

^BEC=i Z.BAC=iA, (Geo. prop. 47), and by (7), cos.^ 

* EC , . 1 . BC , 
A= — , and sm. i A= — ; hence 

' ^ 2EC« 2EC* . AB+AC , , . 

2 C08.HA= ^3i = 2XBiEC = -AB - =^ + ^^'- ^ ' 

... 2BC« SECxCD AB— AC 
and 2 sin. *A= ^^ = gABxEC " Aid =^— ^^^- ^- 





30 PLATTB TSIQOROHSTBT. 

42. Thborrh. Any side of a 
triangle is eqaal to the sum of the 
products of the other two aides, 
into the cosine of the angle in- 
cluded between it and that side. 

For (7) gives AD = c cob. A, 
and CD=(icos.C; 

.-. 6=ceos. A+dcoa. C, and similarlf 
tf=6cos. A+acos. B, 
o=iico8.C+ccos. B. 

43. Multiplying the first of these hy 6, the second by t, 
and tEe third by a, we have 

J'=6ccoa, A-i-ai>coa. C, 
c'^bc COB. A -i-ize COB. B, 
«''=a6co8.C+<wcos. B. 

44. Adding now the Erst and second of the aboTe, and 
aubtracting the third, we obtain the following: 

J=+c=— tt''=2iccoa. A; 
whence changing the sidesj and dividing by 2be, we han 

(1.) COS. A= — 37- — , and similarly 

, and (J.) COS. C,= — — — , 

45. AddiDg I to both sides of the equations in (4^ 
we hate from (1) ,. . . . 



(Aig.,A,..3i.) ^i^a^m^. 

But (41) 1+ COS. A=2 oDB.'|A, and theiefois 

...co.'iA=^^^^t-^-^- 
Ut now S=i{a+b-i-e), then 5— a=J(6+e--), 
S—!>=i(,a+c—b), and jS—c=^(a+b—t}. 

Substituting these Tallies in die ^ove, and extracting 
'« fool, we ohtaia 



TLASm TBIOOIIOMSTRT. 31 

«oi.|As ij j^ '» and simi] 



OOB. 4tt= /^ ■ ■ 



iB=J'- 



16. In the same maimer, from (44) we obtain 



1 — COS. A=l — 



20c 



^ 2be 2bc 

_ <g4-e~A)(<-f 6— c) _ 
•" 26c ' 

.•,9m.'4A='^^^ ~ — — ^, . 



or 



nn.^A= ^ ' ^ — ^^, ftnd timilarlj. 



and 8u,.iC= y <^Xa-*> . 



?• Again« dividing the expressions for the sine in (46), 
Jus corresponding expression for the cosine in (45), Mre 
lin the following expressions for the tangents of half 
angles: viz. 



18. Whenee Tan-iAs ^J^(S^a)(S—b)(S—e). 



m 



PLANE TRlGONOMETRr. 

49. From the expressions in (48), is easilj dertTed the 

illowing very convenient rule for calculating the tkrte 

tgleg of a triangle when the t/iree sidea are i/iven. 

EuLn. Add the three Hides togetber, and take half 
'tbeir sum ; from half the Bum suhtract each side separate- 
ly, then subtract the logarithm of tlie half sum from 20, 
and under the remainder wrile the logarithms of the three 
remainders ; luilf the sum of these will be a eoTistant, from 
which, if the logarithms of the three remainders be succes- 
Birely subtracted, the new rcmaiodcrs will be the logariih- 
mio tangents of half the angles of the triangle. 

NoTR 1 . The above rule has this advantage over all others, 
that the three angles are obtained with little more labour 
than one, and when the three angles are thus found inde- 
pendently, if their sum be IBO", the calculation is corrett; 
if not, it must be examined till it prove. 

NoTB 2. In order to know which of the angles we have I 
obtained, it is necessary to observe, that the rojutanl I 
— (S — a), gives tan. ^A, the coTtstant — (S — fi), givM I 
tan. ^B, and the constant — (5 — c), gives tan. JC. I 

Note 3. The angles might also be calculated from <\\t | 
pipressions in (Arts. 44, 45. 46. or 47); but those in (44) 
are not suited to logarithmic calculation, those in (45 aoJ 
46) do not give the angles with the same accuracy ia pw- 
ticular cases, since for a small angle the cosine TWWt 
slowly, and for an angle nearly JtO" the sine varies slowlji 
and those in (47). as well as all the others, require win- 
dependent calculation for each of the tjjree angles. 

Example. Given the three sides 
of a triangle ABC; viz. AB 340, 
BC 380, and AC 360, to find the 
three angles. 

AB, c= ;140 
BC, a= 380 ^ 

AC, 6= :160 



540 ar.co. Log. +10=17-2G76O6 
160 Log. = 2-204120 

ISOlxig. = 2-255273 k 

200 L"g, = 2-301030 "^ 

2 |24-02802[) 
Constant. =12-0140145 




■ 



^m 



B SKIOOirOlfBTBT. 



Tan. iA=32° 50' 3l-7''=9-S0g8945{con'. — Log.(5— a)} 
Tan. 1B= 29" 50' 46" =9-75874 J 5 [con'. — Log.OS'— b)] 
Tan. ^C=27" 18'42-3"=9-712yS45[con'. — Log.C,S'— 01 
Therefore L\ 05° 41' 3-i", IB 59" 41' 32", and LC 54" 
37' 24-6", the Bum of which is exactly 180°. 



n terms of 



1. "What are the three angles of the triangle ABC, AB 
being 100 yards, BC 150 yards, and AC J20 yards! 
Am. ZA 85" 27' 34", /.B 52° 53' 28", and /.C 41" 38' 58". 

2. The three sides of a triangle are AB 470, BC 398, 
and AC 420 ; what are the angles ? 

Ans. LA 52° 45' 49". IB 5T 9' 22", and iC 70" 4' 49". 

3. The three sides of a triangle are AB 2601, BC 140-4, 
nndAC190'O; required the angles? Ana. ZA3I''47'3r', 
;B 45- 37' 46', and LC 102° 24' 43". 

4. The three sides of a triangle are AB 562, BC 320, 
andACeOO; required the angles? 

Ans. lA 18" 21' 24", LC 33- 34' 47", LB 128" 3' 49". 

50. It is required to find expressions for the sine and 
cosine of the sum and difference of two angles 
the Bine and cosine of the angles themselTes. 

Let CAD=a, BAC=fi, then 
fiAD=a-|-&, and it is required to 
£nd esprcBsions for the sine and 
cosine of (a+ft), in terms of tlie 
sine and cosine of a and b. 

From B draw BC-i-AC, and 
BE-i-AD, and through C draw 
CFu-BE, and CD -i- AD, then. 
FEDC will be a parallelogram, and -^ " 

PE will be=CD, and FC will he ^ED ; also since the 
right angled triangles AOE, BOC, have the right angles 
AEO. BCO, equal to one another, and the angles at 0= 
bdng rertically opposite, the remaining angles OAE and 
OBC are equal, or the /.CBF=a. Now, (6.), 
„. , , ,, BE CD+BF CD , HP- 
Sm. («+6)= — = -r^ = AB + A 








PLASB TBiooaoKniHr. 




k 



, Cos. (a+6) = cos.acoB. 6— : 
_rLet now CAE=(t, and CAB=6, 
then BAE wUl =(a— i) ; draw BC 
J.AC, CD and BE each J-AE, and 
BF J- to CD; then smce iDAC+ 
/.ACD, are together = a I'i, and ACB 
19 a i^L, take away the common L ACD, 
and there remains the Z.BCF=:the 

IZ,CAD; .-.the angle BCF^a, also 
IFD^BE, andFB=DE; h^ce we have 
^ „. , ,. BE CD— CF CD 
% 




fi3. Sin. {a—b)= 






AB' 



AD+FB AD 



; therefore 



.ught. 



54. Cos. (u — 'j)=; COS. a cos. 6' 

Collecting now these four resu 

ference, we have the expressions 

Sin. {a+b)= ain. a cos. i+ cos. a sin. b. 
Sin. (a — b)= sin. a cos. b — cos, a sin. b. 
Cos. (o+6)=^ COS. u COS. h — sin. a sin. b. 
Cos, (a— 6}= COB. a cos. ft+ Bin.ra sin. 6. 
■51) + (S3) 155 Sin.(a + t)+sin.(a— 6)=2sin.tt««t. 
'51)— (63) 56 Sin. (a+6)— sin. («— 6)=2cos.aain.t, 
(54}+f52) 57 Cos.(a— 6)+cos.(a+fi)=2coa.«cM.6. 
(54)— (52) |58 Cos.(a— 6)— co3.(a+6)=2sln.asiii.i.. 
Let now a+t^S, and a — 6=rf, then a=^{S+d), vA. 
h={{S^d). Substituting these Taluea in the last bra, 
they become 

(55) I 53 Stn.^+sin. (?=2sin.^(5+d)o08.J(5-4 
(66) I 60 Sin. .S-- sin. d=2coiX{S-^d)am.k[S—iV 
(57) 61 Cos. rf+ COS. 5— 2 COB. i(.S'+rf) COB. !(«—<;). 
(50) I G2 Cos, d— COS. ,^;2Bin.|(^V+d)sin. !(;;-</> 
The last four expressed in words are as follow:— 

59. The sum of the sines of two angles, is equal to 
twice the sine of half their sum, into the cosine of Itaif 
their diifcience. 

60. The difference of the sines of two angles, is equal 
to twice the cosine of Aa^" their sum, into the sine of Aa!^ 

tbeii difierence. 



yi.*ira TWQOWOMETKT. 



35 



fil. The Bum of the cosines of two angles, is equal to 
twice the cosine of Aaf/" their sum, into the cosine of half 
their difFereDce. 

62. The difference of the cosines of two angles, is equal 
to twice the sine of kalf their sum, into the sine of half 
their difference. 

Note. Since S and d are any two itrtgles, we may writo inBtcad 
of them a and b, by wridog at the same time 4(a-l'i) for i{S-ir-d), 
*nd 4(0— i) for J(S^— ^)i T instead of S write A, and instead of d 
wile B, then l(S+(0=i(A+B), and i{S— rf)=i(A— B); whence 
(59)^(bO)giTesb^^^^_^^_^^^_^^^_^jj^^^^^^_g^. 

^ - — 1/ i iigy By dividing both numerator and denomi- 
nator of the second side hy 2 cos. ^(A+B) cos.4(A— B). 

11.A+ sin. B . w, , i>v 
^^^^^^ = tan.KA+B). 

^B-I!^A-=-'■^^+^> 

B. B+ COS . A _ cot. i(A+B) _ 
*.B— COS.A ~ taii-lCA— B) 
oot.i(A-B) 
(an. i(A+B)" 
In Arts. (51, 52, 53, 54), let the angles a and 6 be equal, 
then 0+6=2(1, and n— 6=0, 



(59)H-{81) giTea 


64 


(69)^(62) giTe. 


65 


(60)+(61) gira 


66 


(60)-h(62) giTes 


67 


(61)+(62) Eire. 


68 



(51) gives 

(53) 

(52) 

(S4) 
(51)-i.(52) 



i.2a=2 6iii.a 
^0. 



m. (.+») _ (.to. . c. 1+ o 



B+Bin-'a^l, {Art. 2]>. 

, . ,, tau.a+ tan-ft 
n.(„ + 4)=— ^--^-. 



\m^<.^) 1 7* 1 "iSr'di 



:a, and (73) giyes | 75 | tan, 2a^ __ 



'ian.&' I 




, _ _ JO 6S) make 6 =0, and (Arts. "0 anJ 
e haTfi the following : 
Bin. A , '- ' 

1—coB.A^'' tan-ii 

\=^:>^=^.=^'^'^^- ^y'"^ 

h,g, (82). 
84. To find the numerical values of the sine, cosine, mi 

tangent of 30°. 

Leta=sin.30"; then (21) cos. 30'=Vl— x*-", and (g ) 
si n. (30° +30°). or sin. 60"= cob. 30" (26)=2jr^ 1-V= 
Vl — X-; dividing both sides of this pquation by •Jl—^ 
gives 2x=-l, or3r=J, conseqaentlj Vi — i— \'|=^^'3= 
; and since — = tan., fan. 30°= — j^ = -= = 
[ 4^3"; .-. sin. 30° = ^. cos. 30° = i ^3, and tan. 30°= 
^. or ^^/3--, congequentljsin. 60°=^^3, cos. CCrri, 

' tan. 60°=V3, &c. 

85. To find the sine, cosine, tangent, &c., of 45 ■ 
Let.r:=ain. 45°. then ^1— i"^"(2l)i= cos. 45°= an, (30 
—45X26) = sin. 45". 

Hence sin. 45°=- 



;e squaring a-^l — x' 
COS. 45"= ^1—^=- 



I. From the valiiea deiQcei vo^^^*'a.u^?.3^^lA^(^ 



i 



PLANE TRIGONOMETRY. 37 

of (51 — 54) numerical expressions for the sine and cosine 
of 75% and for the sine and cosine of 15**. Ans. Sin. 7^** 

= ^^ cos. 750= ^^, sin. 15-= ^^, and cos. 15' 

2^2 2^2 2^2 

2. Prove the truth of the following expressions. 

„. ^ COB. a tan. a cos. a see. a 

Sin. a= COS. a tan. 0= -— — =: = : 

cot. a see a cosec. a 

^ . sin. a cot. a sin. a cosec. « 

Cos. a= sin. a cot. a= : = =: 

tan. a cosec. a sec. a 

— sec. a sin. a cosec a cos. a sec. a cos. a 

Ian. a^ = := r =-: 7=-. 

cosec. a cot. a cot. a sin. a cot.* a 

3. Show that if (A+B+C)=180% or he the three 
angles of a triangle; tan. A-{- tan. B-f- tan. C= tan. A. 
tan. B. tan. C. 

4. Prove that sin. A=sin. (60**+ A)— sin. (60"— A), or 
that sin. A= cos. (30''— A)— cos. (30°+ A.) 

5 Provethatsec 2A-l/' '^' ^"^'^°' A x j^/cos.A-sin.A\ 

I r: 2A ^ai^T^lcos. A— sin. aJ + sI co8.A + 8in.A J 

1+tan.^ A cot.*A+ 1 > ' ^ ^ 

"" 1— tan.«A ~ cot.«A— r 

6. Prove the rule in (art. 40) from (arts. 36 and 59), and 
(Alg.69). 

7* Prove that the difference of the sides of a triangle, is 
to the difference of the segments of the hase made hj a 
perpendicular upon it from the vertex, as the cosine of half 
the sum of the angles at the hase, is to the cosine of half 
tilieir difference. Also, that the sum of the sides, is to the 
Terence of the segments of the hase, as the sine of Iialf 
the sum of the angles at the hase, is to the sine of half 
their difference. 



APPLICATION OF PLANE TRIGONOMETRY TO 
THE MENSURATION OF HEIGHTS AND 

DISTANCES. 

86. The instruments commonly used for measuring 
heights and distances are^ a chain, a quadiaat, ql ^c^^t^, 
and a theodolite. 

A chain is used for measuring tliose distaivce^ ot \\tv^^ 
rjttch are to be given sides of triangles. T\ie Im^^fvssSi 



cbain is G6 feet in lengtli, aad is divided into 100 equal 
links, consequently eiich. link is "^-92 inches long. Every 
ten links from each end to the middle of the chain is dis- 
tinguished by a piece of brass having as many points aa it 
is tens of links from the end of the chain, 

A quadrant is need for determining vertical angles. It 
is made of brass or wood in the form of a quadrant of a 
circle, and its circumference is divided into 90°, and these 
agMn subdivided, as far as the diraensiona of the quadrant 
nill admit. Abo a plummet is suspended by a thread &om 
the centre. A square is used for finding the ratio of the 
sides of a right angled triangle. It is made of the same 
materials. Two of its sides are divided each into 100 
equal parts; and a plummet hangs from the opposite angle. 

A theodolite is used for measuring horizontal, as wellaa 
vertical angles. It is a circle of brass divided into 360 
degrees, having an index moveable about its centre, and ia 
furnished with a telescope, moveable on a graduated semi- 
circle, which is used for measuring vertical angles. It ii 
indispensihle where great accuracy is required. 

87- fo find the height of an accessible object Btaoding 
on level ground. 

Ruti:. Measure any convenient distance from the bottom 
of the object, and there lake the angle of elevation of it( 
top; then state, £ is to the tang, of the angle of elevation, 
as the measured distance to the height above the level of 
the eye, to which add the height of the eye, and the tun 
will be the height required. (See art. 33.) 



H^' 



. 1. Required the height of a tower, whose angle of elec- 
tion, at the distance of 100 feet from the bottom, ii 51° 
30', height of the eye 5 feet. Ans. 200-148 feet. 

2. At the distance of 100 feet from the bottom of » 
tower, on a horizontal plane, the angle of elevation of iti 
top was 47° 30', the centre of the quadrant being fixed 5 
feet above the ground ; required the height of the tower. 

Ana. 114-131 feet. 

3. At the distance of 130 feet from the bottom of> 
tower, and on the same horizontal plane with it, the angle 
of elevation of its top was 50- 43', and the height of UW 
eye 5J feet; required the height of the tower. AnB.164'4!3- 

88. To find the distance of an inaccessible object, fitUD* 

given point. I 

Rule. Set up a mark at the given point, and take an; I 

'ler station^ and meaaure iW &u^ante\n\.we«u the pni I 



PliANE TRIGONOMETRT. 39 

point and the other station^ and at each of the ends of this 
line measure the angle subtended by the object and the 
other station; then the sine of the sum of the observed 
angles^ is to the sine of the angle at either station^ as the 
distance of the stations is to the distance of the object from 
the other station. (See art 39.) 

EXERCISES. 

1. Being on the side of a river^ and wanting to know the 
distance of a house on the other side, I measured 266 
yards in a right line^ by the side of the river, and found 
that the two angles, one at each end of this line, subtended 
by the other end and the house, were 38° 40', and 92'' 4&; 
what was the distance between each station and the house? 
Ans. Distance from one station, 354*38 yards. Distance 
from the other station, 221-67 yards. 

2. From a ship a headland was seen, bearing NE.^N.; 
the vessel then stood away NW.^W. 20 miles, and the 
same headland was observed to bear from her E.;|^N.; re- 
quired the distance of the headland from the ship at each 
Station. Ans. Distance from the first station 19 09, and 
from the second 26'96 miles. 

3. Having measured a base of 260 yards in a straight 
line, close by one side of a river, I found that the two 
angles, one at each end of the line, subtended by the other 
end and a tree close to the opposite bank, were 40® and 
80°, what was the perpendicular breadth of the river ? 

Ans. 190046 yards. 

89. To find the height of an inaccessible object. 

At any two convenient 
points F and G (these being 
in the same vertical plane 
with AB), observe the angles 
of elevation AFE and AGE, 
and measure the distance GF g. 
or CD. Then because the d' 
exterior angle AFE is= ^ 

FAG + FGA, the angle FAG is the difference of these 
two angles. Now the triangles AFG and FAE evidently 
give the following proportions, which will determine AE, 
to which adding EB or FC, the height of the eye, the alti- 
tude will be determined. 

From A AGF, sin. G AF= sin. (F— G) : sin. G : : GF : AF 
From A AFE K : sin. F : : AF : AE 

These proportions may be wrought separately, ot V^^Wet 
being eampounded (Alg. ] 16) they give the ioftomiv^-.'^^aji. 




40 n-Amt TdiGOKflMBTnr. 

(I'— G) : Bin. F. sin. G : : GF : AE, which hy Logarilhmi 
gives Log. EA= [Log. GF+ Log. sin. F+ Log. sin. + 
Log. coscc. {F— G)— 30j. If the distance FE were sought, 
we would hitve in tlie same manner Log. FE^ [Log. coseo. 
(F— G)+ Log. COS. I-+ Log. sin. G+ Log. GF— 30], 

If two atstions ia the same Btraigbt line with the object, 
and in the same horizontal plane with it, cannot be found, 
lake one station as near as possible on a level with the base 
of the object, and select another station such that the first 
station can he seen from it, and also the object; and that 
its distance from the first can be conveniently measured; 
then at the first station measure the angle of elevation o( 
the top of the object, and the angle subtended by the ob- 
ject, and the second station; also at (he second station 
measure the angle subtended by the first and the object; 
then measure the distance between the stations, and state 
the following proportions, in which S is the angle at tkt 
first station, S' that at the second, D the distance betirefn 
the stations, D' the distance of the first station from tkt 
object, E the angle of elevation of the top of the object nl 
the first station, and A the altitude of the object. 
Sin. (S'+S):Hin. S'rrD :D' 
II .- tan. E : : D' : A 

Compounding these proportions (Alg. 116) we hare 
K sin. (S'+S) : sin. S' tan. E : : D : A, hence 
Log. A=; [Log. coseo. (S'-|-S)-|- log. sin. S'+ log. tan. E 
+Log. D.— 30], 

NoTB I. If a station cannot be conveniently found ons 
level with the bottom of the object, the angle of elevation nt 
depression of the top iindbotlotn mtiy be taken, and theparV 
calculated for separately; then if the angles of the top and 
bottom of the object he both angles of elevation, or both 
angles of depression, the difference of the elevatinns et 
depressions of the top and bottom will be the altitude «f 
the object; hut if the lop be elevated, and the bottom d«- 
pressed, their sum will be ibe altitude of the object. 

NoTB 2. If the horizontalangles Sand S' be taken withi 
theodolite, if the base ts not on a level plane it must fM 
multiplied into the cosine of the acclivity or declirity be- 
fore being used in the above calculation. 

Note 3. If the horizontal angles S' and S be measured b; 
a sostont, then the distance from the first station .ind tb« 
ohject must be mull'nilwd into the coaiiie of the angle of 
elevation or depieaaion o^ ^-^lat ■joiiA (>l ^% «i^\«ct, tht 



FLANS TRIGONOMETRY^ 41 

image of which was made to coincide with each of the 
stations in measuring the horizontal angles. 

NoTS! 4. In measuring the height of mountains at a dis- 
tance, allowance must he made for the curvature of the earth, 
by adding to the calculated height 8 inches multiplied into 
the square of the distance in miles, or to twice the log. of the 

distance in feet add the constant log. 8*378641, and the 
result will be the log. of the correction in feet yery nearly. 

EXERCISES. 

1. A person 6 feet high, standing on the side of a 
river, obserred that the top of a tower placed on the oppo- 
nte side, subtended an angle of 59**, with a line drawn from 
his eye parallel to the horizon; receding backwards for 50 
feet» he then found that it subtended an angle of 49^. 
Bequired the height of the tower, and the breadth of the 
mer. Ans. Height of the tower 192*27 feet. Breadth of 
theriTerlll92feet. 

2. To determine the altitude of a light-house, I obserred 
the eleyation of its top above the level sand on the sea- 
shore to be 15° 32' 18'', and measuring directly from it 
along the sand 638 yards, I then found its elevation to be 
9* 56' 26"; required the height of the light-house. 

Ans. 302*463 yards. 

3. It is required to find the height of Arthur's Seat, in 
ilie vicinity of Edinburgh, from the following observations, 
taken with a theodolite on Leith sands, about the medium 
height of the tide, the base being 1410*42 feet; at the west 
end of the base, being that nearest to Leith pier, the angle 
subtended by the top of Arthur's Seat, and the eastern 
extremity of the base, was 50° 44', and the elevation of its 
top 3° 59', whilst the angle subtended at the eastern station, 
by the western and the top of the hill, was 123° 29', and 
the elevation of the top of the hill was 4° 17' 30': Find 
the height of the hill above the medium level of the tide, 
allowing 5 feet for the height of the eye^ and the necessary 
correction for curvature, by (Note 4). Ans. The observed 
altitude at the western station gives 821-174, that at the 
eastern gives 82101 9 feet, the difference of the two being 
less than 2 inches. 

4. Find the height of the top of the cross on the spire of 
Assembly 'Hall, Castle Hill, Edinburgh, from the following 
observations, the height of the eye being on a level with the 
sole of the entrance door; at the first station, the elevation 
of the top of the cross was 62° 36', angular bearing of ih^ 
•pire from the second station 35° 36', aivd l\i^ ^xii^xiSax 



42 



rLAXK TRiaonOMETRT. 



bearing of (he spire and the first Etation, taken at tha 
second, was 3U° 22', tlie distance between the stations heiag 
na-e feet. Ans. 235309 feet. 

90. To find the distance of two inaccessible objects. 
Let the two objects be A and B; ^^ „ 

take any stations C and D, such. 

that the objects and the other station 

can be distinctly seen from each, 

and that the distance CD can be 

accurately Measured; then at the ^ ^ 

station C measure the angles ACB 

and BCD, and at the station D, measure the angles BDA, 

and A DC, then measure accurately the distance CD. 

Solution. In each of the triangles A CD, BCD, wff 
have given two angles, and the side lying between thenif 
consequently (Art. 39) AC, AD, CB, and BD can be found, 
then in each of the triangles ACB and ADB we have two 
sides and the contained angle, consequently (Art. 40) tbe 
side AB can be found from either nf these triangles, and if 
it be calculated from each, and the results be found tin 
same, it is a proof of the correctness of the calculations, i 

91. This problem is of Tery estensive application hi 
many departments, both of civil, military, and naval snM 
Teying, and may be solved in a different manner from llwj 
commonly adopted method which is given above, liti 
method here referred to may be called surveying by Stit 
angular Co-ordinates, and consists in finding an exprotiol 
for the perpendicular distances of the objects from thebtf 
line, and the distance of the iniddle of tlie bate &ov 
foot of the perpendiculars, in terms of the baae udfl 
angles at its extremities. 

92. To find an fsrpremon, for the 
perpendicular AD in tem>» of the iase 
BC and tlie angles B and C, at its 
extremities. By (Art. 37) AC=BC 

by (Art. 6) AD=AC. sin. C = BC 
fiin. B Bin. C ,, ^ 4 T^ ri 

l(bTc)- Hence Log. AD= [log. c 

1. B+ log. sin. C+ log. BC— 30]; or Add togetJitr » 

log, cosecant of the sum of tJie angles, the log. alnea of M«l ' 

the angles at ike base, and the log. of the bate ; the utm, ^^ 

ifi^ iij/30 int}ieindex,vnllbetJielog.oft/ie perpendieiJi 

03. To Jind an ea-pressimi Jot "tX), ftw dnxhutw of 





■•(B+C)+fc| 



PliANE TBIGONOMETBY. 43 

foot of iJvR perpendicular from the middle of the base, in terms 
of the base and the angles at its extremities. 
^ ,A «*,v * ^ ^^ sin. B , ,„ „^ Bin. C 

By (Art 37) AC=BC . ' ^„ and AB=BC . ,'\^ . 
^^ ^ siiL(B+C) sin. (B+C) 

Again, (Art 7), DC=AC cos. C=BC ""^—-^ 
And BD= AB cos. B=BC ?^?4^ ; 

sin. (B+C) ' 

• 2FD DC— BD— ^^i^in. B cos. C — cos. B sin. C| 

' ^ "" sin. (B+C) 

But (Art 52), ^8in.B cos.C — cos.B sin. C] =sin (B — C). 
By substituting this value, and dividing both sides by 2, we 

obtain, ED=iBC^!5^^2' ^^^^^^ 

* sm. (B+C) 

Log. ED= {Log. cosec. (B + C)+ log. sin. (B— C)+ log. \ 
BC— 20]. 

Or, add together the log, cosecant of the sum of the angles 
at the base, the log. sine of their differente and the log. of half 
the base, the sum diminished by 20 in the index will be the 
log. of the distance of the foot of the perpendicular from the 
middle of the base, 

94, The foot of the perpendicular falls always on that 
side of the middle of the base whieh is adjacent to the . 
greater angle at the extremity of the base, and its distance 
from the middle of the base should be marked -f- when on 
one side, and — when on the other, (Art. 28). Also, if in 
an extensive survey some of the objects should lie on the 
one side, and others on the other side of the base, the per- 
pendiculars should be marked 4- ^^ ^he one side, and — 
on the other. 

95. The perpendiculars DG and 
CF being thus calculated, and also 
EG and EF, the distance of the two 
objects C and D can be found, for 
CI being drawn parallel to AB, the 
triangle DIG will be right angled, 
and Dl is the difference of the per- 
pendiculars, and IC is the difference 
of the distances of the perpendicu- 
lars from the middle of the base, 
since EF is — ; and if one of the perpendiculars had 
been measured downwards, DI would still have been th^\t 
difference, ahice the perpendicular meaauiedl dio\^ii\;^^^^ 




i^ 



w 



Tonid be mimis. Xow Log. DI+IO — Log. IC=Log, 
tan. DCI. BatLog. secDCl+IiOg. IC— 10= Log. DC; 
hence we have the following luJe for finding the distance 
of two objects : — From (Ae log. of the difference of the per- 
pendicida.rs, inei'eased iij/ 10, of an iitdea^, eublract tlie log. 
of the distance of the perpendimlar, the remainder wiQ, be 
&i log. tan. of an (mgU; to the log. eeeamt of this angle add 
the log. of the distaaiee of the pei^ieitdieulart, and the turn, 
rejecting 10, from the index, will be the log. of the diffaud 
sought. 

96. If the distance between the objects A and B be 
given, to find the distance between the stations C and D, 
(Art. 90) ; measure the angles at C and D as before, and 
assume CD any length, as 100 or 1000, and from this 
assumed length find AB ; then the calculated lengih of 
' " ■ the assumed length of CD, as the true length of 
the true length of CD. 



1. Find (he distance between the two objects A and B, 
(Fig. art. 90). on the supposition that CD is 300 yardi, 
ZACB^56°, /.BCD=37°, /.ADB=ij5°. and L\T>Q=\\'. 

Ans. AB=341 aSyardi. 

2. Being desirous of finding the distance between two 
. ohjeclB A and B, I measured a base, CD, of 384 yariii 

on the same horizontal plane with the objects A and Bi 
At C I found the angle BCD=48° 12', and ACD=89°lff; 
at D the angle ADC was 46° 14', and BDC 8?° 4'. It ii 
required from the data to compute the distance between A 
and B. Ans. AH=358-5 yard«. 

3. Wanting to know the distance between two inaccM- 
eible objects A and B. (Diag. art. 90), I measured a base 
line CD of 360 yards: at C the horiaontal angle AC6 
was oliserred with a sextant, and found to be 53" 30', anJ 
the angle BCD 38° 45' ; at D the liorizontal angle BD.4 
was 67° 20', and the angle ADC 44° 30'. Required ll« 
distance between A and B. Ans, 548149 yardi 

4. Wishing to know the distances and positions of ■ 
namher of the principal objects in and about Edinbnr]^ 
I took a station on the top of Arthur's Seat, and wiUit 
theodolite meaamed the angular hearings of the following 
objects, with a line drawn through the top of Nelsoa'l 
Monument, on the Calton Hill, viz.:— Spire of AMemWj 
Ball (H). Castle Tower (T), Dome of St George's Chittdi 
(O), Spire of St AnAieVa CVuxiAi V^\ lAcWUle's Mono- 



PLANE TRIGONOMETRY. 



45 



tnent (M), Spire of North Leith Church (L), Inchkeith 
Light-house (K), and Berwick Law (B). I then went to 
the top of Nelson s Monument, and took the angular bear- 
ings of the same objects with a line drawn through the 
former station on the top of Arthur's Seat. The angles 
were as given in the following table, where the single let- 
ters in the first column signify the objects after whose 
names they are placed above : — 



Vertex of 

the triangle. 


Angle at 
Artiiur's 


Angle at 
Nelson's 


Sum of 


Difference 


Perpendicu- 


Distance of 
the middle 


Seat. 


Monument. 


angles. 


of angles. 


lar in feet. 


of the base. 


H 


. 26" 48' 


100'' 8' 


126" 56' 


73" 20' 


315M 


3400-7 


T 


SO" 22' 


105" 35' 


135" 57' 


73" 13' 


3974-5 


3907-2 


G 


25° 40' 


127" 54'i 


153" 34'4 


102" 14'4 


4257-9 


6231-1 


A 


13'' 57' 


1.85" 44'i 


149" 41'i 


121" 47'4 


1892 


4779-1 


M 


11«'23' 


137° 25' 


148" 48' 


126" 2' 


1463-1 


4429-5 


L 


28° 26' 


131" 38' 


160" 4' 


103" 12' 


—5923-7 


8103- 


K 


57*' 49' 


113" 4'4 


170" 53'4 


55" 15'4 


—27913 


14729 


B 


[—n5°5V 


61" 14' 


177" 5' 


54" 37' 


—87983 


—45465 



The base of the above triangles, namely, the distance 
between the top of Arthur's Seat and Nelson's Monument, 
was determined from the third triangle, of which a side, 
namely, the distance from Nelson's Monument and the top 
rf the dome of St George's Church, was previously deter- 
Biined from a similar survey, in which the base was 
measured, and the distance so determined agreed to the 
tenth of a foot with that deducible from data given in 
Wallace's Theorems and Formulae, and hence concluded 
to be very correct. The base thus determined was 5675 
feet. 

• The following is the method of calculating the perpen- 
dicular, and the distance of its foot from the middle of the 
hise in the first triangle; and the others are exactly 
similar : — 



For the perpendicular. 

Jiog. 5675 = 3-753966 

Log. cosec. 126" 56'=10-097271 
Lbg: Bin. 26' 48' = 9*654059 
Lo«r- sin. 1 00" 8' = 9*99317 2 

>. 3151*1 = 3*498468 



For its distance from the middle of 
the base. 

Log. 4(5675) = 3*452936 

Log. cosec. 126° 56'=10-09727l 

Log. sin. 73" 20' = 9-981361 

3400*7= 3*531568 



OTB 1. The foot of the perpendicular falls always on that side of 
niddle of the base which is adjacent to the greater angle. All 
Jars in the ahove &I1 upon that aide ot \k<&'\^«i&i^\<\i\0(v 
Velsoa'a Monument except the \aBt,vA)kS[i\ft^'^'tc£st^ 



niftNG TmoofmntKTRT. 




'OMiked — , and the last throe porpendicuUra are — , beenuBP they 

lie towards the east, or minna side of tlie base. 

Nora 3. The abuvo exercise contaiiiB 20 questions like tho firsl 

three exunples, for each of the 5 plaices ma; be aombined with iQ 

the otiien ; those who wish U> compare tho relntive merits of tin 
I <dd method and that given above, may calculate them both mja ; 

^^^_ the results ooght to be the same. 
^^^^- . Note 3. If the objects observed above w 

^^^^1 Innudai? of a county or an eslate, its area could easily be founl 
^^^^K bom the above, as nill be sbonii in MensuFation or Land &a^ 
^^1 Teying. 

r 97- Given the angular bearings of tliree objects wioM 

distances fram one anothei are known, to find the distanct 
of the station where the angles are taken from each of tftt 
I three given objects. 

I Let the three objects be (A, A', or 

I A"), and B and C, of which B and " 

C are the estreme objects, and A, A', 
or A", the object which has its angu- 
lar position between the objects B imd 
C, as seen from the station S, and let 
CSA, BSA, be the angles formed by 
these objects at the station S. 

CoNSTRUonoN. At the point B, 
In the line joining the extreme ob- 
jects, make the angle DBC=CSA, 
and at C, in the same line, make the 
ZDCB=BSA; then abont the ABDC describe a oirde, 
and It will pass through the station S; join (A, A', or A"), 
and D, and produce the line to meet the circumference in 
S, and S will be the position of the station sought. 

Now in the ADBC there are given the is at B and C, 
and the side BC, hence BD and DC can be foond, 
(Art. 39); and since the three sides of the AABC ate 
^ven, the is can be found, (Art, 49), therefore their diffe- 
rence ACD or ABD can be found; then in the A« ACD 
or ABD, there are given two sides and the contained 
angle, therefore (Art. 40) the is CAD or CAS. and BAD 
or BAS, can he fonnd. Again, in the AACS there are 
given the Is CAS and CS.4, and the aide AC, hence 
(Art. 39) AS and OS can be found, and from the AABS. 
in the same manner, can be found BS. 

1. If the three objects B, A', C, lie in the same stiaigbc 

line, iiaving calculated CD as above, we have in the 

jCiDCA', two sides, DC, A.'C, ami ^ikts co^\^a^'a<:& im^e at 

C, to find (Art. 40^ l\ie LCN!\>, ox ^>^%-, W^enK^-iM 

^BA'S, there are gv^en i\ie Ls^Nl'S.^^A- ^^'S^. ^aA'i 




PI«ANE TRIGONOMETRY. 47 

side A'B, to find BS and A'S, by (Art. 39); and in a 
similar manner, from the A^'^^ ^^^ distance CS can be 
foiind. If the station were taken in the line EC, or EC 
produced, the distance could not be determined. 

2. If ihe third object lie between the station S and the 
line EC, as at A"; to the ZDCE add the LECA", then in 
the Al^^^'' there are given two sides and the contained 
/., hence the LCA"D can be found, and consequently its 
supplement, C A"S ; then in the ACA"S there are given 
two angles, and the side A"C, to find (Art. 39) A"S and 
CS, and in the same manner, from the AEA"S, ES can 
be found. The solution when the three objects form a 
triangle fails, when the points D and A coincide, or when 
the station is in the circumference of a circle described 
about the three objects. If the station S were in the pro- 
duction of one of the sides of the A formed by the objects, 
or in one of the sides themselves, the solution is easy, with- 
out describing a circle. 

3. If the station were within the A 
formed by the three objects, as at S, in 
the annexed figure, then the /.« ASB 
and ASC being measured, construct the 
i^BCD= the supplement of ASB, and B^ 
the angle CED = the supplement of 
ASC, and about the A^CD describe a 
circle, and join DA, and the point S, 
where the line AD cuts the circle, will 
be the station required, for the Z.BSD 
will be = the /.BCD ; .*. BSA, its supplement, will be of 
the given magnitude, and the same may be shown of the 
ZASC. Now, in the A^I^C there are given the angles 
and a side EC, .*. ED and DC can be found ; and since 
the sides of the A ABC are given, the is can be found, 
.•. in the AAED, AB, and ED, are known, and the angle 
ABD, hence the /.BAD, or B AS, can be found, .*. since 
ASB is also known, and the side AB, the sides SB and 
SA can be determined ; and if from the ZBAC, the Z.BAS 
be subtracted, the LCAS will be known, and since ASC is 
also known, and the side AC, the side SC can be found. 

4. The station is without the triangle, upon one of its 
sides, or within it, according as the sum of the two mea- 
sured angles is less than, equal to, or greater than two 
right angles. 

JEXSRCISES. 

J. In diagram, (Art. 97), let BC=290 yaxda^ XC=\^^ 





h 



. yards, and AB=240; iASB^^0°5', and ZASC=2r>M5'. 
It is required to find the distances, AS, BS, and C4.h 
Adb. as 43207 yards, BS 27046 yards, and CS 33634 
yards. 

2. In the same diagran], let the three ohjccts, B, A', C, 
in the same straight line, be distant from each other ai 
follows: viz. BA' 490 yards, A'C 300 yards, and conse- 
quently BC 790 yards; and let the Z,A'SC=43°, and 
BSA'aaMS'. It is required to find the distances SB, 
SA', SC. Ans. SB 782'17 yards, SA' 423'93 yards, and 
SC 390-104 yafds. 

3. In the same diagram, let A"B=500 yards, BC=630 
yards, A"C=540 yards, ^BSA"=3r, and ZA"SC=28' 
24'. It is required to determine the distances SB, SA" 
and SC. Ans. SB 612-523 yards, SA" 137118 yards, 
and SC 656-67 yards. 

4. In the diagram, (Note 3), lot A, B, and C, represent 
three objects in the same horizontal plane, whose distances 
are aa foUows: viz. AC 460 yards, AB570 yards, and BC 
620 yards. At a point S within the triangle formed b; 
these objects, the /.ASB measured by a circle was found 
to be 125" 15', the Z,BSC 124" 15', and the i. ASC I lO'Sff. 
Required the distances AS, BS, and CS. Ans, AS 247-8Sa 
yards, BS 280 804 yards, and CS 310 323 yards. 

5. In the side AB of the triangle ABC, whose ndes 
were AB 1200 yards, BC 1000 yards, and AC 974 yaris, 
I took the angle BSC 74'' 12', Required the distancesof 
the station S from each of the three points. A, D, and C 
Ans. SC 814-274 yards, SB 843-094 yards, and S A 356^ 

6. In the side AC, produced beyond C of the same 
triangle, the LBSC was 37° 25' 53". It is required to find 
the distance of the station S from each of tBe objeett. 
A, B, and C. Ans. SC 1000 yards, AS 1974 yards, and 
SB 158816 yards. 



a EXERCISES. 

1. After observing the elevation of a tower, which i» 
100 feet high, to be tiO", how far must an observer measut* 
back on tlie level plane before its elevation becomes 30*? 
Ans. 11547, or 66^ J3f«et 
S. From the top o? a towei, -wVisk Wx'^vU 108 fert, 
(he angles of depreaaioa o^ tVa \o'e naWm^iwsi. -A iv^*- 
(ical column in the hoiUonVA\ ^\a»'!.. ^^ ^""nA V!>>i 



f <^ 



NAVIGATION. 49 

30° and 60° respectiyely. Determine the beight of the 
colanm. Ans. 7^ feet. 

3. T&e sum of the three sides of a triangle is 2539058, 
snd the angles ate to one another as the numbers 3, 4, and 
5 ; it is required to find the sides and angles. Ans. The 
sides are 707'107> 866*025, 965-926 ; and the angles are 
&, 60°, and 75°. 

4. A person on the top of a tower, whose height is 100 
feet, obserres the angles of depression of two objects on 
she horizontal plane, which are in the same straight line 
irith the tower, to be 30° and 45°. Determine their dis- 
tance from each other, and from the tower on which the 
ibsenrer is situated. Ans. 73*205, 100, and ] 73*205 feet. 

5. From the top of a tower 120 feet high, the angles of 
lepression of two trees on the same horizontal plane with 
she base, was of the nearest 53° 24', and of the other 20° 
W, Required the distance of each of the trees from the 
)8se of the tower. 

Ans. Nearest 8912, and the other 324*982 feet distant. 

6. Wishing to know my distance from two distant ob- 
ects, and their distance fjRem one another, I measured 
beir angular bearings 38° 24^, then advancing in a direct 
ine towards the right hand object 300 feet, their angular 
)earing was 4(y* 13' ; returning to the original station, and 
lien measuring 350 feet towards the left hand object, their 
mgular bearing was then 39° 58'. Find the distance of 
he first station from each of the objects, and their dis- 
ance from one another. Ans. From right hand object 
{22308, from the other 6110*24, and their distance 
;il8-66 feet. 



NAVIGATION. 

98. Navioation is the art of conducting a ship through 
be wide and pathless ocean, from one part of the world 
) another. Or, it is the method of finding the latitude 
nd longitude of a ship's place at sea, and thence de- 
srmining her course and distance from that place to any 
ther place. 

The Equator is a great circle circumscribing the earth, 
▼ery point of which is equally distant from. \.\i& '^^^^'^^ 
lius dMdiBg the globe into two equal parts, CQS\&^\i&Tc^- 
pherea; that towards the North Pole ia called ftifc TLOt?SiKca> 
id that towards the south the Sotttlieiix liem\s^ei^ 



50 SAYtQATtCa. 

The Meridian of any place on ihe earth is a great circle 
passing through that place and the polee, and cutting the 
equator at right angles. The j(rs( meridiaa with us is 
that which passes through the Royal Observatory, at 
Greenwich. 

The Latitude of a place is that portion of its meridian 
which is intercepted between the equator and the given 
place, and therefore never exceeds Q0°. 

The Difference nf LatUwk between two places on the 
earth is an arc of the meridian, intercepted between their 
corresponding parallels of latitude, showing how far one of 
them is to the northward or southward of the other. 

The Longitude of any place on the earth is that arc or 
portion of the equator which is contained between the Jird 
meridian and the meridian of the given place, and is called 
east or west, according as it may he situated with respect 
to the first meridian. Longitude is reckoned east and 
west half round the equator, and consequently may be as 
large as 180°. 

The Difference of Loiiijilude between two places is an 
arc of ihe equator intercepted between the meridians of 
those places, showing how far one of them is to the east- 
ward or westward of the other. 

The Mariner's Con'poM is an artificial representation of 




the horizon. It is divided into 33 equal parts, called 
points, each point consisting nl \V Va'. Hence the ouid- 



NAVIGATION. 51 

ber of degrees in any number of points can be obtained by 
multiplying the points by 11^, or by multiplying 11" 15' 
by the number of points ; ana . degrees can be reduced to 
points by multiplying the degrees by 4, and dividing the 
product by 45. 

A jRhumb Line is a line drawn from the centre of the 
compass to the horizon^ and obtains its name from the 
point of the horizon it falls in with. Hence there are as 
many rhumb lines as there are points in the horizon. 

The Course steered by a ship is the angle contained be- 
tween the meridian of the place sailed from^ and the 
rhumb line on which she sails^ and is either estimated in 
points or degrees. 

The instance is the number of miles intercepted between 
any two places, reckoned on the rhumb line of the course; 
61 it is the absolute length that a ship has sailed in a given 
time. 

The Departure is the distance of the ship from the meri- 
dian of the place sailed from, reckoned on the parallel of 
latitude on which she arrives, and is named east or west, 
according as the course is in the eastern or western hemi- 
sphere. 

If a ship's course be due north or south, she sails on a 
meridian, and therefore makes no departure; hence 
the distance sailed will be equal to the difference of lati- 
tude. 

If a ship's course be due east or west, she sails either 
on the equator, or on some parallel of latitude. In this 
case^ since she makes no difference of latitude, the distance 
Sailed will therefore be equal to the departure. 

When the course is 4 points, or 45^ the difference of 
latitude and departure are equal, and each is the (distance 

X %/*«^) = the dist. x '707 nearly. When the course is 
ess than 4 points, the difference of latitude exceeds the 
leparture; but when more than 4 points, the departure 
.'xceeds the difference of latitude. 

99. The distance sailed, the difference of latitude and 
leparture, form the sides of a right-angled triangle, in 
i^hich the hypotenuse represents the distance, the perpen- 
licular the difference of latitude, the base the departure, 
ind the angle opposite the base is the course, and conse- 
luently the angle at the base the complement of the course; 
lence any two of these five l^eing given^ the others can 
)e found by the rules for right angled triangles, (Arts. 33 
ind 34)| which give the following xelatiOBa ^jsi^u^ ^^ 
Nurts: 



HAVtOATIOIt. 




Bad. : tan. of the c 
Had. : sec. of the c 
Uist. : dif. lat. = Had. : 
Wist, : dept. =; Had. : 
Uif. lat. : dept. = Had. ; 

100. To find the latitude 



w 



= dif lat. : depL 
= dif. lat. : di»t 
COS. ofthccouree. 
Mue of the course, 
lau. of the course, 
md longitude at which a ship 
Las arrived, ivhen those of the jilace vhicli she left, uua 
the difference of latitude and luiigiiude ivhieh she has made, 

KuLB. If the latitude left and the diSetence of lati- 
tude be hoth north or huth south, their sum is the latitude 
uome to ; but if the one be north and the other south, 
their difference is the latitude come to, of the same oane 
as the greater. 

Note. Tho longitudB is olitained in the same way, only when dm 
longitude Ballad from and the difference of Inogitude are of the taisa 
name, and their Bum exceeds lliO", it must befcubtracted from 360", 

ExAiirLK. If a ship sail from lat. 39" 14' N., long. 21° 
17' '^V'., till her difference of latitude be 312 miles N., and 
difierence of longitude 169 E. ; required her latitude aiil 
longitude arrived at. 
Lat, left. 39° 14' K Long, left 21" 17' W. 

Dif lat. 312= a' 12' K Dif. lung, 169= 2*49' E. 
44° 26' N. Long, in lb" 28' IV. 



In the following 
rived at are required 



the latitude and lon^tode ai- 



"■■■•«• "SSif ^'f^ -- 



1. 42M2' N. 17° IB' E. ISft S. 216 W. 3B* 42' K. IBM^E. 

aerars. i9's2'w. loo n. 2row. sb'u's. sraa'W. 

3. ss'iB's. iB-ai'E. aie s. iss e. 36° sea 31* o'E. 

4.12Mi'N. 73°I6'W. 612 N. 724 E. 20° 4G' N. 60'12'W. 

B. ea-lS'N. aS'SB'E. 147 N. 301 E. 31' JS' N. 3U*5S'E. 

e. 52MB'S. 28°27'W. 218 8. 1B6 W. 65° fiti' S. 31*33'W. 



n 



101. Given the latitude and longitude sailed (torn, and 
io those arrived at, to fiad the difference of latitude aiiil 
difference of longitude made. 
'■VLB, If the latitudea or \oiigAuiBa \it \>Ql,h of the saof 



NAVIGATION. 53 

Dame, their difference reduced to miles will be the diffe- 
rence sought; but if they be of different naraes, their sum 
reduced to miles will be the difference sought. 

Note. If the longitudes be of different names, and their sum ex- 
ceeds 180°, it must he suhtracted from 360**, and the remainder re- 
duced to miles will he the difference sought. 

In each of the following exercises it is required to find 
the difference of latitude and the difference of longitude : 





Sailed Atom. 
Lat. Long. 


Answers. 




Lat. Long* Dif.lnt. Dif.long. 

Mileg. Miles. 


1. 


25° 16' N. 150 3' W. 


20° 17' N. 42° 13' W. 299 1630 


2. 


12° 50' S. 40° 34' E. 


35° 42' S. 1 8« 14' W. 1 372 3528 


3. 


50' 17' N. 5' 7'W. 


70» 12' N. 50*' 16' W. 1195 2709 


4. 


18°25'S. 15°35'W. 


0° 0' 350 17' W, 1105 1182 



Plane Sailing. 

102. For the proportions necessary to solve cases in 
plane sailing, see (Art. 99). But Plane Sailing can also 
be solved by the Traverse Table in the following way : 
Damely, when the distance and course are given, find the 
given course, and under the distance, in the proper co- 
iamns, you will have the dif. of lat. and dept. required. 

Note 1 . In the small Traverse Table given in this work, look 
for the given course at the side of the page, and the first figure of 
distance at the top or bottom, where these columns meet, take out 
the dif. lat. and dept., removing the decimal point as many places 
to the right as liie figure is to the left of the units' place ; do the 
Same with each of the figures in succession, and add the results; 
the sums of the differences of latitude and departures thus found 
iHll be the dif. of lat. and dept. sought. In a similar way, if the 
coarse and dif. of lat. be given, may the distance and departure be 
Found. 

Note 2. When the course is not one of the things given, the 
solution is not so conveniently obtained from the Traverse Table as 
by trigonometrical calculation. 



EXKUCISES. 

1 . A ship from lat. 49^ 57' N., sails SW. by W. 244 
miles ; required the latitude she is in, and the departure 
made ? Ans. Lat. in 47° 41'. Dept. 2029 miles. 

2. A ship from 1" 45' north latitude, sails SE. by E.^ 
till she arrives in latitude 2° 46' south ; lec^vui^dL V^x ^v^- 



64 KaVTGATIOlf. 

tance and departure. Ans. Dist. 487-8. Dept. 405(1 
miles. 

3. If a ship sails NE. by E. | E„ from a port in 3' 15' 
BODlh latitude, until she departs from her first meridian 
40S miles ; required what distance she has sailed, and 
ivhat latitude she is in ? 

Ans. Dist. 4491. Lat. in 0° 3' S. 

4. Suppose a ship sails 488 miles, between the south and 
the east, from a port in 2" 52' south latitude, and then bj 
observaiion is found to be in 7" 23' south latitude ; what 
course has she steered, and what departure has she made? 

Ans- Course S. 56" 16' E. Dept. 405'8 miles. 

5. If a ship sail south-eastward from latitude 50° 16' N., 
till her distance is 137 miles, and her departure 112 miles; 
required her course, and latitude come to ? 

Ans. Course S. 54° 50' E., and lat. 48° 57' N. 
fi. If a ship from latitude 32° 37' N. sail south westward, 
till Ler difference of lat. is 114, and her departure W 
miles; required her coursp, distance, and latitude arrired 
at? Ans. Course 8. 40' 24' W., dist. 149-7 miles, and 
lat. 30- 43' N. 

^V Tbavebsf: Sailing. 

* 103. "When a ship is obliged to sail on different coutsci. 
the crooked line which she describes is called a traverse; 
and the method of finding a single course and distance, 
which would have brought the ship to the same place, is 
called resolving a traverse. 

Ilui.B. Make a Table of any convenient size, as that of 
the following example, and divide it into six columns : in 
the first of these place the several courses the ship has 
made; in the second place the several distances she Iiat 
made on each course. The third and fourth columns ate 
to contain the differences of latitude, and therefore to he 
marked N. and S. at the top ; and as the fifth and sixth 
are to contain the departures, ihey are marked E. and W. 
at the top. 

Find by the first and second analofjies, (Art. 99), ot 
from the Traverse Table, the dept. and dif. lat., which place 
in their proper columns, for each of the courses; then the 
difference between the north and south columnB will be 
the dif. of lat. made good, of the same name with the 
greater quantity ; and the difference between the sums of 
the east and west columns will be the whole departure, of 
the same name Kritli the gteateT m.i;i\SvMi Svaaatt. 



nATIOATIOir. 



55 



ExAHPiiB. A ship in latitude 38° 25' N., and longitude 
le** 17' W., sails SW. by W. 56 miles, W. by N. 110 
miles^ W. 95 miles, SE. by E. 50 miles, S. 103 miles, and 
S. ;| W. 116 miles; required the latitude in, and the de- 
parture made ? 



T&IYXBSE l^LE. 1 


Conecbed 
oounea. 


Pointa. 


Dis- 
tances. 


Dif. of latitude. 


Departure. 1 


N. 


s. 


E. 


w. 


SW.byW. 
W. by N. 

West. 
SR by E. 

South. 
S.1 W. 


5 

7 
8 
5 


i 


56 

110 

95 

50 

103 

116 


21-5 


31-1 

27-8 
103- 
115-9 


41-6 


46-6 
107-9 
95 

5-7 






21-5 
Dif. lat. » 


277-8 
21-5 


41-6 


255-2 
41-6 


256-3 


Dept. s= 


213-6 



Now 256' miles of dif. lat. = 4** 16', which being south, 
sabtract from 38° 25' north, and the remainder is 34° 9' N., 
which is therefore the latitude arriyed at. 



EXERCISES. 

1. Suppose a ship from latitude 49° 57' N., sails SSE. 
15 mUes, SE. 34 miles, W. by S. 16 miles, WNW. 39 
miles, S. by E. 40 miles ; required the difference of lati- 
tude, and the departure she has made, and also the course 
»he has made good? Ans. Dif. lat. 65-32 S., Dep. 14-14 
W., and course S. 12° 13' W. 

2. A ship from latitude 17° 58' N., sailed S. | W. 40 
miles, SSE. J E. 97 miles, N. by E. { E. 72 miles, SSE. 
I E. 108 miles, N. by E. | E. 114 miles, SE. by S. i E. 
126 miles, and NNE. \ E. 86 miles ; required the latitude 
she is in, the departure made, and the course made good? 
Ajis. Lat. in 16° 52' N. Dept. 258-87 E. Course made good, 
8. 75° 40' E. 

3. A ship from latitude 51° 25' N., sails SSE. iE.16 
miles, ESE. 23 miles, SW. by W. ^ W. 36 miles, W. | N. 
12 miles, SK by E. ^ K 41 miles ; required the latitude 
urrived at^ and the course and distance TCkaA<& ^<;^qW 



B'AvraA.TIOIT. 

50= 25'. Course S. IS" 15' E. Dist. 62'704 



K Faballel Sailing. 

P 104. Parallel sniling is the method of finding the dis- 
tance between two places ailuatod under the eame parallel 
of latifade; or of finding tlie difference of longitude cor- 
responding to the meridional distance or departure when a 
ship sails due easl or west. 

Since circles are to one another as their radii, the 
distance of two meridians on the equator, or the dif- 
ference of longitude, will he to the distance of the 
fiarae meridians at any oilier latitude as radius ; cos. of 
lat. 

105. Hence (he following prnpnrlions can be derived; 
Dif. long. : raer. disl. = li : cos. of the latitude. 
It: cos. lat. ;^ dif. long. : mer. dist. 
B : sec. lat. = mer. dist, : dif. long. 
Also, cos. of any lat. : cos. of any other lat, ^ the mer. 
dist. on the first : the mer. dist. of the other. 



KXBItCISEB. 

1. If a ship, in latitude 42° 54' N., sail due west 196 
miles; required the difference of longitude made? 

Ans. 4" 27' 36" H'. 

2. A ship from latitude 51° 25' N., and longitude 9° 2? 
"W., sailed due "est 1040 miles; required the longitude 
at which she then arrived ? 

Ans. Long, arrived at 37° 16' W. 

3. A ship, by sailing due west 45fi miles, made a difit- 
rence of longitude of 12" 15', or 735; required the latitude 
on which she sailed? Ans. Lat. 51° 39' 14''. 

4. A ship, in latitude 51° 16', sailed due east for 240 
miles, another sailed due cast 308 miles, and made the 
same difference of longitude as the former ; on what lati- 
tude did she sail ? Ans. Lat. 3G° 35' B". 



MinnLE Latitcijb Sailing. 



^^- rerse 



lOfi. In Middle Laiiiude Sniling, (he longitude is eaky- 
lated from the departure made, either in a single or tra- 
Terse course, by I'aralk! ^ni/inf/, upon the supposition 
Jt is equal to the meridian distance, on the middle ' 
f/eJ bet*¥een that sai\ea£tQTOaQiftvaVwtTC«e4a.t. | 




NAVIGATION. 57 

This method^ though not quite accurate^ Is nevertheless 
efficiently so for a single day's run, particularly near the 
?quator^ or when the ship's course is yearly east or west, 
[n high latitucles^ when the distance run is greats it may 
ead to slightly erroneous results. 

In the annexed diagram, let ABC 
epresent a %ure in Plane Sailing, in 
vhich AB represents the distance 
aOed^ LBAC the course^ AC the 
lifference of latitude made^ and CB 

he departure. At B make the 

.CBD = the middle latitude, and 

iroduce the line AC to meet BD in ha 

). Then if BCD be considered a ,^ 

i^re in Parallel Sailing, in which ,^ 

!)B is the meridian distance^ BD f^ 

rill be the diff. long., and the Z.D 

he complement of the middle lati- 

ade. 
107. Now from the two right angled triangles, ACB, 

ICD, and the triangle ABD, the foflowiug analogies can 

e obtained : viz. sin. D : sin. A : : AB : BD, or 

1. Cos. mid. lat. : sin. course : : dist. : dif. long. 

2. Sin. course : cos. mid. lat. : : dif. long. : distance. 

3. Dif. long. : dist. : : sin. course : cos. mid. lat. 

4. Dist. : dif. long. : : cos. mid. lat. : sin. course. 

dif lat. 
In the last of these analogies, substitute ■ for 

° COS. course 

be distance, and multiply the extremes and means, and 
iere results dif. long, x cos. mid. lat. = sin. course X 

= tan. course X dif. lat. 

». eourse 

5. Hence (Alg. 103) dif. lat. : dif. long. : : cos. mid. lat. 
tan. course. 

6. Dif. long. : dif. lat. : : tan. course : cos. mid. lat. 

7. Gos. mid. lat. : tan. course : : dif. lat. : dif. long. 

8. Tan. course : cos. mid. lat. : : dif. long. : dif. lat. 
These eight analogies are sufficient to solve all the cases 

r Middle Latitude Sailing that ever can occur ; only the 
rinciples of Plane Sailing and Parallel Sailing must fre- 
tiently be applied first to obtain the necessary data. 

Example. A ship from lat. 39° 41' N., and long. 31° 
' W., sailed NE. ^ E. 590 miles ; required the latitude 
ad longitude come to ? 

First to find the latitude come to, we have h^ "PVaxka 
uling B : cos. 4^ pts. : : dist. 590 : dif. lat. ^14l^ tkiA^^'^. 



,^^.., 



»• 41' N. I 



^ 



£8 NATIGATI01T, 

Latitude sailed from, 

Dif. lat 374 = 6- 14' N. 

Latitude come to, 45° 55' N. 

Sum of latitudes, 85" 36' 

i sum, or mid. latitude, 42° 48* 

Applying dow analogy (1), we liave 
Cob. 42° 48'. ar. co. of Log. cos. = 10134464 

la to sin. 4i pta. Log. sin. = 9-888185 

As dist. 590 Log. di8t. =_2'770853 

Is to dif. long. 10" 22' Log. 62I'6= 2-793501 

This dif long, being E., while that sailed from is W., 
their difference is the longitude come lo, which is there- 
fore 20° 41' W. The ship is therefore in latitude 45' 55' 
K., and longitude 20° 41' \V. 

EXERCISES. 

1. A ship from lat. 14° 46' N., and long. 24" 46' W„ 
uiled SE. by S., until by observation she was found ta be 
in latitude 10° 30' N. ; required the distance sailed, and 
her present longitude? 

Ans. Dial, 307-9 miles; long, in 21° 50' TV. 

2. A ship from lat. 49" H' N.. long. 0° 19' W., sailed 
320 miles between the south and west, and then, by ob- 
flervation, was found to be in latitude 45° 8' N. ; required 
the course, and the longitude come to ? 

Ans. Course 8. 26° 39' 0" W. ; long. In 9° 52' V. 

3. A ship from lat 40° 3' R, and long. 3° 52' R, sulcj 
.280 miles between the north and east, upon a direct 
GOnran, and made 186 miles of departure; required the 
course, and the latitude and longitude come to? Am. 
Course, N. 41° 37' 39" E. ; lat. come to, 43° 32' N.j and 
long. 8° 1' E. 

4. A ship in north latitude sailed 500 miles upon 
direct courae, between the south and west, until her diffe- 
rence of longitude was 440 miles ; required the cotirse 
steered, the latitude sailed from, and the latitude come to, 
allowing the middle latitude to be 43° 45' north? Am. 
Course, S. 39° 28' 14" W.; lat. sailed from, 46'58'N.; 
and lat. arrived at, 40° 32' N. 



k 



MENSURATION OF SURFACES. 



59 



MENSURATION OF SURFACES. 

In the mensuration of plane superficies the prohlem to 
)e solved is, — How often is a square of a given magnitude 
ontained in a figure of which the rectilineal dimensions 
re given? The method of solving this prohlem depends 
ipon the form of the figure whose surface is sought. The 
quare employed to measure it may he of any dimensions; 
ut those most commonly used are a square inch^ a square 
sot, a square yard^ a square link, and a square acre, 
lie numher of times that the square is contained in the 
iTen figure is called its area, and the square that is em- 
loyed to measure it is called the measuring unit. 

Table of Lineal Measure. 



Inches. 


Link. 










7-92 = 


1 


Foot 

1 






12 


1-5151 = 


Yard. 






Pole or 
Perch. 




36 


4-5454 


3 = 


1 




198 


25 


16-5 


• 5 = 


1 


Chain. 








Fur. 
longs. 




792 


100 


66 


22 


4 


1 




7920 


1000 


660 
5280 


220 


40 


10 = 


1 

8 = 


Mile. 

1 


53360 


8000 


1760 


320 


80 





Table of Square Measure. 


Square 
Inches. 


Square 
Link. 






Square 
Foot. 




2:7264= 


1 






Square 
Yard. 




lid 


2-2956=: 


1 






It* 


Square 
Perch. 




1S!96 


20.6611 


9a 


1 








Square 
Chain. 




99204 


625 


272*25 


30-25 = 


1 








4356 




Square 
Rood. 




^264 


10000 


484 


16 = 


1 










Square 
Acre. 




ISfiSlSt) 


25000 


10890 


1210 


40 


2-5 = 


1 






Square 


6872640 


100000 


43560 


4840 


160 


10 


4 = 


1 » 


l4489G0(y 


84O0000fi 


Sm8400' 


30976001 


10240(\ 


6400 1 2560 \ 6\0 =\ \ \ 



Problkiu ^^^^H 

To find the area of a Paralelhgram. whetjier ft %ff¥ 

RuLR I. 'When the length and perpend Icular breailti 

are given; Multiply the length by the perpcndicolaj 
hreaclth, and the product will be the area. 

DEKDNSTomoK. Let ABCD lio x it. 


contatna tliE meaBuring lineal Doit 
any niamber of times, as S, and of 
wUch thB pfrpandiculnr hremllh 
All wmtiuiifl the Biuno unit .iny 
numbor of timea, ks 4 ; then, if lines 
















1 
















\ 
















i 
















a 


of each unit in the sides, parallel " 
to the adjacent sides, as in the annexed 
the whole figuro will be divided into as 
nnits in the product of BCxAD; for til 


diagram, it is evident tliJ 
many Bqnaroa as there i«| 


the Brat as tbera are nnits in AB: henee there aro na man; tqnintf 
in tha whole figure as there aro units in the product of the anil 
in BC multiplied into the number of units in AB. 

equal lo a rortanglo having the same hose nnd altitude, or perpendi- 
cular breadth, BBlho parallelogram. 

EsAMPLK. What ia the area of a pamllplngram whnw 
length is 2-JO feet, and perpendicular breadth 160 f«lt 
Ans. 240x 1C0=38400 sq. feGt=426GJ square yards. 

EXRRCISES. 

1. What ia the area of a square whose aide Js ^ 
links? Ans. 124:2562.5 square ]inks=124 ac. 1 ro. f^ 

2. Find the surfiice of a rectangular pane of glass, in 
length being 5 feet 7 inches, and breadth 3 feet 5 inche*. 

Ans. 19 square feet 11 square inch** 

3. How many square yards of plastering are in the Oftf 
ing of a room, its length being 2u feet, and breadth XTM 
6 inches? Ans. 4S\\ \ square yard). 

1 4- What is the area of a field in acres whose length il 
1535 linkg, and whose breadlli is 1270 links? 

Ans. 19 aeres, 1 rood, 39-12 pereJie* 
5. How many roods, of 3G square yards each, an in i( 

1 wall 320 feet long, and 13 feet G inches high; and fihol 

^^1 trill it cost at L.2, 12s. Cd. per rood^ 

^^ft ^Vtis.\3^ roods, oDrtJ^ 



lESumnuTioB or sdbfaccs. q1 

H. "When two sides and the contuined angle are 
lultiply the product of the two sides by the natu- 
if tile included angle, the product ivill be the area. 
I Logarithms. Add the Logiiritbms of the two 
I the log. sine of the contained angle together, tho 
inUhed by 10 in (he iiides will be the Log, of the 

TBiiioN. Let ABCD lie a, iiai-al- p c 

in whidi iii'e givua tb« twu ai Ji:9 I 7 

AB, and the cuotaiueJ angle / / 

iw DE -^ to AB, lliL-Q by lost / / 



3E_ 
ID" 



a. of tho figui'u = A1).UK= L 



:AB.AD. r 



lin. DAB. (Trig. Art. 6). 



PI.B. What IS the area of a parallelogram, two ad- 
lea being 34(iand 210, and tbeeontained angle 62° 
E.sin.62''12'=.8fl4581x34ti>: 210-64273-63546. 

[n this case, whather we use tbe natural Bines carried tu 
f decimala, or tho logairithmic HiiicH wiUi li plicei^ of deci- 
lau ojdy dopond on the firuL fi figures ef tho ajiawer ^m 
•eat, BO that the answer ciui he fuuud witluii Ites than a. 
^ai\ of tho nhele, wbich will be suttidently correct fer ul- 
ractical purposea. If jjivater aociiracy he required, re- 
lat be hod to tables earned to more dodnml places. 

EXEltCISES. 

iat is the area of a field la the form of a pa- 
im, two sides being "JZG and 609 links, and the 
angle 81" 15'? 

Ans. A acres. I rood, 28'fll2 perches, 
parallelogram has two of its sides 360 and 200 
id the angle contained between them 30°. What 
■a? Ans. 30000 sqiiare yards, 

e of tbe sides of a field, in the form of a rhombus, 
links, and one of its angles is 70° 18'. What is its 
1 how much is it leas than if it had been a square 
he same length of side! Ans. Its area 14 acres 

Tches, less than the square 3 roods, 20'53 perches, 
itv many acres are in a rhomboid, whose leas angle 
nd the including sides 2u35, and 1040 liaks^ 

Ans. 13 acres, 29.12 perches. 

Fbobleu IL 

id the area of a Triangle, when there are given, ( ' 
! base and perpendicular altitude, two &\' 
ained aagie, or the three sidus. 



^ 



3 MxiHtntATioH or amwACsa. ■ 

Rule I. Multiply the buse Ly the perpendicular aiti 
tude, and half the product will he the area. 

Rule II. Multiply htilf ihe product of the two sides li 
the natural sine ot' the iucluded angle, and the prodiu 
will be the area. Or, add together the log, of one side, th 
log. of half the other, aud the log. sine of the containe 
angle; the sum rejecting 10 from the index ivill be the Iq 
of the product. 

Remark. The reason of these rules is ohvioufi &ai 
Problem I., and (Geo. Prop. 28), which proyea that a ti 
angle is half of a parallelogram. 

ExAUPLE. What is the area of a triangle, two of vhttt 
sides are 8ii and 90 feet, and the included angle 47° 131. 
Tlie nat. sin. of 47" 13' = -733927 

Multiply by K^fix 90) = 3825 

And the product is the Ans. =280? 2? sg . feet 

Exercises. 

1. What is the area of a. triiingle whose base is 7" 
feet, and perpendicular altitude 48 feet? 

Ana. 1680 feet, or 186| yards 

2. How many square feet are in a right angled tiiangbj 
whose base is (iO, and perpendicular 40 feet? 

Ans. 1200 square feet 

3. What is the area of a triangle whose base is 54, ajA 
perpendicular altitude 29.52 chains? 

Ans. 79 acres, 2 roods, 32-64 pol«» 

4. What is the area of a triangle, two of whose sides "" 
45 and 5675 feet, and the contained angle 3(i° 45'? 

Ans. 763 986 square fee 

5. How many square yards are in a triangle, of nliie 
one angle is 45°, aud the including sides 30 and 24 feetl 

Aus. 28-2842 square yarf 
Rule III. Find half the sum of the three sides, 
from it subtract each side separately; then multiply 
half sum and the three remainders together, and thesqoia 
root of the last product will be the area. Or, add the lo- 
garithms of the half sum and of the three remainders ttf 
f;ether, and half this sum wiU be the 
Dgarithm of the area. 

I)GHaNSiiu.TiO!t. Let AJBC be a triangle, 
of which tlie three sides ore given, and adopt 
tile sanie notAtioa as in Trigonometr}' ; wa 
hare bf Rul e 2, area ={ be Bin. A. K ow 
lia. A M= V(l -Hifa. A) (,V — coi h.~i = 





ME.VSUBATION OF SURFACES. 63 

,J^^^[^^^ (Trig. 45 and 4e) = ^^«(s-^)CV-6)C»=0- 
- 3 a>B area is = i&e x if. ^s(, — a)"(s= 6) (s ^c) = 

. IT iJie bnangle be ei^uilalenil, and have each of its aideB a, 
t wiU becorna 5, (s — a), (s — i), and (s — c), will each become 5 - 
therefore the area of au equUatend triangle isVg x^xfXj = 

Example. How many square yards are in a triangle, 
vhoee three sides are 50, 40, and 30 feet ? 

Here half the sum of the three sides is 60, and the three 
iraneinders are 10, 30, and 30; therefore the a>-ca is 
^60302030= Va6<>i)0O=60O feel;=66^ yaida. 

EXEBCISE9. 

1. "What is the area of a grass-plot in the form of an 
Equilateral triangle, each of its sides being 234 links ? 

Ans. 37936 square poles. 

2. How many square yards are in a triangle trhose three 
aides are 39, 36, and 15 feet ? Ans. 30 square yacds. 

3. Hon many acres are in a field in the form of an is- 
osceles triangle, each of the equal sides being 500 links, 
and the third side 600 links ? Ads. 1 acre, 33 poles. 

4. How many square feet are in the area of a triangle, 
whose three sides are 74, 82, and 90 feet 1 

Ans, 2855-61 square feet. 
Rule IV, WLen there are given a side, and the angles 
ttt its extremities. From twice the logarithm of the given 
tide suhtract '301030 ; to the remainder add the log. cosec. 
of the sum of the two measured angles and the log. sine of 
each uf these angles ; the sum after rejecting 30 from the 
index will be the log. of the area, 

DehONSTRITIOM. (Diagram last case.) Let AC be the giren Bide, 
nhieh call b, and A. and C the two measured angles ; then since IB 
it the supplement ot A+C, sin. B= sin. (A+C), (Trig. Art. 30.) 

HraicesiD.(A+C):sin,A-. :6:BC=^^^|^^,{Trig.Art3(i.) Abo 

R-ain.C: ; { BC =-^?J?iA, Y BD=1" °' - * - .^"'^ , and the area = 
\ lin. (A+C)/ Run-IA+C) 

lixBD= S HBH^IA+C li *'^«1'S'°™ - .in: lA+CI = ""^ (A+C), 

=1 gives the ahove rule. 

, "What is the area of a triangle, OTie dt \)^yow 



'64 



sides is 46 yards, 
and 57" 12'! 

Log. 46=1-662758x2 
Subtract Log. % 

Log. cosec. 129° 30', 
Log. Bin. 72° 18', 
Log. sin. 57= 12' 



nd the angles at its extremities 72" 
Jlere the sum of the angles is 129° 
—3-325516 
■301030 



3-024486 
=10-112534 

= 9-978939 
^ »92457 2 

Log.l097-96aquare yards. Ans.= 3'040591 



I 



1. How maDy acres are in a triungiikr field, ota 
■whose sides is 1046 linka, ;md the angles at its extre 

ies 68° 24', and 71° 1^'? 

Ans. 7 acres, 1 rood, 29-208 pi 

2. How many square yards are in a triangolar field, 
of its sides being 465 feet, and the angles at its extre 

■ s 45", and 78" 15'* Ans. 9944-13 square ya 

3. Two of the angles of a triangle are 76° 13' and ' 
9', and the side betneen them 3475 links ; how iq 
acres are in the field ? Ans. 144 acres, roods, It 

PilOBLEll III. 

To find the area of a trapezoid. 

EuLK. Multiply /tay" the sumof the parallel sides by 
perpendicular distance between them, and the prodact' 
be the area, (Geo. Prop. 33). 

ExAMFLE. What itt the area of a trapezoid, its pan 
sides being 30 and 20 feet, and the perpendicular dJsti 
between them 14 feet t Here the paralleS-Hides beiiu 
and 20, half their sum is 25, which being multiplied 
14, gives the area 350 etiuare feet. 

XXEKCISES. 

1. What is the area of a field in the form of a i 
pezoid, its parallel sides being 536 and 378 links, and 
perpendicular distance between them 418 links ? 

Ans. 1 acre 3 roods 25-6416po 

2. What is the area of a board 20 feet long, ^ brei 
at one end being 2 feet 6 inches, and at the other 1 1 
3 inches? Ans. 36 feet 8 \iA 

3. The side of a roof Is 40 feet at the easing, 24 I 
at the ridge, and the distance from the ridge to the eai 
isJ8feet; what is the urea of the aide of die roof ? 

A.T\a. &1& fc<it, Qc 64 B^oare jti 



< jixbsebatioit of bubiacbs. u 

Problem IV. 

To find tbe area of a trapezium. 

K171.E I. Dinde it into two triauirlea by a diagonal, and 
find tbe area of each of them, and their atun nill be the area 
of the trapezium. 

Note. — The srcas of the triangles are to be found by either of tbo 
rules given in Problem 11., Bccordiag- to the data. 

RiTLB II. Measure each of the diagonals, and the angle 
of their intersection ; then half the product of these dia- 
(^Doals, multiplied into the nat. sine of the angle of their 
intersection, will be the area. Or add together the loga- 
rithms of the diagonals and the log. sine of their intersec- 
tion, the sum diminished bj 10'301030 will be the loga- 
rithm of the area. 

Note. — This mle is kIbo trne of paxallelogramB. 

DsKaHBTiuTioK.— Let ABCD be a. tra- 
pezium, of wliich AC and BD are tlie dia- -' . 
gunalB which iuCerBeecinE; then sin. AED \ 
Ez ain. DEC, (Trig. 30) = sin. CEB = Bio. 
&EB,(Geo. prop.3), which callsiu.E, and 
ietAE=a, EC=.c,B£^,aiid UE=(f; then 
(Prob. IL Rale II), AAED=i ad torn. E, 
d}EC=idc Hiu. £, aBECz=IAc Bin. E.and 
ULEB=ia& Bin. E ; .■. tbe whole figure= 

KoJ+A+ftc+afc) ain E=i{iH-c)(6+tOBin. c 

£=)ACxBD Bin. E, which U the rule. 

Example. If AC=530 links, BD=608 links, and the 
(.AEB=52° 12'; it U req,uired to find tbe area. 

Nat. Bin. 52= 1 2'=-790155 X 530 x ^ =127309-7736= 
1 acre, 1 rood, 3 poles, 21 yards. 

EXEBCISES. 

I. In the trapezium ABCD, if the diagonal AC be 
270 feet, and the perpendiculars upon it from I> and B 
180 and 120 feet; find its area in square yards. 

Ans. 4500 square yards. 
_ 2. In a trapezium ABCD, if the diagonal BD be 1476 
links, and tbe perpendiculars on it from A and C 557 
links and 403 links; find its area in acres. 

Ans. 7 acres, 2 roods, 39-84 poles. 

3. If in the same figure AB^490 links, BC 464 Unks, 
BD 756 links, Z,ABD 74° 44', iDBC 80M7'; what is 
the area of the figure in acres ? 

Ans. 3 acres, 2 roods, 2-496 poles. 

4. If AB=36feat,BC=34, CD=42, AD=44,andthe 
diagonal BD^=62 feet ; what is the area of the tra^eiuBo, 
in yards ? Ana. 16V40ft ss(^Ma.t(i -^aiia. 




F 

^^K 5. If AC:r^7eO links, BD=810 links, and the angle 
^^KAEB be 79° 1 6' ; what is the area of the field in acies? 
^^K. Ana. 3 acres 3864 poles. 

^^K 6. If AD=3aO links, DC=3eO links, BC=filO links, 
^^■AB=534 links, and ihe angle ADC 88° 30'; what is tlie 
^^bma in acres 1 Bt Tiig. At'=4(i(> 834. 
^^r^ Ads. 1 ac. 2 ro. 26'13 poles. 



llraHSOTATIOW (# SiniFACES. 






To find the area of any irregular figure. 

Rule. Draw diagonals dividing the figure info triangles, 
or triangles and trapezioms. Then find the area of all ihese 
separately, and their sum will be (he content of the lybole 
irregular figure. 

1. Find the content of the 
irregular figureABCDEFGA, 
in which are given the follow- 
ing diagonals and perpendi- 
culars, namely 

AC=530 links. GD^424 
FD^4a6 



Bi =134 

GC=394 

^^ DC=182 

^^ To find t 



Ans. 1 




Ea = 



30'2812 poiet 
Problem TI. 

To find the area of a regular polygon, when the leng!'' 
id the perpendicular upon it from the cenlreiiK 
given. 

KuLE I. Multiply the side by the perpendicular, and 
that product by half the number of sides, the last pmdac' 
will be the area. 

Dguonbtbation. Thu polygon consists of 
03 many trionglea as iC hath sideB, whoso 
heights and bases are each equal to thop or- 
pcndlcular, and the side o{ the polygon ro- 
spectively; if » be a side of the polygon, p 
tbe perpendicular, uid n the nnmher of 
Bides, then ^ps will be una triangle, (Prab. 
'IL Rule I.) and iiips all tlio triangles or 
(ha whole polygon. 

' £xAifPi:.B. Find the area of a heptagon, each of n 
mdes is 20, and the perpendicular 20-7^5 feet 




JIEN8URATION OF SUBFACES. 



67 



BXERCI8E8. 

1. Find the area of a hexagon^ ivhose side is 10 feet, 
id the perpendicalar upon it from the centre 8*66 feet. 

8-6g X 10 X 3=259-8 square feet. 

2. Find the area of a pentagon, whose side is 8 inches, 
id perpendicular upon it from the centre 5*506 inches. 

Ans. 110*12 square inches. 

3. Find the area of an octagon, whose side is 15 feet, and 
e perpendicular upon it from the centre is 18*1 1 feet. 

Ans. 1 20*73 square yards. 

4. Find the area of a decagon, whose side is 4 feet, and 
e perpendicular upon it from the centre 6*156 feet. 

Ans. 123*12 square feet. 
Rule II. When the side only is given ; to twice the 
Tarithm of half the side add the logarithm of the numher 
sides^ and the log. cotangent of (180° divided by the 
imber of sides) the sum diminished by 10 in the index 
U be the logarithm of the area. 

RuLfS IJI. Multiply the number opposite to the name 
the polygon in the following table by the square of the 
i^h of the side, and the product will be the area. 
For the demonstration of these rules see Key. 



Jo. of 
ides. 


Names. 


Areas or Multipliers. 


18() 
n 


Log. cot. — 


3 


Triangle. 


*433013 


60** 


9-761439 


4 


Square. 


1-000000 


45° 


10-000000 


5 


Pentagon. 


1-720477 


36*' 


10-138739 


6 


Hexagon. 


2-598076 


30° 


10-238561 


7 


Heptagon. 


3-633913 


25°« 


10-317336 


8 


Octagon. 


4-828427 


22*4 


10-382776 


9 


Nonagon. 


6181824 


20° 


10-438934 


10 


Decagon. 


7-694209 


18° 


10-488224 


11 


Undecagon. 


9-365640 


16°A 


10-532205 


12 


Dodecagon. 


11-196152 


15° 


10-571948 



Example. Find the area of a pentagon, each of whose 
as is 8 feet. 
Rule II. 2 X Log. 4 ^ =1 *2041 20 

Log. cot. (^-5^'=36°'\= 10*138739 

Log. (n=5) = *698970 

Ans. 110*11 = 2041829 

Rule III. Tabular multiplier=l-720477x(8^=64)=ilVQ*UQ^1'^ 

:bxeiicisbs. 
/. Required the area of a regular pentasoTi, o^ ^V\«^kv 
ker-efibe eqmd sides is 15. A.tl^- 'SSTlA^ 



r 



» 



KEKeDXATIOK OV SOBFACKS, 



I 



2. Bequired the area of a regular hexagon, either of 
equal sides is 20 feet. Ana, 115'470 aq. yards. 

a. Jiteqiiired the area of a regular heptagon, either of 
whose equal aides is 3 feet. Ana. 32'705 sq. feet 

4. Required the area of a regular octagon, either of 
.vl.oae equal aides is 10 feet. Aob. 53649 sq. yanft, 

5. Eequired the area of a regular decagon, either of 
iWhose equal sides ia 18 inches. Ana. lySllS aq. feet 

6. Required the area of a regular dodecagon, either of 
Trhbse equal sides is 4^ feet. Ans. 226732 sq. feet 

7- Required (bj Rule 2) the area of a regular polygon of 
16 sides, each side being 5 feet. Ans. 502734 sq. feet 

8. Bequired (by Rule 2) the area of a regular polygon of 
100 sidea, each side being 12 feet Ana. 12728'2 sq. yardi. 

Phoblkm VII. 

To find the diameter and circumference of a circle tin 
one from the other. 

Ca9E I. To find the circumference of a circle when the 
diameter is given. 

Rule. Multiply the diameter hy y, jM or 3'1416,.3iJ 
the product will he the circumference nearly. 

Cask II. To find the diameter of a circle when the at' 
cumfereace is given. 

Role. Multiply the circumference hy ^, ^ JJ, or -31831> 
the product will be the diameter. 

Note. In each of tl)e abovo nilGS the first mnltipIieF is Ibe larf 
Bccnrate; the Bcmnd tho tuoat correct; and tlie third tbe miH 
Tenient, and very Dearly as accnrato aa tho eecond: The c 

of these ruica will be given in tho Key. The answen ted* 
queations will be given as they are deduced from the MirJmnltipW 
in each rule, and the pupil can Hnd the answers hy the other 
pliers also, and thereby judge of tlie accuracy of each of the t 

Example 1. Find the circumference of a circle wbtm 
aiameterisl2inchcs. By Rulel. 31416 Xl2=37-6992Aii« 

EXAMPI.B 2. Find the diameter of a circle, the circumft^ 
encebeing20inches. Bj-Eulell. 31831 >:20=6-3662Aiil 



EXERCISES. 



i 



1. If the diameter of a cylinder he 3 feet, what ia Hi 
rcumference ?' Ans. 9-^248 f«t 

2. If the diameter of a gasometer be 60 feet, what is 
drcumference f Ans. 188'496Mi 

3. What is the circumference of the earth, its diaroeW 
hieing 7912 mllea* Kna.1.i«S56-3392 



HmffiniATiof) OF BirKPACSB. 69 

4. What is the circumference of the planet Jupiter, i(s 
ameter being (19170 miles? Ans. 280 136 472 mile i^. 

5. If the circumference of a rcruini tree be 7 4et 3 
ches, what is its diameter? Ans. 230/7^ feet. 

6. If the circumference of a cylinder be 10 feet, what is 
, diameter? Ans. 5-09296 feet. 

7. If the circumference of a circular pond be 300 feet, 
lat ifl its diameter? Ans. 95493. 

8. If the wheel of a carriage turn 528 times in a mile, 
lat is the diameter of the wheel? Ans, 3'1831 feeb 



Problem YIII. 




To find the length of an arc of a circli 

ItnLB I. When the chord of the whole arc and alt 

E chord of half the arc are given. From 8 times the 

ord of half the arc, subtract the chord of the whole 

;, one third of the remainder will 

the lengih of the arc nearly. 

Thns, in the annexed diagram 

— — - = the length of the arc 

irly; or, if the chord AB and the 

ight CD be given, it becomes 

^}AB'-tCD'— A B 

- 3 

For the demonstration of this rule, see the Key. 
CxAUPLB, If the chord AR be 36, and the height CD 
f; what is the length of the arc ACB f 
Sere AC=N/ia'+(7-5_)'=V3i{0-25=]9-5;andhence 

5x8—36 136—36 ■„ , ,. ^ ., , 

-— ^ — 5 ^40, length of the arc nearly. 

EXERCISES. 

1 . The chord of the whole arc is 50-8, and the chord of 
f the are is 30'6 ; required the length of the arc ? 

Ans, 64'6. 
i. The chord of the whole arc is 45, and the chord of 
f tlie arc is 25'5 ; what is the length of the arc? 

Ans. 53. 
i. If the chord of the whole arc he 16 feet, and the 
eht 4 feet ; what is the length of the arc ? 

Ans. 18 518 feet. 
I. If the chord of the whole arc be 24, and Ike \ie\^\.'i ■, 
at is the length of the arc ? K^tia. %1. - 



n 



I 



i 



RuLK II. When the radius of ihe circle and the nmn- 
Iter of degrees in the lire are given, or con be found from 

the data. Multiply 'be nnmber of degrees in the arc by 
the radius, and by -0174533, the product nill be the length 
of the arc. 

. Rule III. Tuke out from (Table XI.) the numbers cor- 
WBponding to the number of dfffrues, minates. and seconds, 
ia the arc ; their sum multiplied by the radius will be the 
length of the arc. 

Note I. To tlnd tlie radius and number of degrees in an arc, 

leu the chord AB and height CD are given; ni(liua=i' — ^j^ ; 

&nd Iflugant j (btg ACB)= ~ ; for — = long. CAD = lang. CEB, 

(fioo. prop. i7, cor. 1.)= lang. iCOU,(Geo, prop. 47)= lang, iAOB, . 

"which Che aro ACB ia the measure. 

. KoTK 11. The uumbei"0174£S3 given in Rule 3 is thus oblaio- i 
ad; — Since tlis eemicircumfererice of a circle whose itidiua is I n < 
S-UI5a365359, if thi3bedLviJedl)y IBO, Iha uumbHr of degreesin 
1, aemicircle, it will give the atiove number for tlie leugth 5 t is- 
gree when the rndina is L ; and since tlie circumferencEH o( citxdes an 
Ui one another as their radii, the above rule is obvious. 

ExAMPLB. Whatia the length of an arc of 20" 30' 10", 
the riidius being 16? 

Dy Table XI. arc an-loradlns ! =-3490659 
30' „ ='00ST2Cfi 

, 10" „ = ^485 

[ arc 20° 30' 10" =-3578410x 1 6=:5-72345fi. 

EXERCISES, 

J. "What is the length of an arc of a circle contsin- 
iag 49° 30', the radius being 30 inches ? Ana. 25-9itt]. 

2. What is the lengtli of an arc, of which the chord it 
20, and the radius of tlie circle also 20 ? Ans. 20-9439. 

3. Find the length of an arc, of which the chord is 36, 
and the height 12? Ans. 45-8642. 

4. What is the length of an arc, the radius of the circle 
being 40, and its height R? Ans. 51-4a 

5. What is the length of the circular arch of a bridgti 
Ihe span of (vhich is 18 feet 6 inches, and ihe centre of the 
arch aboTe the top of the piers 6 feet 9 inches ? 

Ans. 244923 fed 
Problem IS. 
To find the area of a circle. 
JiULE 1. Multiply half the circumference by the railim. 
* the product will W t\ie aiea. 



71 

RoLB IL Maltiply 3'1416 by the square of the ra^ 
dias, or '78o4 by the square of the diameter, and the pro- 
dnct will be the area. 

RuiB III. Multiply -0795775 by the square of the cir- 
cumfereoce, the product nil! be the area. 

Demonstkitioh. The circle may lia conceived to ba maile up of 
ta infinite namber of small triangles, tlie anm of wIiobq bases i» the 
ctrcumfereace, and the vertices being all in Ihe centre, the altitude 
of each of the trianglos wili be the radiua of Iha eircia; (harefora the 
aea. of all the tnangleg, or of the whole circle, will be (Prob. IL) 
hilf the sum of all the bases, or the soraicircumferHice multiplied into 
thdr common altitude, or into the nidiua of the circle, which ia 
Kale 1. 

Agam, by (Prob. 7), the Bemicircumference ia = to 3-U16r, 
whidi being multiplied by r, according to Rule I., givea 3H16r^, 
which is Rule IL; and the second form ia ^-^ x (9r)" = ■7B54rf'; 
where r ia used for i%dlua, and J for diameter. 

Also, since similar surfacea are to one another as the squares of 
tbeir like parts, (Geo, prop. 71, ear., and prop. 76, cor. 6), aud wnco 
Ihearea of a circle whose circuinference is 3'1116 ia -7S3i, tho-e- 
foM (3-1416)*: I : i-7Bo* : ■7BS*-H(3-U16)'=.-tl7S£776, theanaof 
lotrclewhoseciTCumfeFeucB is 1; hence P-.(circiimf.)'::-()79577£ : 
the area to any other circumference, which proportion bemg wrought, 
^m Role 111. 

ExAMPi.8. 'What IB the area of a circle, its radius 
bdngS? 

By Rule U. 3]4]6x(5'':=25)=78-54, the area re- 
paired. 

EXERCISES. 

' 1. What is the area of a circle whose circumference is 
333 feet, and whose radius is 53 feet ? Ans. 980i sq. yards. 

2. What is the area of a circle, whose radius is 1522 
tDcIieB, and its circumference 9563 inchest 

Ans. 50537-8 sq. feet. 

3. Wliat ia the area of a circle whose radius is 46 ? 

Ans. 66476256. 

4. What ia the area of a circle whose radios is 45 feet? 

Ans. 70^-86 eq, yarda. 

5. Wliat is the area of a circle whose cireumference is 
100 feet? Ans. 795775 sq. feet. 

6. What is the area of a circle whose circumference is 
40 yards? Ans. 127324 sq. yards. 

7. What is the area of a circle, when a chord in it is 32, 
and the height of the arc cut off 8 ? Ans. 1256'64. 

8. What is the area of a circle, when a chord at the dis- 
tance of 5 from the centre measuxes 24? Ans, 53fi'^'Jft\. 

Kors, The tbird and fourth exercises above axe m\eo&«^ ^n ^>^ 1 



uB also the Bevectli and 

I Problem X. 

k To find the area of a sector of a circle. 

Rule I. SluUiply half the length of the arc of the sec- 
tor hy the radius of the circle, the product will be the area 
of the sector. 

Note. The demouatratiou is evident from the firat case of Prob- 

BxAHPLB. What is the area of a eector, the arc being | 

10'2 feet, and the radius of the circle 14 feet ? i 

— xl4=;51xl4=71'4Bq. feet the area. ^ 

EKEBCISES. 

1 . What !a the area of a sector whose arc is 20 j^ inchesi 
the radius of the circle being 14 inches ? 

Ans. 143^ square inches. 

2. Find the area of the sector whose arc is 10 feet, the '' 
radius being 12 feet? Ans. 60 feet j 

3. Find the area of a sector whose arc is 100 feet, audi 
whose radius is 54 feet? Ans. 300 sq. yards. 

Rule II. Multiply 0087266 by the number of degrew . 
in the sector, and by the square of the radius, and the lastj 
product will be the area. 

DEHaNSTRATian. The area of a circle whose radius is 1, is 3'14M| 
tfaerefore the area ofasectorof I degree IB 3-U16-H36e=-0D872fi^ 
wiiicb b«ing multi^ilied by tbe square of the radiaa, will give the uc^ 
of a Bector of one degree to that raJiuB, and thia multiplied bj l(* 
number of degrees iu any other sector, will give the ares of iW 
■ootor; hence the rule is obvioua. 

KoTE. When the number of degrees in the sector are Dot givcHi 
they must be calculated by trigonometry from the data; and whea 
the radius la not given, it must be calculated in like maunei'. 

ExAupLB. What is th<i area of a sector of 45% th«J 
ladiua being 12 feet? 

Uere -0087266x45 gives -392697, and -392697 
(12'=:144) gives 56'048368 sq. feet, the area reqniied. 

EXERCISES. 

1. What is the area of a sector of 35°, the radius bei]i| 
45 feet? Ans. 618-497775 feet 

2. Bequired the area of the sector, the arc of wbirh ii 
30 degrees, and the diameter 3 feet? Ans. -589045 fe*; 

3. What is the area of a sector ivhase arc is a quadrant, 
or contains 90 degrees, the diaiaeteT being 18 feet ? 



4. What is the area of a sector of 60 decrees, the radius 
bdag 12 feetl Acs. 8377536 sq. yards. 

5. What is the area of a sector containing "jS degrees, 
the radius being 5 inches ? Ans. 15'70788 inches. 

6. What is die area of a sector whose chord is 24, and 
whose height is 5 ? Ana. 225'31]. 

j 7- ^tat is the ar*a of the sector of a circle whose radius 
I ii 9, and the chord of its arc 6 ? Ans. 27-5266. 

Pboblkm XL 
To find the area of a segment of a circle. 
Rule I. Find the area of the sector having the same 
Arc with the segment by the last problem. Find also the 
irea contained by the chord of the segment, and the radii 
of the sector, ^en take the difference of these two when 
the segment is lesa than a semicircle for the area of the seg- 
ment, and their sum if it is greater than a. semicircle. 

KuLB II. From the product of the number of degreesin 
the sector, multiplied into -0087266, subtract half the 
natural sine of the degrees in the sector, the remainder 
aaltiplied bj the square of the radius will be the 
the s^ment. d 

DiBionsTBaTtos. Let d= tlio degrees in 
tbe Ugte ACB, then by the secoad Rule in 
last probteni the urea of the sector ACBD is 
■OI)a7266rfr', and by Ruie 11. Problem 2, 
ii' not. sin. ACB = (he area of the triangle | 
ACB ; the difference of these expresaiona, or 
[■0087266J— i DSt, sin. ACB)!" is evidently > 
tlie area of the eegnient ABD; but Ihia is 
Rule II.; and Rule I. must be obvious from 
Ui inspection of the tignre. 

flxAMPLB. Find the area of tbe segment ADBEA, its 
chord AB being 12, and tbe radius AC or BC 10. 
(AC=10) : (AE=6) :; R : sin. ACE=36'' 52' 11-3". 
Therefore ZACB=73° 44' 22-6" =73-7396". Hence the 
area of tbe Bector=-0087266x737396x 100=64-3496. 
Again, the area of the AACB=6x8:^48. Whence the 
area of the segments 64-3496—48= 16-3490. Again, by 
Rule ir. -0087266 x 73 7396— i nat. sin. 73" 44' 22-6"= 
■643496— ■4a0000=-l 63496, which being multiplied by 
t^=100, gtTes 16-3496 for the area of tbe segment. 

1 . Find the area of tbegreater segment o£a c\ic\e,\la ii\0T4. 

being- 34. and tbe radius of tbe circle 20. Ana. \\9\'i'i\T - 

2. Find the area of a segment of whict the aic cciiiVavB* 

^3', tbe radius of the circle being 12. Ana. '2^-Wi\'2B-k, 

■- -_ M 




7* 



MKHetTBATION OP SDBFACXa. 



3. Find the area of a se^ent of a circle, the chord of 
which is 20, and the height 4. Ans. 59-002. 

4. Find the area of a seg'ment of which, the arc conl^ns 
90°. the radius ofthc circle bd:ig IW. Ans. 256-8546. 

5. Find the area of a Begmenl of which the arc contains 
270°, the radius being 15. Ans. 642-64095. 

PROBI-EM XII. 

To fiod the area of a Zone, or the space included betwetu 
two parallel chords, AB and CD. 

RuLK I. Find the area of each of 
the segments DECQ, and AEBP, 
their difference will evidently be the 
area of the zone ADCB. 

Rui.a II. Find the area of the 
trapezoid A BCD, to which add 
twice ihe area of the segment 
AFDS, and the sura will be the 
area. 




t from the previww jn- 
it ia neccBBary to find Iht 



Note I. These rules are so aolf-evidei 
lileros, that they require no demonstratiou 

Note 11. The dnta Ja often such that 
rsdioH of the circle. This can be daue aa follows; when we bin 
given the chords ABoad CD, and The perpendiculiir distance betncB I' 
them, let OP be drswu .>- to AB, then being produced it will tha te f ' 
J- to DC; draw DU |1 OE, and .-. J- to AP; draw also from the cmW | ■ 
O, OS-^ to AD, and let it meet ihe circamference in F, and thioDgli i 
S draw SR \\ AP, or DQ; then tbe As DAG and SRO hare the aA» 
of the one respectivrfy -^ to those of the other, sod are .-. rimilM. 
Now DG— the distance of the || chords, and AG= half the difir- 
eocc uf Ihe chords, whilst SR= oue-fourtli the sum of the two cboris 
AB and DC. Agnin, from the two similar as ADG- aud SRO "e I 
have DG :GA=SR ; R0= '^-, IT now d= the diatauee o£ llw ' 
chords, c= the grettter chord, <■'= the less; SR will =i(c+<j'J sud 
AG=i(i — c'), and DG=i/, stibatituting these values in the»bovo 
value of RO, we have 110= i^ and OP — OR— BP= 

'J±i^=^ _trf„ M^l(_yi-*". HcDce the centre will be wiUi- i 

out or within the *one ncooi-dlDg dh Uic diffrraxx of lie nquant <ifli' 

ckorda u yrealfT or lei' than id^, and the centre will alwajs be a) <1k 

iiatanee "+''"'^'+''^ from tbe less oliiird. 

WoTK III. T he distance OP bebg thm found, we l»«vo Ofa = 
n/ OP'+ PB*, or VOQ,*+CQ»; ulso AD = jDO* + AC*- 
Vi(o — c'j'+ii", from which the lines necessary to find the af« «• 

eusil/ fouod, when the two ehmdaaad the distance between ihenm 



ExAUPLE. Let the ^ealer chord be 96, the less chord 
60, and the distance between them 26; required the area 
of the zone. 

Here substituting the proper values in the formula for 

l.?6x36+4x{36)> _ 2913 



OP found above. 



^ 0P=; 



=14, therefore the radios is =^~(48)^ + (l4)J=50, and 
the angle contained by (he sector of nhieh 60 is the chord 
is 73" 44' 23"=73-73972, which being multiplied by 
■0087266,and by the square of the radius, gives 160K7425 
for the area of the sector, which diminished by 1200, the 
area of the triangle, gives 408-7425 for the area of the less 
segment. In the same manner, we iind the area of the 
^eater sector 3217'485, and its corresponding triangle 
672; hence the greater segment is 3'''45 483, which, dimi- 
nished by the area of the lesser segment, as found above, 
gives for the area of the zone 313(i-7425. 

BXERCISEB. 

:hord 40, and the less 30, and 
1 '35, required the area of the 
»ne. Ans. 158174. 

2. The greater chord of a circular zone is 32, the less 
chord 24, and the perpendicular distance 4 ; what is the 
ureaof the zone ! Ans. ]13'516. 

3. Required the area of a. circular zone, each of whose 
parallel chords is 50, and their perpendicular distance 50! 

Ans. 3213-485. 

4. Find the area of a circular zone, the greater chord of 
which being equal to the diameter of the circle, is 40, and 



the less 20. 






PnOHLBM XHI, 



icluded 



To find the area of a circular ring, ( 
between the circumferences of two 
concentric circles. 

£(;lb I. Find the areas of each 
of the circles, and their difference 
wll be the area of the ring. ^^i 

"lutE II. Mnltiply the product 
nf the sum and difference uf the 
diameters hj '7S54, and the product 
"ill he the area of the ring. 

DEHONHTHiiTmH. The urea uf the circle ABF, diminished bj tfae 
ina of the circle DEG, is evideatly the are^ q[ t\i« in - - - ■ 
lielweeu theii' circuiiiicreuces, uhtch is Rule 1. 







KEMSnRATION OT BDITACKS. 



Id order to show «>e truth of Rule II., \et A.B=d, DEW', then 

Rule I., tho area of the ring ia d'x78S4— if'x'7BS*=(*— d") 

^eSi=(d+d')x[d—dy7SSi, which ia Rule II. 

Example. What is the area of a ring incladed be- 
tnreen two concentric circles, the diameter of the greater 
beitig 12 inches, and that of the less 8 inches ? 

By Rule I. (12)' X 7854=1130976, area of the greater 
circle, and 8'x '7854^50'2656, area of the less circle; 
.-. 1 130976— 50 2656=62 GSS, area of the ring. 

Bj Rule II. (12+8)(12— 8)>C 7854=20x4x-7854 

■.62-832, the area of the ring as before. 

EXEBCISE9. 

3. Required the area of the ring, the diameters of whose 
bounding circles are 5 and 4. Ans. 7'068fi. 

2. The diameters of two coDcentrjc circles are 16 and 

10; what is the area of the ring included between Iheir 
circumferences? Ans. 122-5221 

3. What ia the area of the ring included between the 
circumferences of two concentric circles, whose diameters 
are 24 and 18? Ans. 197-9208. 

4. If the diameters of two concentric circles be 15 bbiI 
12, what ia the area of the space included between their 
circumferences? Ana. 63'61'ii 

Problem XIV. 

To find the area included between two arcs of eirelM 
Iiaving a common chord, when the chord and height oi 
each, of the segments are given. 

Note. If the segments be on the same side of the chord, the sp''^ 
included between the ares ia called a, loae. 



I 



, of each of the segments ACB, 
then their difference will be the 
the same aide of 
:hord, and their aum will be the area when the Kg- 
meols are on different sides oC Ihe chord. 



Rule. Find the aret 
ADB, by Problem XI., 
area required when the segments a 




Note. For the raetViod o^ (uiAin^&eTufims.^wnw* 
Note I,, Prob. VllL, tlio ni\e la ao tf 
Figures 1 and 2, that, it leiviirea ^^ "^ 



HaHSURATIOII OP SmtrACBS. 77 

LK. Find the area included between the circular 
)g the common chord 12, the height of the one 
id the other 2. 

will be found, that the radius of the circle, the 
whose arc is 3, is 7*5. and that the radius of the 
3, and that the aic of the first contains 106° 15' 
that of the second 73° 44' 24"; consequently b; 
SI., the area of the 6rat segment is -0087266 x 
927289--48000O=447289 x (7 ■5)'= 25 160006; 
:ea of the second is -0087266 X 73 74 ='643499— 
:-163499x(10)*=zl63499. Hence if the arcs 
same side of the common chord, the area of the 
be the difference of these areas, or 8-8101; and 
; on opposite sides, the jnclnded area will be the 
ese areas, or 41 -2099. 

EXERCISES. 

1 the area of the space contained between two 
ircles having a common chord 36, the height of 
leing Q, and that of the other 6. 

Ans. 7929054, or 373-58856. 
1 the area of the space contained between two cir- 
s having a common chord 30, the height of the 
[5, and that of the other 3. 

\aa. 41-7127651, or 



r 



I 
I 



3 nearly the . 
straight Hn 



ea of a figure bounded by any curve 
md two other straight lines drawn 
: perpendicular to the 



> extremities of the c 

line. 

, Let the base 

divided into any 

limber of equal 

' the perpend icu- 

6 ; a', V ; a", b", 

:h meet the curve 

ints a, a', a", &c. 

sum of the first 

lerpendicularadd 

s sum of the remaining odd peipe'n<S.c\)\M*, voA.' 

?s the sum of the even perpendVcvAana -, ftaa «>«&, 

/ bf the third part of the common A.\ft\,auce o^ "Oa* 

liars, will be the area nearlj; aud, V\ie ■u.eas.e.T- "^^ 




Wz 



prapeadiculars are taken to one another, the more exaet 
wUl the approximation be. 

EsAiiPLE. Let Ba be 400 links, AB 110, ab 115, 
tt'6'119, a'^" 125, a"'h"' 128, a"h" 13], a'b'138, a-^'*J46, 
and PQ 145 ; what is the area of the figure ? 

Here the area of the figare will be {110+145 + 
?(1I9+128 + 138) + 4(115 + 125+131 + 146)1 X 's" = 
51550 links = 2 roods 2'48 poles. 



1. If the base be 360, and it be divided into 6 eqnal 
parts, and the perpendiculars be in succession 20, 31, 43i 
57, 08, 78, and 90 ; what ia the area of the figure ? 

Ans. 199W. 

2. If the base be 196, and it be divided into 10 equal 
parts, and the perpendiculars be in succession 112, Jilt, 
93, 80, 71, (JO, 52. 41, 30, 18, and 10 ; what is the area 
oftheiigure? Ans. ]208fif 

3. If the base be 366, and it be divided into 6 eqnil 
parts, what is the area, the perpendiculars being bucmi- 
BiTcly 0, 20, 38, 50, (il, 70, and 78? Ana. imBl 



MENSURATION OF SOLIDS. 

In the mensuration of solids, the problem to be soW 
is, how often is a cube of a giren length of side containol 
in a solid whose rectilineal dimensions and form are ipwiil 
and the problem is solved in different vcaye, accoiding Ib 
the Tarioua forms of the bodies whose solidities areaoughb 

The number of times that a body contains a cubic indi, 
a cubic foot, or a cubic yard, &c., is called its solidi^. 

Table of Solid IMeasuke. 



49li7g308SI)0a 



S874960a0 1DG4S000 



/25435806105600o'iU7\Sn95'yiO(k4a\ll6m\i^'n68flOOl 613 



MESSIRATION OF SOLIDS. i'J 

Problem I. 

To find the solidity of a priBm. 

Rule. Multiply the area of the base by the perpendi- 
cular height or length of the prism, and the product will 
be the solidity. 

DeuonstbatioK. If the BuperRcial asua of the boss be foDod, itod 
I height be then taken equal to the length of ihe measuring unit, iF 
will evidently conlain Jia many cubes of tbo required dimensiooB u 
Ihe base containa superficial units of thu same length of aide; and 
fiinoe the dimenHioDH of a prism &re tho same at all diatanoes from 
iW ends, the next uuit of length woald canlain the same number of 
eabes aa the fiiet, and so on throughout ; the number of nHmn'n^ 
eaita in llie wbalo will bo equal to the atiparfidal nnibin its eud mul' 
dplied into tbo Untar unite iu its length, whivh is the rule. 

Vote I. If the surface of u prism be required, it may he found by 
ilie following 

RcLE. To twice the area of its end, add the perimeter 
of the end multiplied into the height or length of the prism, 
and the sum iriil be its sur&ce. 

Note II. Tlie boat raethoii of giving boys a competent knowlcdgo 
of the mensuratjon of solids, is lo have the various iigui-as formeil 
of wood or pasteboard, which they may be mado to measure, and 
Ihenby leam both the theory aud practice at the same time. 

Example. Find the solidity and surface of a triangu- 
lar prism, the sides of the triangular base being 15, \% 
and 9 inches, and the length five feet? 

Here the area of the base =s/lQ%^Gx^= J'ifdi'd, 
(Prob. II-, Mens. Sur,)=54 inches =§ foot; which mul- 
tiplied by 5, gives I J of a foot for the solidity. 

AgaiD, to find the surface, we have twice the area of the 
md ^J foot, and the perimeter of the end is 3 feet, which 
multiplied by 5 feet, gives 15 feet for the surface of the 
«des, to which add the area of the two ends as found 
ahoTe, and we have the whole surface of the prism 15^ 
t^oaie feet 

mXEHCISES. 

I. "Wlint is the surface and solidity of a cuhe whose side 

16 inches ? 

Ans. Surface lOj feet, and solidity 2-J-S cubic feet, 

9. What is the surface and solidity of a cube whose side 
ia 6 feet? Ans. Each 216 feet. 

3, What is the surface and solidity of a parallelopiped, 
*hose length ia 12 feet, breadth 2 feet, and ie^vV \. l«i^ 
6 inches ? Ans. Saxface 90 feet, and ao^Sit^ ^ ieaV 



«aa 



y 



i 



» 



10 

Find the surfiice and solidity of a square ptism, the 
nde of its base being 1 foot 9 inches, and its length 16 
feet? Ans. Surface 116^ feet, and solidity 49 cubic feet 
5. Find the surface and solidity of a triangular prism, 
each side of the base being 2 feet, and its length 12} feetl 
Aub. Surface 7ii'4641, and the solidity 21-6506. 
t>. Find the surface and solidity of a pentagonal prisnii 
each aide of its base being 3j feet, and its length 15 feel 
4 inches? 

Ans. Surface 310-485, and solidity 3231629 feet 

7. Find the surface and solidity of a hexagonal prisnii 
each side of its base being J 1 inches, and its length 3 feet 
9 inches ? Ans. Surfece 24>9912. and solidity 8-18664 feet 

8. Find the surface and solidity of an octagonal priam, 
each side of the base being 6 inches, and the length H 
feet ? Ans. Surface 58'414'J, and solidity 16'89849 feet. 

PltOBLBM II. 

To find the surface and solidity of a cylinder. 

Rule I. Multiply the circumference of the base by titt; 
height, and the product Tvill be the convex snrfoce, tBt 
which add twice the area of the end, and the sum will ba 
the whole surfece. 

Rule II. Multiply the area of the base by the heigbfc. 
and the product will be the solidity. .,' 

Note. The demanstration of the rules are Uie Bams as thond 
Problem 1.; the area of the base U found by Problem IX, " 
ladon of Surfaces. 

Example. Find the surface and solidity 
of a cylinder ABDC, the diameter CD of 
its base being 16 inches, and its height AC 
5 feet 10 inches. 

^ince the diameter of the base is 10 
inches, or 1^ feet, 31416xi(=41888is the 
circumference of the base, which being mul- 
tiplied by 5 feet 10 inches, the length, gives 
2i-4346, the convex surface, whilst ■7854x (lj|)'x23 
2-7925 feet, the area of the two ends; hence tttewbi' 
surface is 27'2271 feet. 

Again, the solidity is -7854 X V* X 5^=8-1448 cubic fe 
the solidity. 



I 



I. Find the snrfitce and solidity of a cylinder, the 3Sl 
meter of whose end is 3 feet, and its length 7i feet ¥ 

■ ,. Surface 8\a^Sl ^eel, oa^i wVdit^ 53-01^ 



ITIOS OF SOLIDS. 81 

3. Find the Bolidity of a cylindrical pillar of 5 feet dia- 
inet«r, its height being 30 feet, and find also its convex 
aur&ce. Ans. 58905 cubic feet, and its convex suifuee 
471-24 feet. 

3, The diameter of a circular well is 4 feet, and its 
depth 36 feet ; what did the digging of it cost, at 10a. 6d. 
per cnbic yard ^ Ans. L.8, 15b. lid. 

4. If a cylindrical stone roller be 5 feet 3 inches long, 
and 1 foot 9 inches in diameter, nhat ta its solidity, and 
how often will it turn in rolling a field containing 4 acres, 
3 roods, and 20poleB? 

Ans. Solidity 12'6278, and wUi turn 697993 times. 

Problem III. 

To find the surface and solidity of a pyramid. 

Rule I. Multiply the perimeter of the base by ka!f the 
ilatit height, to the product add the area of the base, and 
the anm will be the surface. 

RpLE II. Multiply the area of the base by one t!drd 
df the perpendicular height of the pyramid, and the pro- 
duct will be the solidity. 

DcMONHTKATiON. The B[dc9 of the pyramid are evident!)' triangles, 
ihc sum of whoBO bases is tlie perimeter of the base, and the coni- 
moii altitude of the triangles is the slant height at the pjruniid ; 
hence the Ham of all Uie bases, or the perimeter of the base of the 
ptnamid, multipliad by half the common aldtude, that is, /ui^Qie 
>^t height, wUl giro the smfaee of the eidea, to which add the area 
of the baee, and the smn will be tho nbole surface nf llic pyramiii, 
Kbieh ia Rule I. 
The second rule is demonstrated in Geometry, (prop. 1 03, cor. S-) 
Note L It is often neeasaary to find the slant height from the 
porpendicolar height, and the length of a side of tho base, the num- 
ber of sides in the base bemg also given. The slant height ia tlie 
hjpoteDDee of a right angled triangle, of whieh the perpendicular 
be^t. is one of (he sides about the light angle, and llie o^cr ia tlie 
perpendicular fmra the centre of the base upon the side; this side 
may be found by adding together tlio log. cotiui. -— (n being the 
lumber of Hides in the base) to the log. of Aa|/'the side; tlie sum di- 
mtiUHhed by 10 in the index will be Uie log, of the perjieiulicular; 
^See Prob. VI. Mens. Sutf.) henoe it ia evident from Geometry 
[prop. 3S,) how the slant height can be found from the perpendicular 
beight being given, and conversely. 

KoTE ir. If (Area times the Bolidity of a pyramid be divided by the 
*rca of (be base, the quotient will be the perfendlcular altitude. 

Example. Find the surface and solidity of a square 
ppumid, the perpendicular height being 12 feet, aa4 ejw^w 
side of the base JO feet. 



I 
I 




I- 89 MKMITHATIOK OV 

Here, in order to find the perpen- 
dicular frgmthecentre upon the Bide 
cif the base, we divide 180" by 4, 
whioh gives 45°; then C Bis equal to 
cot.45°x^DE=4DE= 5;iiEainAC 
the Blant height ia =v'AB' + BC' 
= ^144+25 = v'l^=J3' h^^nce 
the slant surface is =40 X V = 20 X 
]3=:2(iO: and the area of the base 
is evidently 10x10 = 100. The 
whole surface is therefore 260+ 
100=360 feet, or 40 sq. yards. 
Again, the solidity is ^ the area of the base, 100x4, 
third of the perpendicular lieiglit, or 400 cubic feet. ' 

£SEBCISEB. 

1. Required the surface and solidity of a square p 
mid, each side of whose base is 30, and the slant he 
25. Ans. Surface 2400, and solidity 6 

2. Required the surface and solidity of a hesag 
pyramid, each of the equal sides of its base being 40, 
the perpendicular height 60. 

Ans. Surface J2470765,and solidity 83138- 

3. Find the surface and solidity of a trianguliirpyrai 
each side of the base being 20 inches, and the perpend 
lar height also 20 inches. Ans. Surface 797705 Inc 
and solidity 11547013 inches. 

4. Find the surface and solidity of a pentagonal p 
mid, each side of the base being 2 feet, and the i 
height beiog 4 feet. 

Ans. Surface 268819 feet, and solidity 861 

5. Find the solidity of an octagonal pyramid, «' 
perpendicular altitude is 15 feet, and each side of the 
3 feet 3 inches. Ans. 2550 

Pboblbm IV. 
To find the surface and solidity of a cone. 
Rule I. Multiply the circumference of the base by 
the slant height, to the product add the area of (he t 
and the sum will be the surface of the cone. 

Rule II. Multiply the area of the base by oite-thir 
the perpendicular altitude of the cone, and the pro 
will be the solidity. 

Deuonstrition. If tlie cone be couceiTed to be so oonBtitntec 
ilH surface may be cut by a. atraiglit line pBHHOg from the vertl 
the base, its EorCacQ lieiiig lVi:aU,^en off lutd eKteaded on • [ 
will be a aeotoi- of a circ\E,liie ateti.Ql'«^vitt'«a'\«V!iiuiHi^ v 




■ HeiHtOBATIOK 0¥ BOUD^ 83 

■plying the length of its arc by half the ntdius of the circle; but its 
arc will be the cireamference of the base of the cone, ftnd the radiuB 
of the circle will be the slant height of the cone; to which, if tlie area 
of the base be added, the sum will be the surface of the cone. Hence 
Hula I. ia obvious. 

Kule II. ia dcmDDstrated in the treatise on Solid Geometry (Prop. 
104.) 

Note L To find the slant height from the perpendicular height 
nnd the radios uf the base. To tbs square of the perpendicutor 
beight add the square of the radius of the base, and the square root 
cf the sum will tie the slant height. 

NoTB II. To Bud the perpendicular height from the slant height, 
mnd the radiaa of the base. From the squars of the slant height 
■ubtract the square of the radius of the base, and the square root of 
the remainder will be the perpendicular height. 

Example. Find the surface and ^ 

solidity of a cone BCDA, the per- 
pendicular height AE being 24, and 
the diameter of the base BD 20, 

By Note I. A^=J{^W^<^ 
=26, and the circumference of the 
base wiU he 31416x20=62 832, ^ 
vhicb being multiplied by 13, half 
the slant height, gives the slant sur- q, 

&ce 816816. to (vbich add 31416 
XlO» = 314-16, the area of the base, and the Bum 
1130-976 will be the whole surface. Again, since the area 
ofthe base is 31416, if this be multiplied by 8, the third 
of the height, it giyes 2513*28 fot the solidity. 

XXERCISES. 

1. Required the solidity of a cone, whose perpendicu- 
lar height is 10 feet, and the diameter of its base is 7 feet. 

Ans. 128-282. 

2. Find the surface and solidity of a cone, whose slant 
height ia 25, and the radius of the base IS. 

Ana. Surface 1884-96, and the solidity 4712-4. 

3. Find the slant surface and the solidity of a cone, 
■whose perpendicular height is 5, and the diameter of whose 
base is 24. 

Ans. Slant surface ^90-0896, and the solidity 753-984. 

4. Required the surface and solidity of a cone, its height 
heing 36, and the circumference of its base 94-248. 

Ans. Surface 2544-C57, and its sohdity 8482 32. 
Problem V. 
To find the surface and solidity of a frustum of a pyra- 
i; mid or cone. 
' DEFjNiT/or.-. The irustom of a pyramid oi co-a^ Sa 'iaa 



i 



M 

part that remains after a part has been cut off by a plana 
parallel to its base. The part cut off will be a pyramid oi 
. cone similar to the origiual one. 

I Rule I. Multiply half the buu of the perimeters of tlie 
^ jfeno.ends by the slant height, and the product will be th( 
surface of the sides; to this add the areas of the two endl, 
and the sum will be the whole surface of the frustum. 

Rule II. Add together Ihe square of a side of each enj, 
and the product of those sides, multiply the sum by thp 
tabular area belonging to the form of the base, and the 
product by ^ of the perpendicular height, the last product 
will be the solidity of the frustum. 

Note I. Tho tabnlar area belongiBg to the form of the bueviO . 
be gal a,t Problems VI. or VII. of Meusuraliuu of Surfaces. 

Note II. The solidity c&n also be eilcuLatod by tbe foUowiof 
Tulee: — 1st, To the ureu of each of the ends of tlic fruntura, uW 
a mean pmportioDal between them, then mukiptytfae Bum b^ ^oFlta 
height, and the product will be the solidity, ^d, Flod tbo solidity otillt 
whole pyramid or cone of which the EnistDiu is a, [lajt; tlteu (h^aillt 
of any line in the base, is to the difference between this cube ut 
the cube of the corresponding line in the top, as the soHdlty of ikl 
pyramid or cone to the solidity of the fVostum. 

DkNoKeiEauo^. Since each of the aides is a Irapoioid, ite anavil 
be fooad by multiplying half the sum of its piirallel aiflaa by A) 
perpendicular distance between them, and in each of th^ ud« thi 
perpendicular distance between tho pai'allel sides is the bshk^ lit. 
IheslanCheight of the frustum; also, since the sum of all the pMoM 
aides is evidentiy the aum of the perimelera of Ihe ends of the &»■ 
tum, the whole slant surface, or eurbco of the sides, is half the IHB 
of ^c perimeters, multiplied by the slant beigbt, lu which if A> 
areas of the two ends be added, the sum will be the BUlbm «Cl^ 
frnatUDi, which is Rule I. 

Let ABCD— P be a pyramid, nnd AO 
ft frustum of it conloiacd l>etweeQ the 
pbuies ABCD and KFGH, and Ice 
EFGH— P be the pyramid cut off by tha 
plane EF6H; it is evident tliaC the frus. 
tum AG ia the difference between tho 
wiiola pyramid ABCD — P, and liio pyra- 
mid cut off EFGH— P. Let now AB„fi 
and £F=&', also let the altitude PM be 
represented by a, then since the pyraiuids 
are similar, AB : EF : : PM : PL or 6 ; 6' ; : o -. PI,= ^iMiiKp 

1 of the Iw 
, cor. S) ll* 

solidity of the pyramid ABCD — P a^pl^, and the pynouJ 
EFGH— Pis -pi^ — cs— x|, .'. the ftuBtum AG b^*'- 
fi—j _ — (6*— 6«). How \Xl 4«i iMaula t^ thfl Erattora ii •»- 




mi 




, BubstltDllng Cbis ID tile expresaion hand above for QiB aolidity 

of the frurtum, it beeorars p- ( , ) = p- (l^ + bb' f) = 
(4'+M'+6*) p- , Whicb is Rule II. The ralea given in tlio noKa 

ore easil]' derirabte from the above expression, or from the theorem 
that Gimilar Bolida ore ta one another Jis the cubes of Ihoir like parts. 
I D order to make the proof applicable to a cone, consider b and b' 
the radii Ot the bascB, and put 3'14I6 for;i. 

ExAsiFi.E. Find the surface and solidity of the frustum 
of a cone, the radius of the greater baee being 8 incheB, 
and that of the less 3 inchea, and the altitude of the fruB- 
tum 12 inchea. 

Since the altitude DG— 12 and 
GB=5 DB^ v'1^4 + 25= V'fi9 
=13, and the sum of the circum- 
ferenccB of the bases =(AB+CD) 
X3-I416 = 22x31416=69 1152, 
which being multiplied by half the 
slant height, 6^, gives the slant sur- 
face 4492488 iachee; also the sum , 
of the areas of the circular ends is 
=(8'+3"}x 31416=73x3 1416 
=229-3368, to which add 4492488, and we have the 
whole surface, 678-58o6 square inches. Again, the solidity 
i»= (8'+8x3+3^) x314I6x V ^97 X 3J416 x 4= 
1218'9408 cubic inches. 

EXEItCISKB. 

1. Find the surface and solidity of the frustum of a tri- 
angular pyramid, each side of the greater base being 15, 
and of the less 9, and the perpendicular heiglit of the frus- 
tum 12. An». Sor. 568 978794, and sohdity 763 834932. 

2. Find the surface and solidity of the frnstum of a 
S({iiare pyramid, each side of the greater base being S feet, 
and of the leas 2, and the perpendicular height 12 feet. 

Ans. Surface 1983073, and solidity 156 feet 

3. Each side of the greater end of a piece of squared 
timber is 28 inches, each side of the less end 14 inches, 
uid its length 18 feet 9 iuches; how many solid feet does 
it contain? Ans. 59'5486I1. 

4. Required the surface and solidity of the frustum of a, 
bexagonal pyramid, the side of its greater end beio^ 4. 
feet, and the aide of its less 2 feet, and the "eex^ftiiitfsiiias 



86 MSSBUSATIOV OF 80UDB. 

heJgtt of the fnisfum 9 feet. Ans. Surface 225-547-22' 
and the solidity 2I8'238384. 

5. Fiod tbe surface and solidily of a frustum of a cone, 
the diameter of the greater end beinp; 5 feet, that of the 
less 3 feet, and the perpendicular height 4 feet. 

Ans. Surface 78'5]62. and the soUdity 5I'312a fe.t 

6. What is the surface and solidity of the frustum of n 
cone, the di.imeter of the greater end being 20, that of tbe 
less end 10, and the slant height of the frustum 13 ! 

I Ans. Surface 1005-312, and the solidity 219912- 

■ PSOBLEU YI. 

To find the solidity of a wedge. 

ItuLE. Add twice the length of the base to the length 
of the edge; then multiply this sum by the height of tbe 
wedge, and again by the breadth of the base, and onB'siilli 
of the last product will be the solidity. 

Demonstration. Wlien the length 

of the base BC is equal to that of the Ejp — |F 

edge EF, the wedge U eridently equal /)'. il 

to bair s prism of the biudo base and / i\ :\ 

altitude. / l\ M 

And (iMording as the edge is shorter A / J i T)! \ 

or longer than the baao, Ihe wadge ia V' f V" " ''■ 1 

greater or leas than hnlf a prirnn of VJ— -tr ■. \ 

the Bame height and breadth with the B^ ^C 

wedge, and length equal to that of the 

edge, bjr a pyramid of the same height and breadth at the base *1Hi 
and the length of whose base ia equal to the difference of the lengUn 
of the edge and base of the wedge; putting now BC=L, AB=1. 
EF=/, and EG=y5, we will have t6/A^ziiAx(=tL=}=0=Jfta+JM 
(L— ')=iM(3/+2L— 20 = jM{2L+0, which ia the rule, 

Coa. If t=L, the rule will become ;fi/;(3L)=lML=i a prism o( 
Ihe same tiaae nnd height, as it evidently ought. 

Note. The surface of a wedge may be found bj the nilea for iho 
mensuration of Hurfnces, hy cal culating sepamlcly the arena of (h« 
bnae, sides, and ends, and addmg iheix surfaces logether for tfai 
whole surface. 

Example. How many solid inches are in a wedge, the 
length of whose base is 15 inches, its breadth 8 inchet. 
length of the edge 12 inches, and the perpendicular height 
16 inches? 

Here twice tbe length of the base, (30 in.), added to 
the length of the edge, (12 in.), is 42, which multiply by 
the breadih of the base, (R in.), and by tbe perpendicular 
height, (16 in.), and we bave 5373 inches, which being 
divided by 6, gives 896 solid inches for tbe solidity. Or 
(15 X 2+ 12) X 8 X \6-^6=8Sft feii\\i.\wcVvw. 



HElel 



MMBORATies or ■ouiM. 



e length and breadth of the liase of a vreAge are 
SS and IB inches, and the length of the edge is 53 inches; 
■what is the solidily, supposing the perpendicular height to 
be I7-I45inches? Ans. 3>]006 solid feet. 

2. Required the solidity of n tredge, the length and 
breadth of the base being 9 and 4 iucliea, and the length 
of ihe edge 1 1 inches, and the perpendicular height 10 
inches? Ans. 193^ cubic inches. 

3. R'-quired the solidity of a wedge, the length and 
breadih of the base being 26 iind 18 inches, the length of 
the edge la inches, and the perpendicular height 28 inches? 

Ans. 3 feet 444 inch ei. | 

^^B Problem ^^^H 

^^^KBnd the solidity of a prismoid. ^^^^^M 

Bulb. To the sum of the areas of the two ends aM I 

fbnr times the area of a section, parallel to and equally 
distant from both ends ; multiply this sum by the per- 
pendicular height, and ^ of the product %Till be the so- 
lidity. 



(fasr For the ■ 

NoiE 2, For the demonstration of IUg aliove rule, aee ths Key. 

ExAMPLR. What is the solid content of a prismoid, 
whose greater end measures 12 inches by 8, and the less 
end 8 inches by 6, and the length or height 60 inches? 

Here 12x8+8x6=96-|-4a^I44= the sum of the 
areas of the two ends. 

Also ^-~ X ^- X 4 =10 X 7 X 4 = 280 = four times 

the area of a section, parallel to and equally distant from 
both ends. Therefore C144+2B0)x V=''24x 10=4240 
is the solidity in solid inches, which is equal to 24537 cubic 
feet. 



1. What is the solidity of a rectangular prismoid, the 
length and breadth of one end being 14 and 12 inches, 
and the corresponding sides of the other end 6 and 4 
inches, and the perpendicular 30 feet? Ans. 17^ solid feet. 

2. The length and breadth of a stone pillar at the greater 
end are 28 and 18 inches, and the leng\.H aai \«eaA'&i «^ 



the less end 16 and 10 inches, the perpendicular height 
being 9 feet ; ivhat ia the solidity of the pillar? 

Ana. 19-75 solid feet 

3. How many solid feet of timber are contained in a 
rectangular beam, the length and breadth of the greater 
end being H'i and 20 inches, the length and breadth of the 
leBH end 16 and 10 inchea, and the perpendicular length 
24 feet? Ans. 62| solid feet. 

4. What ia the capacity of a coal waggon, the inside 
diniensionB of which are as follow: at the top the length 
DO, and breadth 56 inches; at the bottom the length 42, 
and the breadth 30 inches; and the perpendicular depth 
48inchea? Ans. 7^^^ solid feet 

I Problem Till, 

I To find the curre surface of a sphere, or any segment or 
Kone of it. 

Rule. Multiply the circumference of the sphere by tb* 
height of the part required, and the product will be the 
curve surface, whether it be a segment, a zone, a hemi- 
Bphere, or the whole sphere. | 

I NoTK I. The height of the vhole sphere is its diameter j bout 
the whale surface of a sphere is the Bquore of the diameter mslli- ' 
plied into 3-14 1 6, which ia foiir times the area of a great cink tf 
the Bphere, or of a ciide having the Buae diameter as the sphsn. ^ 

Note 11. The eurfoce of a hemisphere is twice the area of ( [ 
circle, having the same radios as the hemisphere, j 

Note III. The deiaonstration of the above rule will be givoi i 
the Key. 

Ekahpls. Find the curve surface of a segment of ■ 
sphere, the height of the segment being 5 inches, and th« 
diameter of the sphere 12 inches ? 

Here 31416x12=37-6992, the circumference of the ■ 
sphere, which being multiplied by 5, the height of the palt 
. gives 188*496 square inches, the surface required. 



I 1. If the mean diameter of the earth he 7912 mile^l 
and the obliquity of the ecliptic 23° 28', find the surface ofJ 
the torrid zone, which therefore estends 23* 28', on es^ ' 
side of the equator? Ans. 78314300 square mita 

2. The diameter and obliquity of the ecliptic being sbj*- 
posed the same as in the lust exercise, find the surface m 
the frigid zone! k.na. ftV^^AW «c>ywe inil»] 



MBReTTRATTOI* OT SOUDS. 89 

he diameter and obliquity being the same as abore. 
5 surface of the temperate zone 1 

Ans, 51041500 square miles. 
4. If the earth he supposed a perfect sphere, vrhnse dia- 
meter is "JQIS mileSj how many square miles are in its 
whole surface ? Ana." 196663000 square miles. 

FnoBLEM IX. 

To find the solidity of a sphere or globe. 

RtTLK I. Multiply the surface by ^ of the radioB, or by 
J of the diameter, aad the product will be the solidity. 

Rdlb II. Multiply the cube of the diameter by -5236, 
and the product will be the solidity. 

Note. The demonstratian of tbeae RuleB will be gJTcn in the Tiey. 

ExAMPLS. If the diameter of the moon be 2180 miles, 
and it be a perfect sphere, how many cubic miles of mat- 
ter does it contain? 

Here (Log. 2180)x3=3'336456!<3 =10015368 

and Log. -5236 = 1-719000 

Therefore the solidity is 5424600000 = 9-7U36S 

NoTB. Thia result ie given only to that degree of eisctnees that 
can bo obtained by using tbe Tables pven in tliia work; if the di«- 
taeter be uabed, and the cube be multiplied b; '5236, we will find 
the ■oliditr to be £4S46174T£-2 mUea. 

Tbia rceult differs from the last by less Iban a. three hundred 
thoosandtb part of the whole. The following answers are given as 
thc^ oan be derived by logarithms, ho also were the answers in the 
laM problem. 

BXBBCiaES. 

1. How many cubic feet are in a globe, whose diameter 
is 3 feet? Ans. 14-1372 cubic feet. 

9. How many cubic inches are in a ball of four inches 
diameter? Ans, 33-5104 cubic inches. 

3. How many cubic miles are in the earth, its diameter 
being 7012 miles? Ans. 259333000000 miles. 

4, How many cubic inches are in a globe, whose dia- 
ittetec is 16 inches ? Ans. 214466 inches. 

I 

Pboblbh X. 

LTo find the solidity of a segment of a sphere. 
RniiB I. To three times the square of the ladliw o^ \\» 
tt add the square of its height ; and thia aam m\^\A^\«& 



M 



l 



by the lieinht, and tlie product agaia by -5236, wlU give 
the aolidity. 

Rule IT. From three timps the diameter of the sphere 
mbtract twice the height of the segment ; multiply the re- 
mainder by the square of the height and by '5236. 

Note. For the demouslnitian of the abo\-e Rulea Bee the K»J. 

ExAMPLR. The radius A7t of r> 

tbe base of a segment CAB of a 
'sphere is 5 iDches, and tts height 
Cn 4 iaches, nhat la the solidity o( j^i 
the segment ? 

By Rule I. 5'x3+4"=01x4 \ ■ 

=3(i4x-5236=l90-5904, the soli- \ 
dity. '-.^ _,y 

By Rule II. The diameter of the ' 

circle is (Geo. prop. 68) '^-^ = V = ^^h which beii^ 

multiplied by 3, and tntce the height subtracted, giTM 
22| ; this again being multiplied by IG, (the si^uare of Ike 
height), gires 364. which again being multiplied by 5236. 
giTcB, as before, 190-5904 for the solidity. 

EXBRCIBBS. 

m 1. What is the solidity of the segment of a sfheKi 
I whose diameter is 20 and its height 9^ Ans. ndlWil 
r 2. What is the solidity of the segment of a sphere, tiw 
radios of its base being 10 and its height 9 ? 

Ans. 1790-4344, 

3. What is the solidity of a spherical segment, the n- 
dius of whose base is 16 and its height 8? Ana. 3485-08IB- 

4. What is the solid content of a spherical segment, tb( 
diameter of the sphere being 30, and the height of the wg- 

lent 24? Ans. 126669312. 



Pkoblbm XI. 



I > 



To find the solidity of a frustum, or zone of a sphere. 

RuLB. To three times the sum of the squares of ibe 

radii of the ends, add the square of the height of the 

zone ; multiply the sum by the height, and the product 

again by -5236, and the product will be the solidity. 

NoTK I. If the zone be in the middle nf the sphere, Ihenw 

ids will be equal, aril the solidity mnj' bo found by tlie M 

\le: — To six limes the B<\uR,re of Ihe rndiUB of Ihe eod sdi! 

foare of the height ot t^iicVuew ot Aib -inoe-, niuUk^ly the mud tf 



I MflmmATioir of solidb. 

Ihe beighl of the ione, and that prodnut by '5236, wl 
I the solidity. 

NoTB II. For the deroonstratioQ of those Ruira see the Key, , 

ExAMFLB, "What is the solid 
content of the zone ABCD, whose 
greater diameter, DC, is 24 inchps, ] 
the less diameter, AB, 20, and Ihe 
height mn. 16 inches? 

Here Dn=12, Am=10, and 
mn^ie. 

which being multiplied by 1 6, gives 
15456; thia again being multiplied 
by '5236, giTes the solidity 8092-761 6. 



EXBRCISBS. 

1. What is the solidity of the zone of a sphere, the dia- 
meter of whose greater end is 4 feet, the diameter of the 
less end 3 feet, and the height 2^ feet ? 

Ana. 32-725 solid feet. 

2. What is the solid content of a zone, whose greater 
radins is 24 inches, the less radius 20 inches, and the dis- 
tance of the ends 8 inches ? Ans. I2532'88»6 solid inches. 

3. What is the solid content of the middle zone of a 
sphere, whose top and bottom diameters are each 6 feet, 
and the height of the zone 4 feet ? Ans. 146-608 solid feet. 

4. Required the solidity of the middle zone of a sphere, 
tlte diameter of each of the ends being 8 inches, and the 
hdriLt of the zone also 8 inchest 

Ans. 670-208 solid inches. 




Problbu XII- 

To find the solidity of a circular spindle. 

RpLB. Find the area of the revolring segment ACBE, 
which multiply by half the central distance OE. Subtract 
the product from i, of the cube of A13, half the length of 
the spindle; and multiply the remainder by ]2'56Q4, and 
the product will be the solidity. 

Note I. The Burfaee of a circular spiodla may be found aa fol- 
lows. From the product of the chord of the arc and radius of the 
■arcle, subtract the [iroduct of tbo length of Ihc arc multiplied into 
the oentrol distance 0£, and the renmindor multiplied by 6-2S32 
*ai be the surface. 

Note II- For the demonstratian of theee KoV^, vx "Odr Y^v.'j. 



03 



ME!tSOB4T»(WI OP 9m,W»i 




: 44728800 
intral distance OE 



Example. The length AB of a 
drculnr spindle ADBC is 16, and 
the middle diameter CD 8 ; what A 
is tlie solidity and surface of the 

Here — — = — - — i= Y := 20, 

the diameter CF ; therefore the ra- 
dius CO=]a And AE^AO=8 
-h10=-800000, the natural sine of F 

the angle AOC. Hence the angle AOCs^SS" 7' 48". anJ 
therefore the whole angle AOB=106'' 15' 36"=106-26''. 
10&2G''x '0087266 = -9272885 

i natural sin. KM)" 15' 36" = -4800005 

difference of do. = -447288 

multipljing by R'=100 
area of Che segment 

This being multiplied by half the 

which is 3, gives 134-1864, this product being subtrscteo 
from |of the cube of AE=170'6666, gires 36'4802, whiA , 
being multiplied by 12-5664, gives ^8-42478, the soli- i 
dity. 

Ag^n, to find the surface, we have 16 X 10—2(106^ 
X -0087266 X 10x6) = 160— 1 1 1-27462= 48-72538 X 
6'2832=306-1513. 

EXKBCIBE6. 

1. What is the solidity and superficial content of a cif' 
nilar spindle, whose length is 48, and middle diameter 3B 

^iaches. . (Solidity 17-31299 feet 

■^"^- (Surface 327054 feet, 

2. Find the solidity of a circular spindle, whose length 
IB 15 inches, and middle diameter 8 inches ? 

Aqb. 433-40932 inokei. ' 

13. Required the solidity and surface of a circular epA-' 
He, whose length is 20, and its greatest diameter 15. 
i 



I 



,„ 7 Solidity 2164m 

-*"^t Surface 817-636* 



Fboblgm xni. 

To find the solid content of the middle frustum or zoiK 
, of a circular spindle. 

RuLB. From the sijuare of half the length of the »?»• 
■die, take ^ of ibe G^^u&ie o^ ^^\i 'Cat^\bQ'^\\x aC the luiMl), 



HXKSORATION OF SOLID S- 



93 1 

:tliofthe ' 



zone, miJtiply the remainder by the Baid half lengt 
zone ; from the product subtruct tbe product of ihe gene- 
rating aiea und central distance ; ihun the remainder mul- 
tiplied by 6-2832 trill be the content of the middle zone-. 

ExAMPLB. What is the eo- 
lidity of the middle frustum 
ABCD, of a circular spindle, 
ivhose middle diameter t/ih is 
18, the diameter AD or BC 
of the end is 8, and its length 
07-20? X jj» 

Join DC and Dn; produce \]. 

nm to 0, the centre of the 

arc EDCF, and join EO ; then DC is eTidently =or=20, 
and TM is =^^(m» — AD)=5. Again, Dm'^sDi'-J-jm*^ 
100+25=125, and DK«-§-2rMi= the radius of the circle 
=125-^10^12-5, hence, since nv=:9, the remaining part 
of the radius or central distance ^=3*5, and therefore the 
square of Ei=(12-5)''— (3-5)2=16 j(9=144, Euisthere- 
&ne=12. 

Now, the generating area is the area of the segment 
CDn, together with the area of the paralleloOTam Dr; but 
the area of the segment CDm (Mens. Surf, Prob. 11,) = 
69-88875, and the area of the parallelogram D*- is 20x4 
=80. Hence the generating area =69-88875+80 = 
149-88875. Now, by the rule 12 "-4(10) '=144— 33^ 
=110-6666, this multiplied by 10, half the length of the 
zoae, gives 1106'666; from this But)tract the generating 
area, multiplied into the central distance 3*5^524 6106, 
and there remains 582-053, which being multiplied by 
6-2832, gires the solidity of the zone 3657-174+&C. 



1. If a cask in the form of the middle frustum of a cir- 
cular spindle have its bead diameter 24 inches, bung dia- 
meter 32 inches, and length 40 inches, how many cubic 
inches does it contain? Ans. 27287-90178. 

2, What is the solidity of the middle zone of a circular 

n'.le, whose length is 25, greatest diameter 20, and least 
eter 15 inches ? Ans. 385537 ft-et. 

Problem XIV- 
To 6nd the solidity and surface of a circular ring. 
Btn,B I. To the inner diameter add the thickness of the 
ring, multiply the sum by the square of Ike tbicVaftw, aa.\ 



94 lassuKATttm ta souoe. 

that product again by 24674, the last product will be the 
solidity of the ring. 

RuLR II. To the inner diameter add the thickneSB of 
the ring, multiply the sum by the thickness of the ring, 
and the product again by 0-8696, the last product will be 
the surface of the ring. 

Note. The demoDstration of (he aboTo Rules will be found iii llw 
Key. 

Example. What is the solidity 
and surface of a circular ring, 
whose thickness AB or CD =4, and 
the inner diameter BC 8 inches? 
Here the inner diaraeter + the ^i 
thickness is 8+4=12. which being 
multiplied by the square of the 
thickness 16, and that product by 
2-4674, gi»es the solidity 473-7408. 

Again, for the surface we have 8+4=12x4=48x 
9-86i«3=473-7-108, so that in this example the surface 
and solidity are both expressed by the same number, but 
this can only be the case when the square of the thicfcnea 
is equal to the thickness, or when the thickness of the riijg 
is 4, as in the present example. 




1. What ia the solidity and surface of a cylindrical ring, 
whose inner diameter is 14, and thickness 6 inches? 

.^^ J Solidity 1776 528 inchefc 
■^''^- (Surface 11 84-352 indiM. 

2. Reqnired the solid and superficial content of a cylin- 
drical ring, »hose thickness is 9, and whose inner diame- 
ter is 31 inches. , J Solidity 7994-376 inche*. 

■^''^■\Superiicies 3553-056 inche*. 

3. What is the solidity and Euperficial content of a tj- 
lindrical ring, whose inner diameter is 15 inches, aid 
whose thickness ia 5 inches ? 

, /Solidity 1233-7 incheB. 

P ADS. ^ s^pg^jjigg ggg^g j^y. 

Problem XV. 
To find the sorface and solidity of each of the r^nUr 
bodies. 

Dbp. a regular body is a solid bounded by plane free 
that are similar surfaces. 
The whole number of regular bodies that can be foroicJ 
in tbiB masner U &ie. 



MCMSUBATIOll OF BOLID8. 



96 



] . The Tetraedron, or regalat pyramid, viid lia« fonr 
equal and limilar triangtUar faces. 

% Tbe Hexaedron or cube, irhich has six equal square 
hcea. 

3. Tbe Ootaedrou, which has eight equal triangular faces. 

4. The Dodecaedroo, which has fire equal jfenlagoiial 

5. The Icosaedron, which has twenty equal and similar 
triaagnlat faces. 

If the following figures be drawn on |iastehoard, and the 
lines on the first and third row be cut half througb, and 
then folded together, they will form the five regular solids 
lepreseiited by the figure» in the second and fourth lowi. 




A Q ^ 





The following is a table containing tbe eurfaces and 
scdidides of each of the fiye regular bodies when the Uu%ti^ 
<tf id edge ia one. 



toKtoBJtxwK- OB aauaa. 



S0I.IIHTIE8 aSD Sl-RI'ACHS OF HEUIrt..K BoOtES. 1 


1^^^ 


K.,™. 


Solidiiiei. 


Snrf.™. 


6 

12 
20 


HeiBBdrulL 
Octaedrob. 
Dodecttedron. 


■117861 
l-OOOOOO 

■471+05 
7-668119 
2-181695 


1-732051 

6-000000 
3-464103 
20-645739 
8-660254 



Bole. Since Himilar surfaces are to one another as 
squarcB of their like sides, and since similar solids an 
one another as the cubes of their like sides, the saifaoe 
the table above, multiplied by the square of the lengi 
the edge of anj aimilar figure, will give its surface; 
the solidity in the above table, multiplied by Hie cub 
the length of the edge, will give the solidity. 

ExAMPLs. Find the surface and solidity of a regi 
octaedron, whose linear edge is 5 ? 

The tabular surface of an octaedron is 3-464102, wli 
being multiplied by the square of 5 or 2o, gives 86'609 
the surface required. 

Again, the tabular solidity of an octaedron is '4714 
which being multiplied by the cube of 5, or by 125, gi 
58-925625, the solidity required. 

I!XEBCISEB. 

1. What is the surface and solidity of a regular tetE 
dron, the length of its edge being 6 inches? 

, „ f Solidity. 25-455816 indi 
■*°^\ Surface. 62-353836 inch 
. What is the surface and solid content of a kexi 
dron, the length of whose edge is 4 inches ? 

. „ J Solidity, 64 inch 
■*''^\Surfiice, 98iDch 

3. What is the superficial and solid content of a d« 
caedron, whose linear edge is 3 feet ? 

, (Surface, 185-811561 fs 
■*"^- \ Solidity, 206-904213 fs 

4. What is the solid content and superficies of a regu] 
jctaedron, whose linear edge is 10 inches ? 

.^, (Solidity, 471-405 inche) 
^"^^ i Superficies, 346-4102 inchi 

5. What is the surface and solidity of a regular icosi 
dron, whose linear edge is 8 inches? 

. f Surface, 554-256256 ioehi 
■^^- V^Vi4A^A\\TQ2ria4 iachi 



UiftD 8tniT£7iira: 



LAND SURVEYING. 



Land Surreying is the art of fiading the extent of area 
of a field or an estate. 

Land is commonly measured hy a chain, invented bj 
Edmund Gunter, and hence called Gunter's chain. It is 
66 feet, 22 yards, or 4 poles in length, and is dirided into 
100 equal links, so that each link is 7-92 inches. 

An acre of land is one chain in breadth and ten chains in 
, length, or ten square chains^ and therefore contains 100,000 
sqoare links. 

An acre of land is also 4840 square yards, and is divided 
into 4 roods, each rood is divided into 40 poles or perches, 
and each perch into 30| yards. 

The chain must always be accompanied by fen arrows, 
to the top of which is generally attached a small piece of 
red cloth, to make them visible ; and the surveyor will 
often find it necessary to have a few poles, with smaU flags 
attached, called station staves, to place at comers, or other 
particular points in the field or estate that he may be sur- 
veying, to render these points more visible from a distance. 
The surveyor will also find it very convenient to have a 
staff or rod, divided into 10 eqiial parts, corresponding to 
10 links of the chain, with which to measure off-sets, or 
short distances, between the chain line and some bend or 
coiner of the field. 

A cross staff, which consists of two set of sights at right 
angles to each other, placed on the top of a Staff, to assist 
in finding the point on the main line, from which a per- 
pendicular to any bend or comer ought to be raised. 

The surveyor who has the above instruments is prepared 
to take such dimensions of an estate as will both enable 
him to lay down a correct plan of it on paper, and also to 
find its area, which are the two principal things required 
of him. 

FnoBLBu I. 

To measure a field of three sides. 

At each comer of the field place a station-staff; then if 

the sides be all straight lines, measure thcra, and the area 

j can be found hy(Mens.ofSurf.,Prob.n.); or, by the as- 

siitance of the cross-staff; find where a petpenditwlM. 

from one of the corners would fall on one o^ \,\ievL&^,v^>i^ 



98 TUAim itvHvmttxm. 

measure that side, and the perpendicular upon it, from the 
opposite comer, and the area can still be found by the 
same problem, and a plan of the field can be drann Irom 
either dimensions. 

Note 1, Tbc dimeDBionB of the Egur« cau be laken in both v*js, 
Hnd if the area calculnted fnitii each be the same, the dimeaBtona 
have been carrectly laken ; if not, it should be measured again, t 
find where the mistake has been committed. 

NoTB U. Id order to measure a line, the assistant takes nid 
arrows in his left hand, and an end of the cbsin and one arrow i 
his right; (hen advandug In the place where he is directed, the fol- 
lower makes agaa to him wilh his hand, tUl he is exaetljr in a list 
witli the BtaUon to which he is advancing, and at the end of tbc 
chain he sticks into (he earlh the arrow which lie holds in hia ri^l 
hnnd; they then advance, the leader taking another arrow into his 
right hand, until the follower cume to the Hi^ arnin-, which he taka 
up with his chain Imnd, the leader at the samo time putting down 
second Brraw with his right hand ; they thus proceed till all II 
arrows are in the hauds of the follower. Tliis tslies place at the md 
uf the eleventh chain. The follower then advaucea to tlie leader, 
and counting the arro»s, delivers them again to the leader, 
places one down at the end of the cham, and then advances as be- 
fore, the follower at the samo time marking in his field book the n- 
etuDge tlial has token place, leal, if the line be long, an error of la 



Example. Required the 
lar field, ivhose three sides ar 

Here if we lay down the 
plan on a scale of 10 chains 
to the inch, and then on the 
plan measure the perpendi- 
cular on the side which is 1 4 
chains, it will be found to be 
eiactly 12 chains; hence 
the area of the li eld Is !■«» 
X 600=840000 square links, 
which being reduced to acres, 
gives 8 acres, 1 rood, 24 poles. 
The same result will be ob- 
tained by finding the area fro 



ind plan of a trianga- 
13, 14, and 15 chains. 




I (he three sides. 



EXERCISES. 

1. What is the area of a triangular piece of land, iH* 
three aides of which are 720. 609, and 1044 links? 

Ans. 2acres,Oro.,2I4784r' 

2. What is the area of n field in the form of an equil.i- 
teral triangle, each of its sides being 925 links? 

Aaa. ^ ■AM^s.'i TO.. la-iasa perch* 



-LAND SURVETII^Q. "96^ 

3. What is the area of a triangular field, one of its 
sides heing 546 links, and the perpendicular upon it from 
the opposite angle 432 links, the perpendicular falling at 
the distance of 230 links^ from one extremity of the base, 
and 316 links from the other; required also a plan of the 
field ? Ans. 1 acre> ro,^ 28*6976 perches. 

Note I. When the field is triangular, hut not hounded hy straight 
lines. Fix, as hefore, station-staves in the three corners of the 
field; measore the three lines hetween these three comers, and the 
perpendicular distance of the hends in the sides of the field from 
these lines, at the same time carefully marking the point of the main 
line on which the perpendicular falls; the field will then consist of a 
triangle, and on its sides several small triangles and trapezoids, the 
areas of which must be added to the area of the principal triangle, 
to get the area of the whole field; but if the boundary of the field be 
on the outside of the main lines, such parts as may lie within the 
main line must be deducted. 

Note II. In measuring such a field, it is most convenient touse 
a field-book, consisting of three columns; in the middle column are 
marked the distances along the main line to any comer from which 
a perpendicular is to be measured, or to where the boundary of the 
fidd leaves the main line; then the perpendiculars from the various 
comers or bends in the boundary are to be marked in the other 
eoiunms, observing always to place the perpendicular in the right- 
hand column when it is measured to the right of the main line, and 
in the left-hand column when it is measured to the left of the main 
line. 

NoTB III. Begin at the bottom of the page and mark upwards, 
stating at the beginning what side of the field is first measured, and 
m what direction you proceed, and at each comer state whether you 
turn to the left or right. Mark also where the main line crosses any 
hedge, ditch^ or river, &c. The above directions being attended tu, 
it will be easy to lay down a plan of any field from Uie field-book, 
and also to calculate its contents, but it is sometimes convenient to 
draw a rough sketch of the field in the field-book, and mark .the va- 
rious lengths on this plan^ both of the main lines, and also of the 
perpendiculars upon them, from the various bends or comers in the 
boundary of the field. 

Example. Lay down the outlines, and find the area 
of a field, from the following field-book ? 





^^B* 


AC 


^^^H 


■• 


954 




840 


^^^^H 


^^V 154 


560 


^^^^H 


^^^ 


208 


^^^^H 


^^^^^m 


000 
RoffC 


1 


BC 


^^^^^^t 


694 




^^^^^^ 


436 


^^^^1 


^^^^^t 


000 
RoffB 


^ 


AB 


^^V 


785 




^^H Begin at 


A 


and go west. 


^^B perp. on the left 


base line. 


perp. on the right. 






^^H Area of the triangle 


ABC, n 


^^V ftund from its three sides. /N-^ 1 


■ 78i //N?\ 1 


^B mi f 


'i \N 


^H 


\A 


H 2|24:« / \\ 


^H tog. 1216o=3-085112 / \\ , 


^H „ 431-5=:2-634981 ^ '^ 


^H „ 522-5=2 718086 Offset A on BC. 


H ,. 262-5=2-419]29 347x69= 23943. 


^M 2| f0857308 Offsets on AC. 


^■Xce. 268320 =5-428654 ;0D „^ 


^^W^^ — x74 = J6S)li 


H ^';^"X352_ 40128 


H. ^^^;^^X280_ 29400 ■ 


^H 56x57= »Id9 


^^^^ 1043S9 


^^^^^B AreaofAABC = 268330 


^^^^^V AVLole 372679= i 


^^^^^^^^^^ 3 acres, 2 roods, 36-S8S«fM 



lAWB SUBVBTtXfl. 


^ 


ESRRCISRS. 




Draw a plan of a field, and find its area, froit 


the 


ing field-book. 






CA 









1252 






37 


lUOO 






69 


824 






45 


716 






72 


610 






15 


424 




^H 


< 55 


212 




^H 





000 
EoffC 




1 




BO 




■ 




683 


•* 






636 


40 


^H 


• 


354 


64 -* 


^H 


' 


229 


49 ; 


^H 




000 


^ 






KolTB 




■ 


AB 


■ 





973 


II 




48 


745 




^H 


76 


600 




^H 


56 


495 






25 


256 




^H 





000 






Begin at 


A, and 


go north-east. 





Ana. 3 acres, 3 ro., ]9'0648 perchea. 
Construct the field-book, draw a plan, and find the 
Df afield, from the following description: beginning at 
)Uth comer of a triangular field, where the station was 
: boundary, and proceeding along the west side lo- 
9 the north, at the distance of )50 links, the ofTset on 
oandary to the left was 46 links ; at 210 links farther 

the perpendicular on the boundary towards the left 
tl links ; 1 60 links farther on the ol^act on the same 
iraa 34 links ; and the north-west corner also in the 
lary was 154 links farther on. Then turning to the 
, the main line continued in the boundary to the dis- 

of 209 links, and at a distance of 120 links farther 
in inset to the vertex of a triangulat \«ni Va W"ff 
iarf iras 48 links; 140 links fatlhet oiv V\it \a 



i 



103 •LAtm euRTETniGi 

again coincided with the boundary, and continued 30 to the 
end, which was 113 liiilcs farther on. Turning again to 
the right, the boundary of the field lying towards the left 
of the main line, throughout its whole length, at 100 links, 
the offset was 20 links, at 200, 34 links, at 300, 50 links. 
at 400, 65 links, atiiOn, 30 links, at 600. 12 links, and 

_ the whole distance Co the first station was G64 links. 

■ Ans. 2 acres, ro., 32-47 poles. 

PfiOBLBM II. 

To measure a field of four sides. 

KuLK. Measure a diitgonal and the perpend iculnn 
upon it from the other corners. Or, measure a diagonal 
and the font aides, then a plan of the field t ' 
stcucted, and the perpendiculars measured on the plan, or 
the area of each of the triangles may be found from the 
three aides. 

Note. Id nrdpr 1« prove tlie carrectnesa of the dimenioiii, 
mcBAurc the other diagoiukl; tben if its length on the plao corre^HHid 
with its length as measured on the field, the dimensions have been 
token correctly; if not, there baB been an error cooiniitCed, which 
most be dieeovered liefore proceeding fartlier. 
EXEItClBRB. 

I. What is the area of the field represented in the fol- 
lowing diagram, and whose dimensicws are as stated be* 

AE 238 BE 229 B 

AF 496 DF 388 
AC 670 

Ans. 2 acres, ro., 
10-7I2 poles. 

3. Bequired the 
area of a field, the di- 
mensions of which ta- 
ken as in the abore 
are AE 460, AF 742, 
and AC 984; the per- 
pendiculars being BE 
:W9, and PD 361 
links, and a proof diagonal BD 72? links. 

Ans. 3 acres. 1 ro., 7-424 perchw 

3. Required the area of a field, the dimensions of 

which are as follows, (the point F being in the diagtam 

nearer to A than the point E), AF 280, AE 594, AC JO20, 

and the pcrpendicuIaiB being DF 460, and BE 340. 




I.AND 8DRTETIN0. 



103 



■ARK. When the four sides are not straight lines, it is neces- 

go round the houndary, measuring the four sides, and all the 

1 and insets, from the angles of the boundary on the main lines; 
neasure a diagonal, whic& will enable you to draw a plan of the 
on which the perpendiculars can be measured from Ihe comers 
) diagonal, and the area of the two main triangles calculated, 
idi add the spaces indicated by the offsetSi and deduct those 
ted by the insets, and the result will be the area of the figure. 

Draw a plan, and find the area of a field, from the 
wing field-book ; A being at the south-east comer, B 
e south-west comer, C at the north-west comer, and 

the north-east comer, and the four stations being all 
e boundary of the field. 



- 


BD 






1420 


Dia((onaI 




1000 


proof line. 


am 


toB 






AC 






1500 


Diagonal. 




1000 


• - 




Bo£PA 






DA 






1200 






1000 






750 







580 


94 




318 


60 




000 







RoffD 






CD 







838 




144 


490 




86 


300 
000 
RoffC 






BO 







1156 




40 


1000 




60 


735 




74 


530 







340 
RoffB 


1 



IiAIIB SBSVXTfira.' 



14 



Begin ut 



AB 




922 





806 


24 


710 


36 


540 


Cross 


4S0 




350 




17t> 




000 




A, and 


go west. 



Ans. 10 acres, 3 ro., 36-4096 perehcJ. 



^^H BuLS. Bj means of diagonals, divide the field into 
triangles and trapeziums, and find the areas of eacb of 
these figures separately, and add the results togcthei fui 
the whole area. 

Note. A five-sided tigure can be divided Into s trapezium and ) 
ttioDgle ; a Bix-sided figure cau be divided into Iwo traj>eziuma ; > 
seveu-aidad Rguro can bo divided intu two trspeziuniB and a triuglel 
and gencniU;, if r be the number of udes in a figure, it can be ^- 
vided into j — 1 trapeziums; and it the i^uotient contain a iaH, il 
representa a triangle, 

EXBRCIStS. 

1. Required the plan and area of a field from llit 
following dimensions ? 



Pboblbh III. 




.LARD SUBTBTIFO- 

DA I Disg- 
1756 i 

1280 . *S4 B. 
872 ■ 
R off D I along AD. 



C524 
Prom 



Bonth-east to D. 

Ans. Ifi tWFM, I 10. 25 '376 perchei. 

Required the plan and area of a field from the follow- 
field-book and accompanying sketch ? 



B380 



F({68 
Begin at 



1)80 

600 

E, and 



Diagonal 
440 D. 



along AC- 



120 D. 

go north-west to A. 



Ans. U Bvem 3 nods, 26'592 perches. 




SCRVETTIIO. 



3. R«quired the pliin nnd area of a six-sided field, from 
the following field-book ? 




Phoblkm IV. 

To survey a piece of land in the form of an irregular 
belt. 

RcLB. Select tiro stations one at each end of the belt, 
such that A straight line joining these stations may be 
wholly within the belt, then measure this line and the off- 
sets from it upon the several bends or angles in the boun- 
dary, and calculate the area as before directed; if two sta- 
tions, one at each extremity of the belt, cannot be found, 
such that the straight line between them will fall within 
the belt, divide it into several parts, such (hat each part se* 
parately can be measured as directed above, then the sum 
of the areas will be the area required. 

KXBRCISKS, 

1. Hequired the area of a belt from the following 
£dd'book and sketch. 



fTseta on the 




left of AB 


AB 


230 


1118 


223 


970 




952 


190 


824 




750 




620 


288 


550 


310 


400 


300 


280 


260 


154 


200 


000 


Begin at 


A. and 



offsets on the 
right of AB 



Ana. 5 acres, 3 rooda, 2195^ ^eiiTcLCT. 





Hbe LAKn B«trveit*4. 






B 2. Find the area of a field from the foUowing field-Look. 




V 210 


1940 









W 208 


1800 


12 






■l 202 


1600 


26 






^B^^-. 214 


1400 


30 


! 




^^^^B- 


1200 


48 




^^^^B 


1000 


72 




^^^■l^ 


800 


117 






^^^B 110 


600 


145 






^^H 


400 


180 






^^K 


200 


206 






^^r 


000 


222 






Ana. 4 mhm, 3 roods, 2-72 po'ea 




1 


3. Find the area and draw a plan of an irregular Geld 






from the folloivtng field- 


ook. 









3425 

3200 235 






436 


3150 

3012 350 






*- 576 


2R24 

2720 290 






700 


2400 

2100 500 






642 


1830 

1424 413 






730 


1200 






^ fil4 


956 402 








600 308 






412 


412 

306 218 









000 






Ana. 29 acrws, 2 ro., 7 532 poles. 






PSOBLKM V. 






To measure a lake, mere, wood, or any field without 






entering it. 






RuLB. By the belp of a cross-staff or any other meant. 






fix three poles near to three corners of the figure, so tlat 






lines joining them may contain a right angle, and place a 






fourth pole in aucK a. ^oailitwi that it can be seen from eadi 






of the extreme poles &tsl \.\a.^X«4, mi w^ -oasa m, --^waible 




^ 














1 



LAKP SUETSTlNe. 



Ii0 



to another corner of the figure; then go round the four 
sides of the trape^nm thus formed and measure its sides 
and the insets on the sides of the figure. Construct the 
trapezium and find its area, from which subtract the space 
indicated by the insets, and the remainder wiU be the anea 
of the figure. 

Note I. The right angle in the figure is necessary to enable the 
surveyor to obtain the diagonal without any trigonometrical ealeula- 
tion; but if he has any instruments for measuriBg angles, he may 
place the poles in any oonvenient fMMition, and then measure one of 
the angles, and this with the data taken as above, will enable him, 
by the help of trigcmometry, to find the area of the figure, and to 
inw a phui of it 

NoTB II. The angle contained between any two lines may be 
measured by the chain or any graduated line in the following man- . 
ner. Measure 1 00 links or any other convenient distance along each 
of the lines containing the angle,, then measure very carefully the dis- 
tance between the extremities of these measured distanees, and half 
of this distance divided by the former, will give for quotient the na- 
tval sine of half i&e contained angle. 

Example. Find the area of a wood from the following 
field-book, the angle in the surrounding trapezium at A 
being a right angle. 



BA 






DC 




1150 







950 





990 


110 




450 


120 


850 


100 




000 





eoo 







RoffD 




250 
000 


50 











AD 




RoffB 






1550 
1000 





CB 






650 





1340 







350 


50 


940 


60 




IJO 


160 


640 


50 




000 





290 






Begin at 


A, and 


go east 


000 


, 


RoffC 




Ans* 12 acres, ^ 


> roods, 19 


1 poles. 



PnOBLBM VI. 

To find the breadth of a riyer or other obstacle without 
crossing it. 
^let She adjoining figure represent a river the breadth of 
L is required. 



w 

^^^1 Bare 



LAVt> AFfiVdrt^ffi. 




■From B to C mea- 
gure any distance _^_^ 
BC perpendicular to ^M 
AB. and make CD ^^^^^ ^^ ^J 
=BC, then measure ~ 

from D perpepdicu- 
lar to BD, till it 
meet the line AC 
produced in E,whict 
will be known by (he points A and C, at which ranrh 
must be placed, appearing in Ihe same line; then DE a 
equal to AB. For (lie triangles ABC and EDC have ihit 
angles at C eqaal, being vertically opposite and (he an^es 
at B and D right angles; also BC=CD, therefore (Geo 
prop. 6,) AB=DE. 

A'wOier method. Measure a line from B to C perpen- 
dicular to AB, and mark the point F, where a perpendicu- 
lar to AC meets AB produced, and measure BF; then bt 
(Geo. prop, (il ), FB : BC : : BC : BA, or BA ia equal to 
the square of BC, divided by BF. 

Note. If the impediment on the line were & house w taj other 
such obstacle, which only extended Tor a, short way; we can measun 
a line perpcndiculai' to tha line we want to mcaanre, till we »re bt 
jond the obetacla ; then measure perpendicular lo Iho but line, or 
parallel to Ibe original line, till we are past the obstncle ; if now we 
measure a perpendicular to this last line equal to the lirst perpendi- 
cular, and on the same side of Che BMond, we will have again ratorn- 
ed to the line we were firet mcasaring; and the diataaco of the point 
where we first left it and that at which we again returned to il Bill 
be equal to the second measured peiTcndicular, for Ihey wiJl be op- 
]ioeiIe sides of a pamllelagrani. 

DESCRIPTION OF THE PLANK TABLK. 

The plane table is a 
plane rectagular board. 

bogany, and liaving a 
frame, by means of which 
a sheet of ilrawing paper i| 
may be fised on it. The ^^ 
frame is generally gradu- 
ated to as»>ist in laying olf 
distances. The table is 
placed on a tripod stand, 
and then made perfectly level, by n 
It has also a compnss B aUa.c\ii;4 




IS of adjusting Bcrem. 
it, \o fttvable the sut- 



■ UKs sokmiNG. Til 

tefor to place it in exactly the same direction at rarious' 
stations, where he may require to uae it. It ia also furnish- 
ed with a moveable index, marked I, I, on the diagram, 
commonly about 18 or 20 inches long, made of brass, and 
should be finely graduated, to assist in laying oif distances. 
It has upright sights at each end, also made of brass, and 
about five inches high, through which the poles at each 
station are seen, when the index is directed towards it. 

By means of this instrument, a. plan of a field or an estate 
can be laid down on the ground in such a manner, as Co 
determine the principal points, from which the plan can 
easily be filled up, and the contents calculated with conside- 
rable accuracy, by measuring the necessary perpendiculars 
' and other lines with a scale of equal parts. 

Prdble.u "Vll, 

To survey with the plane table, by going round the field, 
and planting the table at each of the corners of the field. 
Let the table be placed , j-, 

' as at A, having a sheet of ''. 

paper properly adjusted to 

, receive the plan. Place 

j the table level, and turn 

I it round till the compitss 
needle settle over the 

I fiewr-de-Ui, or north point; 

;| then having placed poles 
at each of the corners, lay 
the chamfered edge of the 
index on the point A, and 
turn it round till the pole _ 

I at B be seen coinciding /' 

] with the hair in the other 
sight ; then having first observed that the edge of the in- 
dex is still on A, draw a Hne from A to B along the edge 

' of the index, and make it equal to 328, the measured dia- 

' tance from A to B ; then remove the table to B, and plac- 
ing its centre over the hole where the pole was, adjust the 
table as before, and placing the index on the line AB, 
turn the table round till the pole at A coincide with the 
hair in the Bight of the index, when the needle of tlie 
compass will again settle over ^a fieiLr-de-lis as before, un- 
less acted upon by some local attraction. Apply now the 
index to the point B, and turn it round till the ijole at, C 
coincide with the hair ot the index ; measaxe We iiv4\a.t\'ijs 




113 I^ND BtnmeTIKUIi 

from B to C 310, and drawisf; a line along the edge of the* 
index, make it equal to BC 3J0. Remove the Cable iiLtlie 
same manner to C, and adjuat it as before ; place the in- 
dex ott the point C, and turn the index till the point D 
coincide with the hair as before; draw the line CD 355; 
then remove the table to D, and hnving adjusted erery 
thing as before, make DE 318; lastlj-, remove the table 
to Ej and having adjusted it carefully as before, apply the 
index to the points E and A, then will the pole at A be 
seen coinciding with the hair in the sight of the indes, and 
the distance from £ to A, measured on the scale, will be 
equal to tlie distance measured on the ground, if no mis- 
take has been committed. Bj this means we obtain m 
exact plan of a field or an estate, and at the same time 
prove its correctness, by the two above-mentioned tests. 

If it were required to determine the position of any ob- 
ject, either within or without the plan, if not at too great a 
distance, this can easily be effected, if it be visible from two 
stations ; for we have only to apply the index at each of 
the stations where the object is visible, to the point of the 
plan representing that station, then turn the index till iJie 
required object coincide with the hair in the sight of the 
index, then draw a line along the edge of the index, and 
the object, will be situated somewhere in that line ; do tLe 
same at the other station from which the object is viable, 
and the intersection of the two lines will evidently be the 
position of the abject upon the plan, and its distance fiom 
any other point can then be measured on the plan. It J» 
evident that in this way any number of points can be de- 
termined, if they be visible from any two stations, so that 
the surveyor can determine as many points as he pleuM 
on the plan, and then measure their distances from bb; 
other points that are laid down on it. 

In the above plan we have then AB 328, BC 310, CD 
355, DE 318. EA 450, and the lines CE and CA aa 
measured on the plan, are found to be 526 and 506 n- 
spectively, from which the area of the field can be deter- 
mined ; or the perpendiculars on CE and CA can be 
measured on the plan, and the area more simply detennin- 
ed from them ; and it will be found that the field contuu 
2 acres 13-56 poles. 

The plan of a field or an estate can also be laid down by 
the assistance of the plane table with less labour than tbat 
described above, if two stations can be found either within 
or without it, whose distance can be ascertained, and from 
each of which all iW c(»m«xa «i *.V fisU. at estate can b« 



Ken; for we can place the table at one of tbe statiaos, 
and having made it level, and turned it round till ihe 
needle stand over the feur-de-lis, place ihe index on the 
point representing the station at which the table is placed, 
and tore it round till the pole at the other station appear 
to coincide with the kair in the sight of the index, then 
draw a line along the edge of the index, and, from a scale 
of equal parts, mnke it equal to the distance between the 
stations. The index being still kept over the point which 
represents the station where the table is placed, turn it 
round till it coincide with the direction of each of the cor- 
ners in succession, and dran- indefinite tines through that 
station in the direction of each of the corners ; then remove 
the tahle to the other station, and place its centre exactly 
over the hole in which the pole stood ; adjust the table as 
before, applying the index to the line representing the dis- 
tance between the stations, and, lookiiig back, make th& 
hMr in the sight to correspond exactly wiili the pole in the 
first station. Apply now the index to the point of the 
plan representing the second station ; turn it round till the 
poles at each of the comers appear to coincide with the 
hair cff the sight of the index, drawing at each coincid*nert 
an indefinite line in the direction ot the comer, through 
the station-point, the interaections of the lines drawn 
through each of the stations, towards the comers, will givn 
the proper position of the comers upon the plan, and these 
being joined, the outline of the plan will be obtained ; and 
the other lines which may be necessary for findingthe area, 
or tliose already laid down, may be measured on the plan 
with compasses, on a scale of equal parts. If any other 
points are to he constructed on the plan, their bearings 
should be laid down at each of the stations, in the siime 
manner as the comers. By this means the surveyor ob' 
tains A correct outline of the field or estate which he is 
tarveying, and also the position of any particular point 
reqnireil ; he can then by the chain measure any distances 
that maybe npcessnry for the more minute tilling up nf 
his plan, and then fiuish it off in any style that may be 
necessarj. 

A proof of the accuracy of the plan may be obtained, 
either by measuring certain lines on the field with the 
chain, and on the plan by compasses and a scale of equal 
parts ; if the plan be correct Ibese measurements will be 
the same ; if incorrect, they will be diffei-ent. Or Ihe table 
may be removed to a third station, ivhich is markrd on live 
plan, and the table being hero adjusted aa\)eto\e, v^ >^«i 







114 1>AKS SinVSYISSi 

index be applied to the point which represents tbe Btatioa 
where the table is placpd, and at the same time to the 
point on the plan repreaenting any other station, then the 
pole placed in that station will be seen to coincide with tbe 
Imir in the sight of the index, if the plan be correct ; and 
te same will be the case with every other point laid dona 
itiieplan. 



pKon]:,EM VIII. 



To Burrey with a theodolite. 

The theodolite is an instrument used by BDrreyora fat 
measuring horizontal or veitical angles, and the most per- 
fect instrument for these purposes yet invented. Without 
it, it would be almost impossible to survey a large estate ot 
a county. It consists of two circular brass plates, finely 
adjusted to each other, the lower being graduated arooiid 
the edge, from 1" to 360°, and the upper has a vernier 
attached to it, so as to enable the surveyor to read the 
measured angle to minutes. The upper plate also carries 
a telescope with it, which is attached to a semicircular arc, 
by which the vertical angles are measured. The upper 
plate is furnished with levels, to enable the surveyor to 
place it perfectly level, and also with a compass, to enable 
him to find the bearings of various objects. 

In order to surrey with tbe theodolite, the surveyor 
must place up poles in the various points the position of 
which he wishes to determine, and having selected tvro 
convenient stations, he places the theodolite first in the one 
station, and having set the index of the vernier to 0°, Imm 
the instrument round till the pole in the other station hi 
seen to coincide with the cross hairs of the telescope; then 
fix tbe instrument by means of its screws, at the same time 
slackening the screw which fixes the horizontal plates to- 
gether, BO as to allow the upper plate to revolve freelj 
upon the lower, and the instrument is then fit (or use. 
Turn the upper plate, which carries the telescope along 
with it, till the pole in the nearest position to that station 
be seen to coincide with the cross wires in the telescope, 
and mark down the angle, as read from the horizonlnl 
fixed plate. Then turn tbe telescope round till tbe pole in 
the next station appear to coincide with the cross wires, 
marking the angle as before ; proceed thus tilt tbe ongalu 
bearings be obtained of all the points which you wiih to 
determine, and observe if the angular bearing of the second 
$tiition be ttie same as &»^ e«^, ti^'wik, {<n If it is not, the 



jnuro wwvOT iwc 1 Jo 

instTument must Lave been displftcod during ils reTolution, 
and it will be necessary to measure the angles again. 

HaTiDg got all tlie angular bearings of the points to be 
determined from the buse line at the first station, the theo- 
dolite must then be removed to the second station, and ad- 
justed in the same manner aa it naa at the first, making 
the cross wires in tlie telescope coincide with the pole in 
the first station ; then measure in succession the angular 
bearings of each of the points whose position is to be de- 
termined, as at the first station, observing, when the teles- 
cope is brought back to the first station, Uiat the beating of 
that station appear the same as formerly set down ; if it do 
not, it will be necessary to measure the angles again, until 
this be the case. 

When the distance from the station at which the theo- 
dolite is placed to the various comers, whose bearing it ia 
necessary to take, can be determined, it will not he neces- 
sary to take a second etation, as both the area and a plan 
of the field ot estate can be obtained from one set of 
angles. 

ExBBOisB. Find the 
area of .1 field, of which 
the adjoining figure is a 
sketch ; the theodolite 
havingbeenfirst placed a t 
A, the angles were found 
to measure as follows: 
ZBAC 94- 50', IC\D 
44° 63', iDAE70''28', 
Z,EAF92''2',and/.FAB 

57° 47'. while the distances were AB 504 links of the im- 
perial chain; AC 445, AD 325, AE 201, AF 648. The 
Uieodolite was then removed to B, and the following angles 
were measured : i^FBG 79° 5', iGBH ll-'SO', and ZHBL 
68° 58'; while the distances were BF 572, BG 404, BlI 
417, and BL 140. 

The above is the park in which Craigmillar Castle is 
Bituafed, the Castle being on the north side of BC, and 
east aide of BL. Ans. 5 acres, 2 roods, 7'6 poles. 

rHOBLBH IX. 
To find the surface, and draw a plan, of hilly or sloping 

The rules for finding the surface of hilly or bIq^\ii^ 
ground are the same as those for le^el gtoijjii ■, WvS.ii.Ns^s- 




1 16 I.AWD aOBTSTntO. 

ing down a plan of hilly or sloping gjonnd, every line 
raeaanred on a surface inclined to the horizon, must be tt- 
duced in length, on the plan, by the folloning proportion ; 
R : eosinc of the inclination : : measured length : the length 
on the plan. 

Note. Some iand-BurreyorB ore of (i[iinioii, that in ineo^unn^ 
hilly ground, tbe horizontal surges of the base efaould be given for 
iha are& of the ground ; and in sapport of tb^ opinion tell ua, (hat 
19 many treea or upright atalka of com will grow on the borizonUl 
surface aa on the sloping side of the bill ; but allhough this be per- 
fectly true, it appears to be the busineBs cS the land-aurveyoc to give 
the surface, uid tba business of the cultivator of Ihe soil, to coniider 
what the viktue of the surfaee is for &ny particular purpose. In lay- 
ing doKD a plan of hill; ground, bowerer, since all plans arc drawn 
OD plane aurfaees, it ia necoasary to reduce fhe length of the linas, 
80 as to prevent diatortionH of Iho plan, which would be unavoidRbl» 
if any other method were adopltid. 

ExANFLK. A line of 960 links was measured np a hill, 
whose indination to the horizon was 17" 12'; what is the 
length of the base-tine ? 



r 



B 

Cos. \T 12* 
Surface line 950 k 
Base line 9075 log. : 



EXERCISES. 



1. What is the length of the base-line correspondii^ to 
1560 links, measured on an inclination of 21' 14'? 

Ans. 1454-1 Uak«. 

S, What is the length of the hase-line corresponding to 
SlOO links, the first 500 of which were measured on a 
slope of 9= 1 6', the next 900 on a slope of 18" 33', and the 
last, 700, on a slope of 23° 10' ? Ans. 1990-26 linki 

3. What is the length of the base-line corresponding to 
1800 links, one- half of which is inclined to the hoiizon, at 
an angle of 21°, and the other at an angle of 17° 12'? 

Ans. 1699^7 lint*. 

Problem X. 

To deduce from angles measured out of one of tlie tlk- 
^ions, but near it, the true angles at the station. 

p.When the centre of the instrument cannot he placed in 
^ vertical lino occu^\e4 V^ *.\ie wia nC % signal, the 




LAND B0RTETINO. 

la observed must undergo a 
"** , according to circum. '- 

KC be the centre of Ihe sla- 
P the place of the centre of 

_„_iMtruinent, or the summit of 
P'Ae obeeived an^le APB; it ia 
I nquired to find C, the measure 

of ACB, Bupposing there to be known APB=P, BPC= 

p, CP=d, BU=zL, AC=K. 
j Kbgc thff exterior angle of a triangle is equal to the sum 

ofthetwointerioroppositeangles, (Geo, prop, ]9),wehave, 
I with respect to the triangle lAP, Z.AIB:=iP+/.lAP ; 

aodwith regard to the triangle BIC, i^AIB=Z,C + iCBI. 

Making these two values of iAIB equal, and transposing 

iCBI, ive obtain LC=l.P+LlAP~LCBI, But the A» 

, CBP, giye sin. CAP = sin. IAP= ^sin, APC= 

B?^. riu. CBI= 8iu.CBP= f^. sin. BPC = ^■ 

"And as the angles CAP, CBP, are, by the hypothesis of 
W lite problem, altrays very small, their sines may be Bubsti- 
I tialed for their arcs or measures ; therefore 
_ (ifllnJP+p) dat^ 
R L ' 

Or to have the result in seconds, 

L /• 

The first term of the correction will be positive, if the 
Mgle (P+p) is comprised between 0° and 180°, and it will 
become negative if that angle exceeds 180°. The con- 
trary will evidently be the case in regard to p, since it is 
affected by the negative sign. The letter K denotes the 
distance of Ihe object A to ti>e right, L the distance of the 
object B, situated to the left, and p the angle at the place 
of observation, between the centre of the station and the 
object to the left. 

PnoHLEU XI. 

When a Hne is measured at an elevated level, to find its 
length, when reduced to the level of the sea. 

Let r= the radius of the earth, k= the height above 
Ihe level of the sea, at which the base ia mftaft\Kei.\ 'feevv 
since Ihe circumferences of circles are to oat a.n'i'Co.'a »* 



'^e 



118 T.J 

their radii, r+A : r : : m', {the measured Icngtli), ; m (the 

tnielength= «j=W^l— -+^ — ^ + -_&c.j- ; 

bat the radius of the earth being very great in relation to 
the height of any mountain on its surface, all the terms of 
the series may be neglected after the second, which gives 

the reduced length :=m' — ■ — ■ hence 

Rule. Add the It^arilhma of the measured base, and 
of its height above the level uf the sea, both in feet, ami 
from the Euni subtract the constant logarithm '{■ZIQBQO, 
the remainder is the logarithm of the correction in feet, 
which is always subtraetive. 

ExAwpLB. If the length of a line be 22540 feet, at an 
eleTalion of 14600 feet above the level of the sea; what 
will be its length when reduced to the level of the sea? 

^Log. 22540 =4-352954 

Log. 14600 =4164353 

85 17307 
Subtract constant Log. 7319890 

Correction =1575 =1'197417 

.-. 22540— 1575 =22524-25 feet. Ana. 
EXERCISES. 

1. If the length of a line be 31576 feet, at an elevation 
of 1800 feet above the level of the sea; what will be its 
length when reduced to the level of the sea? 

Ans. 3157a279fe<'t 

2. At an elevation of 82Z feet above the level of the 
aea, a road was 5 miles ; what is ils length when rednceil 

^^_.t9 the level of the sea ? 

^^B Ans. 4 miles, 1759 yards, 1-963 feet. 

Problem XII. 

The Division op Land is one part of the land- 
measurer's profession, he being frequently reij^uired to divide 
a piece of land between suodry claimants, in proportion to 
their fespectiTe claims. In some Instauces, the division 
relates only to the quantities that each claimant is entitled 
to ; in others, the quality as well as the quantity must be 
considered, in doing whiah, the measure of the whole 
must be first accura«:\j astei\.a.\ne4, \i'3 aanu; d^ dw rules 



LAND SURVETIXG. 119 

which have been fonnerlj given in this work, and then 
the division must be made according to the form of the 
figure, and the value of the claims. 

Cask I. To cut off a portion of a square or parallelo- 
gram, by a line drawn parallel to one of its sides. 

Rule. Divide either of the sides, adjacent to that to 
which the dividing line is to be parallel, in the ratio of the 
parts into which the field is to be divided, and through 
the point of division draw the required parallel, and it 
will divide the parallelogram in the ratio required, (Geo. 
prop. 58). 

ExAAiFLB. If a field in the form of a parallelogram 
contain 7 acres, and one of its sides be 200 links ; how 
must the other be divided, so that a line drawn through 
the point of division parallel to the given side may cut oif 
one-fifth part of the field ? 

Since the field contains 7 acres, its w^hole area in links 
is 700000, which being divided by 200, gives the adjacent 
side 3500, one-fifth of which is 700 links, the base of its 
fifth part. 

EXERCISES. 

1. If a parallelogram be. 1200 links in length and 400 
links in breadth ; what length must be taken, so that the 
area cut off by a parallel may be an acre ? Ans. 250 links. 

2. From a square which contains 5 acres, 3 roods, and 
10 poles, there is to be cut off 1 acre, 3 roods, and 5 poles, 
by a line drawn parallel to one of its sides ; at what dis- 
tance firom the corner must the parallel be drawn ? 

Ans. 233*6 links nearly. 

Case II. To cut off a part from a triangle, by a line 
drawn through the vertex. 

Rule. Divide the base in the same ratio as the parts 
into which the triangle is to be divided, and a straight line 
drawn through the vertex and the point of section will di- 
vide the triangle, as required. (Geo. prop. 58.) 

Example. A triangle whose base is 960 links, and 
area one acre, is to be divided into three parts, by lines 
drawn from the vertex to the base, so that the first part 
may be 24 poles, the second 1 rood 36 poles, and the third 
1 rood 20 poles ; required the length of the segments of 
the base ? 

Since the parts into which the triangle is to be divided 
are 24 poles, ^6 poles, and 60 poles, and their sum is 1 60 
poles, by the Rule of Distributive Pro]^ottvo\i, qx ?^\.\^^\- 
Aip, we have 



I 



I.Aia> BOBVIETIira. 



im : 24 : : OHO : 144 links, the first base. 
160 :?<>:: 960 : 4afi links, the second bas«i] 
100 : 60 : : yeO : 360 links, the third bane. 



1 . A triangular field, whose base is 1380 links, is to le 
divided, by Tmes drawn from the vertex to points in (lie 
[riven base, so that the parts may be to one anoibcr in tlie 
proportion of L.3, 12b., L.5, loa., L.7, 5s., and 1..6, Bs.; 
required the segments of the base? 

Ans. 216, 345. 435, and 384 links, 

2. A triangular field, ivjiose area is 2 acres 30 poles, is to 
bedividediiitobveparls, intheratioof 2, 5, 7, 9, andll, 
by straight lines drawn from a given angle to the opposite 
side, which is 1530 links; >vbat must be the length of ibe 
begments of the side ? 

Ans. 00, 225. 315, 405, and 495 links. 
Casb III. To cut off from a triangle any assigned part, 
by a line drawn parallel to its base, another side at leatt 

ituLB. From the area of the whole triangle subtract 
the part to be taken away; then state, as the whole triangle 
is to the remaining pnrt, so is the square of either of the 
sides to the square of the distance from the vertex to ihe 
point nbere the parallel cuts that side ; extract the square 
loot of this result, and the root will be the distance from 
the vertex required ; ihrougli this point draw a parallel to 
the base, and the thing required is done. 

£xAMPi.K. From a triaiigle which contains 4 acres, il 
is required to cut off I aere 3 roods, by a line drawn po- 
i-allel to one of its sides, the other two sides being 300 and 
:j(iO links respectively; at what distance from the veciex 
will the parallel cut each of these sides? 

The whole area is 640 poles, and the area to be cutoff 
is SKO p(>leB ; therefore the area of the remaining triuiigle 
^vill be 3G0 poles ; hence 

fi40 : 360 : : (300)' : 50C25, the square root of which is 
225, the distance from the vertex on tbe side which is 900. 
Again, 

640 : 3fi0 : : (360)' : 72900, tbe square root of wUcb ii 
270, tbe distance from the vertex on the side which iaSGO 

EXRIECISBS. 

1. If tbe base of a right-angled triangle be 2100 liaki, 
and tbe perpendictt\at lOQUWaka-, a\,»W\. di&tauoe from 



LANK SUnVEYING. 



1^^ 



tlie yertes must it line he drawn parailet to the buse, so 
that the area conbiined between the base ani] the parallel 
amy be an acre? Ans. 95! -2 Haks nearly. 

2, la the same triangle, if a line be drawn parallel to 
ibe base, so tliat tbe area of ihe trapezoid may be 1 acre 
2 roods; at what distance from tbe base must it be drawn 7 

Ans. 74'2 links nearly. 

3. If a trapezoid, containing 2^ aerea, be cut off from 
Ihe same triangle, by a line parallel to the hypotenuse, at 
what distance from the right angle will it cut the base 
and perpendicular? 

Ans. The perp. 872'9, and base 1833 links. 

Casb IV. When a given quantity is to be cut off from 
a field, by a line drawn from a point in one of its sides. 

RubB. Dmw a line by trial, cutting off as nearly as 
possible the given space; measure the space cut off, and 
lake the difference between it and the space required ; di- 
Fide tbe remainder by half the trial line, and the quotient 
will be the perpendicular altitude of a triangle, having the 
trial line for its baae, which triangle being added lo lh« 
space cut off if it be too little, or taken from it if it be too 
great, will give the space required. 

ExAMPLB. Suppose 3 acres are to be cut off from a 
field, by a line drawn from a given point in one of its sides, 
and that a trial line of 6'00 links has been drawn, and the 
space cut off is found to be 285000 square links, which is 
too little by liiOOO square links ; divide this remainder by 
300, half the trial line, and the quotient is 50 links ; draw 
a perpendicular to the trial line, on the opposite side of the 
space, because it was too small, and make it equal to AO 
links; if its extremity be in the boundary, the inter* 
mediate part being a straight line, it is the point requir- 
ed ; if not, through its extremity draw a parallel to the 
trial line, and it will cut the boundary in the point requir- 
ed ; join this point and the given one, and the thing re- 
'is done, as is evident from (Geo. prop. 2S). 

MISCFLLANKOUa EXBRCISKS. 

(ill a ridge of grass cost, at 10 guineas per 

49*7 its length being 1 7^ yards, and its breadth (!^ yards ? 

Ans. r..2,9s. 0|d. 

2. What is the area of 18 drills of potatoes, the length 

being 454 links, and the breadth 62 hnks; and what will 

they cost at L.24 per acre V 

Ana, 1 ro, 5 0368 poles; cost US, \Jrt,.\i^*>, 



122 LAso auKvrriNG. 

:). ^Yba^ is the value of 3<> drills of potatoes, at 3s. fid. 
per 100 yuidti, ilie niean leiigtii being 2/2 yurds? 

Ans. L.14, 5b. 7j<l. 

4. What will 60 drills of turnips cost, at L.21. ISs. per 
acre, the menu Irn^tli being 11)6 j'urds, and the diiit<iiii.'e 
between each drill -J Itol ? Ant. L.35, 49. 7^(1. 

5. If the breadth of a ridge be 5^ yards, what leujiib 
muBt lie taten to aiate half an aeref Ans. 440 yank 

tt. To preveat iujuriug the uropn, a land-surveyor mea- 
sured one side uf a iriaiigulnr field, which be found to be 
(i54 links, and the angles at its extremities 5(1° 20' and 64" 
15'; required the area of the field in acres? 

Ans. 1 acre, 3 roods, 313056 pol«. 

7. The annua) rent of a triaogular field is L.43, I 5b., 
its base ineasurea 2500 links, and perpendicular 1400 
liaks; what is it let for per acre ? Ails. L.2, 10s. 

6. A field in the form of a right-angled triangle is to be 
divided between two persons, by a fence made from the 
right angle meeting the hypotenuse perpendicularly, at tie 
distance of 880 link-s from one end ; required the area of 
each person's share, the len^rth of the divisiou fence being 
660 links? Ans. 2 acres, 3 ro., 2464 poles, and 1 nav, 
2 ro., 21-36 poles. 

Q. A genileinan has a rectangular garden 100 feet long 
and 80 feet broad ; ivhat must the breadth of a walk routul 
it be, so as to occupy half the garden ? 

Ans. 12-9844 feet nearlj. 

10. The side AB of a triangular field is 40 chains, lit' 
RO, and C.\ '25; required the sides of a triangle, parted 
oft' by a division -fence made parallel to AB, and proceed- 
ing from a point in CA, 9 chains from the angle A? 

Ans. 16, 19-2, and 256 chaint 

1 1. The side .^B of a triangle is 050 links, and the side 
AC 560 links ; it is required lo bisect it by a line drawn 
from a point in the side AB, 4U0 links from A; find the 
distance from A, where the line will cut AC? 

Ans. 455 links. 

12. What is the annual value of a pentagonal field, it 
L.2, 5s. per acre, the side AB being 926, BC 536, CD 
835, DE 828. EA 587, and the diagonals AC and CE 
1194 and 1223 links respectively? Ans. L.18, lOg. 7jd. 



SPECIFIC GRATITT. 



123 



SPECIFIC GRAVITY. 

ie gravity of a body is its weighty aa compiured 
t of another body of the same size, whose weight 
ered as the unit of measure. The body used as 
iard or unit of measure is a cubic foot of rain or 
.water^ which at a temperature of 40° Fahrenheit's 
leter, weighs exactly 1000 avoirdupois ounces. 
n&c gravities of bodies may be expressed either by 
against them the quotient arising from dividing 
[ght by that of an equal bulk of water, or by writ- 
ist them the weight of a cubic foot of the body^ in 
)ois ounces; the first can be derived from the 
y dividing by 1000, and consequently the second 
first by multiplying by 1000. In the following 
will give the weight of a cubic foot of the sub^ 
I avoirdupois ounces as its specific gravity. 

SFECIFIC GaAVITIES OP BODIES. 



if EARTHS, METALS. 

/ental . . 2590 

nyx . . 2638 

loudy . . 2625 

f oriental white 2730 

, of Piedmont 

, of Malta 

, Spanish saline 

, of Valencia . 

, of Malaga 



Of . m • - 

B, long . 
f short . 
ripe 
, starry 
9 crude . 
, glass of . 
^lass of natural 
lolten 

Giant's Causeway 2864 
>f Judea, . 1104 

2000 
M, . . 3549 
mtal, . 2723 



2693 
2699 
2713 
2638 
2876 
1078 
926 
909 
2313 
2578 
3073 
4064 
4946 
3594 
5763 



Bismuth, molten, . . 9823 

Brass, cast, not hammered, 8396 

Brass, wire-drawn . 8544 

, cast, common , 7824 

Cohalt, molten . . 7812 

hlue, glass of . 2441 

Copper not hammered, 7788 

, the same wire drawn 8878 

Chrysolite of Brazil, . 2692 
Crystal, pure rock of Ma- 
dagascar, . • 2653 
Crystal of Brazil, . . 2653 
Calcedony, common . 2616 

— ' , transparent . 2664 

i veined . 2606 

Camerlian, pale . . 2630 

Chalk, British . . 2784 

, simple . , 2613 

Diamond, white oriental 3521 

, Brazilian . 3444 

Emerald of PerU| . 2775 

Flint, white . . • *15i^V 

'-'■ — > Eg^ptaaxL • ^ti^^ 
Gk>ld, pure, csiftt, \iviX> t^\> 





^ 






19369 


Pktina, crude, in grains. 


15602 




, guinea of Geo. 11., 


17150 








, fuiDeaofGeo. III., 


1763.1 


mered, 


19500 




, trinket standard, 


13709 




20337 




Gamet of Bohemia, 


4189 




21042 




Giniwl, . . . 


4000 




B2069 




Gypsum, opaque . 


2168 


Pearl, virgin oriental 


2684 






230C 


Pebble, Englidi 


2609 




., rhomboidal . 


2311 


Prasinni, . 


25S1 




— — .oimeifonn cry Blallized -2306 


Peat, hard 


1SS9 




Glufls, green . 


2643 


Porcelam, china, . 


2385 




, white . . 


28!)2 


Porphyry, red . 


2;6i 




^~~, bottle . 


2733 


— .green . 


2fi7« 




, Leith crjBtal , 


3169 


Quartz, cij-alallized . 


26S5 


— ; fluid . . . 


3329 


Ruby, oriental 


42!1 




Grsnile, red Egyptian 


2654 


-.Brazilian . 


35 Jl 


Hone, while razor 


287<i 






Hyacinth, conunon . 


3637 


hakened . . 


7B31 




Iron, cast . 


7207 




, bar 


7788 


, do. not hardened 


TBlf 




Jade, white . . . 


29.50 




10471 






2966 




lOSIl 




~;^e". ■ . ■ . 


2983 


. Paris gtandard . 


lOlJi 




Jaaper, clear green . 


2359 


, Bhiliing of Geo. n. 


10000 




, red . . 


2661 


, shilhng of Geo. Ill 


10S34 




, yeUow . . 


E7I0 


, French Poiu 


10408 




— ^ — , veined . 


2696 


Sapphire, oriental 


3991 




.bloody 


2628 


. . Brazilian 


3121 




iaxgon of Ceylon, 


4416 


Spar, white sparkling 


24M 




Lead, molten . 


11352 


■ . red do. 


3111 




, ore of black lead 


4059 


, green do. 


2r« 




, do. do. vitreous 


655S 


, blue do. . 


SCSI 




, ore red lead . . 


6027 




SSH 




Limestone, 


3179 




SBIJ 




■ , white fluor . 


3158 


Sardonyn. pnro 


25M 




. green 


3182 


Stone, paving . . 


SIK 






4756 


, caller'B 


Sill 




Molyhdona, 


4738 


, grind . . 


9u: 




Mercury,soUdoreongea!ed 15632 


,niill . 


24S1 




, fluid . . 


13G68 




SIS 




, natural calj of 


9230 




1981 




_,pm:ipitatered . 


8309 


Sulphur, native 


2»31 




, bro«n cinabar 


10218 


-, molten . 


im 






6902 


Tin, pure Comisli melted, 




Nickel, molten . 


7807 


and not hardened . 


7251 




Marble, green . . 


2743 






,red . 


2724 


Talc of Muscovy 


mi 






2B3B 


, black crayon 


30» 




■ , Pyreuean . 




, do, Germui . 


»4( 






26 95 


. yeflo* 






— , white Grenadan 


2705 


,bh«:k . . 


aS 




, violet Italian . 


2858 


, white . 




, green Egjptiaa 


t&66 VlviB'^". ■ 






, Op»l, '\ . . . 


^■V\t\-«l«Ol'DO, . 




















I 



r BTECIFIC GRAVITT. 


135 


Wolfram, . 


7113 


Wine, Bourdeaos . 


994 


ZinCj molten 


7191 


, Madeira . 


. 1038 






, Port 


997 


LIQCORS, OILS, ETC, 








Acid, aulphuric . - 


1841 


KESIM, QUMS, iND 


ANIMAL 


-, nitric . 


1271 






, inuriatio 


1191 


Aloes, socotrine . 


, 13B0 




1033 




1359 


, white aoetooB 


lOU 




. 1398 




1558 


Bcea wax, yellow 


B66 


, formic 


994 


, white , 


969 




837 


Bono of an ox, . 


16fiG 


, highly rertified 


629 


Butter, 


942 


, mixed with water 




Camphor, 


S89 


7-BlbB alcohol . 


867 


Copal, opaiiue . 


. 1149 


, 6-BthB do. 




, Chinese . . 


1063 


, 5-BlhB do. 


920 


Fat, beef . . 


823 


, 1-aiha do. 


943 




e2« 


. 3-Utl« do. . 


9(i0 


, hog's 


. 937 


, 2-athH do. 


973 


,y^tl . . 


934 


— , l-Blh do. 


9BS 


Gamboge, . . 


. 1232 




897 


, Arabic 


1207 
. 1453 


Ammontac uqum 
Beer, pale . 


!(I33 


, brown . 


ID3-I 


— -, Eapborbia 


H24 


Cjder, .... 


101 B 


, seraphio . 


. 1201 


Ether, gnlphuric 


739 




, Ditric . 


909 


-, do. of A 


oppo 1235 


, nmristic . 


730 




933 


, aoetic 

Milk, woman's . 
.cow's . 




in a loon 


heap, 836 

. 1745 

1450 


1820 




1032 


Honey, . . 


.asses' . 


1036 


Indigo, 


769 


^..e-we'a . 


1041 


Ivory, 


1B2S 


, goat'a . 


1035 


Lard, . 


948 


.Siare'a . . 


1031 


Myrrh, . . 


1368 


, cow's clarified . 


1019 


Opium, . . 


. 1336 


Oi), essential of torpentin 


870 


Spermaceti, 


943 


^, do. of lavender 


894 


Tallow, 


942 


— , de. oTcloTeB . 


1036 


Shoemaker's wax, 


897 




1044 






— , do. of olives . 


915 






— , do. of almonds 


917 






— , do, of lintsatd 


940 


Alder, 


800 


— , do. of whale . 


S93 


Apple-tree, 


793 


— , do. of hempseed 


926 


Ash, the trunk, . 


84.'. 


— , d*. of poppies 


994 


Bay-tree, 


823 






Beach, . . 


. 853 


Tnipentine, liquid 


991 


Bon, French . 


912 


Urine, human . 


]01 


-, Dutch 


, 1328 


Water, ram or distilled 


1000 


■ , Brazilian red 


1021 


,9m (average) . 


1026 ' Campeach)- op log wood, 9\* 


of Dead Sea, 


1340. Cedar, wild 




Wine, Biirgundy 


992 , Paleatino 


■■ ^^^J 





SPBCmC GHATTITT. 


^ 


Cedar, Indian . 


1315 


MMtic-tree, 


349 


, American , 


6BI 


Mahoganj-, 


1063 


Citron, . 


726 


Map4 . . 


761 


Cherry-tree, 


715 


Medlar, , .. 


944 


Cork. . . 


240 


Mulberry. Spanish 


B9T 


Cypress, Spanish . 


Sit 


Oak, eo years old . 


im 


Ebony, Americaa 


1331 


Olive-treo, 


9W 


. .Indian . 


. 1209 


Omnge-tree, . 


7a5 


Elder-tree, 


695 


Pesr-lree, , 


6S1 


Elm, trunk of 


. 671 


PomeBTanate-tree, . 


1351 




eno 


Poplar, . . 


311 


Fa, male . . 


630 


, while Spaniah 


S29 


■ , female . 


493 


Plum-u™, . 


783 


Hazel, . . 


fiOO 


Vino, '. . . 


U27 


lacmin, SpnniBh 


770 


Walnut, 


671 


Juniper-tree, 


. BS6 


Willow, . . . 


iBi 




703 


Yew, Dutch 


781 


Lignum vite, . 


. 1333 


.Spanish . 


HW 




6M 






^K WEIGHT AND BPE 


nnc GKAviTirs of diffekbnt gases. 


^H FallKolleit'B 


Themoqi.5 


=. Barometer 30 mches. 





PSptc tn^vil;. Weight cublcfML 

Atmospheric air, . 1.3 535'0 gn. 

HydrogeD gas, . , 0-1 43.75 

Oxygengaa, . . 1-435 627'813 

Mtrousgoa, . . 1'4544 63fi'333 

Ammoniac gas, . . '731 1 319'B33 

SnlphureouB acid gas, , 2-7611 1207-978 

H Problem I. 

I Given the specific gravity and solidit; of a body, to find 
its -weight. 

RuLR. Multiply the solidity in feet by the tabular spe- 
cific gravity, and the product will be the weight in avoir- 
dupois ounces. 

Example. Wlat is the weight of a block of Pyrenesn 
mntble, containing 7"25 cubic feet? 

Here we find the tabular specific gravity 2726, tberefoK 
272Gx7-25=:197tS3-5 avoirdupois ounces, or 11 cwt,31l» 
3^oz. 

EXERCISES. 

1. IFhat is the weight of a log of mahogany containing 
35| cubic feet? Ans. 1 ton 3 qrs. 9 lb. 9J oi 

2. What is the weight of a piece of French box, mee- 
auring 3| cubic feet? Ans. 1 cwt 3 qrs. 3 lb. 8 □£ 

3. What is the weight of a piece of cast iron containing 
450 cubic inches? Acs. 1 ewt. 5 lb. 5 oz. nearly 

4. What is the weight of a Ic^ of beech which ia 20f«l 
long, and 1 foot 6 incVica at^uiwe? 

Atia- \ ton \ c-jA,. \ cj. \Q ft. 4 m 



^ 



^ 



SPECIFIC GKATITT. 127 

5. What is the weight of a cjlindrical pillar of Egyptian 
CTanite, its diameter being 3 feet 3 inches, and its height 
15 feet 5 inches? Ans. 9 tons 9 cwt. ] qr. 18 lb. 5 oz. 

6. What ia the weight of a hollow cast iron cylinder, 
whose length ia 5 feet, outer diameter 30 inches, and the 
thickness of the metal 1^ inches? 

Ans. 15 cwt 3 qra. 1 Ih. 13 oz. 
Problem II, 

GiTen the specific grayity and tveight of a body, to find 
its solidity. 

KuLB. Rednce the given weight to ounces, and divide 
by the specific gravity of the body, and the C[U0tient will 
be the solidity ia cubic feet. 

Example. How many cubic feet of male fir ore in a log 
which weighs one ton? 

Here 1 ton reduced to ounces is 1 x20>: 112x 16— 
35840, which being divided liy 550, the specific gravity of 
male fir, gives the quotient 65^"^ feet, 

ESERCISES. 

1. An irregular block of red marble was found to weigh 
2 Ions 7 cwt. 13 lb. 12 oz., how many cubic feet did the 
block contain? Ans. 31 cubic feet. 

2. If a mass of molten sulphur weigh 15 cwt. 2 qrs. 
lb. 2 oz., how many cubic feet are in the mass? 

Ana. 14 cubic feet. 

3. How many cubic feet of green porphyry are in an 
irregular block which weighs 7 cwt. 1 qr. 24 lb. 4 oz.? 

Ans. 5 cubic feet. 

4. How many cubic feet must a vessel contain to be 
capable of holding a ton of rain water? 

Ans. 35'84 cubic feet. 

5. How many cubic inches are in an irregularly formed 
Teasel, which contains 5 lb. 7 oz. avoirdupois weight of 
run Tvafer? Ans. 150'33d cubic inches. 

Pboblebi III. 

To find the specific gravity of a body, its weight and solid 
content being given. 

Rule. Divide the weight in avoirdupois ounces by the 
solid content in feet, and the quotient will be the specific 
grarily. 

Note. If the eolidity he in inoheB, multiply the weight in avoirdu- 
paiftoauceB by 1728, and divide llie product by the solidity in cubic 

»FLE. If a mass of 9 solid inches weigh 15 oi, 1 it, 

» the specific ^ javity of the body^ 



Here hi,', ounces, multiplied by 1728, (see note) gires 
26028, whitli bemg divided by 9, gires 2892, the specific 
gravity; hence the body ia white glass. 

EXERCI3K?. 

1. If a mass of 4^ cubic feet of stone weigh 7 ewt 
3c|T8< 261b. 15 oz,, what is the specific gravity of the stone; 
and find from the table what kind of stone it ib. 

i„, J Spec- g"'- 3182. 
\Oreen limestone. 

2. If a mass of Sfeet of wood weigh 1 cwt. 8 lb. 12oz., 
what 18 its specific gravity? Ana. 644. 

3. If apiece of silver of 4^ cubic inthes weigh 1 lb. lOo*. 
41 dr., what is the specific gravity of silver? 

Ans. 10474neariy. 

4. If a cubical piece of iron, whose side is 5 inches, 
ireigh 35 lb. 3 oz. 6 dr., what is the speci'fic gravity of iron! 

Ans. 7788. 

5. A bar of melal 6 feet long', 4 inches broad, and 2 
inches thick, weighed 1 cwt I qr. 22 lb. 4 oz., what wai 
the specific gravity of the metal, and find from the tahh 

m yibat kind of metal it was? 

■ Ans I ^P^^'^" e™''*y 7788. 

■ ' I Hammered copper or bar iron' 

F Pbobi-km IV. 

To find the specific gravity of a body without knowing 
its solidity. 

Cash I. When the body is a solid heavier than water. 

Rule. Weigh the body in air, and also in water, then 
stale ; The weight lost in water, 
Is to its weight in air. 
As the specific gravity of water 1000, 
Is to the specific gravity of the body. 

Note. A body may be weighed in water by suspendiag it bea i 
fine chord attached la the ami of a tiaUnce, eo tUnt the body na} 
deBcend inio the water while the other scale in which the weights aW 
rBmaios oat of tha water. The body descending into the water "ill 
dJsjilacG a quantity of water equal to its own bulk, and will bo 
buoyed or preasod upward with a force equal to tha weight of llm 
water displaced, and hetica will lose a part of its weight equ&l la llw 
same bulk of water; and from this the truth ot the rule is obvioiA 

ExAMPLB. If the weight of a body in air be 40 ob., anil 
in water 10 oz.; what is its specific gravity, and what hodj 
is it? 

Here the weight lost in water is 30 oz,, and hence 30: 
40 : : 1000 : 1333, the body's Byecifio gravity, and by in- 
specting the table, we ?iu4 \l lo \ie ligTvu-m. ■uVI'b, 



vnanc GXATRfr. 



1. If die weight oFa body in air be HO ot., and in rain 
fit distilled water 42 oi. ; what is the specific gravity of 
tfiebody? Abb. 6250. 

2. If the weight of a piece of metal be 1 5 or. in air, and 

14 OS. in water; what ia its specific gravity? Ans. 15009, 

3. The weight of a stone in air is 26 oe., and in water 
16 ox, ; what is its specific graTity? Ans. 2600. 

4. Find from the table what substance that ia, which 
weighs 4 lb. in air, and 3 lb. in water. Ans. GirasoL 

5. What is the specific gravity of » body which weighs 

15 ox. in air, and 14 oz. ] dr. in water. Ans. 16000 
Cask II. When the body ia lighter than water. 
Bulb. Attach to it a piece of another body heavier 

than water, so that the mass compounded of the two may 
sink together. Weigh the denser body and the compound 
body separately, both out of the water and in it, and find 
how much each loses ia water, by subtracting its weight IB 
water firom its weight in air; and subtract the less of these 
remainders from the greater j theu use this proportion : 

Ab the last renainder 

Is to the weight of the light body in air. 

So is the specific gravity of water, 1000 OBuces, 

To the specific gravity of the body. 
Example. Suppose a piece of filbert tree weighs 15 lb. 
in air, and that a piece of copper which weighs 18 lb. in 
air and 16 lb> in water is af&ied to it, and that the com- 
pound weighs (f lb, in water ; required the specific gia- 
Tily of the filbert 1^ 

Here the weight lost by the copper was 2 lb,, whilst the 
mass, which weighed 33 lb. in air only weighed 6 lb. in 
water, it therefore lost 27 lb ; hence the difference of the 
weight lost is 25 lb. ; and therefore by the rule we have 
the following proportion: 

25 : 15 ; : 1000 : 600, the answer. 

EXBRCISKS. 

1. If a body lighter than water weigh 20 lb. in air, and 
a body heavier than water weigh 12 lb. in air and 10^ lb. 
in water, and the two together weigh 4^ lb. in water; 
what is die specific gravity of the lighter body ? 

Ans. 769/b. 

2. If a piece of wood weighing 14 lb. in air be attached 
to a piece of metal which weighs 11 lb. in air and 10 lb. 
in water, while the compound mass weighs 2 lb. in water; 
£ad the specific gravity of the wood ? Ai«, SJft-^^. 

3. A piece of metal whicli weigha \6 \\i, vo. ^lv^ %&^- 



k 



SPECIFIC GRAVITT. 

14 lb. in water, is atdichcd to a piece of wood weighing 20 
lb. in air, and the compound mass is found to weigh only 
one pound in water; what was the specific graTitj of the' 
wood ? Ans. 606 nearly, j 

4. If a piece of metal which weighs 14 lb. in wr and 
12^ lb. in water, be attached to a piece of cork weighing 
3 lb> in air, and it be found that the compound mast 
weighs also 3 lb. in water; what is the specie c giavitj 
of cork? Ans. 240. 

Case III. To find the specific gravity of a fluid. 
BoLG. Take some body of known specific gravity, 
weigh it both in and out of the fluid, and find the loss or 
weight in the fluid ; thea say, 

As the weight in air of the body 
Is to the loss of weight in the fluid, 
So is the specific gravity of the body 
To the specific gravity of the fluid. 
' ExAUPLE. If a piece -of metal whose specific gravity ii 
19500, weigh 9| lb. in air and 9^ lb. in a fluid ; what k 
the specific gravity of the fluid ? 

Here the weight tost in the fluid is half a pound ; and 
therefore by the rule we have 

9| 4 : : 19500 : 1000, the specific gravity of the bodj, 
it was therefore rain or distilled water. 

EXSRCISKS. 

1. If a piece of bar iron weigh 7 lb. 9 oz. II dr. _ ^ 

and in a fluid 6 lb. 11 oz. ; what is the specific gravity «( 
the fluid? Ans. 940, lintseed «L 

2. If a piece of green Egyptian marble weigh 2 lb. 9 oii 
[' 11 dr. in air, and 1 lb. 6 oz, 5 dr. in a fluid ; what is tlw. 

pecific gravity of that fluid? Ans. 12411 

3. If a piece of unhammered copper, weighing 2 lb. 8 ol 

9 dr. in air, lose 5 oz. of its weight in a fluid ; what is ibt 

specific gravity of tho fluid? Arts. 9SX' 

F110BI.BM Y. 

To find the quantity of each ingredient in a givett com- 
pound of two ingredients. 

KuLE. Find the specific gravity of the compound, and 
of each ingredient ; then multiply the weight of the nu 
by the specific gravity of the body, the quantity of whit 
you wish to find, and by the difference between the 



fie gravity of the mass and the other body; divide this nw 
duct by the difierence of the specific gravities of the booii^ 
multiplied into the specific gravity of the compound 
and the quotient wiVl ^ve &e (^aa.^<;i.*.^ qC that body. 



131 

ExAUPLB. A mixture of gold and copper weighs 18 
lb., and its specific gracity is 14520, Ihe specific gravities 
of gold and copper being reapectively J9258 and 77"f ; 
how much of each was id the mixture? 

ix(U530— 77Bt 



(19-250— 77Ba)itU, 

lexTTBOxClHaSB— 146'J0) 

(19258— 7788)xl4S2o' 



= 14-012, the gold, 
= 3-988 the copper. 



1. Amislare of gold and silver weighed fij lb., and its 
specific graTity i*aB 15800; what quanlitj of each metal 
did the mixture coatain^ 

Ans. 62-8177 of gold, 221823 of ailver. 

2. A mixture of silver and copper weighed 100 lb., and 
Its specific gravity was 8530; what quantity of each meial 
did the composilion contain ? 

Arts. 33-92 lb. of silver, 6608 lb. of copper. 

3. A mixture of pure platina and silver weighed JiO lb., 
and its specific gravity was 16420; how much of each 
metal did the composition contain ? 

. I 62-5865 of platian. 
^^^- I 17-4135 ofBilvcr. 



1. Gauging is the art of finding the quantity any vessel can 
Or does uontain, and conipnaea the methods of ineasuring the 
dimeDsioDs, and therefron) calculating the cnntenta. 
* 2. All diniensioDfl ave taken in inches and tentlis of incht-h, 
sod the conti^nts are foimd in the same, by the rules for men- 
nntion of Bolids, whose furma are sitnilar to those of tlie Tea- 
ttta gauged. 

3. These contents are converted into pounds weight, ga!- 
lone, and bushels, by the use of ei-rtam known nnmbeni, 
called divisors, multipliers, and gauge-points, and which arr 
contained in the following 

Table. 




In Lhis table, the divieors for cii'des for flint and plate glass, 
and the divisors for arjuarus for the di^ereat kinds uf saap,ui(l 
far the imperial gallon and bushel, ai'e those legalized, as the 
standard of ineuaurernent, by act of Parliament. All the 
other divisors for circles are obtained by dividing their res- 
pective diviBora for squares by '7S5396, and the other divbors 
for squares by imiltiplying their respective divisors for circiea 
by -7B5308. The divisors are the measures of capacity, 

The multipliers are the refiprocals of the divisors, and are 
obtained by dividing nnity by the relative divisors for sqaaies 
or circles. 

Gauge-points are the square roots of the divisors; they aN 
chiefly used for calculations by the sliding rule. 

4. Tabular divisors, multipliers, and gauge-points may be 
found for any form of regalar superjicies, and for any sUod- 
ard of capacity. In general they are given in treatises of gaog- 
ing; but as they are seldom or never used in practice, they 
have been omitted from this. Sncb, hov>ever^ as may be de- 
sirous of knowing them, and having thetn at command, may 
easily compute them by the following rules. 

For Conical Figures, — Multiply the number of cubic 

inches, in the measure of capacity, by 3, for divisors; thebc- 

tors and gauge-points may be obtained from these as before. 

NoTB. In using this t&ble or rule, the whole depth mast bo 

For Regular Polj/gons. — Divide the measure of capacity by 
the tabular multiplier given in the table of polygona for di- 
visors. (See page 6!l.) Proceed as before for multiplien 
and gauge-points. 

When the middle area of aroj regular frustuia is taim, — 
For the gauge-points. — Take the square root of dx times the 
divisors, for polygons, for squares; and for circles, the gtpisw 
_ nota of six times the circnlar measure of capacity. 

fi. Many calculations in gaaging are made by the sUding- 

P THE Sl:IDIN0-RaLE. 

The sliding-rules generally used are of two forms. One is 
1'2 inch broail, and '6 loch thick; tlie other is 1-7 inch brMiI 
and -2 inch thick; both are 9 or 12 inches long. 

Tho former lias four sides, in each of which is a groo'e 
and a slider. One side is marked A on the upper edn, snJ 
MD on the lower. The opposite side is marked D. Of 
the other two sides, one is marked SS, tha other SL. 

On the first side, the mark A is placed at the end of alinr, 
called the Une of numbers, which is divided into 100 parts. 
ia such a manner as to constitute a scale of graduated loe*- 
rithms, numbered from 1 to 10, and ninningfrom left to rignt. 
In this line is a brass pin, marked 1MB, at 2218']92, Si"! 
wnoiher IMG, at 277"27i- The former is the divisor (m 
imperial bushels, the \a\,let ioi wtv^rLni. ^Utnis. 



The line marked MO on the same aide, is for computing 
ualt-floon ; its numbers stand In inverted order to titoee un 
tKe line A, snd ore aa placed that 10 ia oi>poBitc to 1MB 
n A. 

On the opposite side, which la marked D, there is hut one 
ine, numbered I to 32 on the upper edge, and on the lower 
■2 to 10. Thb line ix the same m construction as the line of 
nnmhera A, but has a double radiua. In the upper portion 
of the line, at 18-7)192, is a brass piu, marked IM.U; this is 
the circuhir gauge-point for an imperial gallon. In tlie lower 
part, at 49'0977, ia another pin, marked MS, wliich is the 
gauge-point for squares, and one at -531441, marked MB, 
the gauge<point tor circlei^for an imperial biiahel. 

The sides marked SS ami SL are used for ullaging, or 
finding; the quantity in ca!<k9 that arc neither full nor empty. 
SS signify segment standing, and SL Begment lying. 
These linea occupy both the upper and lower edges of the 
mde, and their numbers are placed from left to right. 

Of tlie sliders, two are marked B and two C. All tlie 
lines on the bees so marked are similar, and eciiial to the line 
of nuniberg A. On the under face of one slider are marked 
the gaajre -points, divisors, and factors for squares and circles, 
for a, gallon and huahel. Another slider baa on ita under face 
three lines; the first or uppermiist is divided into inches and 
tenths, the second is marked splid., and the third 2d va- 
rietj'. 

These lines are for reducing the 1st and 2d variety of casks 
to cTlinders, hut in practice are very seldom used, 

Aiders having the same letter, are, in use, joined together, 
ao that ttie letters are the extremities, and thus tiiey form 
sntt are rend as one slider having two radii. Either pair uf 
diders mav be used in all computations. 

The other form of sliding rule in common use has hut two 
sliders, which being of the same thickness as the stock, are 
grooved, to slide on tongues on the stock. Both the faces of 
these sliders have on them the lines of numbers. On one edge 
of the stock are two lines of numliers, so placed, that 1 on tho 
upper cuts 1'63 on the lower. On the other edge are also two 
aimilar lines, but placed so that 1 on the upper cuts 1'227 on 
the lower. In all other respects both rules are alike. 

G. To estimate the value of the numbers on the sliding-mls. 

On all sides the value assigned to 1 determines the value of 
the rest of the numbers on the same line. If the assigned 
value be ■!, 1, 10, 100, or 1000, then 2 will be -2, 2, 20, 200, 
or 3000, and the intermediate divisions will have correspond- 
ing values. On the line marked MI), 10 may be read 1, 
and then the others must be proportionately decreased. 

7- To multiply by the sliding-rule. 

Rule. To 1 on A set the lesser factor tn\ ^B, a^vi t^iwisS. 
the greater faclor on A will be the product c 




KxiKPLE 1. Multiply Sii by 7. 

cipiHi^ite to I iin A set 7 on B, (tnJ against 56 on A will In 
31t2 on B. 

2. Multiply !WM hy 48. Ana. 17472 

3. Multijilv 75-0 by 6-S. Ans. *09* 

4. Multiply sag by -2. Aus. 164-6. 
Note, a umple way lo read the product is, after Bxiog the rule, 

to give the nuDiericBl value of tlie Jint figure at the greater bclar 
In Qie vnil fi){ure of the lesser factor, and valuing tlie prodnet le- 
cordiugly. Thus, in the 2d enunple, 6, the uni'f ligure o[ the lesser 
Tactor, after taking the Taluo of ;!, tlio first figure of the greater bc- 
lor, and whiuh in hiadrrdf, bemniea SQ^, imd the tesser fiiclDr tboi 
reads lUOO, wbidi caaily leads tu the correct value of the pradad, 
17*72. 
6, To divide by the a!Ming--riile. 

Agninst tba div'dfind on B eet the divisor on A, flJui 
lite to 1 on A will be the quoticDt on B. 
1. Divide .'WO by 4JS. 

1 B to 45 on A, and beneath 1 on A is 8, the SB- 
Divide 85'5 bv 9. AnB. 9-5. 
Diviile 1470 by 42. Ans. 35. 
4. Divide 1728 by 14-4. Ans. ISO, 
KoTE. WlioQ the divisor ie greater than the same nnmbei <t 
figyiras of the dividend, the quotient coniiats of as many Agnret u 
the dividend hiia more tliwi Uie diviaor; but fthen Icea, the quonsiLl 
lias another figure in addition. 

9. To work Proportion or the Rule of Three on the slidinj- 

Rur,£. Set the first terra on A to the second on B, then 

Bgainat the third on A will be the fourth on B. 

BxAuPLE I. If 14 books cost 35s., how mui-h will 44 bookE 

To 14 on A set SJSs. on B, and 44 on A cub llOs. on B. 
2. Wbatisthefourthproportioniil toSl,27,nuii8l? Ans. 24a 
a, Kequiied the third proportional to 48 and G4'^ Ana, 8d^ 
Note. The third proportional is found by using the aecond M 
for bothia second and third term. 

10. To square a given number. 

RubB. Set I on B to 1 on D; ogainat the given number 
D will he its square on B. 

Example 1. What ia the square of 9! 

Fliice 1 on fi tu 1 on U, und against 9 on tlte lower line •£ 
D will be 81, the answer, on B. 

2, What is the square of 731 Ans. ilB*- 

The s*|narB of s.viy number may he a!?o found by (Art, 7) 
^ Another methud ts to 'w>(iivt Ute '£\&«n^ wdv^t (be gjna 



GAUGINO. 135^ 

number on B to the same number on A, and against 1 on A 
wUl be the square required on B. 

Note. If the given number be on the lower part of the line on 
the side D^ then the square thereof will consist of twice the number 
of figures that are in the given number ; but if it be on the upper 
part, then of one less than twice that number. 

11. To extract the square root by the sliding-rule. 

Rttle. Set 1 on B to 1 on D, and against the given num- 
ber on B will be the required root on D» 

^Example 1. What is the square root of 49? 

Place 1 on B to 1 on D, and opposite to 49 on B will be 7 
oh D, the root required. 

2. What is the square root of 13*69? Ans. 3*7. 

3. What is the square root of 420-25? Ans. 20-6. 

4. What is the square root of 22500? Ans. 150. 

Note. If the given number have 1, 3, 5, 7, &a, figures, the root 
will be on the upper part of the line D; but if of 2, 4, 6, 8, &e., it 
will be on the lower part. 

12. To find a mean proportional between two given num- 
bers. 

Rule. Set one of the numbers on C to the same number 
on D, and opposite to the other number on C will be the re- 
quired proportional on D. 

' Example 1» Required the mean proportional between 4 
and 16. 

To 4 on D set 4 on C; against 16 on C is the answer, 8 
on D. . 

2. What is the mean proportional between *20 and 301 

Ans. 24-49. 

3. What is the mean proportional between 72 and 288? 

Ans. 144. 

4. Required the mean proportional between 2-2 and 250? 

Ans. 23-45. 

13. To cube a given quantity. 

Rule. Set the given number on C to 1 on D, and against 
the given number on D is the cube required on C. 

Example 1. Wh^lis the cube of 3? 

To 1 on D place 3 on C, and beneath 3 on D is 27, the 
answer on C. 

2. What is the cube of 9? Ans. 729. 

3. What is the cube of 7*5? Ans. 421-875. 

4. What is cube of -4? Ans. -064. 

14. To extract the cube root of a given number. 

Rule. Move the slider either way till 1 on the line D and 
the given number on C are opposite the same number^ wvd 
this number is the required root« 



J 3d oacging. 

ExAMPLB 1. What is the cube root of 343T 
Find 343 on C, then move the slider till 1 on D k against T 
1*11 C, and 34.1 on C is against 7 on D, and 7 is the answer, 

■ S. Required the cwbe root of 17S8? Ana. IZ. 

■ 8. What 13 the cube root of -SI Ana. -926. 
I 15, To find the areas of squares, circles, paralledogiams, &qi} 

ellipses, by the pen and the aliding-rnle. 

Note. All nreaa are as soUik, having a depth of ene inch. 

Rdle. 1, Find the surface in inches and tenths, by the role* 
for the " mensuration of sarfacec," then divide or multiply by 
the proper diyiaor oi mnltiplier in the table, (Art. 3.) 

Rdlg 2. Set 1 on C to the proper gauge-point on D, anj 
against the given side or diameter of a square or circle on D, 
will he the answer required on C. 

RuLB 3. To the proper square or circnlar divisor on A, 
set one aide or diameter on B, and opposite the other side m 
diameter on A will be the answer on B. 

Note. In computing the areas of pamllelograma of ellipses \j 
Rule II., the mcaa proportionBl between their «des or diimetsnl 
muBt ba firat found by (Art. 12.) ' 

ExiHTLB 1. What is the area, in bushels, of a eircnlff 
room, whose mean diameter b 200 inches? 

By Rule 1. Now 2824-2003, is the tabular circular divisor, 
and -000354 the tabular circular multiplier for bushels. 

Therefore 260x280-=-2824-2903=23'93 bushels. Am. 

And 2e0x2«0x-000'3fi4 =23-93 do. km. 

By Rale II. The tabular circular gaose-poiot for bBsheli • 
is fi3-1441 ; then set 1 on C to 63-1441 on D, and agtunet 260 ' 
on D is 23-93 on C, the answer in bushels. 

By Rule III. The tabular circular divisor for boaheb \t ■, 
2824-2903. 1 

Then to 2824-2903 on A, set 2(50 on B, and opposite to 260 " 
on A will be 23-93 bushels on A, as required. 

EkavpubZ, Required the area, in gallons, cd a sqnai^, 
whose side is 120-7 inches. 

Bv Rule I, The tabular dirisor for ^usres i> 277-27^ 
and the multiplier is -003606, for gallons. 

Then 130-7x1 20-7H-277- 2 74^ 62-5 gallona. Ani. 

And 120-7xl2O-7X*0036O6=62-fi gallons, Ans. 

By Rule II. The tabular gauge-point for squares b 1R-6S1S 
for a gallon. Tberel'ore to 16-6515 on D sot 1 on C, and 
againat 120-7 on U wiU be 62-5 on C, the answer in gatlonk | 

By Rule III. To 277-274 on A set 120-7 on B, and ow^-- 
site to 120-7 on A, will be on B 52-5 gallons. Am< 

ExAJBPLE 3, There \a b.q e\^^^aa ■«'b«* S.raaT>-<«nie i" "**■ 



oAVonro. 137 

is 154 inches, and its conjugate diameter 55 inches; what is its 
snpeifieifis in poands of plate glass? Ans. 688-68 lb. 

ExAitFLB 4. Giren a frame 45 inches long and 15 broad, 
to fmd its area in ponnds weight of hard soap, hot and cold? 

A«. f 24107 lb. hot. 
, -'^^t 24-87 lb. cold, 

ExAMFLB 5. How many pounds weight of flint glass are 
there in the area of a flint glass pot, whose mean diameter is 
27*2 inches? Ans. 67*25 lb. 

16. To flhd the contents of solids of greater depth than one 
inch. 

Bulb L find the content in inches by the rules for the 
mensuration of similmr solids, tben divide or multiply by the 
tabular divisor or multiplier proper to the required answer. 

Rule II. By the sliding-rule. When the vessel is square 
or circular. To the proper gauge-point on D, set the depth 
on C, and against the given side or diameter on D will be the 
answer on C. Bnt if the vessel be a parallelogram or an 
^pse, find a mean proportional between the sides or the dia- 
meters given, and with this, as a side or diameter, proceed as 
before. 

Example 1. A brewer's cylindrical mash-tun is 144 inches 
in diameter and 82 inches deep; how many quarters will it hold 
wbenl^U! 

Bulb I. By the divis(»r for circles for a gallon. 

144xl44xd2.^d5d-036da.l879 gallons=29 qrs., 2 bush., 7 galls. 

By the multiplier for circles for & gallon. 
144xl44x32x*0028d2=sl879 gallons =29 qrs., 2 bush., 7 galls. 

Rule II. By the gauge-point for circles for a gallon. 

To 18*7892 on D set 82 on C, and against 144 on D is 1879 
on C, which being gallons, and reduced, gives 29 qrs., 2 bush., 
7 galls. 

Example 2. There is a square guile tun whose side is 48*7, 
and depth 27*4 inches; what number of gallons would fill it? 

Rule I. By the divisor for squares for a gallon. 

48r7x48*7x27-4-j-277 274x=234-37 gallons. 
By the multiplier for squares for a gallon. 

48-7x48-7x27'4x-003606=234*37 gallons. 

Rule II. By the gauge-point for squares for a gallon. 
To 16*6515 on D set 27*4 on C, then against 48*7 on D will 
be 234*37 galls, on C, the answer. 

Example 3. A soap franco, 45 inches long and 15 broad, is 
fiUed to the depth of 51*4 inches with soap 8ilicated\ h.^^ 
many pomid^ hot and cold, does it contaixi\ 



138 GAVomo. 

, RULB I. 

n ti. 1- ■ „ i 45X1 6x51 ■4-f-24-n4.=l 443-33 lb. hot. 
By the divBOTB, | 45,^1 5^51 -i-i-ai' 30=1 489-05 lb. cold. 
m.l,».„„I.Uli.,. S45xl5x51-4x04169=1442-fBlbhol. 
Bj themultipliMS, ^ 4.,xl5xar4x-04292-<1489-10 lb. cold. 

Rdlb II. Find the mean proportional (Art, 12), which a 
26 nearly. 

Then to 4-9031 on D, set fil-4 on C, and against 20 on D tj 
14431b. hot on C. Ans. 

To 4-827 on B, set 51-4 on C, and a«ainat 26 on D is 1489 
lb. cold on C. Ans. 

Example 4. Grain to the depth of 14-8 inchea ia pnt ml* 
an elliptical couch frame, whose transverse diameter is IIS-Z, 
and conjugate 111 inches; required the number of bnaheltit 
uontains? 

RiTLE.T. By the divisor for circular bushels. 

140-2x111 x14-Q-h28Z4-2903=84-44 bushels. Ana. < 

By the multiplier for circular buahels. 

145-2xl]lxU'8x'000354=84-43 bushels. Ans. 

RuLB II. The mean proportional between 145-2 andltl, 

»126'96, (Art. 12). 

Tlien to G3'1441 on B, set 14'8 on C, and against 126-95 oa 
D. U 84-43 bushels 0: " ' 

Example 5, IIow many pounds weight of plate glass metal 
will fill the frustum of a cone, whose dinmeler at the top is 
2(1, and at tho base 27 inches, and whose height is •'W incheit 

Rule I, By the proper tabular divisor. 

27'+2^+(27x20)xYH-14-387S=1160 lb. Ans. 

KuLB II. The mean proportional between 27 and 20 is 
■" (Art. 12.) 

1 to ■," on C set 3-703 on D, and opposite 20, 23-2, 27, 

is 278, 375, 607, on C. Total, 1160 lb. Ans. 

ExiucPLE 6. Required the content, in imperial bushels, of 

malt hopper, in the form of a truHtum of a square pyrjmtJ, 

and whose depth is 3(5, aide of the lower haw 14, and of the 

upper 34 inches! 

Rule I. By the proper tabular multiplier. 
34+14=48; and 48*^ (34^04) xy'-WWifi 1=98-93 bushels. 
K II. The mean proportional between 14 and 31 ii 
(Art. 12.) 



1 



Then place y on C to 47-0977 
■ — D ia, lO'O, 26-8, (!2-5, on 



tliegj-eater bo'mg 75 \D\igMi4&ft\viii\^^rtQ»i 



OAUGINO. 



139 



long and 35 broad, and the depth of the solid 30 ; to find how 
many pounds of hard soap cold it can contain. 

RuLs I. By the proper tabular divisor. 
75x60 » 3750 
45x35 » 1575 
iisx!^x4»1020q 

15525 
5=Y 

77625-^27-14=2860 lb. Ans. 

Rule II. The mean proportional between 75 and 45 is 58*^9 

50 35 41-8. 

Then to 27*14 on A set 75 on B, and at 50 on A is 138-2 on B. 



^7-14 
27-14 



99 



W 



581 
45 



» 



And286xV=2866'lb." Ans'! 



>9 



41-8 
35 



» 99 



89-6 
58-2 



286 



17. Gauging open vessels. 

All vessels which admit of having their dimensions taken 
from within are considered open vessels. Some of them are 
regular or nearlv so, others are irregular in form, but all can 
be reduced to the forms of solids. For the purpose of facili- 
tating the taking of account of quantities of goods contained, 
and securing correctness in calculations, many are fixed and 
tabulated; while the ascertained lengths, breadths, or diamcr- 
ters^ and the computed superficial areas of others, are record- 
ed in a dimension book. To fix a vessel is to determine its 
poeition and dipping place, to tabulate one, is to compute* and 
form into a table, the area at every inch and tenth of an inch 
of the depth. 

18. To gauge a malt- cistern or couch-frame, soap-frame, ot 
other rectangular vessel. 

Note. In rectangles the shortest distance from end to end, and 
side to side, is the true length and breadth. 

Jhtle^ Measure with a proper instrument or tape, the in- 
ternal length, breadth, and depth of the vessel, and calculate 
the area in the denomination required. The areas of malt- 
kilns are computed according to their forms. 

19. Precedent of a dimension-book for Maltster's utensils, 
and a Soaper's soap- frame. 

Mr G. H. Maltster. 



Ko. of 

Malt 

fitoiuo. 

9 



Ctotamt. 



I 



41 HI 

37-Oj 



! 

i4 



i08*5 
1910 



6 I 



a 



U>8'2 5-2<> 
110-8 16'57 



Couches. 



I 



a 

27*4 leao 

240 1452 



5 

1 



111-6 
111 



i 



•<i|7-26 



KUns. 



No. of 
KUn 



Length. 



Breadth. 



Area. 



,l»t. \ 
\ 2nd. m. 












OAtroiso. 
Messrs J. T. & Co.'s Hard Soap-frame. 



-SS' 


:^ 


BrodUi. 


„-.^ 


c»,..™. 


1 4fi-0 


150 


21-107 


28-47 



aeUn. Hms IM 



■; large oi 



20. To gauge a cylindrical vessel. 

Note. In circlea the greatest disCaoaa between uiy two poiiUs of 
flie circumfproiico is the true diaroeier. 

Rl'ls. Measure the diameter and perpendicular depth with- 
in, and caluulatu the area iu tlte capacity rerj^iiired. 

21. To gauge & vessel ia the form of a &uatuin of a pyn- 

RuLS. Measure the sides or diameter at the top ontl bot- 
torn of the vessel, and the perpendicular depth; find then 



Nota. In measuring circnlar veasela il 
each end cross diametei-s, und Sod the mi 
fimajl vessels ore gauged in the above man 
and tabnlaled. 

22, When the side or diameter 8,t any aswgned depth of • 
frustum of a pyramid or cone b required; 

Bulb. Divide the difference between the top and botton 
sides or diameters, hy the depth of the frustum, and the quo- 
tient will he a common factor, hy which multiply the yn- 
posed depth; and, if the given depth be reckoned from tlx 
smaller end, tht prodoct must be added to the less side o; " 
meter; but, if the depth be reckoned from the greater end,tt>i 
product must be subtracted from the greater side or diame- 
ter; and the sum or ditfereuce will be the diameter o: 
quired. 

KxiupLB. Given a, frustum of a cone, whose top and bot- 
tom diameters are 28 and 41, and height or depth 62 inches,U 
find tlm diameters at 7, 12, 16, 23, 31, and 44 inchea from tiK 

First *1—2B=1 3, di8erenca of the given diameltOT. 
Then 13-^-5^=-25, the cooinwn factor. 

F«tpr. DqWhu. Prnducli. DIamelcri. 

(■ 7= I'75"| (29-75\ r 7' 

13= 3-00 31-00 .13 

Now2ix- llZ ',Z .-.2Btbetopdiam.= f^lJU^Ul 

31= 7-75 • 35-76 31 

U4=ll-00 US'OOJ L4i. 

Were the products subtracted from the greater diameter thi 

remainders would he \,\ie d\e.mete'nU,\.\^«R£ae distances bun 

tils bottom d[ ilie fruaWm.. 



CAuaina. 141 

gauge, fis, anil tabulate a Tessel n-lticli is nearly cir- 

Rtnai. Find the top diameter, which multiply by -7, and 
the product will be the side of the inscribed aquare nearly. 
By triab find the exact eides, mark the angular points, from 
each let fall a plumb line, mark where the plummet touches 
the bottom, and then, withaehalked line, strike a line each way 
across the bottom, and pussjng tlirough the oiipMite marka. 
Strilce lines up thui^dea, joining the top and bottom points, 
and the veaw! will be quartered. Determine the dipping 
place, from which let Ckll a plumb line; mark where Uie 
plummet touches the bottom, and from this mark take the 
Teasel's perpendicular depth. Divide the depth into frustums 
of 10 inches deep, beginning at the top, if tlie vesspl is to be 
gnuged for dry mches, that is, when the distance between the 
top of the liquor contained and the dipping place, is to he af- 
terwords taken to determine the quantity in the yessel, but if 
gauged for wet inches, which is when the actual depth of the 
liquor is taken, beginning at the bottom. 

At the middle of each frustum take cross diameters on the 
lines on the sides. Find the mean diameter and area of each 
fhtstnm. Multiply the area by the depth of the frualum 
for its content, and these contents added together will be the 
conUnt of tile whole vessel. 

If the bottom be nnlevet, measure water into the vessel till 
the bottom be just covered. Find the deepest depth, and 
o«er it fix the dipping place. 

Should there be any incumbran-ces, calculate their solid con- 
tent, and deduct the proper proportion from the frustum, or 
pan of a frustum, in which situate. 

To tabulate a vessel. Calculate the areas at every inch and 
tenth of an inch, as may be required, of each frastum, which 
areas, if the gauging be for dry inches, must be suceeaNvely 
deducted from the full content, if for wet inches, be similarly 
added to the quantity to cover the bottom. 

When the aides ol a vessel are much curved, the frustums 
ahouid be only 4, S, or 8 inches deep. 

Large vessels, and when great exactness is required, are 
divided into 8 equal parts at the top and bottom, by quMter- 
ing as above, and then dividing each quarter into two equnl 
parts, and striking lines between the ends of each correspond- 
ing diameter. In these cases there are four cross diameters, 
whose sum must be divided by 4 for a moon diameter, by 
which the area must be found. 

ExiUPt^ Of the gauging and tabulating to wet laches of 
a vessel, nearly circular, by the above rule. 



i 



Mr J. P.'s Spirit Receiver, gaiiced 12tli January 1814, 
by J. R. and W. B. 



1 Cr.»l>l^..c~ 


H^n 


=. 




— 


■SSSL- 


1 1 1 ! ^ 1 4 


..r. 


n^tb 


i>.c. 


U1-* lD.ip-perinea.ure 
l^T.Drplh. 


El 


w^we? 




74S7 


1 


S 


_ 






c..,„,. 


I«IU!« 



INCUMBRINCE. A beam across the top, whose length is 67'i 
breudth 03'), and de])th 05-2 inches, and whose ares b •7417 
gallons, which mufit be deducted from the area of the first Wi 
inches from the top. 

Precedent of tabulating to the incli. 



WetincHei 


o.«.. 


i"hl 


Gallon!. 


lnTh"«. 


GMllam. 


Drip 

,% of lowest area 

iowEst area 
3 

5 

B 

10 

-2d ureii 

11 

12 


21-0001)0 


14 
IS 
16 

17 

18 
19 
20 
Sdarea 
21 
22 
23 
34 
25 
25 

ar 

28 


332-74488 
356-86758 
380-59028 
404-51298 
428-43668 
452-36838 
476-28108 
22.53770 


29 
30 
4thar«a 
31 
32 
32-3 
,', Sth area 
32-i 
33 
34 
35 
36 
37 
37-5 


57M3H 
70H6M 

ai-ssm 


24-SSI60 
65-19388 
83-74448 
U4-2!)G08 
138-84768 
163-3!)928 
187-95088 
212-S0248 
237-05408 
23-92270 


723-3148 
745-OSIB 

rsi-sMT 


7SS-7S«3 
766-1981 
TtTiat 

aoe-OBW 

829-0308 

849-97M1 
860-44784 


(98-81878 
521-35648 
543-S94I8 
566-43188 
588-90958 
611-50728 
634-04498 
656-58268 


284-89948 
308-8-2218 



NoTJi — For the sake of cnncipenpsa tho first addition only of tif J 
area of each fmBtuni Le shown. Had the tabulation been to tb* 
tenth of BQ inch, each area niuat have beeo divided by 10, beliiK 
being added. From this table another is conBtrueted in whj^ MEMl 
pilous only are given, and in the conBtruction of whicli all bdsw- 
Hii-teiilhs IB rejected, ani att b,\™-^b VaV™ tn he another gallnk 
Thus, the content in the above «i.Uea,\,iimft \B'Siw*-«wiA,\a *iieiii- 
c-ond table be 2\2 gallona, tni &« co'n\oiiV »*. \^ m*** «>#^>a 
35? gallons. 



OAUOINO. 



143 



24. To gauge a vessel nearly rectangular. 

Rtn:.E. Find the perpendicular depth; divide it into sec- 
tions of 10 inches or less each. Take two lengths and two 
breadths at the middle of each section; add the two lengths 
and two breadths respectively together, and divide by two for 
a mean length and breadth, which multiply together, and re- 
duce the product into the capacity given ^r the area. Mul- 
tiply the area by the depth of its section for the content there- 
of; add all these contents together, and the sum will be the 
content of the whole vessel. 

Example of a Brewer^s mash tun gauged by the above rule^ 

Mr R. H.'s Mash Tun, gauged 10th Feb. 1844, by T. R. and 

J. S. 



IMllS 






Men 
Length. 






Mem 
Breadth. 


Arru in 
Oalloni. 


Areas in 


Contents in 1 


illOIU. 


.I^enguu. 


BraaatDs. 


Qri. 


B. 


Galloni. 


Qrs. 

4 
3 
3 
4 
4 


B. 

4 
5 

7 
1 
2 


Gallons. 


13 
10 
10 
10 
30 


98-6 
100*4 

1027 
I0AH> 
107*3 


99*2 
100-8 
1031 
105*(» 
107*2 


969 
100*6 
102-9 
1060 
107*2 


636 
66-1 
68*0 
69.6 
71-8 


62-9 
66-0 
68*4 
70*1 
72.0 


63*2 
660 
68-2 
698 
71-9 


22-5392 
23*9423 
25*3061 
26-42a3 
27*8038 


• •• 

• •• 
••• 

• •• 


2 
2 
3 
3 
3 


6 5392 
79423 
l*3(>ni 
24283 
3-8038 


5*0096 
7-423«l 
5-0610 
0-2830 
6-0380 


*>. 


Total 


depth. 


i 


:;ont( 


mt. 


20 


5 


7*81'46 



To tabulate the above, proceed as in (Art. 23), but to qrs., 
(ushels, gallons. 

• 25. To gauge and fix a rectangular back or cooler. 

* Rule. Measure the mean length and breadth, and find the 
area. Cover the bottom with fluid; take the depth in various 
p&ces, add the dips together ; divide the sum by the number 
of dips, and the quotient will give the mean depth. Try till 
a depth of the fluid is found, as near the side of the cooler as 
poedble, equal to the mean dip. Fix this as the dipping place^ 
snd opposite to it, on the edge* of the vessel, make a mark, 
and specify also the distance the dipping place is from the 
point marked. Measure the whole depth of the cooler; cal- 
culate the area to every tenth of the depth, after deducting 
therefrom the mean dip, and tabulate, as in (Art. 28.) Should 
the place of the mean dip be inconvenient, any other place may 
be taken, by marking its position, and showing on the vessel 
how much deeper or ebber it is than the mean dip, which dif- 
ference must be subtracted from or added to the mean dip, as 
the case may be. 

A mean dip answers only when the bottom is covered with 
fluid ; under other circumstances it would be better to fix the 
dipping place at the deepest depth. 

Example of the Gidng and gauging ot a Teclaxk^vjA^ax \i^'c?^ c»^ 
cooler. 



* c 

£ , £ 

3 

* i. * 



144 OADOINC. 

Let the fignre ABCD represent m rec- 
tangular back or cooler, whoso mean 
leogth, AB, is 17s inciies, and mean 
breadth, CD, 124 inebes; let the deptha 
b« at <i (t3-6, at b Oa-3, at c 08-8, at rf 03-^ 
at e 033, at /"oaA at g 03'0, at A 03-2, - 
at ■ 02-S, and at A 03-0; to find the area 
and meGui depth. 

Now 175xl24-217'XI, which, if gallons 
be the denomination reqaired, being divid- 
ed bv 277-274, gives 78'2619 galk., or 2 
barrels and 6'2GI0 rbUs. = area. Then 
03-6, +O;i-3+O3'8+03'2+O3'3+03'0>O3'O 
+03-2+02'9+03'0=31-2 and 31-2-1-10, (the number of dipi ta- 
ken), =312«tnean depth. By trial a dip =3-1 ia found at 
0, opposite to it is made the mark^', and it being i inchM 
from the side, a note to that efidct ie also marked ob Um edg» 
of the cooler. 

26. To gauge a copper with a riung ci 

Rdie. With a straight edge or cord, hetd tightly acrori 
the mouth of the copper, aa a diameter, and with a Une * 
plummet, find the perpendicular distance between the tc, 
the crown, which should be ita centre, and the straight edge or 
cord; measure this distance, and it will be the depth of tha 
body of the copper. 

In the same manner measure the perpendicular distance be- 
tween the straight edge or cord and the lowest point of the 
bottom, where the crown rises from the sides. Twice the di»-' 
tanco that one perpendicular ia from the other, will be lb*' 
diameter of the splierical seement forming the crown, anil 
their difference in length will be the altitude of the s^men^ 
or rise of the crown. * 

Quarter the top and bottom, so s,a to touch, but not cut tiia- 
crowTi, and join tne corresponding points by lines, drawn with. 
a straight edge and lung pescil, up the sides. Divide thai 
depth of the body of the copper into frustums of 4, G, 8, &c»} 
inches deep ; take cross diameters in the middle of each fru^j 
tum; find the mean, and comnute its area, which raultipl|l| 
by the proper depth, and add the products together. Calci^i 
late the content of the frustum containing the crown and tbsi 
content of the spherical segment or crown, by the rule fiJt 
computing the solidity of St segment of a sphere, page 9()>i 
From the former subtract the latter content, add the remain-t 
der to the sum of the products or contents of the other ftiw, 
turns, and the total will be the content of the whole copperi 

The most expeditious and satisfactory way of finding thH 
content of the frustum containing the crown, ia to ^i^^g 
the crown with liquor, and then draw it off into ft propelj 



I'he tabulation. m\i.a\.^)e\a'\iaxit\'Q,&[<£ixa, Bud galloni) 



a 



0At7QING. 



145 




for dry inches; hence the areas xnttst be successirely deducted 
from tile fall content. 

Example 6f a copper With a rising crown. 

Let the figure ABCD 
repreeent a copper with 
a rini^ crown, to he ^ 
ganged and in<diedy and 
whose dimensions, ta- 
ken by the above rale, 
are depths BF 48 and 
GC 54 inches; cross 
diameters, at 6 inches 
from the top, OO'O and 
90*1; at 17 inches, 85*3 
and 85*5; at 26 inches, 
89-2 and dOl ; at 83 

inches, 76^4 and 75'8; and 39 inches, 70-8 and 71*4; and also 
at 45 inches, 67*4 imd 67*4 inches; and let the cross diameters 
at the top of the crown bead's and 64*3; and at the bottom of 
the crown 59*8 and B0*2 inches. 

j^QYT <»'8H-g*'3 «64*0» mean diameter at the top of the crown; 
and ^'^^^'^ «a60*0=mean diameter at the base of thecrown. 

Then 642+602+(64x60)x 5^, the height of the crown, 

X •002832=65*3399 gallons = the content of frustum CFDH. 
Again, Y=30= radius of base of the crown, and 64 — 48=6 
ssheight of the crown; whence (3x30^+62)6x*6236x-003606= 
30*9950 gallons = content of the crown, page 90. Therefore 
65*8399— 30*9950 Bd4'3449 gallons required to cover the 
crown. 

Form of Table for Dimension-book. A. B.'s copper, gauged 

16th March 1844. 



f 



19 
10 
8 
6 
6 
6 



48 



lOCllM 

fifom tha 
top. 



6 

17 
86 
33 
39 
45 



Cron 
Obunetera. 



Mean 

Dlaine> 

ten. 



ft^'OWl 



85-3 85*5 
80S dOl 
76«4 75-8 
70-8 71 '4 
e7'4l67-4 

To cover th« Crown. 

Deeth and Content 



90- 
85 
80< 
76 
71 
67 



Area* In 
OaUoos. 



22 9438 
20-6583 
18'1738 
16-4(i40 
14-3192 
12*8676 
34-3449 



Areas in 



B. 



2 
2 
2 
1 
1 
1 



Gallons. 



4*9438 
26583 
01738 
7-4040 
5-3192 
3-8676 



Content in 



B. F. Gallons. 



7 
5 
4 
2 
2 
2 



25 



2 
2 

2 
1 

3 



2 



5-3256 
8*5830 
1*3904 
8-4240 
4-9152 
5-2056 
7-3449 



6-1887 



27. To gauge a copper with a falling crown. 

Rule. Find, as in (Art. 26), the perpendicular depth, from 
the straight edge to the angle where the body and crown of the 
copper are joined. Quarter, and take cross diameters ; find 
the area and content of each frustum ; the content of the 
crown^ by measurement or calculation; add tYi^-wVc^a <wcl- 
tents together, and tabulate, as in the piecedvn^ ^tVa!(^<^« 

1* 



lo gan^ a Teasel \rith a fall or drip, 
'hen a vessel U in an indiaed poaition, tlie quantity d 
oesMry to cover the bottom is aaid to be the fall or drip. 

RuLB. Measure into the veasel water sufficient to cov 
the bottom, or ascertain the quantity required, by the rules 
for finding the solid contents of ungulas. Find the greatnt 
depth of thu fall or drip, to which let fall a plumb line from 
a point level with the lower side of the top of the vessel; 
deduct the depth of the drip, and the remainder will be th« 
depth of the vesael, as positioued. Divide this depth inti 
proper segments, of which find the areas and contents, ae 
cording to (Art. 23), and the sum of the contents, added to the 
quantity for the drip, will be the whole content of the vessel 
in its inclined state. 

Tabulate as before. 

Should the fixing of the dipping place over the greatrrt 
depth of the drip or iall be inconvenient, any other point ma]' 
be taken, if the difference in height between the two paints 
be added to or subtracted from the depth, as the case may U. 

29, To gauge and tabulate an irregular oral vessel, bj 
means of ordinates. 

Let the two following figures represent the hases,OT battoiu 

' 'wo irregular oval V' 
Fig. 1. 




GAUOIKO. 147 

To find the centre, and to quarter the base, of an irregu- 
lar elliptical vessel. Draw any lines, as ab and cd^ Figure 1, 
parallei io one another; bisect them in e and/; through e and 
/draw the line gh^ the bisection of which is O, the centre of 
the vesseL With as a centre, and any distance, cut the 
periphery in h and /, which join, and bisect kl in m. Through 
the point m and the centre draw AB, which is the trans- 
verse diameter. Through the centre O, and perpendicular to 
AB, draw CD, the conjugate diameter ; then the distances 
AD, DB, BC, and CA, win be equal to one another, and the 
figure is quartered. 

To draw the ordinates. 

Parallel to AB, the transverse diameter, Figure 2, draw 
MN, cutting CD, the conjugate diameter in H. On the lines 
AB and MN, and on either side of H and 0, the centre of the 
vessel, set off any even number of equidistant points, as 
Jl, 1'; 2, 2'; 3, 3'; 4, 4'; 6, 6'; 6, 6'; 7, 7'; 8, 8'; 9, 9'; 10, 10'; 
the sum of all the distances of which shall be less than the 
transverse diameter at the top of the vessel. Through these 
points draw lines terminating both ways in the periphery, and 
they will be the ordinates. 

To draw lines from the ends of the ordinates up the sides of 
.the vessel* 

Fix a string or straight edge across the top of the vessel, so 
that it shall be exactly over the respective ordinate on the 
bottom, which will be the case when the plummet of a plumb- 
line, ffidling from any two distant points in the string or 
Btraight edge, shall touch the ordinate. Then with a chalked 
cord strike a line up the sides between the ends of the ordi- 
nates and the points on the top cut by the string or straight 
edge. Draw such lines from the ends of all the ordinates, and 
of the transverse and conjugate diameters. 

Fix and mark the proper dipping place, (Art. 23,) and ascer- 
tun the perpendicular depth of the vessel at that point. 

Divide the depth into proper frustums of 2, 4, 6, 8, or 10 
inches, beginning from the top and leaving the odd inches in 
the bottom frustum, if the tabulation be for dry inches, but 
if for wet, beginning at the bottom and leaving the odd inches 
in the top frustum. Take the ordinates in the middle of each 
frustum, except there be a curb or cover through which the 
gauges must be taken, and if the tabling be for dry inches, then 
deduct the thickness from the top frustum, and take the or- 
dinate in the middle of the difference. 

Should the sides of the vessel be other than perpendicular, 
the points for the ordinates must be marked, so that a line 
drawn between any two correlative points, will cross the per- 
pendicular depth at right angles, and at proper distances from 
the top. 

To find ihegroBB area and contant of eacViitvx^Wxsu^ 



Its 

Rule, To the snm of the first an J Inst ordinateB, adil fiinr 
tiniea the aum of the even ordinatei, and doable the a um of i 
the W9t; then to the product nf the total -multipUad by the 
eqnidiatance of the ordinates, add the prodnct or the aiiiti of 
the extreme ordinatea, multiplied by the sum of the segments, 
that is, the difference between the Ivanaveifle diameter and tlie 
part intiTcepted by the cxlrame ordinates. One-third nt tliis 
Bum divided or multiplied hy the proper divisor or factor for 
B(|iiare8, will be the area in gallons, which multiplied by llu 
height of the frustum will give the content thereof. 

Measure the quantity required to cover the bottom, to whitli 
add the contents of the several frustums, and the si 
the content of the wliole vessel. 

Find the depth of the drip, subtract it from the gross deptli. 
the remainder, deducting the curb as before directed, will be 
the net depth to be tflhulated. 

When there are incumhTsnces, compute their areas and va- 
tents, which deduct from the areas and cuntenta of the r ~ 
tire frustum, 

KxAHFi.B of the ganging- of an irregular vessel. 

Suppose Figure 2, page 141, to repreafat the base of ai 
guiar ovul vessel, on whieli 11 euuidistant urdinates have bcA 
taken, and also in the middle of every frustum, according t< 
rule, and that the dinieunlons, areas, and incumbrances, an I 
in the following scheme; — 




ate|?.J|:!|«.. 

--'blfP 77-" '»■»«■» 



m-smtI 



Incumbrances. Two iieanis inside the lop, each 80 i 
lonit, 04 broad, and Ofi'2 deep. Area 2'4809 gallona, 
deducted from the first 06'2 inches from the top. 



1i 



"*f 



GAUGING. 149 

leraiionsfor finding ihe areas for the several frustums. 





FnutuoM. 1 




1 


8 


3 


4 


5 


6 


'»«'»[£« 


24-7 
SI '3 


253 
84*0 


25*6 
86*5 


27*2 
88-6 


3(H} 
3*7 


317 
827 


eme ordinate*. 


46-0 


49*3 


52-1 


55*8 1 007 


64*4 


. Second 
V Fourth 
jcs, < Sixth 
4 Eighth 
^ Tenth 


47-2 
896 
760 
7«'-5 
490 


48-6 
7«»*4 
76-7 
71*1 
49-5 


■ m 

71*0 
77-1 
71*5 
50*9 


51-4 
71-6 
775 
72-2 
51*6 


52*5 ■ 
72-8 
780 
73-0 
53*2 


72*9 
78*3 
73-5 
54*1 


ordlnates. 


318-3 

4 


316*5 

4 


320*4 
4 


324-3 

4 


328*9 

4 


302*4 

4 


ordin. x 4 =s 


1249.2 


1266*0 


1281*6 


1297-2 


13756 


1329*6 


e Third 
) Fifth 
* 1 Seventh 
C Ninth 


(Ji-4 
74-8 
751 
62-1 


0^*2 
75*3 
75-7 
62*9 


62-8 
75-6 
76 
63-5 


63« 

76-2 

!76*6 

64*4 


64 3 

767 
770 
66*3 


65*4 
77-2 
77*3 
66*0 


irdinates 


27S4 
2 


2761 
2 


2779 
2 


880*8 
8 


283-3 
8 


285-9 
2 


wd. X2s=» 
do. x4s 
do. 


546*8 

1249*2 

46-0 


552-2 

12660 

49-3 


555*8 

1281-6 

52*1 


561-6 

1297*2 

55*8 


566*6 

1315-6 

607 


571-8 

1329*6 

64*4 




1842*0 


1867*5 


1889-5 


1914*6 


1948-9 


1965-8 


lie equidistance. 


18420* 


18676. 


18895- 


19145- 


19499- 


19658- 


nneordnates 
egments 


460 
3-4 


49*3 
4-8 


5-8 


i^*l^ 
6-8 


6)7 
8*1 


64-4 
9-2 


rd.X8umofseg. 


156*40 
18420- 


236*64 
18675. 


30218 
18895- 


379-44 
19146- 


491-67 
19429- 


592-48 
19658- 




18576-40 


18911*64 


19197*18 


19525*44 


19920^ 


20250-48 


c inches 


6192-1333 


6303*8800 


6399-0600 


6508*4800 


6640-2233 


675a-l6'J(J 


77^41 
ans 3 
nbrances 


22-a321 
2-48U9 


227362 


230784 


23-4730 


23-9482 


24-3447 


cum. deducted 


19*8512 


227352 


23-0784 


23*4730 


23-9482 


[24-3447 



5»r wet inches tabulate as in (Art. 23); if for dry, as in 
28), only commence with 1*9 the depth of the curb instead 
ill." 

To gauge a still. 

Is are of various forms. Those in common use consist 
lead and body. The head is generally a conical frus- 
the body a copper with a rising crown, and covered 
I spherical segment. 

le stills are too small to admit of the internal dimen- 
being taken easily and accurately. Such may be gauged 
dng the external dimensions, and dedxiclvsY^ \X\«tAxw«v 
ickuess of the copper or metal of wMch. TCia^"fc\ \s>qX ^^^^ 
\d moat satisfactory method would \)e to ft\V ^^^^^^ Vv^^^ 
nd draw it off into a given measure. 



ISO 



I A usma. 



To gniige those which admit of their dimensions being ia- 
ktii interiiully. 

Role. From the point where the glohular part of ths 
hody heRiaa, extend a, line na a diameter; measure the perpeo- 
liicular dlBlance between the centre of the crown nnd the top 
of the glohaler part, which will be the whole height otlhe 
atill ; from this deduct the depth from the diameter tfrtht 
trrown, and the remainder will he the hei^lit, and the diame- 
ter the huge of the g-lobular portion. The part below tha 
diameter will be a copper with a rising crown, and ita dimen- 
gione mnst be taken and content found by (Art. 26). The glo- 
bular part wilt be n spherical zone, and its content must ba 
found liy the " Rule for meaanring the aolidity of a zone of a 
aphfre, pai^'e 81>. The Bum of the contents thua found will 
he the whole content of the body of the still. 

To gauge a still head, 

Rvi.B. If it be of an irre^lar form, divide it into patti, 
whose foriiiB are regular. Take the internal dimenaions, if 
possible, of each ; if not, the external dimensions, and allow for 
the thickness of the melal, Compute the contents by tlie 
" Rule for the Mensuration of Solids" whose figures are simi- 
lar. Add the contents obtained to the content of the whnle 
of the body, and the sum will be the full content of the »tiU. 

EicAHrLB, Suppose 
the annexed figure to 
representa^til!; Athe 
head, a frustum of n 
I'one; B the breast, a 
ifone of s sphere; and 
C the body, a copper 
with a rising crown; 
and the diuensiona to 
be as follows: of the 
head ab, the height 2ti 
inches, diameters at 1C\ 
tapJfflV& and 12-1, ' 
at bottom lij 23-8 and 
24'2; depth of breast 
br 13'0j diameters at 
A; 238 and 24-2, at A/ 
6!)-« and 60-4; depth ' 
of body 27-0, diame- 
tt-rs at 4'6 from top, c, 
5(i-8 and 6r'2, at 13-5 from c, 50-0 and 60-4, and at 22-8 from 
c, 45 '3 and 45'1; depth of rrown, de, 4, diameters at top, m". 
'" " and 43-2, and at bottom, op, 39-7 and 40-4 incbea, to find 
hole content. 




^^the whi 



=12-0, ■ 



^!i?:r-*JL =niVQ,>C&stt 



GAUGIKO. 151 



12«+24^+.(12x24)VX*002832-21*885e96 gaUs., the content 
of the head. 

2d, ^^ -12-0, and ^^^ -30-0, then 

12a+308+ *^ xl3x2x-002832=81-019744 gaUs., the content 
of the hreast, 

3d, 55f±*?:?=,57.0; then 572x-002832x9«82;81051 2= con- 
tent of Ist frustum. 

^^°*^ - «60-2; then 50-22x-002832x9ifc64-230779=: 2d, 

and^^^^*^«46-2; then 45-22x-002832x9«52W3003= 3d, 

199*114294 galls, 
the content of the hody. 

4th, ^!l±i51=43-0, and ¥1±^=400; then432+402+(43x40)| 
x*002832sil9'518144 gallons, the content of the frustum con* 

taming the crown; and »'7+^-^ ^20; whence (202x3+42)4 

x*5236x*003606» 9-183726 galls., the content of the crown; 
therefore 19*518144— 9-183726=10*334418 galls, to cover the 
crown * conseouentlv 

21-885696 + 81-019744+199114292+10-334418=312-354160 
galls., whole content. 

These dimensions, areas, and contents, should he formed into 
a tahle, as in (Art. 26), and inserted in the dimension-hook. 
When stills are gauged with a view to tahulating, the height 
or depth from the top of the hreast to the top of the crown is 
divided into frustums of three or four inches deep, and cross 
diameters are taken in the middle of each, and the areas and 
contents are calculated as for cylinders. 

Malt Gauoing. 

Malt is made from harley, grain, or pulse, which must he 
covered with water for at least 40 hours in the cistern, and 
then removed to a couch-frame, where it must remain for 26 
hours; afterwards it is put and wrought upon the floor until 
ready for drying upon the kiln. During the manufacture, 
gauges must be taken of it in cistern, couch, on floor, and oc- 
casionally on kiln ; and that gauge which shows the greatest 
net quantity is set forward, and charged with duty. 

» While in cistern or couch, one hundred bushels by gauge 
are estimated as 81-5, and while on floor or kiln, as 50 bushels 
net. 

Cisterns and couch-frames must be constructed with their 
sides at right angles to one another and to the bottom. Cis* 
terns are not to exceed 40, and couclviiaixi^^ ^ vclOcv^ vol 
depth. 






GJlVGISG. 

To gauge malt in ciatetn, coucii-frame, or kiln. 



ItDi.B. Take 5 or 10 dips of the malt, odd them together, 
and divide the suiu by tlie numtipr of dipa, for a meaii depth, 
by which multiply the arsa of the vessel, preTionsly comput- 
ed, for the quajitity contained. 

ExAUPLB 1. How many bnahels of malt are there in a cl»> 
tern, the area of which is 6'67 bushels, and the dips 26-4, 2S'3, 
a4'9, 28-0, 26'8 inches. 

Now 26-4+25'3+24-9+'28-0+26-8=]31-4, which -^3, tha 
number of dips, gives 36'2. This multiplied by 6' 57, produces 
172'13 bushels. Ana. 

By the sliding-nile. 

Place 6-67 on B to 1 on A, then beneath 20-2 on A is 17S-I3 
onB. Ana. 

ExiMPLE 2. The aren of b couch-frame is 8"11 bushels, and 
the dips of the malt therein 22-0, 21'3, 21-9, 22-4, 23-0, 21-^ 
22'0, 22'2, 23'1, and 22'5 inches; how many bushels does it 
hold? 

Now (22'O+81-8+2]-9+22-4+23'0+21-7+22'O+2E-2+23'l 
+22'fi)H-10, the number of dips is 22-2; fhia multiplied by 
8-11, gives 180'04 bushels, Ans. 

By the sliding-rule. 

To 8-11 on B set 1 on A, and against 22-2 on A is 180'04 
on B. Ana. 

32. To gauge malt on floor. 

Rule. Find the depths in various places, as in (31), and 
compute the mean depth. Measure nith a tape the length 
and breadth of the malt on floor. Should the sides or ends be 
slant, take the dimensions from the upper edge of one side or 
cud to the lower edge of the other. 

When tile fonn of the floor is irregular, divide into regular 
forms, measure tlie diniensiona, and calcnlale tiie content of 
each separately; add the contents together for the whole con- 
tent. 

E.'CAMPLB 1. There is a Sooe of malt whose length is 450, 
breadth 372, and mean depth (U-3 inches; what number of 
bushels is there (herein? 

(460x372x04-2)h-2218-1&2, the divisor for bushels, t»31Q-9 
bushels. Answer. 
By the rule. 

To 4-2 on D set 4S0 on C, and against S72 on A is 316'9on 
C. Ans. 

Example 2. An irregular- formed floor of malt is divided 

into two pieces, of wh'ic^ owe laftX-^itW mean depth, 278 tliB 

fcreadfh, and 304 inchea Uic \ew¥jX^', ^^\^n.■0l\'6■^■\axBwsa^^!^ 

OSl, tcngth 553, aMbteaAV^ iflaw^eaatt-, t<<Piiv-H&,'^'-^-^- 

eonteut of the floor inbusWVat 



GAUGING. 153 

Noiir (8©4x2r^08^2)-5-2218«192t« 121-9 
And («6dxl08x05-l)-^2218-192s 137*3 

Tota], 259*2 bushelB. Ans. 

By the hiding rule, 
let, To 3-2 on D set 304 on C, and against 278 on A is 1 21*9 on C 
2d, 6-1 „ „ 553 •„ „ 108 „ 137*3 

Total, 259*2 Ans. 

33. To reduce the quantity of malt injone stage of operation 
to the quantity in another stage. 

Ist, To reduce cistern or couch bushels to net. 

Rui.E. l^fultiply the gauged quantity by 81*5^ and divided 
the product by 100. 

%, To reduce floor or kiln bushels to net. 

Rttijb. Multiply the gauged quantity by 50, and divide 
the prodact by 100 ; or divide the gauged quantity by 2, the 
quotient will be in either case the anstver. 

3. To reduce dry barley to floor or kiln bushels. 
RuuL Multiply the given quantity by 2. 

4. To reduce dry barley to cistern or couch bushels. 
Rule. Multiply the given quantity by 1*227. 

5. To reduce floor or kiln bushels to cistern or couch 
bushels. 

BuitS. Multiply tlie given quantity by -CI 35. 

6. To reduce cistern or couch bushels to floor or kiln 
bushels. 

Rule. Multiply the given quantity by 1*63. 

ExAUPLB 1. Given 250 bushels of malt in couch-frame, to 
determine the con'esponding quantity in floor and net 
bushels. 

Now 250x1 •63=407*5, the floor bushels required. 

And 250x81•5-^100=203•75, the net bushels required. 

ExAMP&B 2. A circular kiln, whose area is 23*93 bushels, is 

covered with malt, of which the mean depth is 7*5 inches ; 

required the quantity net, and corresponding cistern bushels? 

Now 23-93x7*5«179*4 bushels on the kiln. Then 

179.4-i-2=89*7, the net bushels, and 179*4x-6135«110-0 

bushels in cister% as required. 

Cask Gauging. 

% Casks, from the difl^erences in their foTm^,\\«bNfe,^«t 'Ccv'Sk 

mrpoae of ganging, been divided into four Nat\^\.\ft^, «Jgt'fc^^^^5^ 

/rith the solids to which their curvatures Yiov^ W\^ <:i\ft^^^"^ ^^' 



1S4 

eemTilance. Those which are much curred between Ibe 
middle and ends, are considered as the middle frustum of « 
spheroid, and are called the first Tariety; those which are cor- 
respondingly less curved, aa the middle zone of a parabolio 
spindle, the second variety; those whose similar parts tn 
very slightly curved, as two equal frustums of a parabulic 
conoid joined togetlier at the greater ends, the tliird va- 
riety; and those which are straight betweeu the middle and 
the ends, as two equal fjiistume of a cone joiued together St 
the greHter ends, tlie fourth variety. 

35. Of instruments used in cask gauging. 

These are the diagonal rod, head-rod, bnng-rod, and OOH 
and long callipers; they are adapted to casks of the 1st varirtj. 

The diogonul rod is i feet long and '4 of an inch square, nod 
folds by joints to 12 inches in length. On one aide is a diago- 
nal line for from 1 to 240 gallons; on the upper edee of U» 
aide, from left to riglit, is a line for ullaging a half-hogsheld 
lying; and inversely, from right to left is a line for nllaeiDg a 
half-hogshead standing. The second side is divided into inchot 
and tenths. On the third side are two lines for nllaging a 
barrel lying and standing, two for a hogshead lying and ataod- 
ing, and reversely, two for a kilderkin lying and standing, and 
one for a firkin standing. The fourth aide ismarked with tw* 
lines for ullsging a puncheon lying and standing, two for a bolt 
lying and atunding, and reversely, one for a firkin lying ""■' 
principal line is the diagonal line, and is formed thus: I _ 
periment, a cask containing 144 gallons has a diagonal of 40 
inches or very nearly, hence 144 is marked on the rod againsl 
40 inches; then since the contents of casks are as the cubes of 
their diagonals, as 144 : 40^ : : any other diagonal : cnbeof the 
quantity. 

Tlie head-rod oonsista of two parallel parts, joined at tk 
ends, and a slider which moves between them and has t*a 
f^ces. On the end of one face is tixed a piece of brass crooked 
in the form of _|~ which projects one inch; the lower edge of 
this face is divided into inches and tcntha, reading from tb* 
right, on the upper edge is the line of numbers com men cio^at 
25 on the left end; the slider, on this face, has attached anin- 
dex of brass, fixed perpendicnlarly, from which to the right 
end Is continued the line of inches from the rod; on the Wl 
of the brass are two scales, one marked spheroid, the other a 
line of inches; on the upper edge of the slide is the line i>f 
numbers, having a double radiua, and beginning at IS'TSKV 
tlie gnuge point for circles. On the upper part of the rod ani 
on the slide, of the other fuce, are lines of numbers, and on 
the lower part of the rod the line for uliaging, marked 

The bang-rod is hnlf-an-inch square. Two sides are diTidcJ | 
into inches and tenths; another marked, imperial area, ebo** I 
the area in gallons wHea B^^^i^>l \ii ^-Va &\B.uu;t«r of a circN; 



GAUGIRO. 155 

ind the fourth expresses, when measuring the diagonal from 
lie lowest point of the head to the middle of the bung, the 
sontent of the cask. 

The cross callipers are two rods, sliding on one another, and 
laving at one end of each a perpendicular leg; the sliding parts 
lie divided into inches so as to show the distance the 1^ are 
ipart. 

The long callipers are similar to the cross callipers, but with 
horter legs and returned at right angles. 

36. In finding the content of a cask by the diagonal rod; — 
»lace the rod so that its bevelled end shall touch the lowest 
loint of one head of the cask, and observe the number on the 
liagonal line cut hv the centre of the bung; turn the rod to 
he other head and repeat the same; the mean of these two 
[oantities or numbers will he the content sougl^. 

37* When the content is to be found by the callipers, the 
Messary dimensions are the head diameter, the bung diame- 
er, and the length. 

To find the head diameter. Put the crooked brass at one 
nd of the head rod into the chimb of the cask, move the 
Uder, till the piece of brass or index on it touches a point f 
p the opposite chimb, and against the index will be the dia- 
leter in the lower part of the rod, except the index is with- 
at the rod, when the end of the rod will be against the dia- 
leter on the slider. Cross diameters must be thus taken at 
Oth ends of the cask, a mean be found between each pair, and 
x>m these a third mean Ji>e taken, which will be the head dia- 
leter* 

To find the bung diameter. Place the bun^ rod perpendi- 
nlarly in the bung hole, observe the dimension cut thereon 
Y the inner edge of the hole or stave, for the vertical diame- 
t; lay the cross callipers with the legs downwards, over the 
liddle of the cask, which is generally the centre of the bung- 
ole, expand or contract the callipers till the legs touch easily 
)th sides of the cask, deduct two inches, or more or less as 
le thickness of the stEives may warrant, &om the dimension 
town on the callipers, the remainder will be the horizontal 
ameter, and the mean, between it and the vertical diameter, 
ill be the bung diameter. 

To find the length. Apply the long callipers along the cask 
• as to touch both ends; do this in several places, the mean 
f the dimensions shown by the callipers will be the length. 
Should the cask be of a variety other than the first, it is 
istomary to make an allowance in the length, according to 
LB form. 

38. To compute the content of a cask by the head rod. 

Set the index on the slide, to the head diameter on the 
wer part of the rod, on which find the bung diameter, and 
ark the number it cuts* on the scale marked ^\|\ie\ orot. \^^ 






isa 

dtde. Applj this number to tlie lower scale on the slidr, and 
beneath it on the rod is the mpari diameter. Fix the leli end 
of the sUde to the lensth of the cask on the upper part of tlie 
tod; find the mean diamelci- on the upper Ime on the elide, 
and over it, on the up|>er pArt of the rod, is the content iu 
gallons. 

Note. Wlien nnj dimension for finding the content of n out bj 
the head rod is tliroira off tlie role, work with hslt thnt dimenaon, 
Uid doubli! the answer. 

If the Begtnent or inean dLainet«r lie throivn off, double titc 
mean disnieter, and tahe \ the answer. 

If the mean diameter and length be thrown off, donUl 
both,' and take ^ the answer. 

39. To compute the cojitent of a cask by the sliding-nil«. 
Subtract tlie head from the bung diameter, and inuttipl; 

the difference, when it is 

Add the product t-o the head diameter, the sum will bellu 
mean diameter. Then act thelen^h on B tn the gauge-p<nnl 
UK D, and against the mean -diami^r on D will be the conteiil 
onB. 

40. To compute the content of a cask by the pen. 

Find the luean diameter by (Art. 3!)), multiply its aqiure 
by the proper tabular mulliplier, and this product multiplitd 
hy the length will give the content. 

Example I. It is required to find the content of a caalt of 
the first variety, whose dimenaiona are, head diameter W'Z, 
bung diameter S3-0, and length 35'G incliea. 

By the head rod, 
"Set the index on the slide to 282 on the lower part of thft 
rod, and over 83-0 on tl>e same line, on the scale marked spliF- 
roid ia S'3; this applied to the lower scale on the slide, cutt 
on the lower part of the rod 31-6, which is the mean dian»t«r. 
Place the left end of the slide to 3fl-6 on the upper part of U« 
rod, and over 31 '5 on the upper line of the slider, atsodien 
tlie rod 100 gallons, the content. 

By the sliding-mle. 
First, Sg-O—Sa-Z^J-O, and4-8x-68=3-a; hence 283+ 8^3- 
31-fi, mean diameter. Then set 35'6 on B to IS-'USS on D, 
and at 31'& on D cute on B 100, the content in galloua. 
By the pen. 
The mean diameter is 31'5; then 
»J-Jx31-5x-O02832x36G— ACfe^SiieeSiwoft^'OQtwBUnt 



GAUGIKO. 157 

ExAMPLR 2. Let the head diameter of a cask 1)e 22*7, the 
mng diameter 81*3, and the length 50 inches; required its 
intent? 

By the head rod. 

Set the index to 22*79 ^^^ against 31*3 will he 6 on the scale 
ipd.) which found on the lower scale, b against 28*7 on the 
od. Then place the end of the slide to 50, on the upper part 
I the rod, and against 28*7 on the slide will be 116^ gallons^ 
he content. 

By the sliding-rule. 

Suppose the cask to be of the second variety; then 
31.3—22*7 «-6, and 8*6 x'64— 5-5. Again, 22*7x5*5 «»28-2, 

he mean diameter. 
Now set 50 .on B to 18*7892 on D, and beneath 28*2 is 112*6 

^Is., content. 

By the pen, 

The wiean diameter is 28*2 ; then 
28*2x28*2x*002832x50*0— 112*6 gallons, content. 

Ullagtno. 

41. A cask is on ullage when the liquor it contains does not 
1^ ity^ and to compute the quantity contained is uUaging. 

The d^pth of the liquor is called wet inches, the remainder 
»f the bung diameter or length dry inches. A cask on its 
jde is said to be lying, one on its end standing. 

The dimensions necessary for finding the ullage of a cask 
ire^ the bung diameter, if the cask be lying, the length, if it be 
landing, and the wet inches, together with the content. 

Should the content not be given, it must be found hy the 
»receding rules. 

If it be fifiven or ascertained, and the cask be standings 
neasure with a bung rod, or some other proper instrument, 
hrough a hole in one end, the length, and the inches on the 
od, wet with the liquor, will be the wet inches. If the cask 
« lyiuff, measure the vertical bung diameter, and the inches 
ret wiu be the wet inches, for all practical purposes. 

42. To find the ullage of a lying cask by the head-rod, slidr 
ng-mle, or ullage-rule. 

Set the bung diameter on C to 100 on the line marked SL, 
nd against the wet inches on C will be the segment or mean 
rea on the line marked SL. Then set the content on B to 1 
n A, and beneath the mean diameter or segment on A will be 
he ullage quantity on B. 

Note. When the wet inches are less than ^'^ of the bong diame- 
)r, the segment or mean area is found on the upper part of the line 
L, on the first form of Rule (5); and on a line marked Seg. Ly, 
omediately beneath the line marked SL, on the second form of 
Ittle (5). 



II qnot 



43. To ull^e 8 lying cask by the pen. 



Divide the wet inches by the bung diametev ; if th 

Snotient be lees than '500, gulitract ^ of the difference frn 
le quotient; if the quntient exceed '500, add 4 of the diSt 



laotient be less than '500, gulitract ^ of the difference 

Ment; if the quotient exceed '500, add + of the ^i 
the quotient; the remainder or aum nmrtiplied by 
content will give the ullage quantity. 

ExAUPLB. The content of a lying cask is 124 gallon^ 
bnng diameter 3<!'0, and the wet 24'8 inches; how 
gaUona does it holdl 

Cy the Blidiug-rule, 



-} 



Set m-i) on C to 100 on SL, and against 24-8 on C i . 
I SL, the segment; tlieii set 124 on B to 1 on A, and below 
on A is 03 galls. Ana. on B. 

By the pen, 
2i-fi-^m=-SdO; from '630— '500— '190, which -^4='04I 



1. ,.,„..., 

^^^P The rule by tite pen m ubvioualy only an approxlmatioa 
^^^B 44. To 6ud the ullage of a standing cask by the sliding 
^^^^ Ullu^e rule. 

^^^■^ Use the side or line marked SS., instead of the aide or ' 
^^^B , marked SL., and proceed aa in ullaging a lying caek. (Art. 



NotG. When the wet inches are Icfs tlion ,', of the lengtb, 

, segmeat or mean area is roaud on the uiiper pai-t af the line SS| 

the first fol-m of Rule [5); and on a line marked Seg. St., in 

diately below the lice mai'ked SS., on (he second form of Rule (j 

45. To find the ullage of a standing cask by the pen. 

HnLK, Divide tile wet inehea by the length of the cA) 
the quotient be under '500, subtract ,'„ of the diHerence . 
the quotient ; if above 'SOO, add ,'o of the difference ta 
quotient. The remainder, or sum multiplied by the ' 
of the cask, will give the ullage quantity, 

E^jtHFLB, Find the ullage quantity in a standing t 
^ whose content is 120 gellons, length 43-0, and wet 

By the sliding-rule. 
Set 430 on C to 100 on SS, and beneath 17'R on C i« 
an SSi then place 120 on B to 1 on A, and beneath 4«>Di 
I will be on B 48 gallons ullage. 



I 

^^^^ Now 17-6-f-43'a=-401), and -500— ■409=-O9t. Tht* ^ 
^^■'^lO^'OOQl, and-409— '00t)='400, ttnd-400»cl20= 
^^^V the ullage quantity. 



GAUGING. 159 



Miscellaneous questions. 

1. Bequired the divisor, multiplier, and gauge point for a 
Done of nint glass. 

Ans. Divisor 33, multiplier '030303, gauge point, 6*7445. 

2. What are the divisor, multiplier, and gauge point for a 
sentagon in bushelsl 

Ans. Divisor 1289-2889, multiplier '0007756, gauge point 

35-9067. 

3. What is the circular gauge-point for gallons, when the 
niddle area of a regular frustum is taken? Ans. 46*024. 

• 4. Multiply 74 by '027, by the sliding rule? Ans. 1-998. 

5. A malt couch is 7*44 bushels in area, and mean depth 
18*3 inches; required the content in bushels, and in net malt I 

. „ f 136-15 bush. 
^^•\ 110*96 net. 

6. What is the quantity of malt on a floor whose length is 
f66, breadth 164, and depth 03*1 inches. Ans. 175*5 bush. 

7. A soap-frame 45 long and 15 inches broad, has in it hot 
ilieated soap to the depth of 61*4 inches, — how many pounds 
loes it contain? Ans. 1443*2 lb. 

8* Given a plate glass pot whose depth is 21 inches, and 
MBS diameters at 3*5 from the mouth, are 24*2 and 23*9; at 
10*5 ore 25-3 and 25*5; and at 17-5 are 27*4 and 28*1; to de- 
mnine the pounds weight of metal it will holdl 

Ans. 967-44 lb. 

Note. Plate glass pots are tabled for drj inches. 

9. A floor of malt is found to be 250 bushels, and its cor- 
^ponding couch was 158*8 bushels; from which would the 
Inty charge arise, and upon how many bushels net? 

Ans. Couch; 129*42 bush. net. 

10. The length of a cask of the first variety is 43*0, bung 
6*5, and head 28*2 inches; how many gallons is its content! 

Ans. by the rule, 141* I ,, 
by the pen, 140*7 J ^*^^* 

11. Given a cask whose content is 108 gallons, bung diame- 
er 33*5, and wet 24-1 inches, to find the ullage quantityi 

Ans. By the rule, 84*5 ) ,, 
By the pen, 83-6 J ^^^^^ 
.12. Divide 228 by 67 on the sliding rule? Ans. 4. 

13. There is a mash tun whose top diameter is 70, bottom 
liameter 50*4, and depth 40*0 inches; required the content, 
ad tabulation at each inch, in qrs., bushels, and gallons! 

Ans. Content, 6 qrs. 3 bush. 6 galls. 

14. What is the area in gallons, of an ellipse whose diame- 
ers are 72 and 60 inches? Ans. 10*1952 galls. 

15. Required the content in bushels, of a frustum of a 
one, of which the top diameter is 24^ the bottom d\ax£ke\>^x^) 
nd the depth 53 Inches? Aus. VI 'W WsJto^ 



■ . depti 



GAVaiKG. 



IS. The length of a cooler is 212, the breadth 148, and the 
. deptba at different places, 3-8, 3-6, 4'4, 4-4, 3-9, i% 3% i% 
4'3, and 3'9 inches; how many gsllons does it contain? 

Ans. 452-63 giil, . 

17. A copper with a rising crown has to be inched and ta- 
bulated, for dry inulies, in barrels, firkins, and gallons, the 
dimensions of which are — depths 48, frotn the centre of th« 
crown, and S4 froTa tile rising of tlie crown to the top of the 
vessel, the cross diaincters at 6 trom the top 7B'0 and 74'6, at 
17, 74-8 and 74-4, at 26, 69'6 and 69-0, at 33, 66-6 and 65-S, it 
39, 61-6 and 61-2, and at 45, 57-1 and 57-2 inchea, and Ui« 
quantity to cover the crown. 22 gallons: wliat is the conlent 
of the copper? Ana, IS bar, 2 firk, 5'04 galli. 

18. Of a standing cask the length is 40-0, wet 19-0 invlu^ 
and the content 207 gallons; how much doea it conlaiii? 

Ans. 87 i galls. 

19. The depth of a flint glass pot is lfl-5 and its mean dia- 
meter 2^3 inches; find tlie area and content, 

Ans. Area 8i-273G Ih,, conti-nt 1503-5G1S1.. 

20. Suppose the dimensions of a guile tun, in the form of Uie 
fhistum of a square pyramid, to be depth 30 inches; the length 
of a side B inches from the bottom, 32'5, at IS inchee, 9-J, 
and at 26 inches, 42'5 inches; it is required to find the ai 
in gallons of each irnstom. Ans. 1. 3'8094 galbni 

2. B-34S8 do. 

3. 6-5143 do. 

21. Required tlie content of a cask whose head diameter ii 
17-4, bnng diameter 19-6, aad kngth 23-7 mches? 

Ans. 24 veariy. 

22. Ilowmany bushels of malt are there in a floor, of whieh 
the length is 863, breadth 28S,aud depth 1-7 inches? 

Ans. 188-5 biuli. 

23. What quantity of hard soap, hnt, 19 coiitsined in a cy 
linder, whose diameter is 3G'5 and depth 71'4 inches? 

Ans, 2667-81 lb. 

24. Let the altitude of tile globular part oFa still be S inchei, 
and the cross- dial nete is at the top and bottom be 27'2 and 
26'8, snd 55'2, and M'8; aUo the cross diameters oFthe b«df 
of the still at 4-5 inchea down be 59-8 and GO-2; at IS-fiincbH 
fl3'8 and 64-4; at 22-5 inches 64-0 and G4'G, and at 32-5 inclw 
62-0 and 02-4 inches; and let the quantity required to cover 
the crown be 3fi gallons; required the content in callonsl 

Ans. 5iM)-S6Sgall>. 



GENERAL EXERCISES. 

1 the difference between tlie areas of two rect- 
■npnlar fields, the one 560 links bj 426, and the other 280 
Jinks by 213? Ana. I nc. 3 ro. 7 272 poles. 

2. The aides of three squares are 12, J5, and 16 feet 
tespecllTely; find the side of a sq^uare that is equal in area 
to all the three? Ans. 25 feet. 

3. If the side of an equilateral triangle be 99 feet; re- 
l]Uired the side of another, whose area aha!] be one-ninth 
of the former? Ans. 33 feet. 

4. Find the sum of the areas of the two equilateral tri- 
mgles mentioned in the last question ? 

Ans. 523-94573 square yardi, 

5. How many yards of paper Tvill be required to line a 
ihest that is 5 feet 3 inches lon|T, 3 feet 2 inches wide, and 
I feet 6 iiicbeB deep, the paper being Iti inches broad ? 

Ans. IBi yaids, 

6. What quantity of canvass, yard wide, will be required 
] cover the convex surface of a conical tent, the diameter 
f the base being 18 feet, and the perpendicular height in 
lie centre 12 feel? Ans. 47-l24yard8. 

7. What are the three sides of a right-angled triangle, 
hose aides about the right angle are to one another as 4 
1 3, and whose area cost as much to pave it at one shiU 
og per square yard, as the pallisading of its three sides 
I8t at half-a-crown per lineal yard '? 

Ads. Hyp. 25 yards, and the sides 20 and 15 yards. 

8. A roof, which is to be covered with lead, weifjhing 
lb. per square foot, is 30 feet 3 inches long, and 15 feet 
incbes broad ; how much lead will be required to cover 
? Ans. 37 cwt. 2 qrs. 19 lb. 14 oz. 

9. Suppose the plate of a looking-glass is 30 incbes by 
I, and it is to he fraraed with a frame of such width, that 
I aniface shall be three-fourths of the surface of the 
iBs; required the width of the frame ? 

Ana. 4'3I 15 iuches nearly, 

10. How many bricks, each 9 inches long, 4i inchei'^ 
nad, and 3 inches thick, roust he taken to huiid a 
feet long, 20 feet high, and one foot thick ? 

Ans. 28444J.' 

11. If the circumference of a circle, the peiiraetM ot & 
lare, and of an eguiJateral triangle, be eac\i ^ ^eW. 



what IB the area of each of the figuree in square jards, and 
which has the greatest area 9 Ans. Circle II '45916; 
squared; and triangle 69282 square yards; and hence 
the area of the circle is the greatest. 

12. If the diameter of a circle, the aide of a eqnare, and 
the side of an equilateral triangle, be each 100 feet; nhat 
are their areas, and which is the greatest! Ans. Circle 
7854, square 10000, and triangle 4330-13 sqnare feet; and 
hence the area of the square is the greatest. 

13. The rent of a farm is paid in a certain fixed nam- 
her of quarters of wheat and barley; when wheat is at 55*. 
per quarter, and barley at 339., the value of the poriioni 
of rent paid hy wheat and barley are equal to one anothei; 
hut when wheat is at fios., and barley 4l8. per quarter, 
the money rent is increasc^d by L.7- What is the rawn- 
rent ? Ans. 6 quarters of wheat, 10 quarters of barley. 

14. There is a wa^on with a mechanical contrivaiicr, 
hy which the diSTerence of the number of revolntion* of 
the fore and hind wheels on a journey is noted. The cir- 
cumference of the fore-wheel is a (lOi) feet, and of the 
hind-wheel b (12) feet; what is the distance gone over, 
when the fore- wheel has made n (1000) revolutions more 
than the hind-wheel 1 

Ans. — — -. or 15 miles, 7 furlongs, 60 yarfd 
1,1. Tliere are four numbers, such that if each he roulli- 
plied by their sum, the products are 352, 504, 396, and 
144; find the numbers r Ans. 7, 14, 11, and 4. 

16. A man being asked what money be possessed, re- 
plied, that he bad three sorts of coins, namely, half- 
crowns, shillings, and sixpences; the shillings and sii- 
pences together amounted to 409 pieces ; tlie shillinss and 
half-crowns 12,54 pieces; but if 42 was subtracted from 
the sum of the half-crowns and sixpences, there would le- 
main 1103 pieces; what did the man possess in all' 
Ans. 995 half-crowns, 259 shillings, 150 sixpences; in all 
L.14l,ls. 6d. 

17- Find two fractions whose sum is I, and whose pro- 
duct is .,3,? Ans, -jJ, and.*,. 

18. Suppose a ladder 100 feet long placed ai^ainst a per- 
pendicular wall 100 feet high ; how far would the top of 
the ladder move down the wall, by pulling out the boliont 
thereof 10 feet! Ans, 50126 feet, or 601512 laehe^ 

19. A may-pole having been broken hy a blast of wind, 
it was found that the pcirt broken off measured 63 Jni, 
and by falling, the W^'^iiBwiAe Mn'stWatfaa ^oad, at 



the distance of 30 feet from ihe foot of the pole ; it is rc- 
(jnireil to detennine what wns the height of the pnle when 
slandinsr upright? Ana. 118-3085559 feet. 

20. Find four numbers in arithmetical progreasion, 
whose common diflerenee is 4, and I heir continued product 
]76985? Ana. 15. 19, 23, and 2". 

21. Find the distance from Eddystone Light-house to 
Plymouth, Start Point, and Lizard, respectively, frora the 
follomng data: the dislance from Plymouth to Liiiard 
being 60 miles, from Lizard to Start Point 70 miles, and 
from Start Point to Plymouth 20 miles ; also the bearing 
of Plymouth from Eddystone Light-house is north, that of 
the Lizard W.S.W., and that of Start Point E. by N. 
Ans. From Eddystone Light-house to Lisiard 5312 
miles, to Plymouth 14-19 miles, and to Start Point 1713 
miles. 

2'2. At an election foul candidates offered themselves, 
and the whole number of votes was 5219 ; the number for 
the first candidate exceeded those for the second, third, 
lud fourth, by 22, 73, and 130 respeclirely; how many 
roted for each? Ans. For the first 1361, for the second 
1339, for the third ]2!I8, and for the fourth 1231. 

23. What ia the area of an isosceles triangle inscribed in 
k circle, vrhose diameter is 24, the angle included by the 
mual sides of the triangle being 30 degrees? 

Ans. 134-354. 

24. The sides of a triangle are respectively 40, (30, and 
W feet ; find the radius of the vnscribed, and nlso of the 
iicumscribing circle? Ans. Inscribed S^^IS, and the 
ircamacrjhing 10|^lg. 

25. Required the aide of an equilateral triangle, whose 
rcB is just tiro acres ^ Ana. 079(319 links. 

26. A field in the form of an equilateral triangle con- 
lins juat half an acre ; what must be the length of tether, 
xed at one of its angles, and to a horse's nose, to enable 
im to graze exactly half of it ? Ans. 48*072 yards. 

27- The area of a right-angled triiingle ia 60 yards, one 
f the sides is 8 yards ; reijuired the other side, and the 
ypotenuse? Ans. 15 and 17 }'urds, 

28. The area of an isosceles triangle inscribed In a 
rcle is half an acre, and the angle contained by the equal 
des is three times the angle at the base ; required the 
igles of the triangle and the diameter of the circle? 
ns. Yectical angle 108°, anglea at the base 3<i°, diametei 
'the circle 121'367 yards. 



164 CSXBftAI. SXKSCISE8. 

29. Fiad the expense of gilding a spbeie at Hd. pn 
ijuare inch, its diameter being 2 feet? Ana. L.Il,6s. 2j<i. 

30. Find the expense of inclosing a piece of grDunil in 
)f a Bector of a circle, whose arc conUina Sl)°, 
i area is a quarter of nn acre, at the rate of 'i 

lilliags per lineal yard? Ans. L.37, }Qs. 4^d. 

, 31. It 13 required to find the area of a circle vrhoae ra- 

Ib equal to an arc of 70 degrees of another ciitlf 

« diameter is 20 feeta Ans. 468-922 feel 

. If a circle he described Tcith radius one, and a Recond 

nth a radius eijual to the circumference of the first, )ind 

f third with a radius equal to the circumference of Uie 

Kaacond ; what will he the area of the third circle ? 

Ans. ^BflfrSa. 

33. If the vertical angle of a triangle be 100°. the dift 
ftxence of tlie sides 4^, and the dilference of (he segnienU 

of the base made h_T a perpendicular upon it from the tvi- 
tes 5 feet; what are the sides and angles of Ihe triangle, 
and what is itsarea? Ana. y60387, J-IUIS?, and 1»h 
6-03699 i the angles are CI" 69' 7", and 10' (T 53", and 
the area 2-94l.'>9. 

34. What will the diameter of a globe be, when its snt- 
fftce and solidity are both expressed by the Bame nambert 

.All*, a 

35. If a round cistern be 26'3 inches in diameter, and 
L$6 inches deep ; bow many inches in diameter must a 

hold three limes as much, the depth being the 
' RBine as before? Ans. 45553 inchn. 

36. How many spheres 5 inches in diameter will hr 
equivalent to a sphere 20 inches in diameter? An*. 61- 

37- How high above the earth's surface must apenon 
be raised to see a third part of its surface ? 

Ans. To the height of the earth's dinmeht. 

38. Sapposing the ball at the top of St Paul's'Chatth 
to be 6 feet in diameter; what would the gilding of it 
oome to, at 3^d. per square inch ? Ans. L 237. It's. H- 

39. Three persons having bought a sugar loaf, want to 
divide it equally among them, by sections parallel to tlic 
base ; what must be the altitude of each person's shan^. 
supposing the loaf to be a cone, whose height is 18 inchtt! 
Ann. 12-480U the upper part, 32430 the middle part, and 
2-2756 the lower part. 

40. If a cubic foot of metal be dmwn into n wire of Vn 
of an inch diameter ; what will be the length of the wite. 
sJJoiring no loss in the metal i 

hvA. \&;ift\1 feet, or 31 i mile«- 



CENBBAL EXBIlCieBS- 165 

-41. If a sphere of copper, of Due foot in diameter, was 
In be beat out intn a circular plate of ^'^^ of an inoh thick ; 
what woold be iis diameter? Aue. 113)37feet. 

42. If a round pillar 7 inches in diameter have 4 feet of 
stone in. it ; of what dianieler is the column, of equal 
length, that conlains 10 limes as much ? 

Ans. 22136 inches. 

43. What is the area of a triaagiikr field, whoae three 
Ndes are reepectivelj 1717 links, 1515 links, and 808 
liskar Ans. 6 acres, 19 poks. 8954 yards. 

44. The perpendicular, from the verteZi on the base of an 
tquilatexal triangle, is 10 feei<; find (he sides of the triangle, 
and the diameter of the circumscribing ciccle^ Aos. Each 
aide is 61^3) and the diameter of the circle ifl 13^. 

45i Giren the base 160, the vertical angle 100°, and 
the difierence of the sides 8'7365, to find the other sides 
and angles? Ans. The sides 100 and 1087365, and the 
angles 37° 59' 19" and 42° 0' 41". 

46. Given the vertical angle of a triangle 120°, and the 
three sides to one another as the numbers 7. 5, 3 ; also the 
numerical expression of the area equal to ten times the 
perimeter, to find the rest. Ans. The sides are 93J ^3, 
%|V3. and 40^3"; the area is 2000^3 ; and the other 
angles are 38° 12' 48", and 21° 47' 12". 

47. Given the vertical angle of a triangle 80", the base 
100, and the sum of the other two sides 150, to find the 
Bides and angles. Ans. The sides are 92 313, and 57fi8ti8, 
md the angles are 65° 22' 55", and 34° 37 5 '. 

48. The sum of two sides of a triangle is 300, and their 
difference is 100, and the angle contained by these two 
ttdes is 60°; find the other angles, the tliird side, and the 
uca ? Ans. The angles are 90° and 30°, and the third 
lide ia 100^3=173-205, and the urea is 5000^3= 
1B60254. 

49. What is the solid content of a cone, the diameter 
if its base being SO inches, and its perpendicular altitude 
©inches? Ans. I 81805 feet nearly. 

50. What is the solidify of a segment of a sphere, 
irhose diameter is 10 inches, and the height of the seg- 
aent4incheBi Ans. 184-3072 inches. 

51. A gentleman wanted to know the contents of a 
iqilare field, but had forgotten the dimensions, only he 
-emembered that the distance between a large oak which 
;rew within the field and three of its cornets, in & «m<lc(a- 
ive order, were 116, 156, and 166 ja.ida; ie(^\tei iA»a 



f 186 GENEBAL EXEIICISG3. 

eontentsof the Belli and theside of the tiquurn? Ans, Cm 
tent 813585 ncies, and the side 198-437 yards. 

52. How often will the fill of a conical glasg, 2;^ inch, 
ide at ihe top, and 2^ icches deep, be contained in an< 

ther in the form of a t'ruslum of a cone 3^ inches wide - 

the top, 2 inches wide at the bottom, and 4^ inches deej 

Ans. 8,Yj tima 

53. How many apheriotl lenden halls, of half an incl 
dinnicter, could be obtained from a cubic foot of lead, saf^ 
posing no waste of metal in casting? Ans. 2()401ft 

54. How menj cones, of one inch diameter of base, anli 
one inch perpendicular altitude, would be equal in solidity 
to a sphere of 6 inches diameter? Am. 433.' 

55. If the diameter of a lub at the top be 30 inches. uC 
its depth 2 feet, what must the bottom diameter be, fio M 
it may contain 8 cubic feet of water? Ana. 2'0044fut> 

56. If a ctstem. 4 feet long, 3 feet wide, and 2^ M 
deep, were proportionally enlarged in all its dimeniioiu,N 
as to hold 4 times as much ; what would then be its iu< 
mensioDS ? Ass. 6-3406 feet long. 4'7622 feet wide, avt 
3-9685 feet deep. 

57. If the base of a triangle be 30, and the other liilH 
25 and 20 ; find the segments of the base made by a pf 
pendicular upon it from the Tertex, the segmenU msJo 
by a line bisecting the vertical angle, and the leoglb (^ 
a lise drawn from the vertex to the middle of the bw- 
Ans. The segnuents of the base by the perpendicular b« 
1875 and 11-25; the segments by the line bisecting t\it 
vertical angle are 16^ and 12^, andjhe line from the TW-i 
tei to the middle of the base is ^287-5=17 nearly. I 

58. Find the dimensions of a room, in the form of », 
rectangular parallel Dpi ped, the solid content of which iBj 
cubic yards is expressed by the same number as the supe^ 
ficial yards in the walls, ceiling, and floor together, asn 
such that its length may be double of its breadth, and il^ 
breadth double of its height ; find also how many peOJ 

I the room would contain, each having the fourth part ' 
Bquare yard to stand on? Ans. 14 yards long, 7 ) 
broad, and 3^ yards high, and would contain 392 peopli 
39. The area of a triangle is 59(11-16 square yards, 
perimeter 370 yards, and an angle 85° 27' 34"; reqiiii 
the side opposite to it? Ans. 









TABLE I. 

rHMIC SINES, TANGENTS, AND 8ECANT9. TO KVERt POINT 
AND QUAkTEK POINT OF TriK C0MPA8S. 


" 


Co-inev 


ThubbiiI. 


CoUng. Secant. 


Coscc, 


A 
I 


S03 


Itt 000000 0. 000000 

9.909477 aeomg 

9.89790* 8.9933B8 
9.99S274 ai71S*T 


11.808681 la000623 
11.006603 10,002096 
1ft 828763 1ft 001726 


11,309204 
11,008898 
lft833480 
lft700T64 
10614429 
la 637176 

loiiT'ieo 

10 369008 

lo'28«960 

ia-i34973 
10197641 
]ft 172916 


i! 

9" 

W 

ii 


571 
934 


9.991fl74 8,298863 
0.086786 9.308788 
S.US0886 9 4S1939 
0.073641 9.SBSm 


ia701838 
Ift60I31« 
1(1618061 
]a446853 


ia008*26 
10.013314 
ia019116 
ia03SI69 
ib.03B86- 
Ift043e37 

:o.*o66e5o 




0.»0fl«15 
9.0S6163 
9.946430 
9.933350 


a617XH 
0.674820 

B. 737067 
9.777700 


10.383776 
ia326171 
ia272043 

10.233300 


B7 

06* 


9.904S38 
9.888185 
9.860790 


aS7OI09 
9.014173 

9.967296 


lftl7S107 
1ft 139801 
1ft 086837 
lft(M3706 


1ft 0061 72 
limi8]6 
miiMBlO 




9.849185 10.000000 








'1'HiiKene. Conee. 




B TABLE n. 

^T LOGARITHMS OP NLMDERB. 




130 

160 
ITO 


23 
33 

34 

ai 


1361728 
1.380211 
1897940 


44 
46 


L 623240 
1.033468 
L643463 
1,663213 


(a 

(34 
65 


1.7S63aO 
1.792il93 
L799M1 
Leoei80 
L812913 


83 
84 
86 


1,1108486 

1JJ19678 
1,924279 
1,029410 


m 
Da 
m 


2T 
S8 
39 


1.431 
1.44; 


364 
68 

98 

C3" 

14 
479 

<m 


47 
48 
49 
*>_ 

63 
64 

66 


1.672098 
1,681241 

1.690196 
_1.B98»70 

liTieoos 

1,724276 
1.732394 
1.740a63 


67 
68 

7U 

7i 

7a 

74 
75 
■76 
77 
78 
79 


L 826076 
1.8t!2A09 
1838849 
1,846098 
1.861268 

L863323 
],ee!U.f3 
1.876061 
1.^0814 
1,886491 
1.HI2O06 
1,897637 


87 
88 
89 
90 
^91 
92 
03 
94 
06 
96 
07 
98 
99 


1.939619 
1,944483 
1,949390 
1,934343 
1960041 
L9637S8 
1.068483 
1.973128 
1,07772* 
1,082371 
1,086772 
1991226 
1,996635 


HS 
128 

m 


32 

33 
U4 
36 
W 
37 

.19 


.511 
631 

tbbo 

.M8 
l.fiTff 
1. 69: 


MS 

178 


Mi 
202 
f4 


56 
57 

69 


1.J481B8 
1.766876 
1.76342S 

l.'77S15l 


■ 






A 





































;. 4 1 6 




* 1 M ,1 1... 




JoiSus 


T^^ 


^fii-WAWwa 




loHowlw^M 


206666 




g 7090 


T360 


7634 


79041 81731 


8441 


8710 8079 


0247 






; 9783 


sioeai 


210ai9 


ilO««;21(ie53i 


131112 


31138^211664 


211021 






?3124» 


3720 


398G 




3818 


378S 


4049 4314 


4579 






«109 


SBTS 


5038 




6166 


64:10 


6694 6987 








7747 


8010 


8273 




S798 


9060 


9323 9686 


084t 






830370 


asosai 


320893 


321153 




231675 


221036 32211)6 


a234,'ifl 






asii 


3336 


34!KI 


3766 


4018 


4374 


4683 4792 


6051 


2f.9 






B826 


6084 




6900 


6858 


7116 7372 


7030 


258 




tI 61M 


MOO 


em 




9170 


9420 


B683 B«oa'2noi9a|2C6 






330960 


331316 




231724 


3319Tt: 


232234 2324S8ja32743,2S8 


e 32S0 


3604 


3757 


40U 


4364 






5376,2.-i3 




t fiTBl 


8033 


63Sfi 




6789 


7m 


7393 7544 


7795253 




3 8Z9T 


am 


8799 


9049 


9299 


9560 




340300,260 






Ml 048 


241297 


341546 


241795 


242044 


343293 2641 


3790,249 








3783 


4030 


4277 


4628 


4773 6019 


6266248 




3 6769 


eoofi 


6252 


6499 


6749 


6991 


7337 7482 


7728846 




^ 8319 


B464 


8709 


8954 


9198 


9443 


9887 9932 


360176,345 






351S51 
3680 


361396251838 
38231 4064 


261881 


2iH125 2(^ 


3610,343 
6081243 




F»a36n4 2S»7M 


26699626(1337 




2667 li 


266988 367198 


3674;i! 


341 


r9 791i 




83991 8637 




9116 


8366 9594 




239 








3fl(ff87;2(n0S6 


361263 


261601 


361739 361976 


262214 


238 




SlFMBf 




3102 3399 


3636 




4109 4346 




237 








0026^ 876! 


699B 


6232 


6467 6703 


6937 


336 




ra ?'«« 


7641 


787ffl 8110 


8344 


8S7S 


8812 9046 


9879 






IH B74( 


9^8O^70213|2704^6 


370679 


370913 


271144 271377 


271609 








2733(16; asm 3770 




3233 


346* 3690 


39271232 




»r«8! 


4830 ^ «B1 




6S42 




6232230 




Cal «6&3 


eaait 715l| 7380 


7609 


7838 8ltfi7l e3S( 


8r.25!'m 






^79439279(167 


27DB9S 


3eoi2!1.28(X161iL«0578 






2817) 


J81943 


382169 




2623 2849 




237 




^ 3537 


3763 


3879 




4431 




4883 6107 




236 




iffl ff7K 


6007 


6332 




6881 




7130 7354 


7878 






I02l8(|a 


S349 


8473 




8920 


91t 


9366 9689 


9812 








W04aa, 


290703 


3B092( 


391147 


391369 


291691 29iai;i 


293034|222 




lea 2478 


m<J 


3B20 




3363 






4346;321 




ia9 *m 


4907 


6137 


6341 








6446230 




m\ em. 


7104 


7333 


7642 


7761 


7979 


8108 8416 






loal »(i7i 


»2!'» 


9507 


9725 


9943 


300161 :300378i300695 


300613 


318 
"217 




t»iaov2a-Mivn:^\mmvm 


S03114^23;tl 


J02647 303764309980 


196 3*12 




3844 


4069 


43751 4491 


4706 4921 6136;aie 




0] 5dU6 


8781 


69y(, 


6211 


6435 6639 


0884 7068 7383315 




ISW 7710 


Tifii 


8137 


8381 


8504) .8778 


8991 9304 9417313 
311118 311330 3I1542i3ia 




990 9Sd3 


aiO(K6 


310269 


110481 


310603310906 




UM 3111)66 


2177 




2600 


2812 3023 






je? 4078 


43gn 








634(1 6661 6780310 




PV(H msD 


6390 


6600 




7018 7227 


74.W 7646 7884209 




D^ SB: 


S4B1 


8im 




910(i 9314 


8523 9730 9038|208 






380063 


1307(i9 


520977 


S31IS4I321391 


3215981321806 


3220121207 
3240771206 






ia*«y 


3m>^ 








4(»4 


4899 


6105 


6310^ 561(1 


6721 69361 613i;305 




B96 61M 


874^ 


6900 


7166 


736M 7663 


7767 7B72 817620* 




3Ba 8089 




8891 


9194 


9898^ 960 






4I*'330ei 


33081: 


331022 


33122 


3S1427 a31(130;3ai8Sa| 30;M| 2236202 




OSSl 3G4C 








ai47 3649 3S50 40Bll 4353'3Ba 




4fi4^ MM 


486 




8367 


64Ea 566e\ BSW*. W«A WSVBS^-^ 


4ed eeeo' eswy Towf raw 

fcW «i«/ SSfi.^ »»« 9363 


7469 76B9\ 18iB\ Wit®, ^aKllKR^ 


*4 340643l3*08tll3*l<mln4-\237 


341*Wa41G3-2WlK!«\ «ftfe\ tm&v^g^ 


1 i 1 ^ 1 H \ * B \ t, \ 1 \ ^ ^^_^^ 


























ioj3(ii728,attini7|36Mii)s;; 

1 3612 3801 

a 735W 7642 772(H TSlfi 
« B21ffl 9*01 fl5S7 — " 
537100a3713633714;j7U, .^^^ . 

6 3912 3096 3-280^ 3464 

7 1748 4032 JJlia " 

a es77 erm (mal 

ft| 83981 " " 



ffl380ail|380.392|380573|380r&* 

1 aoi7 3197 aa77 2557 

3 38IS -ma 4174 4353 
8 S606 6185] am 6142 
« 789ffl 7668 7746 7!)33 

s ei66 B34a 9.^30 ::.": 

6390035:39111^391388301464 

7 3897 3873 304S 

B 4482 4627 4W2 4977 
8l 9199I 637*1 65481 ((723 



970143 370338 J70513 m 



S30934lieUlS 
3737 2917 
4533^ 

6321 6409 
8101 8379 
8S76f 

391641 — , 
WOO 3875 



66&6| 7071 



^1476 3B16.'>638 
33771 34581 " 
607W B249| 



1993] 31691 33lfi 

37fii aeau uoi i 
5501 rniw sem \ 

734fil7«ffl7awn 



|3981M|3MI , 

9S47 400020kO(ll»3' 

H40137a 1746 1917 

" 3293 3464 3635 

eooa 6176 5346 

67ld 6881 70S1 

841« 8579 8749 

^10102 410371410440' 

I 1956 2124 

34671 aatsi — 




B267 9*26 

I1144113S3U 

3659 3796 



.>..,■.,, ,„„.., ,1166*l[41S8aa|416974|*16Ui;41630a4 

8807 69731 7139 730& 7472 ' -"■' ' 

8467 8639 8798^ 8964 S139 

wfttUoii'nut.tfnaa'i 



1 430I2I 420286 420451 '43061^4201 
176» 19;)3^ 30^ 3361 2426 
34iq 3574^ 3737 3901 4065 
5045^ 6308^ 6371 B534 5697 
667^ 6836 6999 7161 7J"" 
8297 84591 8631 8783 S944 



1431604^ 
334ffl 
4882 
6511 
8135^ 
97521 99141*30075 430286 4303^'*30S59' 



(a0945'431110 *2197Sta 
369U 275* aei8| 
4328^ 4393 45RH 
6860| 6023^ 8186^ 
748^ 7648 7811^ 
9106 9268 042S 



1307201*30881 l4310Mti 




UOABITBUB OF NUNBEHB. S 




1 8 18 


3~ 


r~5- 


« 1 * 1 6 




TT 


1731--) M7468,4478a3|44T778 


147933 




448397 


t&m 


IM 


8861 [lOIS 9170 




9478 


mx 


9787 


91M1 


160095 


154 




iOiCa 4505.-17 ]lfi0711 


lS086r 


151018 


151172 


161336 


151479 


1633 


164 




IJMO 3093 




240O 


3563 


3700 


3859 


3012 


3165 


163 




3*71 3624 


3777 


3930 


40B3 


4235 


4387 


1640 


489: 


163 




4097 5160 


5302 


5454 


6606 


W58 


SOK 


8063 




153 




6B18 6870 


6821 


6973 


7136 


7276 




7670 


7731 


163 




803S BlBl 


8336 




8638 


8789 


8940 


8091 


9243 


151 




96*3 9BH 


9845 


0996 460146:460296 «0*47 460697 


1B0748151 








ssoia^mr 


463847 462097 4U314B 463296:463445 463694.463744 150 


41H2 uai 


4340 4490 


luay 


4788 


1936 


6085| 5231149 




e533 B6S0 


68^9 6977 


6126 


6271 


64231 


6571 8719140 




70W 7164 


7312 7460 


7608 


7766 


7001 


8053 8300148 




84SS 8613 
0968 «0U6 


87»0 6D38 
470263470110 


908; 


93331 
170701 


0380 
170861 


9627| 9675:148 




tn«3S 15S5 


1733 1878 




3171 




"m4|*'2610;146 




2903 3049 


3195 3M1 




36331 


sni 


3836^ 40711*6 




4363 4S0S 


4853 4799 




5090 


6236 


6381 653^146 




B816 6063 


6107 0363 


8397 


6543 


0687 


0S33| 8976 


145 
145 




47726^ 4T7411 


17;55S 17770(1 


477844 


4T7989 


478133 


178278 478133 


Wll 8856 


8899 9143 


9387 


9431 


9576 


97191 Baa 144 




480151180394 




180725 


18080! 


181013 


iaU56 481209 






16B6 1739 


1873 301* 


3159 




2445 


268i 








3018 3159 


3303 3445 


3687 


873( 


3872 










4443 4585 


4737 iaeo 


soil 








5679 






0863^ 6005 


6147 63S< 


8430 




8714 


6855 


6897 


113 














8269 


8410 


141 












9S37 


8677 


9818 


141 




490099490239 


iTKX\-.. ■. .■ ■■... ! -^'11 


i:)0911 19I08i;491222 


140 




149150'2,40ir^,41)17- '1 !'ja34149a4ai492S3iil«) 




3900 


3040 




3737 


3876 


4O15130 




43!M 


4133 




6128 


5287 


6406,130 




MS3 


5323 




6616 


6653 


e79lll30 




9068 


Szoe 






7621 7759 


7897 


8036 


6173 138 




8448 


8586 




8863 


8899 0137 






0550,138 




e824 


KW3 


500090 


600336 


600374600611 




S00785 


600923137 




soiioe 


501333 


1170 


1607 


17441 1880 


3017 


3161 


329i:i37 




E664 


2700 


3837 


3973 


310ft 3346 
447l[ 4807 




3618 


3666136 




S927 


4063 


4100 


4336 
505G93 


4743 


4878 


6011136 




606380505431 


606567 


606828 


505964 606099:506234 






6B40 


6776 


6911 


7046 


J 181 


7310 


7451 7586 








799J 


B12f. 




8395 


8530 


8664 


8799 8934 








9337 


9471 




9710 


9871 


5100M 


610113 610377 


610411 






B10879 


S10S13 










1182 1616 








son 


S151 


3284 




3551 




2818 2961 










3484 


3617 




3883 


1016 


1149 4383 




133 




mi 


4813 


1946 




6311 




5476 5609 


5711 


133 




600(1 


6139 


6371 




6536 




8800 ffi)33 


70e4;132 




7328 


7460 


7592 




7866 7^ 


8119 8251 


8382133 




518040 


618777 


618909,5I004il619171i619303 6I9t)4'519.'.6tl 


519697X31 


9D59 


52OO0( 


52022] 


52B(63 5201&1,620«15 520746 620876 


621007 1 131 




oma 






1661 




1922 




BIKJ 






2576 






2066 


3006 


3226 


3356 


3486 


3618130 




3876 


4006 




426B 


4306 


1626 


4666 




4915 






6174 


5301 


6434 


636.3 


6^3 


5822 


6061 




62!0 


129 




64G9 


6398 


6727 


6856 


6985 


7111 


7343 




7501 


129 




7759 


7388 


8016 


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8102 


8531 


8660 


8789 


129 




9045 


9174 


9303 


9687 


9815 9943'630072 


128 




S3032Sfi304565305B4 


530712630840530068:5310965312331 1351 


128 




1 1 1 3~7~3~T"0~5"~P6~ri~T'ft \ ^"^^^ 





3 92ifl 9401 

5371088 

6 2913 3U06 



3 SWIfi 5765 

* 1aao\ 756S 

b\ Biee 9:)43 

e 330935391113 3912) 

7 8607 3873 - 

8 M63 4627 

9l l)l!)9l 6ST4^ I 



6481 

T30ffl 7483 
91241 93001 



703ffl^3T(W! . _ 
S17A 3360 
40ld 4198 
G84H 6039 
767ffl 7853 
94871 966e 



1|381U« 

2737 29171 309! 

*7ia 4891 

6321 6409 667; 



11381476 38n 
■ 33771 at 
. B07ft ffi 
' 6856 T( 
' 8634 Sf 
3D040S|3SW 

3169^ a 

393ffi « 
____ 5676 R 
734B| 74191 W 






418641 |*lS80a|41BB74]416U114iei 

7473 763a 7SM 7! 

9139 939a 940ffl 9 
!0781 430945 421110 431 

, 3436 359q 3754^ 2 

3901 4I»5 433a 4393 « 

5697 5aent am m 

7334 T4SS1 mitJ 

WM4I fliiMtl Pafsujn 



L0OAS1THM8 OF NUMBBBS. 5 




1 a 1 8 


LJZ 


6 1 It ! Y 1 8 1 9 


IT 


W7313 447*68,447623 


M7778 


*47933:**8088 




1^562 


r55 


8861 fiOlSl 11170 


032* 


9*78 




9787 994: 


450095 












151171. 


16132( 451*79 


iei3 


161 










3706 


2869 


3012 


3166 


163 




3471 3^ 3777 


3930 


4081 


4335 


4387 


15*0 


4893 


162 




*09T 6J6IM 630a 


645J 


660t 


fi75S 


5910 


6062 


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6S1S 6670^ eeai 




7125 


7276 


7438 


7679 








8^ 8181^ 8336 
OM 9694| 984S 


8487 


8638 


8789 


B910 


9091 


9243!l6I 








4S10«8|46II9840I3184SI499: 1649[ 17B» IMS' au98 


2a48'150 




462M8 462697 I162S47I462997 


1631*6 


M]3296;*63**6,*a3694 


463744, 50 




4630 




4936 5085 


533* 19 






6136 


6274 


6*23 6671 


8719 « 




7016 71641 7312 7i60 


7608 


7766 


7904 8063 


8300, 48 




SIM 86*3 8790 S93S 


9085 


9233 


9380 9637 


9676148 




flaffl 170116*70303470410 


i70657 


470704 


170861 *70098|471146;i47 




471438 1585 1733 1878 


2025 


2171 


2318 216* 


2610146 




3903 3IM9 3196 3341 


3*87 




3779 3921 


10711*6 




4362 4£08 4653 4799 


49** 






5536146 




fiSlI 59631 6107 eSK 


6397 




6687 6831 


69761*5 




477S66 I77411[477555 t777D( 


1778K 


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8711 8856^ 8999 914i 






9676 97191 8863 144 






1807aj 




181012181156481299 


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760-!'l' 'ii7:J;i' 't'-r.-.' 'nil.-.. 


1025 


1101 


76 




1176 








1778 


1853 


76 




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axK 




222S| 2;i63' 237rt: 24K 


2629; 3604 






S679 


2754 


2829 29()4 


2978, 30fl3l 3128 32<n 








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696( 


6041 




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? 71fi6 7330 




7379 


7463 






767( 


77* 


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8046 


8i:i0 


8194 


8368 




8416 


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8860 


B9S4 


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8| 837l| MU 




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9716 9B20 


9884 


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770836 770110 


7701S4770SS7lT70631'77mo 


i m 


770663 )70e% 




ni073 


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TTima 


771293 


7713C7 


77144 








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IB81 


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3028 


2103 


217 


1 B 






2S42 


3616 


3668 


2762 


S835 


290 


1 3 


3065 3U8 


3201 


3374 


3Ma 




3494 


3667 


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1 ^ 


3788 8860 


39K 






41M 




4398 


437 


1 <> 


4517 4fi»0 






4809 




1965 




810 


6 


Bate B319 




5465 


SS33 


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6683 


6766 


682! 


7 


607« 60t7 




G1H3 


6265 




6411 


6483' 669 


8 


6701 6774 


6846' 6tH9 


6993 7M14 


7137 


7209 728 






707a' 7844' 7717i 7789 


7862 


7934 B0» 


600 77810l|778m 


nsaot 


778368778441 


7I851ff 


778585 


778658 






8874 8847 


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9091 916a 


823U 


9308 


9360 


816 




9596 9669 9741 


oaial H«e6 


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78003! 


7S010I 


leoiT 




78031- r803ea7BO4lil 










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10S7 1109 


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1999 


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3368 


332 




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3616 


2688 2769 2831 


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3374 


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8189 3360 


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3403 8475^ S6SB 
4118 4189 4361 


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47601 48311 4902' 4874 


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786686 


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6183 


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6893 


69E4 


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7106 


7177 




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6366 


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6713 


6782 


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6921 


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r 1 


' LOQARITHMS Of NUMBERS. IS 




.1 (1 1 1 1 i 1 3 1 4 1 fl 1 fl 1 7 1 H 1 B 111. 






98l»71,88093t 


e8U98f 


8S1043.S81O9d88116d|B81313B8127188133d 6!f 




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14a 


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1613 


1670 172] 


1784 184 


: 1S9H 67 






r 1S55 




S06B 




3183 


3240 3395 


2361 341 


2468 67 






3 aoas 


368' 


263S 


369 


3762 


3809 38681 2923' 2980| 3037| 57 






i soas 


3150 


320 


3261 


3331 


3377 3434 


3491 364f 


3606 67 






B 3aai 


3718 


377. 


3S3J 


388» 


3846 400^ 


4069! 411 


4173 57 






B 4339 


4386 




439S 




4513 4569 


4626 4083 


4739^ 67 






7 4795 


485: 




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502-. 




5183 624 








B 6361 










6644 670(1 




S870: 67 






V 5BB6 




6039 6096 6163 


62091 6266' 63311 6378 


6434 66 








pg^ 


W6604:S86660 


886716 


1886773886820886886 886042 886998 


66 








71S7 7M23 


7380 


7036, 7393 7449; 7606 


766 


66 






^ 7817 


787. 


7730 7788 


7843 


7998 


7966, 8011 


8067 


8123 


66 






» 817fl 




83921 8348 


8404 


8460 


86161 8573 


8639 


8686 


66 






8741 


8791 


8863 8909 


8965 


9031 


9077 


9134 


81901 9246 








9308 


9368 


941*1 9470' 9626 


9682 


9638 


9694 


9750I 98061 66 






9863' 


991S 


9974:890030 890086 890141 


SS019' 


890353 


89030! 


8903U6 66 






890421 89M77;890S33| 0389; 0^; 0700 








0924' 56 






(WeW 10S6| lOBl 11471 1303 1269 
16371 1693' 1649' VW- 17fiO' 1816 




1370 


1426 


1483 56 






1872 


1B38 


19831 3038' 66 






tnoua 




993317 893373 


392429 


892484 


993640 


Wem 


66 




aesi 


37071 370a 2811 


2873 2S29 


2988 


3040 


309b 


aiBi 


66 






8307 




3439 3484 


3540 


3681 


366: 


3706 


66 






H 8763 


3817 3873^ 8921 


3984 4039 


4094 






4261 


66 






2 4ait 




4638 4593 


4648 




4761 


4814 


66 






4870 




6001 6146 






6312 


5367 


66 






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6809 


6864 


6920 56 






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6681 66361 6692 6747 6803 


6306 


6361 


6418 6471 66 






» 6636 


6857 


6913' 6967! 7fti2 6,1 






t T0T7 


71321 71871 7343 7387' 7353' 7407 


7462 7617 7673 56 






[l|aB7637^6BS 3977371897793 


897847 887902 


887967 


898013 


898067 8981321 55 




8178 8331 8386 8341 


8396 8461 


B606 


8661 


8616 8670 56 






87M 87«l 88aa B89( 


8944 8999 




9109 


9104 9218 66 






M 9073 9a2S 9383 »4:)7 


949a 8*47 


960^ 


9666 


97n| 9708 66 






9831 S87S 993q V9Ba 


900039 aootm 


90014U 900203 


800258 900813,66 






»00ae7 H>043'i'90(M76 900631 


0586 0640 


0696 0749 


0804' 0869, 56 






0S13 0908 103-i 1077 


1131 1186 


1340 1296 


1349,' 1404 56 






1468 1613 1667 1633 


1676 J731 


1785 1840 


18941 1948; 54 






a003 3067 Slial 31661 3331 2375 


2339 3384 


2138, 2492I 54 






l! 3647 30O1 2656 2710 376*' 2818 


2873' 3927 


29811 3038. 64 








9033071903361 903416 903470 


903534903678 


54 




3633 3687 3741 


3796 


3849 3904 3968 4013 


4066 


4120 


64 








4337 


4391 4445 4499 4653 


4607 


4061 


64 








4878 


4933 4986 6040 6094 


5148 


6302 








6See 6310 5364 


541S 


6472 5636 6680 5634 




6742 








6796 6850 fiWW 


6968 


6013 6066 6119 6173 




6381 








63SS 03S9 6443 


6497 6661 6604 66S8 6712 




6820 








6874 6937 6981 


7086 7089 7143 7196 7360 




7368 


61 






7411 7466 7619 


7673 7636 7080 773*! 7787 


7841 


7896 


64 






7849 8003 8056 


8110! 8163! 83171 8370 8324 8378 


8431 54| 








908860,908814 908967, 64 




9031 9074 9126 


9181 


9236 9289 9342 


93961 9440 9503 64 






Wi66^ 0610 sees 


9716 


9770 9823 8877 


S930l 9984910037 63 






nMnB1014491D197 


B10361 


910304 910358 910411 


910464910618, 067! 63 








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0838 089] 0944 


0988; 1051 


1104 63 






lui 1211 




1317 


1871 1424 1477 




1637 63 








1797 


1860 






2160 63 






3339 3976 




2381 






2700! 63 






2783 asoa 


S 2913 




Uiaaj 3178 


3331 63 






at!S4' 3337 


3.'19ol 3443 34» 3S49I 3602 


3666' 3708 3761 


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14 LOOABITHMB OP NUMBERS, 






91 407919141321914184 


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450^^ 






46ti(f 


4713 


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3 487:2 492G, 4977 


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6341 


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6611 




6716 


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ma 




4 fi9Z7 SSSi; 6033 


6065 




6191 




6396 


eat 




S 0454 6507 flSSO 


66J3 


6661 


6717 




0833 


687 




e 6980 7033 7085 


7138. 






7298 


7848 


740 




7 760& 7658 7611 
B SOSO: BOSjl 6135 


7863 


77l( 




7820 


7873 


793 










8345 


8307 


845 




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8713 




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8869 


8921 


897 


B30'9ia07S;S19130 91S183 919335 


9i^7|819340|91B392 


UM44t9194S 




1 S301 »eS3l 9T06 9768 


9810 


0861 


9914 


0907 


moi 




a Wias'saOlTO 920328 9a028o|830332»3(l38*!B2Wae!»aO«89 


OM 




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1332 


1374 


1420 










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1843 


1894 


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1098 


2060 


310 




S aSOO' 3268 2310 


S3S2 


3414 


2406 


3518 


3570 


382 




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3881 


3933 3986 


3037 


3080 


314 




8 3244' 3298 3348 


3399 


a«6ll 3S03! »656 


3007 


806 




gl 8702 3814' 3866 


3017 
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3969' 4031' 4073 4134 




M0D24379 


924331934383 


134486 


124638 


934589 9346U 


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1 4796 






4961 


6008 


5054 


5108 




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3 6312 






6467 


6618 


6570 


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672 




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6879 


6981 


5963 


60S4 


0086 


6137 


0188 


634 




4 0342 


0394 




64»7 




6600 


6851 


6701 


676 




6 6857 


6908 




7011 


T06i 


7114 


7166 


7310 


726 




6 7370; 7«3 


747; 


7fi2J 




7627 


7678 


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9295"72 929623 029674 


129736 


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1 0030 9981 


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930063930134030186 


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13038 


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0502 


0643 


0694 


0745 


0796 


084 


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1101 








1306 


136 


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1458 1609 


1560 




1681 






1814 


186 


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1966 2017 


3068 


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2230 


9371 




837 


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2474 3524 


3675 


2620 


3877 


3737 


3778 




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3081 3031 


3082 


£133 


S183 


3334 


3286 




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3487 3638 


3589 


3631 


iteoo 


3740 


3791 


3841 


389 


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9! 3993 4044 


4(104 




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4346 


4396 


4347 


439 


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6154 


6205 


6266 


5306 


6366 


640 


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3 6507 6553 


fi608 


56581 


5709 


67ro 


5809 


5660 


591 


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6111 


6ie2i 6212 


0262 


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f»63 


641 


■, 


t 6614 6564 


C614 


6665 6716 


6765 


6816 


6866 


691 


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7117 


71fi7 7217 


7367 


7317 


7367 


741 


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6! 7618 7568 


7618 


7868; 7718 


7709 


7819 


7860 


791 




71 8019 8060 


8110 


8169 B2I9 


B269 




6370 


843 




a\ S52D 8570 


8620 


8670 8730 


8770 




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9120 


9170 9230. 9270 




9369 


941 


870 

J 




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HOU 




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0616 0666; 0616 0666 0716 


0766 0816 


0865 


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1014 1064 in* 1163 isia 


1363^ 13)3 


1363 


141 




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Ifill 1561 1611 IMO 1710 


178(J 1809 


1869 


190 




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3008 2058 2107 3147 2307 


3266 2306 


3855 


3*0 


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376^ 3801 




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rj— 1 1 1 '2' V 3 I *■ ^ ^ ^ ^ ^ ^ ^ 

































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16 LOGARITHMS OF NUMBBM. 


M. 


1 1 3 


3 1 4 1 6 1 d 1 7 1 8 1 


mmi^ 


BTsiHWiJO 


973266'973313( 




973405 973461 97340719 




1 




3630 


3083 




3774 








a 


40fil 


4097 


4143 


4189 


4335 


4281 


4327 4374 4420^ 




3 


«12 


4668 


4604 




4696 


4742 


4788 4834 «88M 






tBIi 


5018 


6064 


6110 


6166 




6348 6394 EaM 










6524 


S670 


5610 




6707 6763 6708 






6801 


6937 


6983 




6075 


6131 


6167 6313 6368 




7 


0300 


0396 


6442 




6633 


6S79 


6626 6S71 6717 




8 


6808 


6854 


6900 




6992 


7037 


7083 7129 7176 




9 


726( 


7312 


7368 


7403 7449 


7495 


7541 7686 763i 


SfiO 


077724 


977769 


977815 


OTTSaiSmM 




97799f 978043 ff60899 






aiBi 


8226 


8272 


8317 8363 




8454 8500 8M 






3 


8637 


e6»i 


8723 


8774 8819 




8911 8956 9003 






3 


9093 


9138 


8184 


9230 9275 




9366 9412 9467 






4 


9Mf 


9694 


0639 




977( 


9821 9867 99U 






£ 




980049 


880094 


8801+0880186980231 


980376 SSO323Ba0367A 




( 




0603 


0549 


0694 0640 




0730 0776 Offll 




Hli T 




0957 


1003 






1184 1239 1371 




Hf ^ 






1466 


IWl 1647 




1637 I68G 1738 






8 




1864 


1909 


19541 3000 


3041 


309( 213f 31S1 






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0^271 


983316 


9^^ 


982407 982*62 


9S2497i88a543:9S368SiS^8^ 


i 


^M\ 1 


3723 


2769 


2814 


38Se 3904 


2949 


3994 8040 BfiH 




^H 3 




3220 


3266 


3310 3356 


8401 


3446 3491 aSW 




^1 8 


3626 


3671 


3710 


3762 3807 


3863 


3897 S842 aeer 




^H * 


4077 


4133 


«67 


4312 4267 


4309 


4347 4389 4487 




H » 


4827 


4672 


4617 


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ooiaoi 




998W9 S* 




BOl 3700 


110199 


891113 


3717 


103S» 


001311 


J6 


008889^ 




iffil sem 


108579 


893743 


3707 


107268 


001331 


17 


998679^ 




OSS 3080 


H«964 


894366 


2697 


1060341 


001331 


17 


998669 61 




^SWTO 


I06:»7 


806984 


2flS7 


10401^ 


001341 


17 


998659)30 




BU 3660 




1897696 


"1677" 


lL10a4O4 


10.001361 


17 


9,998649 39 


'843 aesi 




89930; 


3667 


100797 


001361 




998639 38 




M33 3641 


100568 




3658 


099197 






098629 27 




W17 3631 


098983 


00239! 


3648 


097603 






998619 26 




ISM W39 


097404 


9«1987 


361)0 


096013 


001391 




998609 a 




UCa 361S 


ooeaai 


905670 


3039 


0944S0 


001401 


17 


098699 H 




fist 3603 


094364 


907147 


3630 


093853 


001411 


17 


998580 33 




IS»7^3n3 




908719 


2610 


09I2SI 


001423 


17 


998S78 33 




I85»3S84 


091147 


9103S0 


36111 


(W971S 






998568 31 




HM OflS 


089596 


911846 


3593 








998668 30 




l«W|3S66 


IL 088061 a 91.S401 


2683 


11.086699 






9.998648 19 


M8^SSS6 




9H9.'J1 


3574 


085019 






098637 18 




1033 3547 


084078 


916496 


3666 


0835061 


00147; 


17 


998627 17 




iB5d3S3S 


083450 






081066 


001484 


18 


998519 16 




»73 3339 


081937 


919568 




0901331 


001494 


19 


998506 16 




M»l 1630 


080409 


B3I098 


2538 


078904 


001606 


181 


99849614 




EI03 3513 


078897 


932619 


3530 


077381 




1^ 


99S486 13 




mo 3503 


077390 


924136 


3521 


075804 


001536 


19 


998474 13 




tlia !4»4 




935649 




074351 


001636 


K 


993464 11 




mo 3486 




937166 








li 


0984S3 10 




■IM 3477 


TTofiSioj 


.938658 


M95" 






[i9 


9.9984^79 


1597 3469 


071413 




3486 


069846 


0015% 




998431 8 




NWS 3460 


069932 






068363 


001679 




998421 7 




Jt44 2tB3 


008456 


OSiHM 




066866 


001600 




998410 6 




615 3443 


066986 


934616 


2461 


065384 


001001 


18 


998,399 5 




■481 343S 


065519 


9^)6093; 2463 


063907 


001613118 


998388 4 




943 3437 


064O.W 


937665 3*W 


062435 


00I623!l8 






■see 3419 




9;(90.33l 3437 


060968 


00163418 


998366 31 




850 3411 




910104: 2430 


089.506 


00164618 


098366 I 




396 3*03 


0fl07O4 


941952 2421 


OAao^ 


«l\«ft\S ■*!»«.' 'is 


\ 


«i 






1 f'-r 


T^s^T-rr-ss-T 


-\ 



















22 (5 DKQRBRS.) 


TABLE OF 


tOOABIT 


l^sS^I 




-CSST- 


T-ng. 1 B. 1 






Ta 




■oiTi^'mi 


11.03970 


5S41953 


"2421 


1L05SO4) 


iaoai66^i8i9.9b 




1 911738 23M 




9*1404 




066.W6 


O01687 


19 


9 




3 M3174 3387 


056831 


94*852 


3*06 


055148 


00167^ 


9 




3 944(100 sars 


053394 


940295 


2397 


033703 


001689 


19 


9 




4 940034 33T1 


(I6^»6I 


94773* 


3390 


032366 


001700 


19 


9 




£ 04;4fi6 2363 




949168 


3383 


060833 




19 


9 




a 948874 23M 


091 131 


330397 




049403 


00179a'l9 


9 




7| BWJ87 «M8 


049713 


932021 


3387 


047979 


001734'19 


9 




S BSiem 3340 

9 S68100 ssaa 

19 asms 9329 


O4S304 


B63441 


9369 


046569 


00174619 


91 




048900 


954868 


9351 


04614* 


001767191 M 




046301 


938367 


3344 


0*.3733 


001768 W li 


ll&9SSeiU 3317 


1LO44106 


8.987674 




lL0*232fl 




191 9 9| 




la. 997384 3310 


042718 


95907B 


3329 


0*092; 


001791 


19 ai 




IS 968070 3303 


041330 


980473 


2323 








14 980093 2395 


039948 


9«I86S 


2314 








IB H1438 2388 


438671 


963255 




0367*3 


001^191 a 




IB 982801 33B0 


0371B9 


964639 






001837 


l^ $t 




1T> 964170 3373 


035830 


966019 






OOlStS-JO^ (i 




18 966334 3266 


034468 


997394 


3396 




ooi86iS9 a 




19 966893 3369 


0331U7 


908766 


3279 


931934 


001873)9 a 






96824913253 


031 7 SI 


970133 


3273 


029867 


00188*1? a 




8.9696001 334A 


IL 030100 8.971496 


■M66" 


U.D28504 


10.00189S 


lo^ajl 




3! 


970947 2338 


039033 


g;3866 


9258 


03T145 


001906 


M^ 9| 




33 




037711 


974309 




923791 


001930 


M a 




M 


973639 3224 


026371 


975580 


2244 


02*1*0 


ooi^ssM an 




SB 


974993 2217 




97®oa 


3337 


03309* 


O019«4 


10^ M 




301 9763931 3310 


03370; 


9T834S 


3330 


02173i 






27 977619 3203 


033381 


979688 


2223 


020*1* 






38 978M1 3197 

39 9S03Ee 3190' 


031069 


990931 


3217 


019079 






019741 


933361 


9210 


017749 






89 981673! 31S3 


018437 


983677 


3204 


018*331 




Ma98a883| 9177 




a9B4899| 3197 








33 9841BBI 3170 


013811 


988317| 3191 


013763 




1 




33 


98H91 2163 


014609 


9876331318* 


013488 


O0904ll31 


1 






986789 9197 


013211 


8688431 3178 
990149^ 3171 


011138 


003060131 
003066*31 


1 




S3 


988083 3160 


0I1S17 


009861 


( 




at 


989.'(74 2144 


010896 


991431 9166 


008649 


O03O78M 






37 


998660 2138 


009340 


9937601 3159 




00300031 






S8 


(IM943 2131 




994046 3163 


005956 


00310331 






a> 


993333* 3196 


006778 


99S337 3148 


00*663 


00311691 






40 


994497I 3119 


006603 


99OO24I 9140 


003376 


00312731 


( 


41 


8.993768, 9113 


11.004333 


8997909, 3134 


U.002093 


10003141 


M 






43 


9970381 2106 


003984 


999188' 3127 


000813 




11 






43 


9983991 3100 


001/01 


9.000466; 3131 


10989035 




31 






4* 


999380' 3094 






998269 


ooaiTf 


U 






459.000816 3088 


10 9991 8< 


eoaoOT 3109 


996993 


003191 


u 


1 




46 003089, 3083 


997B3I 




996728 


003303 


31 


1 




47 8033181 3078 




003534! 2097 




00331691 


1 




48l 004663^ 2070 


993437 


00879*3091 






I 




49i 005806 3064 


994193 


0060*7 3063 




003343 11 


i 




60' 007044' 3058 


009396! 2080 




003365331 an 


019.008378,3063 


10.991722 


9.0106*6 3074 


10.989461 


moo2308t3ii ii.m 




63, 009S10 


3046 


990490 


011790 3088 


988310 


003381^ 






CS 010737 


aoio 


989203 


8130311 3063 


968960 








84 011983 


9034 


989038 


014968' 3066 


986733 


002307 32 






65 013163 


2039 


086818 


0Ifi603 3051 


964498 


003330X3 


1 




66' (11440( 


2033 


986600 


016733 204fl 


983268 


0023^33 


a 




67; 01661) 




SS4387 


017939 2039 


983011 


O03W«'13 




59' 01883*1 




983176 


0191831 2034 


980817 


00336913 


a 


SO 0180.31 


2006 


9S1909 


WltS4l)3'. WJB 


970597 


00337125 




/- 


so/ 019235 


2000 


flSWIBB 


m,\<m -WKl 






9 


//: 


_I.ff«w 


5™v 


iCounti 


V T»«*. 


V'^S^ 'fl 



. TAKGENTS AND SECANTS 


(6 DBGRKKS.) 23 


^ 






Tt-W. 








SS^oBir 


10.9807 6o 


7mm 




WsT^oWimM 


W'9.B97614 






35 19W> 


S-95&, 


02233* 


2017 


977166 0U2.mi 


22 


997601 






33 1980 


078368 


024IH* 




076956, O0;i412|23 


997688 






CU ISSi 






3006 


874719| 00213B 


aa 


807671 


67 




H6 1878 


97398* 


036466 


3001 


0736*6 002430133 


907661 


M 




103 1073 


97479J 


037665 


1906 


S73316 002153 


23 


89T547M 




180 lOffi 


973(11* 


028861 


1990 


971118 002166 


13 


887534^ 




iS7 1M3 


07M33 




1985 


06996*' 002480133 


097630 B3 




U 1057 


871256 




1979 


068763 003403 


33 asnm 


^2 




lia 1S6I 


e700B-J 






807675 00-.i507 


Z3[ 997188 ei 




as iwa 


966911 




11H19 


B66;B1' 002630 


J3 907480 [50 




(67 IMl 


lULeij77*U 


9^034791 


1961 


10.966208 


Ift001i534 


23:8.997466,49 


13 lOM 


966679 


036089 


1968 


0640S1 


002648 


2S| 997452 48 




B2 1030 


965418 


03714* 


1963 


082866 


003561 


33 997439 


47 




r4 10-^ 


96ti59 


03831U 


1918 


061 684 


002075 


28 997426 


16 




186 1920 


96aiw 




1943 


960616 


003589:33, 997411 






HS 1016 


961052 


010G61 


1038 


959318 


00260333; 997307 


14 




lor 1910 


96U803' 041813 


1B33 


9S8187 


002617 23 097383 


43 




HI 1006 


069658' 0429731 lOM 


067027 


00363128' 997369 


42 




ISH ia9» 


953515 0*1130 1923 


fl55870 00'.'64523, 997365 


41 




BSl lB9g 


057^76 (MfilS* 1S18 


9547161 O.i'jesO^a' 0973*1 


40 






10.856338,9-0*6431 


1913 


10,963566 


ia002673 24 9,997327 39 


m\ 1884 


955105 


0*7682 


1908 


952418 


002087 21 007313.38 




»G 1ST9 


953974 




1903 


961373 


00270121 997299 






144 1878 


962816 


0*9869 


1898 


9S0131 


00271624 907286 






(7« 1870 


0517;il 


051008 


1804 


918992 


00272924 897371 






100 ISOfi 


950600 


06211* 


1889 


917856 


00274334 997267 






no\ 1B60 


940481 


063277 


1884 


046723 


00276824 907243 






185 J866 


918385; OH407 


1879 


0*6693: 002773341 887238 






r*B 1880 


0473611 055535 




9*4*65 0027863* 807211 






m 1B48 


MfllUI 056C68 


1870 


913311 O028OIS4I 997190 






KG] 1841 
»T 1830 


10.916034 a06778I| 1666 


10.012310 lft002816 


24 9-907186130 


943929 


068900 1800 


811100 002830 


21 897170 !2S 




1781 


1831 


9*2828 


060016 1866 


039984 0a:.'S44 


24 007166.27 




ffl 




911729 




038870 002859 


21 997111 ,26 




«7 


1833 


840633 


oteMolilie 


8377CO 002873 


2* 097127 ,86 




160 


1B18 


0395*0 


063348 1842 


036668 002888 


34 997112121 




HH 


1813 


838149 


061*63 1837 


835517 002902'31] 987098 23 




BO 


1808 


837361 


065550 1833 


03444* 002017 26 997083 22 




F» 




930276 


066656 1828 


9333161 002932 361 9970G8'21 




WB 1T99 


836181 


0077621 1821 


832218 00^19*7 


26 097053 30 




fSf 


17» 


10.934116 


).068846[ 1819 


10.931151 la 002961 


2619.997080 


10 


m 


J790 


833038 




930062 002976 


25 897034 


18 




136 


1786 


03J061 


071027 1811 


828973 002991 


25 897009 






107 


1781 


930893 


072113 180(1 


937887 oaiooe 


25 !K>6994 


.6 




tT6 


1777 


929821 


073197 1803 


926803 003021 


25 986979 


16 




»a 






074278 1797 


025722 O03D36 


M 996864 


14 




W6 


1768 


827e94 




92464* 003051 


36 996049 


la 




186 


1764 


826634 


076*321 1780 


033568 00306036 996934 
S22496 003081 126 096919 


13 




UK 


17M 


826676 


077605 1784 


.1 




no 


1766 

riwr 


BM3& 0785761 1780 
ma23ie7'0.O79644| 1776 


02142*1 003096 26 
10930360110,003111126 


ooesM 






^ 


9.^8^ 


"0 


IBS 


1740 


823*17 


080710 1773 


919200 003126^36^ 006874 






m 


1743 


921360 


081773 1767 


818327 003143^36 996868 






S76 


17B8 


02013* 




»17ia7 00315725 996843 






'IB 


1734 


919281 


083891 1769 


916100 O03i72'36l 996828 






tea 


1739 


0182*1 


084917 1765 


816063 003188 26, 990812 






rw 


17^ 


917203 


086000 175! 


814000 003203 2ffi 996707 
812960' 00331836^ 996782 






^ 




01S16S 


087050 1747 


2 




«4 1717 




088003 17*;i 


S11902, 00.T.'3*2fl' «W7l« 


1 




m 1713 


911106 


088144' 1739 


910866 «fia«11&^ SfiBlwAtt 


s. 




SSm-t. ,Col»Bir.-T 


-^r.'.: T'c.^«. V *, a«jg 


^ 


'■■' «^-^_:^ 



■ 


i 




^ 






2i (7 D80REB8.) TABLE OP 


.OOARITHMIO StHM, \ 


ri.l Sta. 


ro- 


C»ee. -\ T-nt. nn 








0U.O8MI4 


\Tu 


lUUlill* 9.081.144 


1739 










1 0S«9-^i 


17O0 


91:1078 000167 




909813 




3 










913063 OB 1338 






003380 








3 088070 


1701 


011030 002266 


1737 


907734 


00339S 


a 






4 OeSDBO 


1896 


910010 093303 


irai 


■ B06698 


0033131 








B 0BI008 


1693 


908993 0B4336 


1719 


906664 


003037 










1688 


907976 096367 


1716 


904633 




361 






T 0B3037 


1685 


906963 096395 




903606 




<2N 








1680 


905953 097422 


707 






m 








1677 


90tiJ44 098*46 


703 






m 








1873 


903938 009468 


699 








IJS 


Im 




895 


ia899613 


ia003433 


13719. 






16G5 


901034 101504 


691 


S9S4S6 










1661 


900H36< 103619 


1887 


897481 










1657 


899038 103633 


1884 


896488 










16S3 


898044 104643 


1680 


896458 








18 102048. 


1849 


897953 IftWaO 


1876 


aOM50 




37 1 






IT 1030* 


1645 


896963 106S66 


1673 


893444 




37 1 








1843 


8969761 107669 


1669 


893441 


00353S37I 1 








1638 


S94008I 109669 


1665 


891 440 




37 9 






aol 10S9D2 




1661 


890441 




3l\ 1 




31 9.1O607S 


IM" 


10-S9»»7 


9.nUG66 


"1668- 




la00366S 


loTii 






23 107S51 


1637 


893049 




1664 


686449 


O03600!37 






33 10S937 


1633 






1660 




O03616I3J 






34 109901 


1619 


S00099 


113533 


1647 


886467 


003638197 






2S 110873 


1616 


889137 


114631 


1613 


886479 


003649I3T 






ae 111843 


1613 


88SliS8 


115507 


1B39 










W 113808 


1808 


887191 


116491 


1636 


883509 








38 11377* 


1605 


886336 




1632 


882528 










1601 


885363 




1639 


881548 








at. 116698 


1697 


8443031 119429 


1625 


880671 


00373115 




31 (.118656 


ISIH 


La8S3344 


1.130404 


TSis" 


10.8796W 


10l0037«» S. 






39 117613 


1690 


682387 






87863t 


0081^ 






33 118507 


1687 


881483 


13234( 


1615 


677662 






34 118S10 


1583 


880481 


133317 


1611 


876683 








3d iso*m 


1680 


87953] 


124384 


1608 


876716 








38 131417 


1676 




126249 


1604 


871751 








37 1333d3 






136211 


1601 


873789 








38 133306 


1M9 


876604 




1697 


872838 








39 134348 


1666 


876763 




16M 


871670 








40 136187 


1663 


B74813I 


129087 


1691 


87(«13 






41 1^136139 


1659 


10873476 






10.869961 








ti 1370(10 


1656 


872940 


130994, 




smict 








43 137093 


1553 


8730OT 


181944 




668066 








44 138035 


154B 


871075 


133893 




867107 








46 138S54 
46 130781 


1M6 
1643 


670146 
669319 


13*(39 
134784 


1674 
1671 


866161 
86«il6 


0(^^ 






47 131706 


15^ 


868394 


136726 


1567 


8843741 








48 133630 


1636 


867370 


136667 


1564 










4S 133661 


1539 


86S449 




1681 


86at95 


O04054S9 






flol 134470 


1639 


8655110 






861456 


004O72'3a 




SIB. 135387 




10864613 


B.l^«476 


TfiM" 


10.860624 








63 186303 


isaa 


863607 


14040! 






ooHoSsa 






63 137316 


1619 


862784 






868660 


oo4i»tpe 






64 138138 


1616 


861872 


143261 




857781 








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(IBDESBEKa.) 29 








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078570 


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1036 


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980 


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1023 


666967 




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60 


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fl7a2ao 


333046 




606354 




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090134 [60 1 




m 


rsTT- 


lawBOat 


ai»42.'i9 


T020" 


1*6667*1 


Ki:ou95i; 


9.990107 


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1019 


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67*466 






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990062 


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n 


970 


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1016 


663907 


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073300 


336702 


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837919 


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8*9339 


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30 


(13 


DEORBES.) TABLB OF 


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Sttut ib.r 




TiT 


iaiW7VlMU.3Ba3U4 BOO 


10.636686 








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6473BS; aeawt^ 959 


636060 


01130649 




2 3631S1 


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e4SU9 Wi45IA 968 


63548S 


011334149 




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64H274; 14650901 957 


634910 


01136449 




907 


C457WI »ti56r>4! 95S 


634336 










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64fll86| ■mdi'.ll i^ 


633763 










904 


04484^ SttOaiO 9.53 


6:»190 








868901 


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644099 3«738a !fca 


632618 








3Sa43 


9oa 


643567 367968 961 


632047 


0H6lBiffl 






3fi6W4 


801 


tAMm ae&tii 960 


ftn476 






ID 




899 




630908 




SassoM 


"806" 


io.6ti93(i|g.3di«ui| 948 

641397 370^113 940 










3586U 


897 


630168 










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629201 










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628813 






la; 3«KJlo 


8B3 


tt;l!)785 371933 


943 


628067 






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639^ 372499 


943 


637501 






171 3612C7 


891 


638713 373064 


941 


^6836 






18 3li\Si2 




638178 373GB9 




636371 






19 3Baa66 


8«9 


637644 374193 


939 








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C371U 374756 


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9.3a3422 




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6a»)«8| 376881 91*5 


634119 






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8B4 


635515 37(1442 934 


633668 


011957jBt 




24 3Bflt)lfl 




634984 877003 933 


623097 






36 aGOHfi 


883 


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620 


428113 006773 


608 


393337 


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967116 '69 




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620 


425800 607137 


605 


392803 


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019 


4254SS 007500 


606 


393600 


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967013 67 








425176 00786a 


604 


393137 


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4248A4 603235 


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891419 


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960869 '64 






518 


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603 


801060 


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96 


966808 63 






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423931 eoiiiia 


603 


890688 


033244 


86 


960766 62 




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517 


42.Wai 609674 


803 


390826 


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96 


966705 « 






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602 


389964 




96 


966663 50 


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10.033398 


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603 


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86 


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422ItB8 sum 


601 


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86 


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960293 


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887079 


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420630 613281 


699 


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41S995 S16077 


697 


384923 






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41«aeS 616TB3 


697 




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418076 616161 
417771 616509 


696 


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696 


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607 


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410942 618662 


694 




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416839 619008 


594 


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693 


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414728 620076 


693 


379924 


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40 (33 


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1^308296 ia086566 






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101 


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632053 


581 


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596727 


401 


404273 


632401 




387599 036675 






596021 


490 


40^979 


632760 




367350 036720 






506315 


490 




633098 




366002 036783 




f 


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480 


W(3fll 


633447 


680 


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17 


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4030971 633795 


580 


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1028011 631113, 


579 


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488 




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19.598076 


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10.401926,0,0361861 


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678 


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677 


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577 


362736 037438»1 


38 


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485 


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576 


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399591 637960 


578 


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81 


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484 


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483 


30872d 63^92 


575 


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673 


360683 






601860 


482 


3!t8140^ 639083 


571 


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602150 


483 


307860 640027 


571 


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602139 




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369629 






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181 


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3061061 611717 


573 


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10.3901189.6420911 


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39.'>830 


64243* 


672 


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305513 


642777 


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396256 


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071 


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478 


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614190 


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392108 


646881 


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10.371081 U. 67 3301 


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616 


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370370 673374 


546 


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416 


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1531S1 


984094 421 


IW53(l6 








m aia 


155058 


994947 421 


00.^.1 kVJ 




851997.20 




mrm^ 


iai52t«fl 


9.99519! 










IS9 313 


16280 


99545L 


421 


004543 




'851747 


18 




(a: 318 




995706 








851022 


17 




UM ai3 




9959.'i7 








851497 


16 




»2 213 


iSUh 


090210 




003790 


14aa28|209 


85137i, 






rem 312 


15229 


990463 




0135.37 


14876l'209 


aM24<i 






m 213 


152164 


990715 


121 




148879209 




la 




ia« 313 


163030 


990008] 121 




149001209 


85U9!W 


13 






lawos 


097221 421 


003779 


119130209 


8.50870 


11 




as 212 


151782 




002527 


149255 209 


850715110 




ml 213 


iaisiess 


"TaiiTHB" 


121 


10.0<tfl!7i 


107149331209 


9.830019 




(^ 3U 


161528 


997979 


421 




14960T|210: B50493 






IBQ 311 




998231 




001761 


1496.32:210 


86O30S 






m 311 


151374 

151148 


993431 
098737 


121 


0015J6 
0012^3 


14B7582U 


860242 
85011 






ITS 311 


161031 




121 






B49990 






\M 311 


160894, 


B90342 




000758 




649864, 






15U:«8 




121 


000505 




B19738 






ISS: 211 


150641 


999747 


421 


000263 


1503892lS 849611 
1.50516^210l 849445 










10.<H«)I>W 421 










s«.„i 




l->u^. 




\ 










*(i\ww™- 


^ 





1 


iS^^TABLE IV. A TABLB Oe NATOEAI, SIMM. 




Ul «■ 1 l> A> . *■ 


ITl* I *' 1 4' > *' > *'-™. 




iUM.m,»ln^J.iK»m 0523aB 


iMifrae 1 087i5ti 


1WS38 131869 13917360 






0361!H), 063030 






104818 


133158 13MUISI 




vinM5i» 




03JM81 


063917 




087735 


106107 


133447 1397«,M 




ttWMMTS 


msssa 


oaa77i 


OSXWT 


U70837 


088035 


106396 


1I3T39 I4O037J6T 




400UM 


umu 


OWMJ 


O0S198 


070917 


088315 


lOMHK 


1S9IU4 1 14Utf6W 




At»l45t 


018(H)7 


03031)3 


0S3T88 


071307 


088606 


IU5976 


1SI313I 140613 w 




«'W171S 


0IH1»7 


OMdU 


064079 


071497 


otsaau 


loewM 


133(»l' 140901 M 




KS 


oiiHsa 


OUOUM 


OMaau 


071788 


U891B4 


106S63 


12389U , 141188a 




UIU7TS 




uuUik) 




i>«»4?4 


1U68U 


134179 1414775* 




KWWW 






OS4U60 


ofaaaa 


089763 


107133 


U446r 1 141T<G'a| 




loioujm 


OJuasi 


037806 
038007 


056341 


0736W 


0000531107431 


121766' 43UiaM 




U'OOSWO (M0S5^ 


073948 


09034a, 10771O 


U604ff usaiM 


■ IHWMUIIO'JOW 


038380 




073338 


090633 


10799i) 


1363ilil, tMHMq 


^■m>t«' 


mixa 


aWfi78 


06«U3 


073538 


090933 


106389 


i:iS6^ fiVUffl! 


^■Ruii 


irjl534 


0389B9 


066403 


073818 


091313 


108S78 


11t6910 , U33WM| 
13fU9i» 1 ItUttMH 


HK*^^ 


(WlSlfl 


03y3CO 


066693 


074108 


09ISU3 


108867 


BBIhsh 


O33106 


03!IS»I 


036963 




091791 


109166 












07468U 


093081 


109445 


iMTTtf i4«anm 












074979 


093371 


lumat 


137066 ]448Mi1 






Vims 


oiin33 


057*64 


076:i6M 


09366(1 1 100311 


i37afiti 1 imeSa 




amoMiSKwaiHlU 


040713 


068145 


U76569 


093050' 103ia 


13T6»'l44Mii| 




ui|l«X>llW 1 iKUfiW 






076849 


093339 10603 137990 lluHafl 




sjootBfflB uaaasi 


IMI3tU 


068730 


076139 


093639 10891 ll383U>!l4SM« 




i»W»UIO (UU*1 


(Hi685 


069016 


076439 


093819 lllWO ! 138SU7 








«4187fl 




076719 


094103 1U4H9 


138796 






013168 


069697 


077009 


094398 111768 


139084 


1^^^ 








059^7 


0773BU 


<I94«87 1 113047 


13WT3 






a;[o»T8M l«5.fU5 








0B4977 113336 


139661 






aoW727 0*8177 


in:\fm IHIOKW 077879 

w>^m "«"7.^i iirhioM 


096367 
095556 


113<Ei5 
112914 


139949 
laU338 






04;Wiy FNiLIH!) 07^459 096846 


113303 


lau&M 






■ai|«»B0i7|oa«4fw 


IM^Ii)i0:00l339il':w;49 


096135 


713483" 


laws' 






»j'UlltKIIM (MHTO) 


044-01 


l>Ullii"J,U7SHWit 


U9U435 


113781 


laiioa 






ayuuUciouliKJTim 






07U329 


096714 


1140T0 


laiisi 






3*0.KWW)H)a73i0 


044783 


063310 


079619 


097004 


114339 


131IISD 






MOIUISI 


omoiu 






079909 


097393 


U4I148 


laUMS 






aU 010473 


037U3a 


045363 


063791 


080199 


0B7583 


114937 


193368 






y7 ui07«a 


0Z831a 


0456M 


063081 


oao489 


097873 


US336 


133645 






38U11UM 


02SS03 


046W4 


063871 


080779 


098163 


U5516 




liuiiiw 




^U1IM4 


033794 046335 


063661 


OS1069 






laaiai 


laoawul 




40011U»5 


030085 ' oiossa 


063993 


081369 




iiwws 


ia»4io 


15068U3II 




41011IW6 


039375 


o*B8ie 


064343 


081649 


099U3U 




133898 




43012317 


omm 


047106 


064633 


081939 


099^30 


11667 I 


13U98U 








omsj 


047397 


064833 


0»»3S 


069009 


116960 


134274 






M;oia7i» 




047688 


066113 




099899 


117249 


134663 






«l'013090 
46013380 


030639 


047878 


066403 


08380S 


100188 


U7637 


134861 






030839 


W83e0 


086693 


088098 


100477 


117836 


135139 




47013671 


031130 


048659 


066984 




100767 


USllA 


136m 




4S'|)1396'J 


031411 


048860 


0U63I4 


0B367S 


101056 


11B4U4 








4U0113KI 




019140 


066664 


083968 


0348 


86UJ 


36004 






Sll!ui4fi44 


mm. 


049431 


060854 


084368 
084547 


0836 
9ii 


8»&i 


1303U3 






51,014835 


033363 


oSfsT 


"067145 


119270 aoE&rteii^ 




a3,oism 


033674 


060012 


0S7Ua 


0S483 




11965J 'S'^ 




S3 U1S4I6 


OUl'864 




067735 




*I3 




64 0157OT 








Od£4 








AS 015998 






0«SJ<I6 


0B5 








66 omasa 


033737 




00859(1 


0Su9n 








57 0165S0 


034037 


051464 


0U888G 066 •* 








.« 016871 


034318 


051756 


0691 6| I8>f f> 




m'l\3 




fi'J (117163 


034009 


0530« 




A,m 




60U17i63 


034S99 U533SII 1 IMiQ7fi6 OS ob 


04538 U1389 Ml .1 ■MU.ul S 


J 


"'«"" «»" "*''' ""' 5* ?i — ^^ » i ^,MS 









A TARI.B OF 


NATURAL BINES. 631 




».f 










0,156*34 


TmisT 




'wfesn 


2419M 


268810 276B37 


292972B0 




1 156;a:J 


i7;ro:ts 


imu95!-^l!(li 






2691 Oil 276917 






B 107009 


i74aii 


iBiijM ; lioawi 






26**81 270197 






«1«;2US 


ITi-WS 


laiuuGMosTrt.-. 




2-mti9 




2»;M)li,,W 




4 ixisa* 


174794 


lUiy.'il,2U!>i.5'l 


MliftISS 




■ifmvi 271.768 


29;!4W.W 




6 157471 


17601*) 


IIW237 ! »>itt3+ 


KllkilW 


a4a3.'« 


2eiW:'4 1 277«M 


39:lT{Uft-. 




8158158 


17.5,J67 


iWiWiwratjiy 


uiimii 


2*iei5 


2W15U5 277^10 


201iHO;e4 




71.Wt45 




19:M)7 ! aWiWS 


^J6ftU 


34:eS7 


aeoTss a77oi»4 


2U43IH|63 




P1587y3 




imsmt 1 310187 


337;il8 


344179 


361066 I377S71 


294696163 




» 




H33 (W 




4Ma 


26d47 3- * .»48 6 










iWH W 




JO 0^' 3784.12 2flfifiSlfl0 










93SMJ 




20 9U8 1 ^»&(-J0|4i) 




J4 "M 




9i » 








iJ 




W 










4^604. 


liB 








(6 




« ID 


771M4 








•9 


29 51.46 




( d W 






«>H 




2b. 


U 










•9 




bJJ JflUW* 


29 09 43 




fi 










18 [.28006 






» » 






M 2N 


6464 28094( 






9U- 








B4434I 2) OJOkOB 




i- *^ 


m 








2- 


994 








'^ 










TI6 










\ 30 


























wuea Uwaisa Lffliffja aootsB^i 




■M 






2603«olafi jm'js4<iis aooToeao 








^lir-« \'Mm jmB63si 








m. 1 30 36138 








M loesafl 








M 36 






201 W 


34^ r30B33S 






til- 


J0237 






a* m 


302fl4 38 




w 




1 4H. 1 "B 46 30293 JS 






«+ 


>9U LftJH IttHnSU J86IU6 803303 










1 »n a7i04nlwwn 3'»4mio 






^ U« J8 UtU dU. 6U 








(H4 


T8 


363798 27flAn0 


SST36 3040331 








ja 


ZZI 




3MU39 0880 


287039 


804. 












9304 


Sfl* 


MM 80 


2879 


804687 6 




1 






aoaoiJ 


sum 


337880 


8.W«W 27 440 


288 Dfi 


304864 










»W9a7 


23096 


337908 


BIHftS. 37 M 


368*6 


306 








«H1 


M 


J3L«S 


WLfl 


AS 66 ^jono 


388763 


3064 










imiw 






6644 3723fO 
68 2- 27^660 


289002 

2893 


TOS6B6 
1069 2 




en 












80189 


■iOfi 










-uti-ca 


27 


•mn KHL6|8 








UU 2K» 


273400 


"90 46 aoosoa 8 












27167 


■W4 30-oeol 




w 








■>7T8flfl 


39- 


•» W 6 






IMBfl 








"7 » 


"vm 


30 T3 
















■^ o9 








W 168 








■^ 98 


2B 637 










■/■J*«4 


2*1 :«7 




27.W78 


20181.1 


308464 2 




winaiie 




->?4fl68 




36S6:n 


270368 1 29-2094 


308740 1 




eonam 


inowo aoTQis 


r}*m 


"41 fl-" 


j^Mio * "OMW v^wm!L2«*^Sl^ 




U. BO - 


















H 




















i 






64 


A TABLE Of NATURAL BINES, 










»»• 


■3JHI73T 


40075*^619 4383FPS 








374ffl?r 






l|lO"'M 


ja4 1 -J 3.MAI 


i748-rlj9-P90') 




4. 8S 438833^ 






sLn-i^ 




(jS It 


140 «1 OJ 


4( ^ 






4)i<m7 


Sm. H Ml 








4JB40Q 43n6s3 


1 




4'UOI 3 






37l(^rtfll30l8(^ 




iJami 1394111 I 




mi WOO 

6|jlOC-6 






feOoS 30i070 


4II80U5 


439936 MOffiCT 


1 




i7JIS 




AJOfl 


( 24 aspur 


wAm 


4341)19 iHoeatw 




TJlUBoJ 


JJ 49rf 


34.«.).l 


36U"W 


3 0404 31C200O 


408jOO 


434483 tiimm 






8^311229 


327 68 


344 Ob 


36064. 


j-e76j 


30-8^ 


4W801 


4247.26 «4«l4an 






O^UIBOO 


3280t. 


34449 


360811 




39J140 


40' 127 


4.4900 4ttBM 






ujiaoae 


)2a»7 


144 5 
34.K)25 


361082 


377571 


mu7_ 


40939- 


43,rfSir 44IHS| 




^ai" 


3930 6 4(mfiS8 








3Z8W. 


345 '08 


3fll6B5 


JT7841 


303942 40- "I 










9^141 


340671 


36IS96 


3 8in 


304300 41 -i 


^ 4 ff 






l2siJ8B8 


3-WI6 


345844 


362167 


378379 


J944T7|4I 


w 






19S13I6* 


3:iOU)l 


34611 


362438 


37S649 


304744141 1 


4 U 








■iam^ 


J4a.]90 


J62700 


8-8018 


306011 410's. 


4 >n, 44 « *t 






ITJlSTll 


SJO'K 


346663 


8B2US0 


3~Jia7 


305r8 41 4 






imta-r 


fc)O.H 


3469ai 


303 11 J-J45b 300640,411614 


427J58 <430lJ|| 


1 






tl" « 




ll>W 7 J3jt*13 4I177J 




1 






IJ Hi 




i 1 J094 (06(180 1 4IJ04fl 




1 




i. m A7 ^ 




lKoa<^ ilM H 41 31 


4J»14 '44^^ 


1 






JJ 1 J4* 


3UJJ6 


JSUOJ- MJ 4 4t 7 


4 t4\ 4441 14 ip 






J.J1H87 34a«« 


161600 


jaoa " 4 


4 4 n 


B 




94J15049 


(S-lBl 148573 


364977 








1 issi -oau 


J3243> MStt5 


30j14S 


381'BJ 






jeiifj 1 


33!riO 340117 


366418 


WWi 






37 a« 


dSaOSi J49390 


366689 


381H 1 1 






jsiatBTw 


333258 3JWi' 


31.5900 


3S 40 3 a 15 414104 


429UMI 44f(.7T3l| 




39317020 


JSaBJ,) J4»Jj 


jnfl 31 


W +lu 308482 414429 


43U->4Ui4MS9V^| 




3iajJc« 


333iW7 i> 1 7 


J6.501 JM6S.J 


398^40 4146CU 


4»)6ulMunH 


l«l»l W^S. JJl (HH, 


iOW-lb 414S68 


4.307-4 1 44W)MH 






J.l.J.t.'i. t- V 1 iK, 2 1 Kf g 4 J ^3 
















36 " 


* J 








87 










3a3ij 


tdW«fl « M , JBMUW 3H48,f 4 11 *«■ 1 4 








39 )]U7iJ' 


336i 4 di^tb(td6lil!kH> Ideal 1 4I114U|4I 








« J'.W* 


33im7 3S^931 1 3U9-J0G 1 3^»3U( 41 14IS 


4 


', 


1 


l«a ikj- 


XXSl 


J6MH JUU4 6 




4 1681 






1 4J320fn 


33-095 


3534 360 47 


iHjOOt 


4 1948 4 








td^Mm 


J3730!) 




3h 174 










MdJUtH 


33 J4J 














|45'fil*.W 


JJUl 


WHJJl J7».7 


JHG 11 


4U'2747 4 








4631!1716 


3381 to 


3j4)ft 3 JS 1 


weiffl 


4030U 4 








47 321000 


3;ie4Si 


dfi48J5 371 m 


3K 47 


4a279 4 








48333^66 


3.38738 


366107 


J71J68 


jy 10 










4S322M1 


339(1 


366J79 


^11)38 


3877H4 


403SI1 4 








S03Z381S 


339^ Sfisari 








* 








3395^9, JUiO^J 


37J1 8 


388320 


404344 








aaoeai 300194 


37244B 


0SS5M 


40401 4 








is 


340100 3S(Hbe 
34U38U MB 38 


37^-18 
37 01*8 


3WW, 4 

jsei 4 










34«6Bd(i?i WO 




381 








34uri\367ia\,irjB-B 


3fJ 






5T'a24 43 


341200l357BWl\Kli jn WJlJi-s 






. 


a83''aH8 


34147 


iB78S 


31\(« 


■w J 






->A u:A«. 





3*1 47iar)&*Wi"r4aJl\*JW - - ,-^iv .™,a, 

^ \^^^^\i ^^"'-' «wi^^ w«i^ ^^>'y■!%>^*Ti 




NATIJBAI. BINES. b5 

ai^aa-Xoa "lain ***'"■ 



MUS2 

.46 «S 

ra|*G2366 



i8 54 S3. 


^ 


9fil 


*W ^ 




59 sn 










1)9. 


6433 IS 




























'V 


SK 43 
Ml « 

m4i 



SSflR 


•u 




47S6» 


011904 


6. U 


oaj 


H64 


J06 86 


«M 





fl( U3 




50-^ 




49J4 


50 6JS 



46 jl. 

(■94 { lib BBMmU 
J7l56ii4nfl M 'K 





(HI 226 I <ai5i 
titl*SO 1 654 i 
611073 [ 6M9S 
641896 6Sfi 80 C 
t WiV ft\<!MAW\ fl 

. ..^^[ -w'-^ •>."'-\ ^t'-C S 



P NATCBAL BINea. 



fHP 



7ilSE0 
731780 
7919^ 
733163 



733766 
721967 
783108 



ffil 



ViftMi ' V7n< 



1 75M73 

r TU0S3 

1 756853 

] 7W0U 

) Tfieaa* 



T»3S32 
733730 
733997 

73«m 
73IA30 

73*717 1 

7349U 7 

73511a 7 

73fl309 7 



74M78 
745a7U 
HBBM 
7*6087 



7S7375 
7S7565 
7B7J6S 
7579*6 



76ISe»6 
7«6793 



76a6a(s 

7a88U 
709038 
lOftll* 

76)HiXf 



777329 ; 

777fiia ] 

777685 7 

777878 7 

77S(Wi> 7 



779520 J 

779702 J 

77UBei 7 

780067 7 



7910*6 

791924 
791*01 



5 800731*1 
i 800006*; 
1 801060 *l 
a 801951 *« 
" 801*284. 
80160?*; 
801776*: 
801IM9'4: 



► 73*773 
t 791974 
' 730174 
i 79WT4 



t 730975 
I 736175 
1 7a(375 

1 736575 



1 736U97 

1 736394 

<i 736*91 

" 736S87 

786884 7 

737081 7 

737277 ] 

787174 

737870 

737867 

738063 

738359 

738*55 

738S51 



747806 
747798 
747991 
748134 



758703 7 
761«93 7 
7fiO0H9 7 



7*9148 
749341 
749534 
749726 
749918 
750111 
750303 



771810 
VriBSA 
773179 
T7338* 
779MS 



) 7«)«4 7 

) 740609 7 

} 740806 7 

i 741000 7 

r 741195 ■ 7 



I 729IM13 

J 730169 

) 730961 

I T30MD 



•3iamh* 



7418H6 
741781 
M197e 
743171 

742366 
742561 



'59989 
7A'i372 



) 7GOa»|i7 
i 760317 17 
J 760406 1 

7«0flW 

T6D784 

760973 

761 IGl 

761350 

761737 7 

761915 , 7 
769104 1 7 
76 2292 7 
T6:!*80 . 17 JUSG 



77*677 
174781 
T74B44 

T7519S 
776ffl2 
775498 



7B2789 7 
782970 7 
7B!J161 7 



B 791757 

" 791935 809*70 38 

792112 81.1964437 
799990'80:;8173S! 

709*07 802991'3ai 

T92644 8031S*;34 

J 803337,33 

) a035n'32 

i 80368431 

i aoas.vTiBO 



794061 
794338 

79*415 
79*591 
794768 
79*944 
7951 U) 



TS4921 
765109 
769296 
765^3 



776679 
775863 

77804S 
7762;lO 
776413 



8IV.75(i 10 
8l>rill:;8 18 ; 

'ikwso I mam n 

'951196 ! 80G273 10 , 
'9(l002leiH)44i;i5l 
796178 ' 806617,14 1 
'96351 1 801(788 13 
'g6.V)0 1 806960119 
'96706 !S07139lll! 
BO73l»(10' 



\ ^k^] <tu ^TTT 



, 797057 I S0747SI 9i 

786S7S 797233 1 807617 8i 
786750 797408 1 807818 7 1 
786836 1 7975»4 80799(K 6' 
787114 T97769 806161 6 
787294 797935 1 SOOM*, <' 

.«™M...>*n™iiwma\n'i»Yis>\^»»iWi ^ - 
7ft5&57 \ n«i^V«i'«A"'^£^\'^^-i 
i ^o - I a» - ■, - ^ ^ -^'* '^-^^ 



i 



. ^ 




68 B N R Ea 






iUhjiss 


k II i' MH 


W^ 


-TIT" 


WlfiT 


^■^ir 


V"''"H.a 








m 


84^ 
















ti»8Ue 


4S^ 


eS4B7 








tew 


UU&f 


-W. 






tA 6 








1181 


82MJ38 


3li!li04 


848WM 


65 W> 










8B986 


a^dOO 


SJaMi 


S4t» 


8S 










a2U£J 


SdOU 


(MUtUU 


843U 


*wuua 








rkioaia 


8:0 


830 


83U 


34J 2B 


868- 


SllM 








8:M>48a 


830J;)7 


8JU0.JO 


H4ir 


8D8JI14 


8<l 
















868513 


W 


WAHSO an uH 










840^ 


84lt08(i 


tUMWl 


8i> e 

80-63 


MKM 88I8HIS 




ni810891 


S^^UWd 83082' 34M0e~a)J »a 8S88U 


ft7fiU 






laBiiow Si w 


83W84 840ati 84Ucflt: BaSMU 




«IU7 






13 8UKM 82 


83 4B 


840 >^»m BJJOU 


8b U 


Irti? 






14Bll*wla. « 


aj jwj 


S4utM2 BiM jU as 2fi8 


sasuM 








lGBlla7i,!UtM 


8a 


84 03tf 8j(WS BjIMUM 


aoB uu 


a 






16^L7U » 




84 U 


Mioou Boasso 


U8^ 










3J 1M 








rt 




331 M 




89UI \mmi 








iB'Siaafla ftwa 




84(188 


sjtww sbooo 








MOlSlUMa Sffli 


esa- 


84 aj 


80 I8a04tt 








aiisiMffi 8ia6*i 


8II343B 8411182 


as '691 29 








a:«8ja7ftl R,»U6 


SSmi 84 W 


8fl 






KlSimi i9 


833 UO 84 UO 








wsiaioi 8ja JG 


saaaa 84. 










!«HU82 812 








W;SliM39 -J4U 3J3. U 








a8'8l3778 SAff SAKfaS W 














aiBiaw: saatti rsj (<4j ss 


85 BM BO 48 




austuie e- »hHt> 


85 04 HbbJ9 






aiiaiias* 12a i4046 (H 51 er w, 






XJlSHUa S:,445b Mi A^ >1 JU44 8b 






tSlSl^fi^a 1 MJ20 SUM!? 1 8439 HSJUDb 86 






34 8Ur«l 8i tfO 8J4&f7 8«0 H5. jalS, 






3aS149o9 8:( IMU 8^4088 8M jJJ. ObANMi 






J080 a' S3**) I844.JM B*Hk. 1 J5 ' 






J7H N, AtiOUS W 






a)8 « 
■M 8 






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4J816.JW W 1J M 1 




■^ 


8H&Slill| 




-M81(M74 U4 » 84 8o4 | B(M«<J 




■mj W 






iMlSeti 


W JO (* -t, ^ 8.>4! 8fai8Jb 


IN. 


U8ur» 






4rHiufiua 


JU 63644(1 M,«8. 8&£0li3 H0!ij»82 


u^ 


(« aa 






4 81UU 7 


8:/69 1 JbtiU) 84bOJ8 Slrj 8t>4128 










4881,146 


Bros 8J6 G4 H4 8WJ64 »6t- 








4U817aia 


aa we.m >mM sasa 8M4 








I5OS174S0 


82741 OSJ 84U503 


8561 rfi 8MflU7 








5iaiTW« 


^ «a 8466.58 


.58 8W 






52 817815 


■» 83 8468 8a5S«l 8b4*l« 






.13817082 


b3 WJ 84ai 666 






54818150 


84 1 






























68 818818 




1 




Af 818118a 


-88 


v(. 


84 8J4 










m. a 



US 



t ■'\ o'Tl 






tp-aa- 1 8«~^ 



iTBl 



'JUBKUTU SMiHlt : 90C6H ' 911178:2 

wwuos au!f i7<j I wears i visuoo 

MBM6»1 8DUd(M lloe7W lows 
TISO 686 aJHHi I i»(i9- 3 9 38 

«eu 98 a» "■ -■ " ■ ■■ 







92741^ il 

1197^10 e 

ll!l7fl 9 S 

107 IS S 

1127836 H 



9398111 
It 9:W9D1 
7 MOOBO 
940 89 a 



91ST0eSB 
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04038 946 mSS 

tH04B6 !H(U 5 

IHOeSfi 94<i!i6S» 

IM06W ft464aj60 

HU S3 »4U»A,4tt 

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J 9 JWH 
|] SHSoaS " ■" 
9 903709 
a 9 03831 1 9 Ii4 
S U(I3B58;9 hi: It 

904083:9 ua 9 W 
B 004207 19 5 9 S5> 

1 MMOai 9 043 9 «f 
S 90445S 9 (j I 9 S-y 
I 90467919 ftS 8 9")< 
:i 904TU3I913U0 S 
8 1 904827 912 20 9 
7 904961 B 2"39 

S 90 5075 9 ^3fl8 

3 !>05198|9 

1 905322 1 U 596 

906«fl|y 2" 5 9 

S Mn.'iG'.l I 9 ..S34 9 Jx i 
S i)05C93|9 Jj tiO 
3 90.W10 9 31 9 MO I 

1 'JOdSiSa \0 « 04 , 

\) 9*1608219 109 l»tf> 

OitO I 




5 jaoaa 93" 84 Is 



aCi'Aa'- j'ia -la*' Tia' rtJk^ t a>^ ^ io 



94907'' 1 
1M44 950063 U 
W4«M 9fil Silf 



' nsos 6 6 

tl 50606 fi{H| 

Mm im 

>4 0SD786 3U^ 




3 96K9T:M9801 
9 B66473 1 960BT3 

OGAMS 9%H3 
9 90,5024 OTOnU 

1 9en7oo 97on84 

aHftWlM SUllOl 9e5776l97IH5S 
/ ' eo Safi3(W ' 9612fi3 1 B65S»a6 iWWMlXaHWU 

/ /'itri7--TT>" 



981 J-i* o tewa, 1 1 
■"V -ti.o ' 



A TABLK OF NAT0nAL ilNEfl. 7l| 




' i^' 






11987731 
Ui987779 


ssnwfl 


99^^(110346^2 ft«.lte96-S64|8e8&» 




989a4^fiO|| 


9P03U* 


IWJoHJ 9941 I'.JI, 1j9 »i 


J9«64d 


999401 




9903*0 


991!6 


994 


obn 


99941] 


U9985aS8 




IWUSSS 


99-^ J 


9'H 




99942J 


99986»eT 


ffitOU^ 




























flaosog 


J9 5 


991 




Jf 41 199877 5*11 




[wufiia 


992 92 


994 J. 


tmsisa 




B&yjea 


9B«ei7 


9947h .JUOJ 


<mBsaB3 




99oejii 


99-2862 


994792 99«4 J 




Mril«8!38 




l>92S9a 








900TW 








990-48 


99-296fl 






iai\)S»iT.I 










t*lU885n 


9908^7 


9a.*m 






is'Msaes 


9ffll86G 99306S 






lti!»S40U BldBOa iO.103 






17 9*«-T") DHWM 911 J7|4 












J0 938j«. 


JM JO 


90J ■>« 












4.lMMg583 



n 4U 91J8JJ 
991 7B3 J"!.!!*. 
W1S2I)^ 
P91S57 



61(9'^ JO 
Jta'9l4J94J 
ej989l)8.t 



9^2234 



eeo^ 1 

fl7 99 u 

wnsii- 



VU30 

11)2511 114491 
99'2Mfa 1 994622 



Biuei 

994<r2fi 
994(1* 
9940)48 

gwi^o 

Q9«ai 
994183 
994.14 



09jjU7 997008 



99j6B 1 
99v5H9' 
19,'Hil 



(97063 
99 7fi 
•)■) )r 

wriiu 



998"74 IWgi 
99a»l 8991 
99a'iU8 



9« 

9 9996^98 

99904 29 

J99672a 

, 1W«) 49998937 

I 990ae~ 999971 S« 

.990694 99eg71ifi 

1 999701 99ft97S3« 

> 999709 99997828 

■ 9TO716 9<t99S03» 

M 999722 9999B121 

l9ff)7'«) 99998330 



rf9jS8* 407314 I 998441 '999'^ 
99% 99 Jd6 99845 999274 
J959-r D47a '99W 3 1 9992SS 



999 K 99990619 

999743 99998G18 

999749 99998817 

699T66 99998916 

999703 999S901S 



999-S 999996 



994.'-( 



1E3=I 



I 


n TABLE VI L ] 




Amount Dt L.l laid M compound intcrut roi u; number of fouL ' 




1 i!w.Ww , i,iK>iV^ hkMk^ 


lo^KKXJ 


i.mm. LoAuouo: iMm 




a' 


l.(l6<^2S 


1.06U900 1.0712^ 


1.061600 


1.002035 






3 


1.0768BI 


i.i»a7a7 


1.108718 


1.121861 


1.141166 








1.10U813 


1.136SI» 




1^88869 


1.193619 








1. 131406 


1.160371 


1.187686 


1.316663 


1.948183 


1.278282 ]gM 




e 


1.1S»6»3 


1.I94U52 


1.329265 


1.265319 


1.S03380 


1.4O7100 umOB 




J 


i.iaaaso 




1J172279 


1.315939 


1.360689 




8 


1.3]MI)3 




1.916800 


1.368669 


1.423101 


1.4n4S6 i.«9 




8 


1^48883 


1.304773 






1.486095 


1.M13W IMM 




10 


1.380K6 


1.313916 


i;il0699 


1.480344 


1.562960 


1.6388M i.mM 




11 


1.31 «)87 


1.384331 


1.499970 


1JB9164 


1.6338D3 


l.Tlim J^jH 






1.344889 


1.435781 


IJHIOBB 


1,601032 


1-696891 








1.378611 


1.468531 


1.58:»66 


1«H1074 


1.772196 


ijBsM ^^m 




14 


1-113974 


1,613590 


1.618695 


1.731876 


1,801916 


ijirnn kS9 




IS 


M4K!08 


1.557067 


1.876349 






3.07«M HBS 




16 


1.18t6<>a 


1.604708 


1.733086 


l[87298i 


2;02«ffU 


ajffisn itSM 




17 


IJUIGIS 


1.663848 


1.794876 


1,947000 


3.ll8Bn 


sxim 


^H 




18 


Lsseese 


1.702403 


l.a6748e 


3.035817 


2.208479 


a.tomil 




IS 


1.598690 


1.763508 


1.033501 


9,106849 


3.307880 


s-Bjewi 








i.saMils 


1.806111 


1.960780 


2191133 


3 411714 


26taiM 


g| 






1.679^83 


1.80029S 


3.059131 


JJ7876S 




i7Bg*aa 




a* 


1.731671 


1.916108 


a.i3iHia 


J 360019 


3633669 


sjciun 




33 


1.761G11 


14B3587 


3.i08114 


i464 16 


3 763160 


a071Bl<l 


1 




31 


I,e0871i6 


3.039791 


9.je3J3S 


3563304 


.i 876014 


asasiw 






1.863944 


2.003778 


3.368945 


9 66.1636 


aoosiBi 


3.3a68Si 




38 


1.900eB3 


2.156591 


3.116959 


2 7724701 8140679 


a555e» 






1.947800 


2.831289 


2.6^1567 


9883388 3.282010 


a7a3» 




38 


1.996496 


2^87938 


3.BJW17i 


.^998 03 


atwioo 


asm 




99 


3.010407 


2.366566 


3.7118-8 


3118861 


3.5840^8 


4iims> 




ao 


rrniMS 


2.427382 


S.806794 


3.^3398 


a *MiB 


i.mni 




a: 


>.lflO007 


2.500080 


2.905031 




a913857 


tmxa 




83 


2-3KITfi7 




3.008-08 


3 508069 


4 068961 


4.;«w 




33 


2.368851 


2.052336 


3.111913 


364S3S1 


4 37«M0 


6,ilWlB» 




St 


3.31fiaa3 


2,781908 




3 794318 


4.ieej(, , w 




ss 


3.37330S 


2.813862 




3.916089 








3.432635 


3.898978 


8.450386 


4.103933 








2.493349 


2.086327 


3.571039 


4.^68000 










a074783 


3.608011 








39 


3.618674 


aiS7027 


a826a-2 


4.816368 


BJibS 1 1 




40 


3.685064 


a363038 


S*»2C0 


4 801021 


6.810.*. 


J S iOWM 






3.752190 


3,859899 


4.0978J4 


499J0S1 


6.0-8101 




42 


3.890996 


a4606a6 


4.341.f58 


5109781 


6351615 








3.801620 


asAifii? 


1.380703 


£400196 


6 637138 


8j*«g3™l 




44 


3.963S08 


3.671152 


1.G13312 


5.618615 


e»3<iiJ3 




4S 


3.037903 


3,781698 


4.-na359 


5.811176 


7 248- ) ' «'""■]?; 




46 


B.11386I 


3.S9*W4 


1.868011 


8.071833 


7Ji4-l 1 1 




47 


3.191687 


4.0118!J6 


6.0J7284 


8.317816 






48 


3.371490 


4.133252 


5.213^ 


6 870528 






46 




4,S.^ZI9 


5.39e06u 


6833349 








3^437 109 


4.383906 


5.58192; 








SI 


3ja3036 




5,780a90 






63 


a.6U113 


ijxmfis 


fijiaarid 


7 (Ml 




63 


3.701390 


4.790112 


0.193108 


1J4 1 1 






a7nap2jj 

3.888773 


1.9.tll25 6.**83J 
5.08^149 1 B.fi*!lll 


KM II li 




^■M 


3^86993 


6.334613 1 H.R66J01 




mmLit.06Bm 


6391861. llAOBSSl I li lit 


^^^V 1 .1.1877 H3 


6.KI340V\T364aea U 


^^Bl *.393m 


r,.-ioim' I6\i6«t 1 A 


^^^L«.8997» 1 &.t>ai60a' IXiVKSt^ \ ^^^3 



*■ 




TABLE VIIL 76 






The prwenl nine of L.l due at 


the end of amj Number of Yesri. 




&Oirpircen- 




[ajprcWt. 


,.p««nu 






T^ 


-ftifiHlO 




.9fi6184 




IMm* 


.ysuasi 


.9i>39« 




3 


.951SU 


.9«.'i96 


J);i35!l 


,fi34B.«) 


.015730 


.907039 






3 




.!) 161*2 


.901943 


.883996 


.876397 


.8G!M38 


.839811 




4 


isofissi 


.88S*a7 


JI71443 


.854804 


.838581 




.793094 




S 


.HSSBM 




341973 


.831937 


.803451 


>a3636 


.747368 




a 


.882297 








.767996 


.746216 


.704961 




» 


.8tl2fl5 


.813093 


.785991 


.769918 


.734838 


.710681 


.666067 




8 


.820747 


.789409 


.769413 


.730690 


.703185 


.876839 


.637413 




^ 


.800739 


.786417 


.733731 


.703687 


.673904 


.644609 


.601898 




.T811S8 


.744094 


.708919 


.676664 


.64.3938 


.6i:i913 


.658395 




.T6ai4« 


.723431 


.684946 


.640681 


.616199 




.636788 




|2 


.743556 


.701380 


.661783 


.634597 


Ji89U64 


;5S6837 


.496969 




,72Si30 


.680951 


.839404 


.600574 


JS64273 


.630331 


.468838 




.707737 


.661118 




.577475 


.639973 




.443301 




i** 


.890166 


.641863 


J»6891 


J»6364 


J116730 


:*81017 


.417366 




i« 


.e7.%25 


.633187 


*7870« 


.M390e 


.4»*4e9 




.mm 




^^ 


.857196 


.606018 


.657304 


.613873 


.473176 


!436397 


.S713fl« 




i« 


.841166 


.68739S 


.638361 


.»361S 


.453800 


.416531 


.360344 




is 


.SiSBlS 


.570386 


.«2015S 


.474643 


.433303 


.395734 


.330613 




^ 


.619371 


■663676 


.SOSAGS 


.456387 


.41464.3 


.376889 


.311806 




ai 


.535386 


Ji376«e 


.485671 


.4BS834 


.306787 


.;I68943 


.394166 




29 


.880865 


.631893 


.409161 


.421966 


.370701 




.377605 




3a 


.666697 


J(06S93 


*S3a88 


.405736 


.363360 


!336S7l 


.361797 




94 


A53876 


.4919D4 


.437967 


.390131 


.347703 


.310008 


.346979 




35 




.477606 


.433147 


.376117 


.833731 


.395303 


.293999 




28 


.536335 


.463896 


.408838 


.360689 


.318403 


.381341 


.319810 




27 


.613*01) 


.450189 


.396013 


.340817 


.304691 


.267848 


.307368 




38 


.600B78 


.437077 


.S816M 


.833477 


.391671 




.196630 




39 


.188661 


.434348 


.869748 


.320661 


.»79016 


.343946 


.184657 




SO 


.476743 


.411987 


.366378 


.308319 


.367000 


.331377 


.174110 




31 




.399987 


544330 


.396460 


.366503 


.320360 


.164366 




sa 




.388337 


.333690 




.344600 


.309866 


.164967 




33 


.443703 


.377036 


.331343 


.374094 


.J83B71 


.199873 


-146186 




3* 


.4.51905 


.366045 


.310476 


.363663 


.333896 


.190356 


■137913 




as 


.431.571 


.366383 


.399977 


.353415 


.214364 


.181390 


.130106 




36 


.411094 


.346083 


.389833 


.343669 


JW5D38 


.173667 


133741 




97 


.401087 


-334683 


.3S00S3 


J«14397 


.196199 


.1644.36 


-1167S3 




38 


.391385 


.336236 


.270563 


.335385 


.187760 


.186606 


-109339 




89 


J8I741 




.361413 


.316631 


.179665 


.149148 


-103066 




*0 


.373431 


.306667 


.3626rJ 


.308289 


.171929 


.142046 


«97333 




41 


.363347 


.297838 


.344031 


J»0027a 


J64636 


.136383 


.091719 




43 


.354W5 


.3S8S69 


.335779 


.193676 


.167440 


.128840 


-086637 




48 


.345889 


.380543 


.237806 


.186168 


.160681 


.133704 


-061630 






.337404 


.379373 


.330103 


.178046 


.144173 


.116861 


■077009 






.339174 


.364439 


.313639 


.371198 


.137964 




■073060 




46 


.331148 


.356737 


.305468 


.164614 


.imm 


! 105997 


-068638 




47 


.313S13 


.349369 


.193630 


.168383 




.100949 


-0646r>8 




48 


.806871 


.241999 


.191806 


.163196 


!l30S9S 


.096143 


-060998 




49 


.398316 


.334960 


.186330 


.146341 


.116693 


.091564 


457546 






.300943 


.338107 


.179063 




.110710 


.087304 


■064388 




61 


.383848 


.331463 


.173998 


-13S301 


.106043 


.083051 


4161316 




82 


.378833 






.180097 


.101380 


.079006 


•04831S 




S3 


.870189 




.161496 


.136093 


.097014 


.0^6330 


.0(6581 




«4 


.368679 


.303670 


.168035 


.130383 


mssn 




«]>»««» 




» 


JW7J51 


.196767 


.15»»768 


.115ftM 


\ J8S«M6\ J»KSA\»«* 




S8 1 


.350879 


.191036 


.146660 


.111201 


\ joatft\%\ .'*««\'}.\'«« 




Br / 


.3*4700 




.140784 


.106930 


\ JO»\SEa\ s«v^iAT 




^ / 


338790 


180070 


.136076 


1 .102811 


\ iynwa\K*»ga\J 




iS 1 


.232986 / 


174836 


.131377 


\ .osasA 


i \ S3\VSS\ \ .«*»».S 




0J^ 


337384 /. 


109733 


.126934 


\ .09506 


O \ jtfl\'«S\-'= 


^ 




I 












1 

















^^^=^= 


— - ■ 


76 




TABLE IX. 1 


The aamunl of L.l per ttnnuai in any number of Vea™. 1 






l.UWHMJO l.umiiMJi l.UIMHiu, l.UKJiiUO i.uwxmu 


2 OSOOoS 2 0^ 




a.035000 




9.03500UI S.04UU00. 2.046U0I 




3.975025 




3.106225 


a. 1216001 3.137035 






4.1SMW 






4.31«64, 4.279181 






s.3Masg 


6JW9138 


5.862466 


e.4i8aK 


e.47071< 






ftSSTTW 


8.468110 


8.S5015S 


6.633976 


6.718892 






T.MT4ao 




7.770408' 


iism* 


8.01U15i 






a73mi6 




9.061687 


9-311336 








0.854310 


10.150106 


10.36S49U 


10-583795 








ii.uimsa 




11.731393 


12-006107 


19.a88a« 






13.483480 


12,8077IM 


13.1*1992 


13.486861 


13.8U178 






13.795553 




I4.6(J19B2 


15.025805 








le.inMia 


16.6177(« 


le.ll.'fOSO 


10.626838 


17.'l699i; 






16.618053 




17.670988 




18.9331^;^ ]■.;■:-.- .. .-;« 




lT.Oai027 


1&5989H 


10.ai5USl 


aci;023688 






19.380226 


30-160881 


2o.97in;)o 


21.821631 


2271!)-;^!: j- 




ao,8e4;3o 


21.761588 


33,705018 


2a 697512 






33.386349 


23.414436 


24.499691 


35.046413 


26^S66<Vl .-,.--■ ..VI 




33.946007 


26.116868 


36.357181 


37.671339 


29.0636«ia ■l•).i^i^J^^•nt M.in-jmk 


30 


2G.544658 


26,370374 


38,379683 


39-778079 










27.183274 


V8. 876486 


30.269471 


81.869209 


33-783137 






29 


2S.86286(( 


30.530780 


33.338902 


3*.a47970 


36-303378 






23 


30.584427 


ai.4528S4 


34,460111 


36-617889 


88.937030 






24 


33.349038 


31.426470 


36.666328 


39-083604 


41.689196 








34.157764 


36.459364 


38.949857 


41.646908 


44.565210 














*t 16 


4 .5 0015 






jn 


3 00 


40 09634 


1. 9060 


081il 


50 1331 


H.660 Si 




Jn 


3985980 


13 U933 41, -W "7 


90 688 


5. 9^3333 


68 4(2683 






58290 


46 H8C0 


!•■ 


13 


6 3 






1390 03 




«l 




46 >» 


50J02e7S 


* J 


■i 


48 a a 


S. .502759 


J 




60d54O34 


66 07 


j 




S3IIU885 




1 




M-oasw 


60 


1t> 




R 




69 33948 






6 3^m 


60 




64 S3979 




w 


6 4f 554 






08701 






2S39S0B 


B- 02.196 93 


i. 


66«080i 


86 483899 80 


i 


8 6SJ333 


8904 1 


,« 


81.610131 


9^ 71 .1 


48 


81.55403* 


90 i 




87.607885 


IK .[( 


U 




04 jS 


49 


8l!l310J 






97.184349 


.13 ".t 


Bl 


100.921458 


11-1807 J 13t, S 


53 


101.444491 


m 0931971112 


S3 


oao556oe 


id'mim 


1» 








i 



li6i 111.756996 131 
V6S /lI5.5i5O021 
the /119.439G91 
W /133.425667 
PS /127.611329 
» lai-flgsij" 
FO (135.991590 



163.0634J7\l9GB^^«»* ■ 






! 


5= 
















TABLE X. 




77 






The preacBt i-flliia af L.l per Bnnum for any Nuinbei of YeBrB. 


^ 


'«i«c""- 


aMrc*t.|at.-r™^ 


4p^c.nMl(p„.™tl,M,.,.™l h1p„r™i 


rr 


.(175010 


.970874 


.96618* 


.kifito, kmx-i, k,i.^ lU'Liyiil 


!i a 


1.91742* 


1.913470 


1.893004 


1.880095 


1.872UUh 




: 3 


3.856034 


2B386I1 


2.S01637 


3.775091 


2.748964 




\. * 






a073079 


3.629895 


3.587520 




3.466100 


a 


4,6*6828 


4.679707 






4,389977 


4.329477 


4.312364 


! e 


6.B08135 


B.4I7191 


5.»iAo--ii 




6-157872 


5.075693 


4,917334 




T 


G.31S391 


6,33038.3 


iMHAi 


u.iwms'i 


6.893701 


6.780373 


6.682381 




B 


T.17013T 


7.019692 


0,873058 


0.73^1745 


0,696386 


6.463213 


6.209794 




9 


7.970866 


7.78Gli« 


7.607»S7 


r.43S,i;i2 


7.288790 


7.107823 


6,801693 




ID 


B.7fi3064 




8.316006 


auoawi 


7.912718 


7.721736 


7.360087 




11 


9,614209 


9.:^ h 4 






8^28917 


1106414 


imsas 




19 


10.2677((fi 


9,y540-M 


9 66Ui4 


938.10 4 


91186^ 


8 803253 


8.383844 




13 


10.983186 


10,634B'J6 


1OJ0273S 


9 t&M 1 > 






U 


11.090913 


11.296071 


1J*306J0 


If 1 ^ 1' 




u 


13.381373 


11^07930 


11JU7411 






IS 


13.066003 




13 094117 






17 


13.713iae 


13.166118 


U 651321 






IB 


14.36S184 


13.753513 


13189083 




19 


14.878881 


14.323799 


13 70(H37 


IJIkI 1 


» 


15,589162 


14,877475 


14 12403 


la-wj 1 1 




91 


16.184649 




14 697974 


14.0"911,0 114 1 




SS 


16.T6S413 


16,936917 


]6 167 1J5 


L4 4.11ir 11 1 




33 


17,333110 


16,443006 


16 620410 






34 


17.884986 


18.936543 


I6 05S.WS 






25 


18,434378 


17.418148 


16 4S1 1 11 




96 


18.960611 


17.876843 


IbH^ 'l 




BT 


114e40U 


18.33 031 


17 -s I 1 




aa 


ia,964fi89 


18.704108 


171.1 U 1 




89 


90.453660 


19,188466 


18 0k 


1 1 




80 


30.930293 


19.60O44I 


18 jgjii 


17 ) i. 10 Sh 4 1 i 4 IIJ W83l| 




31 


81.396407 




18 73037H 


17 6884D4 16 544391 






83 


21.849178 


20..'!88766 


19O0I4S66 


17 87355^ 16 788891 


16 802677114 084043 




Ha 


38.291881 


20,763783 
3l,l3fc37 


19 3B02OS 


18 H7B4t. 1? ^Sfl' 


16 00364914 230230 




34 


aa.73378B 


19 7l)08« 


1S41119B 17 246758 


10192904114.368141 




SA 


23.1*5157 


21.487220 


-OOOOOUl 


H 604613 17 46101 


lflJ7419414 4J824b 




36 


33.S6636I 






18908 8- 1 666041 


1654686 14.620987 




ar 


23.967318 






19 14.n 117 %3l4n 


16 71128 14 736781 




38 


34.348603 






111.^ 8C4|]8 04.f990 


Jb^89.il4 846019 




ae 


84,730344 


22.908 15 




lJ6S44S6|lti22'J(>,% 


17 01 04lll4 949075 


M 


35.103775 


33.U4773 


21.,J6a073 


19.. 92,, 4 18,401584 


17.10908615,046297 


1*1 


35.466122 


23.413400 


31.599101 


19.993052 


18.566109 


17,294366 15, l;i80l6 


.43 


36,820607 


33.7<U36» 


31.a«6fl;i 


aul85627 


18.723550 


17.423208:i6.23464S 


13 


36.166446 


83S81903 


32.063689 


20,370795 




17.545912.15.306178 


M 


36.603849 


94.35^4 


33.382791 




19^018983 


17.66277316,383188 


46 


36.833034 


a4ju87ia 


33,496460 


.■0,720040 


19.16(a47 


17.77407015.455833 


,'« 


S71M1 


24.776449 


32 700918 


^884054 


19388371 


179aOI*7 6o4 




37 487483 


36034708 


2^899438 


1M2B36 


19414709 


981016 6 6>Wn 8 


iS 


J7nai54 


35,^66707 


23 091"^ 


^1 195131 


19,535807 




,49 


280 'm 


"6 50 667 






19 661398 


816872- f 2 


fiO 




7") A J4M613 


214SJ185 


19 762008 


1825692jl (,8(1 


'fil 


■« 6 


21 617485 


19867950 


I83J8! 158 T 6 


53 


65 


1 475N 




lH4 8f7 1 S.1J93 


es 




8726 o 


2i 0()6345 


8 41140, I6<>0C974 


6* 




99 m 




■,■66 


«^ W1 ■i&^y.W,! •*«»« 


■0S 1 


JS A- W.W.^M>^. ^ '^ 




SIS T. ;:i^^ 


Wji 





P 

78 TABLE XI. 






^ 


L«nSthB of Circulu Area to Rwliiu 1. J 




-rr 






JS 


1 


.(1174533 


16^ 


^^7936^7 






1 


m>ti 






S 


.08«»8« 


17 


.2W17060 


«j 


.8736046 


3 


5818 


3 




8 


.odiatm 


18 


.3141693 


60 


1.047197B 


3 








« 


.0698132 


19 


.3316126 


70 






11636 






S 


.0873065 




.3490668 




1.3963634 










6 
T 


.I0471BB 
.1W1730 


23 


JW6S101 

.383973* 


BO 
100 


1,5707963 
1.7463293 


6 

7 


17453 






8 


.1398263 




.4014267 


11» 


1.9198623 


a 


33271 






a 


.1070796 


at 


.4188790 


■120 


2.1Ht*30fil 


9 


36180 






10 


.1T4S3Z9 




.436.3333 




3,2689380 




sme 








.IBlSSfla 


36 


.4637856 




£.4434610 






SO 






.aoii*3»fi 




r*7iaa80 




S,(J179939 


30 


8736G 


30 






.asmfis 




.48809^2 


160 




40 


116365 


*o; 




14 


.a4*MGl 


39 


.6061ifi5 


170 


2'9fl70597 


60 


1454M 


60 




IS 


.■jauw* 


30 


.6-23S98fl 


IBO 


3,1415937 


60 










1 


TABLE All. • 1 








1^; 


rHittT 


-KStT 


— TnT |(Ut..| Amt-T 


oiSn 


T 


■nm- 


ooi?e4§li3 


W 


rssT 


olSee^ST 


*+ 


ITOIM 




{ 




006 


0021660618 


'4 


1,096 


0107338664 


^ 




0191 




1 








b| 






4 


l!o476 


0201 








01 


OM.'!ai373a 






013flS79347 




1.0B 


Oil 




11 




OlSfi 


00B395031B 




,0385 


0138B00003 




1,0626 






4 




OIB 


ooeieooiaa 




,035 








aaa 




l| 






00TS3i4J78 




.0376 


0169881064 




l!0575 


rm 








ooseoonia 






017033S393 


6 


1,06 




TABLE XIII. ^^H 




UKrulNumbem.wicbtheii'J'aguithini. ^^^H 




„ ,. . . «* 1 








AreBorcircle (do. do.) .7gM T« 




Conlont of sphere (do. do.) Jme T.TI 




Surface of a sphere (do. do.) 11416 .tf 




Numb^ofeecondsiiiSflD'' USeuOO CU 


Numbor □rarcn of 1° in tlie radius fiT.39fiT8 l.TI 


'BaBeofNnperinnLoBarithms 3.71823 .43 


/jJI!odaliiiolcomramho^^t\'Caai -VHUdM T.6S 


1 Inches inn meUe 'ft^W* TJ« 


//common tropical year in mton «.\m ia'!* m^ffii ^ 


1 Inchea in a pendulum, flhictv VOoTUte* «s<^™4'.-v g,jj(|a