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Piniiiiiiii
6000364648
I'iJlf e. Vl
A SYSTEM
OF
PRACTICAL MATHEMATICS,
PJLIftT I.
CONTAINING
ALGEBRA AND GEOMETRY.
BEING
N^ XVI.
OF
A NEW SERIES OF SCHOOLBOOKS,
BT THE
SCOTTISH SCHOOLBOOK ASSOCTATION.
9ttllfeely tot te 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
3f £ 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 aii.
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(6cf c)— «?(56— c) . . =102
3. (a— l)(6l)(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(Jc)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;^i4ry — 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; 45cj +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" + 8jy+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 4a6 ; 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. 2abc+^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/',
g4cr».
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 dirisor 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^iJ
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 lial/c, by
I Proof. 2ahox7l'=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*L3— 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* — 3af2 .
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. 5641 — rg.
. ^ . 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''+4an=+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"— aSSCj:— 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. 3j> ,— ^.
_ 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 rr7 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^. AniJ
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
ito (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 (at^)'—
*+ 133 ^ ^ 12
!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«'« 43ar' + 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 coeiBcieat of
tie second term h always equal to the power to which the
liinomial is raised, and that the successive coefficieats cai;
be ohtained by multiplying the coefficient 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 coeffieient of the third term in the expansion
pf {o4«)'' cftn be obtained br mnltiplying 5, the eoeffioi
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 jr^ ? 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 2ac426cfc^
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. 3a4fi<
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:J2.
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;+6f4; 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
^xfQ^—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;.
;vf^ 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^ — lOi3x^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 40j32=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^=E3j«'; 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.
(50it?)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, LlS
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. ayard: 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 halfguineas ^
crowns, using in all 202 pieces: how many pieces y
there of each sort? Ans. IBO halfguineas, 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
(5J47
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:=35w419, it
becomes 9a?=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
15y=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^^42j?=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+7ff8i=
(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) 525 14
By subst. in (1) 15
Transposition 16
(16)512 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: (3in2nf4)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?,
ar42y32!=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 onethird of tho other two, and
the third with onetburth 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^=21 ; 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 ayard as there were yards in its length.
fiow many yards were in each ? Ans. 12 yards and \6 ^^i&!&.
7* Tb»pUte of a Jookingglaaa 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 apiece 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, Lalfacrown 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 aynrd 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,
4y4l==*=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:42y=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. yAy7a> ^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 AJ2.
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 {GaJa^l ^^a), diff {GaJ'^'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 aci^bc^zr 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'15, 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*/? 13;+:^ =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 C4aft')*.
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
af 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 — 7jrl(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.+<i2.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)
.(ac)
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=:, 157 =:li; •*•=> 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»aicex^: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.Al\<,g.J!. (2.)
1 log. i;=log. j(_lBg. p. (3.)
log. J— lo g.
(«
Iog.ii=^5j3J. (5,,
121. The interest is generally conrerted into prindpa
yearly, but sometimes halfyearly, 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 halfyearly, 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. IA
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 halfyearly? 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^ halfyearly, 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
= 77 '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
rfiL 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 LICK^
payable halfyearly for 20 years, interest at S per cent pa
annum, also payable halfyearly ? Ans. L.125.5, 2a. 1(^4
4. What is the present value of a perpetuity of L.10
per annum, payable halfyearly, interest at 5 per cent, peg
annum, being also payable halfyearly? 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. 142 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 peipendiculnr 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
cilled 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 foursided 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 foursided 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 selfevident trutL
47 The side opposite to the right angle of a rightangM
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 rightangled, 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 poljgon 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.
SoBoLiun. 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¢
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 aje 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 iBCD 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
1E
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
i8:7the 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 = iEFD, 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, since 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
rZ; 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. 24), 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 = onehalf of DC, (Prop. 28, Cor.) .'. the wbt
figure ABCD is = a iz:d nliose 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 ABCD,
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 ABAF, FH is the
rectangle ABFE, 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 ABAD ; .. the rectangle ABAD is =ABAF4
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 ABBC, 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 ACCB. for
is contained by ACCl", and CF ia =CB, being sides of
square; .. the rectangle ABBC ia =CB*+ACCB.
Q. E.
Cor. Uence also the rectangle ABACz= 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 ABAC=AC'' + ACCR.l ,p „„ .
I also ABBC=BC^ + ACCB. f U"p.Jt..J
P .. AB«=;AC^ + BC«+2.\CCB. (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''+2ACCB.
To eacU of these equals add CB".
.. AB'' + CB'=AC« + 2CB' + 2ACCE. (Ax.a)
But 2ABBC=2CB« + 2ACCB. (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«+2ABBC. (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 2ABHC to both sides of the result in Cor.
gives AB«+BC' + 2ABBC=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 ADDB, 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»+2ADDB.
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 ADDB 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 + CBBD+BD=, (Prop. 34,)
but since AC=CB, substitute CBBD for .\CBft
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, ACCo— i)}.
Cor.I. {fl+ft)«+(i.— 6)»=2i^+2i' {AD^(o— 6).AC=(o+i)}.
Cot.*. («+fr)»{ci4)«=*n6.
Cor. S. «»+i»=2D6+(fl— *)*.
Cor. 6. («ift)'4o6+{a— iV.
Cor. 7. (atiXa— t)+i'=o* AC or CB=a, CD_b.
Cor. S. (2a+i)tla*=i(a+&)* AC=CB=a, BDb.
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 ADDB=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 rightangled 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)(BCUAC).
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'+2ABBD.
For AC*=AD'+CD^ since ADC is a t'L. (Prop. 38)
But AD'=AB'+2ABBD+BD', (Prop. 37).
.. AC'=AB'+2ABBDBI>'+CD^.
And BD'+CD'=CB', since CDB is a t'L (Prop.S
.. AC=AB'BC'+2ABBD.
Otherwise,
AP' AB'+D B'+SABBD; add CD' to both,
and AD'+CIJ=AB=BD'CD'+2ABBD.
.. AC'=AB'1BC''+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"— 2ABAD.
Since BD is the difference of AB and AD, (Prop. 38),
BD>=AB"+ AD'— 2ADAB; 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 4.2ADDE. (Prop. 40;)
CBa^AD'+DC^— 2ADDE, (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+ADDB^CB*.
For CBS— CDs=BES— EDa. (Prop. 42.)
= (BE + ED) (BE— ED) = BDDi
(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::pG^
(Prop. 46, and As. 6,)
Nest let the chord ED he ^GH. FC shall he ^:CI^
for since DF'^4FC»^HI;!tCL=, 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 : CAEF : 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 AEEB— CEED.
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 .. AEEB=CEED. (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, AEEB— 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*= ACAB. (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^
^BAC, 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=ABBe.
or. 1. Since AEB is a rightangled 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^=ABAC, and BE==ABBC ;
AE'+BE^=ABAC+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»=ABAC: 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 etiri
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^=ABAD; .. 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=VAB\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
aaaigbc 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 ~ito 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 rZ*
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
BOttD 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, BCDA, and
aE. 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 BCDA be applied to the pyramid
FGHE, 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. ABCDEP, and GHKLMH,
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 ABCDEFGH, 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 ABCEFG. and ACDEGH, 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.\>
I4I 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 EEGBACs
ABFDCG, (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 ABCDEFGH, and ABCDKLMN
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 ABCDOPQR, 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 ABEP, and
plane ADNO parallel to BCMP. Hence AEO
IR and BFPCGQ 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 ABCDEPQH, 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
>CEHG, ADCKNM, will be equal.
Pro POSITION C, — Theorem.
»araIIelopipeds ABCDEFGH, and ABKLOPQR.
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 ADLEHM,
1 BCKFGN, 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,
{KDEFNH, 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 BCDA 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 EFGHCL. 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 EFGHCL, and
EHKGLD, 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 EFGA, and B!1KE, 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
KLDjG, and .■'. .^^ either of the two prisma. Conse
quently, the solid HCLKEFG is ::^^ the two pyramids
EFGA, BHKE, 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 twothirds 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 onethird of the cylinder, therefore the
hemisphere is twothirds of the cylinder ; .. the whole
sphere will be equal to twothirds 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.
LA^
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 onefourth 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 twothirds 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 twothirds 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.
PHOBIEM 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 rightangled 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
FHEr ■•
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.)
Pbohiem 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
44c, 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 afl, afS, 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;'^+10jy + 7. bya;'' aEi/+4.
Ans. a:*+4A— tiOA''inx— 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;'(plK+(5p+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+aXaf «), 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<)(«+6c).
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— cJrf) (a — ifc+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 (xf fif 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,fif3?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,.(293I),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+ ^r2,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,plyjc5^, 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:^ — 2rJl.
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=bxid, 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
™"»+t7:='
^_ .(.+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 aday, started n days
Itefore another, who goes h miles aday. 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 aday, and
cuiuot finish 70 sheets, which I could wish to do ; but if
1 could wort ten hours aday, 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«jy"=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.422I483.
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
jears 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 LANDSURTETING, 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 SCHOOLBOOKS.
BTTHB
800TT1SH SCHOOLBOOK 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 Coordinatesy 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"
Rightangled Triangles, . . 22
Obliqueangled 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. 5185, . . 34
Mensuration of Heights and Distances, . 37
yy Rectangular Coordinates, , .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 1959041.
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 2659916, 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 lefthand 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 2899547, 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 lefthand 6gures, as
in the last article, to which prefii the proper index bj
(Art. 11); then from the righthand 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 righthand 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=375, therefore ad.i 37
I Gives log. of 34863 =4542364.
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 468375.
Log. of 4683 by the table is 1670524
Diff. 93x75=6973. ■■• add 70
Hence log. of 46 8375 =1670594
Ex. 3. Required the logarithm of •0076452.^.
Log. of 007645 by the table (Art. 1 1) is 3883377
Diff. 57x25=1425, ..add 14
Hence log. of '00764525 =3883391.
EXERCI8K3.
The log. of 4fi is
=1662758
The log. of 97 is
=1986772
The log. of 286 is
=2456366
. The log. of 901 is
=2954725
. The log. of 7569 is
=3879039
. The log. of 2344 is
=3369958
. The log. of 46874 is
=4670932
. The log. of 954625 is
=3979832
. The log. of 356773 is
=0552394
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 =1380211
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 dicitnsl of , ne
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=T4647HK, and the tabular dttTorence is 14!), twothirds 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 =1968483
Take log. of 4
Hence log. of =i% or 2325
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
lefthand 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, 3785fi24
The next less tab. log. is the log. of ol 04= 3785615
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. 5314782, ii
2UfJ)34.
2. The number corresponding to the log. 3290035, ii
■00195.
3. The number corresponding to the log. 2531907, »
34()'335.
4. The number corresponding to the log. 0357912, il
227988.
5. The number corresponding to the log. 6486S05, it
3063410.
6. The number corresponding to the log. 1282169, 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 845.
LoE. ot S1\=.0^of£ftlV
Log. 0^84.5 =\^ie6Vl^
Log. of product, a^^ AS>5='iAamV
Logarithms. 7
Ex. 2. Required tlie product of 56*125, and 743*75.
Log. of56125=l749156
Log. of 74375=2871427
Log. of product, 41 743 =4620583:
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=2551724
Log. of 6385 =1*805161
Log. of 07425 =2*870696
Log. of 842725 =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. 396678.
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. 396256.
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 lefthand 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 743625=287l353 2871353
Log. of 426 125 = 162953 6 ar. co. = 8 370464
Quotient, I74'>09=1'241817 i2418J7.
Ex. 2. Divide 398^ by 7439.
Log, of 398125=2600019 2600019
Log. of 7439 =3871515 ar. co. = 6128485
Log. of quotient, *06351 85 =2728504 2^728504.
Ex. 3. Divide I by 78.>4.
Lc^. of 1=0000000 0000000
Log. of ■78.'j4=l 895091 ar. co. =10104909
Quotient, 12 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 2925.
2. Divide 46328 by 474.
3. Divide 364 x 5475, by 7854.
4. Divide 463x2525, by 31416.
5. Divide 1 by 31416.
fi. Divide 31416 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. 131634,
Ans. 977386.
Ans. 253743,
Ans. 372127.
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 3125 log. 0494850 ar. co. 9505150
I : 730625 log. 1863695 log. IB65695
^:L.l25 log 096910 log. 009f)9IO
:L.29225 =l4fi57a5 1465755.
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!<'225 =
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=1301030
Log. 30=1477121
60=1778151
., 1002000000
20=1301030
„ 15=M760!>1
2=0 301030
3=0477'2I
10=1000000
8=0903090
368]i
41
12=1079181
711 2604
Subtract sum of logs, of the first terms =5681241
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 lagaiithms 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 aday, can reap a field,
nieasurinjt 150 yards by 240 yards, in five days, how manj
men, working twelve hours aday, 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 aday, how many hours aday 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 (7285)'? 
Log. of 7285 =1862430
Log. of 38fl625=55872y0
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 =189509 1
Log. of 886227=7947545
Ex. 4. What is the fifth root of 34625?
Log. of 34625 =1539390
I
J
Log. of 808869=1907878
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 1125 ? Ans. 324736.
3. What is the seventh power of 5125? Ans. 928662.
4. What is the square root of 2625 ? Ans. 512347.
5. What is the cube root of 17? Ans. 257128.
6. What is the tenth root of 538625 ? Ans. 2361 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. 254327
Y 19x52 )
11. Find the value of (l05)i'^ x705 x ? Ans. 361733.
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'
9418615
9434080
1 00 154(15
70= 16'
9973716
10445259
10471543
21 => 54'
9571695
9604223
1003252y
36° 15'
9771815
9865240
10093425
108» 18'
9977461
10480642
10503081
164= 39'
9631593
9675564
10043971
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' =9649602 DifF. 566
^^Correcti
n for 16" =
91
9056
^gangent
of24°3'16"=
9tJ4Hti»3
w
EXERCISES.
Angles.
Sine.
Tangent.
Secant.
36° 12' 51"
9771444
9864670
lU'093227
430 18' 43*
9836305
9974394
10138089
69° 56' 13"
9'972813
10437449
10464637
98= 17' 40"
9995434
10836287
10'840853
52> 26' 19"
9899109
10114057
10'2 14947
1340 45' 6'
9'851359
10'003765
10152405
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"
9922405
9967832
9917072
9927979
9447701
9801384
10398387
10166244
i 0202642
9465476
9'912545
Cosecant.
10261103
10430554
10249172
10'274663
10017775
10111161
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 gieA^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 9496724.
Firsl, By inspection of the columns of tangents, we find
that the nest less than it is 9943732, 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 9496919, 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.
9346373=] 2° 49' 59" 9764825=30= 1 1' 36"
9782599=37= 18' 49" 9978546=43=35' 7"
9952864=630 47' 8" 10 6 35 887 =760 58' 42"
Secants. Coaiaes.
10'468357=70° 6' 55" 9'367285=i76° 31' 43"
10'004763= 8= 28' 1 1" 9846349=45° 24' 3*/'
10I47364=44=34'52" 9 167249=81= 32' 54"
Cotangents. Angles. Cosecants. Angles,
9874639=53= 9' 24" 10438674=21= 21' 27"
10104738=38= 9' 25" 10017643=73= 46' 43"
9964286=47= 21' U" 10114763=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^
f57
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 rightangled 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.
14084
18779
28168
Side adjacent
26488
35318
52977
346 28°
162.4358
3054957
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 righthand 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 49742 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\J3fl5, 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 821295760.
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.'i4445.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.36666i32a 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.75308a2S.
11. Table XL contains the length of circular arcs to ra
diuvl, 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''=:l2217305
4'= 0698132
20'= 0058178
6'= 0017453
40"= 0001939
3"= OOOOUS
Xength of arc 74'" 26' 43"=12993I52 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
105, 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 23fi021I, 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 triangle, 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 thirtysix 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 TrA!»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. 1320).
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, auiplement.
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 conespondiag 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 3567)16= 2^532408
Ar. CO, Log. R =10000000
Log. coaec. A =]ll 130803
Log. CB2fi4 = 2'4^i604
Log. AB 3567^6= 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, 3726J3
2. Given the base 560 feet, and the angle at the biw
required the perpendicular and hypotpnuse.
Ans. Perp. 6487^6' 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 54221. Feip. 341889.
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' 114". 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. az^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=73*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° =10062469
Log. sin. 67= 43' 26"= 9966317
Log. 450 — 2633213
Log. AC=480838 =abB1993
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, 132166 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". OdMr Z.89° 37' 43", or 15° 54' 17". Third side,
99988, or 27401 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 J9l58^.
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 2103, the side
BC i6Q% and the ZC
1I0°1'20"; requiredthe
ii A and B and the side jgg 2
AB.
4(80— /.CllO" 1' 20")=34° 59' 20"= half the Bom of the
2* A and B.
Side AC —2103
Side CB = 1602
Sum of Bides=3705 Ar. co. Log. =7431212
Dif of 8ide8= 501 Log. =1699838
^ sum of the is 34° 59' 20" Log. tan.=984504S
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"= 10306561
Log. sin. C=110*» V 20" = 9972925
Log. BC= 1602 = 2204663
Log. AB=3049 = 2484149
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 326753.
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 992795.
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^ 67 523". and AB 536937.
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' 254", ZB 47° 40' 346", and BC 998161.
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, Aiai>coa. C,
c'^bc COB. A iize 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+bie), 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
_ <g4e~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 TaniAs ^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=172G76O6
160 Log. = 2204120
ISOlxig. = 2255273 k
200 L"g, = 2301030 "^
2 2402802[)
Constant. =120140145
■
^m
B SKIOOirOlfBTBT.
Tan. iA=32° 50' 3l7''=9S0g8945{con'. — Log.(5— a)}
Tan. 1B= 29" 50' 46" =975874 J 5 [con'. — Log.OS'— b)]
Tan. ^C=27" 18'423"=9712yS45[con'. — Log.C,S'— 01
Therefore L\ 05° 41' 3i", IB 59" 41' 32", and LC 54"
37' 246", 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 1404,
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 BCiAC, and
BEiAD, and through C draw
CFuBE, 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 JAE, 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(54
(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(al'i) for i{Sird),
*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).
^BI!^A='■^^+^>
B. B+ COS . A _ cot. i(A+B) _
*.B— COS.A ~ taiilCA— B)
oot.i(AB)
(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+ tanft
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^'' tanii
\=^:>^=^.=^'^'^^ ^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^ 1V=
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. Bf 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 2Al/' '^' ^"^'^°' A x j^/cos.Asin.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. 200148 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. 114131 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, 22167 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 nAmt 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 lighthouse, 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 lighthouse.
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 821174, 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
nae 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 Coordinates, 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 eapressimi 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=3585 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
Lighthouse (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
34007
T
SO" 22'
105" 35'
135" 57'
73" 13'
39745
39072
G
25° 40'
127" 54'i
153" 34'4
102" 14'4
42579
62311
A
13'' 57'
1.85" 44'i
149" 41'i
121" 47'4
1892
47791
M
11«'23'
137° 25'
148" 48'
126" 2'
14631
44295
L
28° 26'
131" 38'
160" 4'
103" 12'
—59237
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 = 3753966
Log. cosec. 126" 56'=10097271
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'=1009727l
Log. sin. 73" 20' = 9981361
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 390104 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 612523 yards, SA" 137118 yards,
and SC 65667 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 2478Sa
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 814274 yards, SB 843094 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, 965926 ; 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
;il866 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 rightangled 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. iBai'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. ealS'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. 4878. 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 southeastward 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. 1497 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
215
311
278
103
1159
416
466
1079
95
57
215
Dif. lat. »
2778
215
416
2552
416
2563
Dept. s=
2136
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. 6532 S., Dep. 1414
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. 25887 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. = 9888185
As dist. 590 Log. di8t. =_2'770853
Is to dif. long. 10" 22' Log. 62I'6= 2793501
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, 3079 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.
792 =
1
Foot
1
12
15151 =
Yard.
Pole or
Perch.
36
45454
3 =
1
198
25
165
• 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
22956=:
1
It*
Square
Perch.
1S!96
20.6611
9a
1
Square
Chain.
99204
625
272*25
3025 =
1
4356
Square
Rood.
^264
10000
484
16 =
1
Square
Acre.
ISfiSlSt)
25000
10890
1210
40
25 =
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 2JO 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, 3912 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, iiaial p c
in whidi iii'e givua tb« twu ai Ji:9 I 7
AB, and the cuotaiueJ angle / /
iw DE ^ to AB, lliLQ 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>: 2106427363546.
[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, 3264 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. 282842 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^)CV6)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 grassplot 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, 285561 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
Rain.C: ; { BC =^?J?iA, Y BD=1" °'  *  .^"'^ , and the area =
\ lin. (A+C)/ RunIA+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=1662758x2
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°
—3325516
■301030
3024486
=10112534
= 9978939
^ »92457 2
Log.l09796aquare 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, 29208 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. 994413 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 paralleSHides 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 256416po
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{iHc)(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 ^ =1273097736=
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, 3984 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, 2496 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 207^5 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.
86g X 10 X 3=2598 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**
9761439
4
Square.
1000000
45°
10000000
5
Pentagon.
1720477
36*'
10138739
6
Hexagon.
2598076
30°
10238561
7
Heptagon.
3633913
25°«
10317336
8
Octagon.
4828427
22*4
10382776
9
Nonagon.
6181824
20°
10438934
10
Decagon.
7694209
18°
10488224
11
Undecagon.
9365640
16°A
10532205
12
Dodecagon.
11196152
15°
10571948
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=l720477x(8^=64)=ilVQ*UQ^1'^
:bxeiicisbs.
/. Required the area of a regular pentasoTi, o^ ^V\«^kv
kerefibe 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=376992Aii«
EXAMPI.B 2. Find the diameter of a circle, the circumft^
encebeing20inches. BjEulell. 31831 >:20=63662Aiil
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«S563392
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. 509296 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'+(75_)'=V3i{025=]95;andhence
5x8—36 136—36 ■„ , ,. ^ ., ,
— ^ — 5 ^40, length of the arc nearly.
EXERCISES.
1 . The chord of the whole arc is 508, 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 <
SUI5a365359, 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 anloradlns ! =3490659
30' „ ='00ST2Cfi
, 10" „ = ^485
[ arc 20° 30' 10" =3578410x 1 6=:572345fi.
EXERCISES,
J. "What is the length of an arc of a circle contsin
iag 49° 30', the radius being 30 inches ? Ana. 259itt].
2. What is the lengtli of an arc, of which the chord it
20, and the radius of tlie circle also 20 ? Ans. 209439.
3. Find the length of an arc, of which the chord is 36,
and the height 12? Ans. 458642.
4. What is the length of an arc, the radius of the circle
being 40, and its height R? Ans. 514a
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 3U16r,
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, thoe
foM (31416)*: I : i7Bo* : ■7BS*H(3U16)'=.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)=7854, 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. 505378 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 3U16H36e=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. 618497775 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. 275266.
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' 113".
Therefore ZACB=73° 44' 226" =737396". Hence the
area of tbe Bector=0087266x737396x 100=643496.
Again, the area of the AACB=6x8:^48. Whence the
area of the segments 643496—48= 163490. Again, by
Rule ir. 0087266 x 73 7396— i nat. sin. 73" 44' 226"=
■643496— ■4a0000=l 63496, which being multiplied by
t^=100, gtTes 163496 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\'2Bk,
■ _ 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. 59002.
4. Find the area of a seg'ment of which, the arc conl^ns
90°. the radius ofthc circle bd:ig IW. Ans. 2568546.
5. Find the area of a Begmenl of which the arc contains
270°, the radius being 15. Ans. 64264095.
PROBIEM 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 aolfevidei
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= ouefourtli 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 ncooidlDg 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"=7373972, 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 4087425 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(i7425.
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. 3213485.
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=20x4x7854
■.62832, 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. 1225221
3. What ia the area of the ring included between the
circumferences of two concentric circles, whose diameters
are 24 and 18? Ans. 1979208.
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
92728948000O=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 88101; 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 37358856.
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. 417127651, 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 iiguias 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 2JS 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 216506.
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 310485, 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 818664 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
2i4346, the convex surface, whilst ■7854x (lj)'x23
27925 feet, the area of the two ends; hence tttewbi'
surface is 27'2271 feet.
Again, the solidity is 7854 X V* X 5^=81448 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^ 5301^
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
47124 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 oitethir
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» = 31416, the area of the base, and the Bum
1130976 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. 128282.
2. Find the surface and solidity of a cone, whose slant
height ia 25, and the radius of the base IS.
Ana. Surface 188496, and the solidity 47124.
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 ^900896, and the solidity 753984.
4. Required the surface and solidity of a cone, its height
heing 36, and the circumference of its base 94248.
Ans. Surface 2544C57, 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 coa^ 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)
X3I416 = 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
=2293368, to which add 4492488, and we have the
whole surface, 67858o6 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 22554722'
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 1005312, 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 I7I45inches? 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=964a^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. 1975 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 314 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=376992, 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 = 1719000
Therefore the solidity is 5424600000 = 97U36S
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. 141372 cubic feet.
9. How many cubic inches are in a ball of four inches
diameter? Ans, 335104 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(i4x5236=l905904, 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, 1905904 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. 17904344,
3. What is the solidity of a spherical segment, the n
dius of whose base is 16 and its height 8? Ana. 348508IB
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 8092761 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. 32725 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. 146608 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. 670208 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 62S32
*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"=10626''.
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 1341864, this product being subtrscteo
from of the cube of AE=170'6666, gires 36'4802, whiA ,
being multiplied by 125664, gives ^842478, the soli i
dity.
Ag^n, to find the surface, we have 16 X 10—2(106^
X 0087266 X 10x6) = 160— 1 1 127462= 4872538 X
6'2832=3061513.
EXKBCIBE6.
1. What is the solidity and superficial content of a cif'
nilar spindle, whose length is 48, and middle diameter 3B
^iaches. . (Solidity 1731299 feet
■^"^ (Surface 327054 feet,
2. Find the solidity of a circular spindle, whose length
IB 15 inches, and middle diameter 8 inches ?
Aqb. 43340932 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 817636*
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 62832 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
0720? 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'Jjm*^
100+25=125, and DK«§2rMi= the radius of the circle
=125^10^125, hence, since nv=:9, the remaining part
of the radius or central distance ^=3*5, and therefore the
square of Ei=(125)''— (35)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,) =
6988875, and the area of the parallelogram D* is 20x4
=80. Hence the generating area =6988875+80 =
14988875. Now, by the rule 12 "4(10) '=144— 33^
=1106666, 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 582053, which being multiplied by
62832, gires the solidity of the zone 3657174+&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. 2728790178.
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 ftet.
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 08696, 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
24674, gi»es the solidity 4737408.
Again, for the surface we have 8+4=12x4=48x
986i«3=4737108, 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 84352 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 7994376 inche*.
■^''^■\Superiicies 3553056 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 12337 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 SlRI'ACHS OF HEUIrt..K BoOtES. 1
1^^^
K.,™.
Solidiiiei.
Snrf.™.
6
12
20
HeiBBdrulL
Octaedrob.
Dodecttedron.
■117861
lOOOOOO
■471+05
7668119
2181695
1732051
6000000
3464103
20645739
8660254
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 3464102, 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
58925625, 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. 25455816 indi
■*°^\ Surface. 62353836 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, 185811561 fs
■*"^ \ Solidity, 206904213 fs
4. What is the solid content and superficies of a regu]
jctaedron, whose linear edge is 10 inches ?
.^, (Solidity, 471405 inche)
^"^^ i Superficies, 3464102 inchi
5. What is the surface and solidity of a regular icosi
dron, whose linear edge is 8 inches?
. f Surface, 554256256 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 792 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 offsets, 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 stationstaff; 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 crossstaff; 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.. laiasa 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, stationstaves 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 fieldbook, 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 lefthand 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 fieldbook,
and also to calculate its contents, but it is sometimes convenient to
draw a rough sketch of the field in the fieldbook, 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 fieldbook ?
^^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 224:« / \\
^H tog. 1216o=3085112 / \\ ,
^H „ 4315=:2634981 ^ '^
^H „ 5225=2 718086 Offset A on BC.
H ,. 2625=2419]29 347x69= 23943.
^M 2 f0857308 Offsets on AC.
^■Xce. 268320 =5428654 ;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, 36S8S«fM
lAWB SUBVBTtXfl.
^
ESRRCISRS.
Draw a plan of a field, and find its area, froit
the
ing fieldbook.
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 northeast.
Ana. 3 acres, 3 ro., ]9'0648 perchea.
Construct the fieldbook, 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 northwest 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., 3247 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.,
107I2 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., 7424 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 fieldbook ; A being at the southeast comer, B
e southwest comer, C at the northwest comer, and
the northeast 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., 364096 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 fivesided tigure can be divided Into s trapezium and )
ttioDgle ; a Bixsided figure cau be divided into Iwo traj>eziuma ; >
seveuaidad 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
Bontheast to D.
Ans. Ifi tWFM, I 10. 25 '376 perchei.
Required the plan and area of a field from the follow
fieldbook and accompanying sketch ?
B380
F({68
Begin at
1)80
600
E, and
Diagonal
440 D.
along AC
120 D.
go northwest to A.
Ans. U Bvem 3 nods, 26'592 perches.
SCRVETTIIO.
3. R«quired the pliin nnd area of a sixsided field, from
the following fieldbook ?
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 fieldLook.
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, 272 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 crossstaff 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
fieldbook, 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 fiewrdeUi, 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 fieiLrdelis 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 abovementioned 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 1356 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 feurdelis, 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 stationpoint, 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 diffeient. 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 iandBurreyorB 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 landaurveyoc 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 basetine ?
r
B
Cos. \T 12*
Surface line 950 k
Base line 9075 log. :
EXERCISES.
1. What is the length of the baseline correspondii^ to
1560 links, measured on an inclination of 21' 14'?
Ans. 14541 Uak«.
S, What is the length of the haseline 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. 199026 linki
3. What is the length of the baseline 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 =4352954
Log. 14600 =4164353
85 17307
Subtract constant Log. 7319890
Correction =1575 =1'197417
.. 22540— 1575 =2252425 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, 1963 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
onefifth 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, onefifth 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
iallel 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 rightangled 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. 0d.
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 landsurveyor 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 rightangled 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 links 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., 2136 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. 129844 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, 192, 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, wiredrawn . 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, cijalallized .
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
7BlbB alcohol .
867
Copal, opaiiue .
. 1149
, 6BthB do.
, Chinese . .
1063
, 5BlhB do.
920
Fat, beef . .
823
, 1aiha do.
943
e2«
. 3Utl« do. .
9(i0
, hog's
. 937
, 2athH do.
973
,y^tl . .
934
— , lBlh do.
9BS
Gamboge, . .
. 1232
897
, Arabic
1207
. 1453
Ammontac uqum
Beer, pale .
!(I33
, brown .
ID3I
— , 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
^..ewe'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
Appletree,
793
— , do. of hempseed
926
Ash, the trunk, .
84.'.
— , d*. of poppies
994
Baytree,
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
MMtictree,
349
, American ,
6BI
Mahoganj,
1063
Citron, .
726
Map4 . .
761
Cherrytree,
715
Medlar, , ..
944
Cork. . .
240
Mulberry. Spanish
B9T
Cypress, Spanish .
Sit
Oak, eo years old .
im
Ebony, Americaa
1331
Olivetreo,
9W
. .Indian .
. 1209
Omngetree, .
7a5
Eldertree,
695
Pesrlree, ,
6S1
Elm, trunk of
. 671
PomeBTanatetree, .
1351
eno
Poplar, . .
311
Fa, male . .
630
, while Spaniah
S29
■ , female .
493
Plumu™, .
783
Hazel, . .
fiOO
Vino, '. . .
U27
lacmin, SpnniBh
770
Walnut,
671
Junipertree,
. 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, . , 01 43.75
Oxygengaa, . . 1435 627'813
Mtrousgoa, . . 1'4544 63fi'333
Ammoniac gas, . . '731 1 319'B33
SnlphureouB acid gas, , 27611 1207978
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
272Gx725=:197tS35 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 \ cjA,. \ 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
3cT8< 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 Pbobikm 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 ligTvum. ■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
(19250— 77Ba)itU,
lexTTBOxClHaSB— 146'J0)
(19258— 7788)xl4S2o'
= 14012, the gold,
= 3988 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. 628177 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. 3392 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 625865 of platian.
^^^ I 174135 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 inchth,
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 eirtam known nnmbeni,
called divisors, multipliers, and gaugepoints, 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.
Gaugepoints are the square roots of the divisors; they aN
chiefly used for calculations by the sliding rule.
4. Tabular divisors, multipliers, and gaugepoints 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 gaugepoints 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 gaugepoints.
When the middle area of aroj regular frustuia is taim, —
For the gaugepoints. — 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:IDIN0RaLE.
The slidingrules generally used are of two forms. One is
1'2 inch broail, and '6 loch thick; tlie other is 17 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
ualtfloon ; 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 187)192, is a brass piu, marked IM.U; this is
the circuhir gaugepoint for an imperial gallon. In tlie lower
part, at 49'0977, ia another pin, marked MS, wliich is the
gaugepoint 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 slidingmls.
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 slidingrule.
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 750 by 6S. Ans. *09*
4. Multiply sag by 2. Aus. 1646.
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!Mingriile.
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. 95.
Diviile 1470 by 42. Ans. 35.
4. Divide 1728 by 144. 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 muih 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 slidingrule.
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 42025? Ans. 206.
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. 2449.
3. What is the mean proportional between 72 and 288?
Ans. 144.
4. Required the mean proportional between 22 and 250?
Ans. 2345.
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. 421875.
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 alidingrnle.
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 gaugepoint 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 28242003, is the tabular circular divisor,
and 000354 the tabular circular multiplier for bushels.
Therefore 260x280=28242903=23'93 bushels. Am.
And 2e0x2«0x000'3fi4 =2393 do. km.
By Rale II. The tabular circular gaosepoiot for bBsheli •
is fi31441 ; then set 1 on C to 631441 on D, and agtunet 260 '
on D is 2393 on C, the answer in bushels.
By Rule III. The tabular circular divisor for boaheb \t ■,
28242903. 1
Then to 28242903 on A, set 2(50 on B, and opposite to 260 "
on A will be 2393 bushels on A, as required.
EkavpubZ, Required the area, in gallons, cd a sqnai^,
whose side is 1207 inches.
Bv Rule I, The tabular dirisor for ^usres i> 27727^
and the multiplier is 003606, for gallons.
Then 1307x1 207H277 2 74^ 625 gallona. Ani.
And 1207xl2O7X*0036O6=62fi gallons, Ans.
By Rule II. The tabular gaugepoint for squares b 1R6S1S
for a gallon. Tberel'ore to 166515 on D sot 1 on C, and
againat 1207 on U wiU be 625 on C, the answer in gatlonk 
By Rule III. To 277274 on A set 1207 on B, and ow^
site to 1207 on A, will be on B 525 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. 68868 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 2487 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 slidingrule. When the vessel is square
or circular. To the proper gaugepoint 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 mashtun 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.^d5d036da.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 gaugepoint 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*7x274j277 274x=23437 gallons.
By the multiplier for squares for a gallon.
487x487x27'4x003606=234*37 gallons.
Rule II. By the gaugepoint 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 ■4f24n4.=l 44333 lb. hot.
By the divBOTB,  45,^1 5^51 iiai' 30=1 48905 lb. cold.
m.l,».„„I.Uli.,. S45xl5x514x04169=1442fBlbhol.
Bj themultipliMS, ^ 4.,xl5xar4x04292<148910 lb. cold.
Rdlb II. Find the mean proportional (Art, 12), which a
26 nearly.
Then to 49031 on D, set fil4 on C, and against 20 on D tj
14431b. hot on C. Ans.
To 4827 on B, set 514 on C, and a«ainat 26 on D is 1489
lb. cold on C. Ans.
Example 4. Grain to the depth of 148 inchea ia pnt ml*
an elliptical couch frame, whose transverse diameter is IISZ,
and conjugate 111 inches; required the number of bnaheltit
uontains?
RiTLE.T. By the divisor for circular bushels.
1402x111 x14Qh28Z42903=8444 bushels. Ana. <
By the multiplier for circular buahels.
1452xl]lxU'8x'000354=8443 bushels. Ans.
RuLB II. The mean proportional between 1452 andltl,
»126'96, (Art. 12).
Tlien to G3'1441 on B, set 14'8 on C, and against 12695 oa
D. U 8443 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)xYH14387S=1160 lb. Ans.
KuLB II. The mean proportional between 27 and 20 is
■" (Art. 12.)
1 to ■," on C set 3703 on D, and opposite 20, 232, 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=9893 bushels.
K II. The mean proportional between 14 and 31 ii
(Art. 12.)
1
Then place y on C to 470977
■ — D ia, lO'O, 268, (!25, on
tliegjeater 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
77625^2714=2860 lb. Ans.
Rule II. The mean proportional between 75 and 45 is 58*^9
50 35 418.
Then to 27*14 on A set 75 on B, and at 50 on A is 1382 on B.
^714
2714
99
W
581
45
»
And286xV=2866'lb." Ans'!
>9
418
35
» 99
896
582
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 couchframe, soapframe, 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 dimensionbook for Maltster's utensils,
and a Soaper's soap frame.
Mr G. H. Maltster.
Ko. of
Malt
fitoiuo.
9
Ctotamt.
I
41 HI
37Oj
!
i4
i08*5
1910
6 I
a
U>8'2 52<>
1108 16'57
Couches.
I
a
27*4 leao
240 1452
5
1
1116
111
i
•<i726
KUns.
No. of
KUn
Length.
Breadth.
Area.
,l»t. \
\ 2nd. m.
OAtroiso.
Messrs J. T. & Co.'s Hard Soapframe.
SS'
:^
BrodUi.
„.^
c»,..™.
1 4fi0
150
21107
2847
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 diameteis, 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" (2975\ r 7'
13= 300 3100 .13
Now2ix llZ ',Z ..2Btbetopdiam.= f^lJU^Ul
31= 775 • 3576 31
U4=ll00 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 nlticli 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 incumbrances, 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.ipperinea.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 052 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
210001)0
14
IS
16
17
18
19
20
Sdarea
21
22
23
34
25
25
ar
28
33274488
35686758
38059028
40451298
42843668
45236838
47628108
22.53770
29
30
4thar«a
31
32
323
,', Sth area
32i
33
34
35
36
37
375
57M3H
70H6M
aissm
24SSI60
6519388
8374448
U42!)G08
13884768
1633!)928
18795088
212S0248
23705408
2392270
7233148
745OSIB
rsisMT
7SS7S«3
7661981
TtTiat
aoeOBW
8290308
84997M1
86044784
(9881878
52135648
543S94I8
56643188
58890958
61150728
63404498
65658268
28489948
30882218
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
Hiiteiilhs 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
cond 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
986
100*4
1027
I0AH>
107*3
99*2
1008
1031
105*(»
107*2
969
100*6
1029
1060
107*2
636
661
68*0
69.6
718
629
660
68*4
70*1
72.0
63*2
660
682
698
719
225392
23*9423
25*3061
2642a3
27*8038
• ••
• ••
•••
• ••
2
2
3
3
3
6 5392
79423
l*3(>ni
24283
38038
5*0096
7423«l
50610
02830
60380
*>.
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 (t36, at b Oa3, at c 088, at rf 03^
at e 033, at /"oaA at g 03'0, at A 032, 
at ■ 02S, and at A 030; to find the area
and meGui depth.
Now 175xl24217'XI, which, if gallons
be the denomination reqaired, being divid
ed bv 277274, gives 78'2619 galk., or 2
barrels and 6'2GI0 rbUs. = area. Then
036, +O;i3+O3'8+03'2+O3'3+03'0>O3'O
+032+02'9+03'0=312 and 312110, (the number of dipi ta
ken), =312«tnean depth. By trial a dip =31 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 raultipll
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 remaint
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,
892 and dOl ; at 83
inches, 76^4 and 75'8; and 39 inches, 708 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 <»'8Hg*'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*6236x003606=
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 Dimensionbook. 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
853 85*5
80S dOl
76«4 758
708 71 '4
e7'4l674
To cover th« Crown.
Deeth and Content
90
85
80<
76
71
67
Area* In
OaUoos.
22 9438
206583
18'1738
164(i40
143192
12*8676
343449
Areas in
B.
2
2
2
1
1
1
Gallons.
4*9438
26583
01738
74040
53192
38676
Content in
B. F. Gallons.
7
5
4
2
2
2
25
2
2
2
1
3
2
53256
8*5830
1*3904
84240
49152
52056
73449
61887
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. Onethird 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" '»■»«■»
msmtI
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
'»«'»[£«
247
SI '3
253
84*0
25*6
86*5
27*2
886
3(H}
3*7
317
827
eme ordinate*.
460
49*3
521
55*8 1 007
64*4
. Second
V Fourth
jcs, < Sixth
4 Eighth
^ Tenth
472
896
760
7«'5
490
486
7«»*4
767
71*1
495
■ m
71*0
771
71*5
50*9
514
716
775
722
51*6
52*5 ■
728
780
730
53*2
72*9
78*3
735
54*1
ordlnates.
3183
4
316*5
4
320*4
4
3243
4
328*9
4
302*4
4
ordin. x 4 =s
1249.2
1266*0
1281*6
12972
13756
1329*6
e Third
) Fifth
* 1 Seventh
C Ninth
(Ji4
748
751
621
0^*2
75*3
757
62*9
628
756
76
635
63«
762
!76*6
64*4
64 3
767
770
66*3
65*4
772
77*3
66*0
irdinates
27S4
2
2761
2
2779
2
880*8
8
2833
8
2859
2
wd. X2s=»
do. x4s
do.
546*8
1249*2
460
5522
12660
493
555*8
12816
52*1
5616
1297*2
55*8
566*6
13156
607
5718
1329*6
64*4
1842*0
1867*5
18895
1914*6
19489
19658
lie equidistance.
18420*
18676.
18895
19145
19499
19658
nneordnates
egments
460
34
49*3
48
58
i^*l^
68
6)7
8*1
644
92
rd.X8umofseg.
156*40
18420
236*64
18675.
30218
18895
37944
19146
49167
19429
59248
19658
1857640
18911*64
19197*18
19525*44
19920^
2025048
c inches
61921333
6303*8800
63990600
6508*4800
66402233
675al6'J(J
77^41
ans 3
nbrances
22a321
248U9
227362
230784
234730
239482
243447
cum. deducted
19*8512
227352
230784
23*4730
239482
[243447
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 globular 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 121, '
at bottom lij 238 and
24'2; depth of breast
br 13'0j diameters at
A; 238 and 242, at A/
6!)« and 604; depth '
of body 270, diame
ttrs at 4'6 from top, c,
5(i8 and 6r'2, at 135 from c, 500 and 604, and at 228 from
c, 45 '3 and 45'1; depth of rrown, de, 4, diameters at top, m".
'" " and 432, and at bottom, op, 397 and 404 incbea, to find
hole content.
^^the whi
=120, ■
^!i?:r*JL =niVQ,>C&stt
GAUGIKO. 151
12«+24^+.(12x24)VX*00283221*885e96 gaUs., the content
of the head.
2d, ^^ 120, and ^^^ 300, then
12a+308+ *^ xl3x2x002832=81019744 gaUs., the content
of the hreast,
3d, 55f±*?:?=,57.0; then 572x002832x9«82;81051 2= con
tent of Ist frustum.
^^°*^  «602; then 5022x002832x9ifc64230779=: 2d,
and^^^^*^«462; then 4522x002832x9«52W3003= 3d,
199*114294 galls,
the content of the hody.
4th, ^!l±i51=430, 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» 9183726 galls., the content of the crown;
therefore 19*518144— 9183726=10*334418 galls, to cover the
crown * conseouentlv
21885696 + 81019744+199114292+10334418=312354160
galls., whole content.
These dimensions, areas, and contents, should he formed into
a tahle, as in (Art. 26), and inserted in the dimensionhook.
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 couchframe, 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 815, and while on floor or kiln, as 50 bushels
net.
Cisterns and couchframes 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, couciiframe, 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 264, 2S'3,
a4'9, 280, 26'8 inches.
Now 264+25'3+249+'280+268=]314, which ^3, tha
number of dips, gives 36'2. This multiplied by 6' 57, produces
172'13 bushels. Ana.
By the slidingnile.
Place 667 on B to 1 on A, then beneath 202 on A is 17SI3
onB. Ana.
ExiMPLE 2. The aren of b couchframe is 8"11 bushels, and
the dips of the malt therein 220, 21'3, 219, 224, 230, 21^
22'0, 22'2, 23'1, and 22'5 inches; how many bushels does it
hold?
Now (22'O+818+2]9+224+23'0+217+22'O+2E2+23'l
+22'fi)H10, the number of dips is 222; fhia multiplied by
811, gives 180'04 bushels, Ans.
By the slidingrule.
To 811 on B set 1 on A, and against 222 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 (U3 inches; what number of
bushels is there (herein?
(460x372x042)h22181&2, the divisor for bushels, t»31Q9
bushels. Answer.
By the rule.
To 42 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<<PiivH&,'^'^^
eonteut of the floor inbusWVat
GAUGING. 153
Noiir (8©4x2r^08^2)52218«192t« 1219
And («6dxl08x05l)^2218192s 137*3
Tota], 259*2 bushelB. Ans.
By the hiding rule,
let, To 32 on D set 304 on C, and against 278 on A is 1 21*9 on C
2d, 61 „ „ 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 couchframe, 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 2393x7*5«179*4 bushels on the kiln. Then
179.4i2=89*7, the net bushels, and 179*4x6135«1100
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, headrod, bnngrod, 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 halfhogsheld
lying; and inversely, from right to left is a line for nllaeiDg a
halfhogshead 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 headrod 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 bangrod is hnlfaninch 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 slidingnil«.
Subtract tlie head from the bung diameter, and inuttipl;
the difference, when it is
Add the product to the head diameter, the sum will bellu
mean diameter. Then act thelen^h on B tn the gaugep<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 S30, 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 830 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 316, which is the mean dian»t«r.
Place the left end of the slide to 3fl6 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 slidingmle.
First, SgO—SaZ^JO, and48x68=3a; hence 283+ 8^3
31fi, 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
»JJx315xO02832x36G— 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 slidingrule.
Suppose the cask to be of the second variety; then
31.3—22*7 «6, and 8*6 x'64— 55. Again, 22*7x5*5 «»282,
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 headrod, slidr
ngmle, or ullagerule.
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 Blidiugrule,
}
Set mi) on C to 100 on SL, and against 248 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,
2ifi^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 pait af the line SS
the first folm 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 430, and wet
By the slidingrule.
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 176f43'a=401), and 500— ■409=O9t. Tht* ^
^^■'^lO^'OOQl, and409— '00t)='400, ttnd400»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 12892889, multiplier '0007756, gauge point
359067.
3. What is the circular gaugepoint 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. 1998.
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 13615 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 soapframe 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 253 and 25*5; and at 175 are 27*4 and 28*1; to de
mnine the pounds weight of metal it will holdl
Ans. 96744 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 241 inches, to find the ullage quantityi
Ans. By the rule, 84*5 ) ,,
By the pen, 836 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, 38, 36, 4'4, 44, 39, i% 3% i%
4'3, and 3'9 inches; how many gsllons does it contain?
Ans. 45263 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, 748 and 744, at 26, 69'6 and 690, at 33, 666 and 65S, it
39, 616 and 612, and at 45, 571 and 572 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 400, wet 190 invlu^
and the content 207 gallons; how much doea it conlaiii?
Ans. 87 i galls.
19. The depth of a flint glass pot is lfl5 and its mean dia
meter 2^3 inches; find tlie area and content,
Ans. Area 8i273G Ih,, contint 15035G1S1..
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, 9J,
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. B34S8 do.
3. 65143 do.
21. Required tlie content of a cask whose head diameter ii
174, bnng diameter 196, aad kngth 237 mches?
Ans. 24 veariy.
22. Ilowmany bushels of malt are there in a floor, of whieh
the length is 863, breadth 28S,aud depth 17 inches?
Ans. 1885 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, 266781 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 45 inchea down be 598 and GO2; at ISfiincbH
fl3'8 and 644; at 225 inches 640 and G4'G, and at 325 inclw
620 and 024 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 oneninth
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. 52394573 square yardi,
5. How many yards of paper Tvill be required to line a
ihest that is 5 feet 3 inches lonT, 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. 47l24yard8.
7. What are the three sides of a rightangled 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 halfacrown 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 lookingglass is 30 incbes by
I, and it is to he fraraed with a frame of such width, that
I aniface shall be threefourths 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 433013 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 forewheel is a (lOi) feet, and of the
hindwheel b (12) feet; what is the distance gone over,
when the fore wheel has made n (1000) revolutions more
than the hindwheel 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
halfcrowns 12,54 pieces; but if 42 was subtracted from
the sum of the halfcrowns and sixpences, there would le
main 1103 pieces; what did the man possess in all'
Ans. 995 halfcrowns, 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 maypole 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. 1183085559 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 Lighthouse 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 Lighthouse is north, that of
the Lizard W.S.W., and that of Start Point E. by N.
Ans. From Eddystone Lighthouse to Lisiard 5312
miles, to Plymouth 1419 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. 134354.
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 rightangled 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. 468922 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, JIUIS?, and 1»h
603699 i the angles are CI" 69' 7", and 10' (T 53", and
the area 294l.'>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. 12480U the upper part, 32430 the middle part, and
22756 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=173205, 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. 1843072 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 198437 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. 63406 feet long. 4'7622 feet wide, avt
39685 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 1125; the segments by the line bisecting t\it
vertical angle are 16^ and 12^, andjhe line from the TWi
tei to the middle of the base is ^2875=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(1116 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.
"
Coinev
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
iai34973
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
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47
48
49
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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
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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
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W
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1. 69:
MS
178
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■
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10.966208
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082866
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33 997439
47
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28 997426
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10.856338,90*6431
1913
10,963566
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955105
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100 ISOfi
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940481
063277
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0*6693: 002773341 887238
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0473611 055535
9*4*65 0027863* 807211
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1870
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10.916034 a06778I 1666
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943929
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21 897170 !2S
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1813
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107133
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037806
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073948
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03y3CO
066693
074108
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091791
109166
07468U
093081
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074979
093371
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137066 ]448Mi1
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oiin33
057*64
076:i6M
09366(1 1 100311
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076849
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193368
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A TARI.B OF
NATURAL BINES. 631
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0,156*34
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268810 276B37
292972B0
1 156;a:J
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2W15U5 277^10
201iHO;e4
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337;il8
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64
A TABLE Of NATURAL BINES,
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40075*^619 4383FPS
374ffl?r
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31.5900
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A TABLK OF NAT0nAL ilNEfl. 7l
' i^'
11987731
Ui987779
ssnwfl
99^^(110346^2 ft«.lte96S648e8&»
989a4^fiO
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994 J.
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999749 99998817
699T66 99998916
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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
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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
1696891
1.378611
1.468531
1.58:»66
1«H1074
1.772196
ijBsM ^^m
14
1113974
1,613590
1.618695
1.731876
1,801916
ijirnn kS9
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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
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1.663848
1.794876
1,947000
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18
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1.702403
l.a6748e
3.035817
2.208479
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1.598690
1.763508
1.033501
9,106849
3.307880
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1.806111
1.960780
2191133
3 411714
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1.679^83
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3.059131
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3 763160
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3563304
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1.863944
2.003778
3.368945
9 66.1636
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3.3a68Si
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1.900eB3
2.156591
3.116959
2 7724701 8140679
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1.996496
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3.31fiaa3
2,781908
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3.37330S
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3.916089
3.432635
3.898978
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2.086327
3.571039
4.^68000
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3.608011
39
3.618674
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3.890996
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4.341.f58
5109781
6351615
3.801620
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1.380703
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6 637138
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3.671152
1.G13312
5.618615
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3.037903
3,781698
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7 248 ) ' «'""■]?;
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3.S9*W4
1.868011
8.071833
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47
3.191687
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6.0J7284
8.317816
48
3.371490
4.133252
5.213^
6 870528
46
4,S.^ZI9
5.39e06u
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3^437 109
4.383906
5.58192;
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3ja3036
5,780a90
63
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3.701390
4.790112
0.193108
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5.08^149 1 B.fi*!lll
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3^86993
6.334613 1 H.R66J01
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r,.ioim' I6\i6«t 1 A
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TABLE VIIL 76
The prwenl nine of L.l due at
the end of amj Number of Yesri.
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4
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341973
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8
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.703185
.876839
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^
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.T811S8
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.6i:i913
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.723431
.684946
.640681
.616199
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2
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.661783
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Ji89U64
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JS64273
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J»6891
J»6364
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.e7.%25
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.453800
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is
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.474643
.433303
.395734
.330613
^
.619371
■663676
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.456387
.41464.3
.376889
.311806
ai
.535386
Ji376«e
.485671
.4BS834
.306787
.;I68943
.394166
29
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.409161
.421966
.370701
.377605
3a
.666697
J(06S93
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.405736
.363360
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.361797
94
A53876
.4919D4
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.390131
.347703
.310008
.346979
35
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.376117
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.395303
.293999
28
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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
9311336
0.854310
10.150106
10.36S49U
10583795
ii.uimsa
11.731393
12006107
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
30160881
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.injmk
30
2G.544658
26,370374
38,379683
39778079
27.183274
V8. 876486
30.269471
81.869209
33783137
29
2S.86286((
30.530780
33.338902
3*.a47970
36303378
23
30.584427
ai.4528S4
34,460111
36617889
88.937030
24
33.349038
31.426470
36.666328
39083604
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
Moasw
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
111807 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
» laiflgsij"
FO (135.991590
163.0634J7\l9GB^^«»* ■
!
5=
TABLE X.
77
The preacBt iflliia 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, kmxi, 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.»iAoii
6157872
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,y540M
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
lawj 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
lJ6S44S6lti22'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
.623S98fl
IBO
3,1415937
60
1
TABLE All. • 1
1^;
rHittT
KStT
— TnT (Ut.. AmtT
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