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

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About Google Book Search Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers discover the world's books while helping authors and publishers reach new audiences. You can search through the full text of this book on the web at |http: //books .google .com/I Piniiiiiiii 6000364648 I'i-Jlf e. Vl A SYSTEM OF PRACTICAL MATHEMATICS, PJLIftT I. CONTAINING ALGEBRA AND GEOMETRY. BEING N^ XVI. OF A NEW SERIES OF SCHOOL-BOOKS, BT THE SCOTTISH SCHOOL-BOOK ASSOCTATION. 9ttllfe|ely tot t|e 9iMoti&tUm, ts WILLIAM WHYTE AND CO., BOOftSJEELLJiBB TO THB QUKXN DOWAOXR, 13, GEORGE STREET, EDINBURGH. HOUXSTOH AND STONEMAN, LONDON ; W. GRAPEL, AND G. H. AND J. SMITH, UTERPOOL; ABEL HETWOOD, MANCHESTER; J. ROBERTSON, DUBLIN. MDCCCXhY. ••-I \ -'\K\ \ PREFACE. The following is the First Pari of a Treatise on Practical Mathematics, and comprehends that por- tion which does not require the use of Tables. In adding another to the many existing Treatises on this subject, it may be proper to state the objects that have been kept in view in- its composition. These have heen^ Jirst^ To exclude all useless matter, and thereby to keep the work within a small com- pass; secondly, To make it as entirely demonstra- tive as possible, without reference to any other work on Mathematics. IFor this purpose, as well as for its own intrinsic usefulness, a Treatise on Geometry is introduced, in which, by adopting a different order of the propositions from that used in Euclid's Ele- ments, and using symbols for certain expressions of frequent occurrence, an unprecedentedly large quan- tity of geometrical truths is presented, without in any instance detracting from the fulness of the de- monstrations, which are always given at length. The article on Algebra, it is hoped, will be found to be sufficiently extensive for most practical purposes; and the pupil that has thoroughly studied it, will find himself well prepared for entering on the study of larger works. In many instances exercises have been introduced of such a nature, as not only to iUnstrate the rules, but to assist in reducing cq^iAaaxl DEFINITIONS. Art. 1. Algbbba is a biancL of mathcmatica in nhich ealcnlations are performed by means of letters wkich de- note Dumbera or quantities, and signs which, indicate ope- lationB to be performed on them. 2, The first Jetters of the alphabet, as a, b, e, &c., are lUed to denote known quantities, and the latter letters, as t,f, t, &c., to denote unknown ones. _ 3. The sign + (named phis), indicates that the quanti- lia between which it stands are to be added together ; thus 3-f £ denotes the sum of the quantities a and b. 4. The aign — (named minun), indicates that the num- ber or quantity placed after it is to be subtracted from that pited before it; thus a — h denotes the remainder left by fsloag the quantity b from a. 5. Theaign x (^aB,Taed multiplied 177(0), indicates that the ^UntitieB between which it stands are to be multiplied the one by the other; thus aXcdenotesthata is to be taken as rflen as there are units in c, or that c is to be taken as often u there are imita in a^ This symbol is however seldom )<%d, as a.e, or simply ac written as the letters of a word, iidicales the same thing. 6. The sign -^ (named divided by), indicates that the 'IMntJty before it is to be divided by that placed after it ; tliua u-^c denotes that a la to be divided by c. This sym- ^1 is also seldom used, as division is more commonly de- iDicd by placing the dividend above a line as the numera- toi of a fraction, and the divisor below it as its denomina- tor; thus ~ is the same as a-i-i. 7. The sign ^ (read equal, or i» equalto), indicates that i^ quantities before it are er[ual in value to those after it; thus 4x3+7=9x2+1, for each is equal to 19. 8. The quantities before and after the eigti = are to- Ktilet called an equation; that portion which stands before ~ ' 1 ^ being called the first side of the efYaatitm, a.iA * aafierit the second. 9. The symbol J denotes that tlie nmnber oyer which it is placed is to have its square root extracted; thus Vl6 "'•^''^'^^ '^^ ^1'''"^^ rootoflfi, which is 4, and the t/^ denotes the square root of a, that is a number that, be- ing multiplied into itself, would produce a. 10. In the same manner, the cube root of a number as a is denoted by ^a, the fourth root, by Jl/o, and so on. 11. A number placed before a letter or combination of letters is called a coejicieiii; thus 3tt denotes three times a, and 3 is called the coefficient of a. The first letters of the alphabet are frequently called the coefficients of the latter letters; thus, in the expression 3cx, 3c is called the coeffi' cient of jr. 12. When the same letter enters several times as a mul- tiplier into an expression, instead of repeating the letter it is only written once, and a figure written after it to indi- cate the number of times it enters as a multiplier; thus a", «', «■*, &c., denote respectively the second, third, and fourth powers of ti, and tte smaU figures, 2, 3, 4, &e., placed after the letters, are called the exponents or indices of the letters. 13. Fractional exponents are also osed to indicate roots; thus, instead of ^j; x^ , is written, for ^ x^ 3.3, for ^ ^^ sc*, and so on to any extent; fractional exponents, where the numerator is not nnr', are also used; thus ,r', x^, &c,, the former of which denotes that x is to be raised to the second power, and then the third root of this power ex- tracted, and the latter denotes that j: is to be raised to the fifth power, and then the square root of this power extract- ed; or generally the numerator of the fractional exponent denotes a power to which the quantity is to be raised, and the denominator indicates the root of this power which is to be eitracted. 14. When ^ = J, it is frequently written thus, a:h:i e : d, and read, a is to t as c is to d, and the four qnantitiea are said to constitute a prop'irlion or analog!/; the terms a and d are called exlreme>i, and and c j/ieans. 15. The symbol -■. ia used instead of the words Cherefore ortOTtatvptrti^iy, which occur very ftequently in mathemalioal leasoning; and the symbol ■,■ instead of became. Ifi. Litte quantities are such as are expressed by means of tie game ietters, and llie same powers of these letters, ALGEBRA. 11 and unlike quantities are expressions which contain dif- ferent letters or different powers of the same letters; thus 3a^c^ and Ja^c^ are like quantities, whilst Sbx'yand Wxy* are unlike. 17* A simple quantity consists of one term, as Acx; a compound quantity consists of two or more terms connected hj the signs -f* o' — > ^^s 16a^(;-|-a5 and ISc'o?' — icd are compound quantities. 18. A vinculum^ bar , or parenthesis ( ), is used to collect seyeral quantities into one; thus a + ad or (a+x)d denotes that the sum of a and ^ is to he multiplied into d; also ^4ac — h^ or {4ac — P)^ indicates the square root of the remainder left by subtracting the square of b from four times the product of a multiplied into c. 19. The reciprocal of a quantity is the quotient arising from dividing unity by that quantity; thus - is the recipro- cal of o^ and can also be written a"^ ; in the same man- ner the reciprocals of a^, x^, 2", are -g- or a"^,~5 or x" , ~ or 2"^, where n may represent any number either whole or fractional, and is used as a general symbol for any expo- nent. 20. Find the numerical values of the following expres- sions, when a=8, 6=4, c=3, fl?=2, c^l5,/^0. 1. ac-\'(b — d)e — bed . . . = .30 2. a(6c-f c)— «?(56— c) . . =102 3. (a— l)(6--l)(c— l)(c;+c) . . = 210 4. anJce-\-b — CtJce-^-b . . =: 35 5. a\be-^e)'^f^2^d+b . . = 960 6. a^(2b+e+4)^+bi(e+c^d)^ . =14 In the following equations, the first and second sides will always give the same numerical value, if the same value be given to the letters on each side : verify this. 'J.f^^x^ + xy+y^. 8. (a-^aj)(a— a;)=a^ — x^. 9. (a + b—c)(a^b+c) =a2— 6'-*— c'* + 2bc. 10. x^^y*=(x^y){x^+x^y+xy^+y^). ALOKBBA. ADDITION 21. Is commonlj dirided into three cases; — \st. When the quantities arc like, and have like signs; 2rf, When the <]uaiititie9 are like, but have unlike signs; 3d, When the quantities are not nil like, and have unlike signs. " ' Rui,E. — Add the coefficients together, and to anex the literal part. tiQa^'bcxi,' I tbe ^^H 3a — oc 5V^^+? ^^M a — 7«c (ir"+/)^ ^^1 Sa — 9ac 19 (x'+y)^ ^^M 8a — 4ViH? ^^^ _5a — 3^ fl 3 (a'^+y')^ Sum, 26a — 29ac 35 (a:'+y')^ Ei. I. Find the sum of ax^+^aa:^+3ax^+4^(u:s^ |a«*. Ans. Q^ax^. 2, FindtheBttmof2a»w:'+4awu:'+7^anij;'+4Jaj7t:c'. Ans. ij^amx'. 3. Find th e sum crf^V«^+F+V^N^ + V^Hy* Ans. 44^x--\-i/*. Find the sum of 12(j:«_r)3^(3;»_:^3^7(,^4 j^i Ans. 4H(j'^— 3)i 22. Cabe II. When the quantities are like, hut hare unlike signs. RiTLE. — Add the coefGoients of the plus quantities inti> one sum, and those of the miiiua quantities into another, thrar difference is the coefBcient of the sum, and is plus if tarn of the plus coefficients be the greater, and minus ALGEBRA. 13 EXAMPLES. If^ 2d. Zd. 7acx 4a^T+P («+ft)(a?'+y*)^ 4acx —4aJT+p 10(a+6)(a?*+y»)^ -^ — 7aV r+^ 8(a+bXx^ + y«)i i. Find the sum oia,Jmx — ^afjmx-^ la^mx — xkijmx -^-iajmx —jjmx. Ans. 4^a^nix. 2. Find the sum of ^mpx^y^ — Ampx^y^ — Qmpx^i/^ + ^ah/px^y^ — Qmpx^y^ + "^mpx^y^ . Ans. — Smpx^y^ , 3. Find the sum of 3(a;«— y«)^+.7(a;«— 3^«)^— 4(a;«— Ans. (x^—y^)i 4. Find the sum of 5(6— c)a?^— 7(6— c)a?^+4(J--c)a;^ -4(6-.c)a;^+7(6--^>^--3(6— c)A Ans. 2(6— c>^* 23. Case III. When both the signs and the quantities are unlike, or some like and others imlike. Rule. Find the sum of each parcel of like quantities bj the last rule, and write the seyeral results after each other, with their proper signs. Note. The above rules will be obvious from the following consi- derations: — The first rule is simply this, that any number of quanti- ties, as 4, and 5, and 7, of the same kind, will make 4+5+7, or 16 quantities of the same kind. In the second, it must be remembered, that minus quantities are svbtractive, while plus ones are additive, and that to add and then subtract the same quantities, is the same as to perform no operation at all; therefore to add a greater quantity, and then subtract a less, is the same as to add their difference; and to subtract a greater, and then add a less, is the same as to subtract their difference, which is the rule. Again, it is evident that the third role just enables us to combine several accounts in the second into one sum. w GXAiiPLE. Here we begin ivlth tlie fin 3x1/ +4a: — Sac terni,whithcontainBay. Wi — 4<ij;^-i-4ry — 3az therefore collect all the (iByV: "J ex — Soa" — 2xy into one sum; then find th( — 5ni +4(a; ^-aa^ sum of the {azYs, and so oi 3ji; 4-5cj +3aj:' till wo hare collected all thi 8i:i/—4az+ ]3fj;— 5aj-' terms. 1. Find the snm of 4,t'3+8jy+4/, 3j-^+4j;y+y', 2.1:' — 4j-(/+2/, andj:'^— 7-. Ans. 10i" + 8j-y+6j^^ 2. Find the sum of 3a6+4u«— c,/, — 66v'+8crf+4a^ -|.6&e — 4ut+5e(i, and 46i.--|-2cii. Ans. a(iJ+4ae+12cii+4fia 3. Find the sum of ia'+lOub+ll'', 3a-+4dl+i', 2a'+Gul,+4b',ania'—6ab—7IA Ans. 10(i'+14a/i+56'. 4. Find the sum of a—3ah^+2/,, Z<t—iah^—b, ■7a+ 14ah^—3l<, and a~4ah^+'^. Ans. — 2<(+3A^+26. SUBTRACTION. 24. RtTLE. Change the aigns of the quantitiea to he sub- tracted, or suppose them changed, and then collect the quantities, as in addition. Note. This rule will appear evident from the foUi rationa : — If we are roiiiiired ta Bubtracl 3a — b Cram So, this meaitt that we are to take away (3a— A) from oa; if liien wa take AWayS^ the remninder will ba io— Su ; but it ia evident we have hero l**— avay too muub by & ; we must then add b, and we will have for true remninder So — 3a+b=2a+b, where it is evident that the K_ Iaf the Bnbtrahend have besn changed, and tlien the quantitieB ■»!• kfited by addition. The sune proccBS of rcaaooing will apply to [ From 7a 4- 5nc^— Sic— 4ic', Take Ha^Sac^ + iljc —QUK Rem. 5a+8ac'— 7te+aic*. 1, From 'i'' + 2iib+b'', take n* — 2<il,+b'. An«. S. From x'^+Sj^j/ -j- ary* + ifi, takea-'— Sa'^y + 3j?y'- Ans. feV+SJ'*. 3. From x^—a^ take a^—3ax+x\ *' Am. *:i-^»4.2ojr— a*— «*, '4. Prom 3bx'—4cx+5A take Sjt'— 46j»+3m. Ana. 7^*— 7< ALGEBRA. 15 MULTIPLICATION. 25. In algebraic multiplication three things are to be attended to; first, the sign; second, the coefficient; and third, the literal part of the product. Rule. When the signs of the factors are like, the sign of the product is plus, and when the signs are unlike, the sign of the product is minus. The product of the coeffi- cients of the factors is the coefficient of the product. And the letters of both factors, written after each other as the letters of a word, form the literal part of the product, the letters being commonly arranged in the order of the alpha* bet KoiB 1. That like signs give plus, and unlike give minus, can be shown in the following manner : — When +6 is to be multiplied into +a, the meaning is, t£at +6 is to be added to itself as often as there are units in a, and therefore the product is 4-a6 ; if now (6 — 6), which is evidently =0, be multiplied by a, the product must be s=0; bat +&x+a has been shown to be +a6, and that the whole product may be =o, the other part must be — ab; therefore — 6x+a= — ab. Again, since the product of two factors is the same, whichever be considered as the multiplier, +ax — b= — ab, and if (a — a), which equals 0, be multiplied into b, the product must be ; but it has been shown, that +ax — 6= — ab, therefore that the product may be 0, -—ax -—6 must be equal to -{-ab. Hence like signs give plus, and unlike tigns give minus. Note 2. When the same letter appears in the multiplicand and multiplier, it will appear in the product with a power equal to the sum of its powers in each factor ; for a^xa^ denotes (aaa)x(aa)=:. (uma=a^=a^^'^'^\ that is, its power in the product is the sum of its powers in the multiplier and multiplicand. Multiplication naturally divides itself into three cases. 26. Case I. When the multiplicand and multiplier are both simple quantities. Rule. Multiply the coefficients together for the coeffi- cient of the product, and the letters together for the literal part, and prefix the proper sign. Example. 5(ic x 3a6^= 15a'6^c. 1. Multiply da^c, by 7ah''(?. Ans. 2laW. 2. Multiply — 5(icd, by 3bcd, Ans. — I5al)c^d'^. 3. Multiply — 4aV, by —7bcd. Ans. 2Sa^(^d. 4. Multiply 7 OCX, by — Socy. Ans. — 2la^c^xj/. 27. Case II. "When the multiplicand is a compound and the multiplier a simple quantity. Rule. Multiply each term of the multiplicand by the multiplier, and write the several products after e?Le\i o\5cL'ei, mib their proper signs. 16 AU»BItA. ExAMPLB. Multiply 3ac+2bd+5c'', by 4,ib. Multiplicand, 3ae+2l>d+Sa''. Multiplier, 4alt. Product, 12d'bc+8ub''d+m^. 1. Multiply 7a''4.4o6*^-6^ by Sac. Ana. 2]d^c+12a^b\-i.3ab*e. 2. Multiply 3a=+4a=6—7<»c, by 5a'6, Ans. \5a^b+20a*b^~-3Wbc 3. Multiply iafl+§a'6—|ac«, by 4ai.". Ans. 2a-b-c+^a^b^—3d'b^e'. 4. Multiply fa'+lafi+SJ", by 3«6. Ans. 2(i^i+4a'6*+|ai'. 5. Multiply 7a'c' — 56c^ — 4ni, by faic. Ans. 5ia36c3— 3ja6V— 3a'6V 28. Case III. When both multiplicand and multipLiei arc compound quantities. RuLK. Multiply al! the terms of the multiplicand by each term of the multiplier, and collect tbe several pro- ducts into one sum by addition. Example. Multiply 3(t'— fti6+36', by a~b. Multiplicand, 3a' — 6iib-\-3b''. Multiplier, a — 6, Product by a, 3a'—6d'b+3ab\ Product by b. —3a %+6 <ib^— W. Total product, 3a»— 9a^6+9a6'— 3&». 1. Multiply o+ 6, by a+6. Ana. a^ + Sab + S*. 2. Multiply a—b, by o— 6. Ana. a''—2ab+b* 3. Multiply a+i, by 0—6. Ans. a-— 6' 4. Multiply a*+a6+6*. fcy «— *■ Ana. a'— fc>. 5. Multiply a' — ax+x^, hy a+r. Ans. e^^t^. 6. Multiply 7ab + 3ae+4d, by 3a— 2c. I Ans. 2]o''6+9o=c+]2rtrf— I4a6e— 6ac^— 8«/. L 7. Multiply 5fl+3r+3/,ty3a + 2r. ^ Ans. i5a''+l9ar+3ai/+6x*+23y. 8. Multiply «+iy— 2, by |:ot.3-/. « ., ,., Ans. :ij.=+3ir^-4^+liv'- _,. 9. Multiply js+3^j/+xi/^+,iA by K— J/. Ans. ar* — y*. 10. Multiply^*— a<i+a;"— / + !, bya" + s-— 1. Ans. i«— 3:4+xS— j:'+2jr — 1. . Multiply ax-\-bx^ + ax», hy l + j-+j;* + j/', g4-cr». PB. Tu KonBM I. The square of the mun of tn'o quanti- ^ eqitat to tbe sum uf the B^uuies o{ «ttcli of the qoan- ALOEBBA. 17 tities, together with twice their product. (See Art 28^ Example 1.) 30. Thbobbm II. The square of the difference of two quantities is less than the sum of their squares hj twice ^eir product. (See Art. 28, Example 2.) 31. Theobem III. The product of the sum and diffe- rence of two quantities is equal to the difference of their squares. (See Art 28, Example 3.) Note. The above theorems are veiy important, and should be committed to memory. 12. Write by Theorem I. the squares of the following quantities, and verify the results by multiplication, (o;-}-^) , 13. Write by Theorem II. the squares of the foUowinj^ quantities, and Terify the results by multiplication, (a — d?) , (2a— ^)^ (3a;— 2y)^ (a^— 60^ {aG—yy, (3a*— 4c)^ Ans. {a^—2ax^x% (Aa^^Aah+h'')X^x'—\2xy+Ay*), (a4_2a262+6*), (a«c*— 2a<jy+y«), (9a*— 24a*c+16c0. 14. Write by Theorem III. the product of the follow- ing quantities, and yerify the results by multiplication, (a+a;)(a— a;), (2a+x){^a—x\ (3a:+2y)(3a:— 2y), (a*— h^W + h% (4a2— 9/)X(4a^+9c2). Ans. {a^ — x% (4a2 — a;*), {9x'' — Ay% (a* — b^), (16a*— 81c*). 15. Find the continued product of (a — x) (a-^-x) (a^+o?*) (a*+a?*). Ans. a® — a^. 16. Find the continued product of (2a + a;) (2a — x) (4a* +a:*) (16a*+a?*). ' Ans. 256a8— a;8. 17. Find the continued product of (a?* +xy+y^) {x — y) {a^+y^). Ans. x^ — /. 18. Find the continued product of (a^ — x^y+xy'^ — y^) ix+y) (x^+t^). Ans. x^^ — x^y^+x^y^ — y^^. DIVISION. 32. Division being the converse of multiplication, natu- rally gives the following Bides, 1st, If the signs of the dividend and divisor are like, the sign of the quotient is -{- ; and if unlike, the sign of the quotient will be — . 2d, Divide the coefficient of the dividend by that of the divisor, for the coefficient of the quotient. 3d, 8i9ce anf quantity divided by itself giYea 1 fox cjvxo- 18 tient, and a quantity multiplied by 1 gives that same quan- tity for product, it follows, tliat the letters which are com- mou to the di-risor and dividend, with the same exponent in each, will not appear in the quotient. 4th, The letters which aie in the dividend, and not ii the diyiBor, will appear as multipliers in the quotient while letters which are in the divisor, and not in the divi- dend, will appear as denominatora of a fraction u quotient, 5th, "WTjcu the same letter is in both dividend and di^i-J sor, with different exponents, it will appear in the quotient with an exponent equal to the difference of its exponents, and in the denominator of a fraction, when the exponent of the divisor is the greater; thus, a-^a =:a >or—,a ~a =a 33. Cask I. When the dividend and divisor are both simple quantities. Uui.K. Place the dividend as the numerator of a frac- tion, and the divisor as its denominator, then divide by the I above general rules. I ExAJiFLE. Divide lia-l/c, by I Proof. 2a-hox7l'-=lia^i'c. I 1. Divide Sa'x^, bj — 2oa''. Ans, — 4ax. I 2. Divide — ZJp^nu-*, by Opvix. Ans, —3ja^. ^^^ 3. Divide —84j;yi^ by— 12^3/j. Ana. 7^^ ^^^^ 4. Divide 120aV/, by 30<i.e^^. Ans. ^^^f 5. Divide —iOx^^/h, bysJff'^z^. Ans. — 5«^ I 6. Divide 36^Vj. by 20j-«//Jii Ans. IJj^M^^* 34. Casb n. When the dividend is a compound, and the divisor a simple quantity. EiTLB. Divide each term of the dividend by the divisot, as in the last case, and the sum of the separate quotients will be the answer. Example. Divide 3a'b+4u%-:, by 4ab. Sa'b+ia'bc 4a& Jfiride J^a'^'f— 24a6*L-3— 10<i^6=c^,hy 3aie. ;^2a^<r— SiaiV— loA^ _ ^^^^ _^|,t*— '^3^ , ALGEBRA. 19 1. Divide 3a4— 6a3+4a«— 8a, by 2a«. 3 4 Ans. -a* — 3a-f2 . 2. Divide 12a3—36a26+36a62— 12^.3, by 4a6. Ans. -7 9a -4- 96 . 3. Divide ^Sa^h^X'-^6a^l»jc^+14al?x*, by 2ahx. Ans. 2ia3b—lSabx+7b^Ji^. 4. Divide 4()p2^y2—3ap*a:V—60i?2:i:»^, by —10j5^^ Ans. — ixif+3j3»3i^if+6a^i/^. 5. Divide 7a36^c^— 12a*63o-.14a62c*, by 4aMA Ans. -j^ aM— SftM— S^aM. 6. Divide 4aV— 8a2ca:+4a2^, by 4a2. Ans. (^—.2cjc+a^. 35. Oasb III. When both dividend and divisor are com- pound quantities. RuLB. Place the quantities as in division in arithmetic, arranging both dividend and divisor according to the powers of the same letter. Divide the first term of the di- vidend by the first term of the divisor, and put the result vfith its proper sign for the first term of the quotient. Multiply the terms of the divisor by this quotient, and sub- tract the product from the dividend; to the remainder bring down as many terms of the dividend as may be ne- cessary, then divide as before, and so on till the work is finished. bxamples. Divisor. Dividend. Quotient a+x)a^ + Sa^x+3ax^+x^(a^+2ax+x^ a^+a^x 2a^^x+3ax^ 2a^x+2ax^ ax^^x^ ax^+x^ Divisor. Dividend. Quotient a -'— 2ax + a;« )a4— 4a5ar+6a V— 4aa?5 + a;4 (Gt2__2aic + a;^ a^—2a^x+a^x^ — 2a^a;+ oa^x^ — Acix^ a'^x^ — 2ax'^ ^x^ Here it ia evident the -quotient ■would but go on to Infinity, the power of .r incn each step. This quotient givea several interesting Beries, by subsl tuling fractional values for ir ; for csample, if we make Buccessirely equal to ^, ^, ^, ^, |, we niU have the folloi ing results : — I ri=srri=2=l + i+i+J+TV- &c. to infinity. , - ._, =j^j=4=l + J+A+H+A'. + &o->oi«l t) /^(i— 2nd Igiyes |— ^+^+4TTV\\.-V&ft.toial ^^ ALGEBBA. 21 1. Ditide a* + Sa^x+iOa^x" + lOa'x^ +5ax* +x> by fl'+aar+o;''. Ans. a^ +3a''x+3ax' + x^. 2. Dmde a^— 5a* +\Oa'—lOa'+ 5a— I by a'— 3a« +3a— 1. Ana. «"— 2«+ 1. 3. Dmdear* — a* hy x'-\-x^a-^xa' + a'. Ana. x — a. 4. Divide j:^+j)* by x*— j:'y+arV"— ^^i'^+i'''- Ans. x+i/. G. Dmde 9j;6_46x=+95*« + 150j! by ^2—4^— 5. Ans. 9.e*— 10.r' +oj-»— 30x. 6. Diride 25a;9— «»— 2j;'— &f" by 5:c=— 4j:". Ans. Sx^+'ir'+Sx+Q. 7. Dindejc*— ^x'+iT'+la;— 2 by |j^— 2. Ana. Jj:' — Jar^ + l. 37< Veri^ by diTision the truth of the follofring ex- preuJDiu : — »*+fl =x*—x'a+x^a''—xa'' + a*——^. The eight expreaaions given above are particular illua- tatioiiB of the four follovruig TIteorema: — 38. Thborbm I. The difference of the same powers of 1*0 quantities is alwaja divisible by the difference of the quantitiea themselvea, whether the Mponents be even or odd. See Ist and 3d. 39. Theorem U. The difference of the same powera irf two quantities is divisible by the sura of the quantities "ithoat a remiuader, if the exponent be even, but not if Ihe exponent be odd. See 2d and 4th. 40. Tbrorku HE. The Bum ofthesamcpovfet^oH^o quantities is nerer dmsihh by the difference of tW <\aB.u< ^ titles without remainiler, whether the exponents be odd or even. See 5th and 7th. 41. Theorhm IV. The sura of the same powers of two. quantities is divisible without remainder, by the sum ' ' quantities, if the exponent be odd, but not if it be r ALGEBRAIC FRACTIONS. 42, The Rules for the management of Algebraic Frac- tions are the same with those in common aritlimetic. Case I. To reduce a mixed quantity to the fonn of » fraction. Rule. Multiply the integral part by the denominator of the fraction, and to the product annex the numeratoi with the same sign as the fraction; under this sum place the former denominator, and the result is the fraction re- quired. Example, ReduceSa + j; to the form of a fraction, 2nxi+j inb+x I ~ 6 ■ Here 2(/, the integral part, is multiplied hy i, and .v, the numerator, is added to the product, the sum of which forma the numerator; under which we write the former denomi- nator b, which gives -^ — -j the fraction required. 43. If the numerator be a compound quantity, and ibe fraction be preceded by the sign minm, the signs of ( " '" terms must be changed by the rule for subtraction, (Art. 24); thus, (i+.r — ~ — = I. Reduce3a-{ — to the form of a fraction. 3a»+ja " ^^ Ana, , Reduce « — ''+"j' to the form of a fraction. d — , Reduce I5a*+j: — to the form of a fraction. Ans. _-7 ALGEBRA. 23 5. Reduce x—a to the form of a fraction. Ans. . X 6. Reduce a-}- a:- to the form of a fraction. a — X 44. Case II. To reduce improper fractions to whole or mixed quantities. Rule. Divide the numerator hj the denominator, as far as an integral quotient can be obtained ; then write the remainder for the numerator, and the former denominator for tlie denominator of the fraction. Example. Reduce -. . 4a =3a+l_j-. Here 12a* divides by 4a, and gives an integral quotient, so also does 4a divided by 4a ; but 3c will not give an in- tegral quotient when divided by 4a, therefore it is written as a fraction at the end with its proper sign. 1. Reduce to a mixed quantity. Ans. 3j: . 2. Reduce to a mixed quantity. Ans. 4x+ — . ^ ^ . 10a«6+2a8— 36« . . , 3. Reduce ^ to a mixed quantity. gra Ans. 56-4-1 — r-g. . ^ . 12d*— 6a3+9a» ^ • . , r 4. Reduce ^ to an mtegral form. . ^ , *»+3A+3«»+a» ' . , Ans. 4a«-2a+3. 5. Reduce — ^— ^ — -rri *o an mtegral form. Ans. a; 4- a. 45. Lemma. To find the greatest common measure of two algebraic quantities ; that is, the greatest quantity that will divide them both without leaving a remainder. Rule. Arrange the quantities according to the powers of the same letter, as in division^ then divide that which contains the highest power of that letter by the other> and the last divisor by the remainder continually, till there be BO remainder, the last divisor used is the greatest common measure. Note 1. If all the terms of one of the quantities be divisible by any simple quantity which does not divide every term of the oMher, nnce it cannot form m part of the common measure, it may \>q «\x\lc^i^ r 94 out of all the tfrmH ly dividing by that aimpte qiULQtity before (ha ganeral division ia perfomted. Note 2, If tho coefficient of the leading term of the dividend bs not divisible by the coefficient of the leaJiag term of the divisor, every tBrm of the dividend may be multiplied liy Buuli a nnmbOT will make its coefficient diviaible by lliat of the iliviBor without i mainder; by this meoiiB fmctions are avoided. Note 3. Place the i^uantitics in two coIudhib, bo thai the a. which ia to be the divisor may aland in the left hand column, k the other in the right; then divide the right hand colnma by the le and place the quolieut ailer the right hand column ; then divide t left band column by the Tenmlnder, (reduced, if necessary, bf Note 1 ), and place Uie quotient before the left hand column ; pro- ceed in this manner till there be no remainder. Example. Required the greatest the qoantities x" +2ax-\-a^, and a;'— ;c+a r+a x'+2ax'+ t by dividing by — 2««, x+a 46. The process of finding the greatest commoD measaro can often be much more elegantly performed, by reducing the quantities into factors by the theorems, at the end m multiplication and division ; then the product of all the factors that are common to both quantities is the greatest common measure sought. The above example wrought ii this way tvonld be as under. Thence x+a being the only factor common to both, ii the greatest common measure. 1. Find the greatest commou measure of a' — x*, and' o>— 2a«Ji+oj;«— 2a:=. Ans. a»+«*. 2. Find the greatest common measure of x^+i/^, and 3!*+x^y — xy' — >/^. Abb. x+y. 3. Fiad the greatest common measure of 4Rr^ + IBj- — 15, and 24j;^— 22^^ + 173^5. Ans. 12i— S. 4. Find the greatest common measure of x" — a% and iI!*-ftr'o+6j;'a''+4a-n=+n*. Ans. j; + . 47. Case HI. To reduce fractions to their lowest terms. Edle. Divide bofh numeralor and denominator by their greatest common meosuTe, and the quotients will be the fraction in its lowest terms. Example. Reduce - -3i'a+3lo* -1 to its lowest terms. ALGEBBA. 25 Ans. Here it will be found, that x — a is the greatest common measure ; therefore we hare the following : — jr3— ai;*a+3«o"— aS-S-Cj:— a) ~ x'^2ax+t^' 1 . Reduce ^r to its lowest terms. Ans. -^-^ . oox—cx 3^ — f 2. Reduce -52^2^^:es to its lowest terms. Ans. ^i?. 3. Reduce — -z — . » to its lowest terms. "^-^^ Ans. Jtt?^. 4. Reduce -j — — -=-= — . to its lowest terms. * m"+ii" 5. Reduce — » - ^>, - — to its lowest terms. Ans -2*^^ 4x"+6ar+9a^ 48. Case IY. To reduce fractions with different deno- minators, to equiyalent ones having a conmion denomi- nator. Rule 1. Multiply both numerator and denominator of each fraction by the product of the denominators of all the others ; the resulting fractions will haye a conmion deno- minator. Rule 2. Multiply the terms of each fraction by the quotient arising from dividing the least common multiple of all the denominators by its denominator ; the resulting fractions will have the least common denominator. 49. Note. The least common multiple of several quantities is the least quantity that is divisible by each of them, without leaving a remainder. It may be found by resolving each of the quantities into its simplest factors, and then multiplying together all the sepa- rate factors that appear in the whole, using the highest power of each that appears in any one quantity. For example, let us find the leastcommonmiiUtipleof (or' — a^W(ar+a:a+a^)(jr — a),(^ — 2ar+a*), ss.{x — a)*, 3^ — €?=z{x-\-a){x — a), it will be the product of (a:^+a:a+a") (f — a)'(j:+a), because these are all the different factors which ap* pear in the whole, and {x — a) is used in the second power, because it appears in that power in one of them. « n Sample. Reduce to a common denominator i— , ztt* 4a iJtX A ^ ^3^- By Rule 1, Zx \2x 3jf Aa ^ \2x ^ Zx 2a ^ 4a ^^ 3x 12x ^ 4a ^ 3x 108:e» Uiaa^ 24(^x w B7 Rule S. The least common multiple of all the de- nominators is Hirjj; therefore the fractions redaced to their least common denominator aaa 3x 3j _ Sj* <■ ^ 3* ~ 13ot' 2a a _ 2fl" fi ^ a ~ ISol' a» 4a 4a' ^ X 7- = 1^- , . , 3* 3a , Se denoremator ^, — , and — . a'3x' X . 9j« 2b* lii , . , 0-* a+i , . denominator —-7, — -.. and - d ' a+b' a—i' 3. Bedace ton 4. Reduce to their least common denominator , 4(Kr . 7a'— To J 3aj+3i* fmd-— -_ Ans. --3--j-> -,— ^. _ 7« 3« &. Reduce to their least connuon denominator - 6. Reduce to a common denominator i ADDITION OF ALGEBRAIC FRACTIONS. I £0. RuLK. Reduce (he fractions to a common denomiaa- I tor; then add their numerators, and under I their common denominator; the fraction so formed will b* I tlie fraction required. ■~» J^bcAM pi«E. Add together the fbllcMiuifir fiaciioM e ^. , -V and t;^^ Here the least common denominator of 9U the fractiQns is a«— $« ; henqe the fr^tionp If^qme WM?' ^SIS*^ ^ i^^ ' ^ - "^"^ 'V™ ^ tfiewfore 1. Add iogether ^ , 7 , and i-. Ans. ^^^^-^. 2. Add together ---^, and ^--^. Ans. -^— rj • 3. Add together -,j^, ;^^, and ^. Am. ' fl*__^ • i Add together ^^ and ^-?-- Anfl. j^__\q ' 5. Add together ^, g, j, and ^. Ans. ^, or Ji«» SUBTEACTION OP ALGEBRAIC FRACTIONS. 51. Rule. Reduce the fractions to a common denomi- nator, then take the difference of the numerators, under which write the common denominator, and the result will be the fraction required. ExAMPiiB. From -, take -r-. Here the fractions re- duced to a common denominator are rr-7 and 7iri.> there- fore their difference is ,^ . . 1. From -jr-, take -— . Ans. x, 2. From -7- take — . Ans. ^q- I 1 2c 3. From — take — r- -A:ns. g 4 . II * 4. From , — take ~. — -». Ans. -: . «■. 1 — a- 1 — x" 1 — X . ^ jn+n . , m — » ^ Aimu 5. From take —rr- Alia. -^^ — ^*« LTIPLICATION OF ALGEBEAIC FRACTIONS. RuLB. Multiply the numerators together for the numerator of the product, and the denominalots together for the dtnominator of the product. ExAuPLE. Multiply -^ by -— -. Here we hare 3a x 6 =3a6 for the numerator, and (a-{-h) X (_a — 6)=a' — 6' foi' the denominator j hence the product required is — ,- r,. Note. Cancel faotore whioli Eppcur iu bqth termB. 1. Multiply -£ by 3^. fi. Multiply ^^ by ^^. 3. Multiply^ by ^. 4. Multiply ^:^ by ^j. a b , 6. Multiply ^, —^< and 6. Multiply |, -, 2- e' °°' 2? 19oJ ns, ^,_^. Ans. — ;. 2a+6 ■'^■{^/- ah An^. DITISION OP ALGEBRAIC FRACTIONS. 53, Rule. Multiply the dividend bj the divisor verted, and the product ^ill be the quotient required. Note. The reason of Uie above rule will appear from the follov- jng ejiample: — Lot it be proposed to divide -j by j- : for -^ write i^ then it will be V, wbere llio valae will not be altered by iaalti{i1p[ig omnefator and denominator by d, which gives -^=nx- ; bat n i eijna] to ?-, therefore nx -=-^ x - , which 19 tlie rule. 1. Divide j^ by ^. Ans. ^ Divide ~ by^. Ani-J lDmd.i=ikj^. i,..i!f=i. 5- DiTide - Am- — -. 6. Dmde -^- by y. Ans. — I WTiat is the sum and difference of "-^ and "— ? Ans. Sum, a; difference, i!. What is the aam and difference of - g-' and -"""^I Ana. Sum, a^+it'; and difference, 2(Ke. 9- What is the sum and difference of — and ~» J+M, i— » Ans. Sum, J^ ■ difference, -^^zx- 111. ffliat is the product and quotient of — and ^^? Ans. Product, ; quotient, . II. Iteduce to their simplest forms the following ^fc prmonB:— 1st. — j±^ 2d, 2. |"^+^; 3d, -^-^ Ads. 1st, =?.; 2d, =:x; 3d, ='-^; and 4th, : INVOLUTION AND EVOLUTION. 54. Involution is the raising of powera, and Evoi noN the extraction of roots. loTolution is performed by the continued maiti plication cf the quantity into itself, till the number of factors amount Id the number of units in the index of the power to which He quantity is to be iuTolved; thus, the square of 'iXo=a'; the Sth power of x is xxxxx-=.3:,^ ; ^iie power of I2a) is 2x2x2x2x2 X aaaaa=32a5 . H I 3 16 25 36 7 27 125' 216 343 HI 256 625 I2fl6 2101 3'2 •2i3 1024 3125 1 7776 16807 Ais' 7» 4096 1 6561 Iloots, . Squorea, . CnbeS, ■ 4th pDwora, Stfi powers, 55. In raising simple algebraic quantities to any power, obsetrc the following rules: — Ist, Raise the nnmerieaJ oo- eflicient to the given power for the coeffieient, 2d, Multi- ply tiie exponents of each of the letters bjthe power to whiolv' the quantity is to be raised. 3'^, If the sign of the givdii' quantity be plus, the signs of all the powers will be plmj but if the sign of the given quantity he minus, all the even powers will be plus, and the odd powers will be These rules are exemplified in the following Table, 'which the puinl should fill up for himself; BOOTS AND Boots . . . Squu'es . . Cubes . . . 4 th powers £th powers MPLR ALGEBRAIC QUANTITIES. . +'• + 4^ 2o "35 4fl» 1 4^' a' — «^ 8a' 27r> ii^ .. It' + X' + Jb? 816* ii 2«'„ he four! ~32i» h power. 2435' Ans. 1 "33^ Ans. 343a>»i»i». Ans. „^a*6*c'V: 9. Eaise Ja^px^ to the third power. 3. Kaise ^abc^x to the fifth power. 4. Raise ^a'b*e to the eighth power. Ans. oTgO*^'^ 2401o'fcV i-to (he fourth Ans. - S. Raise 7 56. When it is required to raise a compound qnantl^ tb any power, it can always be effeeted by raulliplying the quantity successively by itself; but (here is another meAtit of finding the powers of a binomial, commonly called thV binomial theorem, first given in all its generality by Sr' Isaac Newton, by which the power required can be writtM at CBce, m'thout going through aU tbc mtenoedUte This ^eorem will now be gircn, and the atndeDt can xerify its results by actual multiplication in the mean lime, as Iha ^eral proof would oat be undeiBtood *.t thi» Btage of the piqiU'a advanoement. SINOIIIAL TDEDREM. 07- Let it be required to raise {it-\-x) to tbo nth power, ithere n may be any number, and a and x any quantities, eithK aimple or cnnpound. The first term nil! be a'. And die Kcond will be obuined frcim it, by multiplying by — , and will therefore be na"~^x. The third will be ob- labed ftom the second, by multiplying hy -^ — , and in the nme manner, the 4tb, 5th, 6tb, 7th, will be obtained by multiplying the 3d, 4tb, 5th, and 6di sueceuJTely by -r -, -T- -, -T- -. and — -, and go on, so that the teiulting series will he (a-t-^)'— *+ 1-3-3 ^ ^ 1-2 !lH )(.-3)(-») (.-4) it!^^3?-\-Scr:., where, by substituting cr of 1 isUeod of n, 2, 3, 4 5, sucaessively it becomes (fl+ j;) 3 =a» + 3«'« 4-3ar' + a;' . From the above, it is evident, (1st), That the power n in the first term is the same as the power to which the hinomial is raised. (2rf), That the powers of a decrease hy unity in each successive term, whereas those of x in- ctBise by unity, till the last term, where it is equal to the power to whi<Ji the hinomial is raised. (Stf), That the number of terms in the expansion is always one more than the power of the binomiah (4'/i), That ihe co-eiBcieat of tie second term h always equal to the power to which the liinomial is raised, and that the successive co-efficieats cai; be ohtained by multiplying the co-efficient of the previous leriB into the power of a in that term, and dividing by the nnmhcr of terms from the beginning of the cxpajiaion; thus 10, the co-effieient of the third term in the expansion pf {o-4-«)'' cftn be obtained br mnltiplying 5, the eo-effioi- Mit of the second term, into 4, the power of a mlWt tetro, §tAiinSagbyS, the Bumb^ ot' one teimfioi%t]b.c\»^'&^ ning of the cspansion. In the same manner, may all the other CO- efficients he oh tain ed, K the Bign of the second term of the biaoiiiial were minus, since the odd powers of a minus quantity are minus, and the even powers are pins, the terms which contain an odd power of the second term will he minus, (Art. 55), and those which contain even powers of that term will he pins. Thus, If it were recjuired to cspand (a+b — e)", it might be effected by first considering (6 — c) as one quantity, anA then raising it to the power denoted by its index in eacli term, and separating into single terms. Thus, 3a»(i— c)= +3a^b—3a^c 63_3i'c+36c=— c 36'c+36c'— 6aJc. J . What is the 4tL power of (u— 6) ? Ans. a*— 4o%+6a=6'— 4n6'+e*i 2. What is the 3d power of (4a— 2j-) ? Ans. 64a'— g6a''ic+48aa:s— &(». 3. "What is the 9th power of ^/x+v t Ans.3:a+3^V+3^'+y'- I 4. What is the Sth power of (2« — x) ? Ans. 32^^— BOtt''j;+80«^^_40aV'+10tw:* 5. What is the 3d power of {a—{x+!/)] f Ans. a' — a:' — i/^ — Sa^x—da'^Jz+Sax" +30^" — 3*"| EVOLUTION. 58. Cask I. When the given quantity is simple. Rule. Estract the given root of the numerical c cieot for the coefficient of the root, then divide the eip<^ uents of eacli of the literal factors by the name of the root, nod the several results connected by the sign of mxiltiplicft- tion will be the root sought. E xample. ■ Extract the 4th root of Sla^a*. "e 4th root of 81 is 3, the 4th root of a' is a'=a, end biaotof x' is x^=3fl, .•.3x<J.V,i^-=3aj:* is tli* ALGEBRA. 33 loot sought. In the same maimer may the roets of the following quimtities be foond. 1. What is the 2d or square root (^ Ida^'b^ifi ? Ans. 4a6'A 2. What is the 3d or cube root of ?^? Ans. ^. a What is the 4th root of ^'? Ans. ^. • ar z 4. What is the 5th root of ^^? Ans. ^. 5. What is the 4th root of -j-r^ ? Ans. -j — , 59. Casb II. When the given quantity is compound, and it is required to extract its square root. HuLE. Arrange the terms according to the powers of the same letter, so that the highest power may be first, and the next highest in the second term^ and so on. Extract the square root of the first term by Case I., and place its root for the first term of the root. Subtract its square from the first term, and there will be no - remainder. Bring down the next two terms for a diyidend, and for a divisor take twice the part of the root already found. Divide the first term of the dividend by the divisor, and place the quo- tient both in the root and in the divisor. Multiply the divisor thus completed by the term last placed in the root, and subtract the product from the dividend; to the re- iBainder, if any, bring down the next two terms, and pro- ceed thus till the root is found. The above rule will be obvious, by observing that the square of a+x is a^+2ax+3o^, and that consequently the square root of a^+2(zx+x^ will be a+x; but after we hare subtracted the square of a, the remainder is 2ax+x^ ^{2a-}-x)Xf the first term of which remainder, if divided bjr 2a, will give the quotient x ; and x being added to 2a, and then the sum multiplied by x, will leave no remainder. £xAMPLE. Extract the square root ofa^+2ah+b^ + 2ac+2bc+c\ a^-+2ab+b^+2ac,+2bc+c^\a+b+c root. Divisor 2a+ 6 I 2ab'\'b^ I 2ab + b'' 2a+2b + c 2ac4-26c-fc^ 2ac+2bc-^c^ I Jlere, after (a+b) lias been obtained in tLe root, it u evident that the remainder can be written 2(i»+6Jc+<^=s {2(a+i) + c]c; whetB a + i takes the place of ain ttQ remainder, and is of the same form as {2a-\-x)x. 1. Extract the scjuare root of Qa'+Gai+b'. Ans. 3a4-fi< S. Extract the square roct of a^-f 8aj-+lfiE^ Abb. a-|>4a!. 3. Extract the equare root of a''c^-\-4a^c.r-\.4x-^. Ans. a'c+S* 4. Extract the Bquare root of 4a*+12a6+96^ + 16fM!+ 24fe+16<:^ Ans. 2a+36+4o. 5. Extractthe square root of a-^+4j-'+2,H+9.r' — 4x+^ AnB. a^+2j;=— a:-J-2. 60. Case III. When the piven quantity is compound and it is reqoirei to extract its cube root. Bui.B. Arrange the terms as in Case II. Find tht cnbe root of the first term, '\Thtch will be the first term of the root. Subtract its cnbe from the first term, ivbicli leave no remainder. Bring down the next term, and divide it by three times tJie square of the root already found ; ttw quotient v«ill bo the eecond term of the root. Raise tlww two terras to the third power, and subtract the result from the given quantity; if there be a remainder, divide its fiat term by three times the square of the first part of the ntt as before, and thus proceed till the work is finished. The third power of {a+x) is a^+3a^i(;+3a*'+j^J hence it is evident, that the cube root of tr'-f Sa'x^-So^ +ie' IS a+jr ; taking away the cube of a, the first term <f the root, the remainder is3a'ie+3(u:''+j?3; the first term of which divided by 3a', gives the quotient 2. ExAMPLK. Extract the cube root of a^+6Ei^.^l2aV ««+fla*j:+12aV+8r-\ | a''+2x. Root. The first term of the remainder is 6«^j-, which diri^ej by 3t^, which is three times the square of «", gives 2Kf«f quotient; and (i'+2j; raised to the third power, gives flit "idnantity whose root was to be extracted, and no remainderir ^ that a''-\-2x is the root eought. li. Extract the cube root of 27a'—^'J<i^x+9a!c' — a». Ans. 3a — f> . Extract the cube root of a'^+6(('6= + 12a'i*+86'. Am, a'+3i». 3B 3L BBdmct the iX^iotit^m^^ik^yJ^VJxf^^f. Am. « — df. 61. Any toalt lAoAem may be eztnMttd by the follow- in? foimula : let n be the nmiae ef the root, which will be i Imp the eqoare root> 3 for the cube root, and so om ; then haying arranged the ttnni at in the Sfnarc and cube rooti^ extiact ihe toot of tbe first term by Case L> and let a re- present that roOft^ then the second term dirided by n(f^\ w31 give A0 aecond term of the root ; the first and second terms of the root being raised to the nth power^ and sob* tnstsd from the given quantity, the remainder, if there be any, will be such, that its first term divided by the same divisor wiB givo tfao third term. The exercises in the square and cube root may be wsoaght by this role. BQtTATlOKS. h Am EqvaUon. is a pr<^siti(m which declares the o^iiaiity of two quantities expressed algebraically. This is done by writing the two quantities, one before md the other after the fiigB(=): thus 44^- je=3jc — 4 is an tquadon, which asserts the equality of 4+«; and Sa? — 4. 8. A jSim^ EquatUm iS one which, being reduced to its simpiest fetrn^ contains otaly the first power of the unknown fwott^. 3i A Qtimdrailic E^fucstkn is one which, being reduced ts its simplest form, contains the squajre of the unknown foantity. 4* When ah Equation, after being reduced to its simplest bmii oontains the third power oi the unknown quantity, it isoslled a Ctibio Equation, 5. A Pure QtutdraUc is one into which only ihe square of the unknown quantity enters. 6L An Actfected Qttadratic is one which contains both the first and second powers of the Unknown quantity. 7* Thk ^esoiuHoh of EquaHone is the determining from •ome giten quantities the value or values of those that are unknown* The resolution of equations is effected by the application of one ^r more of ihe following axioms :— « AXIOHB. (1.) If equal quantities be added to equal quantities^ thc( sums will be equa). (2.) If equal quantities be taken from equal quaatideE, the remaindeTS will be equal. (3.) If equal quantities be multiplied by the same or equal quantities, the products will be equal. (4 ) If equal quantities be divided by the same or equal quantities, the quotients will be equal. (5.) If the same qaantiljbe added to and snbtracted from another, the value of the lattei will not be altered. (6.) If a quantity be both multiplied and divided by tha same quantity, its value will not be altered. (7-) The same powers and roots of equal quantities are From the above axioms the following rules for the re- solution of equations can be derived: — 63. Rdlb 1. Any quantity can be taken from one ade of an equation to the other, by changing its sign. This rule is derived from asioms 1st and 2d, as will ap- pear evident from the following example; Let 3ie — 4=2a!+6; if 4 be added to both sides the equality will still exist by axiom first, but the equation will become 3j;^2j;+6-f-4; where the — 4 has been taken to the other side and its sign changed; so that tak- ing a minus quantity from one side to the other, and changing its sign, is equivalent to adding that quanti^ to both sides; if now 2x be taken from both sides, the equa- tion will become 3x — 2x^10, where by taking 2a; ftom both sides of the equation, it has disappeared from the second side, hut has reappeared on the first; hence, taking a plus quantity from one side, and placing it on the other, with its sign changed, is equivalent to subtracting equals from equals, and it has just been shown, that taking a minus quantity from one side to the other, and making it plus, is equivalent to adding equals to equals. The solu- tion of the above equation will now stand as under : 3x — 4=2x+6, the given equation. Sr:=z2x-^10, by transposing — 4 and adding 4 and 6- 3x — 2ij;^IO, by transposing 2x. .■. x^lO, by performing the subtraction on the first side. 64. Rdlr 2. If, after all the unknown quantities are msposed to the first side, and the known ones to the 1, the unknown quantity have a coefficient, it may be away by dividing both sides of the equation by it. is rule is evidently the same as axiom 4th. MPLB. Given ir + 27=48— 3j-, to find the value ^x-f-Q^—iS — 3x, given ecjiial^on. ALOEBBA. 37 4^-f.3jr=48 — 27» by transposing — 3x and 27- Tx=2\i by collecting the tenns. •*. a;=:3, by applying the role. 65. Bulb 3. If there are fractions in any of the terms, they may be taken away by multiplying all the terms by each of the denominators in succession ; or by multiplying ail the terms at once, by the least common multiple of aU the denominators. It is eyident that this rule is merely an application of axiom 3d, and points out when that axiom may be applied. ExAMPi«B. ^+2 "^7 +^=2^> ^ ^^^ ^® value of a;. ;v-f-^ -f- J +4=2x, the given equation. 4x'{'2x+x+ l6=zSx, by multiplying both sides by 4. Jx-^-I^^lSx, by collecting the like terms. .-. 16=0?, by subtracting 7x from both sides. 66. BuLE 4. If the value of any root of the unknown quantity can be found from the equation, raise both sides to the power denoted by the root, and the value of the \mbiown quantity will be found. This is evident from aiiom 7th. I Example. Given a:* + ix=z^ + ^ > ^^ fi^d *be value of x, I x*+ia:=:| +4, the given equation. j 2a;^-f. arzrar+S, by multiplying by 2. Rule 3d. 2a;^=8, by taking x from both sides. ap*=4, by dividing by 2. Rule 2d. .'. x=:l6, by the Rule. 67. Rule 5. If^ after the equation has been reduced to its simplest form by the preceding rules, the value of some power of the unknown quantity is obtained, its value may oe found by extracting the corresponding root of both sides. This is also evident from axiom 7th. Example. Given — -— =a?+12; to find the value o£x. — — =a?+12, the given equation. o a^+3a;=3ar+36, by multiplying by 3. Rule 3d. x^=:36, by subtracting Sx from both sides. .*. ;r=6, by extracting the square root. The previous rules will be found sufficient for the solu- tion of equations containing only one unknown quantity; the following solutions are added as examples of t\ieii ^i^^- plication. w 1. GiTen 5j^^+34=4r+36 ; to find the vnlof of j-. By transposition, 5x — 4a;=36 — 34. Ilule 1st. .*. by collecting the terms, x=2. 2. Given 4aj! — 5i=3(ic+2(;; to find the vahitatjr. By transposition, 4a<e — 3djr=2c+&b. Rule Ut By collecting, (iaSd)x=2c+5b. 9. Given -^+2=12 — -; to find (he Tslne rfar. By multiplying all tlie terms by 6, the leait comma multiple of the denomiuittors, it becomes 2^ — lO-i-3x^7^ -aB+20. Bale 3d. And by transpoMtion, 2j: + ir + 2.r=72 + 20 + JO, Sale Iftt. Hence, by collecting the tenus, it becomes Jx^lOQ, .: by division, by Rale 2d, j;=I4|. 68- Scholium 1. If the felotion between ,b ,a»d tin- known quantities be not giren in the form of an eguatioiif but of a proportion, it may be changed into the form of an equation, by making tbe first term divided by the second. ^ibe third divided by the fourth, — see detinitiooa ; — M by making the product of the eicCremcs^that of the means. For when a : b :: c: d,hy definition, t ^=^-}, and multiply- ing both sides by irf, it becomes ad^bc ; or the product of the extremes is equal to that of the means. Given -h— : —7 — :: 7 : 4 ; to find the ralne of x. Sj multiplying extremes and means, 10j:-{-ii^~ Multiplying by 4 it becomes 40j-|-32=126— Jj. by transposition, 47j'=94. Kule 1st. nd bence, a-=2. Rule 2d. 69. 80HOLIDM 2. When nn equation appears under the form of the equality of two fractions, it may frequently be solved with much elegance, by making the sum of the nu- merator and denominator on the first side, divided by tlieir ditfercncc, equal to the sum of the numerator and denomi- nator on the second, divided by their difi'erenue. If only one side be a fraction, it may be reduced to the abon fotiD ' &r RntJn^ 1 for a denominator on tbe integral side. Th« tthore principle may be denwnatiated l\wa ■. j— 1=^—1. Ax. 2. .-. -y-=:-j^. (3.) And by diTidiB^ i?) we obtain ^zJr^^lld' "^^ "^'^ S"'^"* aboTe. ExufPiB. Given 4-— S^' = - ; tD find •Ja — ^a—x " 'WBy Bqwmng, ^— ^,^„ -, BylnTerting, — — ="iT^^+°i' Brfncing the first side, 1 — -:=. .' j. Imspasing, '—1^3"+^-;- Stdnoing tbe first iide, . , ^-. 1. Given 3!— 6+25— 36; to fintl a. a. ©Ten 3«— 5=23— *; to tind x. 3. GiTcn 7*— 3=5j+ 13 ; to find a-. * Given 3jr+5=10j.— 10; to find a;. 5. Given 2r+ll:^7j-— 14; to find 3:. 8. Given 15a'— 24=20 +^:c; to find a 7. GiTen i+~ — ^rz4j-— 17; to find ^,. Ans!; tafe 1 ^^H 8. Given*— 2 + 3 — ^=7; to find a:. Ans. x=ia ^^m 9. G:ren|+{+| + |+~=46;fofind«. Ans.a;=^ ^^H 10. Given^+I — |=J;tofindj-. Ans. x=^, I: 11. Given -g- +'^-=14+^; to find r. Ans.iF=13. 12. Given-— H — — =16' ■■ ; tofindg. Ans.3;=7. 13. Given « ^^5| '^+1' toGaix. Ans.a-=S. ; to find .r. Ana. . 15. Given - + j— <^; *" fi"*^ ^' -Ana. ,r=- 16. Given (tr+6^=E3--j-«'; to find x. Ans. a'=(i+fi. 17. Given bx-^-2x—a=3i:+Se; tofind.r. Ans.a:=:f±?", 18. Given 3j;— | +c:c=-^_— ; to find x. PK0BLKM9 IN SIMPLE EQUATIONS. ExAUPLB 1. What number is that to wliicli, if its faaff id its fifth part be added, the sura will he 34 1 Let a- represent the number sought; then its half will Iw ^, and ita fiftli part vrill be r Therefore we will have by tbe question, i+-^ + |= 34, and multipl.vinp! by 10; Jac+&r+2^=:340, Kule 3d; tence by collecting the terms, 17if=340, .-. x= 20, by Rule 2d. 2. What number is that whose third part exceeds ita I seventh part by 4 ? j Let 3; represent the number sougbt; then its tbiid part ] will be 3 , and its seventh pajt ■= . Therefore by the q oiaJiiptj^'ng both sides by 21, ^J^-Sa=84, Eule 3d; ACTce coUectine the terms, \x=SA, ALGRBBA. 41 3. A cistern can be filled by two pipes.in 12 bours, and bytbe first of tbem alone in 20 bours: in wbat time would it be filled by tbe second alone? Put X for tbe time requiredl 12 Tben — would be tbe quantity supplied by tbe second * 12 in 12 bours; and — would be tbe quantity supplied by the first in 12 bours. But in twel?e bours tbe two running together filled tbe 12 12 , astern, .-.—+—= 1, and multiplied by 20a?, 240+12a:=20a:, Rule 3d; by transposition, &c. 240= Sx, Rule 1st; .-. 30= a?, Rule 2d. 4. How mucb rye, at four shillings and sixpence a bushels must be mixed witb 50 bushels of wheat, at six shillings a bushel, that tbe mixture may be wortb five shillings a bushel? Let X be tbe number of bushels of rye. Then 9a?=: tbe price of the rye in sixpences. 600=: the price of the wheat in sixpences. (50-|-it?)10= the price of tbe mixture. .-. 9x+ 600=500 + 1 Oo: by the question. Hence 100=a; by transposition. 70. Scholium. The transferring of problems into algebraic equations will be facilitated by studying carefully the fol- lowing remarks: — 1st, If the sum of two numbers be given, and one of the numbers be represented by x, then the other will be tbe sum minus x. 2dj If the difference of two num- bers be given, and the less be represented by x, the other ^ be a? plus the given difference; and if the greater be represented by x, the other will be x minus the difference. ^, If the product of two numbers be given, and one of them be represented by x, the other will be the product di- vided by X. 4th, If the quotient of two numbers be given, and one of them be represented by x, the other will be the quotient multiplied by x. 5tk, If the sum of two numbers be represented by «, and the less number by x, then the difference of the numbers will be s — 2x ; and if the greater be represented by x, their difference will be (a»— 5). 19. Wbat number 13 that which being incxeaa^3L\i^ \V% half and itB third part, the sum will be 154? Alia. ?k^. ^k£ 20. What numbei U that to TCliich, if its third and foartl pBitB be added, the sum will exceed its sislh part by 17 1 Amis 31. At a. certain election/ 311 persons voted, and tbi successful candidate had a majoritj of 79: liow many Tof< for each? Ars. 195 and llfl 22. ^That number is that from which, if 50 be sal tracted, the remainder tviU be equal to ils half, logethi with its fourth and sixth parts ? Ans. 60(1 23. A hnsband, on the day of his marriage, irtn Ari as old as his wife, but after they had lived together 1 years, he was only twice as old: what were their ages on tl marriage day? Ans. Husband's, 45; wife's, ll 24. It is required to divide L.3(M> between A, B, and < BO that A may have twice as much as B, and G as man as A and B together: what was the share of each? Ans. A's, L.lOOr Bs. L.50,- and C's, L-lS 25. A cistern can be filled with water by one pipe in 1 hours, and by another in 8: in what time would it he fillfl bj both running together? Ans. 4^ honn 26. Two pieces of cloth, which together measared 4t yards, were of equal value, and the one sold at 3s., and tb other at 78. a-yard: how many yards were of each? Ans. 12 yards at 78., and 38 at 3 27. A has three times as much money as B, and if give him L.50, A will have four times as much as B: fin the money of each. Ana. A'a, L.750; Bs, LSM 38. After 34 gallons had been drawn from one of tn equal casks, and 80 from the other, there remained thiii as much in the first as in the second: what did each eo) tain when full? Ans. 103 galloa 29. A person paid a bill of L.lOO with half-guineas --^ crowns, using in all 202 pieces: how many pieces y there of each sort? Ans. IBO half-guineas, 22 30. There is a cistern which can be supplied wit from three different pipes; from the lirst it can he filled i 8 hours, from the second in 16 hours, and from the this in 10 hours: in what time will it be tilled if the three pi] bo all set running at the same time? Ans. 3 hours 28^f mjnntol 31. Solve the above question generally on the suppoi^ — that the first pipe can fill the cistern in a hours, tl^ •ndja b, and the third in Ana,- AtOEBRA. 48 SIMULTANEOUS EQUATIONS. 71. When two or more lyiknown quantities are to be determined, there must always be as many independent equations as there are unknown quantities ; and since the values of these unknown quantities must be the same in all the equations, the values are said to subsist simulta* neouslj, and the equations are called simultaneous equa- tions. Case I. To determine two unknown quantities from two inde- pendent equations. 72. Rule I. Make the coefficients of one of the un- known quantities the same in both equations, then bj add- ing or subtracting these equations, there will result an equation containing only the other unknown, whose value may be found by the previous rules. Note 1. The equations are to be subtracted when the quantity whose coefficient is rendered the same in both equations, has the same sign in each, and added when it has opposite signs. Note 2. The coefficients of either of the unknown quantities may always be rendered the same in both equations, by multiplying the first equation by the coefficient of the unknown quantity, which is to be made to disappear in the second equation, and the second equation by the coefficient of the same letter in the first. By this means the coefficients of that quantity will evidently be the same in both, for it will be the product of its two coefficients in the original equations. 73. Rule II. Find a value of one of the unknown quantities in terms of the other from each of the equations, and make these values equal to each other, which will give an equation containing only the other unknown, from which its value can be found by the previous rules. 74. EuLE III. Find a value of one of the unknown quantities in terms of the other from one of the equations, and substitute this value instead of it in the other, from which there will again result an equation containing only one unknown quantity, which may be solved as before. 75. Rule IV. Multiply one of the equations by a con- * ditional quantity n, then add this product and the other equation together : let n fulfil the condition of making the coefficient of one of the unknown quantities 0, tVi^ii VJckfc equation will onljr contain the other unknown ; del^imvck^ i£f value of n that fuMls the above condition, and ^tjJa^^Lv- ^ tbiB ralue instead of it in the resulting eq\xal\OTi, ^kA. the ralue of the other will be determined. or on. tated I WlUB ^^ (3)- leUtion to one of llie unknoHii ouantiliea, and then in relation to other, whidi will give each J or the tt, of one ol the unknown qnantitiea being found, its vulnecanboBuli. tated instead of it in either of the given equations, from which Example Given 3j;+5tf=35, and 7.^-4y=19: find the values of ;c ind y. By Hule 1st. 1 3xJf fii/= 35. -^^ 2 7^- 4y= 19. <I)X 7 3 2U+35y=245. ^^^1 (2)X 3 4 21^— I2y— 57. ^^^1 (3)-C4) 5 47^=188. ^^^1 (5J-47 6 ■■■ v= 4. ^^H C1)X 4 7 ]2x+20^=I40. ^^H (2)x 5 8 35j:— 20(/= 95. (7) + (8) 9 Alx =235. ^^H ■(9)^47 10 .-. « = 5. By Eule 2d. M From (1.) by tiaD sposttion aud division ' ^ 3 * And from (8) X———. -by _ 19+45 Ax. iBt. _35— Sj_ 245— 35y=.->7+J2;/ by mult, by 21. 188=^47y by irnnsposition. .■- A=.y by dividing by 47. And substituting Ibis value of y ia (1.) tve oblaio 3ir+20=35, Hence 3j-=15, And jr=5, as before. By Eule 3d. It bas already been found from (I), that a:=: ftfltitdting this value, instead of jr in (2) it becomes 7(??=i')-4,=.9. 245— 35^— 12i/=57, by mult, by 3. ]88=47i', by transposition. .-. 4=y. In the same manner may x be found from either eqi fion, hy dubsCiluting a \aiue of y found from (he otl equation. ALGEBRA. 45 By Rule 4th. Multiplying (1) by n, and adding (2), we obtain (3»+7)a:+(5«— 4)y=357i+19, which, if the coefficient of y be 0, becomes (Sn'\'7)xz=z35n+19 ; and since the coefficient of y is =0, 5n — 4=z0; hence n=J ; substitnt- ing this Talae of ti in the equation (37i+7)a:=35w4-19, it becomes 9|a?=47. Hence 47a;=235. And .*. ir=5. Next, let n be such as to render the coefficient of ar=0, then the equation will become (5n — 4)y=357i+19; and 7 since 3n+7=0 .•. niz — « , substituting this ralue instead of 71 in the equation^ and changing the sign8> it becomes 15|y=62f. Hence 47^=188. And .*. y=4. 1 p. f 3a:-}-2y=56 ) to find the values of x and y. 1. ijiven S2a:+3y=54j Ans. a;=12, y=10. 2 r*vp i^^ — 7y= 8l to find the ralues of a? and y. \^4x — y=34j Ans. ii?=10, y=:6. 3 0* J 3^+ 2^=^^ ) to find the values of x and y. J Jip-}- .y=9 J Ans. a?=9, j^^H. 4 p. J 3a;4- iy=38 \ to find the values of x and 15^. ^iven ^ i^^2.y=12 f Ans. x=\2, y=4. e p. J 0?- — ^y*^=20\ to find the values of a? and y, ^ ^^ \x +y =10 J Ans. a:=6, y=4. g p. j x-\-y=8 \ to find the values of a? and y. (a; — j/=d f Ans. xz=z^(s'\'d), yz=^{s — d). 1 -^-r-^^4-2j?=16 / to find the values of x 7. Given < ^_, ^^^^„ > and 3^. Ans. ic=6, y=3. ( , , 1 to find the values of x and y. The above equations may all be solved, by substituting in the answer to the (8) the proper values of a, h, c, a\ b', and c', with their proper signs; only (7) would require to he reduced to the proper form before the substitution can be made. Case II^ 76, To determine the values of three or more unknown quantities from as many independent equationa ?l^ VScietfe are unknown quantities. ^^ Rule I, Multiply eauh of the equations by sucli multi- pliers as will make the resulting coefficient of one of the unknown quantities the same in all thg equations] then b^ adding or subtracting these equations, a new set of equa- tions, one less in number, and containing one unknown quantity leas than before, will be obtained; with which proceed as before. 77. Rule II. Find a value of one of the unknown quan- tities, from each of the equations, in terms of the other ujiknowiiBj then equa.te these values, and a new aet of equations will be obtained, containing one unknown quaii> tity less than formerly; with which proceed as before. 78. KuLE HI. Find a value of one of the unknown quantities from one equation, and substitute this tbIvs, instead of it, in the others, which will evidently give one equation less than formerly, containing all the unknown quantities, except that whose value was found, with which equations proceed as before. 79. Rule IV. Multiply all the equations except one bv some conditional multipliers, as w, ji, p, &c., then add all the multiplied equations and the unmultiplied one to- gether, and a new equation will be obtained, in which if we make all the cocflicients of the unknown quantities ex- cept one equal to U, a set of equations among the condi- tional multipliers will he obtained, which will determtns their values, and these values substituted in the resulting equation will give the vuluc of the other unknown quantity whose coefficient was not considered to be 0. The above rules, as well as tliose given in Case I. are all' evidently true from the axioms. EXAMPLE. > find the values of x, y» Given J2x+7ff--8i= (4^+ y+ 2= — J Multiplying the first equation by 4, the seoond by 6 &e third by 3, they become by Rule 1, 1 1 I".'— 8i<+24j=88 , ^2x+7i/~8r=24[- " (2)-Cl) (2)_(3) (4).^ 2 (5)-;- 3 (6)xl3 (7JX25 12j.'.f42y— 48. 50'/—72z=56. 39tf— 51.'=.';4. 25v— 36j=28. 13>— 17^=18. 325;/— 4«Hit=364. 144. AliGSBSA. 43z=86. 25y=100. 12a:-l32+48=88. 12jp=:72. 47 (»M8) 10 (10)^43 11 Bf subeC. in (6) 1^ Transposition 13 (13) -5-25 14 By subst. in (1) 15 Transposition 16 (16)-5-12 1 17 i.-.ar=6. The ts<rfation bj Rules 2d and 3d are left as exercises for the pnpils. The solution bj Rule 4th is as follows : sraltipljing the 'first of the given equations by m, the Beooiid bj n, and adding these products to the 3d of the given equations^ we obtain the following: (3in-|-2n-f-4)a; -(2iii^7n— 1> + («m— 8n+ l)=22wt + 24*+30. Let now m and n be such as to make the coefficient of y and r each equal to 0, and the following equations are obtained^ in which m and n are the unknown quantities. 1 1 2?/i— 7n— 1=0. 2 6w^— 8n+l=0. 3 6w— 21/1— 3=0. 4 13/^+4=0. 5 13w=— 4. 6 n=—fs.^ In the same manner^ or by substitution^ we obtain m= 15 — rr; and since the coefficients of y and z are each=0, we We (3wi+2n+4)a:=22m+247i+30, in wiiich^ bj substituting the values of m and n, there results, f ^^ ^ ,A\ _ 380 96 ^ ^"^-"18" + ^r -"""26 ""13 +^' hence 14 J2r=r9||. 43ar=258, by mult, by 26. In the same manner, if the coefficients of x and z be equated to 0, we can find .the value of y ; or if the coeffici- ents ofx and y be equated to 0, we can find the value of 2. (^+y — z=10^ to find the values o£ x, y, 1. Given < x — y+2= 6 J- and e. yy+z — a?= 2J Ans. a;=8, 2^n6, 2=4. *+ y+ 2=35'! to find the values of a?, ar4-2y-|-32!=66 >- y, and z. i^+§^y+i2;=13j Ans. a?=:12, y=15, z=8. (1)X3 (2)-(3) Transp. (5)-13 2. Given 3. Given '^^_2g I to find the values of ar, y, and z. I'll^QQ I ^^^' ^=16, l/=\^> «\^- ^y+z=22) a -J 2s ■+y4-«=9(> \ to find the valaes of ST, ■«— 20=3./— 40 J- jj, and z. +20=2;4.5 } Aas.x=aa,i,=30,i=55. ISOBLEMS PRODUCING GQDATIONS WITH TWO OR MORE UNKNOWN QUANTITIBB. 1. Find two numbers fluch that the first with twice the second shall be 2] , and twice the first, with half the second, shall be equal to 14. Ans. 5 and 8. 2. Find two numbers such that one half the first, with i one third the second, may be 7, B.ni one third of the first, i with one fourth of the second, may he 5. Ans. 6 and 12. I 3. Find two numbers such, that if 5 he added to the 1 first, the sum will be twice the second, and if four times the second be increased by 3, the sum will be three times the first. Ans. 13 and S. 4. Find a fraction such, that if its numerator be h ed by 1, and its denominator diminished hy 2, its tbIuq will I be ^, and if its numerator be increased by 3, and its de- fl nominator diminished hy 2, the raluc will GtiU be |. I Ans.A.| 5. Find a number, consisting of two digits, the sum of whose digits is equal to ^ of the number, and the product of whose digits is equal to ^ of the number, Ans. 36. 6. There is a certain number consisting of two figures, which is equal to 6 times the figure in the unit's plaee, and if 27 be added to the number, the digits will be inTerted: what is that number? Ana. 36. 7. There is a certain number consisting of three figures, the sum of the digits is 7i twice the sum of the extrcnu digits is equal to 5 times the meau, and if 297 he subtract- ed from the number, the digits wiU be inverted: what 18 the number? Ans. 421. 8. Find three numbers, so that tjie first, with lialf the other two, the second with one-third of tho other two, and the third with one-tburth of the. other two, mnv each be equal to 34. Ans. 10, 22, 36. 9. Find a number consisting of three figures, whose digits are in arithmetical progression, such that if this number be divided hy (he sum of its digits, the quotient . will be 41^; and if from the number 193 he subtracted, the J digits will be inserted. Ans. 432. I 10. If A and B together can perform a piece of work in I 8 days, A and C together in 9 days, and B and C together ' iaIOdajB, in what time will e.ich of them perform it alone? Ans. A in 14s J, B in \7U' a»i ^ '^^^5\^^- QUADRATIC EQUATIONS. 80. Qdadratic Equations may he divided into two kinds, namely, snch as contain only llie square of the un- known qaanlity, and those which contain both the square and first power of the unknown quantity; the first are called Pwff Qvadralke, and their solution may be effected hy Rule V. of Simple Equations, with one unknown qnantity; the second are caWei Ailfidfd Quadratics, and ihtysie of one of the four following Ibrms: — a^' + 6x=+c, I Wf -~^=~- s JUl Hiese are included in the general fonnula. ax^:±:lix ■ ^:tc; and we proceed now to solve this equation. If the first terra were multiplied by 4a, it would become the square of 2iw; let both sides be multiplied hy this quan- tity, viz, 4ni and the general equation becomes ia'x'dtz4abx ^:t4ac; the first side of this equation evidently wants something of beiof; the square of a binomial, of which the first term is 2(w^; let this qunnlity be p^, then adding this ^oantity to hoth sides, the equation becomes 4a''x'^=4a/ix +p"=p'^:4ae : now if the first side he a complete square, its square root can he no other than 2'iie=t=p, (Art. 29 ;) hence squaring this, its square must be identical with the first ride of the hist equation, but its square is 4a'x':±: iapx^p'^ .-. to 4aV-'^4aJ,E+/)^; hence taking away the common terms from both sides, 4apx^4abJ.; and divid- ug both sides hy 4ax, we have p=b, and therefore the ^ quantity which must be added to both sides of 4a''x'' I ^i(thx^':^4ac, so as to makeit a complete square, is b^, I Hat is, the square of the coefficient of it. If now the root of liotli sides be extracted, weobtain 2ax^=b=2*^ ^ b'^4ac, 2a " fwtoola in which, if we insert the proper values of a, b, Ind e, with their proper signs, (the sign of a being always +), we will have two values of a:, both of which fulfil the wnditions of the algebraic equation. From the above inves/ig'alion the following iu\e\& ia- tired.- — w. wiU Hen I Rule. 1. Transpose all tlie terms containing the {[uantity to one Gitle of the ei^uation, and so tliut tlie tei containing the square of tlie unknown q^uantity may be po sitire, and the known tenns to the other. 2. Multiply both sides of this equation by four times tk coefficient of x'. 3. Add the square of the coefficient of x in the fin equation to both sides, then will the first side be a complt Kquare. 4. Extract the root of botb sides, and the result wifl I a. simple equation, which may be solved by the previw rules. 82. ScHOLimu. When the equation, after being trsii posed aa in the Rule, can he divided by the coefficient j:'', without introducing fractions, it may be solved convi niently aa follows : — Perform the division, then add tl gq^uare of half the new coefficient of a: to both sides, wild wdl make the first side a complete square; then the square loot of both sides, and the equation will be a duced to a simple one, which may be solved as before. EsAUPLB 1. Required (he values of j^ in thi 363:*— 4fe=]it20, by multlpljlng by 12=(4x3), 36*'' — ifiLC+16=1936, by adding (4/ to both sides, Gx — 4^^44, by extracting the square root, 63;=r4it44, by transposition, r^8, or — (>§, by dividing by (i. Either of these values substituted in the given equatii will make the sides equal, and they are therefore both nxi of the equation. Example 2. Find two numbers whose sum is 100, i whose product is 2059. Let j:= the one ; then since their sum is 100, the ol may be represented by 100 — x; and hence their prodn will be x(IOO— ^), which by the question is equal to 20S Hence IOOj: — a:^=2059. x'' — 100j-= — 2059, by changing the signs. ie»_100j^+25O0=441, by scholium. X ^-50=^^1, by extracting the root. .-. «=50±21=7I, or 29, which are the ti( required, and therefore x has come out either tb r or the less part. The solution by the rule is lei Hie eiercise of the pupil. AIiGBBBA. 51 1. OiTen x'^zix+iS ; to find x. Abb. x=9, or 2. Giyen 5x^+7 +4x=^5; to find o^ Ana. x=:2, or -— 2|« 3. Oiren ^'»^=:5-^-dr ; to find x. j^_^ Ans. a?=r5, or — 11. 4. GiTen 4x — 14= — - ; to find x. Ans. ar=:4, or — J. 5. GiT€iL 3« 4-^=2-1 ; to find x. 6. GiTen 4— — r- = — r-^; to find x, Ans. a:=6, or X, 7. Oiven i^'=?i^4-^ ; to find x. Ans. <e=l^, or — |. & Giyen Sst;^ +^+6^=301 ; to find ^. J2Q Ans. a:=:3, or — 4. 9» Giren a?— 1=5 — •; to find x. Ans. orizll, or — 10. 10. Giyen - z=24jt; — TOO; to find a?. Ans. a:=70, or 50. U» Giyen --7= -7o;tofinda^ Ans.;r=:4f0r— *4. 12. Giyen — - + "ZI^^ I * *^ ^^ ^• 13. Giyen r + -tt- =9i to find a?. Ans. xzsl2, or 6. QUESTIONS PBODUCING QUADRATIC EQUATIONS. 1. What two numbers are those whose difference is 15, and half of whose product is equal to the cube of the less I Ans. 3 and 18. 2. What two numbers are those whose sum is 100, and vluMe product is 2059 ? Ans. 71 and 29. 3. Find two numbers, so that their difference may be 8, ind their product 240. Ans. 20 and 12. 4 Haying sold a piece of cloth for L.24, I gained as mach per cent, as the cloth cost me ; what was its prime C08t? Ans. L.20. 5. A grazier bought as manj oxen as cost him L.480, and retaining 6 to himself, sold the remainder for the same sum, bj which he gained L.4 a head on those sold. How many oxen did he buy, and what did he pay for each ? Ans. 30 oxen, at L.16 each. 6. A labourer dug two trenches, one of which was 4 yards longer than the other, for L.20, and each trench cost as many shillings a-yard as there were yards in its length. fiow many yards were in each ? Ans. 12 yards and \6 ^^i&!&. 7* Tb»pUte of a Jooking-glaaa i» 24 inches by \6 \ it \fr to be framed by a fraine of uniform width throaghoat whose surface aball be equal to the surface of the glaaSi Required the breadth of the frame. Ans. 4 inches. 8. There are three numbers in geometrical progression. The sum of the first and second is 10, and the difference of the second and IJiiid is 24. AVhat are the numbers? Ana. 2, 8, E y. A and B set off af the same time to a place at t distance of 300 miles. A travels at the rate of one roilci Lour faster than B, and arrives at his joumey'a end hours before him. At what rate did each travel per honrt Ana. A travelled 6 miles per Lour, and B travelled 3. ]0. A and B distribute L.1200 each among a certioii number of persons. A relieves 40 persons more than B, and B gives L..5 a-piece to each more than A. Howmanj persons were relieved by A and B respectively? Ans. 120by A. andSObyE U. A person bought cloth for L.33, )5s., which he soli again at L.2, 8s, per piece, and (gained as much by the bargain as one piece cost him. liequired the number of pieces. Ans. 15i 12. A company dine together at an inn for L.3, ISi One of them was not allowed to pay, and the &haTe of cad of the rest was, in consequence, Lalf-a-crown more than i all had paid. How many were in the company. Ans. t 13. A draper bought two pieces of cloth for L.3i ft The one was fi yards longer than the other, and eachli them cost as many sbilllDgs a-ynrd as there were yaidili the piece. What was the length of each? Ans. 2 yards an 14. Two girls carry 100 eggs to market. One of then had more than the othei-, but the sum which each re- ceived was the same. The first says to the second, if had had as many eggs as you, I should have received li pence. The other answers, if I had had your number, I should have received 6^ pence. How many eggs hu each, and what did each ivceive ? Ans. The first girl had 40, and the second 60 received 10 pence. 15. Find three numbers having equal differences, to that their sum may be 9, and the sum of their fourtti powers 707- Ans. 1, 3, and ' OVJDBATIC EQUATIOVS, WITH TWO WM^KNO^IA 1^ ».MTITIB8. S3. In solving quadratic equalioTis -wM^i Wo ^tJmvwww gaaadties, it is necessary, frequents, to &Da.a.-^5i'*e«'t«* ALGEBRA. 53 »f the unknown quantities in terms of the other unknown, ind then suhstitute this valae, instead of it, in the other equation, and then solve this equation for the other un- mown quantity ; then the remaining unknown quantity irill easily be found, either by substituting this value, or otherwise, as may be found most conyenient. If the sum, or the sum of the squares of the two quantities be given, and their product be either given or can be found, the sum and difiEerence of the quantities can be found, and then their values determined by the solution of a simple equa- tion. ^ Example 1. Given a?+y=21, and — =2; to find x and y, ^ From the second equation we find xzz2y^ ; substituting this value instead of x in the first it becomes 2y^+y=2l, Hence 16y^+8y+l=169» ^7 comp. sq, 4y4-l==*=13, extract the root . \ y =3, or —3^. These values substituted instead of y in the first equa- tion give a;=18, or 24 J. DxAMPiiE 2. Given a;+y=10, and ii?y=24. ' Squaring the first gives x^ + 2xy + y ^ = 1 00. Pour times the second 4iy=96. Hence by subtraction, x^ — 2xy^y^z=.4. By extracting the root, x — ^y=2 (a). By the first equation, a?+y=10 (6). .-. 2x=\2{a) + {h). and a?=:6. /. 2y=8 (6)-C«) and y=4. Example 3. There are two numbers, the diflference of whose cubes is 117) ^nd the difference of the numbers themselves is 3, what are those numbers ? Let «=the greater, and y=:the less ; then by the ques- tion, x^ — y'=117. And X — y=3. Divide the first equation by the second, and there results x^^xyJ^y'^z=!3>d (a). Square of the 2d x^^2xy^y^=id (h). ,\ 3a;y=30, by subtraction, and xy=ilO (c). a?+y=z7, by extract. lOOt. But x—y=:'^^ Hence x=5, and yz=:2. 194 ALGEBRA. Example 4. Wliat two nnmbers are tbo§e whose multiplied hy the greater is 77i and whose difference tijJiedby thele8siBl2? Lei a;:=the greater, and //^tbe less. Then {.T+,^yT=j^''+^i/=77. And {x—;/}y=xi/—;/'=12. Assume x-=.vy, then by substituting (his value insti t£xia each of the equations, they become And w/'J— y«=12. The first of these being divided by the second, gii ■Co Hence ]2eif + ]2c=77t^77- .-. ]2««— 65,.=r— 77- I 'Completing the square, 5761'"— 31 20f + 422n =529. \ Extracting the root, 2\v — 65:=^::23. Either of these values will fulfil the conditions of i question. Thefirst giveaa'=— ^2, and i/=-=J2, the second gives ir=7. and y=4. BXBRCI8GS. a'+y^IS \ to find x and ?/ 21 t V Ans. ir=:lf, ws 7>'to find .rand;/. ^ ( ar^=18f Ans. 3-=9, y: '3;=+j/^=169 J to find j: and »/. X—y='j\ Ans. x=12, y; jr' — s(^=;72\ to find x and y. k''+V=108( Ans. «;=9,y: fl:4-2y=I5l to find jt and -/. a^^+4/=113 ]■ Ans. :r=8 or 7, ,/=3i 01 *+Jf=^5^) tofindrand,/, ^=-^\ Ans, x=16,y=] :[• Ana. fl;=8, t/=i of two numbeis ina\V\B^i\i3 iVc pwi AIX>£BBA. 55 240, and their sum multiplied by the less is 160; what are hese numbers ? Ans. 12 and 8. 10. What two numbers are those, of which the sum nultiplied by the ^eater is equal to 220, and the difiference nultiplied by the less is equal to 18 ? Ans. 11 and 9, or 10^2 and ^"2. 11. Find three numbers such that their sum multiplied bj the first may be 48, their sum multiplied by the second may be 96^ and their sum multiplied by the third may be 112. Ans. 3, 6, 7. 12. Fmd two numbers such that five times their differ- ence may be equal to four times the less^ and the square of Ihe greater, together with four times the square of the less, may be 181. Ans. 9 and 5. 13. What two numbers are those, the sum of whose squares is 34, and the sum of whose fourth powers is 706 ? Ans. 5 and 3. 14. What fraction is that which is double of its square, aad whose numerator increased by one, and the sum mul- tiplied by its numerator, is equal to the denominator dimi* Hfihed by one, and the remainder multiplied by the deno- ninator ? Ans. \, SURDS. 84. Those roots which cannot be expressed in finite terms, are called surds or irrational quantities, 'Rius the V2, X/a^y s/xT or 2*, a% a;", are surds, for their values cannot be expressed in finite terms. . All surds may be expressed by means of fractional expo- lents, in which it is eyident that the value of the surd will Aot be altered by multiplying or dividing both numerator and denominator of the fractional exponent by the same -umuci , vttuo M =8^=(8^)^, or d^zza~^ for a "raised to mr *nr the rth power, is a"* , and the rth root of this is a "'^ ; but a quantity raised to the rth power, and then the rth root extracted, is the quantity itself; hence a surd is not altered in value by multiplying both numerator and denominator of its exponent hy the same number, and since a**^ \ia% ^il- fi» vadf been shown to be equal to a"*, the surd is not ?\l«feA. ralue by diriding both numerator and denomiualoi o? W^ lexponent by Ihe same number. The following operations npon surds depend upon this principle. 85. 1st, To reduce a quantity to thefonn of a given intrd. BuLG, Ruise the quantity to the power denoted by th< exponent of the surd, and indicate the extraction of same root. Example. Reduce 2(i to the form of the cube toot. Here 2a raised to the tbii'd power becomes 8a^, and dicating the extraction of tlie cube root, (8a')^, the form required. 2o lo* fii 1. Express each of the quantities, ax, Zay, — , ^, tj and ^y^ separately in the form of the square root. An& 3. Express each of the quantities, 2a*c, -^, 4ay^ 3aVi and — ■ separately in the form of the cube root- Anfc 3. Express each of the quantities, — 2(i,— , ^, — -j, and —J- separately in the form of the fifth root. Am Note. IF a surd hsTe a coefficient, the whule may be expRB ui Uie foriu of a sud, by rulHuig the cuefficicut lo the p«w<ff d noted by the sui'd, aud multiplying this power intu the sari, tb placing the eymbol af the root over tbo whole product. IMi 4. Express 4ja, 3^^, ^{ax^)^, and 3(3:»y')i,inft* form of simple surds. Ana. ^'iGa, (27<jj;»)3, (625<w«)*i and (Sli'y^)''. 8ti. 2d, To reduce a stird to its eimplmt forra. Rule. Resolve, if possible, the quantity into two ^ tors, one of whicb shall be a complete power, the root of which is denoted by the surd; place the root of this fatUlt before the symbol, and it will be the form required. If th» Burd have a denominator, multiply both numerator and de* BotBiBator of Ihe fraction by such a quantity as will malu the denominator tlie power deuotci'b^ V\\e ^xud, then ex- tmct its root, and place it wu\iout Xte s'jmWV. ■KxiMp/.E, Reduce ^ 27a? to i^ Min^XeA S.-ina. ALGEBRA. 57 Here 27a'=9a'X3a, the first factor is a square, ex- tractiiig its root^ and placing it without the symbol, we ha?e SatjSof the form required. 1. Reduce V32a^, l/sTa^, ^12^, and (180a'a;«^)i to their simplest forms. Ans. 4a^ V2a, Sa^^Sa, 5,Jb, and 2. Reduce 4/1250^, (Oda^x^y^, and (72a?*y')*, to their simplest forms. Ans. Bxy^lOxy^, 2af6aa;*)'*,and 3. Reduce ^-y* v ""V^' ^^ (~27"J ' *^ *^®"^ ^^°^'' plest forms. Ans. -y-Ay7a> ^4/3^, -—a/So. flieir simplest forms. Ans. i(x^z)^, y^^^^)^' X^21ay)*, and |(10a^«z)*. 87. 3d> ^<> reduce surds having different indices to other eguivalent ones having a common index. Rule. Reduce the fractional indices to a common de* nominator, then inyolve each quantity to the power denoted \fj the numerator of its fractional exponent, and over the rkiilts place for exponent one for a numerator, and the common denominator for a denominator. Example. Reduce (2a)^ and (3a^)3 to equivalent suds haying a common index. Since i=B> ^^^ 3=1 » *^® quantities are equivalent to (2a)^ and (3a®)®, raising each of these quantities to the power denoted by the numerator of its fractional! exponent, they become (8a' )^ and (9a*) 6, which is the form re- quired. 1 1 1. Reduce (ac)^ and 5^ to equivalent surds having a common index. Ans. (a^c^y and (125)6. 2. Reduce 4^ and 3* to equivalent surds having a com- mon index. Ans. (256)^^^ and (243^^. 3. JReduce 4/4a^a; and ^/So^^ to equi\a\eT\V. swcSi^ faring a common index. Ans. ^^25Qa^x^ and^^^Ta'^x^ w 4. Keduce (-3-) and (—Y '" •'q'^'^'alent surds haying index. Ana. ("^ 2" I and f --J- 1* to equivalent Hurda Lst^ ing a common index. Ana. ("i^) ^^^ (3^) • I 8fi. To ai.ld or sahtrad, surd*. I Edlh. Reduce the surds to their simplest form. j if they have the same radical quantity in each, the sum of, \ the coefficients pre6Ked to this radical will he their stUn; and the diiFerence of the coefficients prefixed to the la^' eal will be their difference. But if they have different ibA' cal quantities, their sum can only be indicated by placing the sign 'phiA between them, and their difference by ph the sign miiiug between them. ( The reason of this rule is obvious, for the radical qasn- tity may he represented by a letter, and then the rule will I be identical with that of addition and subtraction in algt' I Example. What is the sum and difference of -JM I :md VIM Here 'v'2a8= VTi4x2=12v'2,_aBd V^ I ='/Qix2=W2; hence their sum is 20^2, and QA' (difference is A-J2. I _ 1. Find the sum and difference of 3v'32 and ^Vm (3€ Ana. s«m30V2. diff. eja' j 2. Find the sum and difference of 3^/54 and ,^250. Ans. mmU^2,i\^.iiJi \ 3. Find the sum and difference of ^^Aa^ and^'H'So. Ans. sum (2a+4),^3a, diff. (2a— 4)4&. . Find the sum and difference of Vao and -J^ Ans. sum I'/o, diff, ^] 6. Find the sum and difference of (Sfi.i^)^ and (98a)i Ans. sum {Ga-Ja^l ^^a), diff {Ga-J'^'jJ^)'' 6. Find the sum and difieteucc -at (^\0W1«,«'^^ uia| (300as)i. AliSKBEA. 59 Abs. ram (lOaWjM+lOa'/^), ftnd diff. lOa'^/TO^ 10a VSa. TO MtTLTIPLY 8UBDS. 39. Rule. Reduce the surds, if necessary, touoommon iex^ then moltiplj the coefficients together for a coeffici- ;, and the surd quantities together for the eurd^ oyer ich place the common index. Example. Midtiply S^/lO hy 2 4/l§. :2=:6, the coefficient, and ^10x^/12=^/120. 6v I20is the result; which, however, can he simplified; ! 4^120=4^8 X 15z=2,yi5; hence the quantity in its ^k8tformisl24/l5. Examlk2. Multiply Vo'by 4^6. ere Ja^a^=J=ia^)K ojidj^b=(b)^=h^=:ih^y^ .-. Ty ^b=z ^a^)^ X (b^y^ia^V'yk \. Multiply 5 V5 hy 3V8. Ans. 30\/lO. 2. Multiply (18)* hy S^/i Ans. lO^^. 8. milii^lj VTO by ^i5. , ^ Ans. (233^65)* or jy225000- i Multiply 5a^ by |ai. Aas-^a^^. t Multiply (a +6)^ by (a— 5>i Ans. (a»— 5')*. m r , _^ 0. Multiply a by a . Ans. a ^ * TO DIVIDE SURDS. 90. Rule. Reduce the quantities, if necessary, to a tinmon index, then divide the coefficients and the surd lantities- separately as in rational quantities. •Example. Divide ahs/<^<^ hy b^/be. are ab-^b=ia, the coefficient, and ac-i^bc^z-r the surd 3/a the quotient is a^ 6 . EbiUMPLE 2. Divide ^^ac by 2sjhc. re the quantities reduced to a common index \)ecoTCL^ 9^^, and 2 (3V')s ... the coefficient of tlxe cjao^xcuX 60 xiOBWtii ■ is - J and the aurd {i^]^=-{j3yi which reduced to its _3(A0i aimpleet form is - (a^i^c)*, and hence the quotient is II. Divide 10 v'27 by Sv^S. Am. 15. Ls. Divide ^4/6 bj |4^a Am. J. |3. Divide ii/iShj 2^Q. Ans. 2:^/2. m4. Divide ^^^a by f^^nt. Ana, — ^^_ |5. Divide I V'^ by J^o*- Am. {J. quir . Divide (a'— 6*)^ by (_a—l)k Ans. {a+6)i INVOLUTION OF SURDS. ^i- 91. Bulb. Raise the coefficient of the surd to the re- quired power, find then multiply the exponent of tbe sori by the exponent of the power. Example. Find the third power of Sv'ac. Here we raise 2 to the tliird power, which gives 8, sinl then nmltipiy ^, the exponent of the surd, by 3, the expo- Bent of the power, which gives- .'. the third power of I/^is8(<ic)^=8(«Vx'^)^=8((c(„c)^ or 8ac-/^ 1. Rjiinn ^Ini-^s to iTip BPcfind Dowpr. Ans. 4f(wl'' I Raise 2{ac)3 to the second power. Am. 4(iK)' Raise 4(60^)"^ to tlie third power. Ans. 646cj^*'M 8. Raise ^v6 to the fourth power. Am. jV 4. B^se a*b* to the sixth power, Ans. 0*6'. 5, R^se 1 4- ^■'' to the third power. Ana. l+3•/i+ar^-a■^'*. RaiKP /.l^Sv'a) to tlie BccDnd jowct. ALOEBSA. 61 EVOLUTION OF SURDS. 92. Rule. Extract the required root of the coefficient, md then multiply the fractional exponent of the surd by ;he fractional exponent of the root Example^ Extract the square root of Qv^oT. Here the square root of 9 is 3, and the fractional expo- Qent of the surd is i, which we are to multiply by ^ the exponent of the root, which gives ^ ; hence the quantity sought is 3(ab)*' 1. Extract the square root of 9^3* Ans. dJ^/s 2. Extract the square root of 36 v^2. Ans. &i^2 3. Extract the cube root of 84/5. Ans. 2^* 4. Extract the cube root of 27>s/7« Ans. 3x7^. 5. Extract the fourth root of 644/4. Ans. 2 x (256)^^ EQUATIONS CONTAINING 8UBDS, ETC. 93. In equations containing surds, before the solution can be effected, the surds must be cleared away; to effect this, transpose all the terms which do not contain surds to (Hie side of the equation, and the surds to the other, then laise both sides to a power denoted by the index of the BQid, and if there was only one term containing a surd, the nird will be cleared away, if there be more than one, the operation must be repeated. If an equation appear under the form xz±za^ x±:h, or 3^''z±zaaf*^Ci it can be solved as an adfected quadratic, by solving first for.the power in the second term, and then for the quantity itself. Example. Given Va;+9=v^a?+]. Squaring both we have a; + 9=ir + 2 v^ -^ L Bj transposition 2V^a;=ft .-. V^=4. And squaring a;=16. Example 2. Given !x^•\'0^:sl^2, to find the value of a?. Here the equation comes under the form ic^'+ar^rzc, since the exponent of the first term is double its "poY^et \tv \\i<^ second; hence we mv^t solve for x^. Tlie opeiraXVoTV X«*^ teas follows: ' . f ^^ im. a*+3!*-}- J=— , by completing Bquare. a'^+^;=:t— , by extracting tlie toot. .'. x^^8 or — 9, by franspositioij. and a:'=G4 or 81, by squaring, hence a-— 4 or 34/3, by extracting the cube root. KXEBCISES. 1. Gi¥en ■/3jH-'1-5, to find j-. Ans. jtsjl 2. Given v'4+5.t=2+. JSJ; to find *. Ans. «=;12, a GiTen ^2.c+10+4=8i to find J. Ans. »-^27. 4. Given ^— +5=7; to find x. Ana. ar=l 5. Given J4j:+17+G^x+S=8^^+3; to find «. Anfcic=l& 6. Given. •J'ic-^t/j: 7 =—7= : tafind^ Ans.ir=l6. I. Given*/? -1-3;+-:^ =-7^} to find k 9. Given ^x:;:^+|=Xl±i> ; to find x. Ans. j=8 ,_ , 9a a 10. Given V«+ Vir+ar; 'J-^, tofind a^. Ana. ar= -. 11. Given^ie — V^=\'aj;, tofindj;- Ans. 3:= /j "^jy 12. Given V'x+a+A/a— x=&, to find x. , • ____^ Ana. :.= I C4a-ft')*. 13. Given vT+W*^+^=l+a',tofind3r. Ans.iPsS. Oiven — - — = — ^_, to find x. Ana, !e=.^ Given ~7^~/t — 5- *•* fi"*^ J^- Ana. ^=tf ^^) ALGEBRA. 63 18. CBven m;+2,J»zz24, to find x. Ana. ;rszl6, or dd. 19. Given a:* — 2x^zzx, to find ar. Ans. x:=4, or 1. 20. Given a^+x^zzS, to find a?. Ans. a?=32, or —243. 21. Givep a»--2aii=rl33, to find a;. Ang. a:z=49, or '-|-\ 22. Given (a;+ 12)*+ (a? +12)^=6, to find a?. Ans. a7=:4, or 69. ARITHMETICAL PROGRESSION. 94. Definition. If a series of quantities increase or de- crease, by the constant addition or subtraction of the same q[iiantitj, then the quantities are said to be in arithmetical proffresdon ; and the quantity thus added or subtracted is called the common difference. Thus, 2, 5, 8, 11, &c., is an increasing arithmetical progression, where the common diflbence is 3; and 19, 17> 15, 13, &c., is a decreasing arithmetical progression, in which the common difi'erence U2. In general, if a represent the first term, I the last term, i the common difference, n the number of terms, and s the mm of all the terms^ the progression may be thus ex- pressed : — 1st term, 2d, 3d, 4th, last. a, a+c?, a+2d, a^-^d, L or a, (3^— d?, a — 2c?, a — 3fl?, h Where the first represents sm increasing, and the second a decreasing series. In each of the series it may be observed, that the coeffi- cient of d is always one less than the number of that term in the series; hence the Tith or last term is equal to a-f n — \d, that is, Z=a+.w — \d> 95. To find s ; observe that the series may be written hj beginning with the last term and subtracting d; thus I, l^ lr^2d% I — 3dt I — 4c?, I — 7j-r-l(Z, where it is obvious, tka^ Ir-.^^fi.^ld^a; writing the series thea in both forms, and then adding ; thus, «=Z+^— c?+^— 2rf+; — M,.. + l-^n — Ic?, where the number of terms on the second side ia eVAeii^'^ f/ .-. 2s=(a+0n, and s=z(a+l) |. In which a\i\i!iVA.tvA^ ing instead of I its talue a4-(n — 1)(/, we obtain Fiom the cqunlions found above, namely, l=a-^n^d, and g=[2a+ii— Irf]|, by substitution and reduction the following tbeorenn may be deduced, from which it will appear, that any thretf of the five quantities, a, d, I, n, s, being giren, the remain- ing two may be found. r ].t. P+.X'-.) 28 2d, 4th, a= 6th, != 8th, 1-. 10th, rf 12th, i Hth, 71 «=,'-V''- 3d, 5lh, =Jv'(2i+<i)--&i.+l* _ 2s Jlh, llih, 131h, 25— Gun -.(.-I)' 161b, IBll, 17lb, «=i{i'(2»-j)'+ai.+<i-2.i. 19th, s='_^+J(;+a). 20lh, s=[2f~«— ]rfl|. The above 20 theorems are suf&cient for the solution of any question that ean be proposed in arithmetical progres- sion; the pupil should deduce the theorems from tbetwi given equaiions, it being one of the best exercises in liUnt' analysis that can be given. KXBBCIBEB. 1. The first term of an arithmetical series is 5, the o iDoa difference 4, and the number of terms 12; find lilt Jast term, and the sum of the sericB. -Afipljr Theorem 5th to fiud t, attAT:\\eOTem\?tVVVoSwA< ALGEBRA. 65 2. Giyen the fint tenn 3, the last term 51, and the common difference 2, to find the number of terms and the som of the series. Substitute in Theorem 14th to find n, and in Theorem 19th to find 8. Ans. n=25^ and 8=675. 3. Given the sum of an arithmetical series 12^100^ the first term I, and the common difference 2. Find the last term, and the number of terms. Substitute in Theorem 8th to find If and in Theorem 15th to find n. Ans. Z=:219, and n=110. 4. A person was forty years in business, and spent the &nt year L.80, and increased his expenditure annually by L4. What did he spend the last year^ and how much during the whole forty ? Aiu. He spent the last year L.236^ and during the whole forty jears L.6320. 5. The first term of a decreasing arithmetical series is 9, the common difference ^, and the number of terms 21; find &e sum of the series. Ans. 119. 6. A man is to receive L.300 at twelve several pay- ments, each payment to exceed the former by L.4. He is willing to bestow the first payment on any one that can tell him what it is. What must the arithmetician have tor his pains ? Ans. L.3. GEOMETRICAL PROGRESSION. %. Definition. If a series of terms be such that each can he produced from the immediately preceding one, by multiplying by the same number, the series is called a geometrical progression ; and the series is an increasing or decreasing one^ according as the multiplier is greater or jess than unity. Thus, I, 2, 4, 8, 16, &c., is an increas- ing geometrical series, where the common multiplier is 2 ; a&d 243, 81, 27> 9^ 3, &c., is a decreasing geometric^ Knes, in which the common multiplier is ^. The common multiplier is called the ratioy and is com- monly represented by r; and if a be put for the first term, the general representation of a geometrical series will be the following: a, ar^ ar^^ ar^, ar\ &c. ; and if n be put for the number of terms, and s for the sum of the series, we will have Mult^lxhotb aides of(l) by r, and it becomes c B AZMSBRJi. Subtract (1) from (8), and thi IT — g^ar' — a. The other terms destroj one another. Hence ()^l>=a(r"— 1). ... ^^'^^jfi, by dividirg by ;^. (3.J Thia IB the formula for s, when r is greater than unity but if r is less than unity, the first term of the series wil he the greatest, and the proper espression for s is obtained by subtracting (2) from ( 1 ), which gives I — w^« — ar". Hence (]—)■>= 3(1—1-). ('--). . . , .:*" If now we represent the last tenn by ^ itia(I) that l=-ii-^K From these two equations, namely, l=ai*~*, and |3 ■ ■! or= -r-- ', the following theorems may be ik rived, The above theorems are given as exercises in Htvtl analysis, and should bM he deduced from the fith and fllll| which were previously proved. When r is less than'I] tlie term r" may be rendered less than any (juantit^ th^ can he named, however small, by taking » sufficiently great; so that (4) in the caso of n=infinity, will becoiu le expression for the sura ( BB continued to infinity. EXERCISES. 7. Gtren the first teim of a geomeV.n'sA »ftTie%\, tlw 'toinoa ratio 2, and the -nttm^w* ol tenaa W, ^ fesA. "'^ ftena and the mm of the 8ou«s- GbbtUvte in !]nieoreiii9 9tli and 6&, and we hare Ana. lzs&l% «=1023. 2f. Tbe sum of a geometrical progression, whose first tefm 18 1, and hoM term 128> i9 255. What is the com* von. ratio? Ans. Theorem 4th girea r=2. 3. Find the sum of the geometrical series, I9 i> i> j» &c., QOntSmied to infinity. Ana. 2. 4U Find the sum of the geometrical series, whose first tan 13 If and common ratio §, when oontinned to infinity. Ans. 5. 5u Find the sum of the geometrical series whose first term is m and commooi ratio ^ when continued to in- Ibltj. * Ans. mn. (k A servant agreed with bis master to serre him for tvelre months, upon this condition, that for his first month's serrioe he should receive a farthing, for the Kcond a penny, for the third fourpence, and so on. What lid his wages amount to at the expiration of his seryice ? Ans. L.5825, 8s. Sjd. 7* There are three numbers in geometrical progression vLose sum is 53, and the sum of the first and second is to the sum of the first and third as 2 is to 5. Required the imnbers. Ans. 4, 12, 36. 8. There are three numbers in geometrical progression. Che sum of the first and second is 15, and the sum of the list and last is 25. What are the numbers ? Ans. 5, 10, 20. GEOMETRICAL RATIO. 97. The geometricsd ratio between two numbers is de- ennined by dividing the one number by the other. The pu^tient is &e value of the ratio. The number divided is oiled the arUecedenU and the divisor the conseqtcent of the itio. Thus the ratio of 9 to 6 is §=1^, in which 9 is the itttecedent and 6 the consequent, and the value of the atip is 1^, Ratio may therefore be considered as a fraction, the nu- nerator of which is the cmUcedentt and the denominator he consequent of the ratio. When the antecedent is greater than the consequent, it is called a ratio of greater ineqiuiliiy^ and when the aiA.^- iedent Is less than tbe xonaeqneiA, it is called a ial\Q ol 96. SSnce ratios can be expressed by fractions, Wve^ ^«ii be compared with each other by reducing the fractions tv a common denoininator ; then that nill be the greater ratio which haa the greater numerator. Ratios are com- monly written by placing two points between the antece- dent and consequent; thus a: h expresses the ratio of a to if and is read a is to h. 99. Proposii'mn ls(. A ratio of greater Inequality minished by adding the same quantity to both its terms; whereas a ratio of leeser inequality is increased by adding the same quantity to each of its terms. For — is a ratio of greater inequality, and if c he addd to each of its terras, it becomes — -— . Eeducingtheseratiol to a common denominator, the first becomes r and the second — — , which is evidently lesa tlmn lis first by Y — :■ Again, let be a ratio of lesser inequality; addetO each of its terms, and it becomes ; reducing t) to a common deoomlnator, they become . — r — , — p- -TT — , where the second is evidently greater than the ; 100- Fropoiition 2d. A ratio of greater inequality il increased, and a ratio of lesser inequality diminished, li} subtracting the same quantity from each of its terms. Let — be a ratio of greater ineqniJity, take c from eaCl of its terms, and it becomes ; reducing these to i common denominator, the first becomes — ^ — . — , anJ the second — ^ — , which is greater than the former b] Again, let — be a ratio of lesser inequality; tabs i from each of its terms, and it becomes ; redodl) .(a-c) these to a common denominator, they become - and — ' — — — , wLicli U efideaWj \e«a V\iM:i the (ami iOJ. Frop. ScL A ratio is not altered bj multiplying or diyiding its terms by the same quantity. Let a : 5 be any given ratio, then it is identical with r=^ = r- Q- E. D. ' MO O PROPORTION. 102. The equality of two ratios constitutes a proportion; lius if a : & be equal to c: d, the two constitute a pro{>or- ion^ and are written thus; a:h: : c: d, or a: hz=c : d, md read^ aisto&ascistoc?; consequently^ since the ratio of a to & is 7 ^ and the ratio of c to c^ is -^^ we have r =: J, in which a and c are called antecedents, and 5 and d sonseqnenCs : also a and d are called extremes, and c and h means. Art. 14. 103. Prop, 1. In every proportion the product of the Extremes is equal to the product of the means. For if a : 6=c : (f, then -=-, multiplying both sides by hd we have ad=zhc. Q. E. D. NoTB. If a : bsJ) : c, then b is called a mean proportional be- tween a and c, and c is called a third proportional to a and b; and [by Prop. 1) it is evident that l^=ac; hence b=^ac, or a mean proportional between two quantities, is the square root of their piodnct. Prop. 2. Two equal products can be converted into a proportion by making the factors of the one product the extremes, and the other the means. For if a€?=6c, dividing both sides by hd, we have -=- ; kence aihzicid. Q. E. D. * "^ 104. Prop, 3. K four quantities be proportional, they are also proportional when taken inversely; that is, the second is to the first, as the fourth to the third. Since 7=-:, 1-5-7 =:l-i--; •*•-=-> and hence hiaszdic, Q. E. D. a o a a c 105. Prop. 4. If four quantities be proportional they are also proportional when taken alternately; that is, the first is to the third as the second is to the fourth. Eot mce-=^, if both Bides he maltipiied by - and t\ie co\sv- moD factors cancelled from the numerator aixd denomVn^- y 70 tor on both sides, we hare -=- ,: a: e:^h:d. Q. K. D. 106. Frtip. 5. Wlieii four quan^ties are proportional, they are also proportional Iry c07/iposilioji; that is, the EQU of the first and second ia to the second, as the sum of ths third and fourth is to the fonrth. „. c a c , , , a+b c+d Since r=: J I r+^~Z + ''> nance -^=^— -. b d b ' d ' b d Therefore (i+t:6=c+(;:d. Q. R Dj 107- Pvop. 6. When four quantities are proportional, they are also proportional b>/ division; that is, the difier- ecce of the first and second is to the second as the differ*- ence of the third aod fourth is to the fourth. Sincere-;, 7 — 1=-: — 1; hence -7—=— 3-. b d b d b a Therefore a—b : b=c—d .d. Q. E. D, 108. Prop. 7- When four quantities are proportional,' fhey are also proportional by mixing; that is, the Gam of^ the first and second is to their difference as the sum of th third and fourth is to their difference. For Prop. 5th ^-=^. and Prop. 6th ^-^. ...2±-^X — =— X— ■ b a — b d e — d' and therefore a+t:((—5=c+d:i>—rf. Q. E.1 109. Frop. 8. Quantities ore proportional to tiu equimultiples. Let a and h represent any quantities and ma and < an^ equimultiples of them., then a : b=ma : ini/. For-=-^; therefore a: h=ma: m6. (%.KI Where m is any quantity, whole or fractional. 1 10. Prop. 9. The like powers and roots of propoi quantities are proportional. Since ~ =2 . ^'= J .-■ a' •■ 6"=c- : if . Q. E. Where n may be either whole or fractional, and quently represent either a power or a root 111. Prop, 10. If two proportions have the sames cedents, another proportion may beformcd, having the seqnents of the one for its antecedents, and the consequoil of the other for its consequents. For if a : 6 :: c : t£, and tt : e : ; c :/, lien ^ = J 1 and by inversion. AI^OBBSA. 71 l»aice|x^:i=axf .'.5 = {; wherofore e : 6=/: di where e and /, the conse* pients of the one, are the antecedents, and h and d, the xmseqnents of the other, are consequents. Q. E. D. 112. Prop. 11. If the consequents of one proportion be the antecedents in another, a third proportion will arise, liaTing the same antecedents as the former, and the tome Donsequents as the latter. Let a:&=c:d^ and l:<=:(i:/; then a:e=zc:f; for from the first ^ = j> and from the second - = ^ ; henoe mtdtiplying these e^aals tdgel^eir, -gyc -^^^^ f' *** " =: p ; hence a : e=ze :/. Q. E. D. 113. Prop. 12. I^ there be any number of propor- fionals, as one antecedent is to its consequent, so is the mm of all the antecedents to the sum of all the conse- {bents. Let a : h=zc : d=e :fi=g : h ; then For ah=:bay and Prop. 1st ad=bc, afzzhe, and ahzzhg; therefore by adding equals to equals, we have ab-{'ad+af '{'ah^ba-{'hc+be-{'hg ; hence a(J>-^d-\-f+h)=zb(a+c ^e-\'g) ; and therefore by Prop. 2d we obtain aib=ia+o+e+g:b+d+f+h. Q. E. D. 114. Definition. When any number of quantities is n continued proportion, the first is said to have to the Uid the duplicate ratio of the first to the second, and the list is said to have to the fourth the triplicate ratio that lie first has to the second. 115. Prop. 13. The duplicate ratio is the same as the "atio of the squares of the terms expressing the simple ratio ; md the triplicate ratio is the same as the ratio of the cubes )f the t^rms expressing the simple ratio. Let a: h^b : €=c : d, then a : c=a^ : h^. And a : cP=a : &*. A h t a h a a a c^ lence a : c::=a^ : b^* abtaaaac? ft ^c ^5 ""6 ^6' ^b •*• " "6»* »d hence a : d=€^ : ^. Q^ TL T>. n 116. Prop. 14. The product oF the like terms ofanfl mmetical proportions are themselves proportional. ta:b=e: d, thea^=^. J n : lc=l: m, then 7=-. and consequently ~ ^ X 7 =4 xf X -■ o y^J h a h m hence aei : hfk=cgl : dJtm. Q. E. D. 117- i'njp. 15. If there he threemagnitudes, n, 6j e.anJ other three, rf, e,/, Buch that a : 6=rf : e, and 6:c=e:/, then a : c=d : f. For 7 =~ and - =- , hence -- x - =- X -,. be c / bee/ and therefore 7=-;,. consequenlly a : c=d : / Q. R D. 1 the turn amc INTEEEST. 118. Interest is the allowance given for the loan or forhearance of a sum of money, which is lent for, or bf- comes due at a certain time ; this allowance heing gens- rally estimated at so much for the use of L.lOO for a year. The money lent is called the principal, the sum pud foe its use is called the interest, the sum of the principal aiA interest is called the amownt, and the interest of L.100 fw one year is called the rate per cent. Interest is either Simple or Compound. Simple interest is that which is allowed upon the ongt' Dal principal only, for the whole time or forbearance. 119. PfionLBM I. To find the simple interest of aD/ sum for any period, and at any given rate per cent. Let r=the interest of one pound for a year, jd= principal or sum lent, (=the time of the loan in yean, the interest of the given principal for the given time, the amount of the givenprincipaland ils interest Ibr^ le ( ; then we will obviously have the following relatioO' among the quantities ; 1 : pt=;r : i .", i=pii. (I.) and hence a=p+prt=p{l+rt). {2.) ALGEBRA. 7^ ' ' ' ' and r=^=^ = - (3.) Bj means of the above five formul2e all the circumstances connected with the simple interest of monej are readily determined. But as the rules for the calculation of simple interest are generally given in reference to the rate per cent. , instead of the rate per pound, as above, the formulae may lie all changed into those relating to rate per cent., by mak- ing r represent the rate per cent, and substituting -— - in- stead of r throughout all the formulae; and the student is requested to write in words the rules which the five for- ibulse contain, by which he will be led to see the advan- tage of algebraic formulae. 1. What is the interest of L.560 for 3 years, at 4^ per cent.? . Ans. L.75, 128. 2. What is the amount of L.420 for 6 years, at 3 per cent.? Ans. L.495, J 2s. 3. What principal laid out at interest for 5 years at 4 per cent, will gain L.60 ? Ans. L.300. 4. What principal laid at interest for 10 years at 3^ per cent, will amount to L.607, 10s. ? Ans. L.450. 5. In what time will L.500 amount to L.800 at 4 per cen,t. ? Ans. 15 years* 6. At what rate per cent, will L.200 amount to L.344iii 18 years ? Ans. 4 per cent. COMPOUND INTEREST. 120. In compoimd interest, the interest is added to the principal at stated intervals or periods, and this amount is made the principal for the next period. Hence if H repre- sent the amount of one pound at the end of one period, nnce this is the sum laid at interest during the next period, we will evidently have the following proportions to find the 4Uxiount of L.1 at the end of any number of periods. 1 : R-=iR : R'^, amount at the 2d period. 1 : R=:R^ : R^, amount at the 3d period. 1 : R:=zR^ : R^, amount at the 4th period. 1 : jB=jff"~* : R^, amount at the nth period. Prom which it appears that the amount of one pound at the end of any number of periods is R raised to the power denoted by the number of periods, and it is plam l\idX. \\i^ amount of /? pounds will be p times the amoxwvt o? otiei pound; bence^ representing the amount of p pouixis >a'S A., and the number of periods by t, we will have I -y^ irlog. jl=lfig.;> + nog. ^. (1.) log.p=l«e.A-l\<,g.J!. (2.) 1 log. i;=log. j(_lBg. p. (3.) log. J— lo g. (« Iog.ii=^5---j3J. (5,, 121. The interest is generally conrerted into prindpa yearly, but sometimes half-yearly, and aomelimea even quarterly. If r represent the simple interest of L.l for* yeai', and » the numW of years for which the calculating is to be miide, then whea the inlerCBt is conTertiMe inb^ principal half-yearly, Ji and ( will have the following viJues: Jl-=l-\- ■=, (=;2re; and when it is convertible ([Qartedff ^=1+ ^, t=4n; also J?=l+ ^, and t=mn. when tl interest is convertible into principal m. times per amran. nxEncises. 1. tVhat will be tlie amount of L.1000 in ten years, kt H per cent, compuutid interest? Ans. L.1628, 17s. 9^4 2. What principal laid at compound interest will amousl to I<.700 in eleven years, at 4 per cent. 1 Ans. L.454, Us. I|A 3. In what time will L.365 amount to L.400, at 4 pM cent, compound interest? Ans. 2 years 122 da» 4. At what rate per cent, compound interest will LJO amount loL.63,lGs.3jd.injive years? Ans. 5 per ceni f). In what time will a sum of money double itself, 4 a per cent, compound intetest? Ans. 14'3yeHl 6. In what time will n sum of money double ilsetfi B 4 per cent, compound interest. Ana, 17"67yeaM 7- In wlmt time nt compound interest, reckoning 5 pc cent, per annum, will L.IO amount to l/.lOO? Ans. 4719 yeai*. a. What will be the compound interest of L.lOO for f weKe years at 4 per cent., if the interest be payable year- ly? what if payable half-yearly? and what if payablt quarterly ? Ans. L.60, 2a. OJd., L.60, 16a. lOid., and L,61, 48. 51ill ALGK&HA. 7d ' ANNUITIES. 122. Ammuitxes signify anj interest of money^ rents, or pensions, payable from time to time, at particular pe- riods* The most general diTision of annuities, is into an- nuities certain^ and annuities contingent ; the payment of the latter depending upon some contingency, such, in particular, as the continuance of a life. Annuities have also been divided into annuities in ^jo«- Bodon, and annuities in reversion, the former meaning Mch as have commenced, or are to commence immediately, and the latter such as will not commence till some particu- lar future event has happened, or till some given period of time has expired. Annuities may be farther considered as being payable yearly^ half-yearly, or qtiarterly. The present value of an annuity is that sum, which being improved at compound interest, will be sufficient to piy the annuity. The present value of an annuity certain, payable yearly, md the first payment of which is to be made at the end of I year, is computed as follows : — Let the annuity be supposed L.1; the present value of ^ first payment is that sum in hand, which being put to nterest, will amount to L.l in a year; in like manner, the iRsent value of the second payment, or of L.l to be re- vived two years hence, is that sum, which being put to nterest immediately, will amount to L.l in two years, and io on for any number of years or payments ; and the sum )f the values of all the payments will be the present value if the annuity. 123. Let the interest of L.1 for one year be represented >7 r, then L.l will amount to 1+r in a year, and the sum kat will amount to one in one year, which call x, will evi- leatly bear the same proportion to L.l, that L.l bears to i+r; hence we have the following proportion: — x: 1=1 : l + r; .*. ar= — , value of 1st payment. In the same manner, that sum which in two years will imoont to L.1, is evidently that sum which in one year wU amount ^o -—-. Stating the proportion so lliat Wi^ oaDtitf sought majr stand last, we have :1 : — , present value of lat pajmentl = 7-7 'cij:"-.!' preaent value of 2 = T- — ^ : 7T~~ra. present value of Zd payment. J24. Tlie present vnlue of an annuity of L.l for n yea y. IS therefore the sum of the series. This is evidently a geometrical series, in fvhicli the fii term is :^ — . and the common ratio is also — - ; heiicefini ing its sum as in geometrical progression, and putting ji_ the sum (hat is the present value of the annuity, vfb hart ^= f^r + O+ry + (l+if + ■'■ (f+0-- (i (I.)-(2.) C30x(l+M (3 125. If the annuity is to continue for ever, then »fc leomes infinite, and also (1-|- ?■)■"; hence t— — loaybecffl I Bidered as 0, and therefore we have for the present toIi t of an annuity of L.] , payable for ever, p= -, value of a perpetuity of L.l. It is plain, that if the annuity be a pounds instead ofM '■ will just bi! a times aS great as before; and therefore l! esent value of an annuity of a pounds, payable for » je* jiriilbe ALGEBRA. 77 iuid that of a perpetuity of a poonds will be a I 126. When an annuity is only to commence n yean hence, and then continue for t years, it is called a deferred annuity, and it is plain that its present value will be the difference between the present value of an annuity to con- tinue for ^ 4- n years, and another to continue for n years ; but we have seen (5) that the present value of an annuity of L.1 to <;ontinue t+n years, is ,, ^ , and that • '' r(l+r)i+n the present value of an annuity to continue n years is rfi-L v> * *^® difference of these expressions is therefore ^e present value of an annuity commencing n years hence sad continuing afterwards for t years; reducing these ex- pressions to a common denominator and subtracting. We ™^® V^ — ^: — > and therefore the present value of an an- unity to commence n years hence, and afterwards to con- tinue for t years, is p=z j- — t^t~* ^md if the annuity be a pounds per annum, instead of one, it is plain that the whole result will be a times as great ; therefore the present value "of an annuity of a pounds per annum deferred for n years, and then payable t years, is p— °*-V ^^ i r(l+ry+» If the annuity be payable for ever after n years, then t, and consequently (l+r)' become infinite, dividing both numerator and denominator of the above expression by (1+r)', and observing that — — r- becomes 0, we have for tke present value of a perpetuity of a pounds deferred for n years, »= -j- -.-. ^ ^ ^ r(l+r)» 127. To find the amount of an annuity left unpaid any number of years, at compound interest. Let A be the an- nuity, then the amount of the first payment which is fore- borne for n — 1 years will be ^(l + r)**"^; of the second for «— 2 years will be -4 ( 1 + r)"-'* ; &c. .•..the whole amount=^(l + (l+r)+(l+r)« + , &c.) to n terms; ^/ \ or the amount = 7 { ( 1 + r^—l h Art 96. {^?>.) k Ex. 1. What is ttio present Tfllue of apentlon of L.IOI^ payable yearly, for 20 years, at 5 per cent, compound in- terest ? Ana. L.1246, 4s. SJd, 3, What is the present value of a perpetual annuity a L. 100, payable yearly, interest al 5 per tent. ? Ana. L.2OO0; 3. What is the present value of an annuity of L-ICK^ payable half-yearly for 20 years, interest at S per cent pa annum, also payable half-yearly ? Ans. L.125.5, 2a. 1(^4 4. What is the present value of a perpetuity of L.10 per annum, payable half-yearly, interest at 5 per cent, peg annum, being also payable half-yearly? Ans. L,20u(t 5. What is the present value of an annuity of I^lOOtfl commence JO years hence, and then continue for 30 yeai^ interest at 4 per cent. ? Ans. L.l 1 68, 3s. 7^0. 6. What is the present value of an annuity of L.50, 1 commence 8 years hence, and then to continue foi A years, interest at 5 per cent. ! Ans. L.589, 12s. 89 7. In what time will a pension of L.50 amount to l,.100O, interest at 5 per cent. ? Ans. 14-2 yearf. 8. To what sum will an annuity of L.24 amount in f* Tears, tvhea improved at 5 p?r cent.? Ana. L.793, llg.8 L 1. It is required to divide each of the numbers 11 a 17 into tvta parts, go that the product of the first parted each may be 45, and of the second 48. Ans. 5, 6, and^ft 2. Divide each of the number321 and BO into twopaiU so that the first part of 21 may be three limes as great c the first part of 30, and that the sum of liie squares of fV remaining parts may be 585. Ans. 18, 'd, and 6, ft ^^ 3. A gentleman left L.210 to three servants, to bedividid in continued proportion, so that the first shall have L4f^^ more than the last : find their legacies. ' Ans. L.120, I..60, and h» 4. Tliere are two numbers, whose product is 45, ani & difiercnce of their squares is to the square of their diSiE! euue HB 7 is to 2: what are the numbers! Ans. and &■ 5. A and B engage in partnership with a cnpitai ct L.lOO: A leaves his money in the partnership for3 monlbh and B for 2 months, and each takes out L.99 of cspitaiasf profit; determine the original conlribution of each, ui Aaa.Ah.4&,i • ' PLANE GEOMETRY. GsoMETBY is that branch of Mathematics which treats of the properties of measurable magnitudes. Magnitudes are of three kinds, viz. lines having length onljy surfaces having length and breadth^ and solids having length, breadth, and thickness. Thai branch of Geometry which treats of lines And sur- £ieet is called Plane Geometry, and that which treats of the properties of solid bodies is called Solid Geometry. DEFINmONa 1. A point is that which has position but no magnitude. 2. A line is length without breadth. 3. The extremities of a line are points. 4. A straight line is that which lies evenly between its extreme points. 5. A w,perficies is that which hath only length and beadth. 6. The extremities of a superficies are lines. 7- A plane superficies is that in which any two points Mng taken, the straight line between them lies wholly in (hat superficies. 8. A plane rectilineal angle is the inclination of two ttraight lines which meet in a point, but are not in the tame straight line. Note. When there A in levaral angles at one punt, as at B, each of ^ angles must be ouned by three letters, and the letter at the angular point must be fbeed between the other two ; thus, the sngla formed by the lines AB and BD meeting in the point B, is called the angle ABD or DBA; also the angle foi*med by the straight Jmes DB and BC^is called the angle DBC or CBD; but when there is only one angle at the point, as at "E, it may be caWed ^Irn^V^ >^« angle at £1 9. When a straight line Etanding on ati- >ther straight line makes the adjacent angles equal to one andihcr, each of the angles is called a riffkl angle; and the Rlriiifiht line which stands on the other is - calk'd a pei-pendiculnr to it. 10. Aa dbluse angl greater than a right ai 5 that which i 11. An acute angle is that which is less than a right angle. 12. A term or hovnd'iry is the extremity of any thing. I.S. A_;^nisthat -which is enclosed by one or inei boundaries. 14. A circle iit a plain figure hounded hy one line, which is c-illed the clrcurafeience, and is such that all straight lines drawn from a certain point within it to the circunifweuce are equal to 15. And ihis point is called the r,enirt. 16. The tfinmettr of a circle is a straight line dni through the centre, and terminated both ways by the tj cumference. 17. A nemicirale is the figure contained by a H meter and the part of the circuiu fere ace cut off by ti diameter. IS. A straight line drawn from the centre to the drcmi* ference of a circle ia'callrd a iWi'ms. 1 19. A straight line which is terminated both wajtU Ihe circumference, hut docs not pass through the centce^,^ called a chord. , 20. The part of the circumference cut off by the choid 14 lied an «rc. 31. The figui y21. The figure hoi mail. ga. Jteefiliiieatji-^i ^ht tiaes. hounded by the chord and arc is called A those which are contained hj N.AJU OEOMXTBT* 81 23. Trilateral figures, or triangles, are contained bj three straight lines. 24. Quadrilateral figures are contained by four straight lines. . 25. MtUiilateral figures, or polygons^ are contained bj more than four straight lines. 26. An equilateral triangle has all its sides equal. 27^ An isosceles triangle has two equal ades. 28. A scalene triangle has three un- cqualsides. 29. A right angled triangle is that which has one right angle. 30. An obtuse angled triangle is that which has one obtuse angle. 31. An acute angled triangle has ftU its angles acute. 32. Of four- sided figures, a square is that which has all its sides equal, and all its angles right angles. 33. A rectangle is that which has all its angles right angles^ but all its sides are Bot equal. 34. A rhomhus is that which has all its sides equals but its angles are not right sngles. 35. A rhomboid is that which has its opposite sides equal to one another^ hut ^ its sides are not equal, nor are its angles right angles. V 36. All other four-sided figures besides these are called trapeziums. n PliARK GBOMETRr. 37- Faral/rl etraifffil lines are sach ae, being in the same plane, and being produced ctct bo far both ways, do not meet. 3B. A pnralfflnffjvim is a four-sided figure whose oppo- site sides are parallel. ^!l. A ]>o«l?tIate requires us to admit tlie possibiliiy of doing Bomefhing, witliout feeing sboirn how to do it. 40. A pfoposition is a distinct portion of science, and ik eilher a prMeni or a llieoyem. 41. A prdtlrm, is an operation proposed to be perfoimcA 42. A theorem is a truth which il is proposed to prore. 43. A lemnia is a preparatory proposition to render whit follows more easy. 44. A corollary is an obvious consequence resulting fno a preceding proposition. 45. A scholium is an observation, or remark upon some- Ibing preceding it. 46. An ariom is a self-evident trutL 47- The side opposite to the right angle of a right-angM triangle is called the h>/pQtemi*e ; one of the sides alniutths right angle is called the basf; and the remaining side ii called the perpendicular, 48. In a triangle vrhich is not right-angled, any aAe may be called the ba«f; the intersection of the other ivo sides is called the vertex; and the angle at that point tbi vfTtk'ol angle. 49. The space contained vrithin a figure is called ill mrfaee; and in reference to that of another figure vriA which it is compared, is called its arta. 50. A polj/gon is a figure contained by more than fiiur itraight lines; when its sides are all equal, and also ill angles, it is called a ifffidar polygon. 51 . A polj-gon of five sides is called a petdagotx; thai df six sides, a /ie.ca30M; that of seven sides, fi AiTifajon; ihst of eight sides, on octagon; that of nine sides, a noivagott; ■hat of ten sides, a decagon; that of eleven aides, an unitr cagon; (hat of twelve sides, a dodecagon; (hat of fifiew sides, a qnindfcaijon. POSTULATKB. 1. I*t it he granted tha( a alraight Hoe way be Jraini from any point to any other poinl. 2. Let it be granted thnt a terminated straight line naf 'e produced to any length in a straight line. PLANS QEOMMTBY. 83 3. Let it be granted that a circle maj be described from Dj centre^ and with anjr radius. AXI01I8. 1. Things that are equal to the same thing are equal to ach other. 2. If equals be added to equals, the sums are equals. 3. If equals be taken from equals, the remainders are qwla, 4. If equals be added to unequals^ the sums are un- guals. 5. If equals be taken from unequals, the remainders are nequals. 6 Things which are double of the same, or equal things. He equal to one another. 7* Things which are halves of the same, or equal things, le equal to one another. 8. Magnitudes which coincide with one another^ that I, which exactly fill the same space, are equal to one another. 9. The whole is greater than its part, and equal to all its arts taken together. 10. Two straight lines cannot inclose a space. 11. All right angles are equal to one another. 12. If two magnitudes be equal, and one of them be reater than a third, the other is also greater than the third. 13. If two quantities be equal, and one of them be less lan a third, the other is also less than the third. 14. If there are three magnitudes, such that the first is reater than the second, and the second greater than the lird, jnuch more is the first greater than the third. 15. If there are three magnitudes, such that the first is !ss than the second, and the second less than the third, rach more is the first less than the third. 16. Through the same point there cannot be drawn two laight lines parallel to the same straight line without co- iciding. EXPLANATION OF SYMBOLS. . means angle. s „ angles. I „ right angle. '/.« „ right angles. = „ equal to. II „ parallel to. ^ „ triangle. ■ . „ inaDgles, j means parallelogram. I „ straight line. ::^ „ greater than. ^ „ less than. 4. „ perpendicular. „ because. ,, therefore. 0^8 w Phoposition I.— TnKoHBM. Tlie angles ACD and DCB, vrhicli one straight line, DO, makes with another. AB, on one side of it, are c;ithcr two right angles, or are together equal to two right angles. If the U ACD and DCB be equal, each of them is a j'i, (Def. 9.) ; but if they are not equali coDceive CE to be drawn ->- to AB, then the La ACE and ECB are two t'Ls; but tbe three is ACE. ECD, and DCB. are together = the two La ACE and ECB, and qIm to the two Is ACD and DCB; .-. the two is ACD and DOB are together = the two La ACE and ECB, but the two Ls ACE and ECB are two I'Ls; .: the two La ACD and DCB are together equal to two r'/.s. Q. E. D. Cor. 1. AH tbe angles that can be formed at the point C, in the straight line AB, on one side of it, are together equal to two right angles. For the L ACD ts = the two Ls ACE and ECD, K that tbe Eum of the Ls is not increased hj drawing the { EC, and in the same manner it ma;^ he shown the sun of the Ls would not be increased by drawing anj numbei of \s through the point C> Cor. 2. All the angles formed at the point 0, on tbe other side of tbe line, by any number of lines meeting ii C, will also be equal to two right angles. Cor. 3. Kence all the angles formed round a point ij any number of lines meeting in it, are together equal '' two right angles. SoBoL-iun. For tbe purposes of calculation, tbe ea- cumference of every circle is supposed to be dirided 3t)0 equal parts, called degrees, and each degree is posed to he divided into 60 equal parts, called min and each minute into 60 equal parts, called seconds. Tit grees, minutes, and seconds, are dislingulsbed by the fff" lowing marks : — 7" 3' 24", which is read 7 degrees, 3 Wl" DUles, aud 24 seconds. In the same manner all the are divided into the same numlier of di and seconds. Since then the circle entirely surrounds its centre, and is similarly sitn- Bted to it in every direction, tbe portion of the circumference intercepted between *ifo Jines druivn from tbe centre to the m/rura/erence, is the meaaute oS \\\e a."n^< s round about a, pMil , minul* ©• PliAKX OKQUBTRY, 85. t the centre; thus the angle AOB is measured hj the itercepted arc AB^ and the angle COB is measured hj le arc CB. Since all the angles round a point are (Cor. 3.) equal to- e^er to four right angles^ and also to 300*, the numerical teasure of a rignt angle is 90^. Proposition II. — ^Thborem. B; at a point B^ in a laight line AB^ two ther straight lines^ CB ad BD^ on opposite sides f AB^ make the adjacent Bgles ABC and ABD pgether equal to two \^t Angles^ these two traig^t mies are in one and the same straight line. For if BD he not in the same | with CB, let BE he in lie same | with it ; then since CBE is a |» and AB makeii J with it, the two Ls ABC and ABE are together = two *U; hut the two Ls ABC and ABD are also together = iro r^Ls hj supposition ; .*. the two Ls ABC and ABE re = the two Ls CBA and ABD; take away the common ngle ABC, and there remains the Z.ABE = the L ABD, be less = the greater, which is impossiMe ; .'. BE is not a the same | with CB, and in the same manner it can he hown that no | can be in the same | with CB, except BD, rhieh therefore is in the same | with it. Q. E. D. Proposition III. — ^Theoresi. If two straight lines, AB and CD, ^ !Ut each other in the point E, the ver- ical or opposite angles, AEC, DEB, ^^ -e iie equal. For the two Ls AEC and AED, fihich the | AE makes with the | CD, are together equal to :wo r^Ls; and the two Ls AED and DEB, which the DE makes with the | AB, are also together equal to two ^Ls; .*. the two Ls AEC and AED, are together equal to the two Ls AED and DEB ; take from each the com- mon Z.AED, and there remains the LAEC=DEB; in the same manner it may he demonstrated that the two U AED and CEB are equal. Q.. E. D. Proposition To make a triangle AEB, ivhose three sides shall be equal to the three given straight lines, AB, C, and D. From the centre A, with n radius equal to G, describe the circle EFII, (Post. 3), and from the centre B, with a radius equnl to D, describe a circle EGII; and from the — ^ ■ point E, where the circles cat each other, draw the |« AE and EB. (Post. 1), AEB will be the triangle required. Because AE is the radius of the circle EFH, and it wat described with a radius ^C, •'. AE is =C, and because BE is the radius of the circle EGII, and it was described with a radius =D, .-. EB is =D. Hence the a AEB has its three sides = to the three {s AB, 0, and D. Cor, I. If the \s C and D were equal, the triangle woall be isosceles, Cor. 2. If the given lines, AH, C, and D, were att' equalj the triangle would be equilateral. Cor. 3. If C- and D were together less than AB, tba circles would not intersect, and the construction would b« impossible ; hence any two sides of a triangle are together greater than the third, which is established in a di¢ manner in (Prop. 13.) PflOPOsiTioN V. — Theohku. If two triangles, ABC and DEF. have two sides, AC and CB, and the contained angle ACB in one, respectireljr equal to two sides, DF and FE, and the contained ai^ DFE in the other, the triangles are equal in every i^iped. For conceive the point C to be laid on the point F, and the |CA on the [FD, then, ■.■ these lines are =, the point A will coincide with the point .^ D. And since CA coincides with FD, and the iC is = the LF, the line CB wiO fall on the line FE ; and ■.• CB and FE are =, tdr point D will coincide with the point E. Htiice sinM (be pointf A and B coincide with the points D and E, the line AB will coincide with the line DE^ and the a iBC will coincide with the aDEF; .*• the two a« are squal, and have all the parts of the one equal to the cor- responding part^ of the other, namely the side AB = the tide DE, the LA = the LD, and the LB = the Z.E. Q. E. D. Pbo?osition VL — ^Theorbm. In any two triangles, ABC, DEF^ if two angles, A and B in the one, he respectively equal to two angles, D and E in the other, and the sides AB and DE, which lie be- tween these equal angles be also equal, the triangles are ^oai in all respects. For conceiye the point A to be lud on the point D, and the side AB on the side DE, then, *.* these Knes are equal, the point B will ^ / \; Br coincide vrith the point E. And » AB and DE coincide, and the AA is = the LD, the ride AC will coincide with DF, and for a like reason BC will coincide with EF. .•. since AC fails on DF, and BC im EF, the point C must coincide with the point F ; and •'.the two A^are in all respects equal, having the other ides, AC, BC, = the two, DF, EF, and the remaining LO s= the remaining Z.F. Q. E. D. Pboposition VII — Theorem. In an isosceles triangle, ABC, the iDgles A and B, opposite the equal does BC and AC, are equal. For, conceive the Z.C to be bisected )y the line CD, then the two a« ACD Bid BCD have ACzzBC, and CD iommon to both, and the LACD = ^' ie ABCD ; .*. they are equal in every respect, (Prop. 5), uid have the Z. A = the LB, *.• they are opposite to the »mmon side CD. Q. E. D. Cor. 1. If the equal sides AC and BC be produced to B and F, the angles E AB and ABF on the other side of ^e base will also be equal. For, the two Ls CAB, BAE, are together = two /Ls, iProp. 1), and the two Ls CBA and ABF are also toge- ler = to two t^Ls; .•. the two Ls CAB, BAE, are =: the ^0 Ls CBA, ABF, (Ax. 1) ; and it was proved i\i^\ l\v<^ nf pt iN^' «ft9uiiMr. » id CAB and CBA are equal ; .-. the angles EAB a ABF are aha equal, (As. 3.)- Cor. 2. The line tliat liisecls the vertical angle of I iaoscelps triangle also bisects the baae. and cuts it at * anftles. For, the as ACD and BCD w^re shown to be e in every respect ; .-. AD is = BD, and iADC is = BD( and tbey are adjacent Ls ; hence (Def. 9) each of them ; a right angle. Cor. 3. Every equilateral triangle is also equiangular. pKOPOSiTioN VIII, — Theorem. If two angles, CAB and CBA, of a triangle be equal, t sides, CB and CA, opposite them nill also he equal. For, if AC be not = CB, let AC ^ be ::^CB, and let AD be the part of AC that is = BC, join BD, (Post. 1 ) ; then the ah ADB. CBA, have AD =CB, and AB common; and the £DAB contained by the two sides of the one is = the iCBA contained by the two sides of iheothec; .-.(Prop.C) Ai__ the aABD is = the aABC, the less := the greater, which is impossible ; .-. AC is not :?^I and in the same manner it may be shown it ib not k Lence AC is = BC. Q. K ? Cor. Every equiangular triangle is also equilaleraL Proposition IX. — Theoeksi. If two triangles. ABC and DEF, have Ibe three ndca' i respectively equal to the three sides of the otT ingles shall be equal in all respects, and haTe d equal, that are opposite to equal sides. Let ABC and DEF be two having AC=Dr, CB=FE, and AB=DE, and let AB and DE be the sides which aj-e not &-■ less than any of the olhcis. Con- ceive the side DE to be applied to the side AB, so that the point D roay coincide with A, and Ihc Uae DE with AB, then the point E will coincide with] jDE is = AB ; but \et t\ie levte^^ ^a\\ va vlw Q^poii rtion, from C as at G, jtim GC -, \Ui:Ti-.- CV,vfcT;-1! .A PLANS OEOMBTRY. 89 and FE is = "BG, being the same line in a different posi- tion, CB is =BG, (Ax. 1); .-. the Z.BCG is = LBGC, (Prop. 7); again, ••• AC is=DF and DF is =AG, .-. AC is =AG-, and hence the L ACG is = the Z. AGC ; .-. also the whole AACB is = the LAGB, (Ax. 2); but LAGB is the LDFE in a different position; /. the LACB is =: the LDFE ; and since AC, and CB, and the LC, are re- spectively = DF and FE and the LF, (Prop. 5), the LA. is = the LD, and the LB is =: the Z,E. Hence the angles are equal that are opposite to the equal sides. Q. E. D. Cor. 1. The areas of the triangles are also equal. Cor. 2. To make at the point F, in the straight line DF, an angle equal to ACB. Construct a A^FE, (Prop. 4.) having its three sides equal to the three straight lines AC, CB, BA, namely FD=AC, FE=CB, and ED=AB; the LDFE will be = the LACB by the pro- position. Proposition X. — ^Theorem. If a side BC of a triangle ABC be produced to D, the exterior angle A CD" will be greater than either of the in- terior opposite angles CAB or ABC. Conceive AC to be bisec- ted in E, and BE joined, and the line BE produced to F, so that EF may be =BE and join FC, and prodace AC to G ; then the A« AEB and CEF have AE and EB, and the contained LAEB in the one = CE and EF, and the contained iCEF in the other.; .-. (Prop. 5) the Z.EAB is = the iECF; but LACD is greater than LECF ; .-. (Ax. 13) the Z,ACD is greater than the ABAC. If the side BC were bisected, and a similar construction made below the base, it might be shown in the same man- ner, that the iLBCG, which is = the LACD, (Prop. 3), is 7^ the LABC; /. the LACD is ;:^ either of the LsCAB or ABC. Q. E. D. Cor. Any two angles of a triangle are together less than tiFo right angles. For the LACD is -z^ the LBAC, and if the LXCB \i^ added to each, the two IsACD and ACB (thatia, Ivjo ^« Prop. l.)are:^ the LsBAC and ACB ; and l\ie sam^ ^Jght be shown of any other two angles. PnoPosiTtoN XI. The grealer side of e?erj triangle has the greater angle opposite lo it. Iftheside ACof the AABC be :^ the side AB, tlie i.ABC willbe^tiie i.ACB. For make ADrrAB, anil join BD. then tlie /.ADB is = the Z,ABD(Prop.7); but the /.ADB is z:^ the iACB (Prop. 10) ; .-. the i.ABD is z^ tbe IaCB (As. 13), still - whole /.ABC z:^ the Z.AC'B. Q. EB the graatei Cor. The greatest aide of any triangle hai angle opposite lo it. Proposition XII Theo«em. Ifthe angle ABC of the triangle ABO be greufer llian the angle ACB, the side AC opposite the grealer angle will be greater than the side AB oppnsile llie Ices. ■_ Or the greater angle of every triangle has the greater ai4f opposite lo it. For if AC he not z^ AR, it most either be =: it less; AC is not = AB, for then the i.B would be = i LQ (Prop. 7). which it is not; neither is AC -:iAB,J then the IV, would be ^i: the Z.C, nhieli it is not {PB II)..-. ACisi^AB. Cor. Tlie greatest angle of eyery triiingle has tbe gresh Bide opposite to it. pKOPosiTioN XIII. — TnaoaBM, In any triangle ABO, the snmofany two of its aides, as AB and AC, is greater than the Temainiog side BO. Produce AB to D, so that AD may be = AC, and join DC; then ■.■ AD is = AC (by Const.), the L ACI> is=the i. ADC (Prop. 7), .-. in ihe AOBC tbe i-BCD is z:^ Ihe /.BDC. hence the side BD is::^ tbe side BC (Prop. 12); but BD is = AB and AC, •■■ i. is = AC, .-. BA and AC are together zp- BC (As. 12.) Q. E Cor. Tbe differenee of two sides ofa triangle is leMtl the third side. For,sinceAB and ACare:^BC,ifACbc taken from« there remains AB z^ the ditfctence of BC and AC, (Al. FLAKE GEOMETBY. 91 1-E Proposition XIV. — Theorem. If two triangles, ABC, DEF, have two sides, AB, BC, of the one respectively equal to, DE and EF, two sides of the other, hut the angle ABC, included hy the two sides of the one, greater than the angle DEF, includ- ed by the corresponding sides of the other; then the side AC is greater than the side DF. Let ABG be the part of the Z. ABC, which is z= DEF, and let BG be = EF or BC. Then the A^ ABG, DEF, are equal in all respects, (Prop.5), and have the side AGr= DF. And as BC and BG are =r, the L BGCis = the L BCG (Prop.7); but the Z.BCG is^^-theZACG, .-. also the LBGC i8:7-the LACG, (Ax. 12); much more then is the LAGC T^ the L ACG, and hence (Prop. 12), the side AC is 7:^ AG, and .-. also z:^ its equal DF (Ax. 13). Q. E. D. Proposition XY. — Theorem. If two triangles, ABC, DEF, iave the two sides AC, CB of the one respectively equal to two sides DF, FE of the other, but the re- maining side AB of the one greater than the remaining side DE of the A.^ other; the angle ACB will be greater than the angle DFE. For, if the L ACB be not ^^ the ZDFE, it must either he equal to it or less; the ZACB is not ■= DFE, for then (Prop. 5), the base AB would be = DE, which it is not ; neither is the ZACB .^ the ZDFE, for then (Prop. 14), the base AB would be ..^ the base DE, which it is not, .-. the ZACB is z;^ the ZDFE. Q. E. D. Proposition XVI. — Theorem. If a straight line, HF, fall upon two Other straight lines, AB^ CD, and make the alternate angles AEF, EFD, equal to one another, the straight lines AB and CD are parallel. B D 92 PLANE GBOMETBY. For, If AB and CD lie not || , they will meet ^hea p duced, either towards B, D, or A, C; suppose that the meet towards B, D, in the point G, then EGF is a ^, an itseilerior /.AEF is ^:^ the iEFGyrop. JO); but Z.AK is = /.EFG, .■. AB and CD do not meet towards B, T and in the Siime inauuer it mny be shown that they do n meet towards A, O, .-. (Def.-37), AB is parallel to CD. Q. E. I Cor. 1. If the esterior anffle HEB be equfd to the ii rior angle EFD, AB is jiarallei to CD. ^_ For, since iHEB it = i-EFD, and also to Z.AEi (Prop. 3), .-. /.AEF is = Z,EFD, and they are alteiaat angles; therefore (Prop. 16), AB is parallel ti Cor. 2. If the two angles BEF and El'D be togethi equal to two right angles, AB and CD are jiarallel. For, since the two U BEF and EFD are = two i*.U by (SopO. an* Ilie t™" ^* BEF and AEF are tngelher = two r'L» (Prop. I), the two U BEF and EFD are =: ' the two Ls BEF and AEF, and takinc; away the comBU angle BEF, there remaina the iAEF = Ibe iEFD.JU^ they are alternate angl^; hence AB is parallel to "CD tProp. 16.) Proposition XVIT.— Thkorem. If two parallel straight lines, AB ir. and CD, be cut by another line EF, in the points G and 11, the alternate angles AGH and GHD wilt be tqual, lh(^ exterior angle EOB vrili be equal to the interior opposite angle GHD; and the two interior angles BGH and GHD on the Mil aide of the line will be together equal to two right angl« If the ZAGII he not = the ZGHD, let LG be d«fH making the /.LOU = the Z.GHD, and produce LG to St. ■.■ the LUiU. is = the ^GUD, and they are altenuW Id, .-. LM is II to C D, (Prop. 1 6). But AB is given || toCD- .'. through the same point G there are drawn tw straight lines LM and AB, {| to the same straight line CDi whicli is imposBihIe. (Ai. 16.) Hence the Z.AGH. " unequal to the ZGIID, tbnt is, it is equal to it. Then, sinc-e the /.AGH is = the lGIID, and also t» the iEGB (Prop. 3), the /EGB is = the lGUD. .: t eilerior Is equal to the interior opposite angle on the sal jade of the line, UAgitiD, since the ^EGB is = the Z.GIID, to eacb e equals add the i.lJOH, Xben ■will the two It ^ t\vo i, BGII and GIID ; bat the i are t-.^-ether = two t'/b (Prop. 1) »d GIID are togetiier equal to two right h,(v.^. , Q. E. J>. -""■■"gilt lines. LM and CD. being cut by a ■* "le t«,o interior angles LGII and GHC, « «ae of iiig (,^nj^g |jj^^_ logelher less than two ™?e two lines will meet if produced far' " "(le where the angles formed are less than. y do not meet on tha* side, they must eitbi f will meet being produced on the other side , \, ftr then the two La LGII and GHC would = two rL't, whith they are not: neither do g produced towards M and I>, for then the E and GHD would be two /,« of a A. !"»d .-. " > (Prop. 10); but the font Lf LGH. GHC, 1), are together = fonr I'Ls (Prop. 1). of , I.GH and GHC, are together leas than i two, MGH aud GIID, are ^^ two I'Ls, t CD do not meet towards M aud D, aud » been shown that tlwy are not 1|, they must (rartia L aud C. Q. E. D., pBOPOsiTiON XVIII, — Theori A a poi"* A to draw t line parallel to a given ine BC. lake any po'"' L>, jmn at llie iioiiit A make ] P^o 9). tIi«^D'\'i = C ^d prodLiue KA to F, ■.- the iEAD = thj bid they are alternate angles, .•■ (Prop. ]6J,^ I bence through the p een dravvn II to BC. e BC of o BC be P"""" ). the e»te- ACD "i" ** ana tl.e ighC Un^^ ■ angles oi ^ Je are togethei c(iu,a\ lo t^io u^ ^\\\. ■Mi'^fi. Throngh C draw (Prop. 18), CE || to AB. Then V Vis II to CE, and AC mecls tliem, Ihe alleraale /.sBACa ACE arc= (Pror- 17); and .' AB is 1| to CE, and I fells upon lliem, the exterior /.ECD is = the interioi ABCbuttheilACEis^ the iUAC,.. the whole iAC is = the two li CAB and ABC. To each of these equals add the ZACB, ,■. (be two ACD, ACB are = the three la CAB, ABC, and BC but the two Z» ACD, ACB are together = three angles CAB, ATl^;, and BCA arc togetlier equal two right angles. Cor. 1. If two angles in one triangle be equal to t angles in another triimgle, the remaining angles ofthi triangles are equal. Cor. 2, If one angle in a triangle he equal to an aiq in another triangle, the sum of ibc remaining angleS' each triangle are'equal. "■■ Cor. 3. If one angle in a triangle be a riglit angle, 1 l^her tTvo angles are together cqnal to a right angle; il \ence each of them is an acute angle. I Cor. 4, Every triangle has at least ttvo acute angles. > Cor. 5. Hence from this proposilion, and (Prop. S), 'angles hare two angles iu Ihe one equal to tf ingles in the other, and it corresponding side equal id eac e equal in al! respects. Prop( >; XX.— Thi If two lines AB and BC meeting in a point B, be respcc- ti»rly parallel to DE and EP meeting in a point E, the included angles ABC and DEF are equal. For join B, E, and produce BE to G. Since AB i* >E, and GB falls nn them, the Z,<'IiA is = the /.OB ~ op. 17): foralikereasonthe ZCBG is :r the ZFH aking equals from equals, there remains the iABC LVEF. Q. E. pBOPosiTioN XXI. — ^Theorem. Pill figure he produced, I so forjued will be togelli buaJ to four right angles. FLANE GXOMBTJIT. 9i Take any point o, and draw oh \\ to EK, oM II to AF, oN II to BG, oP II J to CH, and oQ || to DJ. Then since KA and AB are re- spectively II to ho and oM^ the La is = the La' (Prop. 20) ; for a like rea- fton the Lb is =z the Lh\ Lc =z L<f, Ld = Ld\ and Le = U\ .'. the sum of an Ae Z« a, h, c, </, ^; are equal to the wm of all the Z# a', 6', c', d\ and «' ; but the Ls a\ h\ c\d\ aad t' are together equal to four 7^ is (Prop. J, Cor. 3), .-. all the exterior Ls a, h, c, d, and e, are together equal to four right angles. Q. E. D. Proposition XXII. — Theorem. All the interior angles of any rectilineal figure are to- gether equal to twice as many right angles as the figure has sides, wanting four right angles. For (Figure to Prop. 21) erery interior Z.EAB, together with its adjacent exterior LBAK, are together equal to two ^Ls; ,\ all the interior, together with all the exterior, are €qaal to twice as many /Ls as the figure has sides ; but ail the exterior Ls are = four r^Ls, (Prop. 21) ; .-. all the interior are = twice as many right angles as the figure has sides, wanting four right angles. Q. E. D. Cor. 1, All the interior angles of any quadrilateral figure are together equal to four right angles. Cor 2. If the sum of two angles of a quadrilateral figure be equal to two right angles, tlie sum of the remaining angles is also equal to two right angles. Proposition XXIII. — Theorem. Of all straight lines, drawn from the point A to the straight line BC, the perpendicular AD is the least; AE, which is nearer to the perpendicular, is less than AF, ,which is more lemotej and there can only be Atclyjiv \.nso L 9S VEotMB GSOMKVKT. equal strtiiglit lines, as AE and AD, one on each side of the per- pendicular. Since AD is -u Ic CU, the LADE is a y'L; .-.the /.AED is ^ a 7'L rProp. 19 Cor. 3); and .-.the side AD is ^: A E. Again, since tl.e Z AED is ^::r a I'l. the AEFia^a^'i, (Prop. 1), but tlie iAEU is ^ the L] AFE, (Prop. 10); .-.the ZAEF is p' the ZAFE, anil .'. the side AF is ^' the side AE. In the same it mny be shown, that AC is z^ AF. Again, if DB b» = EIJ, ABwillbe=AE; forinthe two A« ADB ' ADE, the two sides BD and DA are = the two sides and DA, and they contain equal is, for each of tbem r-Z; therefore AE=AB, (Prop.5): and besides AB,lhB« cannot be drawn any other line from the point A^AE|. for if it were nearer to the -I- it would be less, and if mo» remote it would be greater. Q. E. B Proposition XXIT. — Theorem. The opposite sides and i], angles of any parullelo- gram are equal to each other, and the diagonal divides it info two equal triangles. That is, DC=AB, BC = AD, ZDAB=ZDCB, ZADC=ZABC, and AADB=ADCR ■■■ DC is II to AB, and DB meets them, the ZCDBii the ZABD, and :• AD is || to CB, and DB meets thelii the ZADB is = the ZCBD, .-. in the two A« ADB ani DCB, there are two Z» CDB and CBD, respeclively = two Is ABD and ADB in the other, and the uAe Dft Ijinp between the equal angles, is common to bodl. .-. (Prop, (i) DC is = AB, AD is = CB, ZDCB w=i DAB, and the ADCB is = the A^AD : also equal Z^ ADB and CBD, there be added the equal it CDB and ABD, the whole ZADC will he = the whoh ZABC. Q. E. D. Cor. I. The lines which join the extremities of two equal and parallel straight lines towards the same pEii1% are ibemsvives equal and parallel. For if CD and AB be equal aai ^aro\W, the A» CDB and ABD have two sides, CD and DB of (he one = fwo AB and DB in the olher, and Ihe contained Z.CDB is = the contained /ABD. (Prop, 17), -■, the base CB is = the base AD, and the iCBD is = the iADB, (Prop, 5), .-. AD and CB are ||, (Prop. 16,) Cor. 2. If one angle of a parallelogram be a right angle, all the other angles will he right angles, and the figure niU be a rectangle. Cor. 3. Any two adjacent angles of a parallelogram are together equal to two right angles. PaoPoaiTioN XXV.— TnEOiiK: Every quadrilatpral ABCD ivhich n has its opposite sides AB and DC equal, and also AD and CB equal, is a parallelogram, or has its opposite , sides parallel. For join DB. and then the two As DCB and BAD have fte side DC=AB and CB=AD, and the siiilt- DB com- mon ; .-. (Prop, 9) the A" '"'^ equal in all rt^pecls, and , have the Ls equal that are opposite to equal sides; .'. the LABD is = the Z.CDB. and hence DC ia || to AB; also I the /.CBD is = the Z.ADB, and hence AD is || to OB, (Prop, le), and ABCD is therefore a parallelcgram (Def. 38). Q, E. D. P pROPOsJTroN XXVI. — Theorem. A B Parallelograms. DABC and FABE, Mid also triangles, CAB and FAB, ' itonding on the same base AB, and be- tween the same parallels, AB and DE, ue cqnal to each other. For, since DA and AF are re5pectively || to CB and BE, the i. DAF is = the Z.CBE (Prop. 20), also since ED Mb on the two \\b AD and CB, the Z.FDA is = the iECB, (Prop. 17); .■- the A« FDA and EOB hare two ^in the one = two Ls in the other, and the aide AD = lie corresponding side BC, (Prop. 24); .-. the A^'DA is = tlie AECB, (Prop. 6); if now each of these = Ashe liiien separately from the whole figure DABE, there will ttmaia in the one case the iZZ?ABCD, and in the other 4b £:Z7FABE, these i — 7 x are .-. equal, (Ax. 3); and since ttie ACAB is half of the one. (Prop. 2-4), and l\ie C^fiiiS lulf of the other, theaa ^4 are also equal, (,Ait. T)- i w Proposition XXVII. — Theorem. PamHeloRranis.AnCDandEFGII, and also triangles, CAB and EFG. standing upon equal basps AB and EF, and lying between ihe same pa- rallels AF and DG^, are equal to each other. For, since AB is = EF, (Hyp.), and EF is = HQ (Prop. 24), .-. AB is = HG, and Bince AB and HQ at joined towards the same parts by AH and BG, ■'■ ABGl is a CZ7. (Prop. 24, Cor. 1), and it is equal to (b ZZZ7ABCD, -.■ Iliey are on tlie same base AB, and b tween the same ||s AB and DG, (Prop. 26); it is si equal to the CZJEFGH. ■-■ tliey are on the same ba«eH{li and between the same |ls HG and AF, (Prop. 2ti), .-.th* djABCD is = the C^fEFGH ; again, since the A* AW and EFG are halves of these equal [=js. (Prop. 24), tiu AABO is = the AEFG, (Ax. 7-) Q. E. D. Proposition XXVIII. — Theorem If a parallelogram ABCD, and a triangle EBC, be upon file same Lase BC, and be- tween (be same parallels, the parallelograo) is double of the triangle. u C For, join AC, and tbe AABC is = the AEBC, (Prop. 2r>), but tlie c^ABCDis double of tl AABC (Prop. 24), .-. tlie cuABCD is also double Ihe AEBC, (Ax. 7). Q- E- ^■ Cor. If ihu base of a parallplogram be half thai. of' triangle, and they lie between the same parallels, tlietl nJlelugram will be e<^ual to the triangle. Proposition XSIX. — Theorem. Equal triftngles, ABO and DEF, up- a the same side of equ-.il bases. BC and EF, that are in tbe some straight line, / \ 1 '^• between the samo paralkU, UF Z \ , ,1 .l. .A and AD. ^ C e I For, if AD be not || to BF, let AG be drawn throoithj II to BF, and join GF. then the A« ABC and GEF « upon = bases BC and EF, and between t]ie same ||*Bi and AG, .-. the AABC is = the AGEF, (Prop. 27J. bi the AABC is = tbe ADEF,(Hyp.); .-.tbe AGEFis: ihe ADl^. the less = the greater, whieh is imporaibl '■ AG is not II to BG, and ihe sainft 'm&'ij be sbown vf u| PLANE OROAIETBY. 99 other line imssiiig through A^ except AD, Ivhich therefore 18 II to BF. • Q. E. D. Gor. 1. Eqnal triangles on the same base and on the same side of it, are between the same parallels. Cor. 2. In the same manner it might be shown that equal triangles between the same parallels are upon equal bases. Cor. 3. Since parallelograms are the doubles of triangles on the same base, and between the same parallels, (Prop. 28). this proposition and its corollaries are also true of parallelograms. Scholium. The preyious four propositions, with the corollaries of the last two, are also true, if instead of the words *' between the same parallels," we substitute " hav- ing the same altitude," for the altitude is the perpendicu- lar distance between the parallels, which is everywhere the ftune, since two perpendiculars would be opposite sides of a parallelogram, and therefore equal, bj (Prop. 24). Proposition XXX. — Theorem. If two triangles, as ABC and DEF, have two sides AB, BC of the one, equal to two ttdes DE, EF of the other, and the con- ^ E tained angles ABC, DEF together equal to two right angles, the triangles are equal. For, conceive the point C to be applied to F, and the line CB to FE, the point B will coincide with E, and lot A fall as at G, •/ the La ABC, (that is, GEF), and DEF are to- gether = two /Is, GE and ED are in the same straight line, (Prop, 2). Again, *.• GE and ED are equal, the A« CrEF and DEF' are equal, (Prop. 27). Q. E. D. Proposition XXXI. — Theorem. Parallelograms ABCD, EFGH, are equal, which have two sides AB, BC of the one, equal to two aides EF, FG of the other, and the contained angles ABC, EFG also equal. For, draw the dijigonals AC, EG, the A« ABC, EFG have the sides AB, BC, and the xmtained L ABC in the one = the sides EF, FG and con- ained Z.EFG in the other, ,\ the As ABC, TEEG ^\fe ' an equal in al! respects, (Prop. H); consequently the ABCD, Et'GH, which ate doubles of these A«. (Prop. 24 are also equal to each other, (Ax. (i ) Q. E. Cor. 1. KectanglcB contained by equal Btraigtit Iji " e equal to each other. Cor. 2. The squares JeGcribed on equal lines are eqi each other. PaopoBiTioN XXXII. — Theorem. n If ABCD he a parallelogram, and FH and GE parallelograius about its diago- nal DB; and CO and OA the remain- ^ ing parts which make up the whole figure, and are therefore called comple- ments; the complements CO, OA are ■^ equal to each other. ^^ For, since a i::r7 is bisected by its diagonal, the ADCI is = the ADAB, the A^GO is = the ADKO, andtl AOFB is = the AOHB, .-. the two As HGO and OF are together = the two A« I>EO and OHU; taking thei pquals from the equal A" DCB and DAB, there remi' the complement CO = the complement OA. Q. £, Qef. A quadrilateral £gure which has two of its H parallel, but the remaining sides not parallel, is called trapezoid. Phoposition SXXIII.^ — Theobem. A trapezoid ABCD is equal tallelogram of the same base is equal to half the lallel sides AB and DC. Join DB, then the AADB is = a i^^ haTingthesM altitude, and its base = one half of AB. (Prop. 28, Ca also the AEDC is = a i — 7 having the same ijlitude,* its base = one-half of DC, (Prop. 28, Cor.) .'. the wbt figure ABCD is = a iz:d n-liose base is half the 8un«. AB and CD. and its altitude =. the distance bctweeof parallels AB, DC. Q. B. ; Def. A rectangle is said to be contained by two of i adjacent sides. Def. The rectangle contained by two lines, is a tangle nhich has these lines, or lines equal to them, fi two adjacent sides. Def. The rectangle contained by two lines AB and C written for brevity thus: the rectangle AB-CD, e Jeeci'ihed on a line AB, is ^suVVea AB*. is equal to a pn- n i ahiiude, whose /\ I le sum of its pa- / N.I • Ai J( PLANE GEOMETRY. 101 Proposition XXXIV. — Theorem. The rectangle contained by two lines, A6 and AD, is equal to the several rec- tangles contained by AB, and the parts ISLE D ] ? E AlF, FE, ED, into which the other AD B & u c is divided. Make the rectangle AC, having AB and AD for its i^acent sides, and through F and E draw FG, EH || to A.B, then each of the figures, AG, FH, EC, as also AC, is a rectangle, and the three, AG, FH, and EC, are toge- aier =AC. But AG is the rectangle AB-AF, FH is the rectangle AB-FE, for it is contained by FG and FE, and FGis=AB; and EC is the rectangle AB*ED, for it is contained by EH and ED, of which EH is =FG= AB, (Prop. 24) ; and it is evident that the whole is the rec- tangle AB-AD ; .-. the rectangle AB-AD is =AB-AF4- iBFE+ABED. Q. E. D. Proposition XXXV. — ^Theorem. If a straight line, AB, be divided into two a c B arts in the point C, the rectangles AB*AC ~ f AB'BC, i^hall be equal to the square of IB. For on AB, describe the square ADEB, ffld through C draw CF || to AD or BE, ben each of the figures, AF, CE, is a rectangle, and they le together =AE. But AF is the rectangle AB'AC, for t is contained by AD and AC, and AD is =AB, being ides of a square ; also CE is the rectangle AB-BC, for it is ontained by EB and BC, and EB is =rAB, being sides of isquare; .*. the rectangles AB'AC+AB'BC, is =AB^. Q. E. D. Proposition XXXVI. — ^Theorem. If a straight line, AB, be divided into wo parts in the point C, the rectangle )ontained by the whole, AB, and one of he parts, BC, is equal to the square of hat part, BC, together with the rectangle iC'CB, contained by the two parts. Let CE be a square, described on CB, produce EF to [), and through A draw AD || to CF, then AF and AE ire rectangles, and AE is =CE+AF. But AE is the rectangle AB'BC, for it is contained by AB, BE, and BE a zrCB, hewg sides of a square ; CE is bj coTV&\.i\xc.\AaTL the squiire on CB, anil AF is the rectangle AC-CB. for is contained by AC-Cl", and CF ia =CB, being sides of square; .-. the rectangle AB-BC ia =CB*+AC-CB. Q. E. Cor. Uence also the rectangle AB-ACz= AC + AC-( Phoposition XXXVII.— Thbobbk. The square of tlie sum of two lines. AC, _ CB, ia equal to tlie sum of the squares of the linos, together with tivice the rectangle contained hy the lines. That is, AB''= AC-+BC+2ACCB. For AB3=AD'AC+AT?-BC. (Prop. 35.) L and AB-AC=AC'' + ACCR.l ,p „„ . I also AB-BC=BC^ + AC-CB. f U"p.Jt..J P .-. AB«=;AC^ + BC«+2.\C-CB. (Ax. 2.) ' The diaKRim shows how the Bqnivvo described upon i.^— sum of the lines may he divided into the several sqiun and rectangles. Cor. 1. If AB be considered as a line divided into ti parts in the point C, the proposition becomes the followia If a straight line he diviJed into two parts, the squinvi the whole line is equal to the sum of the squares of W parts, together nith twice the rectangle contained by a Cor. 2. If AB and CB be equal, it is plain thai til. rectangle AC'CB will be a squ.irc. IIi?nce tlie squareH a line is equal to four times the square on half tiic line. ■ Proposition XXXVIII. — Theorem. P The square of the difference AC, of two lines AB, BC, is less thnn the sum A ? i of their squares, (AB' + BC'^), by twice the rectangle contained by tiie lines AB, BC. Let AB he one line, and BC another, then AC ia thiil difference, and .'. from (Prop. 37, cor.) AB«=:AC''+CB''+2AC-CB. To eacU of these equals add CB". .-. AB'' + CB'=AC« + 2CB' + 2ACCE. (Ax.a) But 2ABBC=2CB« + 2AC-CB. (Prop. 36.) Take these equals from the former, and ne have AB"+CB»— 2ABBC— ACS. (At. 3.) Hence the squnrc of the difference is less than theiun of the squares by twice the lecUn^Ve. Q. E. tt PLAKX OBOMETRV. lOS Dor. 1. The square of the sum, te- ther with the square of the difference - — ? 1 ^ two lines, is equal to twice the sum the squares of iSie lines. Let AB be one line, and BC another, then AC is their m; make BD=BC, then AD is their difference. AC«=AB« + BC«+2AB-BC. (Prop. 37.) id AD«=AB« + (BC«=BD2)— 2ABBC. (Prop. 38.) iding these equals together, we obtain iC« + AD«=2AB« + 2BC«. Cor. 2. If a line AB be bisected 1 C, and divided unequally in D, the '^ 1 — ? — ? am of the squares of the two unequal arts, AD, DB, are together equal to twice the square of lalf the line AC, and twice the square of CD, the line be- ween the points of section. For AC may be considered as one line, and CD as ano- ber, then AD will be their sum ; and since CB is = AC, )B will be their difference ; .-. AD^ + DB^ is =2AC2 + CD«. (Cor. 1.) Cor. 3. If a straight line AB, be iiected in C, and produced to D, the "^ 5- |oare of the whole line AD, thus pro- nced, and the square of BD, the part loduced, will be together equal to twice the square of alf the line AC, and twice the square of CD, the line Mide vLp of the half, and the part produced. For AC can be considered as one line, and CD as ano- her, then AD will be their sum; and since AC=CB, \D will be their difference; .-. AD^+DB^ is =2AC« |.2CD«. (Cor. 1.) Cor. 4. The square on the sum of two lines is greater ban the square on their difference, by four times the rec- tngle contained by the lines ; for the square on the sum f the lines is greater than the sum of their squares, by mce their rectangle, (Prop. 37), and the square on their lifierence is less than the sum of their squares by twice keir rectangle. (Prop. 38.) Cor. 5. The sum of the squares of two lines, AB, BC, B eqtbl to twice the rectangle contained by the lines, to- [ether with the square of their difference. For (Prop. 38) AB^ + BC^— 2ABBC=:AC2; add to ach of these equals 2AB'BC, and we hare AB2 + BC^= JABBC + AC«. Cor. 6. The square on the sum of two lin^B, K^^^C, I Fot mis equal to four times the rectangle contained by<tlie Un , together with the square of the difference of the lines. Far adding 2AB-HC to both sides of the result in Cor. gives AB«+BC' + 2AB-BC=4ABBC + AC«. The first of these equnls is (Prop, 37); the square I scribed on a line AD, (Cor. .5), which is the sum of snd BC, and the second is four times the rectangles tained by the lines, together with the square on , (Cor. 3), which is the difference of the hnea- Cor. 7. If a straight line AB be bi- ■ected in C, and divided unequally in D, ■ " the rectangle AD-DB, together with the N N of section is equal to the square on (AC or CB) half the JIi For AB"=AD3 + DB^+2ADDB. (Prop. 37). but AB«=r4AC'', (Prop. 37, Cor. 2). and AD^ + DB^— 2ACJ+2CD".{Prop.38,Cot. Substitute these Talues of AB" and AD^+DB' in first, it becomes 4AC«=2AC''+2CD»+2AD-DB. Talte 2AC" from each of these equals, and take halves of both sides, and we have AC"=CDi' + ADDB. If a straight line AB be W- , ^ Bected in C, and produced to D, the 1— rectangle AD-DB contained by the whole line thus produced, and the part produced, together " : on CB half the line, will be equal to the sau on tbe line CD, made up of half the line and part produced. For ADDB=ACBD + CB-BD+BD=, (Prop. 34,) but since AC=CB, substitute CBBD for .\C-Bft- and add CB' to both sides, and it becomes AD•DB+CB^=2CB■BD+BD^+CB■'=CD^(P^op.37 Scholium. The last four propositions, with their co» lories, are also true arithnielically or algebraically, if pi dnct be substituted for rectangle, number or quantllj line, and increased for produced, and the student BflO prove iheir truth arithmetically by taking a number * subjecting it to ihe processes described in each of the M positions and corollaries, and proving the idenlilyDt t results. The following are the algebraic processes tlui equivaleot to these propositions: — Let AC=a, and CB=6, then AB=a+t. (Prop. 35.^ a{a+h) + b(a+h)={a+b)''. .(Prop. 36.) a{a+b)=a''+ab. miProp. 37.) (a+b)3=;a'-V4abJrl>''. rLAm oxomTBT. 10r> Cot. t. The ume aJgebnucallj as tbe proposition. Cor. 2. (2a)''=4a''. (?^o^88'.) {a— S)»=o'— 2o6+6» |aB=o, CB=6, AC-Co— i)}. Cor.I. {fl+ft)«+(i.— 6)»=2i^+2i' {AD^(o— 6).AC=(o+i)}. Cot.*. («+fr)»-{ci-4)«=*n6. Cor. S. «»+i»=-2D6+(fl— *)*. Cor. 6. («-i-ft)'-4o6+{a— iV. Cor. 7. (a-t-iXa— t)+i'=o* AC or CB=a, CD_b. Cor. S. (2a+i)t-l-a*=i(a+&)* AC=CB=a, BD-b. Cw. 9. The rectangle under the sum and difference of two ILnes is eqoal to the difference of their squares. For ADDB+CB'=CD', (Cor. 8), take CB' from both Biei,wd we hare AD-DB=CD'— CB'; now AD is th« nnof CD and CB, for AC is = CB, and BD is their diflcrence. Jn the aboTe it will be remarked that several of the co- nflaries assnme the same algebraical forms; this arises fana the different wa^a in which a line made np of two DBrtsmaybe considered: either the parts of the line may te considered as gepamte lines, and the whole line as their nun, or the whole maj be considered as one line, and one if the parts as another, then the remaining part will be the Itflerence of the two lines ; or when a line is divided into IWD equal and also into two unequal parts, half the line nay be considered as one line, and the distance between lie middle of the line and the point of unequal section as mother, then tbe greater of the two unequal segments wiU w tbe sum, and the lets will be tbe difference of the two PnoPoaiTioN XXXIX. — Theorrh. The square BQ described on t ^potennse BC of a right-angled t ogle is eqnal to the sum of the ^v iqoares LB and KC described on AB IM AC, the sides that contain the ight angle. Draw AE ]| toBF or CG, join AF, HO, AG, and BI, '.• the Li LAB and BAC are t'U LA and AC are in the nme straight line, (Prop. 2); for a like reason KA and AB tre in the same straight line. Again, each of the Is ABH, DBF are r*/*, being Zsinasqnate; to each add the ^ABC, dien the whole ^HBC is s: (he Z ABF; also A&-^%,«iu^ BC=BF. .-. HB, BCare = AB, BF.andiHBCliasIie proTed = the ZABF, ■. the AHBO is = the AAB (Prop. S). But the square LB is double of the AHB and the rZjBF. is double of tiie AABF, (Prop. 28) the square LB is = the cz;BE, (Ai. 6). In the s mauncr it caa bo shown that the square KC is = r — j CK, .'. the whole square BG is ^ the two squares I and KC, that is, the square on the hjpotemiee BC is : the sum of the squares on AB and AC, the sides that ci tain the right angle. Cor. 1 . Hence the square of one of the sides of a rig angled triangle is equal to the square of the hjpotonnse minished by the square of the other side, or equal to ( rectangle under the sum and difference of the hypotenn and the other aide, (Prpp. 3S, Cor. 9), which is thns e pressed, AB'=BC'— AC"-{BC+AC)(BCU-AC). Cor. 2. If the hypotenuse and one side of a, rightaoj^ triangle be respectively equal to the hypotenuse and a liii in another, the remiuning side of the one is equal to thel maining side of the other, and the triangles are equd i eTery respect. Pbopobition XL. — Thborem. In any obtuse angled triangle, ABC, the equate of the side AC subtending the ob- tuse angle ABC, is greater than the sum of y" the squares on the other sides, .AB, BC, y^ which contain the obtuse angle, by twice ^ ^ the rectangle contained by the base AB and the distance BD of the perpendicular from the I angle; or AC'=AB'+BC'+2AB-BD. For AC*=AD'+CD^ since ADC is a t'L. (Prop. 38) But AD'=AB'+2ABBD+BD', (Prop. 37). .-. AC'=AB'+2AB-BD-|-BI>'+CD^. And BD'+CD'=CB', since CDB is a t'L (Prop.S .-. AC=AB'-|-BC'+2ABBD. Otherwise, AP' -AB'+D B'+SAB-BD; add CD' to both, and AD'+CIJ-=AB=-|-BD'-|-CD'+2AB-BD. .-. AC'=AB'-1-BC''+2ABBD. Proposition XLI. — Tbeobbk. In any triangle ABC, the square on a side CB tabteoi iag an acute angki is leu t\iim XW «\3l«l of th« PIiAMB GBOMBTRT. 107 n the sides AC^ AB^ that ODtain the acute angle, by wice the rectangle contained )j the base AB and distance (D of the perpendicular from he acute angle CAB. Or CB«=AB"+AC"— 2AB-AD. Since BD is the difference of AB and AD, (Prop. 38), BD>=AB"+ AD'— 2AD-AB; add CD* to both, and BD'+CD*=AB>+CD*+AD«-.2ADAB. /. CB«=AB«+ AC*— 2ADAB. Scholium. If in the last two propositions the sides op- lonte to the angles at A, B, C, be denoted bj a, 6, c, ?ropodtion 40 will be b^^a^+c^+2cBD, which, by nuuposing a^ and c^, and dividing by 2cy gives BD= — , from which^ if the three sides be given, the dis- nce of the perpendicular from the obtuse angle may be nmd. Similarly from Proposition 41, by the same substitution, "BD— ^i^I_" " 2c Proposition XLII. — ^Theorem. The difference of the squares of the sides AC, CD, of a iangle, is equal to the difference of the squares of the stances, AD, DB, of the perpendicular, from the extre- ities A, B, of the third side AB. Or, AC«— CB«=AD«— DB«. For AC2=AD2 + CD«, (Figure to last Prop.) andBC«=BD2 + CD2. .-. AC«— BC"=AD2— BD2; since CD^ from CD^ ivesO. Cor. The rectangle contained by the sum and difference ' two sides of a triangle, is equal to the rectangle con- bed by the base and the sum or difference of its seg- mts, according as the perpendicular falls without or ifhin the triangle. For AC«— BC«=AD«-^BD«, by the proposition. .-. (AC + BC) (AC~BC) :=: (AD+BD>(AD— BD.) hop. 38, cor. 9). Or, (AC+BC)(AC— BC)=sAB(AD+DB.) Figure 1. and(AC+BC)(AC— BC)=AB(AD^DB.) ¥\gQi«i^, I » The sum of the squares of two sides, C AC, CB, of a triangle ACB, is equal to y^ twice the square of hulf the base AD, and / / iX twice the square of the line CD, joining / j j \ the vertex and the middle of the base. ^ BB Or AC'+CB2=2AD»+2CD2. Draw CE perpendicular to AB, and consider ADC ooe triangle, and CDB as another; then AC2=AD2+DC2 4.2AD-DE. (Prop. 40;) CBa^AD'+DC^— 2AD-DE, (Prop. 41), AD W =DB. .-. ACi!+CB2=2AD«+2DC», (As. 2.) Pbopobition XXiIV. — Throrbh. In any isosceles triangle ABC, th( ■j^uare of a line CD, drawn ftom the ver- tex C lo any point D in the base, togethei with the rectangle contained by the seg- ments AD, DB, of the base, is equal to — - the square of one of the equal sides of the triangle. Or CDa+AD-DB^CB*. For CBS— CDs=BES— EDa. (Prop. 42.) = (BE + ED) (BE— ED) = BD-Di (Prop. 38, cor. 9.) .-. CB==CD»+BDDA, by adding CD= to both sidw Proposition XLV. — Thborkw. ^ In any parallelogram the diagonals bisect each other, and the sum of the squares of the four sides is equal to the iquares of the two diagonals, V AD is II to BC, the Z.ADE is = ^_ the /.CBE. and the Z.DAE is = the iECB ; .-. since A! is =BC, (Prop. 6). AE is =EC, and BE is -ED i wnce AB is =DC, and AD is =BC, (Prop. 24), I squares of AB and AD are together = the squares oFI and BC, but because BD is bisected in E, BA'+AD 2BE»+2AE», (Prop. 43); .-. DC+BC are also sSf^ +2AE»; and hence BA»+AD*+DCHBC« are =41 +4AE=BI>i+AC». (Prop. 37, cor. 2.) 109 PLANS OEOMBTBT. Any portion of the circnm- of a circle^ as ACB, is called tiaigbt line AB^ which joins the des of an arc^ is called the clwrd irc. pace contained bj the arc ACB chord AB is called a segrnenL figure CAD, bounded by two C, AD, and the intercepted arc ailed a sector. 3 radii be perpendicular to each A AC, AB, the sector is called ngle in a segment is fiontsaned by dght lines drawn from any point re of the segment to the extre- f the same arc. -^ Dgle on a segment or on on arc is contained by two< lines drawn from the extremities of the segment or my point in the remaining part of the circumfe- thus the angle ABC is said to be in the segment r on the segment or arc AC. ngle at the circumference of a circle, is one whose point is in the circumference ; and an angle at the 8 one whose angular point is at the centre. 'ur segments are those that contain equal angles, ctilineal figure is said to be inscribed in a circle ts angular points are in the circumference of the rcle is said to be inscribed in a rectilineal figure le circle touches all the sides of the figure. Pboposition XL VI. — ^Theoiiem. (traight line CD, drawn through tre C of a circle, bisect a chord kich does not pass through the it will cut it at right angles ; and it it at right angles it shall hi- , let AB be bisected in D, and L, CB. I 110 PLANE CEOMETHT. Th«i in the As ADC, BDC, AD is =DB, and DC ci man to both, also AC ia =;Cfi, being radii of the a circle ; -■. the /ADC ia =z the £BDC, (Prop. 9), and th( :e adjacent Is, hence each of them is a right angle. Nest let CD he at right angles to A B, AB is hiaectt iD, For since AC ia =CB, and CD common, and the ADC and BDC j' Ls, .: AD is =DB, (Prop. 39, cor, 2. A straight line which bisects a chord at angles passes through the centre. Proposition XLVII Theorkm. ITie angle ACB, at the centre of a cirule, is double of the angle ADB at the circumfe- rence, standing on the ^ Kame arc AB. Join DC, and pro- duce it to F, then -.- AC is =CD, the /CAD is = the ZCDA, -■- the two i CAD and CDA are together double of the /.ADC, butti ZACF is also = the two /.a CAD and CDA, (TWf, V .: the /ACF is double of the /.ADC ; in the same maa it uan be shown, that the /FCB is double of the a CDB; .-. the whole, or the remaining /ACB, is douM» the whole, or the remaining /.ADB. (The ■ applies to the first figure, and rfmaining to the seeond.] Cor. 1. The angles is the same segment of a (urcle, I the angles standing on equal arcs, are equal to each oths Cor, 2. An angle at the circumference of a circle measured by half the arc on which it stands. For by tl proposition it is half the angle at the centre, and the an| at the centre is measured bj the arc on which it ataa (Prop. 1. cor. 3.) Pbopobition XLVIIl. — Tbgorrn The opposite angles of a quadrilateral I figure inscribed in a circle are together equal to two right angles. For the three /s of the A ABC, name- ly ABC, BCA, and CAB, are together lal to two right angles, (Prop. 19), J the /BCA is = the /BDA, (Prop. ; cor. 1), also the /.BAC w = the ThAXim OEOBfSTBT. Ill BDC ; .*. the ZADC is equal to the two angles B AC and €A; and adding the ZABC to each, the two is ADC Bd ABC are together = the three Is ABC, BAC, ACB, lat is, to two / is. Otherwise, The angle ABC is measured bj half the arc ADC, Prop. 47« cor. 2), and the Z. ADC is measured by half the t9 ABC; therefore the two Is ADC and ABC are leasured by half of the whole circumference, and are . Iierefore together equal to two r* Ls. Cor. If a side CB of a quadrilateral figure, inscribed BL a circle, be produced, the exterior angle ABE is equal the interior angle ADC ; for each of them, together nth the angle ABC, makes up two right angles. Pboposition XLIX. — Theorem. Equal chords, AB, CD, in a circle, are ignaUj distant from the centre; and if M distances, EF, EG, from the centre. It equal, the chords, AB, CD, are equal. For draw the radii, EB, ED, and ••• IF and EG are -^ to AB and CD, these iMords are bisected in the points F and X (Prop. 46.) Since then the chords ItB and CD are equal, their halves FB and GD are also nal, and the radii EB, ED, are equal, and the angles at r and G are r* Z«; .*. the A< EFB, EGD, are equal in fwy respect, (Prop. 39, cor. 2); hence EF is =EG. In the same manner, if £F be given zzEG, in the 'I' A« FEB, GED, we have EB and EF in the one =ED and EG in the other; .-. FB is =GD, (Prop. 39, ar. 2), but AB is double of FB, and CD is double of GD ; '. AB is =CD, (Ax. 6.) Proposition L. — Theorem. The diameter AB is the greatest chord n a circle, and that chord ED, which is learer to the centre, is greater than the iiord GH, which is more remote; and ionverselj the greater chord is nearer to lie centre than the less. For since C is the centre of the circle, EC is =CA, and CD is =CB, .-. EC and CD are =AB, Imt EC and CD are greater than ED, (Prop. 13), .-. AB Again, since the chord ED is nearer to the cenlx^ Wv'dXL 118 GH, CF^CL, lut CF^+FD= are =CL'+LH». for ei is equal to the square of the radius ; .■. since GF'.c^CL*', FD^r^LH*, and -■. FD is t^JM, and hence ED::p-G^ (Prop. 46, and As. 6,) Nest let the chord ED he ^GH. FC shall he ^:CI^ for since DF'^4-FC»^HI;!-t-CL=, of which FD^^LH* .-. FC'^CL^ : hence FC^CL. Cor. 1. The angle at the centre of the circle, aubtendcj hy the greuCer chord ED, is greater than the angle ivb tended by the less, GH. For in the two A* ECD, GCH, the two sides EC, Cft in the one, are equal to the two sides GC, CH, in tbi other, hnc the hase ED is ^^ the base GH ; .-. the ^EOS ia^theZGCH. (Prop. 15.) Cor. 2. The angle EUD is tneaaured hy the are Eft and the angle GCH by the arc GU. Therefore the greater chord cuts off the greater arc. Cor. 3. Equal chords cut off equal arcs. For if the chords were equal, the angles at the C would he equal, and consequently the arcs which meaauil I them. ^^^L Proposition LI. — Theor; ^^^H A straight line AB, drawn perpendi- P ^^H colar to the extremity of a radius CA. ia j' ^(T a tangent to the circle at that point; that / /^N is, it touches the circle without cutting it. { C^ U For take any point E in AB, and join V / EC. thensince ACis -L- to AB.itisless N. 7 than CE. (Prop. 23), but CF is =CA, -^ .-. CEp»-CF, and hence the point E is without the circle, and E beiug any point on the line ABi .■- AB is without the circle. Cor. A Etniight line drawn perpendicular to the U gent AB, from the point of contact A, will pass through ll* centre. For it must coincide wilh CA, which is J- to AR Proposition LU. — Tseobem. An angle ADB in a semicircli right angle, and an angle DBA in ; ment greater than a semicircle i than a right angle, aad an angle AED M in a segment less than n semicircle is greater than a right angle. Join DC, and produce AD to F, then ■-■ AC=CD, the iCAD = ihe LCDi, rJLAHX QXOMBTRT. 113 and -.- DCzuCB, the ZCBD = the ZBDC, .% the whole LADB is = the two Ls DAB and DBA, but the ZFDB k also = the two Ls DAB and DBA, (Prop. 19); /. the £ADB is = the ZBDF, and they are adjacent is, .*. each of them 18 a /Z, (Def. 9); hence the ZADB is a r*l, .•. &e ZABD (which was shown to be =: the ZCDB), is less than a /Z. And again, since AEDB is a quadrilateral in- Kribed in a circle, the two Ls AED and DBA are together = twoVZ*, (Prop. 48); .*. since DBA has been prored leiB than a /Z» AED must be greater than a /Z* SoHOLiUM. The proposition might have been demon- rtiated thus: — Since the ZADB is measured, (Prop. 47» Oor. 2), bj half the semicircle AGB, it is a r'Z; since the [DBA is measured bj half the arc AED, which is .^ a MDudicle, itia^^^Af^L; and since the Z AED is measured by half the arc ABD, which ib^p^ sl semicircle, it is :::^ a /Z. PiKOPOSITION LIII. — ^ThBOREM. Parallel chords AB, CD in a circle, btercept equal arcs AC, BD. For join CB, then ••• AB is || CD, md CB meets them, the Ls ABC, BCD ue equal, but the Z ABC is measured bj Wf the arc AC, (Prop. 47, Cor. 2), and the ZBCD is measured by half the arc BD, .•. the arc AC is = the arc BD. CJor. If the adjacent extremities A, B, and C, D, of two iqual arcs AC, BD, be joined, the chords AB and CD will K parallel. Proposition LIV. — Theorem. If a tangent ABC be parallel to a 4. ^ord DE, the intercepted arc DBE is Msected in the point of contact B. Draw FB from the centre to the point sf contact B, then FB is -J- to AB, and •'•it is also -L to DE; hence, since FG i« -L to DE, DE is bisected in G, (Prop. 46); .-. in the A« DGF, EGF, DG is = GE, DF is z= FE, and GF is common, .-. the ZDFG is = the ZEFG, (Prop. 9); hence the arc DB, which is the measure of the one, is =: the arc BE, which 1*8 the measure of the other, .•. the arc DBE is bisected in B, the point of contact. w E OBOllETRT. Proposition LV. — Theokem. igle ABD, formed by a tangent AB, and a chotA DB drawn from the point of contact, is equal to the angle in the alternate aegment BED, and is measured by half Sie intercepted arc BU. For (figure laat PropOj let DE be |{ to AB, then the an BD is = the arc BE, (Prop. 54), .-. the /.BDE, meaawed by half the one, is =; the Z.BED, measured by half the other; but the iBDE is = the iDBA, (Prop. 16),) .-. abo the Z.ABDis = the /.BED; hut the Z.BED is measured bj half the arc BD, .'. the Z.ABD formed by the tangent B^ and the chord DB is also measured by half tbe arc BD. Cor. Tbe angle DBC fonned by the tangent CB, and the chord DB drawn from the point of contact, is measured by half the intercepted arc BEHD. For the part DBE is measured by half the arc DHE, (Prop. 47, Cor. 2), and the part CBE is measured by half the arc BE, .-. the whole /DBC is measured by half itw arc BEHD. Proposition LVI. — Tbeobem. If two chords AB and CD intersect within a circle, the angle AEC formed at their point of intersection is measured by half the sum of the intercepted AC, DB. Join DA, then the /AEC being the exterior angle of the AAED is = to the _ two L» DAE and EDA, but the /DAE is measured by half the arc DB, and the /EDA is IB*«" Bured by half the arc AC, (Prop. 47, Cor. 2); .-. the /AEC is measured by half the sum of the arcs AC and DB. Proposition LVIL — Theorem. If two chords, AB, CD, meet when produced in a point E, the angle AEC formed at their intersection is measured by half the difference of the intercepted arcs AC and BD. _ -For, join BC, then the /AEC isequalto I ttie diflerence of the ie ABC and BCD. I fProp. 19); but the /ABC is measi^ed by f half tbe arc AC, and the /BCD by half the arc BD. .% A J /AEC is measured by half the difference of the arcs A( fjuid BD. PLAVE OEOMBTRT. 115 Cor. If a chord CD produced, meet a tangent £F in &e point E^ the angle C£F is measured hj half the differ- ence of the arcs CF and FD. For the LGEC is = the difference of the Z« GFC and FCE, (Prop. 19), and the ZGFC is measured by half the arc CF, (Prop. 65), and the ZFCE by half the arc DF, (Prop. 47» Cor. 2); .'. the LCEF is measured by half the diftrenee of the arcs CF, FD. ScHouusr. In the application of proportion to geo- metrical quantities, since there is no geometrical principle bj which it can be ascertained how often one line is con- tained in another, the algebraical definition of proportion win not apply to geometrical magnitudes, and therefore it is necessary to substitute another, which is as follows: — De£ The first of four magnitudes is said to have the nme ratio to the second, that the third has to the fourth, wlien any equimultiples whatever of the first and third be- ilif taken, and any equimultiples whatever of the second and fourth — ^if the multiple of the first be less than that of ^ second, the multiple of the third is also less than that cf the fourth; or if the multiple of the first be equal to that of the second, the multiple of the third is also equal to that of the fourth; or if the multiple of the first be greater than tfaatof the second, the multiple of the third is also greater Oaii that of the fourth. Note. A multiple of a quantity is the result of repeating that quantity a certain number of times ; and equimultiples are the re- xiliB of repeating several quantities the same number of times. In order Uiat the propositions on proportion that were demon- rtnted in the treatise on Algebra, may be made available for the derdopment of geometrical properties, it will be necessary to prove, fliat if quantities are proportional by the algebraical definition, they tte also proportional by the geometrical; and conversely. Let now a, 6, c, d, be four quantities that are proportional by the tlgebraieal definition, that is, such that j- = -5, they are also propor- tbnal by the geometrical definition, that is, if ma ^^ nfe, mc Z^ nd; if aa =s aft, iwc = nd, md ii ma^^nb, mc^^nd; since ^=-^, -^ = jjj, for the second is obtained from the first by multiplying both sides rfth« equation by - ; now it is evident that if wia be greater than *> -^ is greater than unity, but "^ ^ = ^^> therefore -^ is greater Aan unity, which can on]|r be true by mc being greater than mf ; Iirace, when the multiple of the first is greater than that of th« se- omd, the multiple of the third is also greater than the multiple of the fourth. w FLAn QSOHETBT. In the Bamo manner, if ma be = ni, ~ = unity, but then ^ liich can only be by mr being =m nd; henco, when tho, multiple of [Jie firtit in equal to tlie luultipio of the second, the tiple of the third in also equal to that of tJiu fourth. Again, if ma be lees than nb, —^ ia less than unity, therefore —r IB aiso less than unity, which can only be by tnc being lesa than aJ} hence, when the muIUple of the first is lees than that of the (he multiple of the third is also lees than that of the fourth. Therefore, if four quantities be proportional by the algebrwcal fiaition, thuy are also praportional by the geometricnL GtDversely, let a, b,c,d,hB four magnitudes, such that whitlevtf nuiubeis n oud n may be, ma is greater than, equal to, or lea ' nfr, according as mc ia greater than, equal to, or less than ad, j-= J, for if not, let j-^-j therefore . — — , and hence na k greater than, equal to, or leas than tib, according aa wc than, equal to, or less than ne; that is, nd aud ne poaseHi properties, or i/ = e, and consequently ^= ^. Therefore, if four quantities he proportional by the definition, they are also proportional by the algcbi^ical. Coutioqueutly, whatever was ilentonatrated to be true of quantitw that are propDrtional by ooe dcfinitioti. is also true of those that Ut proportional by the other, for each defiuidon has been showa to odd- tain the other. Lemma. Quantities that liare the same ratio to the ttat quantitjnre equal to one another; and equal qoanti^- have the same ratio to the same qunntitj. Let a and h have the same ratio to c, thea a=6; Ex since a has the same ratio to c that h has to c, -^ -. Mul- tiplying hy c, we have a=6. Nest, let 0=6, and c be a third quantity, \)itaa-.e=h:t\ for since a^xh, if each he divided by e, we hare ' — {' Def, Similar rectilineal figures are those which haw their several angles equal each to each, and the sides abod these equal angles proportionals. Def. Reciprocal figures, viz, triangles and paralleln- grains, are such as have their sides about tvro of ihtir angles proportional in such a manner, that a side of tht first is to a side of the second as the remaining side of ''' ■econd is to the remaining side of the first. Def. A strjight line is said to be cut in extreme and n ratio, when the whole is to the gteati ^ greater segment to the less. PLAHK OEOMETRT. ll? Def. The altitude of any figure is the straight line drawn from its yertex perpendicular to the hase. Pboposition LVm. — Theobem. Triangles ABC, ACD, and paral- ^ ^ Wograms EC, FC, haying the same altitude, are to one another as their bases BC, CD. Produce the base both "ways, and ^ „ ^ t, ^ take BG, GH, HI, any number of "^ ^ ^ -^ ^ lines, eac^ equal to BC, then IC is a multiple of BC, take DK, KL, any number of lines each equal to CD, then CL 18 a multiple of CD ; join AG, AH, Al, AK, AL, then *.* BC, BG, GH, HI, are aU equal, the A« ABC, ABG, AGH, AHI, are all equals (Prop. 27); .'. what multiple Merer the base IC is of the base CB, the same multiple is the AAIC of the A ABC; again, since CD, DK, KL, are afl equal, the A« ACD, ADK, AKL, are all equal, (Prop. 27); .*. what multiple soeyer the base CL is of the base CD, the same multiple is the AACL of the A ACD. Now, if the base IC be greater than the base CL, the AAIC is greater than the AACL; if equal, equal; and if less, less; .*. there are four magnitudes, namely, the two bases BC, CD, and the two A« ABC, ACD, and of the Vase BC, and the A ABC, the ist and 3d, there haye been taken any equimultiples whateyer, yiz. the base IC and the AAIC, and of the base CD and the A ACD, the 2d and 4th, there haye been taken any equimultiples whateyer, yiz. the base CL and the AACL, and it has been shown that as the base IC "p^, =, or -*i:r the base CL, so is the AAIC z^^ =, or ^^^ the AACL, .-. A ABC : A ACD = base BC : base CD, and •.• ^=I7CE = 2AABC, and £Z=7CF z= 2AACD, .-. AABC : AACD = £Z=7CE : £Z=7CF, (Alg. 109.) and •.• AABC : AACD = base BC : base CD, .-. ZZZ7CE : £=7CF = base BC : base CD. Cor. Triangles and parallelograms haying equal alti- tudes are to one another as their bases. Proposition LIX. — Theorem. If a straight line DE be drawn parallel to one of the ndes BC ef a triangle ABC, it will cut the sides AB, AC, or these sides produced proportionally; and if th^ tides AB, AC, or these sides produced, be cut ]^ro]^ortioiL- ally, in the points D, E, the straight line Y7\i\c\i ^o\ii*& ^^ ¥ 318 FLAJTB flBOJIETBT. points of section will be parallel to the third side BC of the triangle ABC. First let DE be drawn II to BC, a side of the AABC; thenBD:DA b'= = CE:EA;fc ■ * "" DC; Then ADBEzzADCE, ■.- thej are on tbe aane htm DE, and between the same ||s DE and BC, (Prop. 26.) And ■." A^l^li is another mi^nitude, .-. ABDEr AADE=ACDE: AADE. (Lem.,p.ll6. But ABDE: AADE=BD:DA,1 p^^^ „ and ACDE : AADE^CE ; EA. P"P- ***■ .-. BD:DA=CE: EA. (Alg. 111.) Secondly, let the sides AB, AC, of the AABC. or tlua flidesproduced, be cut in the points D, E, bo that BD;DA3 CE:EA; then DE is II to CB. The same construction being made, V BD.-DA=CE:EA, and BD : DA=AJ3DE : AADE,\ „^ -„ and CE : EA= ACDE : AADE, f ' P" ** ABDE ; AADE=ACDE t AADE ; that ia. A« BDE, CDE, have the aarae ratio to the AADE, and .-. ABDE=ACDE. (Lem., p. 116.) and they are on the same base DE, and oa the un ide of it; .-. DE is || to BC. (Prop. 29.) ^^^H PRorosiTioN LX. — Theorem. F If the vertical angle BAC of a triangle be bisected by a straight line ^ \ AD, which also cuts the base, it will ^}Cl divide the base into segmenls BD, ^■"''W'^V / ' DC, which have the same ratio aa the ^ I V othersides, BA, AC; andif the seg- B d C ments of the base have the same ratio which the other Ejdea of the triangle have to each other, ihi ' right line drawn from the vertex to the poiot of 1 divides the vertical angle into two equal parts. First, let the Z.BAC be bisected by the straight liJ« I AD; then BD:DC=BA:AC. For draw CE || DA and let BA produced meet CE in E. Then ■-• A D is II to EC, the exterior iB AD = the il tenor /.AEC. and the alternate b DAG, ACE, « equaX; but the ^BAD is = the ZDAC ; .-. the ZAEC also = th e Z A C E ; hence the s\4e k^ w ■=. t\a uda AC TI.A.NE 6E0METBT. 119 md •.• AD is || to EC, a side of the ABCE, BD : DC= U : AE. (Prop. 59.) but AE=AC ; .-. BD : DC=BA : AC. Secondly, let BD : DC=BA : AC ; join AD, then the BAG is bisected bj AD, that is, the ZBAD = the DAC. The same construction being made as before, vBD:DC=BA:AC, and that BD : DC=BA : AE, (AD being || EC.) .*. BA : AC=BA : AE; hence AC= AE. (Lem., p. 116.) md .-. the LAEC=: the Z ACE, But the ZAEC= the ZBAD, and the ZACE=ZCAD. Prop. 17.) •.theZBAD=the LCAD. Proposition LXI. — ^Tifeorem. Ilendes about tiie e^al angles of [niaogidar triangles are propor- mk^ the sides opposite to the [oal angles being the antecedents consequents of the ratios. Letthe A«AEB,BDC, having the UB= the ZDBC, the LABE= eZBCD, and the LAEB=: the Z.BDC, be so placed that B and BC maj be one continuous straight line. Then *.* the HAB, together with the ^BCD, (which is =the ZABE), less than two r^is, AE and CD, will meet, if produced, Vop. 17, cor.); let them meet in F, then since the iBA=the LBCB, and the ZEABzz the ^iDBC, (Prop. , cor. 1.), FD is II EB, and EF is || BD, •. also )=EB, and EF=BD, (Prop. 24.) Now since EB is || the side FC of the AAFC, AE : EF=: AB : BC, (Prop. ); but EF=BD, .-. AE : BD=AB : BC, and by alternating, (Alg. 105.) AE: AB=BD:BC. ^^in, since BD is || AF a side of the A-^^C, AB : BC=FD : DC, (Prop. 59), but FD=EB, •. AB : BC=EB : DC, and hy alternating, AB:EB=BC:DC. •. also AE : EB=BD : DC. (Alg. 112.) 2lor. Equiangular triangles are similar. Pkoposition LXII. — Theorem. tf the ffldes of two triangles about each of ilieit ^ii^<^ jwgot^xauik, the triangles shall be eqiuan^xilox, ^tA ..^.™ M ^ M lIQO FIiAHE OKOSOtTBT. hare those angles equal that are oppo- site to the sides that are the antecedents or consequents of the ratio. Let the A« ABC, DEF, have theii Bides proportionals, so that AB : BC=DE : EF; BC : CA =EF:FD: and consequently (Alg. 112,) BA: AC=ED:DF, then /,A=/.D, /.B=Ze, and LC^IV. Let the A^FG on the other side of the base EF equiangular to the AABC, so that the Z.ABC may be the /.GEF. the iBCA= the ZEFG, and consequently d ZBAC= the LEGF ; then since the A^ ABC, GEF, are equiangalax EC : CA=EF ; FG. but BC : CA-EF : FD, (by hjp, .-. EF ; FG=EF : FD ; hence FG=FD, (Lem„ p. liB again, BC : B A=EF : EG, and BC : BA=EF : ED ; .-. EF : EG=EF : ED ; hence EG=ED. (Leni., p. lli Wherefore in the two triangles EDF and EOF, I three sides of the one are Tespectively equal to the &I sides of the other; -■- the angles are = that are oppowtc the equal sides, (Prop. 9); that is, the /.DEF =( Z,GEI-, but the £GEF= the ZABC; .-. the IDEVizi /ABC. In the same manner it can he proved, that the iBAC the /.EDF, and the iACB to the /DFE. Proposition LXll I. — Tdk o hem. ■ Triangles which have an equal angle included between propor- ^ tional sides are similar. Let the /.a A and D be equal ; if we have AB : AC=DE : DF, the A« ABC, DEF, are similar. ■_ Take AG=DE. and draw GE \[ to BC, then (Prop. 17) the /AGE= the /ABC, a the /.AEG= the /ACB; .-. the A» ABC. AGE, J equiangular, hence (Prop. 61), AB : AC= AG : AE ; hutAB: AC=DE:DF; .-. AG : AE=DE : DF, (Ax. 1); and since AG H DE, AE is =DF. .-. the two A* AGE and imFlii two sides, AG, AE, of the one, = two sides, DE, DP, ither, and the LA is = the Z.D. .-. the AAGE is eqi every respect to the ADEF, (Prop. 5) ; .■. the IM = (he ZDEF, and the /AEG is = the /DFE. Hot sAown, that the /.AGE \a = tiic LKBC. and that A PJUANS GBOaiETRy. 121 :AEG is = the ZACB; .-. also (Ax. 1) the ZABC is = he ZDEF, and the ZACB is = the ZDFE. Hence the t^s are equiangular^ and .*. similar. (Prop. 61, cor.) Pboposition LXIV. — Thborsm. Equal parallelograms^ AB, BC^ rMch hare one angle FBD of lie one, equal to one angle £BG of be otherj hare the sides ahout the qnal angles reciprocally propor- bnalj and parallelograms which ATe. one angle of the one, equal one angle of the other, and the ides about the equal angles reciprocally proportional, (viz. OB : BE=GB : BF), are equal. Let the sides DB, BE, he placed in the same straight Joe; then since the LDBF = the LEBG, add the ZFBE )cach; .-. the two L$ DBF+FBE, are = the La GBE+ BE; but DBF+FBE, are = two /Za, since DBE is a Bught line, (Prop. 1); .-. the Z« GBE + FBE, are to- ither equal to two t^Is, and hence (Prop. 2) GBF is a nnght fine. Now since ZZI7AB=ZIZ7BC, and FE is another ZZZ7, dzAB r zz=7FE=zz=7BC : ZZI7FE. (Lem. p. 1 16.) But ZII7AB : zz=7FE=DB : BE, and ZZZ7BC : ZZZ7FE :GB : BF, (Prop. 58), and .-. DB : BE=GB : BF. Next, let the sides about the = Ls be reciprocally pro- rtional, that is, let DB : BE=:GB : BF, the n=7AB will = the ZZI7BC. For ••• DB : BE=GB : BF, and since by (Prop. 58), B : BE=ZZI7AB : £=7FE, and GB : BF=ZZ=7BC :/=7 E; .-. /ZZ7AB : £iZ7FEz= zz=7BC : ZZZ7FE. (Alg. 102.) lliat is, the ZZZ7 a AB and BC have the same ratio to eci7FE; .-. zz=7AB=z=z7BC. (Lem. p. 116.) Cor. 1. Since triangles are the halves of parallelograms, <m the same base, and having the same altitude,, equal angles, which have one angle of the one equal to one l^e of the other, have the sides about these angles reci- xally proportional ; and if two triangles have one angle the one, equal to one angle of the other, and the sides out these angles reciprocally proportionsJ, the triangles > equal. Cor. 2. If the angle DBF were a right angle, the pa- Idogiams would be rectangles; hence the «\de% <)i Vn^ I ISS equal rectangles can be conTerted into a propDrtian, bj makiiig the sides of the one the extremes, and the sides is the other the means; and if four straight lines lie propor- tional, the rectangle contained by the estremes is equal to the rectangle contained by the means. Cor. 3. If the means of the proportion be equal, the rectangle formed by them nill be a square. Henro, " when three straight lines are proportional, the rectangle L contained by the extremes is equal to the square on ibe ^^K mean;" and conrersely. ^^1 PsOFOaiTioN LXT — Theorem. If two chords, AB, CD, cut one ano- n ther in a point £, vithin a circle ACD, the rectangle contained by the segments of the one shall be equal to the rectangle contained by the segments of the other. That is, the rectangle AE-EB— CE-ED. Join CA and BD, then the LCAE is = the Z.BDE ; for they stand on the same arc CB, and the ZCEA is = the iBED, (Prop.3); hence the As AEC, BED, ate equiangular, (Prop. 19, cor. 1); hence AE:EC=DE:EB, (Prop. 61.) _ And .-. AE-EB=CE-ED. (Prop. 64, cor. 2.) ^^H Proposition LXVI.^Theorem. ^^^ If from a point E, without a circle ABC, two straight lines be drawn, cut- ting the circle, the rectangle contained by the whole AE, and the external seg- ment, EB of the one, will be equal to the rectangle contained by the whole, EC, and the estemal segment, ED, of the other. That is, AE-EB— CEED. Join AC and BD, then the i^EDB is = the /.EAC. (Prop. 48, cor.), and the LE Is . .-. the As'EBD, ECA, are equiangular, (Prop. 19. cor. Ilj I hence AE:EC=DE;EB, {Prop. 61.) ^^^ And .-. AEEB=CEED. (Prop. 64, cor. 2.) I PnoPoaiTiON LX V II, — Theorem. ' If feom a point A witliottt a. circle, there be drawn PIiANS GBOMETBT. 123 secant AB, and a tangent AD, Hie ctei^le contained hj the \rh6ie secant B, and its external segment AC, will i equal to the square on the tangent D. That is, ABAC=AD^ Join DB and DC, then the ZADC ontained hj the tangent AD and the Old DC) is = the LDBA, (Prop. 55), ;d the ^A is convmon to the two A< DC, ABD. .'.these A» are equiangular, (Prop. 19, cor. 1); and nee BA : AD=AD : AC. (Prop. 61.) .-. AD*= AC-AB. (Prop. 64, cor. 3.) P^iOFOsiTioN LXVin.— ^Th^orbm. Ji a point' £ he taken in the circnm- ence of a circle, and straight lines A£, { be drawn to the extremities of the ^/ meter AB, and also EC perpendicular the diameter, then A£ is a mean pro- lional between the diameter and the scent segment AC ; EB is a mean portional between the diameter and adjacent segment BC ; and EC is a mean proper tional ween the segments of the diameter AC, CB. Or AE^ ^B-AC, EC'rrAC'CB, and BE^=ABBC. Por the A« AEB, ACE, are equiangular, since the Ls B, ACE, are r^Ls, and the ZA common, (Prop. 19, 1); also the ^s AEB, ECB, are equiangular, since rhaye the La AEB, ECB, t^Ls, and the LB common, op. 19, cor. 1); hence also the ^s ACE, ECB, are iaugular, since each of them has been shown to be iangular to A^EB. Now, since the A« BAE, EAC, equiangular, A : AE= AE : AC ; .-. AE^iziABAC. Jid since the A« BCE, EC A, are equiangular, C : CE= CE : C A ; .-. CE^zzBCCA. jnd since the A« ABE, EBC, are equiangular, B : BE=BE : BC ; .-. BE2=AB-Be. or. 1. Since AEB is a right-angled triangle, and EC :awn from the right angle perpendicular to the hypo- ise; " the triangles on each side of a perpendicular cm hypotenuse, drawn from the right angle of a right ed triangle, are similar to the whole, and Ic^ on^ ^AiQ- I Qter; each aide is a mean proportional between the hj| tenuse and its adjacent segment, and the pcrpendicnlai proportional between the segments of the hyp tenuae." Cor- 2. This furnishes another proof of Proposit for since AE^=AB-AC, and BE==ABBC ; AE'+BE^=AB-AC+ABBC=AB=. (Prop. 35- Proposition LXIX. — Theorem. ^<-^ ^ Triangles ABC. DBE, which have one angle, ABO, of the one, equal to one angle, DBE, of the other, are to one another in the ratio compounded \ / of the ratios of the sides about the \ / equal angles. That is, AABC : V ADBE=ABBC : DBDE. « Let the A» ABC. DBE be so placed that AB and I may he in one Straight line, then DB and BC will also in one straight line; and join AD. then (Prop. 58). AABC .- AABD=CB : BD and AABD : ADBE=AB : BE ; .-. compounding theM' tios, and omitting froro the first and second terms A^l we have AABC : AI>BE=ABBC : DBBE. (Alg. 11 Cor. Parallelograms which haye one angle of the 4 equal to one angle of the other, are to one another the ratio which is compounded of the ratios of the t about the equal angles. For they are double of the of which the property has just been demonstrated. Proposition LXX. — Trkohbm, Similar triangles ABC, DEF, b are to one another as the squares of their corresponding For by last Theorem ' ■-■ the A* fe equiangular, the Z.A being ^ Ou i} AABC : ADEF=ABAC: DEDF, but AB ; AC=DE : DF, (Prop. 61). .-, AB : DE=AC : DF, (Alg. 105), and AB : DE=AB : DE, from equality. .-. AB' ! DE»=AB-AC: DEDF, (Alg. 1]6). I .-.AABC: ADEF=AB";DE", (Alg. 102). PLAIIS OBOMETRT* 125 Oor* EquJangglar parallelograms are to one another as e Benares of uieir corresponding sides, since thej are the rabies of equiangular triangles. Pboposition LXXI. — Theorkm. milar polygons ABCDEI, urHIKj maj he divided to the same number of milar triangles, haying le same ratio that the )l7gons have; and the djgons are to one an- ther as the squares of their corresponding sides. Let E and K be corresponding angles, AB and FG cor- ssponding sides, draw the diagonals EB, EC in the one, ad KG, KH in the other; and let A and F, B and G, and H, and D and I be the other corresponding angles. Ince the polygons are similar, and A and F equal Ls, AB : AE=FG : FK, and hence (Prop. 63), the A« BE, FGK are equiangular; .-. the Z.ABE = the ZFGK ; ke away these equal Ls &om the equal U ABC, FGH, kd there remain the equal U CB£, HGIC Now since the 1 ABE is similar to the AFGH, .-. AB : BE=FG ; GK, and le polygons being similar, AB : BC=FG : GH; .•• (Alg. 11), EB : BC=KG : GH; but the Ls CBE, HGK were toved equal, .•. the ^BCE is similar to the A^HK; mce the Ls BCE, GHK are equal: but the Ls BCD, HI are also equal ; .*. the remaining Ls ECD, KHI are [oaL Again, since the A^ BCE, GHK are similar, .-. C : CE=GH : HK; and since the polygons are similar, C:CD=GH:HI; .-. (Alg. Ill) EC:CD=KH:HI; It the Ls ECD, KHI are equal, .-. the ACDE is simi- X to the AHIK, (Prop. 63.) Since the A« ABE, FGK are similar, we have AABE : AFGK=BE2 : GK^ (Prop. 70); and since the A« BCE, GHK are similar, we have ABCE : AGHK=BE2 : GK^ (Prop. 70). /. AABE : AFGK= ABCE : AGHK, (Alg. 102;. By the same mode of reasoning we should find A^CE : iGHK=ACDE: AHIK. And from this series of equal ratios, (Alg. 113), we con- ^ude that the sum of the antecedents, that is the A^ ^BE, BCE, CDE, or the whole polygon ABCDE is to je sum of all the consequents, that is the A* ^^^^ 'HK, HJK, or the whole polygon FGHIK a& onei «ftXft- ^IF cedent, ABE is to one consequent FGK, or (Prop, 70) .AB'':FG^; -■. tLe polygons are to one anoiber as tl squares of their corresponding sides. Cor. In like manner it may be proved that similu figures of any number of aides are to one another as thff squares of their corresponding sides ; therefore, universally similar rectilineal figures are to one another as the square* of their corresponding sides. Proposition LXXII. — ^TaEonKM. Regular polygons, in circles, having the same diu of sides, are similar figures. Let ABCDEF, KLMNOP, be two regular polygons, in- Bcrihed in circles hav- , ing the same n ber of sides, they similar to one Since the chords AB, BC, i^c., are all equal, the audi at the centre are all equal, (Prop. 50, cor. 3); .-, ea(£c ihem is a sixth part of 3^0°, for the same reason eachll the Ls at Q are ^^ a sixth of 360°, .-. the remaiaiag {j ( the As AGB, KQL are togelher equal, (Prop. 19) ; an being isosceles, they are equal to one anoiher ; and nu each of the angles of the polygons is double of the ^ at ik base of the A' ^^ polygons are equiangular ; and since BJ the sides of the polygons are equal, it is evident that th( sides about the equal angles are proportional, .-. the pol)' gons are similar. Cor. 1. The polygons are to each other as the squares the radii of the circles in ivbicb they are inscribed. For they are to each other (Prop. 71) as AB^ : KL und the As AGB, KQL being similar, AB' : KU: AG' : KQ^ Cor. 2. The polygons arc to each other as the Bqnarw< the perpcndicuhirs GR, QS from their centres up^ tl BJdea. For the As AGE, KQS are evidently similar, .-. AG': KQ'=GR'' : QS-'; but it was shown (Cor. J) that " polygons are to one another as AG* : KQ' ; .-, tiey u one anoiher na GR' : QS-. Cor. .?. The perimeters, or sums of all the sides, are tfl oae another as the perpendicuWa horn \.\te centres. ThAKlR OEOHETRT. 12? For since the /^s AGR, KQS are similfir, GR : RA=: »8: SK, and al^ematelj, (Alg. 105), GR : QS=RA : SK; at RA is the half of AB, and SK the half of KL ; /. if n i the number of sides of the polygons, 2n X AR will be \e perimeter of the first, and 2n x KS the perimeter of le second ; but (Alg. 109) AR : KS=2n x AR : 2n X KS, (Alg. 102) GR : QS=2n x AR : 2/i x KS. Pkoposition LXXIII Leuma. If from a quantity there be taken its half, and horn the ifiainder ks half, and so on continually, there will at last smsin a quantity less than any given quantity, however nail that quantity may be. Let A be the quantity on which the operation is to be fidTonned, the seTeral remainders will be -—, — =-rj» '^-^9 and generally the remainder after n subtractions 31 be "-jp. Let also B be the quantity than which the It remiunder must be less, and mB a multiple of B aiUer than A; then A ..^i. niB, .*. — .^^B. m A A A ike noyif n such that 2** zp^ m, then—- .^ — , but — -c:! 2" TO m A I •-. also much more is— .^ B ; hence a remainder may obtained less than B, however small B may be. Cor. If the quantity taken away at each subtraction be nre than a half, for a stronger reason, the quantity ulti- itely left will be less than any given quantity. Paoposition LXKIY. — ^Thsorbm. If a series of polygons be inscribed in nrcle, each having double the number sides of the preceding, the difference tween the radius and the perpendi- lar from the centre upon any side, is s than half the difference between e radius and the perpendicular from e centre on a side of that polygon lich has half the number of sides of the former. Let AB be a side of a polygon inscribed in a c\ie\e> ^oA C a ade of that which has double the numbei ot ^\j^> iSB FLAKK GEOMCTRT. then GI, the iJifference between the radius and lie per- pendicular DI, is .^ the half of EC, the difference betnecti, the radius and the perpendicular from the centre, nn tf side of lha.t vhich has half the number of sides. For the A» AEC, AIF are equiangular, Kince the LA common to both, and the icat E and I are j'Z", ■•■ AE;EC =AI : IF; hut since AEC is a »■■/., AC ^ AE, .•. AI tb» half of AC (Prop. 46) is ^^ the half of AE ; /. since the third term is :^ half the first, the fourth is also ^=- half the second, that is, IF p^ the half of EC. Ag^n, since DEF is a /i, DF ^ DE, but DC=DG, .-. the remainder EC ^ FG ; hence, from a quantjtj FG .^ EC there has been takep FI z^ the half of EC, .-. the remainder IG ^^ half EC. In the same mannerj if AG were joined, and a perptn- dicular drawn from the centre I> upon it, the diSetenci between this perpendicular and the radius would be kil than the half of IG, and still more less than the fourth of EC. Cor. By continually doubling the number of sides of tlw polygons, the difference between the radius and the perpen- dicular ftom the centre, on a side of the ultimate polygoDi may be made less than any given line- For it has been shown in the proposition that this il equiyalent to taking away from EC more than its half, ani from the remainder more than its half, and so on contunir ally; .'. (Prop. 73, Cor.) the quantity ultimately leflUlOi than any given quantity. Proposition LXXV Theoi If there be two similar poly- ^ ins, the one inscribed in a circle, and the other described about it, the perimeter of the inscribed is to that of the described as the' perpendicular from the centre on a side of the inscribed is to the radius of the circle; and the polygons are to each other as the squares of these lines. u p a Let ABCDEF be a regular polygon iiiii:riLei in » circle, and GHIKLM a polygon descdbud about the circle, hafing its sides parallel to the sides of the in- scribed, then (Prop. 20) these polj'gons will he et|iri- tngular, and since their sidea mc ei^viaV, ihej aie uoulit PLANE 6EOMETST. 129 bo, let ON be drawn from the centre perpendicular to lB« and produced to P^ OP will be perpendiculfur to GU, nee AB 18 II GH. .'• (Prop. 7% Cor. 3) the perimeter of the inscribed is > the perimeter of the described as ON is to OP; and Prop. 7^9 Cor. 2) (the inscribed polygon) : (the described dygon)=ON^ : OP^ Cor. 1. If the number of sides of the polygons be con- naally doubled, their perimeters will become more nearly ^uid dian by any giyen difference. For thej are to one another as ON : OP, which is ulti- mtdy a ratio of equality. (Prop. 7^* Cor.) Cor. 2. The polygons diemselyes, upon the same supposi- ion, will become more nearly equal than by any given Oor. 3. Since the circle is always greater than the in- cribed, and less than the described, polygon, each of them nO ultimittely become more nearly equal to the circle than J any giyen difference. Cor. 4. The perimeters of each of the polygons will Itimately coincide with, and consequently be equal to, iie circumference of the circle. Cor. 5. Circles are to one another as the squares of kir radii. For (Prop. 72, Cor. 1) the polygons inscribed in them le to one another as the squares of the radii of the circles 1 which they are inscribed, and these polygons are ulti- late equal to the circles ; .*. the circles are one to another 8 the squares of their radii. Pbofosition LXXVI. — Problem. Haying given the perpendicular from lie centre of a circle upon a side of a gore inscribed in it, to find the per- endicular upon the side of a figure aving double the number of sides. Let CD be the given perpendicular, nd OF the required one. Join AE ; then because AB is a dia- oeter, AEB is a semicircle ; .-. the ZAEB is a t^L (Prop. >'2); but since CF is perpendicular to EB, AE and CF are I; .-. the A» AEB, CFB, are similar; and since AB is buble of CB, .-. AE is double of CF; and since AD is imendicular to EG, ADE is a /L, and .-. = the LAEB, M the angle at A is common to each of the ^s> "BklL^ ^ iSD Kiuaoaomnnr. EAD, hence they are equiangular; .'. AB : AE=AE:, AD, (Prop. 61), and AE-=ABAD; but AE was proTcd to be double of CF, and (Prop. 3?, cor. 2), AE'^=4CF'A nibBtituting this value, we haTe 4GF^=AB-AD; .-. CF* =ziAB'AD, and conaeiuently CV-^^AliAD. Scholium. If a numerical result be required, let EG Iw the Bide of an equilateral triangle, and let the radias of the. circle be 1 , then CD is equal to ^ ; for if EC were joined, EC6 fTOuld evidently be an equilateral triangle, and the perpendicular on CB would bisect it, (Prop. 7. cor. 2.) .'. AD would be (l+i)::zS, and AB would be two; pu^ ting these valueS in the expression above, it becomes CF= Generally, if p he the perpendicular from the centre, upon a side of a figure inscribed in a circle, p, the perpen- dicular upon a aide of one having double the number "i jhj the perpendicular on a side of one having four the number of aides, and ;i„, the perpendicular on " ide of one having 2" number of aides as the first; then Pi= J^-f. P,= y ^ . and p.= J'—--- Also, {Prop. 68), EB=V-AB\BD=Va{I— p), la^iii being one ; in the same manner, if s be a side of the o^ ' nal polygon g,, aside of that of double the Dumber of sidi . and generally a„ a side of one having 2*" the number rf rides as the first, we will have «„+i— */2(l— p„), and if n be the number of sides of the original polygon in& which we commence, the whole perimeler of the polygon will be =n2-+>j2il—p„). Hence, {Prop. 72, cor. 3), if this operation be repeated often enough, the result ffitl be the circumference of the circle to radius one. Proposition LXSVII. — Peoblkm. To divide a given line AB into two parts, such that the greater part, Ah', shall be a mean proportional between the whole line, AB, and the re- maining part, BF. Let BC be drawn J- to AB, ~ ' » from the extremity B. of the aa-aigbc Une AB, and =: the half of AB; join AC, wJ produce it to E, and from l\\e ccnue C, Vv'CB.\lifi wdirf OXOMXTBY OF PLANXS. J 31 ^ describe the semicircle DBE« and from AB cut off JF^ AD ; then since AB is perpendicular to CB, it is a ine;ent to the circle at B ; .*. (Prop. 67)t EA : AB= AB : D ; but since DE and AB are each double of CB, they re equal, and AFs AD ; .-. (Alg. 107), EA— AB : AB= B— AD : AD, which, from the aboTe mentioned equali- es, giyes AF : AB=BF : AF; hence, (Alg. 105), AB : AF :AF : BF. Cor. Since AB=DE, the first proportion gives AE : ED :ED : AD ; •*• the line A£ is also divided in the same amner. Scholium. A line divided as in the proposition, is said ) 1)6 divided in extreme and mean ratio, and also to be It in medial section, or to be divided mediidl j. GEOMETEY OF PLANEa Dt^FiNiTiovs. — ^I. A solid is that which hath length, tsdth, and thickness. II. The boundaries of a solid are superficies. IIL A straight line is perpendicular, or at right angles • a plane, when it makes right angles with every line hich meets it in that plane. IV. A plane is perpendicular to a plane, when the might lines drawn in one of the planes perpendicular to « common section of the two planes, are perpendicular the other plane. Y. The inclitiaiton of a straight line to a plane is the fate angle contained by that straight line, and another awn from the point in which the first line meets the ane, to the point in which a perpendicular to the plane Bwn from any point in the first line, meets the same ane. VJ. The inclination of a plane to a plane is the acute igle contained by two straight lines drawn from any the me point of their common section, at right angles to it, le upon one plane, and the other upon the other plane. VII. Two planes are said to have the same, or like clinations to one another, which two other planes have, hen their angles of inclination are equal to one another. YIII. Parallel planes are such as do not meet qiv^ (Other though produced. IX. A Htraiglit line and a plane are parallel, if they do I not meet when produced. ] ■". The angle formed liy two interaecting platiea « called a dihedral angle, and is meaaurei XI. Any two angles are said ( ' when they are either both greatt a right angle. The same or equal circles, when they i both not greater than a quadr; Def. of the same affeeliim, both not greater than ! term is applied to arcs of the e either both greater or I PRorosiTiON LXXVIII.^ — Theorem. Any three atrmght lines which meet one another, not in the same point, are in one plane. r^t the straight lines AB, BC. CD meet one another in the points, B, C, and E, AB, BC, CD are in the same plane. Let any plane pass through the Straight line AB, and let the plane, produced if necessary, be turned about A B, till it pass through the point C. Theni because the points B and C are in the plane, the straigbl line BC is in the plane, (Def. 7) ; and because the pointt C and E are in the plane, the | CE or CD is in the plane, and by hypothesis AB is in the plane ; -■. the thr» straight lines AB, UC, CD are all in one plane. Cor. 1. Any two straight lines that cut one anothtr are in one plane. Cor. 2. Only one plane can pass through three point^- or through a straight line and a point. Cor. 3. Any three points are in one plane. PflOPDBITION LXSIX.— ' If two planes cut o ^^^B> If two planes cut one another, th( P common section is a straight line. Let two planes AB, BC cut another, and let B and D be ' points in the line of their section. From B to D draw the straight line BD, then since B and D are points JD tilt plane AB, the line BD is in that plane, (Det^ 7) : for th* same reason it is in the plane CB ; .-. being in each of lli> I planes, it is their common section ; hence the comuiiis cecfioit of the two planes ia a sWa\^t\me. OSOM KTRT OF PLANB8. 133 Proposition LXXX. — ^Thbobbm. If a straight line stand at right angles to each of two tnight lines in the point of their intersection, it will also leat right angles to the plane in which these lines are. Let PO he -i- to the lines AB, DD, at their point of intersection ), it is -i- to their plane. For take OB=OD, and join PB, PD, and BD, and draw any ine FE, meeting BD in "E, and join P£. Tben ".- BO=:DO^ and PO common to the two A« POB^ POD, and the contained angles r^ls, the base PB= PD, .*. the A^BD is isosceles, and the AOBD is isosceles hj construction. Now •/ POD is a r'Z, PO^ + OD' = PD*, bat OD«=OE'»+DEEB, (Prop. 44), and PD«=PE*+ DEEB, .-. PO^+OE«+DEEB=PE'+DEEB; take the rectangle DE'EB from both, and there remains P0^4- OE*=PES .-. the ZPOE is a right angle, and PO is at r^Ls to EO ; in the same manner, if the line EO had been in any if the other angles BOG, CO A, or AOD, it could be demon- teted to be at /Za to PO ; .•. when PO is at r^is to two taight lines at the point of their intersection, it is at 0|^t angles to ererj Ime in that plane. Cor. 1. If a plane be horizontal in any two directions, it b so in eyery direction. Cor. 2. The perpendicular PO is less than any oblique Kne, as PB, (Prop. 23), and therefore the perpendicular measures the shortest distance from the point P to the plane. Proposition LXXXT ^Theorem. If three straight lines meet all in one point, and a straight line stand at right angles to each of them in that point, these three straight lines are in one and the same plane. Let the | AB stand at r^ U to each of the \s BC, BD, BE, in B, the point where they meet ; BC, BD, and BE are in one and the same plane. If not, let, if possible, two of them, as BD, BE, be in the same plane, and BC aboTe it, and let a plane pass dnongh AB^ BC, and cut the plane in w\i\c\i'&T> vcA^SS^ are in the [ BF, (Prop. 79) ; then since AB is at r* each of the \a BD. BE, at the point of their intersection, il is also at / Li to BF, (Prop. 80), which ' " plane ; .-. the ^ABF is a r-/ ; but by hypoiheBis, the lABO is also a t'L; hence the ZABF=: the iABC, and thej are hoth in the same plane, nhieh is impossible; .'> th« I BC is not above the plane in which are BD and BE, and' in the same manner it may be shown (hat it is not below it. Wherefore the three [s BC, BD, BE, are in one mi the same plane. PnoPosiTioN I, XXXII, — Theorem. If two straight lines be at right angles to the same plani they eball be parallel to one another. Let the [g AB, CD be at i-' Ls lo the tame plane; AB is || CD. Let them meet the plane in the points B. D, and draw the | BD, to which draw DE at r" Zs tn the same plane; and make DE=AB, and join BE, AE, AD. Then ■.■ AB is -J- to the plane, it shall make r* is with erery | which meets it and is in that plane, (Def. 3) j '■ BD, BE, which are in that plane, do each of them n AB ; .-, each of the b ABD, ABE is a r'Z- For i same reason each of the Ls CDB, CDE is a /Z ; aw AB=DE, and BD common, the two sides AB, BD ■= the two ED, DD, and they contain r* L/. .: the l AD=BE. (Prop. 5). Again, :■ AB=DE, and BE=AD, and the base AE common to the Aa ABE, EDA, the ZABE = the ZEDA, (Prop. 9); but ABE is a /i, EDA is also a i-'Z, and ED is J. DA ; but it is also to each of the two BD, DC : wherefore ED is at r* Z* t* each of the three |« BD, DA, DC, in the point in whidi they meet, .-. these three \s are nil in the same planer (Prop. SI ) ; but AB is in the plane in which are BD, DA, (Prop. 78). since any three |s which meet one another «t in one plane. -■. AB, BD. DC are in one plane; toA each of the U ABD. BDC is a ^'Z ; hence (Prop. Ifl, cor. 2) AB is II CD. pBOTOsmos LXXXIII Theobem. If two straight lines be parallel, and one of them ia rf right angles Co a plane, the other shall also be at i^l angles to (he same plane. ' OSOMBTBT OF PLANES. 135 Let AB» CD be two || straight lies, and let one of tfaem AB be at ^« to a plane ; the o^ier CD is at Ls to the same plane. For, if CD /be not -L. to the plane • which AB is -l., let DG be J- to . Then (Prop. 82) DG is || AB; DG and DC are both || AB, and are drawn through le same point D^ which is impossible. Pboposition LXXXIV. — Theorem. If two straight lines be each of them parallel to the one straight liney though not both in the same plane with , ihey are parallel to one another. Let AB; CD be each ^ j^ ^ wm II EF, and not in the ^ r -b me plane with it ; AB shall k^ -\o .y HI CD. . • / " In EF take any point G, ^ ^ » m which draw, in the plane. passing through EF, AB, e I GH at r^ £« to EF ; and in the plane passing through F, CD, draw GK at r*. Ls to the same EF. And be- Qse EF is -L- b6th to GH and GK, EF is -i- to the me HGK, passing through them ; and EF is || AB ; .-. B is at 7^ ^« to the plane HGK, (Prop. 83). For the BMi reason, CD is likewise at t* Le to the plane HGK. Bnce AB, CD are each of them at / Z« to the plane SK. .% (Prop. 82) AB || CD. PnoposiTroN LXXXT. — ^Theorem. If two straight lines AB, BC, ieting -one another, be pa- lei to two others DE, EF, i it meet one another, though t in the same plane with the it two, the first two and the ler two shall contain equal angles. Take BA, BC, ED, EF, all equal to one another ; and n AD, CF, BE, AC, DF ; •.• B A is = and || ED, AD is both = and || (Prop. 24, cor. 1) to BE; for i same reason CF is = and || BE ; .*. AD aod CF i each of them = and || BE. But |« that are || the ne I aire || one another, (Prop. 84); .*. AD is ^ C¥ *, 1 it ia eguai to it, and AC, DF, Join tkem tAHvax^ tbi le parts, and .-. AC is = and || DF ; and ••■ AB, = DE, EF, and the base AC to tiie base DF, .-. le ZABC = the ZDEF, (Prop. 9.) Propobition LXXXVI.- ■Problbm. plane, fion k To draw a straight line perpendicular to ; a given point above it. Let A be the given point above the plane BH ; it is re- quired to draw from the point A a I -^ to the plane BH. In the plane draw any | BC, and from the point A draw AD J- to BC. If AD be also -i. fo the plane BH, the thing required is done; but if it be not, from the point D draw, in the plane BH, DE at i' /.s " to BC ; and from the point A draw AF -^ to DE, sod through F draw HG |] BC; and •.- BC is at t' ti lo ED and DA, BC is at r' Is (Prop. 80) to the plane passise through AD, DE ; and GH ia || BC, .-. (Prop. 83) Gl is at r* is to the plane through AD, DE ; and is .*. J to every 1 meeting it in that plane. But AF, which ti in the plane through AD, DE, meets it; wherefore GBii -L to AF, and DF is also -i- to AF ; hence AF is -»- te Mch of the Is GF. FD, and ■.■ AF ia -i- to the plane ia which GH and ED are, (Prop. 80), that is to theplue BH. Hence from the given point A, above the plaoe BH, the I AF is drawn J- to that plane. Oor. !• If it be required from a point C in a plau to erect a perpendiculaT to that plane, take a point A above the plane, and draw AF perpendicular to the plane; if from C a line be drawn parallel to AF, it will 1« the perpendicular required. For, being parallel to AF, it will be perpendicular to tbe same plane to which AF is perpendicular. Cor. 2. From the same point, whether without or in ■ plane, there can only be drawn one perpendicular to tin' Proposition LXXXVII. — Tbeorbm. Planes CD, EF, to which the same straight line AB !■ perpendicular, are parallel to one another. If not, they shall meet one another when produced; W them meet their common section, will be a \ GH, i" which take any point K, and join AK, BK ; then, v AB w ^ to the plane EF, il is ^ (Oet V), Ka the | BK, whidl GKOIIBTKT OP rLAKES. a thatpIoBe; .-. ABK ia a t'L; tor ■ame reason, BAK ia a r'Z ; wher»< I the two /j ABK, BAK. of the \BK, are together equal to two r'lt, ich la impomble, (Prop. 10, cor.) ; .'. {danea CD, EF, thoneh ^rodoced, do meet one another ; UuU ia, thej aie illel, (Def. &> PsoroeiTioN I/XXXYIII.— Thborbh. f two atroight lines AB. BC, which meet one another, MfwctiTel; parallel to other twoDE, £F, which meet one diet, but are not in the same plane with the first two; phM which passes thtDi^;h the first two is parallel to t wUdi passes through the latter. ' ^ Ora^h the pcrint B draw BG -«- to b_— — ffc^r^'" -Jj^--~^ I^MW which, passes through EF, ^lIVopi 86), meeting it in G; and «gh 6 dnw GK if EF, and GH D. And -.■ BG is -I- to the plane it is a. (I>e£ 3) to GK and ^ which meet it in that plane ; ■'• of the Ls BGH, BGK ia a r'L. And since AB and are each || ED, they are || one another, (Prop, and BGH ia a r'L, ■■■ GBA ia also a r'L Agwn is II BC, for each of them is || EFi .-. since KGB r'L GBC is also a I'Z, (Prop. I?)- Since .-. BG is 'Z« to each of the linea BC and BA, it ia ~i--to the le AC, and it was drawn -J- to the plane DF. Since 1 BG is -<- to each of the planes AC, DF, ihese planes parallel, (Prop. 87). PnopoaiTioiT LXXXIX, — Tbeobem. r two parallel planes AB, CD, be out by a third GH, their common sectiass EF, GH, with it are dlel. 'or the \s EF and GH are i: e plane, namely EFGH, which I the planes AB and CD ; and f do not meet though produced : the planes ia which they are do meet : .*• EF and GH are parallel .Ml" SSaKETBT OK PUHKB. "^ &. " ^^ " ^ PflopOBixroN XC— Tbkobem. If two parallel planes AB, CD be cut by a third plan* EH, they have the same inclination to that plai Let the 1* EF and Gil bo the common BecHons of the plane EH, with the two planes AB and CD ; and from K, any point in EF, draw in the plane EH ' the I KM at / Le to EF, and let it meet Gil in I. ; draw also KN at / l» tff' EF in the plane AB; and through the |s KM, KN, lets plane be made to pass, cutting the plane CU in the line LO. And ■■■ EF and GH are the common sections of th» plane EH, with the two parallel planes A B and CD, EF i» tl GH, (Prop. 89). But EF is at r-^s to the plai passes through KM and KN, (Prop. 80) ; ■.■ it ia at r-Z* to the lines KJt and KN ; .-. GH is abo at r>la to ths same plane, (Prop. 83) ; and it ia .-. at r'Zir to LM, LO, which it meets in that plane, .*. since LM and LO are it T^Ls to LG, the common section of the two planes CD aod. EH, the Z.OLM is the inclination of the plane CD to ll plane EH, (Def. 6). For the same reason the Z.NK1 is the inclination of the plane AB to the plane EH. Bi V KN and LO are parallel, being the common sectioDsof&a parallel planes AB and CD, with a third plane ; the interiw iNKM is = the exterior /.OLM ; that is, the inclinatia« of the plane AB to the plane EH is =: the inclination d ^ plane CD to the same plane EH. PnoFosiTiON XCI. — Theorem. AB, CD be cut by parallel planH points A. E, B ; C, F, D ; the; liime ratio : that ia, AE : EBa If two straight GH. KL, MN, in shall be cut in CF : FD. Join AC, BD, AD, and let AD meet the plane KL in the point X; and join EX, XF. 7] -.' the two parallel planes KL, MN, are cut by the plane EBDX, the common sections EX,BDareparallel,(Prop.89). Forthesame reason, -.■ the two parallel planes GH, KL, ^f are cut by the plane AXFC, the common ii sections AC, XF ate parallel. And ■■• EX w // BD, a Bide of the A^^Ti. KE-.EB=AX: op.59). Again, v XFis [| AC, aside of the AADC, :!XD = CF : FD. .-. AE : EB=CF : FD, (Alg. ] tpROPOSITIOM XCIl, — Theohem. ight line AB be perpendicular to a plane Gti plane CE toucliing it will be perpendicolai to (" « plane CD. 'or let CBG be their line of com- 1 section, and from any point G, CG, let EG be drawn -i- to it in plane OE. Also, let BF. Gil be < t« CQ in the plane CD. Then LB 18 J- to the plane CD, ABF i /L ; and Bince EG is (| AB, I GH is II BF, the ZEGH ia also a v'l, {Prop, fl t the Z.EGH is the inclination of the plane CE to the le CD, (Def. 6) ; .-. the plane CE is at lY* to tlwj le CD. "^ PaoposiTioN XCIII. — Theokkm. ' two planes AB, BC, cutting one her, be each of them perpendicular k third plane ADC, dieir common ton BD will also be perpendicular be Bame plane. h>m D in the plane ADC, draw -L to AD, and DF -l to DC. v is X to AD, the common section of ^ ilanea AB and ADC,and -.- the plane is at I'La to ADC, DE is at I'ls be plane AB, and .-. also to the | BD in that plane- the same reason DF is at r'ls to DB. And since is at t'Is to both the lines DE and DF, it is at i^Li e plane in which DE and DF are,' that is, to the plane -JProp. 80). ijProi SOLID GEOMETEY, .^golid angle ia that which is formed by more tliid plane angleB at the Bame point, but not in iVe aax k BOt-tD eSUUETBT. I solida bounded by planes are Eimtlar, when theii solid angles are equal, and tbeir plane figures similai^ each to each. III. A parallelopiped is a solid bounded by six planei, of which the opposite ones are parallel. If the aojaceat sides be perpeni^culai to one another, it is a reclangnlar parallelopiped. IV. A ciile is a rectangular parallelopiped, of wIucIl six sides are squares. Y. A prism is a solid of which the sides are paislldo grams, and the ends are plane rectilineal figures. VI. A pyramid is a solid of which the sides are triangle^ having a common vertex, and the base any plane re»' tilineal figure. If the base be a triangle, it is a triangoltt pyramid. If a square, it is a square pyramid. VII. A cyUndei- ii a solid described by the rerolntioK of a rectangle about one of its sides remaining fixed ; whiA side is named the axis ; ajid either of the circles describe^ by its adjacent sides the base of the cyhnder. VIII. A cone is a solid described by the revolution of i right angled triangle about one of its sides remaining fixe^ which is called the axis ; and the circle described by Ihtf other side is the base of the cone- • IX. Cones or cylinders are similar when they are d« scribed by similar figures. Pboposition XCIV — Theohem. If a solid angle A be contained by three plane a ^ BAG, CAD, DAB, any two of these are together greatlt than the third. ^ If all the Ls be =, or if the two greater be =, the proposition is evi- dent. In any other cose, let BAC be the greatest /., and' let ISAE be cut off frota it, = DAB. Through any i point E in AE, let the | BEC be drawn in the plane of the Z.BAC to meet its sdes il B and C. Make AD=EA, and join BD, DC. •.- ih A* BAD, BAE have AD=AE and AB common, an the included Ls BAE, BAU=, the base BE is = BE (Prop. 5). But the two sides BD, DC are IP' BE, EC, . DC is :^ EC ; and since the side AE is = AD, and A( common to the two At ACD, ACE, but the base DCj" EC, .: (he ^CAD is ^^ EAC, (Prop. 14). Conseqiwnll the sum of the is BAD, CM) \f.^ '^'i LBAC. ^ St*n» BWHTBTBI'.' Pbopo9ition XCV. — Teieohem. II tlie plane angles which form any soUd e, are together leas than four right angles, et there be a solid L at the point A con- ed by the plane U BAC, CAD, DAE, 3, these L», taken together, ale ,^ four For, through any point B in AB, let a .e be extended to meet the aides of the , [ ^ in the lines of common section BC , DE, EB, therehy forming the pyramid BCDE — A^^ from any point O within the rectilineal figure BCDI^^ ch is the base of the pyramid, let the \» OB, OC, OD, , be drawn to the angular points of the figure, nhich I thereby be divided into as many ^$ as the pyramid aides. Then '■■ each of the solid Ls at the base of the imid is contained by three plane i«, any two of them together ^ the third, (Prop. 94.) Thus, ABE, ABC, together z^' EBC. .■. the Z« at the bases of the A* ch have their vertex at A, are together ^^^ the L> le bases of the As which have their vertex at O. But he Li of the former are together = all the Zs of ihe rr. .'. the remaining L» at A are together -^ (he re- nins L» at 0, that is, -^ four right angles, (Prop. 1^ _ i Proposition XCVI. — Theobem. ~ ' two solid angles he each of them contained by three e angles, and have these angles equal, each to each, alike fiiluated, the two solid angles are equal, et there be a solid angle at A contained by the three e U BAC, CAD, DAB, and a solid angle at E, con- ed by the three plane U PEG, GEH, HEF, = ler, each to each, and alike situated, these solid a ^or let the [» AB, AC. . EF, EG, EH, b ■■ tl, and let their e es he joined by the i BC. CD, DB, EG, , HE; and thus there formed two isoscele unids, BCD-A, and a-E. Vpon the bases BCD. FGH, let tte - 143 souiD oaoHZmT. EL, fall from the Tertices A, E. Then ■■■ the I'Ld A' AKB, AKC, AKD, have equal hj^oteniues, and the Hde AK common, their other aides,KB,KC,KD, (Prop. 39, cor, 2). are also equal ; .-. K is the centre of the circle that cir- cumecribes the ^ BDC> In like maaner, L is the centre rf the circle that circumBcribea the A FHQ, But these A* are equal in every reapect. For the sides BC, FG, are- equal, '.■ they are the bases of equal and similar A* BAQ EFG; and for the same rnnsou, CD=:GH, andBD=HF. If. .■. the pyramid BCD-A be applied to the pyramid FGH-E, their bases BCD and FGH will coincide. Al» the point K will fait on L, ■.■ in the plane of the circh about FGH, no oilier point than the centre ia equaUjr distant from the three points in the circumfereDMi the perpend ionlai KA will coincide with LE, (Prop. 8% cor. 2), and the point A will coincide with E, ■■• the t'Li A» AKB, ELF, have equal hypotenuses, (Prop. 39, cor. 2), andBK=FL. But the points B, D, 0, coincide wi* F, H, G, each with each ; .-. the |s AB. AC. AD, coineife with EF. EG, EH, and the plane U BAG, CAD, DAft with FEO, OEH, HEF, each with each. Consequeallf the solid i» themselves coincide and are equal. Cor. 1. If two solid angles, each contained by thrrt plane angles, have their linear sides, or the planes that hound them, parallel eaeli to each, the solid angles an equal, (Props. 84 and 89.) Cor. 2. If two solid angles, each contained by the samfl number of plane angles, have their linear or plane ai» parallel, each to each, the solid angles are equal. For eftcK of the solid angles may be divided into solid is, each oon- tained by three plane Li, and the parts being eqTuJ, and alike situated, the wholes are equal. I PaoposiTioN XCVII. — Theorem. If two triangular prisms. ABC-DEP, and GHK-LMH, have the plane angles BAG, CAD, DAB, and ^GKj, L KGL. IXJH, and ale GH, GK, OL, about two of their solid angles at A and G equal, each to each, and alike situated, the prisms are equal and similar. For, since the Bolid Ls A, O, are each contained by three plane Za, wbicb ait T Bides AB, AG, AD, »U' *Liit*L MUD OEOMETBT. 1 43 lal eacli to each, these solid ig beinj; applied to each er, coincide, (Prop. 96.) .-. the U AB, AC, AD, coiii- e with the \g GH, GK, GL, each with each. But AB nciding nith GH. and the point D with L, the |DK at feifupon LM, (AS.16J, and BE on HM; .-. the nt £ coincides with M, and the point t' with X. Con- [oently the rectilineal figures which bound tlie one sm, coincide with, and are equal and similar to the tiliDcai figares which bound the other, euch to each ; ! solid Zs of the one coincide with and nre equal to the id l» of the other, each to each ; and the sulida them- ves wte equal and aimilar. Cor. 1. If two triangular pyianiids have the plane glea and linear sides about two of their sulid angles ual. esefa to each, and alike situated, the pyramids aie nal and Gimiinr. Cor. 2. If two triangular prisms have ihe linear sides init two of their solid angles both equal and parallel, ;h la e»ch, the prisms are equal and simitar. Cor. 3. If two pyramids have the linear sides about lir vertical angles both equal and parallel, each to I jvjraniids are equal and similar. Proposition XCVIII. — Theokkm. ""^^ the opposite sides, as ABCD, EFGIl, of a parallelopi- I ABCD-EFGH, are similar and equal parallelograms, 1 the diagonal plane divides it into two equal and simi- pnsinB. Por, since the opposite planes, AF, ir.areparallel.Andcutbya third plane )) the common sections AD, UC, are ^ 'allel, (Prop. 89) ; for a similar reason > and BC are parallel ; hence the lire ABCD is a i — ? . [n like manner, all the figures which and the solid are i — i s. Hence AD EH, and DC=HG, and the iADC is =EHG, (Prop. ); .-. the As ADC, EHG, (Prop. 5), and conse- ently the c=?s ABCD, EFGH, are equal and similar. Again, -.- AE, CG, are each — and || DH; they arc and II one another; .*. the figure ACGE is a c=7,(Prop. , cor. I), and divides the whole parallelopipeil AG into o triangular prisms ABC-EFG. and ACD-EGH, and ese prisms are equal and similar, (Prop. 9T)-, •.- ftva ine If and iiaear sides about the aoliil i.s a^. V n.i\i.\> I4-I aClLID OEOMETBT. are equal eacli to each. For the plane Ls EFG, ADC, aw. := one another, each being =^ ihe ZEHG, and the linear sides FG, DA, ate := one anotber, each being =EH, and. so of the others. Cor. I. The opposite solid angles of a parallelepiped Cor. 2. Every rectangular parallel opiped is bounded bf tec tangles. Cor. 3. The ends of a prism are similar and equal figures, and their planes parallel. Cor. 4. Eyery parallelopiped is a quadrangular prisnii of nbich the ends are parallelograms, and conversely. Cor. 5. If tno parallelepipeds have the plane anglei and linear sides about two of their solid angles equal, eacli to each, and alike situated, or the linear sides, about two of their solid angles, both equal and parallel, each to eaeb, the solids are equal and similar. Cor. 6. If from the angular points of any rectilineal figure, there be dravrn straight lines above its plane, all equal and parallel, and their extremities be joined, Ihs figure BO formed is a prism. If the rectilinear figure bes parallelogram, the prism is a parallelopiped. If the recti- lineal figure he a rectangle, and the straight lines be per- pendicular to ila plane, the prism isarectangular parallelo- piped. Cor. 7- Every triangular prism is equal to another triangular prism, having its base equal to the half of ona of the sides of the prism, and its altitude the perpendicnlu distance of the opposite edge. The prism EEG-BACs ABF-DCG, (formed by the diagonal plane AFGD), which has for its bitse half the side ABFE, and for its alti- tude the perpendicular distance of the edge CG ; for thej '. are each half of the parallelopiped FD. I Proposition XCIX. Parallelepipeds ABCD-EFGH, and ABCD-KLMN upon the same base ABCD, and be- __ tween the same parallel planes AC, /]_ EM, are equal. //y Because the \» EF. GH, KL, MN, 'fr^—/Mf are || AB, CD, they are i| one A i\///n,/ another, and being all in the same \\y/ /// plane, NK, ML, being produced, will \^Em/-''' .-. meet both EF and HG ; let them WW' meet the former in 0, P, an4 Xbe\aV- * SfWitB OEOSfliTBT. 149 in R. Q, and let AO. BP, CQ, DR, be joined. Tlie re ABCD-OPQR, is a parallelopiped. For, by hypo- lis. the plane Ea !s || AC, the plane of the paralleh :-IIQ is II the plane of lUe parallels AB-EP, and plane AD-NO parallel to BC-MP. Hence AEO- IR and BFP-CGQ are two triangular prisma, whicL e (he linear sides AE, AD. AO. about the solid /A, 0. equal and || the Hnear sides BF, BC\ BP, about solid Lli, (Prop. 98), each to each. Consequently se prisms are equal, (Prop. 97), and each nf them being en away from the whole aoHd ABCD-EPQH, the re- iaders, the parallelopipeds AG, AQ, are equal. In the le manner, the parallelopiped s AM and AQ may be )Ted equal, .'.the parallelepipeds AG, AM, are equal one another. Cor. Triangular prisma, upon the same base, and be- een the same parallel planes, are equal. For, if two planes be made to pass, the one through !, EG, and the other through AC, KM, they will bisect paratlelopipeds AG, AM, (Prop. 98) ; .-. the prisms >C-EHG, ADC-KNM, will be equal. Pro POSITION C, — Theorem. »araIIelopipeds ABCD-EFGH, and ABKL-OPQR. i parallel pb L equal bases, and between the s :, EO, are equal. I^ASB I- When the bases have ) side AB common, and lie be- !eo the same parallel lines AB, Z, these paraUelopipeda AG) I, are equal. ?0T let the planes EAL, FBK, Bt the planeHC of the former, the lines of common section j [, KN, and the plane HF in lines EM, FN. Then since EA, AI„ are || FB, BK irpianesareparallel,and the plane EB ia || the planeHK; he figure AN is a parallelopiped. But ADL-EHM, 1 BCK-FGN, are two triangular prisms, which have linear sides, AD, AE, AL, about the solid Z.A, both = 1 II the linear sides, BC, BF, BK, about the solid /.B, h to each ; .-. these prisms are equal, (Prop. 97, cor. 2) ; I each of them being taken away from the whole solid, {KD-EFNH, the remainders, the paraUelopipeda AG, *, are equal. But AN is =AQ, (Prop. 99) ; .•• ^G^a 4Q. ^ SOLID OBOMETBT. Cabs II. When the bases ABCD, CEFG, are equiangular, haying the /DCB = the IGCE, place DC, CE, iu one |, then GC, C13, are also in one ]. Also let their sides be produced to meet in the points II, K. Since ^ the c^s AC, CF, are equal, they are uompleraentsof thei^Z? AF, and the |HCK is itsdiagonuL Upon the base AF let a parallelepiped be erected, of ihs same altitude with those upon AC, CF, and let it be cat hy planes || its sides, and touching the lines DCE, GCB| these planes divide the whole parallelepiped into four otha paralleiopipeds, upon the bases DG. AC, BE, CF. But the diagonal plane touching HK divides each of the poruUelo- pipeds upon AF, DG, BE, into two equal prisms, (Piop> 1)8). .-. the prisms upon HGC, CEK, are together =: the priBins upon HDC, CBK; and these heiag taken &«s; from the ^ prisms upon HFK, HAK, the reniainderh tlw parallelepipeds upon AC, CF, are equal. Gonsequentlj any two paralleiopipeds upon these bases, and betneeu (ht same parallel planes, are equal, (Prop. 99). Cabb III. When the bases are neither equiangular, nor have one side common, a parallelogram van be described on the base of the one equal to it, (Prop. 26), and eqi^ angular to the other, by which it is reduced to the secood case. Therefore, nniversally, paralleiopipeds upon equl bases, and between the same parallel planes, are equaL Cor. i. Paralleiopipeds of equal bases and equal alti- tudes are equal. Cor. 2. Prisma, Upon equal bases, which are either bolk triangles, or both parallelograms, and of equal altttodeii Cor, 3. Two prisma, upon equal bases, the one a angle, and the other a parallelogram, and having the S altitude, are equal. Cor. 4. A prism upon any rectilineal base ts equal parallelepiped having an equal base and the same aldluds.' Proposition CI. — Ti Paralleiopipeds EK, AI„ of equal altitudes, are to oai another aa their bases EFGH, ABCD. Produce EF' both ways to N and R. and make the liM FQ such that the altitude of the £r=7 EG is to the allitntto of the c=7AC as AB is to FQ, and complete the / — i¥ft aitd the paralleiopiped FS, Hence the cjFW is = tin 'iZZ'AC, (Prop. ti4) ; and \\ie YM&\\e\o'g\^i» VS aad Al aoioD omoiMSfTKr. W A' / -^r- • /t , — t 1 / / V I — lT ^ * ' ting equal bases and altitudes, are equal. Take FA aaj iltiple of FQ, and FN any multiple of FE, and compleU t l=ts Wa, EM, and MN, and the solids QT, EP, and <■ ; it is manifest, tbat since FtJ, QB, are equal, the unit JV, WR, are equal, (Prop. 2?) ; and that, since FW, WR, i eqoal, (he solids FS, QT, are also equal, (Prop. 100); what multiple soerer the base Gli is of FVV, the mum ultiple is the solid FT of the solid FS. In the same maif iT it may be sbovrn, that wh.-kt multiple soever the bma N is of the base EO, the same multiple is the solid KN the solid EK. Now, if the base NQ be :::^ the base GE, e solid NK will be ;:^ the solid FT; if equal, equal; aad less, less. -■. EK : FS = the biise EG : the base FW; d since the base FWz=AC, and the solid FS=:AL; the id EK : the solid ALz: the base EG : the base AC. Cor. 1. Prisms standing on their ends, and of equal alti- des, are to one another as their bases. Cor. 2. Parallelepipeds, npon the same or equal bases, I to one another as their altitudes. For the parollelopipeds KN, KR, on the same base KF, •■to OSM another as NG ; GB, that is, as FN : Fit. ^P^fBi^Ilslopipeds DE, KL, wbicb bave a solid angle «f the one equal to a solid angle G of the other, are to e another in (he ratio compounded of the ratios of the ear sides BA. UC, BE of the one, to the linear sides .•', GH, GL of the other, each to each, about these solid gles. For, let AR, CB, EB be jduced to M, N, O, 80 that a=GF. BN=GH and BO GL. With the lines EB, M, and BN, let the parallel- iped EP be completed; and th the lines OB, BM, and lV, the parallelopiped OP. ten -.- the ratio DE : OP is Pboposition CII. — Theoheii. i 148 90i.it> a^oMenrri , compounded of the ratios DE : EP, and EP : PO, of whicH the former ia = AC : MN, and the latter ia = EM : MO. (Prop. ]0I), or EB : BO, (Prop. 101, cor.2)j the ratio of DE : OP is the aanie nitli that nliich is compounded of AC : MN, and EB : BO, or of AB : EM, CB : BN, and EB : BO. But OP is = KL, (Prop, 98, cor. 5) ; -.■ these paral- lelopipedB have the plane is and linear aides about their solid is B and G respectively equal and alike situated. Consequently, the ratio DE : KL ia the same with that which is compounded of AB : BM, CB : BN, and EB : BO ; or of AB : FG. CB : HG, and EB : LG. Cor. I. Two rectangular parallelopipeds are to one ' another in the ratio compounded of the ratios of the linear aides of the one to the linear sides of the other, each to each. And anj ratio compounded of three ratios, (whose terms are straight lines), is the same with the ratio of the rectangular parallelopipeds under their homologous terms. Cor. 2. Two cubes, or, in general, two similar parallel- --■ -' 'o one another in the triplicate ratio of their Cor. 3. Similar parallelopipeds are to one another at the cuhes of their homologous linear sides. Cor. 4. If four straight lines be in continued proportion, the first is to the fourth as. the cube of the first to the cube of the second. Cor. 5. The rectangular parallelopipeds, under the cor- responding terms of three analogies, are proportional. Cor. 6. If four straight lines be proportional, their cubes are also proportional ; and conversely. Cor. 7. Rectangular parallelopipeds, and cOQSequentljr any other parallelopipeds, are to one another in the ratio compounded of the ratios of their bases and altitudes. Cor. 8. Parallelopipeds whose bases and altitudes an reciprocally proportional are equal, and conversely. Cor. 9. Prisms are to one another in the ratio com- pounded of the ratios of their bases and altitudes. There- fore the 2d cor., prop. 101, and 8th Cor. of this, may be applied to prisms. _ PROPOBITION cm. — TheoBEM. Every triangular pyramid may be divided into two equal prisms, which are together greater than half the irhole pyr,iaid, and two eqaal pyramids, which are nmilu to tho whole and to one auothet, T tOLID GXOHBTBr. 4^ Let BCD-A be any triangular pyra- mid. I*t its lioear sidea, AB, AC, AD. be bisected in E, F, G, and the points of section joined ; and let the sides of the base CB, BD, DC, be bi- sected in 11, K, L, and the points of section joined. Also let EH, EK, and ^^ LG, be drawn, v the two sides, AB, AC, of the AABC, are bisected in EF, the |EF is = and II CU, or HB, (Prop. 59), half the remaining side BCj and so of the other \s that join the points of section. Hence EC, CG, and conseqaently EL, are i m , and the planes EFG, BCD, are parallel. .-. the solid EFG-HCL. upon the triangular base HCL, is a prism. And the solid EHK- OLD is a triangular prism = (Prop. 98, cor. 7) a prism on the base, (IIKL=HCL). and having the same altitude as the former prism ; .■• the two prisms EFG-HCL, and EHK-GLD, are equal to one another, and together = a prism on the base HCL, and having the same altitude as the pyramid BCD^A. .-. since tbe base HCL is the fourth part (Prop. 70J of the base BCD, the whole solid HCDK- EFG is the fourth part of a prism on the base BCD, and having the same altitude as tbe pyramid. The two re- maining solids EFG-A, and B!1K-E, are triangular pyramids. They are equal and simitar to one another, and also similar to the whole, because the ^s which bound the one, are equal and aimiliir to the A^ which bound the other, and also similar to those which bound the whole, each to each, and alike situated. But either of these pyramids is, fur the same reason, = the pyramid KLD-jG, and .■'. .^^ either of the two prisma. Conse- quently, the solid HCLK-EFG is ::^^ the two pyramids EFG-A, BHK-E, and :^ half the whole pyramid. Cor. 1. By taking from the whole pyramid the two equal prisms, and from each of the remaining pyramids two equal prisms, formed in like manner, there will re- main at length a magnitude less than any proposed mag- nitude. (Prop. 73). Cor. 2. Since the magnitude taken away from each of the remaining pyramids is equal to a prism on a fourth part of its base, and having the same altitude as tbe pyra- mid ; the solids thus taken from both will be equal to a prism on the fourth part of their base, or the sixteenth ' part of the original base, and its altitude equal to that of the original pyramid. Cor. 3. For ibe same reason the soUda laVcu Uokv ^% 150 SOUD OKOU8TKT. 1 four remaining pyramids will be equal to a prism taTlng its base a, aixty^fourth part of the original base, and its altitude equal to the altitude of the original pyramid. Cor. 4. Therefore all the solids thus token away, that is, the whole pyramid, is = (^+^'g + Bij+-^j+&c.) of a prism, having the same base and altitude as the py- ramid. But C+ + t\t + A + t4¥+ "'C'. to infin.) = *s (Alg. 96). Cor. 5. A triangular pyramid is equal to the third part of a prism on the same base, and haviDg the same alti- tude. Cor. 6. Since (Alg. 109) quantities are proportional to i their equimuKiples, whatever has been proved of prisms, in regard to their ratios, will also be true of the pyraraidi ou the same base, and having the same altitude. Cor. 7. A polygonal pyramid is equal to the third part of a prism on the same base, and having the same alti- tude. For it can be divided into as many triangular pyra- mids as there are sides in the polygon, and earii of these pyramids will be the third part of a prism ou the sane base, and having the same altitude ; .'- the sum of all the pyramids, that is, the polygonal pyramid, vrill be the third part of a prism upon the sum of all the triangles, that ii< on the whole polygon. I • Pboposjtion civ. — Theorku. A cone is the third part of a cylinder on the same base, and having the same altitude. For, if a series of polygons be inscribed in the circular base of the cone or cylinder, each having its number of sides double of the former, and on each of these a pyramid and prism, having the same altitude as the cylinder, lie erected ; the polygon will ultimately be ^ the circle, (Prop. 75, cor. 3}, while at the same time the prism will be equal to the cylinder, and the pyramid to the cone ; but the pyramid is the third part of the ptism ; .■. also the cone is the third part of the cylinder. H Proposition CV. — Tbeoreu. A sphere is two-thirds of a cylinder, having its altitude and the diameter of its base each equal to the diameter of the Bpbere. ' SOLID GBOMBTRT. Let CDLB be a quadrant of a „ circle, DCBE a square described about it, and DCE a right angled triangle, having the side DE="" and oonsequeotlj anj line, as I DE, will be =K0, and GN win be=:GC. If the irhole figure thus formed revolve about DC, as a fixed axis, the figure DCBE will generate a cylinder. (Def. 7), the ACDE will generate a cone, (Def. 8), and the quadrant will generate a hemisphere ; now these figures may be conceived to be made up of an infinite num- ber of lamina, represented in tbickncBS by the line GI or the line KM, the solids generated by the several parts of the line GI will be as the squares of their generating lines; bat the generating lines are in the cone GN, in the circle GH, and in the cylinder GI. Now the square of GI is = the sum of the squares of Gil and GN, for GN is equal to GC, and the squares of CG and GH are equal to the square of the radius ^ the square of GI. .-. the lamina thus added to the cone and sphere are together equal to the lamina added lo the cylinder, and the same is evidently true at any other point. Hence the cone and hemisjihere together are equal to (he cylinder ; but the cone was shown (ftop. 104) to be one-third of the cylinder, therefore the hemisphere is two-thirds of the cylinder ; .-. the whole sphere will be equal to two-thirds of a cylinder, having its altitude double of the line DC, that is, equal to the diame- ter of the sphere; and it is evident, that the diameter of the base being the double of CD, is also equal to the dia- meter of the sphere. Cor. 1. The portion of the sphere, together with the portion of the cone lying between the lines CB and Gr,are together equal to the portion of the cylinder lying between ' the same lines. Cor. 3. Any portion of the sphere, together with the corresponding portion of the cone, is equal to the corrc- tponding portion of the cylinder. L-A^ J GEOMETRICAL EXERCISES. L 1. If a straight line bisect unnther at right aogles, every point of lie first line wiU be equally diataat from the two extremities of the secDDd line. 2. If Etraight lines be drawn, biaecting two sides of a triaogtc at rigbt angles, and from the point of their intersection a. pErpendicukr be drawn to the thicd side, it will bisect the tliird side. 3. If two angles of a triangle bo bisected, and from the point where the bisecting lines cut one another, a str^glit liue he draon to the third angle, it will bisect the thii'd angle. j I. The difieroQce of aay two aides of a triangle is less thui Ili« I third side. I 5. The aum of two aides of a triangle la greater than twice the straight line drawn from tho vertex to the middle of the base. 6. If the opposite aides of a quadrilateral ligore be equal, Iha figure ia a parallBlogrQjn. 7. If tho opposite angles of a (quadrilateral figure be equal, tlu figure is a paralldugrain. 8. If a. straight line bisect the diagonal of a, parallelogram, It will i bisect the parallelogram. 9. The diagonals of a rhombus bisect one another at right angles. 10. If a straight line biaect two sides of a triangle it will be p»- rallel ta the third aide, and equal to the half of it, and will cut off * triangle equal to one-fourth part of the original triangle. I I . The diagonals of a right angled paraUElogram. are equal IQ. From a given point between two iudefiuite straight lines gtvca iu position, but not parallel, to draw a line which aboil be ter Dated by the given Imes, and bisected in the given point. 1 3. If the sides of a quadrilateral figure be bisected, and the jaoent points of blsectioii joined, the figure so formed will be a rallelogram, equal to half of the quadrilateral figure. 14. If a point be takeo either within or without a rectangle, uri etmight Hues drawn from it to the angular points, the som a! Ibt equares of those drawn to the extremities of one dii^^iuU wilt beeqiul to the sum of the squares of those drawn to (he extremities of tU 15. In any quadrilateral figure, tlie sum of the squares of llx diagonala, together with fonr times the square of the line joEni tlieir middle points, is equal to the aura of the squares of the «da 16. If the vertical angle of a ti'iangle be two-thirds of two ri^ angles, the sqnare of the base will be equal to the sum of thesqnxM of the aide, iucreased bj the rectangle contained by the aides; in* if the vertical angle bo two-thirds of one right angle, the squire <i I the base will be equal to the sum of the squares of the ude^ dim- I niahed by tho rectangle contained by the sides. I !7. To bisect a triangle hyaline drawn from B given point in OM I of Its aides. I 1 3. A perpendicular drawn from an angle of an eqnilaMnl I triangle hi the opposite ^de, is equal to three times the radius at (tir inacribed cii'clei. GCOMETRICAIi EXEBCISE8. 153 1 9. If perpendiculan be drawn from the extremities of a diame- ' to any chord in the circle, they will cut ofif equal segments. 20. If, from the extremities of any chord in a circle, straight lines drawn to any point in the diameter to which it is parallel, the n of their squares will be equal to the sum of the squares of the ;ments of the diameter. 21. If, from two angular points of any triangle, straight lines be kwn, to bisect the opposite sides, they will divide each other into pients, having the ratio of two to one, and, if a line be drawn ongh the third angle and their point of intersection, it will bisect third side, and divide the triangle into six equal triangles. !2. If two triangles have two angles equal, and other two angles ether equal to two right angles, the sides about the remaining rles will be proportional S3. If a circle be described on the radius of another circle, any Bight line drawn from the point where they meet to the outer comference is bisected by the interior one. 24. The rectangle contained by two sides of a triangle, is equal to i rectangle contained by the perpendicular from the contained ^le apon the third side, and Ihe diameter of the circumscribing sle. f5. If a straight line be drawn bisecting the vertical angle of a ingle, the rectangle contained by the two sides will be equal to the are of the bisecting line, together with the rectangle contained by segments of the base. :6. A straight line drawn from the vertex of a triangle to meet base, divides a parallel to the base in the same ratio as the J. 7. If the three sides of one triangle be perpendicular to the three 8 of another, the two triangles are similar. 8. If from any point in the diameter of a circle produced, a tan- t be drawn, a perpendicular from the point of contact to the dia- er will divide it into segments, which have the same ratio which distances of the point without the circle from each extremity of diameter have to each other. 9. A straight line drawn from the vertex of an equilateral ogle, inscribed in a cirde to any point in the opposite circumfe- 9e, is equal to the two lines together which are drawn from the emities of the base to the same point. 0. If the interior and exterior vertical angles of a triangle be cted by straight lines, which cut the base, and the base produced, ' will divide it into tluree segments, such that the rectangle con- 9d by the whole \)ase thus produced, and the middle segment, I be equal to the rectangle contained by the two extreme seg- itSb (A line divided in this manner is said to be divided barmo- Jly.) 1. The side of a regular decagon inscribed in a circle, is equal to greater segment of the radius divided medially; the side of a reg- hexagon is equal to the radius ; and the side square of a regu- pentagon is equal to the side square of a regulai* hexagon, tu- ter with the side square of a regular decagon. PRACTICAL GEOMETRY. Problem I. I I J 1 To diTide a given straight line AB into tvro equal parts. From the centres A and B, with the Bame radius ^- half of AB, de- j Bcribe arcs intersecting in D and E, and draw the pCE, it will bisect AB in the point C. I PsoaLEM II. To divide a given angle ABC into two 'equ:il parts. From the centre B with any radius, de- scribe the arc AC, and from the centres A and C with the same radius, describe a intersecting in D, and join BD; the anglti ABC will be bisected by the |BD. PHOBI-EM III. To trisect a right angle ABC) that is, to divide it inM fliree equal parts. From the centre B ^ith any radius, describe the arc AC; and from the centre C with the same radius, cut the arc in D, and from the centre A with the same ra- dius, cut the arc in E, and join BD and BE, and they will trisect the angle as required, Pkoblem IV. To erect a perpendicular from a given point A, in* given line AC. CiSB I. When the point is near the middle of the line. On each side of the point A lake any two equal distances. Am, An. Fromthecentreswi, «, with any radius greater thai describe two arcs intersect- in D. Through A and D draw the itraight line AO, t it will be the perpendicular required. ^ASK II, When the point is near the end of the line. •"roni the centre A with any ra- 9, descrihe an arc mnr. From | centre m with the same radius, 1 the compasses twice o*er on ,-'^ "'''? arc at n and r. Again, from centres n and r with the same B ^g ^O .us, describe arcs intersecdng in Then draw AD, and it will be the perpendicular re- Anoth^ Method. rrom any point n ae a centre, with idins ■=n\, describe an arc. not less n a semicircle, cutting the line in m A. Through m and n draw the aeter mnr, cutting the arc in r, and »■ Ar, and it will be the pcrpendicu- Kqoired. 'or. If the point from which the perpendicular is to be TO were (be eitremity of the line C, it would only be ^ssary to take iiC as a radius instead of nh., and the I parts of the construction would be the some. Anotfier Method. rom any scale of equal parts take stance equal to 3 divisions ; then L the centre A, with a radius ^ risions, describe an arc, and from centre tr, with a ladiua =: 5 di- ins, describe another arc, cutting b— Former in D, and join DA, and Jl be the perpendicular required, Another Method, ith a marqnois square, which is a right-angled triangli )ut of ivory or wood, apply the right angle to the point tnd make one side coincide nith AB, a line drai g the other side will be the perpendiculai required. B» given ^npeipeni Pbobleh V, given point C| without a straight line AB. I ] w 156 ntACTCOAL GIDUBTKr. Case I. When the point is nearly opposite to the middle of the line. From the centre C describe an arc, cutting the straight line AB in m and n. From the centtes m and n describe arcs, with any radius greater than ^mn, inter- A V^.^_J^_^_^ b secting in s. A straight line drawn through the points C and X'' a will be perpendicular to AB. Or, apply one of the sides containing the right angle of a. marquois equare to AB, and slip it alonj; the line, till the point coincide with the other side ; then a line drawn along this side wilt be perpendicular to AB. Ca3b II. When the point is nearly opposite to the end of the line. From C draw any h'ne Cm, and bisect it in w. From the centre n with the radius Cn, describe thi semicircle CDm, cutting the line AI P ; join CD, and it will be the "" »»■ trpendicular required. Aiw/ker Melliod. From A, or any point in AB as centre, describe an arc through the point C. From any other point m, describe another arc through C, cutting the former in n. Through C and n draw the line CD«; and CD will be perpendicular to AB. N.B. — This can also be done as described in Case L, by the marquois square. Problem VI. At a given point E in the line ED, to make an ao^ equal to a given angle ABC. From the point j, B with any radius, ^c _ ^^ describe the arc ^^C^ mn, cutting AB ^■^ \ and CB in m and 8-=^=- is — : n. Prom ihe centre E with (iae radius Bm, describe die PRACTICAL GEOBfETRT. 157 \e mn^ and lay it from r to «, and through E« draw the [ght line Est; then the angle FED will be = the BC. r the angle be gi^en in degrees. Draw a straight line ID, and using a scale of chords from the centre E, with dius == the chord of 60^^ describe an arc rs; then take the chord of the number of degrees required from the i, and from the centre r, with the chord of the giren e as a radius, cut the arc in s, and through the points nd 8 draw the |EF, and the IFED will contain the n number of degrees. If the angle be greater than 90^, m be divided into two parts, and one part laid off, and I the other part, from its extremity ; it is generally t conyenient to lay off the chord of 60^, and then the rd of the excess aboye 60^. Problem YII. draw a line parallel to a given line AB. kSK I. When the parallel Hne is to be at a given dis- e C, firom the given line. rom any two points m and n ^ g le line AB, with a radius ^ '^^ ^"^ 1 to C, describe the arcs r . 8. DrawDE to touch these ^"^ ^""^^ without cutting them, and c 11 be the parallel required. kSE II. When the parallel line is to pass through a 1 point C. om any point m in the line with the radius mC, describe D 7 r"B arc Cn, From the centre C / the radius Cm, describe the A. — 1 -1 — b Tw, Take the arc Qn in the >asses, and apply it from m to 0. Through C and DE, and it will be the parallel required. B, — ^This problem is more easily effected with a pa- . ruler. Problem VIII. \ divide a given line AB into . 5 ^» )ropo8ed number of equal parts. trough A and B draw h/m and tarallel to each other. In each le lines Km, Bn, beginning at d B, set off as many equal parts .e Mne AB is to be divided into. in>s TK ACTIO AL 0£OHBTBT. Join A, 5 ; 1, 4; ed as required. ,3; 3,2; &c; and AB will be dirid* Another Method. % Through the point B draw tte indefinite line CD, making anyanglenithAB. Take any point D in that line, and through it draw DE || to AB, and set o£f from D as many equal parts DK, KH, HG, &c., DS the line AB is to be divided w ir a. ^ info. Through the points E, A, draw the hnc EA, and produce it to meet CD In C ; ll* lines drawn from C, through the points F, G, H. &C., will divide the line AB into the required number of parts. Phoblbm IX. ^_ lin To find the centre of a given circle, o described. Let ABCD be the given circle ; take any three points, A, fi, C, and from the centres A and B, with a radin* p^ half the distance AB, describe arcs intersect- ing in j- and 3, and from the centres B snd C, with a radius :;^ half the dis- . tance BC, describe arcs Intersecdng in m and n. Then through the points ar and mn, dratv straight lines intersecting be the centre required. Pkoblem X, To describe the circumference of a circle through anj three given points A, B, C, provided they are not in lli^ same straight line. Use the three points the same as in the last problt then a circle described from the centre <>, with a radial equal to the distance oA, will pass through the other tnv and will be the circle required. Cor. Since the three points are not in the same stra line, they may be the three angular points of a triaii "snce this problem also shows how a cucle may bB {bed about a given triangVe. fbactecaij geomethy. Phoalbm XI. To draw a tangent to a given circle, ihat shall pass through a given ]ioint A.*' Case I. When tie point A is in the circumference of the circle. From the point A, to the centre of the circle, draw the radius oA. Then through the point A, draw BC -^ to o.4, and it will be the tangent rec[uired. Cask II. When the point A is without tli To the point A, from the centre draw the straight line oA, and bise it in m. From the centre m, with the radius mA or mO, describe the semi- circle ABO, cutting the given circle in h. Then through the points A and B draw the line AB, and it will be the tangent required. Frobledi Xn. To two given straight lines A, B, to find a third ] poitional. From any point C draw \«, making any iFCG, and make CE= A, CD and CG each =B. Join ED, andc draw GF || to ED, then CF will be the third proportional required. For (CE=A):(CG = B) = (CD=;m: e CF. y.S.—Ii EG had been made =B, then DP, by the same construction, would have been the third propor- Pboblem XIII. To three given straight lines. A, H, C, to find a fourth proportional. From any point D, draw two straight lines, making any_ ZFDH, and make DE=A, EF = B, - DG=:C. Join EG, - and through F draw FH||Er ■• PKACTICAI. eKOHKTBT. Pboblbh XIV. Between two giren straight lines. A, B, to find a mean proportional. Draw any straight line g CD, in which take CE= /^T^X A, and ED=B. Bisect ^ / \ CD in F, and from the B J ' V cenlre F, with the radius *^ FD or FC, describe a semicirek. From E draw EG J- to DC, DieelJDg the semicircle in G, then EG is the mean proportional required. 2^.£. — Another method may readily be deduced from (Prop. 68) Geometry. PnODLEM XV". To describe an equilateral triangle on a given line AB. From the centres A and B, with the radius AB, describe arcs intersecting in C. Draw AC and BC, and ACB will be an equilateral triangle on the re- j quired line. NoiE. An isoBcelea iriangla may lie deacribod on AB, by takiat as mdtua the length of one of the equiil sides. Far the method et diacribing a. triangle, having its sides equal to three given lines, Wt Geometry, (Prop, i.) Pbohi-em XYI. given To describe a square up( line AB. Draw BC -i- to and =iAB, and from the centres C and A, witha radius =AB, describe area intersecting in D, and join DC, DA, and the figure ABCD wiU be a aquare. Another Method. From the centres A and B, with the radius AB, describe arcs crossing at E. " Bisect EA in wi; then from the cenlre E, with the radius Em, cross the arcs BD, AC, in C and D. Join AD, DC, md BC ; the figure so formed will be PRACTICAL GEOMETRY. 161 Problbm XVII. t describe a rectangle^ whose length and breadth shall espectivelj, equal to two given lines, AB and C. t the point B, in the given line erect BD=C. From the point rith a radius =AB, describe an and from the centre A, with a LS :=C, describe another arc, Dg the former in £ ; join ED A.E : the figure so formed will le rectangle required. TB. Any parallelogram may be described in nearly the same ler, by drawing BD so as to make the proper oblique angle, may be effected, by taking the chord of SO"" from a scale of Is, and from the centre B, with the chord of 60°, describe an nd with the chord of the number of degrees that the angle is to In, cut off the portion required, and draw the line BD through tremity. Problem XVllI. inscribe a circle in a given triangle ABC. sect any two of the Lst as A C, by the lines Ao and Co. from the point of intersec- 9, let fall the -t- on upon r of the sides, and it will be idius of the required circle. Problem XIX. a given circle, to inscribe an equilateral triangle, a ;on, or a dodecagon. )m any point A in the circumfe- , as a centre, with a distance equal ! radius Ao, describe the arc BoF. AF, AB, and BF, also bisect AF en BF laid three times round the £' , will form an equilateral triangle. )r AF laid six times round will a regular hexagon ; and AG laid e rimes round will form a regular dodecagon. R. The side of a regular hexagon is equal to th^ t%i- )f the circle in which it is inscribed. FB A OTTO At. OXOMBTST. I PROBLEU XX. To inEcribe a sqaaie in a given circle. Draw two diameters, AC and 13 B, 4- to each other, and join AB, BC, CD, and DA; then ABCD will be the square required. Cor. A circle may be dessribed about a given square ABCD, by draw- ing the diagonals, AB, CD, and their point of intersection, O, will be tlie *- centre ; from which, with the radius OA, describe a circle, and it will be the circle required. Problem XXI, To describe a square about a given circle. Draw any two diameters, AB, CD, _ „ at ■>'Ls to one another; then through the points C and D draw FG, EH || to AB, or -L to CD, and through tlie points A and B draw FE and GH |t to CD, or -L- to AB; and the figure FEHG will he a square described about the »l Problem XXII. To inscribe a regular polygon of any proposed nomlin of sides in a given circle. Divide 360° by the number of sides of the polygon, and make the /.AoB at the centre, containing the number of de- grees in the quotient. Then if the points A, B, be joined, and the chord AB ap- plied to the circumference the proposed □umber of times, it will form the polygon required. Pros L KM XXIII. On a given line AB, to form a reguli proposed number of sides. Divide 180" by the number of sides of the figure, and subtract the quolient from 90°. Make the Z,sOAB and DBA each equal to the difference fio found ; and from the point of intersection O, vith the radius OA or OB. describe a circle, and the chord AB \»ei.\i^ a^^\wi □13 m; if anj to the circumference the proposed number of times, will form the polygon required. Pboblbm XXIV. To make a triangle et^ual to a given trapezium ABCD. Draw the diagonal DB, and „ thiough C draw CE || to DB, meeting AB, produced in E, and join DE; the AADE will \ leequaltothetrapeziumABCD. For the ADEB is = the ADCB, (Gee Problem XSV. To make a- triangte equal to any right lined figur ABCDEFG. Produce the side AB both _ irajs at pleasure, and draw the diagonals EA, £0, and by the last problem make the AE1A= tie trapezium AGFE, and the ^EHB=: the trapezium BCDE. Draw IK || to AE, and HL ij _ to EB ; then join KE and EL, K aad the ^KE]. will bo equal ti the figure ABCDEFG. PnoBLEM xxvr. To describe an oral or figure resembling giren straight line AB. Take any points, C, D, at equal distances from A, B, and on CD describe two isoaceles A*i CED, CFD, and produce the sides to llie points p, o, n, in ; then from tbe centres C and D, with the ndiusAC or DB, describe area bounded by the sides of the is- oweles A* produced ; and from the centre F, willi radius Fm or Fn, describe the arc mii, and fro radios Ep or Eo, describe the ar formed will be on oval. 1 ellipse o le centre E. with the 1, and the figure thus J r^ GENEBAL EXERCISES IN ALGEBEA. I 1. Find the sum of 3a* + 4alj + b\ 4a'— 406+*=, '+4ai+4i', 7a"— 36', and 5o- + 10a6 + 56^ Ans. 18a" + 14a6+ffi'. 2. Find the sum of 16<ic+]6c' + 4ai'', 12c'+7ac+a\ 6a'+4c% l3ac+lUa'+c^ Ua^'+^ac+lc^ and3a''+ 12ac+lV. Ans. ie9a''+70ete+i9<:'. i 3. Findjhe sum of I4a+jac—5c, lSa^5J^+8e, | 7a— 14V^, 3c+5a, 7c— 6Va^+5a, 13n+14Vac+3c, and4^ac— 6a+4c. Ans. Slo— 6Vac+20r, ' 4. Find the sum of 4a + 76— 3c, a + 4o6— c, 5a— 3aS 4-4c, 3a—alj+c+3d, ia+7c—id, and 5d— c+4a— 2i. I Ans. 21a+56+7c+4d. 5. Find the difference of 1^+ i2ab+13b^, and 7a'+ 116". Ans. 8a=+12a6+2i'. 6. From 17((c+5a'+6c°, take 17(ic— 5q"+3c' — id. Ana. I0(t'^+3tfH*i ' 7. From the sum of 3a'-f2(wr + 2a:^, and 5n" + 7(ij:— r. take the sum of 7a=— 7aj;+a;', and 12aj;— 3a»— 4**. Ans. 4a"+4(W+lr- 8. From a^ + b^ +e' + 2al>+2ac+2bc, take 2a6+2ao- 26c— a'— 6"— c*. Ans. 2a'+26' + 2c"+4ic. 9. Multiply a+6 by c — d. Ans. ac+be — ad— M. 10. Multiply a— &+C— <; by ffl+6—c—i;. Ans. a'—b'^ — c'+rf^ — 2ad+2he. 11. Find the product of (a:— 10)(«+l)(j: + 4). Ans. it"- 5j-^— 46x— 40. 12. Find the product of a'+oc+e'^, and a" — ac+e'- Alls. a'+oV+f'. 13. Find the continued product of a-f-l, a-f-S, a+S, and« + 4. Ans. (x'+]0o=+35a=+S0a+2t H. Find the continued product of a — x, a-\-x, fl'+M + a:", anda'- «j; + a:^. Ana. a^—if. 15. Multiply j;'^+10j-y + 7. bya;''- aEi/+4. Ans. a:*+4A— tiOA''-i-nx-— ary+aa 16. Divide fi^- 3it:=+12*:'+14:c, by 3x. Ans. 2^^- a:*+4iB+4|. J7. Divide ^'— 81, by r~3. Ans. r' +-ix-- +9x+^. i8. Divide a'+6^ by a+d. Ans. a''^al-\-h'. 19. Divide a^+y^ by 3-+?/. Ans. x*—3^i/+^'V^ — ^y^+i/*- 20. Divide 3:=+;EV+a'/+/. by *'+/■ Ans.a.^+y^ 21. Divide j^ — f>x*+qa^ — qii'-^px — l,\iyx — I. Ana. 3;'-(p-lK+(5-p+J>^_(p_l>+l. 22. Divide^_4.'+'i^-lf _^i^+27.b7;-- 1+3. Aw. -^'^-Sji'-f j+9. Projuscuous Exehcises. 23. Find the sum of (he products (a+!i+c)(a+b~c), („+6+c){(.^+^), and (a+b+c)(^i+h + c). Ana. a'-\-b-^-i~c'+2(a+Sac+2bc. 24. Find the sum of the products {ci+x){a+x), {a — x) (a^r}.and (a+:e){a^-x). Ans. 3a^+A 25. Find the sum and difference of (a+aXa-f «), and (a_x) X (a^-x). Ana. Sura 2a'+2:c", diff. 4ax. 26. Find the sum and difference of (a+b+c)la+l + c). aBd(a+5-<-)(«+6-c). Ans. Bum 2a''' + 2i' + 2c''+4a6, and diff. 4ac+4bc. 27. Find tiie sum of the products of {a+b+c+d) (a+b+c—d), and (a+J— c-J-rf) (a — i-fc+rf), and alau Ibeit difference. Ans. S\if» 2a''+2ab+2ac+4bc+ 2ad, and their diff. 2ab+2ae—2ad+2b^+2c^—2d^. 28. Uo«v much doea the product of (x-f fi-f 2c)(a4-£ +2c), exceed the product of (a — 6 — 2c)(a — b—2c). Ans. 4ab+ai 29. Divide a'^— «*, by a^—r'. Ans. a+a^^i'+j-. — & by a* — 1/ Ans. a^+aH^+ah'^+b^l 31. Find the greatest common measure of a* +ai — 12i-, and a* — 5a6+fK-^ Ans. a— 36. 32. Find the greatest common measure of 6a'.\.'Jai — 3i',aiid6u- + H,fi-f-3?A Ana. 2a+3J. 33. Find the leaat common multiple of sf* — a\ and /'—a*. Ans. x^ + ji^a—xa^—a^. 34. Find the least common multiple of a — I, a^—i, a— 2, and a^ — 4. Ana. a^ — 5(t^ + 4. 35. Find the least common multiple of 2^—1, 4a'— 1, and4a'+l. Ana. 16a*— 36. Reduce to its lowcfit terms .77^ — :. Xws. — . jy — d) Tib — t EXEaCISES IN ALGEBRA. 37- Reduce to its lowest teTms i8. Reduce to its lowest terms ■— ~ *" . !9. Reduce to its lowest terms ac+ax+bx+bc n'+P+c'+2a6+5ge+at(r f ^^H 40. Reduce to its lowest terms ■ 41. Prove Iha. 1+ "itg=' = (wiwK.tt -l) H 42.P,.„thatl_ia^ = &±J=9fc£:!0. ^B-'«.Pri)T6tli.lbjArt,.(29-3I),AJg.J/j'_pt!f]'|, " o^bemi«c.dloth.fom&i^i*±^]ft£:«f±t?l 44. Prove that if. In the list example, s= ■■, Qieex- ■pression may be reduced to the form sfs — (t)(s — 6)(j_o). 45. Show that a -i- ^ 1^^ ~ + t ' ^^f^^pt a^6. 46. Show that a?+l:^a'+ffl, except a=\. 147. If fl.^i:a:, show that {x.iraf—^.^i<i3?. 48. Multiply a;— ^ by ^ + ^. Ans. f — ^. 49. Mdt,ply;j+ ^r-2-,-T^. ^y?-^- +2^- , a* 4c»d« , He'd* *W .n IT 1^ 1 0*— 91+20 , o*— 13n+42 , a^— ll.+SS SO. M,Jt,ply-jc5^, I7 ^^j— . A»..-^j— ' EXEBCISES IN ALGEBBA. IB? 5.T Find the sum and difference of |- — ^](^ + j^ )■ , [I a\/m ii\ , Sim 2an , Sac 2flm ^-1 (a + ^}(n- Sj- ^'- -^ - ST- ^^-^ ^ - li^- 54. Show that — -—— - - ib equal to ^. Involution and Evolution. 55. Bequired the square of a^^ — j:+l. Ans. a;'— 2jr'+3j:'— 2x+l. 56. Required the cube of \+2x+33:'', 57. Show that the cube of a — -is a' 5— 3f« ). 58. Extract the square root of 4jc* — '12j^+25x'' — 24a: f 16. Ans. 2x^~-3x+4. 59. Extract the square root of a' — 2a^4- V — o +^'!1' Ans. n'— ^+f 60. Extract the cube root of -^ — ^ +3abc^~bV. '' ' , "=' ^ Aub- -t — he. 61. Extract the cube root of «*— fb^+lSj;*— 203^+15a:^ — 6a;+l. Ans. a:^ — 2r-J-l. 62. If two numbers differ by 1, prove that the difference of their squares is the sum of the two numbers. 63. If two numbers differ by any number (a), prove that the difference of their squares is the sum of the num- bers multiplied by their difference (a). 64. If the sum of two fractions =1, prove that their, difference is equal to the difference of their squares. B5. Prove that the square of any odd number diminished by 1 is divisible by 8. 66. Prove that the product of two odd numbers la odd ; Md that the product of two even numbers, or of an even and an odd number is even. tl EXEBCISGS IN SlUFLE EHUATIONS. ^. GiTen 8a;— 4=13— |;r, to find x. Ans. x=2. te.Givcn2K+7+|-=6j:— 23,tofinaiE. Xua.ai^V^, EXERCISES IN ALGEBRA. Ang. x=U 71. GiTeii| + |+ j=7j:— 734+1, to find*. o„4.T K»j.o Ana. x=:i» 72. Giren -^ — ^ =36, to find a:. Am. x^ 73. Giren —— 7""= -— , tofinda:. Ans. ;i:=| 74. Given 1- - — ■ =19 , to find x. ^ ^ * Ana. J^=J 76. Given aj:-\-e=bx-i-d, to find x. Ans. ^= — j 76. Given tM;+6'=o''+6*, to find x. Ans. j;=a+J 77- Given(«+^)(6+.r)— u(J + p)=^'+'^'. tofiudj AnB. x=^ 70. If f _1_ '^ =Sab. x=: ^}^ 79. Il^=ic+d+-. ^ t«+J ™-"»+t-7:-='- ^_ .(.+a.->| '"■"^+lr+:^ + ^— /■■ '- U/„, ^■«{^:jg:lJ?} ..=16 J J-=35. t «-"{i:TgsS} .= 1« J=!l& 84. If-;x+2=38j- (j,+,=46j .= 6 ,,=14 1 ;=32. 85. IfJ*||==20il Ij+J«=34 J ,=32 ■=10. (2x+5,-70=- 86. If-! 6i_ y+3.= I7I+6J+ ! = 288) 227!- 297) .=13 . PBOBLEMS PRODUCING SiMPLE EqI 87- How much money have I in my pocket, when tfce I fomrlt and fifth of the same together amount to 1 Is. 3d. Ana. L.1, as, K. In a company of 266 persons, consialing of officers. merchantB, and students, there were four times us many I merchants, and twice as many officers, as students. How many were of each class ? Aus. 38 students, 152 merchants, and 76 officers. 89. Divide the number a into three such parts, that the second may be m times, and tho third u times as great as the first. . _ a ma na l+m+n' l4ni+n' l+m+n' 90. A father gives to his five sons L.IOIW, which is to be divided among them according to their ages, so that «ach of the four elder may receive L.20 more than his next younger brother. How much will the youngest re- ceive? Ans. L.KiO. 91. A courier who goes a miles a-day, started n days Itefore another, who goes h miles a-day. In what time will the second overtake the first? , "" , Ans. days. 92. A person paid the sum of L.2, ISs. in crowns and shillings, using ]? pieces in all. How many of each sort "Was used ? Ans. 9 crowns and 8 shillings. 93. One of my acquaintances is now 40 years old, and lis son 9 years. In how many years will ihis man, who is now more than four times as old as his son, be only twice as old ? Ans. In 22 years. 94 A transcriber was asked how many sheets he wrote treekly: he answered, " I only work four hours a-day, and cuiuot finish 70 sheets, which I could wish to do ; but if 1 could wort ten hours a-day, then I should write weekly exactly as many above 70 sheets as I now write less ttiau tkat number." How many sheets did he write weekly? Ans. 40. 95. Find two numbers, such that if the first be multi- Jlied by 2, and the second by 5, the sum of the products ■"ill be 31; and if the first be multiplied by 7, and the wlher by 4, the sum of the products will be 68. Ans. First 8, and second 3. 96. It is required to find a fraction, such that if 3 be 'kubtracted from the numerator and denominator, tlie valut ef the remaining' fiaction will be ^, and if ^ be aA4ei \o r. What is the ftactionl Am. ^. 97- The Bum of two numbers is =a, the difference of ! Aeir squares =6, "What are the numbers ? , il»+& , a— • Ans. -5—, and -^. . A, B, and C, one L^lf)0 amongst them, and u one of them can paj this sura alone. But when tliej unite, it can be done in the following ways: by B putliar ^ of his property to all A's; or by C putting § of bn property to that of B ; or by A putting | of his piopeitf to tbat of C. How much did each possess ? Aas. AL.1530, BL.15^0, CLllJft 100. Required to find three numbers, which possess ih following properties: — If fi be added to the first ani second, the sums are to one another as 2 to 3 ; if 2 h added to the first and third, then the sums are to another as 7 to 11; but if 36 be subtracted fiom the « and third, the remainders are to one another aa GkT What are the numbers? Ans. 30, 48, aodSO Quadratic Equations. 101. Ifa:"+6«=27. ^-S, or— 9. 103. I{x'—7x+3i=0. x=6^oi^. 103. If a;*— SJic^lB, x=Q, or — 2j 104. USx''— 3^=65. 3;=5,or— 4|! 105. If -^ = ~. :c=14, or — 1 1+60 3:1^5 106. If «+ -~ =3a^4. r=5, o 107. l(a\l+bV)=b(2a^:i+b). X-. x=: — , er —• 109. If(a— SV— (a+6>+26=0. x=i,ia^. Hi. If»"_7i'=8. x=^,i>t— 1. •m A%aKBKk. m 114. If|"' ^^tul i"-2ij-/= 31 I ^K" ti»"+2«-j-y"=ioi I ^^^Tboblems producing Quai "boblems producing Quadratic Eqitations. U7« What number is h, whose half multiplied by its Ihird part, gires 864 ? Ans. "^2. 118. It is required to find a nmnlier, such thut if we Ent add it to 94, then subtract it from 94, and afterivards moltiplj this remainder by the former sum, the prcduci: may be 8512. What number is it? Ang. 18. 119. What two numbers are those whose product is 144, and whose quotient is 9 ? Ana. 3ti and 4. 120. What two numbers are those whose product is =1, and whose quotient is =?/^ . <— r ■ /7i ^ Ans. V ah and ^ - . 121. A gentleman left L.2I0 to three servants, to be divided in geometrical progresaion, so that the first shall We L.90 more th^ the last. Find their legacies. Ans, L.120. L.60, L.30. 122. A certain capital yields. 4 per cent. ; if we multi- ply the number of pounds in the capital by the number of pounds in the interest for fite months, we obtain 1 1 7041 ^. What is the capital ? Ans, 2650. 123. There are two numbers, one of which is greater than the other by G, and whose product is 240. What uamberB are they? Ans. 12 and 20. 124. It is required to find a number, whose sqiiiue ex- ceeds its simple potrer b/ 306? Xtvs, Vft ■3' ' F^ EXBRCrsSH IN ALGEBBA. 125. It ie required to find a number, Bxich that iTwe multiply its third part by its fourth part, and to the pro- duct add five times the number required, this sum eicesdi the number 200 by as many as the number sought is !e« than 280? Ans, &. 126. A person buys some pieces of cloth, at equal prices, for L.60. Had lie got 3 more pieces for the same buNi each piece would have cost him a pound less. ITow mai^ pieces did he buy? Ans. 12, 127. A person dies, leaving cliiidren, and a fortune of L.46,800, which, by the will, is to be divided equBliy among them. It happens, however, that immediately aftfi the death of the father, two of his children also die. Ili consequently, each child receives L.1950 more than he or she was formerly entitled to by the will, bow many chil- dren were there ? Ans. Eight. 128. Two retailers jointly invest L.500 in business, M which each contributes a. certain sum ; the one let hii money remain five months, the other only two, and eaeh received L.4.'iO capital and profit. How much did each advance ? Ans. One L.200, the other L30B. i2Q. What two numbers are those whose sum is 4\. and the sum of whose squares !^01 ? Ans. 2f> and I^- 130. A capital of L. 5000 stands at 4 percent, compoiud interest. "VThat will it amount to in forty years? Ans. L.24005, 2s. l^d. 131 . How long must L.3600 remain at 5 per cent, cob- pound interest, so that it may become as much as L.5(KI0< at 4 per cent, for twelve years? Ans. Ifiyears, ]36daji 132. What capital, at 4 per cent., will fifteen yean hence be equal in value to L.4500, at 6 per cent, for nine years? Ans. L.422I-483. 133. A town contains 20,000 inhabitants, and we kno" that the population has regularly increased -jjn yearh. What was its population ten years ago ? Ans. 14,882, 134. In how many years will the population of a place become ten times as great as it is at present, if the yeailj increase amount to three persons in a hundred? Ans. 78 years nearlj. 135. What is the present value of an annuity of L.20, to continue for forty years, reckoning interest at the rale of 6 per cent, per annum. Ans, L,300, 18s. 66, 136. What annuity, m^rovcd o,t the rate of 4 per cent. per annum, compound inteieBt, "nVi lA &e coi w iwelTt j-ears amount to L.500, l^s. 'i\&.'< Kto,- \.3a,"w.-Sfe. A SYSTEM or PRACTICAL MATHEMATICS, II. OONTAININO LOOABITHMIC ASITHllEnC^ TRIOONOMETBT, MENSURATION OP AND DISTANCES^ NAYIOATION, MENSURATION OF SURFACES AMD aOLIDSy LAND-SURTETING, SPECIFIC ORAYITT, AND GAUGING. WITH A COPIOUS AND HIGHLY ACCURATE SET OF STEREOTYPED LOGARITHMIC TABLES. BEING PAET SECOND OP N*>. XVI. or A NEW SEEIES OF SCHOOL-BOOKS. BTTHB 800TT1SH SCHOOL-BOOK ASSOCIATION. WILLIAM WHYTE AND CO. B0OK8BLLBR8 TO THB QUBBN DOWAOKR, 13^ GEORGE STREET, EDINBURGH. BOVLBION AND STONEMAN, LONDON ; W. GRAFEL, AND G. H. AND J. SMTTHy LIVERPOOL; ABEL HETWOOD, HASCTaili&TlS.'B.^ FINLAT AND CHARLTON, NEWCASTLE. HDOOCXIiVI. bdinbuboh: ARDBSW JACK. PBWTBB* PREFACE Iv publishing ** Part II. of a System of Practical Mathe- inatics,'' the Committee beg to state, that they have used e^eij efibrt in their power to secure perfect accuracy in the Answers to the questions, which are all given €u they can be obtained from the Tables at the end of the volume. This is a very important point to both Teacher and Student ; and yet it is one that has very generally been overlooked in Treatises on Practical Mathematics. The article on Looartthiis contains all that is necessary* to be known in regard to their practical applications to com- : i&on numerical calculations. r The article on Tbioonombtbt embraces the modem im- provements in the science. By adopting the definitions of the . trigonometrical ratios, now used in all theoretical treatises on the subject, it became necessary to demonstrate all the Rules in a difiFerent manner from that followed in other practical treatises on the subject ; and the consequence is, that the de- monstrations have thereby been greatly simplified. The method of surveying by Rectangular Co-ordinatesy here given for the first time, will be found very useful in extensive sur- veys^ and also in Marine Surveying. The rule for finding the angles of a triangle when the three sides are given, is new. The articles on Mensuration of Surfaces and Solids, Und Surveying, and Specific Gravities, contain clear and perspicuous Rules, illustrated by suitable Examples, and a eollection of Exercises, sufficiently numerous to render the Papil expert in performing the various calculations, and the Practical Measurer acquainted with the moftt tt^ij^T^'^^^ methods of takings dimensiona» V FBEFACK. In order to procure a practically useful treatise on Gauoinc the article on this subject has been entirely written by a prac tical man, an intelligent Supervisor of Excise. All the other articles, both in Parts I. and 11., have beei written by an eminent Mathematician in the Scottish metro polls, who is also distinguished as a public Lecturer on tli< science. For the conyenience of Teachers, and the use of such Stu- dents as do not enjoy the benefit of an instructor, a Key has been published simultaneously. with, the work. Besides the solutions at leogth of all the Exercises in the book, there is appended to it a valuable Supplement, containing demonstra- tions not given in the work, and much other useful matter. CONTENTS OF PART IT. inns, . • . . • .1 I of Tables of Logarithms, . . 2 ic ArithmiBtic, . . • .6 Q of Tables of Log. Sines, Tangents, &c., . 1 1 of Latitude and Departure, . . .15 nterest and Annuities, . . . 16 Trigonometrt, . . 19 f ' • • « • •J" Right-angled Triangles, . . 22 Oblique-angled Triangles, , . .24 tion of useful Theorems, . . 29 I Angles when the sides are given, . . 32 8 -for the Sine and Cosine of the sum and difference jigles, - . ' . . . 33 Trigonometry, Art. 51-85, . . 34 Mensuration of Heights and Distances, . 37 yy Rectangular Co-ordinates, , .42 aduced from the angular bearings of three distant rhose distance is known, • . 46 JUS Exercises, . . . .48 Nayigation, . . - 49 ng, ailing, Wng, . tude Sailing, 53 • ^A. k ^^ • \\i. ■ s b^ w. I SURPACKS, To find the Area of a Parallel agram, . To find tlie Area of a Triangle, To Hod tho Area of a Trapezoid, To find the Area of a Trapeziara, To find tho Area of any Irregular Figure, To find the Area of a Regular Polygon, To find the Dioiueter aud CircumfereDoe of a Circle, the o from the other. To find Che length of an Arc of a Circle, To find the Area of a Circle, To find the Area of a Sector of a Circle, To find the Area of a Segment of a Circle, To find the Area of a Zone, To find the Aroa of a Circular Ring, . To find the Area included between two Arcs of different Circle^ having a common Chord, . . . i ^B* find the Area of a Figure bounded by a curve and atraight To To i or Solids, , ) Ta find the Solidity of a Priam, . . . ] To flad the Surface and Solidity of a Cylinder, . , I To find the Surface and Solidity of a Pyramid, . < To find tlie Surface and Solidity of a Cone, , . 1 To find the Surface and Solidit]' of the FruBtutu of a Pyramid or Cone, . . . . . ( To find the Solidity of a Wedge, . , . i To find the Solidity of a Priamoid, . . .1 To find the Curve Surface of a Spliere, or any Segment or Zone of it, . . . , ,1 To find the Solidity of a Sphere, ... 1 To find the Solidity of a Segment of a Sphere, . . I To End the Solidity of a Zone of a Sphere, . To find the Soliditj- o( a Circular Spindle, To find the Solid Content of the Middle Frustum or Zone of ■ Circular Spindle, .... To find the Solidity and Surface of a Circular Ring, . To find the Surface and Solidity of each of the Regolar Bodw% M a Field ot (hree aiea, ueasure a Field ot four aiiw. C01ITB1IT8. TU P«ge toiiieasiize»Fiflid«f oMratib^nfoiirsidMy 1<M .To survey a pieee of Land in the form of an irreg^ular beU^ 107 To measure a Lake or Wood without entering it, .106 To find the breadth of a River without croaBing it, . IM BcBcription of the Plane Table, . . • 110 To survey with the Pkne Table, . . Ill To survey with a Theodolite^ • . . 114 To find the SurfiMse, and draw a Phoiy of hiUy or sloping ground, . • . . 115 To deduce the true angle at a station from angles measured near it, . . . . .116 To find lite leogiSityf a line ^oh ^fte level of the sea ^onesponding to a line measured at an elevated level, . . 117 On flie Division of Land, . . . . 1 1)B Misedluieoiis Ezer^bes in Ltod Bltfteyilig, l!3i Snworic Gmajoi^ 138 ^tMe of Specific Gravities of Bodies, . . ib. Oiren the Specific Gxftvity and !ISkAidity <^ a Body, to find its Weight, ..... 126 CKrcn the Specific Gravity and Woigitt Of a Body, to find its Solidity, . . • . .127 To find the Specific Gravity of a Body, its Weight and Solid Content being given, . . . . ib. Td find the Specific Gravity of a Body without knowing its So- li^, . . . . . . 128 To find the Specific Gravity of a Fluid, . ' 130 To find the quantity of each ingredient in a given compound of two ingredients, • . . ib. Gaugihg, . • 131 Deseription of the Sliding Rule, . • . 132 To find the content of Solids of greater ^epth than one inch, 137 Ganging open Vessels, . . 139 To Gauge a Malt Cisterii, &c., ... ib. To Gauge a Vessel in the form of a FHistuin of a Pyramid or Gone, ..... 140 To Gauge, Hx, and Tabulate a Vessel, which is nearly circular, 141 To Gauge a Vessel nearly rectangular, • . \V^ To Gmx^ » Copper with a riaing crown, • , \^4 To Qat^ a Copper with a &Umg crown, • . . \i^^ . T0 Oai^ a Veaa^ with a £aU or dnp, . \K^ 'o Gauge and Taliulaie a,\ OrdinateE, To Gange a. SlM, Matt Gauging, Cask Gaugiug, . Ullaging of Casks, Miacellaiieous Questions is General ExHrciaea, . Loganthmic Sines, Tangeata, and Secanls, for every point ani quarter point of the Compaaa, IjOgarithniB nf all niunbors, iraui 1 1o 100, , LogBrilhmBofallDQmbere, from 100 to 1,000,000, Logarithmic Sines, Tangents, and Secants, for every degree and minute of the Quadrant, Natural Sines and Coainoa for every degree and Minute of the Quadrant, . . . • . Difference of Latitude and Departure, Compound InterSBt and Aiinuitiee, , . . Length of Circular Arcs, &a, . . . it. OF LOGARITHMS. 1. Logarithms are a kind of artificial numbers, invented bj Lord Napier, of Merchiston Castle, near Edinburgh, to racilitate certain calculations, such that while the natural numbers increase in geometrical progression, their loga- rithms increase in arithmetical progression. By this arti- fice the operation of multiplying numbers is reduced to that of adding their logarithms ; that of dividing numbers to that of subtracting their logarithms ; the raising of powers to that of multiplying the logarithms ; and the extracting of roots to that of dividing the logarithms of the numbers. 2. If n:=a*, x is called the logarithm of the number n to the base a ; or the logarithm of a number to a given base is that power of the base which is equal to the given num- ber. 3. The base may be any number whatever except 1, but file most convenient base is 10, which is the base of the common logarithms given in this work ; but a simple illus- tration of the nature of logarithms may be given, by taking ike base 2, or 3, or 10, and only using those numbers of the natur^ series whose logarithms are whole numbers. Thus to the base 2. 0. L 2. 3. 4. 5. 6. 7. 8. log. 1. 2. 4. 8. 16. 32. 64. 128. 256. nat. num. Or to the base 3. 0. 1. 2. 3. 4. 5. 6. 7. 8. log. 1. 3. 9. 27. 81. 243. 729. 2187- 6561. nat. num. Or to the base 10. 0. 1. 2. 3. 4. 5. 6. 7. 8. log. 1.10. 100. 1000. 10,000. 100,000. 1,000,000.10,000,000.100,000,000 nat. num. In either of the above series it will be found, that if two logarithms be added together, such as 3 and 5, their sum S points oat the number which is the product of the num- kefs of which 3 and 5 are the logarithms ; or if the diiFe- mnce of two logarithms, as of 7 and 5, be taken, the re^ aiainder, 2, will point out the quotient arising from divid- iiK the number of which 7 is the logarithm, b]f tha^t q^ iUck 5 is tfad h^nthm; hence. _ I of the lor/arithna of two itumbera it the foy-- «W(Am of their product, and the difference of the logarithm of two nv/mheri ie the logar Uhm of their quotient ; wfiicli mt^' be demonstrated generally as follona, bj the notation in(Art.2); for let ii=a', and n'=a'', then their produrt. ^»ea }in'=a'xa''=a"'+ ^>, where a:-f-*', or the snm of the logarithms of the numhers, is evidently the ]ogarilhin({ their product: also their quotient gives -^ — .^a'**^, where x — x', or the difference of ihe logarithms of the nuj hers, is the logarithm of ibeir quotient. 5. The logarithm of a. potver or root of any number the logarithm of the number multiplied by the espoaent which indicates that power or root; for let «=a'repreB«it any number having ils logarithm x to the base o, and 1*1 m represent the exponent of any power, (m^ an inleget)i or root, (m^ a fraction) ; then we have 7(™=;(a') "=: therefore if x be the logari tbm of n, »w is the logarithi its With power or root ; hence by giving proper values t we have the following niles for raising numbers to power, or extracting any root. 6. To square any number by logarilbms, multiply Oil logarithm of the number by two, and the product wiUbt the logarithm of its square. 7. To find the cube of any number by logarithms, tiply (he logarithm of the number by three, and the prodncl will be the logarithm of its cube. 8. To find the square root of any number by logarilhiDS, divide the logarithm of the number by two, and the quo- tient will be the logarithm of its square root. 9. To extract the cube root of a number by logarilbmii divide the logarithm of the number by three, and the quO' tient will be the logarithm of its cube root. k Description of the- Tablks of Looabithnj 10. Since the base of the logarithms here used U 10. any number will be equal to IC, where x is the logarilia of the number; if j = 0, then 10"=!, therefore the loj- ariibmof lis 0; if ^=1, (hen ]0'=10, the logarithm of 10 18 1, bence the logarithms of all numbers between J UmI W are greater than ani \eM \Wti\, V\«rt is, they tn LOGABITHMS. O decimal fractions. In the same manner, by making a;=2, it is evident that the logarithm of 100 is 2, and therefore that all numbers between 10 and 100 will have their loga- rithms 1, with a decimal fraction annexed ; but all num- bers between 10 and 100 are written with two figures, bence the int^pral part of the logarithm is one less than the integral places in the number, and this rule may in the same manner be proved universal. For this reason the decimal part (or mantissa) only of the logarithms is in- serted in the tables^ and the integral part (or index) is pre- fixed by the following rule : 11. Find how many places from the unit's figure the first significant figure of the number is^ that prefixed to the dedmal part found from the tables will be the true loga- liUun. If the first significant figure be to the left of the unit's place, the index is plus, but if to the right, the index is, n^nus, and the sign ( — ) must be placed over it; thus the logarithm of 150 is 2176091, while that of 0015 is I 2*176091, where the ( — ) applies only to the index, not to - the mantissa, it being always ( + )• ' 12. To find the logarithm of any whole number under 100. Look for the number under N, in the first page of the logarithms, then immediately on the right of it is the loga- rithm sought, with its proper index. Thus the log. of 56 18 17481&, and the log. of 91 is 1-959041. 13. To find the logarithm of any number between 100 ' and 1000. iind the given number in some of the following pages ' of the table, in the first column, marked N, and imme- diately on the right of the number stands the decimal part ti the logarithm, in the column marked 0, at the top and bottom, to which decimal prefix the proper index, (Art. 11). Thus the log. of 457 is 2-659916, and the log. of 814 is i^l0624. 14. To find the logarithm of any number consisting of four places. Find the first three figures in some of the left-hand co- himns of the pages, as in the last case, and the fourth figure at the top or bottom of the table ; then directly un- do the fourth figure, and in the straight line across the S, from the first three figures of the number will be d the decimal part of the logarithm sought, and the index must be supplied by (Art. 11). Thus the log. of 7^384 is 0*868292, the log. of 793*5 is 2-899547, au^Xk^X efHMa38iff«»3E39A _ ' 1&. To find tbe logarithm of any number consistiog ol fire or six placee. Find the logarithm of the first four left-hand 6gures, as in the last article, to which prefii the proper index bj (Art. 11); then from the right-hand column, marked D, (meaning tabular difference), take the number oppoaite to that logarithm, and multiply it by the remaining f^rea ol the natural number ; point off from the right-hand side d the product as many figurea as there are in the multiplier; then add the rest of the product to the logarithm befoie found, and the sum is the logarithm required. Ex. 1. Required the logarithm of 348G3. Log. of 34860 by the table is 4.54232? Diff. 125 X 3=37-5, therefore ad.i 37 I Gives log. of 34863 =4-542364. Note. It maj be remarked here, that wheu the firat flgm pointed off in the product nf the difference is S, we may tUber tab the remnining iigureE, or tiie reiaaJnin); Rgtires increased by one, M in either caao the crrer is |, in the last fi^re or the logarithm; Iwl if the figure or figures pointed ofl" exceed "5, the remaning Sgora must slwaya be iucreaaed by unity, as the error by this meara vU be leBB than j, ju the last figure of Clje logarithm, whereas by onJt- tiag then, it wuuld be greater than a half. Ex. 2. Required the logarithm of 46-8375. Log. of 4683 by the table is 1-670524 Diff. 93x75=69-73. ■■• add 70 Hence log. of 46 8375 =1-670594 Ex. 3. Required the logarithm of •0076452.^. Log. of -007645 by the table (Art. 1 1) is 3-883377 Diff. 57x25=14-25, .-.add 14 Hence log. of '00764525 =3-883391. EXERCI8K3. The log. of 4fi is =1-662758 The log. of 97 is =1-986772 The log. of 286 is =2-456366 . The log. of 901 is =2954725 . The log. of 7569 is =3-879039 . The log. of 2344 is =3-369958 . The log. of 46874 is =4-670932 . The log. of 954625 is =3-979832 . The log. of 3-56773 is =0-552394 To find the logarithm of a nilgar fraction, or number. r iiDOAHiTaHB. G Reduce the volgar fraction to a decimal ; then £nd the decimal part of its logarithm hy the preceding rules, and preiis the proper index, as found by (Art. II.) Or, from ibe logarithm of the numerator subtract the logarithm of the denominator, and the remainder will be the logarithm of the fractioa eought, (Art. 4.) A mixed number may be reduced to aa improper fracdon, and its logarithm found in the 81 Es. 1. Required the logarithm of ^■^, =-2916. From log. of 7 by the table ' =0845098 Take log. of 24 =1-380211 Hence log. of a\, or -29 1 (J, is =1 ■464887. NoTB. If (he logarithm of thiR example be taken out &vm ths dednul froetion, since it repeats G, which is the di-citnsl of |, n-e must add | of the t&bulBT diifereDce to the logarithio of the first Imit figuTSB, in order to obtmn the true logarithm ; thus log. of '3916=T-4647HK, and the tabular dttTorence is 14!), two-thirds of ■hich ia 98, which beiug added to the log. foniierlj found, ^vea H6itta7, the some as baforc El. 2. Required the logarithm of 231= 'V''=23'25. From log. of 93 by the table =1-968483 Take log. of 4 Hence log. of =i% or 23-25 17. To £ud the natural numher answering to any given Look for the decimal part of the given logarithm in the different columns, until yon find either it exactly, or the next lesE. Then in a line with the logarithm found, in the left-hand column marked N, you nil! have three figures of the number sought, and on the top of the column in which ibe logarithm found stands, you have one figure more, nhich annex to the other three ; place the decimal point to that the number of integral figures may be one more llian the units in the index ot the logarithm, (Art. 11), and if the logarithm was found exactly, you have the number (cquired. If the logarithm be not found exactly in the table, sub- tract the logarithm next less than it found in the table from tlie given one, and divide the remainder, with two ciphers annexed, hy the tabular difference, which will give other Iwo figures, (if the quotient give only one figure, the first is a cipher), which annex to the four figures found by in- apection, and it will be the number sought, eice^t \Ve \q.- dex show (hat fienumfaer consists of more ihaivavifeaAMts, * I in which case annex ciphers to the number formerly Found, till it contain the required number of integral places. ThU will be the number sought wilb as great accuracj aa it can be obtained by logaritbma carried to six decimal places, and will be always nearer to the true number than by a mil- lionth of itself, however many integral places the numbtf nay contain. Required the number answering to the log. 3785624. The given logarithiu, 3-785fi24 The next less tab. log. is the log. of ol 04= 3-785615 Remainder, ^ 9. Tab. diff. =71)900(13 to the nearest unit. Hence the number sought is 6I04']3, since the index 3 shows that it consists of 4 integral places. In like manner (be natural number corresponding to Ibe fulloiving logarithms may be found. 1. Tlie number corresponding to the log. 5-314782, ii 2UfJ-)34. 2. The number corresponding to the log. 3-290035, ii ■00195. 3. The number corresponding to the log. 2-531907, » 34()'335. 4. The number corresponding to the log. 0-357912, il 2-27988. 5. The number corresponding to the log. 6-486S05, it 3063410. 6. The number corresponding to the log. 1-282169, l> 1^1915. LoaAniTUMic ARirrtiuBTic. 18. To PERFORM Multiplication by Looarithhs. EuLK. Add together the logarithms of all the factors. and the sum will be the logarithm of the product. If flome of the indices be negative and some positive, add the positive indices and the carriage from the decimal part Into one sum, and the negative indices into another, and their difference will bf the index, which will have the same sign as the greater suia. Ex, 1. Required the fioduct of 371 an<i 84-5. LoE. ot S1\=.0-^of£ftlV Log. 0^84.5 =\-^ie6Vl^ Log. of product, -a^^ AS>5='i-AamV- Logarithms. 7 Ex. 2. Required tlie product of 56*125, and 743*75. Log. of56125=l-749156 Log. of 74375=2-871427 Log. of product, 41 743- =4-620583: Note. In finding the number corresponding to the logarithm of the prodact in thiB example, we obtain 41742*9, with a remainder of 84, which is greater than half the divisor 104; we therefore increase the last figure by unity, which giyes the result in the text. The same should be done in eyery case where the remainder is greater than half the divisor. Ex. 3. Required the continued product of 356*225, •6385, -07425, and 8*42725. Log. of 356*225=2-551724 Log. of -6385 =1*805161 Log. of -07425 =2*870696 Log. of 8-42725 =0*925686 Log. of product, 142*32 =2*15326?. NoTB. In the above example we add the decimal part of the log- arithms, which are all positive, and having 3 to carry, we add it to* the positive index 2, which makes 5, then adding the negative in- dices, we find their sum 3, which being taken from the positive in- dex 5, leaves 2, which is the index required, and is positive, because the greater sum is positive : if the sum of the negative indices had bem greater than the sum of the positive, we would have subtracted fte sum of the positive from that of the negative, and marked the lemainder negative: if the sum of the positive and negative indices vere equal, of course the index would be 0. BXBRCISBS. 1. Required the product of 75*825 and 84 75. Ans. 642616. 2. Required the product of -75, -0625, and 846*25. Ans. 39-6678. a Required the product of 4*825, -0225, and 0145. Ans. 00157415. 4. Required the product of 7{, 2}, 31^ and 20 A. Ans. 1012*45. 5. Required the product of '7345, 734*5, and 73*45. Ans. 39625-6. 6. Required the product of -25, -7325, and 01725. Ans. -00315891. JP. To Pemfobm Division by Looabitbl^^. BuLE. From the logmthm of tte di^denSi «vi\i\x«^cX. the logarithm of the diviaor, and the reminder will be the logarithm of the quotient. Or, take the arithmetical complement of the divisor, and to it add the logarithm of the dividend, the sum, rtjictiny ten frmn Ike index, will be the logarithm of the qnoiient. 'llie arithmetical complement of a logariCbm is the diffe- rence between it and 10 of an index, and is contracted into (nr. co.). It is most conveniently found hj beginning at the left-hand side of the logarithm, and salitracting the index and the succeeding figures from 9, except the lait significant figure, which must be subtracted from JO. Ex. Divide 74^-625 by 42 6125. Log. of 743-625=2-87l353 2-871353 Log. of 42-6 125 = 1-62953 6 ar. co. = 8 370464 Quotient, I7-4'>09=1'241817 i2418J7. Ex. 2. Divide 398^ by 7439. Log, of 398-125=2-600019 2-600019 Log. of 7439 =3-871515 ar. co. = 6-128485 Log. of quotient, *06351 85 =2728504 2^728504. Ex. 3. Divide I by -78.>4. Lc^. of 1=0-000000 0-000000 Log. of ■78.'j4=l 895091 ar. co. =10-104909 Quotient, 1-2 7324=0- 104909 Note. When we have to subtract a mlnaa ini by the rules of Algebra, Jjeeausa Bubtraetiog u n adding a plus quantity; and if the minuend ba> must Le eubtracled. EXBRCISES. 1. Divide 385 by 29-25. 2. Divide 463-28 by 474. 3. Divide 364 x 54-75, by 7854. 4. Divide 463x25-25, by 3-1416. 5. Divide 1 by 3-1416. fi. Divide 3-1416 by 180. To Solve a Pkopoetion bv Logaritbus. Hulk. From the sum of the logarithms of the second md third terms subtract the logarithm of the first, and (he ■emainder will be the logariihra of the answer, in the same 'enomination as tbc third term. Or, take the arithmetkil implement of the first leim, Kadi \o \t &dd the loganthnu 0- J 04909. ex, we most add il inus is the samcu Ans. 13-1634, Ans. 9-77386. Ans. 25374-3, Ans. 3721-27. Ans. -318309, Ans. -017453, of the second and third terms, and the sum, rejecting 10 from the index, wilt be the logarithm of the answer. If the proportion be compound, add the logarillims of all the second terms and the third together, and from the sum aulitract the sam of tbe logorilhms of the first leims, and tbe remainder will be the logarithm of the answer. Or, take the arithmetical complement of each of the first terms, and the logarithms of the second terms and the third, then add them all together, and the sum will be the logarithm of the answer, after rejecting as many 10s from the index as there were arithmetical compleraeuta taken. Ex. 1. If 3125 yards cost L.l, 5b., what will 73«62!> jud» eost 1 B As 3-125 log. 0-494850 ar. co. 9-505150 I : 730625 log. 1-863695 log. I-B65695 ^:L.l-25 log 096910 log. 009f)9IO :L.29-225 =l-4fi57a5 1-465755. The liTst method ia wrought in one line, by kilding the ■emd and third togurithma together, &nd froin the suiu Bubtnivtiii); Ibg first, and doing bo in eacli column spparately as we prucend ; nd the second is wrought by ulding tliu three liaes together, nnd abtracting 10 from the index. The answer obtained is L.3!<'-2-25 = 1.S9, 4s. 6d., which may be verified by conunoD Britlimetii!. Ex. 2. If 30 masons build a wall 60 feet long, 20 feet Ugh, and 2 feet thick, in 12 days, working 8 hours per day, law many days should 20 masons take to build a wall 100 ket long, 15 feet high, and 3 feet thick, when they work 10 hours per day? ; 12 days. I 60 ft, 100 ft. 20 ft. 15 ft. 2 ft. 3 ft 10 ho. 8 ho 20=1-301030 Log. 30=1-477121 60=1-778151 ., 100-2000000 20=1-301030 „ 15=M760!>1 2=0 301030 3=0477'2I 10=1-000000 8=0-903090 3-68]i 41 12=1-079181 7-11 2604 Subtract sum of logs, of the first terms =5-681241 The answer. Log. 27=r43r:fli3 Rbiurk. In this eiMinpIe tbe logarithms ot tbe ftrtl, Uivra* tttfc patdovo by tbiuaaelveB, and added inJo one aum.and llic\<igB,ivftwv« Bfti/eteeondandiliird by theniseJves, and added inW, aauVXiM ', *>"■>■ U Bra sum IB auboscted from the second, and tUc remamiet \^ ^ 10 LosAWTSItS. tliB lognritbro or the anawpr, »-hich, bowBver, is less than tlie loga- ritlim of 37, hy 1 in the Ust plaice. Thifi urkeB from the loniiuii- aUutee, that tUo lagai-ithms of 30, S, and IS, are loo small in lb» last figure bj ,%, and tliat of 12 also too Email by ^'g, the i * j'l,; bat the logarithm of GO in tbc first terms is al. L, whieh being taken from the former, leaves ,'g toi lit, whieh in the tables is made lo increaw the last tigiu* )>y unit}', liocnuec it is greater than a lialf. 1. If 25 labourers can dig a trench 220 yards long, 3 feet 4 inches wide, and 2 feet 6 inches deep, in 32 dajs, of nine hours each; how many would it require to digs trench half a mile long, 2 feet 4 inches deep, and 3 Ktt <i inches wide, in 26 days, of eight honrg each ? Ans. 98 labourem 2. If 3 men, working ten honrs a-day, can reap a field, nieasurinjt 150 yards by 240 yards, in five days, how manj men, working twelve hours a-day, can reap a £eld meaaur- ing J92 yards by 300 yards, in four days? Ans. 5 ntta. 3. If 27 men can do a piece of work in 14 days, wott- ing 10 hours a-day, how many hours a-day must 24 boy* work, in order to complete the same in 45 days, the work of a. hoy being half that of a man ? Ans. 7 boon. 4. If 120 men, in 3 days of 12 hours each, can " trench 30 yards long, 2 feet broad, and 4 feet deep ; muny men would be required to dig a trench 50 yatdf long, 6 feet deep, and 4^ feet broad, in 9 days of 15 boiu* each? Ans. 180 21. Involution and Evolution bv Logabithms. RuLB. Multiply the logarithm of the number by lie I exponent of the power or fractional esponent of the tool to be eitracted, and the result will be the logarithm of the power or root required. (See Art. 5.) I Ei. 1. What is the third power of 72'85; or wbl « I (72-85)'? - Log. of 7285 =1-862430 Log. of 38fl625-=5-5872y0 Ei. 2. What is the fourth power of -05326 T Log. of -05326 =2726320 Log. of ■oooooaowv\=?.'2!ssa'«a «E«r*H.K. In this emmv>6, *<: ^ni«^^™^,^'«^,'^>-Zii3 «ddwg the 2, ,hieh w« cans irotu <.\« is=v«>^\ ,«vv.>i»^.*» lOGABITHMS. 1 1 of the index, we subtract it, because it is the index only that is nega- tive, the dednuJ part being always positive. See Art. 10. Ex. 3. Find the square root of 7854 ? Log. of 7854 =1-89509 1 Log. of -886227=7947545 Ex. 4. What is the fifth root of -34625? Log. of -34625 =1-539390 I J Log. of -808869=1-907878 Remabk. In the last two examples, where the index is negative, it most be remarked, that in dividing by the denominator of ^e in- dex of tiie root, the negative index is increased by as many as will make it divisible without a remainder, and then the same number 18 considered as a positive remainder, and taken in with the decimal part of the logarithm; thus the same quantity is first subtracted by considering it negative, and then added by considering it positive; and hence the value of the quotient is not altered. This must al- ways be done when a logarithm with a negative index is divided. EXERCI3ES. 1. What is the sixth power of 1055 ? Ans. 1 -37883. 2. What is the tenth power of 1-125 ? Ans. 3-24736. 3. What is the seventh power of 5-125? Ans. 928662. 4. What is the square root of 2625 ? Ans. 51-2347. 5. What is the cube root of 17? Ans. 2-57128. 6. What is the tenth root of 5386-25 ? Ans. 2-361 18. 7. What is the cuhe root of f ? Ans. -658633. 8. What is the fifth root of J? Ans. 72478. 9. What is the fourth root of i^^^^^^? Ans. 1 05967. 180 10. Find the value of /5^^H1M?\I? Ans. 25-4327- Y 19x52 ) 11. Find the value of (l-05)i'^ x705 x |? Ans. 361-733. 12. Find the value of ( !f l^^'lf o^ ^*? Ans. -849433. EXPLANATION OF TABLES. Of ILoGABiTHMic S1NB8, Tangbnts, and Skck^t^. J. To £nd the logarithmic sine, tangent, ox aec^aX.^ ^l aajr number of degrees and minutes. 19 B.DLE. If the Dumber of degrees be less than 45", wek them at the top of the page ; then in a line with the given number of minutes in the left hand marginal column, marked M at the top, aud under the word sine, tangent, or secant] you have the logarithmic bidc, tangent, or secant, of the proposed number of degrees and minutes. If the number of degrees he above 45°, and under 90°, seek them at the bottom of the page; then in a line wilh the minutes in the right kand taatgvaaX column, marked M at the bottom, and above the ivord sine, tangent, or secant, you have the logarithm sought. Exactly in the same manner is the logarithmic cosine, cotangent, or cosecant of any number of degrees and mi- nutes less than 90° found, in the columns marked cosine, cotangent, or cosecant, at the top or bottom; the name being alwai/B found at the top, when the degrees are found ^ the top, and at the bottom, tvketi the degrees are found at iht bottom. When the degrees exceed 90°, subtract the angle from 180°, and take out the sine, tangent, secant, &c., of the remainder ; or subtract 90° from the angle, and instead of the sine, tangent, secant, &c., of the angle, take out the cosine, colangent. cosecant, &c^ of the remainder, awl conversely. Angle. Siue. Tangont. Secwil. 15= 12' 9-418615 9-434080 1 0-0 154(15 70= 16' 9-973716 10-445259 10-471543 21 => 54' 9-571695 9-604223 10-03252y 36° 15' 9771815 9-865240 10-093425 108» 18' 9-977461 10-480642 10-503081 164= 39' 9-631593 9-675564 10-043971 2. To find the sine, tangent, or secant, of any number of degrees, minutes, and seconds. Rule. Find the sine, tangent, or secant, corresponiliiig to the given number of degrees and minutes, as befolti and from the adjacent column, marked D, take out itis tabular difference, corresponding to the same angle, vrhicli multiply by the seconds in the given angle ; cut off t*" decimal places from the right of tbe product, and add lh« Temaiuiug figures to the logarithm furmerly found, and i< will be the logarithm reijuiied. KEFJCAITATIOII OF TABLES. ^niied the logarithmic tangent of 24° 3' 16"? Tangent of 24° 3' =9-649602 DifF. 566 ^^Correcti n for 16" = 91 9056 ^gangent of24°3'16"= 9tJ4Hti»3 w EXERCISES. Angles. Sine. Tangent. Secant. 36° 12' 51" 9-771444 9-864670 lU'093-227 430 18' 43* 9-836305 9-974394 10-138089 69° 56' 13" 9'972813 10-437449 10-464637 98= 17' 40" 9-995434 10-836287 10'840853 52> 26' 19" 9-899109 10-114057 10'2 14947 1340 45' 6' 9'851359 10'003765 10-152405 The cosine, cotangent, and cosecant of any number of degrees, minutes, and seconds, are to be found in the same manner; except that the proportional part for the seconds most be subtracted. EXERCISES. Angles. 330 14' 24" 21° 46' 52" 34° 17' 32" 32" 5' 35" 73° 43' 8" 129° 16' 11" 9-922405 9-967832 9-917072 9-927979 9-447701 9-801384 10-398387 10-166244 i 0-202642 9465476 9'912545 Cosecant. 10-261103 10-430554 10-249172 10'274663 10-017775 10-111161 3. To find the angle corresponding to any giTen loga- lithmic sine, tangent, 01 eecant; cosine, cotangent, or co- eecant. R111.E. Find in its rcspectiTe column the sine, tangent, or secant nearest to that given, but less, if it be not found exactly in the table ; take out the degrees and minutes which correspond to this logarithm ; then subtract it from the given logarithm, and to the remainder annex livo ciphers, nnd divide it by the tabular difference adjacent to the nearest logarithm found in the table ; the quotient will be (he seconds of the angle, which annex to the degrees and minutes formerly foiind, and it will be the angle ■ought But if it he a cosine, cotangent, or coB&can^, ^n.&^n\^ respectire column the nearest to it, tut gie-A^eT,\^ \*.« ■n.'A, I 14 KXPLAJIATION or TABLES. ^ found cxnctly, and taking out the degrees and minntea correspontJing lo it, subtract the given logarithm from it; to the remainder annex two ciphera, and divide bj the tabular diffotence opposite the nest greater minute ; the quotient will be the aeconds, which annex to the degrees and minutes formerly found, and it will be the angle sought' Example. It is required to find the angle correspond- ing lo the logarithmic tangent 9943765, and to the logs- | rithmic cosine 9-496724. Firsl, By inspection of the columns of tangents, we find that the nest less than it is 9-943732, which correspondi to 41" ]8', and is less than the given one by 13. To this | remainder we annex two ciphers, which mnkca 1300; this I we diTide by the tabular dilference 427, and the quotient I is 3, which is therefore the seconds; and hence the angle ] sought is 41° ] 8' 3". Second, By inspection of the columns of cosines, we find that the next greater cosine is 9-496919, which corre- sponds to 71° 43', and is greater than the given one by 19a; to this we annex two ciphers, which makes 19500; ihii we divide by the tabular difference, corresponding to 43', viz. 637. which gives 31" nearly; hence the angle sought is 71° 42' 31". Sin Re. Angles. Tutigcnts. Ai]gli?B. 9-346373=] 2° 49' 59" 9-764825=30= 1 1' 36" 9782599=37= 18' 49" 9-978546=43=35' 7" 9-952864=630 47' 8" 10- 6 3-5 887 =760 58' 42" Secants. Coaiaes. 10'468357=70° 6' 55" 9'367285=i76° 31' 43" 10'004763= 8= 28' 1 1" 9-846349=45° 24' 3*/' 10-I47364=44=34'52" 9 167249=81= 32' 54" Cotangents. Angles. Cosecants. Angles, 9-874639=53= 9' 24" 10-438674=21= 21' 27" 10-104738=38= 9' 25" 10-017643=73= 46' 43" 9-964286=47= 21' U" 10-114763=50= 9' 18" Note. If the erne, taDgeiit, &c, be knsmt to be tliat'ofatiehsN angle, the correapoDdmg ajigte, as round above, must be subbwltd from IBO% to gel tbo true angle. ^^'1 To find the natural sine, tangent, secant, &c.j of wf .fcer of degrees, minatea, and seconds. ^p Kt«>AITjLTI01I or TABLES. 15 RuLB. Find the logarithmic sine, tangent, secant, &c„ of the given angle, subtract 10 from the index, and take out the numher corresponding to the remainder from the tables of logarithms of numbers, and it will be the sine, tangent, &c., required. The natural sine and cosine of any angle may be more readily found from the tables of natural sines and cosines, the arrangement of ivhich is the same as that of the loga- rithmic sines, &c., except that the differences are oat given. These must be found by subtraction, and the corrections made by the following proportions: 60' is to the given number of seconds, as the diiFerence for 60'' is to the cor- rection required ; and the difference for 60" is to the given difference, as 60" is to the seconds required. Example . What is the natural sine of 12" 34' 12"? Natural ineofl2''34' 217575 Difference, -2178.59- -217575=284 60:12: 284 : 57, o 284 XA^ -f-57 Natural ineofl2°34'ia" 217632 Ex. 2. Find the angle corresponding to the natural co- tiae -843647? -843704=c sine 31- 28'. The differe ce between this and the sive n cosine is 57, Md the difference belwee n COS. 3P 28', an d COS. 31= 29'! iil5fl; hence 156:57: 60:23 nearly; therefore the mgle sought is 31° 28' 22 BXERCISES. HakSnes. Angles. Nat. Coainea. 1 Angi™. ■354638= =20° 46' 17' -576438=54 47' 58" •459637= =27° 21' 49' ■274869=74 2' 45" ■M763g= =57° 57' 21' ■542873=57 7' 14" ■646739= =33= 8' 37' ■984733=10 V 23" ■>. DiFPERBNCE Ol' LATITUDE AND DePABTUHE. In this table are inserted the hypotenuse, sides, and angjea of ail right-angled triangles, whoso acute angles are either whole degrees, points, or quarter points of the compass. The hypotenuse is represented by the distance, placed at the top and bottom of the page, the side opposite the angle by the column marked Dep., (Departure), and the side adjacent the angle by the column mavked. ^£1.^.1^ (Difference of Latitude^. It is principuWy \isei ^ot ^ivi- i ^ \ ing the sides of triangles when the hypotenuse and an angle are given, but may be used for other purposes. Wlien the angle ia found on the left side of the page, the name of the sides must be read at the top, and when the angle is found on the right hand side of the page, the name of the side is found at the bottom. The numbers, 1, 2, 3, &c., at the top or bottom of the page, may be ac- counted 10, 20, 3(1, &c., if the decimal point in the co- lumns under them he removed one place to the right; oi they may be accounted 100, 200, 300. &(;., by remoring the decimal point two phices to the right. Example. Find the sides of a triangle, the bypotenuae being 346, and one of the acute angle 28°? Hypgteliuiie. Angle 300 28° 40 28- 6 38° Side opposite. 140-84 18-779 2-8168 Side adjacent 264-88 35-318 5-2977 346 28° 162.4358 305-4957 6. Another important use may he made of this table, m follows: — In that column marked 6, at the fop and botton, if the decimal point be removed one place to the right, the difference of latitude will express the length of a degree of longitude in nautical* miles at that place whose latitude u the same as the course. Example. Find the length of a degree of longitude on the parallel of 57°! and also on the parallel of 34°f Since 57° is found at the right-hand side of the page, the name must be found at the bottom ; hence the lengdi required is 32678 miles. Again, 34° being found at the left hand side of the pag^i the name is found at the top ; and opposite to 34°i anil under 6, in the proper column, is found 49-742 miles, the length required. I Tablbb of Interest and Annuities. 7. Table VII, gives the sum to which L.l will amoaBt, if improved at compoimd interest, for any number of yeaw not esceeding 60, at 2i, 3, 3|, 4, 4^, 5, and 6 percent The amount of any other sum may therefore be found, bj multiplying the amount of L.l for the given time and ret* Lf the principal, espressed in pounds, ( Alg. 120,) Thus the amount ot T.,30, \0b., M\J3fl-5, for 17 jvoh at 6 per cent., is Ia82'1295765 ; for under 6 per cent, and opposite to 17 years, is found 2()92773. wbich being mul- tiplied by SO.'J. gives 82-1295760. 8. Table Vlll, pves the sum, irhich being improved at compnond interest, will amount to L.l in sinv (fiven mini' ber of years not exceeding fiO, or tbe sum which should be paid down immediately, as an equivalent for L.l, to be paid at the eipiration of tbe given number of years. The present value of any given sum is found by multiplying the present value of L. 1 by that num expreiised in pounds. Thus the present value of L.2.'). due 14 years hence, nt 3^ per cent., is tbe sum found in the table under H^ per wnt., and opposite to 14 years, via. ■6177f!2 multiplied by 25, orL.l.'i-4445.i. 9. Table IX. gives the amount of an annuity of L.I. *ben deferred for any number of years not exceeding &}. nnd improved earh year as it becomes due, at compound intereat, at 2i, 3, 3^, 4, 4^-, fi. and C per cent. Tbe amount of any annuity is found by multiplying the tabu- lar amount, corresponding to the given rate and time, by the annuity cEpressed in pounds. Thui the amount of an annuity of L.RO per annum, deferred for 24 years, and improved at :H per cent., is L.36-666i32a x 80 =L, 2933 32234. 10. Table X. gives the present value of an annuity if 1*1, payable tor any number ot years not exceeding 60, when money can be improved at tbe rate of 2^, 3, 3^, 4. 44. 5, or 6 per cent., the first payment being due one year hence. The present value of any annuity is found by multiplying the tabular value, corresponding to the given time and rate per cent,, by the annuity expressed in Thus the present value of an annuity of L.3fl, to be paid at the end of each year, for 40 years, when money can be improTcd at compound interest, at the rate of 2^ per cent, i» the tabidar value, viz. L.25102775x30=L.753-08a2S. 11. Table XL contains the length of circular arcs to ra- diuv-l, for any number of degrees from 1 to 30, and for every JO degrees from 40 to 180; also the fourth and fifth ulunins contain the lengths of arcs for any number of BiinateB or seconds, the figures given being tbe last of 7 tic^inial places, so that tbe necessary number of ciphers moat be prefixed to put them in their proper -pVaces, vit^B. liBed aloDe. The Jength of an arc to any ot\ieT raA\\i6 \* fituad hf Srst Sniiing the ienplh to radiua \ , ani ftiftvi. la/fipfyiag bj- tbe given radius. I 18 EXPLANATION OF TABtES. Thus, let it be required to find the length of an arc of 74" 26' 43" to radius 12. This is eridentty equal to tli< sum of the lengths of the ares of 7p°. i", 20', &, 40", and 3"; hence length of 7O''=:l-2217305 4'= -0698132 20'= -0058178 6'= -0017453 40"= -0001939 3"= -OOOOUS Xength of arc 74'" 26' 43"=1-2993I52 to radius 1. 12 Length of arc74°26'43"=15'3917824 to radius 12. 12. Table XII. contains the logarithms of the amount of L.l at the end of one year, to ten decimal places, for eTeiy fourth per cent., from \ to 6, it is a necessary addi* tion to the ordinary logarithmic tables, for obtaining cor- rect results in c^uestions on compound interest and annui- ties, -when the number of years for whicli the calcutatioa is made is great. Thus, for example, to £nd the Brntmnt of one penny, improved at compound interest, at the ntt of 5 per cent., since the beginning of the Christian era, or for a period of 1845 years. We take the logarithm of thi amount of L.l at .5 per cent, that is, the logarithm «f 1-05, and multiply it by 1845, then subtract the logantbrn of 240, the number of pence in a pound, to find tlu amount in pounds; hence L.105=-02lia92991 x 1846= 39'094256a'195, from irhich subtract 2-3fi021I, the loga- rithm of 240, and the remainder is 36'7I404(>, which ii the logarithm of the .tmount in pounds, n'hich therefbs consists of 37 places of figures, the first six of which on^ we can find correctly from logarithms carried to six deoi- mal places, the other mast therefore be filled up vHh ciphers, as we have no means of knowing what &tj are; L.5 1 76620,000000,000000,000000,000000,000000. il therefore the amount required. Rehibji . The valoe of a cubic inch of pure gold is about L.4a'434i mod if wo consider the earth as a, globe, whose diameter is TSll miles, we will find that the above sura would be equal in ralouM about 194135001) globes of pure gold, each as large bb our (Mtkl wbilo tha simple interest for the same time, and at tlxi aamewlik would only amount to /s. Old. 13. Table XIII. is insttted to facilitate the talcing oH of the logarithms of Be^eraV TivHft\«i«, <A b«Qji«x,v«coiP- r*nce in calculation. lU uw Sa saSiciKAXi *«%««», ■«*<■ out further cxplaaatioD. E TRIUUNOMCTBI. PLANE TRIGONOMETRY. Artiglb 1. Plane trigonometry Lns for ita object the ilotion of the following problem : — Of the three sides and :ree angles of any plane triangl-e, any three (except tha aee angles) being given to determine ihe other three. 2. This is effected by means of trifjonoraetvical tables, liich contain the ratios of the sides of a right angled tri- )gle to every minute of the <^uadmnt. 3. In trigonometry all the angles round a point are Tided into 360 equal parts, called degrees, each decree to 60 equal parts, called minute*, and each minute into I equal parts, called stcondx. Degrees, minutes, and se- nds, are respectively designated by these characters — ', "; thus the eipression, 36° 1 4' 32", represents an arc angle of thirty-six degrees, fourteen minutes, and thirty- o seconds. 4. In the trigonometrical tables, each side, in succession, oaed as the measuring unite, (called radius), and tha otients arising from dividing the other sides by it. are t in tlie tables under the following names, in referenca _ one of the angles ; viz. sine, cosine, tangent, ci »at, and cosecant, which are contracted thus : e 5. Let ABC be a right angled tri- gle, right angled at C, and let the les opposite to the angles A, B, 0, designated by the letters a, b, and c, rpectively; then the ratios put into - i tables for every minute of the angle , are the following: — 90 Tr-A!»B Ti 11. Cosec.A= . 'i ^T"*' m siile uppoeile' 13. Badias contracted R= - = 13. (e)+(7)l!i.»'^ = = x;- = ^ = tan.(8), ... £ = tan. I h J4. (7)+(6)Biw."^^ = 'x;; = ^=col. (9); .-. j; = <:ot. 1 A ). 1 J5. 1 H-{S) gira ^ =1 X ; = ; = oot. (8); ■■• j^ = V l+(9)Bir.. J^=1X J = J=1«,.(S); .-. i = , 17. i+(iO)g;™__i^=ix|=j=<»..(7)i.-.-i= , _I8.^1+(11) Bive. ;ji.^ =1 X J= = = ■m.(6); .-.^ ; 39. l-^(6) gives r — - =1 X -= -^eosec. (11); .-. — I = COaec. "'" on aa. I ^SO. l+(7)Bi...^-^=lx; = f=.,c.(10); Erop.39.) ^'' -^ ■^ Hence cos- A=; J 1 — »in.'' A, and 22. The relative mafpiitmies of the trigonometrical ratios may lie rpprespntpd as in the annexed dia- I pram, ivhere rudius = AO, OU. I OC, or OE. and in reference to the r angle AOB, or the arc ^^. which is its measure ; ]iD= the sine, OB or 0D= the cosine, AF= the tangent, CI= the cotangent, 0F= the secant, and 01 = 23. AD or R— cosine, is called the verted tine. 24. CG or R— sine, is called the covened sine. 2i>. ED or R+ cosine, is called the iuversed «n«. 26. The difference between an angle and 90* is calW its coraplemenC; thus in the ilii^ram above, COA being a right angle, the iOOB 19 the complement «f A* £AOfi. The names, cosine, cotangent, and cosecant, we contractions I'or sine of the complement, tangent of the •wrapleinent, and eecanl o? l\\e coTo^VmeTv*.. SI 27- The difference between an angle and 180° » its supplement. In the above diagrani, LHOJi ii plement of the IBOA. 28. If XX' and YY' be two lines at right angles to each other is the point of their intersec- tion O; then itll lines measured along XX', or on lines parallel to ^ ihem, Ijing towards the right of YY', are called +. whilst those lying towards the left are called — ; thus OC and OF are +, whilst OE and OD are — ; also those measured n YY', above XS', as BC. B'D, are called - below XX'. as B"E, B"T, are called — . 29. If a line, as OB, revolve Ihroueh all the four quad- rants, XOY, YOX', X'OY', and Y'OX, and in every position have a -J- drawn from its extremity B, upon the line XX', the sine of the angle through which it has passed, beginning in the position OX, is that perpendicular with its proper sign, divided hy OB, and the cosine is the distance from O to the foot of the perpendicular, with its proper sign, divided by OB, Ilence the sine will have the same sign as the peTpeudicular, and the cosine will have the some sign as the line intercepted between O and the foot of the perpendicular. The sine will therefore he + in the first and second quadrants, since BC and B'D are -|- , and — in the third and fourth, since B"E and B"'F are — ; whilst the cosine will he + in the first and fourth quadrants, since 00 and OF are -f, and — in the second and third, since OD and OE are — . The signs of the other trigonometrical ratios can be determined from these, since each of them can be expressed in terms of the sine and cosine, (Art. 13-20). 30. The Bine of an angle w = the sine of Us supplement, and Ifie cosine uf an angle m = — tlie cosine o/ its supplt- For if the ZBOX be = the iBOX', then the Z.B'OX being the supplement of B'OX', will also bp the supple- ment of BOX ; but if the Ls BOX and B'OX', be equal, the triangles BOG and BOD will be similar; .-. ^ will be = 1^, or sin. BOX = sin. BOX = sin.. box:. -Alao ^ will be = — — _, or the cosine at a.ii aa'^t "»a gua/to — the cosiae of its, au-iplement. S2 PLANE TRIGONOMETBT. 31. In the tables of logRritlimic sines, tan^nts, and KcantB, &c., it is the Ingaritlims of these ratios multi- plied by 10.000.000,000 ihat is inserted, and R is the pre- ceding multiplier, it is plain that the ratios are all increas- ed in the same proporlion ; and if we divide each of them by R. vie will have the same values as before ; hence ne obtain ain A cos A 6 a 11 These being all pairs ;en in the form of proporti , thej may be writ- ,B follows :- a(=hyp.:perp.), B : COS. A=e : 6(= hyp. : base), ^- R; tan. A=£:a(^ base : perp.), ^b B : cot. A:=a : £(= perp. ; base), ^" B : sec. A^l/ : i-(^ base : hyp-), '■ E: cosec. A^a; t'(^ perp. : hyp,). Note. From the above six proportions olher six may be obluii»ir by making tha third term of eacli tho first, the fuurth the second, tho first tho third, and the second the foartb. 32. If the side used as the denominator in the Taloeof each of these ratios be called Badins, and the name of the ratio arising from dividin] ■written on that side, we tions included in the two cieut for solving all the ca 33. Cask I. Given ii and one of the acute angli third angle. Bulb. Make the given side radius, responding names on the other sides, thi ing propott I The n the c given side riven side, (Badius), the required side, each of ihe other sides by it k 1 have all the above propor- owtng rules, which are suffi- of right angled triangles that right angled triangle a siJ« o find the other sides and the quired side. 34. Case 11. Given in a right angled triangle, two to find the ancles and the third side, \uLti. Make either of ttie ^\\en eidea radius, and -niit' -— ■ J K.AMX SUOOVOmTKT. S3 the con-espondiag names on the other aides ; then state the folloniDg prnportion : The side made ritdius. Is to the other given side. Is to the name on that side. This result being found in the trignnometricat (ahlea, will determine the vnlue of the angle to which the names were referred ; and since all the angles of every triangle are equal to two right angles, and one of the angles is a right angle, the other will be the complement of that just found. The angles being thus found, the third side can he calculated hj (33). EXAMPLE. The base AG of a right angled triangle is 240, and the perpeDdicuIar BO is 264. Required the angles and the Construction. From a scale of equal parts, lay off AC=240, and -J- to it draw BC=264, join AB, and ACB iviU he the triangle required. Rule (34) gives either ^ AC:CB:3lt:tan. A; or, BC:CA = R:cot. A, ortan. B. 35. In calculating a proportion by logarithms, it is most convenient to take the avit/imeticaC complement of the logarithm of the first term; that is, subtract it from 10 or 20, according as it is the logarithm of a number or of a trigoDonietricBl ratio; to this add the logarithms of like aecond and third terms, and reject 10 or 20 from the index for the logarithm of the answer. The following examples will aU be wrought in this way: Ar. CO. Log. AC 240= 7'6197a9 Log. CB 264= 242Ifi04 I Log. R =20000000 Log. tan. A=47° 43' 35"=I004l3y3 Or ar. co. Log. CB 264= 7 57fi396 I^g.AC240= 2 380211 Log. R = 10000000 >>g.tan.B=42' I6'25"1 _ (..o^o^m angles being thus obtained, the \i5^oVeia\i&e I thu I PLANE TRIOONOHKTBT. e found, (33), by making either AB or CB radiuB; L : sec. A=AC : AB. or E : cobm. A=:BC : AB. Ar. CO. Log. R =10000000 I^g.scc.A =l0\72i97 I^y. AB 240 = 2;^380211 Log. AB 356-7)16= 2^532408 Ar. CO, Log. R =10-000000 Log. coaec. A =]ll 130803 Log. CB2fi4 = 2'4^i604 Log. AB 356-7^6= 25.'J2407 Note. The hjpotenuae might alao be found without findi^ angles, for (Geo., prop. :13). AB*=A(?+BC; .-. AB =jA^+BC', Similarly AC=^AB°— BS, imd ItC=.^ A.B'~/iC. ^^ 49= 1 . In the right angled triungle ABC, right angled at C given AC 300, and BC 221, to find the acute Hnglfannd the hypotenuse. Ans. Z A 36° 22' 40", i.B 53° 37'20". and the hypotenuse, 372-6J3- 2. Given the base 560 feet, and the angle at the biw required the perpendicular and hypotpnuse. Ans. Perp. 648-7^6' feet. Hyp. 857027 feeL ren the hypotenuse 641, and the angle at the base to find the Lase and perpendicular, and the le- maining acute angle. Ads, Base 542-21. Feip. 341-889. Other acute I. 57" 46'. 4. The perpendicular of a right angled triangle is aOOi and the angle at the hase 49"; required the reinwiiin|> parts. Ans. Hyp. 662 306. Base, 434643. Third. LH'. 5. The hypotenuse is 100, and the perpendicular fiO; requited the base and the acute angles. Ans. Base. 60. Angle at the base, 36=52' 11-4". Vertical angle. 53» 7* 486". 6. Given the hypotenuse 580, and the base 361, to. find the other pints, Ans. Angle at the base. al'Sff 27". Vertical Hngle, 38° 29' 3:¥'. Perp. 453939. 36. Thkohkm. In any plane tri- nngle. the sides are to one another . the sines of the opposite angles. Let ABC heanv phine triangle, dra BD -t- to AC. Call the sides oppo- w(o the la A, B, C, tea^i:i;U.v:\i, a, t, ■^' PI.ANB TRIGONOIIETRT. 25 and c, (a notadon which will be frequently adopted), and the perpendicular p. Then froio the right angled A< ABD and CBD^ we have (6.) Sin. A= -, Qiud sin. C= ^ ; .•. sin. A -fc- sin. C= c • a' Or . ' =^ X - = -,or8in.A:;8in.C=:a:c,(AIg.l^). In the same manner it can be proved^ that sin. A : sin.'Bzza : 6, and that sin. B : sin. 0=6 : c; hence g^„ sin. A , sin. A . . • r>i<f^ • x« ^ 37. a-z^c - — 77 =6 - — - ; sin. A=: sm. C - = sin. B r . am. G (prn. B c o 6=c -: — - =a - — r ; sm. B=: sm. C -=: sm. A -. 8121.C sin. A c a sin. C ,8in.C .^ . .<? .t><? c:=za - — 7 =6 -r— :=; ; sm. C= sm. A •«- = sm. B 7 . sin. A sin. B a u 38. Rbmark. The above theorem enables us to find the sides and angles of any triangle^ when there are given either two sides and an angle opposite to one of them, or two angles and a side opposite to one of them. If, how- ever^ the data be two sides, and the angle opposite to. the less, there are two. solutions^ and hence this case is called umbigtuyus. Let AB be the greater side, and make the angle BAD = the given angle, which must neces- sarily be acute, since it is oppo- site to the less side ; then from the centre B, with a distance == j^^ the less side, describe an arc^ ^ ^ and it will either cut the line AD in two points, or meet it in one ; in the last case the triangle would be right angled, and there would be only one solution. In the tirst case» let the arc cut the line AD in the points C and D, and join BC, BD; then each of the triangles ABC and ABD have the given data ; therefore the third side may either be AC or AD, and the angle opposite to the greater side may either be ACB or ADB. But since BC=BD, the ZBCD= the ZBDC, and the ZACB is the supplement of the Z.BCD; .•• the Z.ACB is the supplement of the ZADB. But (30), the sine of an angle is = the sine of its supplement ; .*. both the angle corresponding to the tine in the tables and its supplement mutit be laVeu^oxX^ii^ I 26 FLANR TRIGonOMETBX. angle opposite the greater side; then there will also )>« found two values of ihe third angle, viz. ABC and ABD. one oftvhich gives the tliird side AC, and the other AD. Hence the aides and angles of both triangles are deter- mined. 39. The following rules are evidently deducihle from the theorem (3fj), the hrst being that by which we find an ogle when we know two sides and an angle opposite lo ae of them ; and the second that by which we find a side, 'ticn there are given two angles and a side. Rule I. The side opposite (he given angle. Is to the side opposite the required angle. As the sine of the given angle Is lo the sine of the reijuired angle. BuLE II. The sine of the angle opposite the given aide, Is to (he sine of the angle opposite the required side, I As the given aide i Is to the required side. 1 Example. In the triangle ABC, given AB=450, 1 BC=4I1, and the /,C=60% to find the remaining anglei and ihe third side. Eule 1 . Gives 450:411=: sin. 60" ; sin. A j Ar. CO. Log. AB 450=7-3*6787 Log. BC 41 1=2(513842 Log. sin. 60°=z!l'937531 Log. sin. A=52= 16' 34"=9^8y8160 The third angle Bz;67o 43' 26", (Geo. prop. lit). Again, by Rule 2, sin. 60° : sin. 67" 43' 26' .-AC. Ar. CO. Log. G I. 60": i;(AB=450) : Log. cosec. 60° =10-062469 Log. sin. 67= 43' 26"= 9-966317 Log. 450 — 2-633213 Log. AC=480-838 =a-bB1993 EXfiRcrsEs. 1. Givenoneside215, another248, and iheangleonpo- ^te the latter 74°, to find the remaining angles and tb« ide, Ans. I. opposite, 215=50' 26' 40', Other, 1^49° 33' 20". Third side, 196343-. iven one side 215, another 169, and the angle oppo- e to the former 72°; to find the remaining angles and the Uiirf side. Ans. Z. opposite. lefc^iS* 22' 51". Other, i=5&- 37' 59". Third Bide. 159(»22. 3. GiTcn one angle 64° 13', another 45" 27', and the side opposite the latter 1046 links; required the third angle, and the other two sides. Ans. Third ^70° 20'. Other sides, 1321-66 and ]3J)2'17 links. 4. Gi»eii one angle 49° 15', another 70° 18', and the aide lying between them 5230 feet ; required the sides o[>- pOGite to the giren angles, and the third angle. Ans. bides, 566014 and 45545. Third, LGty 2?'. 5. Oiven one side 800 feet, another 605, and the angle opposite ta the latter 37° 1^'; to find the other paria. Ana, / opposite the greater side, 53° 8' 17". or 12b° 51' 43". OdM-r Z.89° 37' 43", or 15° 54' 17". Third side, 999-88, or 274-01 feet. 6. GiTen one angle 90° 33' 26', another 39° 39^ 20". Skod the side opposite to the former :H002i to find the third angle and the other sides. Ans. Third Z49° 47' 14 ". Other sides, 22926 and J9l-58^. 40, When two sides and the contained angle are given, use the folloiving JitUe : Xhe anm of the tno given sides, l9 to their difference. As the tangent of half the sum of the other two angles. Is to the tangent of half their difference. Dbvonst RATION. Let ABC he a triangle, of which the side BC(a) i8:^AC(6): produce HC to D, so that CD may =AC, and AC to E, M that CE may be =BC ; join BE and AD, and produce AU to F. BD ia=(a+6),and AE is=(a— 6). Because the two /.s CBE and CEB aretogether=CAB and CBA, eachs^ of them beiag^lhe iACD, (Geo. prop. 19); .'. since CBE and CEB are equal to one ano- ther.eachofthemishalf ihesumof the^aAand B. Again, theiCABis^=-theiCEB,oritsequalCBE,bytheiABE, while the ^CBA ia .^ the LCBE, by the same iABE; .'. the /.ABF = half the difference of the Li A and B. AU9 the A« BpF and EAFare equiangular, for CD=CA, .'.the /.CDA = theiCAD = the iEAF, (Geo. prop. 3); also the iDBF = the LAEF, consequently the /DFB = the lAFE, .: each of ihem is a right angle; and since the ^DBF and AEF are eqiiiaBgiOu, BD : DP = ¥.k k 28 PLANB TRICON OMF.TRT. : AF, (Geo. prop. 61), and alternately, (Ale. 105). BD:EA=DF:AF _DF AP P*' ~bf'bf I' =tan.DFB:tan. ABF; that ii, {a+b) : (,a—b)= tan. i(A+B) : tan, K*— B). NoTK 1 . Since the angle C ia given, 4( A + B ) can be found, for it is ^{180°— C)=(90'— iC). The first three terms are therefore known, consequenily the fourth can be found ,' then l(A+B) + J(A~B)=A, the greater angle, which is always opposite to the greater aide, (Geo. prop. 11); and i(A + B)— i(A— B)=B, the less angle. The angle* being thas Found, the third side can he calculated fay (39.) Note 2. The third side can be calculated without find- ing the angles A and B; for !n the /v,ABE, we hare, fay (39), sin. ^(A— B) : sin. ^A + B) ^(a— 6): c = AB. And in the ABDA, since sin. BAD = cos. ABF = cos. ^(A— B), and sin. BDA= cos. DBF= cob. i(A+ B). Cos. i(A— B) : COS. ^(A + U)^(a + 6) : c=AB. NoTB 3. The third side may also fae found without calculating the angles hy the following formula: c= Vt'+6" — ^"b COS. 0, which is easily deducihte from (Geo. prop. 41) and (7); but the form is not suited to Example. Let the side AC of the triangle ACB be 210-3, the side BC i6Q% and the ZC 1I0°1'20"; requiredthe ii A and B and the side jg-g 2 AB. 4(80— /.CllO" 1' 20")=34° 59' 20"= half the Bom of the 2* A and B. Side AC —210-3 Side CB = 160-2 Sum of Bides=370-5 Ar. co. Log. =7-431212 Dif of 8ide8= 50-1 Log. =1-699838 ^ sum of the is 34° 59' 20" Log. tan.=9-84504S 5°_24'_24'^ Log. tan.=8^976aB .-. iB 40"° 23' 44" I aniiA 29° 34' 56" I TLAHS TRIOONOBfBTRT. 29 And by (39), sin. A : sin. C=CB : A B. Log. cosec. A=29» 34' 56"= 10-306561 Log. sin. C=110*» V 20" = 9-972925 Log. BC= 160-2 = 2-204663 Log. AB=304-9 = 2-484149 XZBRCI8R8. L Ohen in the AABC, the i\de AC 241, the side BC 73, and the inclnded angle C 103^, to find the remaining ngles, and the third side AB. Ans. LA 31 o 3' 23", Z B 45° 5& 37", and AB 326-753. 2. Given in the triangle ABC, AC=79, BC=67, and he indnded LC 85* 16', to find the remaining it, and the hiid side AB. Ans. IB 52* 28' 6", and LA 42* 15' 54", and AB 99-2795. 3. In the AABC, given the side AC 3450 links, the ide BC 4025 links, and the LC 91o 30'; to find the re- nainingi^^ and the third side A B. Ans. LA 48° 3*2' 77", IB 39^ 67 52-3". and AB 5369-37. 4. In 4e AABC, given the side AB 800, the side AC 749, and the Z.A 80^ 10'; to find the remaining Is and the third side BC. Ans. LC 52* 9' 25-4", ZB 47° 40' 34-6", and BC 998-161. 41. Theorem. Twice the cosine square of half an mgle is equal to one plus the cosine of the whole angle ; ind twice the sine square of half an angle is equal to one ninns the cosine of the whole angle ; or 2 cos. ^ ^A=rl-|- sos. A, and 2sin.^^A=l — cos. A. Let BAC be the ZA, draw BC ^ ^B JL to AC, and from centre A, irith the radius AB, describe the lemicircle EBD; produce AC [K>th ways to £ and D ; then EC ]£' =AB + AC,andCD=AB— AC, Hso (Geo. prop. 68) EB^z=EDxEC=2AB (AB+AC), and BC2=ECxCD=:(AB+ACXAB— AC). Also the ^BEC=i Z.BAC=iA, (Geo. prop. 47), and by (7), cos.^ * EC , . 1 . BC , A= — , and sm. i A= — ; hence ' ^ 2EC« 2EC* . AB+AC , , . 2 C08.HA= ^3i = 2XBiEC = -AB - =^ + ^^'- ^ ' ... 2BC« SECxCD AB— AC and 2 sin. *A= ^^ = gABxEC " Aid =^— ^^^- ^- 30 PLATTB TSIQOROHSTBT. 42. Thborrh. Any side of a triangle is eqaal to the sum of the products of the other two aides, into the cosine of the angle in- cluded between it and that side. For (7) gives AD = c cob. A, and CD=(icos.C; .-. 6=ceos. A+dcoa. C, and similarlf tf=6cos. A+acos. B, o=iico8.C+ccos. B. 43. Multiplying the first of these hy 6, the second by t, and tEe third by a, we have J'=6ccoa, A-i-ai>coa. C, c'^bc COB. A -i-ize COB. B, «''=a6co8.C+<wcos. B. 44. Adding now the Erst and second of the aboTe, and aubtracting the third, we obtain the following: J=+c=— tt''=2iccoa. A; whence changing the sidesj and dividing by 2be, we han (1.) COS. A= — 37- — , and similarly , and (J.) COS. C,= — — — , 45. AddiDg I to both sides of the equations in (4^ we hate from (1) ,. . . . (Aig.,A,..3i.) ^i^a^m^. But (41) 1+ COS. A=2 oDB.'|A, and theiefois ...co.'iA=^^^^t-^-^- Ut now S=i{a+b-i-e), then 5— a=J(6+e--), S—!>=i(,a+c—b), and jS—c=^(a+b—t}. Substituting these Tallies in die ^ove, and extracting '« fool, we ohtaia TLASm TBIOOIIOMSTRT. 31 «oi.|As ij j^ '» and simi] OOB. 4tt= /^ ■ ■ iB=J'- 16. In the same maimer, from (44) we obtain 1 — COS. A=l — 20c ^ 2be 2bc _ <g4-e~A)(<-f 6— c) _ •" 26c ' .•,9m.'4A='^^^ ~ — — ^, . or nn.^A= ^ ' ^ — ^^, ftnd timilarlj. and 8u,.iC= y <^Xa-*> . ?• Again« dividing the expressions for the sine in (46), Jus corresponding expression for the cosine in (45), Mre lin the following expressions for the tangents of half angles: viz. 18. Whenee Tan-iAs ^J^(S^a)(S—b)(S—e). m PLANE TRlGONOMETRr. 49. From the expressions in (48), is easilj dertTed the illowing very convenient rule for calculating the tkrte tgleg of a triangle when the t/iree sidea are i/iven. EuLn. Add the three Hides togetber, and take half 'tbeir sum ; from half the Bum suhtract each side separate- ly, then subtract the logarithm of tlie half sum from 20, and under the remainder wrile the logarithms of the three remainders ; luilf the sum of these will be a eoTistant, from which, if the logarithms of the three remainders be succes- Birely subtracted, the new rcmaiodcrs will be the logariih- mio tangents of half the angles of the triangle. NoTR 1 . The above rule has this advantage over all others, that the three angles are obtained with little more labour than one, and when the three angles are thus found inde- pendently, if their sum be IBO", the calculation is corrett; if not, it must be examined till it prove. NoTB 2. In order to know which of the angles we have I obtained, it is necessary to observe, that the rojutanl I — (S — a), gives tan. ^A, the coTtstant — (S — fi), givM I tan. ^B, and the constant — (5 — c), gives tan. JC. I Note 3. The angles might also be calculated from <\\t | pipressions in (Arts. 44, 45. 46. or 47); but those in (44) are not suited to logarithmic calculation, those in (45 aoJ 46) do not give the angles with the same accuracy ia pw- ticular cases, since for a small angle the cosine TWWt slowly, and for an angle nearly JtO" the sine varies slowlji and those in (47). as well as all the others, require win- dependent calculation for each of the tjjree angles. Example. Given the three sides of a triangle ABC; viz. AB 340, BC 380, and AC 360, to find the three angles. AB, c= ;140 BC, a= 380 ^ AC, 6= :160 540 ar.co. Log. +10=17-2G76O6 160 Log. = 2-204120 ISOlxig. = 2-255273 k 200 L"g, = 2-301030 "^ 2 |24-02802[) Constant. =12-0140145 ■ ^m B SKIOOirOlfBTBT. Tan. iA=32° 50' 3l-7''=9-S0g8945{con'. — Log.(5— a)} Tan. 1B= 29" 50' 46" =9-75874 J 5 [con'. — Log.OS'— b)] Tan. ^C=27" 18'42-3"=9-712yS45[con'. — Log.C,S'— 01 Therefore L\ 05° 41' 3-i", IB 59" 41' 32", and LC 54" 37' 24-6", the Bum of which is exactly 180°. n terms of 1. "What are the three angles of the triangle ABC, AB being 100 yards, BC 150 yards, and AC J20 yards! Am. ZA 85" 27' 34", /.B 52° 53' 28", and /.C 41" 38' 58". 2. The three sides of a triangle are AB 470, BC 398, and AC 420 ; what are the angles ? Ans. LA 52° 45' 49". IB 5T 9' 22", and iC 70" 4' 49". 3. The three sides of a triangle are AB 2601, BC 140-4, nndAC190'O; required the angles? Ana. ZA3I''47'3r', ;B 45- 37' 46', and LC 102° 24' 43". 4. The three sides of a triangle are AB 562, BC 320, andACeOO; required the angles? Ans. lA 18" 21' 24", LC 33- 34' 47", LB 128" 3' 49". 50. It is required to find expressions for the sine and cosine of the sum and difference of two angles the Bine and cosine of the angles themselTes. Let CAD=a, BAC=fi, then fiAD=a-|-&, and it is required to £nd esprcBsions for the sine and cosine of (a+ft), in terms of tlie sine and cosine of a and b. From B draw BC-i-AC, and BE-i-AD, and through C draw CFu-BE, and CD -i- AD, then. FEDC will be a parallelogram, and -^ " PE will be=CD, and FC will he ^ED ; also since the right angled triangles AOE, BOC, have the right angles AEO. BCO, equal to one another, and the angles at 0= bdng rertically opposite, the remaining angles OAE and OBC are equal, or the /.CBF=a. Now, (6.), „. , , ,, BE CD+BF CD , HP- Sm. («+6)= — = -r^ = AB + A PLASB TBiooaoKniHr. k , Cos. (a+6) = cos.acoB. 6— : _rLet now CAE=(t, and CAB=6, then BAE wUl =(a— i) ; draw BC J.AC, CD and BE each J-AE, and BF J- to CD; then smce iDAC+ /.ACD, are together = a I'i, and ACB 19 a i^L, take away the common L ACD, and there remains the Z.BCF=:the IZ,CAD; .-.the angle BCF^a, also IFD^BE, andFB=DE; h^ce we have ^ „. , ,. BE CD— CF CD % fi3. Sin. {a—b)= AB' AD+FB AD ; therefore .ught. 54. Cos. (u — 'j)=; COS. a cos. 6' Collecting now these four resu ference, we have the expressions Sin. {a+b)= ain. a cos. i+ cos. a sin. b. Sin. (a — b)= sin. a cos. b — cos, a sin. b. Cos. (o+6)=^ COS. u COS. h — sin. a sin. b. Cos, (a— 6}= COB. a cos. ft+ Bin.ra sin. 6. ■51) + (S3) 155 Sin.(a + t)+sin.(a— 6)=2sin.tt««t. '51)— (63) 56 Sin. (a+6)— sin. («— 6)=2cos.aain.t, (54}+f52) 57 Cos.(a— 6)+cos.(a+fi)=2coa.«cM.6. (54)— (52) |58 Cos.(a— 6)— co3.(a+6)=2sln.asiii.i.. Let now a+t^S, and a — 6=rf, then a=^{S+d), vA. h={{S^d). Substituting these Taluea in the last bra, they become (55) I 53 Stn.^+sin. (?=2sin.^(5+d)o08.J(5-4 (66) I 60 Sin. .S-- sin. d=2coiX{S-^d)am.k[S—iV (57) 61 Cos. rf+ COS. 5— 2 COB. i(.S'+rf) COB. !(«—<;). (50) I G2 Cos, d— COS. ,^;2Bin.|(^V+d)sin. !(;;-</> The last four expressed in words are as follow:— 59. The sum of the sines of two angles, is equal to twice the sine of half their sum, into the cosine of Itaif their diifcience. 60. The difference of the sines of two angles, is equal to twice the cosine of Aa^" their sum, into the sine of Aa!^ tbeii difierence. yi.*ira TWQOWOMETKT. 35 fil. The Bum of the cosines of two angles, is equal to twice the cosine of Aaf/" their sum, into the cosine of half their difFereDce. 62. The difference of the cosines of two angles, is equal to twice the sine of kalf their sum, into the sine of half their difference. Note. Since S and d are any two itrtgles, we may writo inBtcad of them a and b, by wridog at the same time 4(a-l'i) for i{S-ir-d), *nd 4(0— i) for J(S^— ^)i T instead of S write A, and instead of d wile B, then l(S+(0=i(A+B), and i{S— rf)=i(A— B); whence (59)^(bO)giTesb^^^^_^^_^^^_^^^_^jj^^^^^^_g^. ^ - — 1/ i iigy By dividing both numerator and denomi- nator of the second side hy 2 cos. ^(A+B) cos.4(A— B). 11.A+ sin. B . w, , i>v ^^^^^^ = tan.KA+B). ^B-I!^A-=-'■^^+^> B. B+ COS . A _ cot. i(A+B) _ *.B— COS.A ~ taii-lCA— B) oot.i(A-B) (an. i(A+B)" In Arts. (51, 52, 53, 54), let the angles a and 6 be equal, then 0+6=2(1, and n— 6=0, (59)H-{81) giTea 64 (69)^(62) giTe. 65 (60)+(61) gira 66 (60)-h(62) giTes 67 (61)+(62) Eire. 68 (51) gives (53) (52) (S4) (51)-i.(52) i.2a=2 6iii.a ^0. m. (.+») _ (.to. . c. 1+ o B+Bin-'a^l, {Art. 2]>. , . ,, tau.a+ tan-ft n.(„ + 4)=— ^--^-. \m^<.^) 1 7* 1 "iSr'di :a, and (73) giyes | 75 | tan, 2a^ __ 'ian.&' I , _ _ JO 6S) make 6 =0, and (Arts. "0 anJ e haTfi the following : Bin. A , '- ' 1—coB.A^'' tan-ii \=^:>^=^.=^'^'^^- ^y'"^ h,g, (82). 84. To find the numerical values of the sine, cosine, mi tangent of 30°. Leta=sin.30"; then (21) cos. 30'=Vl— x*-", and (g ) si n. (30° +30°). or sin. 60"= cob. 30" (26)=2jr^ 1-V= Vl — X-; dividing both sides of this pquation by •Jl—^ gives 2x=-l, or3r=J, conseqaentlj Vi — i— \'|=^^'3= ; and since — = tan., fan. 30°= — j^ = -= = [ 4^3"; .-. sin. 30° = ^. cos. 30° = i ^3, and tan. 30°= ^. or ^^/3--, congequentljsin. 60°=^^3, cos. CCrri, ' tan. 60°=V3, &c. 85. To find the sine, cosine, tangent, &c., of 45 ■ Let.r:=ain. 45°. then ^1— i"^"(2l)i= cos. 45°= an, (30 —45X26) = sin. 45". Hence sin. 45°=- ;e squaring a-^l — x' COS. 45"= ^1—^=- I. From the valiiea deiQcei vo^^^*'a.u^?.3^^lA^(^ i PLANE TRIGONOMETRY. 37 of (51 — 54) numerical expressions for the sine and cosine of 75% and for the sine and cosine of 15**. Ans. Sin. 7^** = ^^ cos. 750= ^^, sin. 15-= ^^, and cos. 15' 2^2 2^2 2^2 2. Prove the truth of the following expressions. „. ^ COB. a tan. a cos. a see. a Sin. a= COS. a tan. 0= -— — =: = : cot. a see a cosec. a ^ . sin. a cot. a sin. a cosec. « Cos. a= sin. a cot. a= : = =: tan. a cosec. a sec. a — sec. a sin. a cosec a cos. a sec. a cos. a Ian. a^ = := r =-: 7=-. cosec. a cot. a cot. a sin. a cot.* a 3. Show that if (A+B+C)=180% or he the three angles of a triangle; tan. A-{- tan. B-f- tan. C= tan. A. tan. B. tan. C. 4. Prove that sin. A=sin. (60**+ A)— sin. (60"— A), or that sin. A= cos. (30''— A)— cos. (30°+ A.) 5 Provethatsec 2A-l/' '^' ^"^'^°' A x j^/cos.A-sin.A\ I r: 2A ^ai^T^lcos. A— sin. aJ + sI co8.A + 8in.A J 1+tan.^ A cot.*A+ 1 > ' ^ ^ "" 1— tan.«A ~ cot.«A— r 6. Prove the rule in (art. 40) from (arts. 36 and 59), and (Alg.69). 7* Prove that the difference of the sides of a triangle, is to the difference of the segments of the hase made hj a perpendicular upon it from the vertex, as the cosine of half the sum of the angles at the hase, is to the cosine of half tilieir difference. Also, that the sum of the sides, is to the Terence of the segments of the hase, as the sine of Iialf the sum of the angles at the hase, is to the sine of half their difference. APPLICATION OF PLANE TRIGONOMETRY TO THE MENSURATION OF HEIGHTS AND DISTANCES. 86. The instruments commonly used for measuring heights and distances are^ a chain, a quadiaat, ql ^c^^t^, and a theodolite. A chain is used for measuring tliose distaivce^ ot \\tv^^ rjttch are to be given sides of triangles. T\ie Im^^fvssSi cbain is G6 feet in lengtli, aad is divided into 100 equal links, consequently eiich. link is "^-92 inches long. Every ten links from each end to the middle of the chain is dis- tinguished by a piece of brass having as many points aa it is tens of links from the end of the chain, A quadrant is need for determining vertical angles. It is made of brass or wood in the form of a quadrant of a circle, and its circumference is divided into 90°, and these agMn subdivided, as far as the diraensiona of the quadrant nill admit. Abo a plummet is suspended by a thread &om the centre. A square is used for finding the ratio of the sides of a right angled triangle. It is made of the same materials. Two of its sides are divided each into 100 equal parts; and a plummet hangs from the opposite angle. A theodolite is used for measuring horizontal, as wellaa vertical angles. It is a circle of brass divided into 360 degrees, having an index moveable about its centre, and ia furnished with a telescope, moveable on a graduated semi- circle, which is used for measuring vertical angles. It ii indispensihle where great accuracy is required. 87- fo find the height of an accessible object Btaoding on level ground. Ruti:. Measure any convenient distance from the bottom of the object, and there lake the angle of elevation of it( top; then state, £ is to the tang, of the angle of elevation, as the measured distance to the height above the level of the eye, to which add the height of the eye, and the tun will be the height required. (See art. 33.) H^' . 1. Required the height of a tower, whose angle of elec- tion, at the distance of 100 feet from the bottom, ii 51° 30', height of the eye 5 feet. Ans. 200-148 feet. 2. At the distance of 100 feet from the bottom of » tower, on a horizontal plane, the angle of elevation of iti top was 47° 30', the centre of the quadrant being fixed 5 feet above the ground ; required the height of the tower. Ana. 114-131 feet. 3. At the distance of 130 feet from the bottom of> tower, and on the same horizontal plane with it, the angle of elevation of its top was 50- 43', and the height of UW eye 5J feet; required the height of the tower. AnB.164'4!3- 88. To find the distance of an inaccessible object, fitUD* given point. I Rule. Set up a mark at the given point, and take an; I 'ler station^ and meaaure iW &u^ante\n\.we«u the pni I PliANE TRIGONOMETRT. 39 point and the other station^ and at each of the ends of this line measure the angle subtended by the object and the other station; then the sine of the sum of the observed angles^ is to the sine of the angle at either station^ as the distance of the stations is to the distance of the object from the other station. (See art 39.) EXERCISES. 1. Being on the side of a river^ and wanting to know the distance of a house on the other side, I measured 266 yards in a right line^ by the side of the river, and found that the two angles, one at each end of this line, subtended by the other end and the house, were 38° 40', and 92'' 4&; what was the distance between each station and the house? Ans. Distance from one station, 354*38 yards. Distance from the other station, 221-67 yards. 2. From a ship a headland was seen, bearing NE.^N.; the vessel then stood away NW.^W. 20 miles, and the same headland was observed to bear from her E.;|^N.; re- quired the distance of the headland from the ship at each Station. Ans. Distance from the first station 19 09, and from the second 26'96 miles. 3. Having measured a base of 260 yards in a straight line, close by one side of a river, I found that the two angles, one at each end of the line, subtended by the other end and a tree close to the opposite bank, were 40® and 80°, what was the perpendicular breadth of the river ? Ans. 190046 yards. 89. To find the height of an inaccessible object. At any two convenient points F and G (these being in the same vertical plane with AB), observe the angles of elevation AFE and AGE, and measure the distance GF g. or CD. Then because the d' exterior angle AFE is= ^ FAG + FGA, the angle FAG is the difference of these two angles. Now the triangles AFG and FAE evidently give the following proportions, which will determine AE, to which adding EB or FC, the height of the eye, the alti- tude will be determined. From A AGF, sin. G AF= sin. (F— G) : sin. G : : GF : AF From A AFE K : sin. F : : AF : AE These proportions may be wrought separately, ot V^^Wet being eampounded (Alg. ] 16) they give the ioftomiv^-.'^^aji. 40 n-Amt TdiGOKflMBTnr. (I'— G) : Bin. F. sin. G : : GF : AE, which hy Logarilhmi gives Log. EA= [Log. GF+ Log. sin. F+ Log. sin. + Log. coscc. {F— G)— 30j. If the distance FE were sought, we would hitve in tlie same manner Log. FE^ [Log. coseo. (F— G)+ Log. COS. I-+ Log. sin. G+ Log. GF— 30], If two atstions ia the same Btraigbt line with the object, and in the same horizontal plane with it, cannot be found, lake one station as near as possible on a level with the base of the object, and select another station such that the first station can he seen from it, and also the object; and that its distance from the first can be conveniently measured; then at the first station measure the angle of elevation o( the top of the object, and the angle subtended by the ob- ject, and the second station; also at (he second station measure the angle subtended by the first and the object; then measure the distance between the stations, and state the following proportions, in which S is the angle at tkt first station, S' that at the second, D the distance betirefn the stations, D' the distance of the first station from tkt object, E the angle of elevation of the top of the object nl the first station, and A the altitude of the object. Sin. (S'+S):Hin. S'rrD :D' II .- tan. E : : D' : A Compounding these proportions (Alg. 116) we hare K sin. (S'+S) : sin. S' tan. E : : D : A, hence Log. A=; [Log. coseo. (S'-|-S)-|- log. sin. S'+ log. tan. E +Log. D.— 30], NoTB I. If a station cannot be conveniently found ons level with the bottom of the object, the angle of elevation nt depression of the top iindbotlotn mtiy be taken, and theparV calculated for separately; then if the angles of the top and bottom of the object he both angles of elevation, or both angles of depression, the difference of the elevatinns et depressions of the top and bottom will be the altitude «f the object; hut if the lop be elevated, and the bottom d«- pressed, their sum will be ibe altitude of the object. NoTB 2. If the horizontalangles Sand S' be taken withi theodolite, if the base ts not on a level plane it must fM multiplied into the cosine of the acclivity or declirity be- fore being used in the above calculation. Note 3. If the horizontal angles S' and S be measured b; a sostont, then the distance from the first station .ind tb« ohject must be mull'nilwd into the coaiiie of the angle of elevation or depieaaion o^ ^-^lat ■joiiA (>l ^% «i^\«ct, tht FLANS TRIGONOMETRY^ 41 image of which was made to coincide with each of the stations in measuring the horizontal angles. NoTS! 4. In measuring the height of mountains at a dis- tance, allowance must he made for the curvature of the earth, by adding to the calculated height 8 inches multiplied into the square of the distance in miles, or to twice the log. of the distance in feet add the constant log. 8*378641, and the result will be the log. of the correction in feet yery nearly. EXERCISES. 1. A person 6 feet high, standing on the side of a river, obserred that the top of a tower placed on the oppo- nte side, subtended an angle of 59**, with a line drawn from his eye parallel to the horizon; receding backwards for 50 feet» he then found that it subtended an angle of 49^. Bequired the height of the tower, and the breadth of the mer. Ans. Height of the tower 192*27 feet. Breadth of theriTerlll92feet. 2. To determine the altitude of a light-house, I obserred the eleyation of its top above the level sand on the sea- shore to be 15° 32' 18'', and measuring directly from it along the sand 638 yards, I then found its elevation to be 9* 56' 26"; required the height of the light-house. Ans. 302*463 yards. 3. It is required to find the height of Arthur's Seat, in ilie vicinity of Edinburgh, from the following observations, taken with a theodolite on Leith sands, about the medium height of the tide, the base being 1410*42 feet; at the west end of the base, being that nearest to Leith pier, the angle subtended by the top of Arthur's Seat, and the eastern extremity of the base, was 50° 44', and the elevation of its top 3° 59', whilst the angle subtended at the eastern station, by the western and the top of the hill, was 123° 29', and the elevation of the top of the hill was 4° 17' 30': Find the height of the hill above the medium level of the tide, allowing 5 feet for the height of the eye^ and the necessary correction for curvature, by (Note 4). Ans. The observed altitude at the western station gives 821-174, that at the eastern gives 82101 9 feet, the difference of the two being less than 2 inches. 4. Find the height of the top of the cross on the spire of Assembly 'Hall, Castle Hill, Edinburgh, from the following observations, the height of the eye being on a level with the sole of the entrance door; at the first station, the elevation of the top of the cross was 62° 36', angular bearing of ih^ •pire from the second station 35° 36', aivd l\i^ ^xii^xiSax 42 rLAXK TRiaonOMETRT. bearing of (he spire and the first Etation, taken at tha second, was 3U° 22', tlie distance between the stations heiag na-e feet. Ans. 235309 feet. 90. To find the distance of two inaccessible objects. Let the two objects be A and B; ^^ „ take any stations C and D, such. that the objects and the other station can be distinctly seen from each, and that the distance CD can be accurately Measured; then at the ^ ^ station C measure the angles ACB and BCD, and at the station D, measure the angles BDA, and A DC, then measure accurately the distance CD. Solution. In each of the triangles A CD, BCD, wff have given two angles, and the side lying between thenif consequently (Art. 39) AC, AD, CB, and BD can be found, then in each of the triangles ACB and ADB we have two sides and the contained angle, consequently (Art. 40) tbe side AB can be found from either nf these triangles, and if it be calculated from each, and the results be found tin same, it is a proof of the correctness of the calculations, i 91. This problem is of Tery estensive application hi many departments, both of civil, military, and naval snM Teying, and may be solved in a different manner from llwj commonly adopted method which is given above, liti method here referred to may be called surveying by Stit angular Co-ordinates, and consists in finding an exprotiol for the perpendicular distances of the objects from thebtf line, and the distance of the iniddle of tlie bate &ov foot of the perpendiculars, in terms of the baae udfl angles at its extremities. 92. To find an fsrpremon, for the perpendicular AD in tem>» of the iase BC and tlie angles B and C, at its extremities. By (Art. 37) AC=BC by (Art. 6) AD=AC. sin. C = BC fiin. B Bin. C ,, ^ 4 T^ ri l(bTc)- Hence Log. AD= [log. c 1. B+ log. sin. C+ log. BC— 30]; or Add togetJitr » log, cosecant of the sum of tJie angles, the log. alnea of M«l ' the angles at ike base, and the log. of the bate ; the utm, ^^ ifi^ iij/30 int}ieindex,vnllbetJielog.oft/ie perpendieiJi 03. To Jind an ea-pressimi Jot "tX), ftw dnxhutw of ■•(B+C)+fc| PliANE TBIGONOMETBY. 43 foot of iJvR perpendicular from the middle of the base, in terms of the base and the angles at its extremities. ^ ,A «*,v * ^ ^^ sin. B , ,„ „^ Bin. C By (Art 37) AC=BC . ' ^„ and AB=BC . ,'\^ . ^^ ^ siiL(B+C) sin. (B+C) Again, (Art 7), DC=AC cos. C=BC ""^—-^ And BD= AB cos. B=BC ?^?4^ ; sin. (B+C) ' • 2FD DC— BD— ^^i^in. B cos. C — cos. B sin. C| ' ^ "" sin. (B+C) But (Art 52), ^8in.B cos.C — cos.B sin. C] =sin (B — C). By substituting this value, and dividing both sides by 2, we obtain, ED=iBC^!5^^2' ^^^^^^ * sm. (B+C) Log. ED= {Log. cosec. (B + C)+ log. sin. (B— C)+ log. \ BC— 20]. Or, add together the log, cosecant of the sum of the angles at the base, the log. sine of their differente and the log. of half the base, the sum diminished by 20 in the index will be the log. of the distance of the foot of the perpendicular from the middle of the base, 94, The foot of the perpendicular falls always on that side of the middle of the base whieh is adjacent to the . greater angle at the extremity of the base, and its distance from the middle of the base should be marked -f- when on one side, and — when on the other, (Art. 28). Also, if in an extensive survey some of the objects should lie on the one side, and others on the other side of the base, the per- pendiculars should be marked 4- ^^ ^he one side, and — on the other. 95. The perpendiculars DG and CF being thus calculated, and also EG and EF, the distance of the two objects C and D can be found, for CI being drawn parallel to AB, the triangle DIG will be right angled, and Dl is the difference of the per- pendiculars, and IC is the difference of the distances of the perpendicu- lars from the middle of the base, since EF is — ; and if one of the perpendiculars had been measured downwards, DI would still have been th^\t difference, ahice the perpendicular meaauiedl dio\^ii\;^^^^ i^ w Tonid be mimis. Xow Log. DI+IO — Log. IC=Log, tan. DCI. BatLog. secDCl+IiOg. IC— 10= Log. DC; hence we have the following luJe for finding the distance of two objects : — From (Ae log. of the difference of the per- pendicida.rs, inei'eased iij/ 10, of an iitdea^, eublract tlie log. of the distance of the perpendimlar, the remainder wiQ, be &i log. tan. of an (mgU; to the log. eeeamt of this angle add the log. of the distaaiee of the pei^ieitdieulart, and the turn, rejecting 10, from the index, will be the log. of the diffaud sought. 96. If the distance between the objects A and B be given, to find the distance between the stations C and D, (Art. 90) ; measure the angles at C and D as before, and assume CD any length, as 100 or 1000, and from this assumed length find AB ; then the calculated lengih of ' " ■ the assumed length of CD, as the true length of the true length of CD. 1. Find (he distance between the two objects A and B, (Fig. art. 90). on the supposition that CD is 300 yardi, ZACB^56°, /.BCD=37°, /.ADB=ij5°. and L\T>Q=\\'. Ans. AB=341 aSyardi. 2. Being desirous of finding the distance between two . ohjeclB A and B, I measured a base, CD, of 384 yariii on the same horizontal plane with the objects A and Bi At C I found the angle BCD=48° 12', and ACD=89°lff; at D the angle ADC was 46° 14', and BDC 8?° 4'. It ii required from the data to compute the distance between A and B. Ans. AH=358-5 yard«. 3. Wanting to know the distance between two inaccM- eible objects A and B. (Diag. art. 90), I measured a base line CD of 360 yards: at C the horiaontal angle AC6 was oliserred with a sextant, and found to be 53" 30', anJ the angle BCD 38° 45' ; at D the liorizontal angle BD.4 was 67° 20', and the angle ADC 44° 30'. Required ll« distance between A and B. Ans, 548149 yardi 4. Wishing to know the distances and positions of ■ namher of the principal objects in and about Edinbnr]^ I took a station on the top of Arthur's Seat, and wiUit theodolite meaamed the angular hearings of the following objects, with a line drawn through the top of Nelsoa'l Monument, on the Calton Hill, viz.:— Spire of AMemWj Ball (H). Castle Tower (T), Dome of St George's Chittdi (O), Spire of St AnAieVa CVuxiAi V^\ lAcWUle's Mono- PLANE TRIGONOMETRY. 45 tnent (M), Spire of North Leith Church (L), Inchkeith Light-house (K), and Berwick Law (B). I then went to the top of Nelson s Monument, and took the angular bear- ings of the same objects with a line drawn through the former station on the top of Arthur's Seat. The angles were as given in the following table, where the single let- ters in the first column signify the objects after whose names they are placed above : — Vertex of the triangle. Angle at Artiiur's Angle at Nelson's Sum of Difference Perpendicu- Distance of the middle Seat. Monument. angles. of angles. lar in feet. of the base. H . 26" 48' 100'' 8' 126" 56' 73" 20' 315M 3400-7 T SO" 22' 105" 35' 135" 57' 73" 13' 3974-5 3907-2 G 25° 40' 127" 54'i 153" 34'4 102" 14'4 4257-9 6231-1 A 13'' 57' 1.85" 44'i 149" 41'i 121" 47'4 1892 4779-1 M 11«'23' 137° 25' 148" 48' 126" 2' 1463-1 4429-5 L 28° 26' 131" 38' 160" 4' 103" 12' —5923-7 8103- K 57*' 49' 113" 4'4 170" 53'4 55" 15'4 —27913 14729 B [—n5°5V 61" 14' 177" 5' 54" 37' —87983 —45465 The base of the above triangles, namely, the distance between the top of Arthur's Seat and Nelson's Monument, was determined from the third triangle, of which a side, namely, the distance from Nelson's Monument and the top rf the dome of St George's Church, was previously deter- Biined from a similar survey, in which the base was measured, and the distance so determined agreed to the tenth of a foot with that deducible from data given in Wallace's Theorems and Formulae, and hence concluded to be very correct. The base thus determined was 5675 feet. • The following is the method of calculating the perpen- dicular, and the distance of its foot from the middle of the hise in the first triangle; and the others are exactly similar : — For the perpendicular. Jiog. 5675 = 3-753966 Log. cosec. 126" 56'=10-097271 Lbg: Bin. 26' 48' = 9*654059 Lo«r- sin. 1 00" 8' = 9*99317 2 >. 3151*1 = 3*498468 For its distance from the middle of the base. Log. 4(5675) = 3*452936 Log. cosec. 126° 56'=10-09727l Log. sin. 73" 20' = 9-981361 3400*7= 3*531568 OTB 1. The foot of the perpendicular falls always on that side of niddle of the base which is adjacent to the greater angle. All Jars in the ahove &I1 upon that aide ot \k<&'\^«i&i^\<\i\0(v Velsoa'a Monument except the \aBt,vA)kS[i\ft^'^'tc£st^ niftNG TmoofmntKTRT. 'OMiked — , and the last throe porpendicuUra are — , beenuBP they lie towards the east, or minna side of tlie base. Nora 3. The abuvo exercise contaiiiB 20 questions like tho firsl three exunples, for each of the 5 plaices ma; be aombined with iQ the otiien ; those who wish U> compare tho relntive merits of tin I <dd method and that given above, may calculate them both mja ; ^^^_ the results ooght to be the same. ^^^^- . Note 3. If the objects observed above w ^^^^1 Innudai? of a county or an eslate, its area could easily be founl ^^^^K bom the above, as nill be sbonii in MensuFation or Land &a^ ^^1 Teying. r 97- Given the angular bearings of tliree objects wioM distances fram one anothei are known, to find the distanct of the station where the angles are taken from each of tftt I three given objects. I Let the three objects be (A, A', or I A"), and B and C, of which B and " C are the estreme objects, and A, A', or A", the object which has its angu- lar position between the objects B imd C, as seen from the station S, and let CSA, BSA, be the angles formed by these objects at the station S. CoNSTRUonoN. At the point B, In the line joining the extreme ob- jects, make the angle DBC=CSA, and at C, in the same line, make the ZDCB=BSA; then abont the ABDC describe a oirde, and It will pass through the station S; join (A, A', or A"), and D, and produce the line to meet the circumference in S, and S will be the position of the station sought. Now in the ADBC there are given the is at B and C, and the side BC, hence BD and DC can be foond, (Art. 39); and since the three sides of the AABC ate ^ven, the is can be found, (Art, 49), therefore their diffe- rence ACD or ABD can be found; then in the A« ACD or ABD, there are given two sides and the contained angle, therefore (Art. 40) the is CAD or CAS. and BAD or BAS, can he fonnd. Again, in the AACS there are given the Is CAS and CS.4, and the aide AC, hence (Art. 39) AS and OS can be found, and from the AABS. in the same manner, can be found BS. 1. If the three objects B, A', C, lie in the same stiaigbc line, iiaving calculated CD as above, we have in the jCiDCA', two sides, DC, A.'C, ami ^ikts co^\^a^'a<:& im^e at C, to find (Art. 40^ l\ie LCN!\>, ox ^>^%-, W^enK^-iM ^BA'S, there are gv^en i\ie Ls^Nl'S.^^A- ^^'S^. ^aA'i PI«ANE TRIGONOMETRY. 47 side A'B, to find BS and A'S, by (Art. 39); and in a similar manner, from the A^'^^ ^^^ distance CS can be foiind. If the station were taken in the line EC, or EC produced, the distance could not be determined. 2. If ihe third object lie between the station S and the line EC, as at A"; to the ZDCE add the LECA", then in the Al^^^'' there are given two sides and the contained /., hence the LCA"D can be found, and consequently its supplement, C A"S ; then in the ACA"S there are given two angles, and the side A"C, to find (Art. 39) A"S and CS, and in the same manner, from the AEA"S, ES can be found. The solution when the three objects form a triangle fails, when the points D and A coincide, or when the station is in the circumference of a circle described about the three objects. If the station S were in the pro- duction of one of the sides of the A formed by the objects, or in one of the sides themselves, the solution is easy, with- out describing a circle. 3. If the station were within the A formed by the three objects, as at S, in the annexed figure, then the /.« ASB and ASC being measured, construct the i^BCD= the supplement of ASB, and B^ the angle CED = the supplement of ASC, and about the A^CD describe a circle, and join DA, and the point S, where the line AD cuts the circle, will be the station required, for the Z.BSD will be = the /.BCD ; .*. BSA, its supplement, will be of the given magnitude, and the same may be shown of the ZASC. Now, in the A^I^C there are given the angles and a side EC, .*. ED and DC can be found ; and since the sides of the A ABC are given, the is can be found, .•. in the AAED, AB, and ED, are known, and the angle ABD, hence the /.BAD, or B AS, can be found, .*. since ASB is also known, and the side AB, the sides SB and SA can be determined ; and if from the ZBAC, the Z.BAS be subtracted, the LCAS will be known, and since ASC is also known, and the side AC, the side SC can be found. 4. The station is without the triangle, upon one of its sides, or within it, according as the sum of the two mea- sured angles is less than, equal to, or greater than two right angles. JEXSRCISES. J. In diagram, (Art. 97), let BC=290 yaxda^ XC=\^^ h . yards, and AB=240; iASB^^0°5', and ZASC=2r>M5'. It is required to find the distances, AS, BS, and C4.h Adb. as 43207 yards, BS 27046 yards, and CS 33634 yards. 2. In the same diagran], let the three ohjccts, B, A', C, in the same straight line, be distant from each other ai follows: viz. BA' 490 yards, A'C 300 yards, and conse- quently BC 790 yards; and let the Z,A'SC=43°, and BSA'aaMS'. It is required to find the distances SB, SA', SC. Ans. SB 782'17 yards, SA' 423'93 yards, and SC 390-104 yafds. 3. In the same diagram, let A"B=500 yards, BC=630 yards, A"C=540 yards, ^BSA"=3r, and ZA"SC=28' 24'. It is required to determine the distances SB, SA" and SC. Ans. SB 612-523 yards, SA" 137118 yards, and SC 656-67 yards. 4. In the diagram, (Note 3), lot A, B, and C, represent three objects in the same horizontal plane, whose distances are aa foUows: viz. AC 460 yards, AB570 yards, and BC 620 yards. At a point S within the triangle formed b; these objects, the /.ASB measured by a circle was found to be 125" 15', the Z,BSC 124" 15', and the i. ASC I lO'Sff. Required the distances AS, BS, and CS. Ans, AS 247-8Sa yards, BS 280 804 yards, and CS 310 323 yards. 5. In the side AB of the triangle ABC, whose ndes were AB 1200 yards, BC 1000 yards, and AC 974 yaris, I took the angle BSC 74'' 12', Required the distancesof the station S from each of the three points. A, D, and C Ans. SC 814-274 yards, SB 843-094 yards, and S A 356^ 6. In the side AC, produced beyond C of the same triangle, the LBSC was 37° 25' 53". It is required to find the distance of the station S from each of tBe objeett. A, B, and C. Ans. SC 1000 yards, AS 1974 yards, and SB 158816 yards. a EXERCISES. 1. After observing the elevation of a tower, which i» 100 feet high, to be tiO", how far must an observer measut* back on tlie level plane before its elevation becomes 30*? Ans. 11547, or 66^ J3f«et S. From the top o? a towei, -wVisk Wx'^vU 108 fert, (he angles of depreaaioa o^ tVa \o'e naWm^iwsi. -A iv^*- (ical column in the hoiUonVA\ ^\a»'!.. ^^ ^""nA V!>>i f <^ NAVIGATION. 49 30° and 60° respectiyely. Determine the beight of the colanm. Ans. 7^ feet. 3. T&e sum of the three sides of a triangle is 2539058, snd the angles ate to one another as the numbers 3, 4, and 5 ; it is required to find the sides and angles. Ans. The sides are 707'107> 866*025, 965-926 ; and the angles are &, 60°, and 75°. 4. A person on the top of a tower, whose height is 100 feet, obserres the angles of depression of two objects on she horizontal plane, which are in the same straight line irith the tower, to be 30° and 45°. Determine their dis- tance from each other, and from the tower on which the ibsenrer is situated. Ans. 73*205, 100, and ] 73*205 feet. 5. From the top of a tower 120 feet high, the angles of lepression of two trees on the same horizontal plane with she base, was of the nearest 53° 24', and of the other 20° W, Required the distance of each of the trees from the )8se of the tower. Ans. Nearest 8912, and the other 324*982 feet distant. 6. Wishing to know my distance from two distant ob- ects, and their distance fjRem one another, I measured beir angular bearings 38° 24^, then advancing in a direct ine towards the right hand object 300 feet, their angular )earing was 4(y* 13' ; returning to the original station, and lien measuring 350 feet towards the left hand object, their mgular bearing was then 39° 58'. Find the distance of he first station from each of the objects, and their dis- ance from one another. Ans. From right hand object {22308, from the other 6110*24, and their distance ;il8-66 feet. NAVIGATION. 98. Navioation is the art of conducting a ship through be wide and pathless ocean, from one part of the world ) another. Or, it is the method of finding the latitude nd longitude of a ship's place at sea, and thence de- srmining her course and distance from that place to any ther place. The Equator is a great circle circumscribing the earth, ▼ery point of which is equally distant from. \.\i& '^^^^'^^ lius dMdiBg the globe into two equal parts, CQS\&^\i&Tc^- pherea; that towards the North Pole ia called ftifc TLOt?SiKca> id that towards the south the Sotttlieiix liem\s^ei^ 50 SAYtQATtCa. The Meridian of any place on ihe earth is a great circle passing through that place and the polee, and cutting the equator at right angles. The j(rs( meridiaa with us is that which passes through the Royal Observatory, at Greenwich. The Latitude of a place is that portion of its meridian which is intercepted between the equator and the given place, and therefore never exceeds Q0°. The Difference nf LatUwk between two places on the earth is an arc of the meridian, intercepted between their corresponding parallels of latitude, showing how far one of them is to the northward or southward of the other. The Longitude of any place on the earth is that arc or portion of the equator which is contained between the Jird meridian and the meridian of the given place, and is called east or west, according as it may he situated with respect to the first meridian. Longitude is reckoned east and west half round the equator, and consequently may be as large as 180°. The Difference of Loiiijilude between two places is an arc of ihe equator intercepted between the meridians of those places, showing how far one of them is to the east- ward or westward of the other. The Mariner's Con'poM is an artificial representation of the horizon. It is divided into 33 equal parts, called points, each point consisting nl \V Va'. Hence the ouid- NAVIGATION. 51 ber of degrees in any number of points can be obtained by multiplying the points by 11^, or by multiplying 11" 15' by the number of points ; ana . degrees can be reduced to points by multiplying the degrees by 4, and dividing the product by 45. A jRhumb Line is a line drawn from the centre of the compass to the horizon^ and obtains its name from the point of the horizon it falls in with. Hence there are as many rhumb lines as there are points in the horizon. The Course steered by a ship is the angle contained be- tween the meridian of the place sailed from^ and the rhumb line on which she sails^ and is either estimated in points or degrees. The instance is the number of miles intercepted between any two places, reckoned on the rhumb line of the course; 61 it is the absolute length that a ship has sailed in a given time. The Departure is the distance of the ship from the meri- dian of the place sailed from, reckoned on the parallel of latitude on which she arrives, and is named east or west, according as the course is in the eastern or western hemi- sphere. If a ship's course be due north or south, she sails on a meridian, and therefore makes no departure; hence the distance sailed will be equal to the difference of lati- tude. If a ship's course be due east or west, she sails either on the equator, or on some parallel of latitude. In this case^ since she makes no difference of latitude, the distance Sailed will therefore be equal to the departure. When the course is 4 points, or 45^ the difference of latitude and departure are equal, and each is the (distance X %/*«^) = the dist. x '707 nearly. When the course is ess than 4 points, the difference of latitude exceeds the leparture; but when more than 4 points, the departure .'xceeds the difference of latitude. 99. The distance sailed, the difference of latitude and leparture, form the sides of a right-angled triangle, in i^hich the hypotenuse represents the distance, the perpen- licular the difference of latitude, the base the departure, ind the angle opposite the base is the course, and conse- luently the angle at the base the complement of the course; lence any two of these five l^eing given^ the others can )e found by the rules for right angled triangles, (Arts. 33 ind 34)| which give the following xelatiOBa ^jsi^u^ ^^ Nurts: HAVtOATIOIt. Bad. : tan. of the c Had. : sec. of the c Uist. : dif. lat. = Had. : Wist, : dept. =; Had. : Uif. lat. : dept. = Had. ; 100. To find the latitude w = dif lat. : depL = dif. lat. : di»t COS. ofthccouree. Mue of the course, lau. of the course, md longitude at which a ship Las arrived, ivhen those of the jilace vhicli she left, uua the difference of latitude and luiigiiude ivhieh she has made, KuLB. If the latitude left and the diSetence of lati- tude be hoth north or huth south, their sum is the latitude uome to ; but if the one be north and the other south, their difference is the latitude come to, of the same oane as the greater. Note. Tho longitudB is olitained in the same way, only when dm longitude Ballad from and the difference of Inogitude are of the taisa name, and their Bum exceeds lliO", it must befcubtracted from 360", ExAiirLK. If a ship sail from lat. 39" 14' N., long. 21° 17' '^V'., till her difference of latitude be 312 miles N., and difierence of longitude 169 E. ; required her latitude aiil longitude arrived at. Lat, left. 39° 14' K Long, left 21" 17' W. Dif lat. 312= a' 12' K Dif. lung, 169= 2*49' E. 44° 26' N. Long, in lb" 28' IV. In the following rived at are required the latitude and lon^tode ai- "■■■•«• "SSif ^'f^ -- 1. 42M2' N. 17° IB' E. ISft S. 216 W. 3B* 42' K. IBM^E. aerars. i9's2'w. loo n. 2row. sb'u's. sraa'W. 3. ss'iB's. iB-ai'E. aie s. iss e. 36° sea 31* o'E. 4.12Mi'N. 73°I6'W. 612 N. 724 E. 20° 4G' N. 60'12'W. B. ea-lS'N. aS'SB'E. 147 N. 301 E. 31' JS' N. 3U*5S'E. e. 52MB'S. 28°27'W. 218 8. 1B6 W. 65° fiti' S. 31*33'W. n 101. Given the latitude and longitude sailed (torn, and io those arrived at, to fiad the difference of latitude aiiil difference of longitude made. '■VLB, If the latitudea or \oiigAuiBa \it \>Ql,h of the saof NAVIGATION. 53 Dame, their difference reduced to miles will be the diffe- rence sought; but if they be of different naraes, their sum reduced to miles will be the difference sought. Note. If the longitudes be of different names, and their sum ex- ceeds 180°, it must he suhtracted from 360**, and the remainder re- duced to miles will he the difference sought. In each of the following exercises it is required to find the difference of latitude and the difference of longitude : Sailed Atom. Lat. Long. Answers. Lat. Long* Dif.lnt. Dif.long. Mileg. Miles. 1. 25° 16' N. 150 3' W. 20° 17' N. 42° 13' W. 299 1630 2. 12° 50' S. 40° 34' E. 35° 42' S. 1 8« 14' W. 1 372 3528 3. 50' 17' N. 5' 7'W. 70» 12' N. 50*' 16' W. 1195 2709 4. 18°25'S. 15°35'W. 0° 0' 350 17' W, 1105 1182 Plane Sailing. 102. For the proportions necessary to solve cases in plane sailing, see (Art. 99). But Plane Sailing can also be solved by the Traverse Table in the following way : Damely, when the distance and course are given, find the given course, and under the distance, in the proper co- iamns, you will have the dif. of lat. and dept. required. Note 1 . In the small Traverse Table given in this work, look for the given course at the side of the page, and the first figure of distance at the top or bottom, where these columns meet, take out the dif. lat. and dept., removing the decimal point as many places to the right as liie figure is to the left of the units' place ; do the Same with each of the figures in succession, and add the results; the sums of the differences of latitude and departures thus found iHll be the dif. of lat. and dept. sought. In a similar way, if the coarse and dif. of lat. be given, may the distance and departure be Found. Note 2. When the course is not one of the things given, the solution is not so conveniently obtained from the Traverse Table as by trigonometrical calculation. EXKUCISES. 1 . A ship from lat. 49^ 57' N., sails SW. by W. 244 miles ; required the latitude she is in, and the departure made ? Ans. Lat. in 47° 41'. Dept. 2029 miles. 2. A ship from 1" 45' north latitude, sails SE. by E.^ till she arrives in latitude 2° 46' south ; lec^vui^dL V^x ^v^- 64 KaVTGATIOlf. tance and departure. Ans. Dist. 487-8. Dept. 405(1 miles. 3. If a ship sails NE. by E. | E„ from a port in 3' 15' BODlh latitude, until she departs from her first meridian 40S miles ; required what distance she has sailed, and ivhat latitude she is in ? Ans. Dist. 4491. Lat. in 0° 3' S. 4. Suppose a ship sails 488 miles, between the south and the east, from a port in 2" 52' south latitude, and then bj observaiion is found to be in 7" 23' south latitude ; what course has she steered, and what departure has she made? Ans- Course S. 56" 16' E. Dept. 405'8 miles. 5. If a ship sail south-eastward from latitude 50° 16' N., till her distance is 137 miles, and her departure 112 miles; required her course, and latitude come to ? Ans. Course S. 54° 50' E., and lat. 48° 57' N. fi. If a ship from latitude 32° 37' N. sail south westward, till Ler difference of lat. is 114, and her departure W miles; required her coursp, distance, and latitude arrired at? Ans. Course 8. 40' 24' W., dist. 149-7 miles, and lat. 30- 43' N. ^V Tbavebsf: Sailing. * 103. "When a ship is obliged to sail on different coutsci. the crooked line which she describes is called a traverse; and the method of finding a single course and distance, which would have brought the ship to the same place, is called resolving a traverse. Ilui.B. Make a Table of any convenient size, as that of the following example, and divide it into six columns : in the first of these place the several courses the ship has made; in the second place the several distances she Iiat made on each course. The third and fourth columns ate to contain the differences of latitude, and therefore to he marked N. and S. at the top ; and as the fifth and sixth are to contain the departures, ihey are marked E. and W. at the top. Find by the first and second analofjies, (Art. 99), ot from the Traverse Table, the dept. and dif. lat., which place in their proper columns, for each of the courses; then the difference between the north and south columnB will be the dif. of lat. made good, of the same name with the greater quantity ; and the difference between the sums of the east and west columns will be the whole departure, of the same name Kritli the gteateT m.i;i\SvMi Svaaatt. nATIOATIOir. 55 ExAHPiiB. A ship in latitude 38° 25' N., and longitude le** 17' W., sails SW. by W. 56 miles, W. by N. 110 miles^ W. 95 miles, SE. by E. 50 miles, S. 103 miles, and S. ;| W. 116 miles; required the latitude in, and the de- parture made ? T&IYXBSE l^LE. 1 Conecbed oounea. Pointa. Dis- tances. Dif. of latitude. Departure. 1 N. s. E. w. SW.byW. W. by N. West. SR by E. South. S.1 W. 5 7 8 5 i 56 110 95 50 103 116 21-5 31-1 27-8 103- 115-9 41-6 46-6 107-9 95 5-7 21-5 Dif. lat. » 277-8 21-5 41-6 255-2 41-6 256-3 Dept. s= 213-6 Now 256' miles of dif. lat. = 4** 16', which being south, sabtract from 38° 25' north, and the remainder is 34° 9' N., which is therefore the latitude arriyed at. EXERCISES. 1. Suppose a ship from latitude 49° 57' N., sails SSE. 15 mUes, SE. 34 miles, W. by S. 16 miles, WNW. 39 miles, S. by E. 40 miles ; required the difference of lati- tude, and the departure she has made, and also the course »he has made good? Ans. Dif. lat. 65-32 S., Dep. 14-14 W., and course S. 12° 13' W. 2. A ship from latitude 17° 58' N., sailed S. | W. 40 miles, SSE. J E. 97 miles, N. by E. { E. 72 miles, SSE. I E. 108 miles, N. by E. | E. 114 miles, SE. by S. i E. 126 miles, and NNE. \ E. 86 miles ; required the latitude she is in, the departure made, and the course made good? Ajis. Lat. in 16° 52' N. Dept. 258-87 E. Course made good, 8. 75° 40' E. 3. A ship from latitude 51° 25' N., sails SSE. iE.16 miles, ESE. 23 miles, SW. by W. ^ W. 36 miles, W. | N. 12 miles, SK by E. ^ K 41 miles ; required the latitude urrived at^ and the course and distance TCkaA<& ^<;^qW B'AvraA.TIOIT. 50= 25'. Course S. IS" 15' E. Dist. 62'704 K Faballel Sailing. P 104. Parallel sniling is the method of finding the dis- tance between two places ailuatod under the eame parallel of latifade; or of finding tlie difference of longitude cor- responding to the meridional distance or departure when a ship sails due easl or west. Since circles are to one another as their radii, the distance of two meridians on the equator, or the dif- ference of longitude, will he to the distance of the fiarae meridians at any oilier latitude as radius ; cos. of lat. 105. Hence (he following prnpnrlions can be derived; Dif. long. : raer. disl. = li : cos. of the latitude. It: cos. lat. ;^ dif. long. : mer. dist. B : sec. lat. = mer. dist, : dif. long. Also, cos. of any lat. : cos. of any other lat, ^ the mer. dist. on the first : the mer. dist. of the other. KXBItCISEB. 1. If a ship, in latitude 42° 54' N., sail due west 196 miles; required the difference of longitude made? Ans. 4" 27' 36" H'. 2. A ship from latitude 51° 25' N., and longitude 9° 2? "W., sailed due "est 1040 miles; required the longitude at which she then arrived ? Ans. Long, arrived at 37° 16' W. 3. A ship, by sailing due west 45fi miles, made a difit- rence of longitude of 12" 15', or 735; required the latitude on which she sailed? Ans. Lat. 51° 39' 14''. 4. A ship, in latitude 51° 16', sailed due east for 240 miles, another sailed due cast 308 miles, and made the same difference of longitude as the former ; on what lati- tude did she sail ? Ans. Lat. 3G° 35' B". MinnLE Latitcijb Sailing. ^^- rerse lOfi. In Middle Laiiiude Sniling, (he longitude is eaky- lated from the departure made, either in a single or tra- Terse course, by I'aralk! ^ni/inf/, upon the supposition Jt is equal to the meridian distance, on the middle ' f/eJ bet*¥een that sai\ea£tQTOaQiftvaVwtTC«e4a.t. | NAVIGATION. 57 This method^ though not quite accurate^ Is nevertheless efficiently so for a single day's run, particularly near the ?quator^ or when the ship's course is yearly east or west, [n high latitucles^ when the distance run is greats it may ead to slightly erroneous results. In the annexed diagram, let ABC epresent a %ure in Plane Sailing, in vhich AB represents the distance aOed^ LBAC the course^ AC the lifference of latitude made^ and CB he departure. At B make the .CBD = the middle latitude, and iroduce the line AC to meet BD in ha ). Then if BCD be considered a ,^ i^re in Parallel Sailing, in which ,^ !)B is the meridian distance^ BD f^ rill be the diff. long., and the Z.D he complement of the middle lati- ade. 107. Now from the two right angled triangles, ACB, ICD, and the triangle ABD, the foflowiug analogies can e obtained : viz. sin. D : sin. A : : AB : BD, or 1. Cos. mid. lat. : sin. course : : dist. : dif. long. 2. Sin. course : cos. mid. lat. : : dif. long. : distance. 3. Dif. long. : dist. : : sin. course : cos. mid. lat. 4. Dist. : dif. long. : : cos. mid. lat. : sin. course. dif lat. In the last of these analogies, substitute ■ for ° COS. course be distance, and multiply the extremes and means, and iere results dif. long, x cos. mid. lat. = sin. course X = tan. course X dif. lat. ». eourse 5. Hence (Alg. 103) dif. lat. : dif. long. : : cos. mid. lat. tan. course. 6. Dif. long. : dif. lat. : : tan. course : cos. mid. lat. 7. Gos. mid. lat. : tan. course : : dif. lat. : dif. long. 8. Tan. course : cos. mid. lat. : : dif. long. : dif. lat. These eight analogies are sufficient to solve all the cases r Middle Latitude Sailing that ever can occur ; only the rinciples of Plane Sailing and Parallel Sailing must fre- tiently be applied first to obtain the necessary data. Example. A ship from lat. 39° 41' N., and long. 31° ' W., sailed NE. ^ E. 590 miles ; required the latitude ad longitude come to ? First to find the latitude come to, we have h^ "PVaxka uling B : cos. 4^ pts. : : dist. 590 : dif. lat. ^14l^ tkiA^^'^. ,^^.., »• 41' N. I ^ £8 NATIGATI01T, Latitude sailed from, Dif. lat 374 = 6- 14' N. Latitude come to, 45° 55' N. Sum of latitudes, 85" 36' i sum, or mid. latitude, 42° 48* Applying dow analogy (1), we liave Cob. 42° 48'. ar. co. of Log. cos. = 10134464 la to sin. 4i pta. Log. sin. = 9-888185 As dist. 590 Log. di8t. =_2'770853 Is to dif. long. 10" 22' Log. 62I'6= 2-793501 This dif long, being E., while that sailed from is W., their difference is the longitude come lo, which is there- fore 20° 41' W. The ship is therefore in latitude 45' 55' K., and longitude 20° 41' \V. EXERCISES. 1. A ship from lat. 14° 46' N., and long. 24" 46' W„ uiled SE. by S., until by observation she was found ta be in latitude 10° 30' N. ; required the distance sailed, and her present longitude? Ans. Dial, 307-9 miles; long, in 21° 50' TV. 2. A ship from lat. 49" H' N.. long. 0° 19' W., sailed 320 miles between the south and west, and then, by ob- flervation, was found to be in latitude 45° 8' N. ; required the course, and the longitude come to ? Ans. Course 8. 26° 39' 0" W. ; long. In 9° 52' V. 3. A ship from lat 40° 3' R, and long. 3° 52' R, sulcj .280 miles between the north and east, upon a direct GOnran, and made 186 miles of departure; required the course, and the latitude and longitude come to? Am. Course, N. 41° 37' 39" E. ; lat. come to, 43° 32' N.j and long. 8° 1' E. 4. A ship in north latitude sailed 500 miles upon direct courae, between the south and west, until her diffe- rence of longitude was 440 miles ; required the cotirse steered, the latitude sailed from, and the latitude come to, allowing the middle latitude to be 43° 45' north? Am. Course, S. 39° 28' 14" W.; lat. sailed from, 46'58'N.; and lat. arrived at, 40° 32' N. k MENSURATION OF SURFACES. 59 MENSURATION OF SURFACES. In the mensuration of plane superficies the prohlem to )e solved is, — How often is a square of a given magnitude ontained in a figure of which the rectilineal dimensions re given? The method of solving this prohlem depends ipon the form of the figure whose surface is sought. The quare employed to measure it may he of any dimensions; ut those most commonly used are a square inch^ a square sot, a square yard^ a square link, and a square acre, lie numher of times that the square is contained in the iTen figure is called its area, and the square that is em- loyed to measure it is called the measuring unit. Table of Lineal Measure. Inches. Link. 7-92 = 1 Foot 1 12 1-5151 = Yard. Pole or Perch. 36 4-5454 3 = 1 198 25 16-5 • 5 = 1 Chain. Fur. longs. 792 100 66 22 4 1 7920 1000 660 5280 220 40 10 = 1 8 = Mile. 1 53360 8000 1760 320 80 Table of Square Measure. Square Inches. Square Link. Square Foot. 2:7264= 1 Square Yard. lid 2-2956=: 1 It* Square Perch. 1S!96 20.6611 9a 1 Square Chain. 99204 625 272*25 30-25 = 1 4356 Square Rood. ^264 10000 484 16 = 1 Square Acre. ISfiSlSt) 25000 10890 1210 40 2-5 = 1 Square 6872640 100000 43560 4840 160 10 4 = 1 » l4489G0(y 84O0000fi Sm8400' 30976001 10240(\ 6400 1 2560 \ 6\0 =\ \ \ Problkiu ^^^^H To find the area of a Paralelhgram. whetjier ft %ff¥ RuLR I. 'When the length and perpend Icular breailti are given; Multiply the length by the perpcndicolaj hreaclth, and the product will be the area. DEKDNSTomoK. Let ABCD lio x it. contatna tliE meaBuring lineal Doit any niamber of times, as S, and of wUch thB pfrpandiculnr hremllh All wmtiuiifl the Biuno unit .iny numbor of timea, ks 4 ; then, if lines 1 \ i a of each unit in the sides, parallel " to the adjacent sides, as in the annexed the whole figuro will be divided into as nnits in the product of BCxAD; for til diagram, it is evident tliJ many Bqnaroa as there i«| the Brat as tbera are nnits in AB: henee there aro na man; tqnintf in tha whole figure as there aro units in the product of the anil in BC multiplied into the number of units in AB. equal lo a rortanglo having the same hose nnd altitude, or perpendi- cular breadth, BBlho parallelogram. EsAMPLK. What ia the area of a pamllplngram whnw length is 2-JO feet, and perpendicular breadth 160 f«lt Ans. 240x 1C0=38400 sq. feGt=426GJ square yards. EXRRCISES. 1. What ia the area of a square whose aide Js ^ links? Ans. 124:2562.5 square ]inks=124 ac. 1 ro. f^ 2. Find the surfiice of a rectangular pane of glass, in length being 5 feet 7 inches, and breadth 3 feet 5 inche*. Ans. 19 square feet 11 square inch** 3. How many square yards of plastering are in the Oftf ing of a room, its length being 2u feet, and breadth XTM 6 inches? Ans. 4S\\ \ square yard). 1 4- What is the area of a field in acres whose length il 1535 linkg, and whose breadlli is 1270 links? Ans. 19 aeres, 1 rood, 39-12 pereJie* 5. How many roods, of 3G square yards each, an in i( 1 wall 320 feet long, and 13 feet G inches high; and fihol ^^1 trill it cost at L.2, 12s. Cd. per rood^ ^^ft ^Vtis.\3^ roods, oDrtJ^ lESumnuTioB or sdbfaccs. q1 H. "When two sides and the contuined angle are lultiply the product of the two sides by the natu- if tile included angle, the product ivill be the area. I Logarithms. Add the Logiiritbms of the two I the log. sine of the contained angle together, tho inUhed by 10 in (he iiides will be the Log, of the TBiiioN. Let ABCD lie a, iiai-al- p c in whidi iii'e givua tb« twu ai Ji:9 I 7 AB, and the cuotaiueJ angle / / iw DE -^ to AB, lliL-Q by lost / / 3E_ ID" a. of tho figui'u = A1).UK= L :AB.AD. r lin. DAB. (Trig. Art. 6). PI.B. What IS the area of a parallelogram, two ad- lea being 34(iand 210, and tbeeontained angle 62° E.sin.62''12'=.8fl4581x34ti>: 210-64273-63546. [n this case, whather we use tbe natural Bines carried tu f decimala, or tho logairithmic HiiicH wiUi li plicei^ of deci- lau ojdy dopond on the firuL fi figures ef tho ajiawer ^m •eat, BO that the answer ciui he fuuud witluii Ites than a. ^ai\ of tho nhele, wbich will be suttidently correct fer ul- ractical purposea. If jjivater aociiracy he required, re- lat be hod to tables earned to more dodnml places. EXEltCISES. iat is the area of a field la the form of a pa- im, two sides being "JZG and 609 links, and the angle 81" 15'? Ans. A acres. I rood, 28'fll2 perches, parallelogram has two of its sides 360 and 200 id the angle contained between them 30°. What ■a? Ans. 30000 sqiiare yards, e of tbe sides of a field, in the form of a rhombus, links, and one of its angles is 70° 18'. What is its 1 how much is it leas than if it had been a square he same length of side! Ans. Its area 14 acres Tches, less than the square 3 roods, 20'53 perches, itv many acres are in a rhomboid, whose leas angle nd the including sides 2u35, and 1040 liaks^ Ans. 13 acres, 29.12 perches. Fbobleu IL id the area of a Triangle, when there are given, ( ' ! base and perpendicular altitude, two &\' ained aagie, or the three sidus. ^ 3 MxiHtntATioH or amwACsa. ■ Rule I. Multiply the buse Ly the perpendicular aiti tude, and half the product will he the area. Rule II. Multiply htilf ihe product of the two sides li the natural sine ot' the iucluded angle, and the prodiu will be the area. Or, add together the log, of one side, th log. of half the other, aud the log. sine of the containe angle; the sum rejecting 10 from the index ivill be the Iq of the product. Remark. The reason of these rules is ohvioufi &ai Problem I., and (Geo. Prop. 28), which proyea that a ti angle is half of a parallelogram. ExAUPLE. What is the area of a triangle, two of vhttt sides are 8ii and 90 feet, and the included angle 47° 131. Tlie nat. sin. of 47" 13' = -733927 Multiply by K^fix 90) = 3825 And the product is the Ans. =280? 2? sg . feet Exercises. 1. What is the area of a. triiingle whose base is 7" feet, and perpendicular altitude 48 feet? Ana. 1680 feet, or 186| yards 2. How many square feet are in a right angled tiiangbj whose base is (iO, and perpendicular 40 feet? Ans. 1200 square feet 3. What is the area of a triangle whose base is 54, ajA perpendicular altitude 29.52 chains? Ans. 79 acres, 2 roods, 32-64 pol«» 4. What is the area of a triangle, two of whose sides "" 45 and 5675 feet, and the contained angle 3(i° 45'? Ans. 763 986 square fee 5. How many square yards are in a triangle, of nliie one angle is 45°, aud the including sides 30 and 24 feetl Aus. 28-2842 square yarf Rule III. Find half the sum of the three sides, from it subtract each side separately; then multiply half sum and the three remainders together, and thesqoia root of the last product will be the area. Or, add the lo- garithms of the half sum and of the three remainders ttf f;ether, and half this sum wiU be the Dgarithm of the area. I)GHaNSiiu.TiO!t. Let AJBC be a triangle, of which tlie three sides ore given, and adopt tile sanie notAtioa as in Trigonometr}' ; wa hare bf Rul e 2, area ={ be Bin. A. K ow lia. A M= V(l -Hifa. A) (,V — coi h.~i = ME.VSUBATION OF SURFACES. 63 ,J^^^[^^^ (Trig. 45 and 4e) = ^^«(s-^)CV-6)C»=0- - 3 a>B area is = i&e x if. ^s(, — a)"(s= 6) (s ^c) = . IT iJie bnangle be ei^uilalenil, and have each of its aideB a, t wiU becorna 5, (s — a), (s — i), and (s — c), will each become 5 - therefore the area of au equUatend triangle isVg x^xfXj = Example. How many square yards are in a triangle, vhoee three sides are 50, 40, and 30 feet ? Here half the sum of the three sides is 60, and the three iraneinders are 10, 30, and 30; therefore the a>-ca is ^60302030= Va6<>i)0O=60O feel;=66^ yaida. EXEBCISE9. 1. "What is the area of a grass-plot in the form of an Equilateral triangle, each of its sides being 234 links ? Ans. 37936 square poles. 2. How many square yards are in a triangle trhose three aides are 39, 36, and 15 feet ? Ans. 30 square yacds. 3. Hon many acres are in a field in the form of an is- osceles triangle, each of the equal sides being 500 links, and the third side 600 links ? Ads. 1 acre, 33 poles. 4. How many square feet are in the area of a triangle, whose three sides are 74, 82, and 90 feet 1 Ans, 2855-61 square feet. Rule IV, WLen there are given a side, and the angles ttt its extremities. From twice the logarithm of the given tide suhtract '301030 ; to the remainder add the log. cosec. of the sum of the two measured angles and the log. sine of each uf these angles ; the sum after rejecting 30 from the index will be the log. of the area, DehONSTRITIOM. (Diagram last case.) Let AC be the giren Bide, nhieh call b, and A. and C the two measured angles ; then since IB it the supplement ot A+C, sin. B= sin. (A+C), (Trig. Art. 30.) HraicesiD.(A+C):sin,A-. :6:BC=^^^|^^,{Trig.Art3(i.) Abo R-ain.C: ; { BC =-^?J?iA, Y BD=1" °' - * - .^"'^ , and the area = \ lin. (A+C)/ Run-IA+C) lixBD= S HBH^IA+C li *'^«1'S'°™ - .in: lA+CI = ""^ (A+C), =1 gives the ahove rule. , "What is the area of a triangle, OTie dt \)^yow '64 sides is 46 yards, and 57" 12'! Log. 46=1-662758x2 Subtract Log. % Log. cosec. 129° 30', Log. Bin. 72° 18', Log. sin. 57= 12' nd the angles at its extremities 72" Jlere the sum of the angles is 129° —3-325516 ■301030 3-024486 =10-112534 = 9-978939 ^ »92457 2 Log.l097-96aquare yards. Ans.= 3'040591 I 1. How maDy acres are in a triungiikr field, ota ■whose sides is 1046 linka, ;md the angles at its extre ies 68° 24', and 71° 1^'? Ans. 7 acres, 1 rood, 29-208 pi 2. How many square yards are in a triangolar field, of its sides being 465 feet, and the angles at its extre ■ s 45", and 78" 15'* Ans. 9944-13 square ya 3. Two of the angles of a triangle are 76° 13' and ' 9', and the side betneen them 3475 links ; how iq acres are in the field ? Ans. 144 acres, roods, It PilOBLEll III. To find the area of a trapezoid. EuLK. Multiply /tay" the sumof the parallel sides by perpendicular distance between them, and the prodact' be the area, (Geo. Prop. 33). ExAMFLE. What itt the area of a trapezoid, its pan sides being 30 and 20 feet, and the perpendicular dJsti between them 14 feet t Here the paralleS-Hides beiiu and 20, half their sum is 25, which being multiplied 14, gives the area 350 etiuare feet. XXEKCISES. 1. What is the area of a field in the form of a i pezoid, its parallel sides being 536 and 378 links, and perpendicular distance between them 418 links ? Ans. 1 acre 3 roods 25-6416po 2. What is the area of a board 20 feet long, ^ brei at one end being 2 feet 6 inches, and at the other 1 1 3 inches? Ans. 36 feet 8 \iA 3. The side of a roof Is 40 feet at the easing, 24 I at the ridge, and the distance from the ridge to the eai isJ8feet; what is the urea of the aide of die roof ? A.T\a. &1& fc<it, Qc 64 B^oare jti < jixbsebatioit of bubiacbs. u Problem IV. To find tbe area of a trapezium. K171.E I. Dinde it into two triauirlea by a diagonal, and find tbe area of each of them, and their atun nill be the area of the trapezium. Note. — The srcas of the triangles are to be found by either of tbo rules given in Problem 11., Bccordiag- to the data. RiTLB II. Measure each of the diagonals, and the angle of their intersection ; then half the product of these dia- (^Doals, multiplied into the nat. sine of the angle of their intersection, will be the area. Or add together the loga- rithms of the diagonals and the log. sine of their intersec- tion, the sum diminished bj 10'301030 will be the loga- rithm of the area. Note. — This mle is kIbo trne of paxallelogramB. DsKaHBTiuTioK.— Let ABCD be a. tra- pezium, of wliich AC and BD are tlie dia- -' . gunalB which iuCerBeecinE; then sin. AED \ Ez ain. DEC, (Trig. 30) = sin. CEB = Bio. &EB,(Geo. prop.3), which callsiu.E, and ietAE=a, EC=.c,B£^,aiid UE=(f; then (Prob. IL Rale II), AAED=i ad torn. E, d}EC=idc Hiu. £, aBECz=IAc Bin. E.and ULEB=ia& Bin. E ; .■. tbe whole figure= KoJ+A+ftc+afc) ain E=i{iH-c)(6+tOBin. c £=)ACxBD Bin. E, which U the rule. Example. If AC=530 links, BD=608 links, and the (.AEB=52° 12'; it U req,uired to find tbe area. Nat. Bin. 52= 1 2'=-790155 X 530 x ^ =127309-7736= 1 acre, 1 rood, 3 poles, 21 yards. EXEBCISES. I. In the trapezium ABCD, if the diagonal AC be 270 feet, and the perpendiculars upon it from I> and B 180 and 120 feet; find its area in square yards. Ans. 4500 square yards. _ 2. In a trapezium ABCD, if the diagonal BD be 1476 links, and tbe perpendiculars on it from A and C 557 links and 403 links; find its area in acres. Ans. 7 acres, 2 roods, 39-84 poles. 3. If in the same figure AB^490 links, BC 464 Unks, BD 756 links, Z,ABD 74° 44', iDBC 80M7'; what is the area of the figure in acres ? Ans. 3 acres, 2 roods, 2-496 poles. 4. If AB=36feat,BC=34, CD=42, AD=44,andthe diagonal BD^=62 feet ; what is the area of the tra^eiuBo, in yards ? Ana. 16V40ft ss(^Ma.t(i -^aiia. F ^^K 5. If AC:r^7eO links, BD=810 links, and the angle ^^KAEB be 79° 1 6' ; what is the area of the field in acies? ^^K. Ana. 3 acres 3864 poles. ^^K 6. If AD=3aO links, DC=3eO links, BC=filO links, ^^■AB=534 links, and ihe angle ADC 88° 30'; what is tlie ^^bma in acres 1 Bt Tiig. At'=4(i(> 834. ^^r^ Ads. 1 ac. 2 ro. 26'13 poles. llraHSOTATIOW (# SiniFACES. To find the area of any irregular figure. Rule. Draw diagonals dividing the figure info triangles, or triangles and trapezioms. Then find the area of all ihese separately, and their sum will be (he content of the lybole irregular figure. 1. Find the content of the irregular figureABCDEFGA, in which are given the follow- ing diagonals and perpendi- culars, namely AC=530 links. GD^424 FD^4a6 Bi =134 GC=394 ^^ DC=182 ^^ To find t Ans. 1 Ea = 30'2812 poiet Problem TI. To find the area of a regular polygon, when the leng!'' id the perpendicular upon it from the cenlreiiK given. KuLE I. Multiply the side by the perpendicular, and that product by half the number of sides, the last pmdac' will be the area. Dguonbtbation. Thu polygon consists of 03 many trionglea as iC hath sideB, whoso heights and bases are each equal to thop or- pcndlcular, and the side o{ the polygon ro- spectively; if » be a side of the polygon, p tbe perpendicular, uid n the nnmher of Bides, then ^ps will be una triangle, (Prab. 'IL Rule I.) and iiips all tlio triangles or (ha whole polygon. ' £xAifPi:.B. Find the area of a heptagon, each of n mdes is 20, and the perpendicular 20-7^5 feet JIEN8URATION OF SUBFACES. 67 BXERCI8E8. 1. Find the area of a hexagon^ ivhose side is 10 feet, id the perpendicalar upon it from the centre 8*66 feet. 8-6g X 10 X 3=259-8 square feet. 2. Find the area of a pentagon, whose side is 8 inches, id perpendicular upon it from the centre 5*506 inches. Ans. 110*12 square inches. 3. Find the area of an octagon, whose side is 15 feet, and e perpendicular upon it from the centre is 18*1 1 feet. Ans. 1 20*73 square yards. 4. Find the area of a decagon, whose side is 4 feet, and e perpendicular upon it from the centre 6*156 feet. Ans. 123*12 square feet. Rule II. When the side only is given ; to twice the Tarithm of half the side add the logarithm of the numher sides^ and the log. cotangent of (180° divided by the imber of sides) the sum diminished by 10 in the index U be the logarithm of the area. RuLfS IJI. Multiply the number opposite to the name the polygon in the following table by the square of the i^h of the side, and the product will be the area. For the demonstration of these rules see Key. Jo. of ides. Names. Areas or Multipliers. 18() n Log. cot. — 3 Triangle. *433013 60** 9-761439 4 Square. 1-000000 45° 10-000000 5 Pentagon. 1-720477 36*' 10-138739 6 Hexagon. 2-598076 30° 10-238561 7 Heptagon. 3-633913 25°« 10-317336 8 Octagon. 4-828427 22*4 10-382776 9 Nonagon. 6181824 20° 10-438934 10 Decagon. 7-694209 18° 10-488224 11 Undecagon. 9-365640 16°A 10-532205 12 Dodecagon. 11-196152 15° 10-571948 Example. Find the area of a pentagon, each of whose as is 8 feet. Rule II. 2 X Log. 4 ^ =1 *2041 20 Log. cot. (^-5^'=36°'\= 10*138739 Log. (n=5) = *698970 Ans. 110*11 = 2041829 Rule III. Tabular multiplier=l-720477x(8^=64)=ilVQ*UQ^1'^ :bxeiicisbs. /. Required the area of a regular pentasoTi, o^ ^V\«^kv ker-efibe eqmd sides is 15. A.tl^- 'SSTlA^ r » KEKeDXATIOK OV SOBFACKS, I 2. Bequired the area of a regular hexagon, either of equal sides is 20 feet. Ana, 115'470 aq. yards. a. Jiteqiiired the area of a regular heptagon, either of whose equal aides is 3 feet. Ana. 32'705 sq. feet 4. Required the area of a regular octagon, either of .vl.oae equal aides is 10 feet. Aob. 53649 sq. yanft, 5. Eequired the area of a regular decagon, either of iWhose equal sides ia 18 inches. Ana. lySllS aq. feet 6. Required the area of a regular dodecagon, either of Trhbse equal sides is 4^ feet. Ans. 226732 sq. feet 7- Required (bj Rule 2) the area of a regular polygon of 16 sides, each side being 5 feet. Ans. 502734 sq. feet 8. Bequired (by Rule 2) the area of a regular polygon of 100 sidea, each side being 12 feet Ana. 12728'2 sq. yardi. Phoblkm VII. To find the diameter and circumference of a circle tin one from the other. Ca9E I. To find the circumference of a circle when the diameter is given. Rule. Multiply the diameter hy y, jM or 3'1416,.3iJ the product will he the circumference nearly. Cask II. To find the diameter of a circle when the at' cumfereace is given. Role. Multiply the circumference hy ^, ^ JJ, or -31831> the product will be the diameter. Note. In each of tl)e abovo nilGS the first mnltipIieF is Ibe larf Bccnrate; the Bcmnd tho tuoat correct; and tlie third tbe miH Tenient, and very Dearly as accnrato aa tho eecond: The c of these ruica will be given in tho Key. The answen ted* queations will be given as they are deduced from the MirJmnltipW in each rule, and the pupil can Hnd the answers hy the other pliers also, and thereby judge of tlie accuracy of each of the t Example 1. Find the circumference of a circle wbtm aiameterisl2inchcs. By Rulel. 31416 Xl2=37-6992Aii« EXAMPI.B 2. Find the diameter of a circle, the circumft^ encebeing20inches. Bj-Eulell. 31831 >:20=6-3662Aiil EXERCISES. i 1. If the diameter of a cylinder he 3 feet, what ia Hi rcumference ?' Ans. 9-^248 f«t 2. If the diameter of a gasometer be 60 feet, what is drcumference f Ans. 188'496Mi 3. What is the circumference of the earth, its diaroeW hieing 7912 mllea* Kna.1.i«S56-3392 HmffiniATiof) OF BirKPACSB. 69 4. What is the circumference of the planet Jupiter, i(s ameter being (19170 miles? Ans. 280 136 472 mile i^. 5. If the circumference of a rcruini tree be 7 4et 3 ches, what is its diameter? Ans. 230/7^ feet. 6. If the circumference of a cylinder be 10 feet, what is , diameter? Ans. 5-09296 feet. 7. If the circumference of a circular pond be 300 feet, lat ifl its diameter? Ans. 95493. 8. If the wheel of a carriage turn 528 times in a mile, lat is the diameter of the wheel? Ans, 3'1831 feeb Problem YIII. To find the length of an arc of a circli ItnLB I. When the chord of the whole arc and alt E chord of half the arc are given. From 8 times the ord of half the arc, subtract the chord of the whole ;, one third of the remainder will the lengih of the arc nearly. Thns, in the annexed diagram — — - = the length of the arc irly; or, if the chord AB and the ight CD be given, it becomes ^}AB'-tCD'— A B - 3 For the demonstration of this rule, see the Key. CxAUPLB, If the chord AR be 36, and the height CD f; what is the length of the arc ACB f Sere AC=N/ia'+(7-5_)'=V3i{0-25=]9-5;andhence 5x8—36 136—36 ■„ , ,. ^ ., , -— ^ — 5 ^40, length of the arc nearly. EXERCISES. 1 . The chord of the whole arc is 50-8, and the chord of f the are is 30'6 ; required the length of the arc ? Ans, 64'6. i. The chord of the whole arc is 45, and the chord of f tlie arc is 25'5 ; what is the length of the arc? Ans. 53. i. If the chord of the whole arc he 16 feet, and the eht 4 feet ; what is the length of the arc ? Ans. 18 518 feet. I. If the chord of the whole arc be 24, and Ike \ie\^\.'i ■, at is the length of the arc ? K^tia. %1. - n I i RuLK II. When the radius of ihe circle and the nmn- Iter of degrees in the lire are given, or con be found from the data. Multiply 'be nnmber of degrees in the arc by the radius, and by -0174533, the product nill be the length of the arc. . Rule III. Tuke out from (Table XI.) the numbers cor- WBponding to the number of dfffrues, minates. and seconds, ia the arc ; their sum multiplied by the radius will be the length of the arc. Note I. To tlnd tlie radius and number of degrees in an arc, leu the chord AB and height CD are given; ni(liua=i' — ^j^ ; &nd Iflugant j (btg ACB)= ~ ; for — = long. CAD = lang. CEB, (fioo. prop. i7, cor. 1.)= lang. iCOU,(Geo, prop. 47)= lang, iAOB, . "which Che aro ACB ia the measure. . KoTK 11. The uumbei"0174£S3 given in Rule 3 is thus oblaio- i ad; — Since tlis eemicircumfererice of a circle whose itidiua is I n < S-UI5a365359, if thi3bedLviJedl)y IBO, Iha uumbHr of degreesin 1, aemicircle, it will give the atiove number for tlie leugth 5 t is- gree when the rndina is L ; and since tlie circumferencEH o( citxdes an Ui one another as their radii, the above rule is obvious. ExAMPLB. Whatia the length of an arc of 20" 30' 10", the riidius being 16? Dy Table XI. arc an-loradlns ! =-3490659 30' „ ='00ST2Cfi , 10" „ = ^485 [ arc 20° 30' 10" =-3578410x 1 6=:5-72345fi. EXERCISES, J. "What is the length of an arc of a circle contsin- iag 49° 30', the radius being 30 inches ? Ana. 25-9itt]. 2. What is the lengtli of an arc, of which the chord it 20, and the radius of tlie circle also 20 ? Ans. 20-9439. 3. Find the length of an arc, of which the chord is 36, and the height 12? Ans. 45-8642. 4. What is the length of an arc, the radius of the circle being 40, and its height R? Ans. 51-4a 5. What is the length of the circular arch of a bridgti Ihe span of (vhich is 18 feet 6 inches, and ihe centre of the arch aboTe the top of the piers 6 feet 9 inches ? Ans. 244923 fed Problem IS. To find the area of a circle. JiULE 1. Multiply half the circumference by the railim. * the product will W t\ie aiea. 71 RoLB IL Maltiply 3'1416 by the square of the ra^ dias, or '78o4 by the square of the diameter, and the pro- dnct will be the area. RuiB III. Multiply -0795775 by the square of the cir- cumfereoce, the product nil! be the area. Demonstkitioh. The circle may lia conceived to ba maile up of ta infinite namber of small triangles, tlie anm of wIiobq bases i» the ctrcumfereace, and the vertices being all in Ihe centre, the altitude of each of the trianglos wili be the radiua of Iha eircia; (harefora the aea. of all the tnangleg, or of the whole circle, will be (Prob. IL) hilf the sum of all the bases, or the soraicircumferHice multiplied into thdr common altitude, or into the nidiua of the circle, which ia Kale 1. Agam, by (Prob. 7), the Bemicircumference ia = to 3-U16r, whidi being multiplied by r, according to Rule I., givea 3H16r^, which is Rule IL; and the second form ia ^-^ x (9r)" = ■7B54rf'; where r ia used for i%dlua, and J for diameter. Also, since similar surfacea are to one another as the squares of tbeir like parts, (Geo, prop. 71, ear., and prop. 76, cor. 6), aud wnco Ihearea of a circle whose circuinference is 3'1116 ia -7S3i, tho-e- foM (3-1416)*: I : i-7Bo* : ■7BS*-H(3-U16)'=.-tl7S£776, theanaof lotrclewhoseciTCumfeFeucB is 1; hence P-.(circiimf.)'::-()79577£ : the area to any other circumference, which proportion bemg wrought, ^m Role 111. ExAMPi.8. 'What IB the area of a circle, its radius bdngS? By Rule U. 3]4]6x(5'':=25)=78-54, the area re- paired. EXERCISES. ' 1. What is the area of a circle whose circumference is 333 feet, and whose radius is 53 feet ? Ans. 980i sq. yards. 2. What is the area of a circle, whose radius is 1522 tDcIieB, and its circumference 9563 inchest Ans. 50537-8 sq. feet. 3. Wliat ia the area of a circle whose radius is 46 ? Ans. 66476256. 4. What ia the area of a circle whose radios is 45 feet? Ans. 70^-86 eq, yarda. 5. Wliat is the area of a circle whose cireumference is 100 feet? Ans. 795775 sq. feet. 6. What is the area of a circle whose circumference is 40 yards? Ans. 127324 sq. yards. 7. What is the area of a circle, when a chord in it is 32, and the height of the arc cut off 8 ? Ans. 1256'64. 8. What is the area of a circle, when a chord at the dis- tance of 5 from the centre measuxes 24? Ans, 53fi'^'Jft\. Kors, The tbird and fourth exercises above axe m\eo&«^ ^n ^>^ 1 uB also the Bevectli and I Problem X. k To find the area of a sector of a circle. Rule I. SluUiply half the length of the arc of the sec- tor hy the radius of the circle, the product will be the area of the sector. Note. The demouatratiou is evident from the firat case of Prob- BxAHPLB. What is the area of a eector, the arc being | 10'2 feet, and the radius of the circle 14 feet ? i — xl4=;51xl4=71'4Bq. feet the area. ^ EKEBCISES. 1 . What !a the area of a sector whose arc is 20 j^ inchesi the radius of the circle being 14 inches ? Ans. 143^ square inches. 2. Find the area of the sector whose arc is 10 feet, the '' radius being 12 feet? Ans. 60 feet j 3. Find the area of a sector whose arc is 100 feet, audi whose radius is 54 feet? Ans. 300 sq. yards. Rule II. Multiply 0087266 by the number of degrew . in the sector, and by the square of the radius, and the lastj product will be the area. DEHaNSTRATian. The area of a circle whose radius is 1, is 3'14M| tfaerefore the area ofasectorof I degree IB 3-U16-H36e=-0D872fi^ wiiicb b«ing multi^ilied by tbe square of the radiaa, will give the uc^ of a Bector of one degree to that raJiuB, and thia multiplied bj l(* number of degrees iu any other sector, will give the ares of iW ■ootor; hence the rule is obvioua. KoTE. When the number of degrees in the sector are Dot givcHi they must be calculated by trigonometry from the data; and whea the radius la not given, it must be calculated in like maunei'. ExAupLB. What is th<i area of a sector of 45% th«J ladiua being 12 feet? Uere -0087266x45 gives -392697, and -392697 (12'=:144) gives 56'048368 sq. feet, the area reqniied. EXERCISES. 1. What is the area of a sector of 35°, the radius bei]i| 45 feet? Ans. 618-497775 feet 2. Bequired the area of the sector, the arc of wbirh ii 30 degrees, and the diameter 3 feet? Ans. -589045 fe*; 3. What is the area of a sector ivhase arc is a quadrant, or contains 90 degrees, the diaiaeteT being 18 feet ? 4. What is the area of a sector of 60 decrees, the radius bdag 12 feetl Acs. 8377536 sq. yards. 5. What is the area of a sector containing "jS degrees, the radius being 5 inches ? Ans. 15'70788 inches. 6. What is die area of a sector whose chord is 24, and whose height is 5 ? Ana. 225'31]. j 7- ^tat is the ar*a of the sector of a circle whose radius I ii 9, and the chord of its arc 6 ? Ans. 27-5266. Pboblkm XL To find the area of a segment of a circle. Rule I. Find the area of the sector having the same Arc with the segment by the last problem. Find also the irea contained by the chord of the segment, and the radii of the sector, ^en take the difference of these two when the segment is lesa than a semicircle for the area of the seg- ment, and their sum if it is greater than a. semicircle. KuLB II. From the product of the number of degreesin the sector, multiplied into -0087266, subtract half the natural sine of the degrees in the sector, the remainder aaltiplied bj the square of the radius will be the the s^ment. d DiBionsTBaTtos. Let d= tlio degrees in tbe Ugte ACB, then by the secoad Rule in last probteni the urea of the sector ACBD is ■OI)a7266rfr', and by Ruie 11. Problem 2, ii' not. sin. ACB = (he area of the triangle | ACB ; the difference of these expresaiona, or [■0087266J— i DSt, sin. ACB)!" is evidently > tlie area of the eegnient ABD; but Ihia is Rule II.; and Rule I. must be obvious from Ui inspection of the tignre. flxAMPLB. Find the area of tbe segment ADBEA, its chord AB being 12, and tbe radius AC or BC 10. (AC=10) : (AE=6) :; R : sin. ACE=36'' 52' 11-3". Therefore ZACB=73° 44' 22-6" =73-7396". Hence the area of tbe Bector=-0087266x737396x 100=64-3496. Again, the area of the AACB=6x8:^48. Whence the area of the segments 64-3496—48= 16-3490. Again, by Rule ir. -0087266 x 73 7396— i nat. sin. 73" 44' 22-6"= ■643496— ■4a0000=-l 63496, which being multiplied by t^=100, gtTes 16-3496 for the area of tbe segment. 1 . Find the area of tbegreater segment o£a c\ic\e,\la ii\0T4. being- 34. and tbe radius of tbe circle 20. Ana. \\9\'i'i\T - 2. Find the area of a segment of whict the aic cciiiVavB* ^3', tbe radius of the circle being 12. Ana. '2^-Wi\'2B-k, ■- -_ M 7* MKHetTBATION OP SDBFACXa. 3. Find the area of a se^ent of a circle, the chord of which is 20, and the height 4. Ans. 59-002. 4. Find the area of a seg'ment of which, the arc conl^ns 90°. the radius ofthc circle bd:ig IW. Ans. 256-8546. 5. Find the area of a Begmenl of which the arc contains 270°, the radius being 15. Ans. 642-64095. PROBI-EM XII. To fiod the area of a Zone, or the space included betwetu two parallel chords, AB and CD. RuLK I. Find the area of each of the segments DECQ, and AEBP, their difference will evidently be the area of the zone ADCB. Rui.a II. Find the area of the trapezoid A BCD, to which add twice ihe area of the segment AFDS, and the sura will be the area. t from the previww jn- it ia neccBBary to find Iht Note I. These rules are so aolf-evidei lileros, that they require no demonstratiou Note 11. The dnta Ja often such that rsdioH of the circle. This can be daue aa follows; when we bin given the chords ABoad CD, and The perpendiculiir distance betncB I' them, let OP be drswu .>- to AB, then being produced it will tha te f ' J- to DC; draw DU |1 OE, and .-. J- to AP; draw also from the cmW | ■ O, OS-^ to AD, and let it meet ihe circamference in F, and thioDgli i S draw SR \\ AP, or DQ; then tbe As DAG and SRO hare the aA» of the one respectivrfy -^ to those of the other, sod are .-. rimilM. Now DG— the distance of the || chords, and AG= half the difir- eocc uf Ihe chords, whilst SR= oue-fourtli the sum of the two cboris AB and DC. Agnin, from the two similar as ADG- aud SRO "e I have DG :GA=SR ; R0= '^-, IT now d= the diatauee o£ llw ' chords, c= the grettter chord, <■'= the less; SR will =i(c+<j'J sud AG=i(i — c'), and DG=i/, stibatituting these values in the»bovo value of RO, we have 110= i^ and OP — OR— BP= 'J±i^=^ _trf„ M^l(_yi-*". HcDce the centre will be wiUi- i out or within the *one ncooi-dlDg dh Uic diffrraxx of lie nquant <ifli' ckorda u yrealfT or lei' than id^, and the centre will alwajs be a) <1k iiatanee "+''"'^'+''^ from tbe less oliiird. WoTK III. T he distance OP bebg thm found, we l»«vo Ofa = n/ OP'+ PB*, or VOQ,*+CQ»; ulso AD = jDO* + AC*- Vi(o — c'j'+ii", from which the lines necessary to find the af« «• eusil/ fouod, when the two ehmdaaad the distance between ihenm ExAUPLE. Let the ^ealer chord be 96, the less chord 60, and the distance between them 26; required the area of the zone. Here substituting the proper values in the formula for l.?6x36+4x{36)> _ 2913 OP found above. ^ 0P=; =14, therefore the radios is =^~(48)^ + (l4)J=50, and the angle contained by (he sector of nhieh 60 is the chord is 73" 44' 23"=73-73972, which being multiplied by ■0087266,and by the square of the radius, gives 160K7425 for the area of the sector, which diminished by 1200, the area of the triangle, gives 408-7425 for the area of the less segment. In the same manner, we iind the area of the ^eater sector 3217'485, and its corresponding triangle 672; hence the greater segment is 3'''45 483, which, dimi- nished by the area of the lesser segment, as found above, gives for the area of the zone 313(i-7425. BXERCISEB. :hord 40, and the less 30, and 1 '35, required the area of the »ne. Ans. 158174. 2. The greater chord of a circular zone is 32, the less chord 24, and the perpendicular distance 4 ; what is the ureaof the zone ! Ans. ]13'516. 3. Required the area of a. circular zone, each of whose parallel chords is 50, and their perpendicular distance 50! Ans. 3213-485. 4. Find the area of a circular zone, the greater chord of which being equal to the diameter of the circle, is 40, and the less 20. PnOHLBM XHI, icluded To find the area of a circular ring, ( between the circumferences of two concentric circles. £(;lb I. Find the areas of each of the circles, and their difference wll be the area of the ring. ^^i "lutE II. Mnltiply the product nf the sum and difference uf the diameters hj '7S54, and the product "ill he the area of the ring. DEHONHTHiiTmH. The urea uf the circle ABF, diminished bj tfae ina of the circle DEG, is evideatly the are^ q[ t\i« in - - - ■ lielweeu theii' circuiiiicreuces, uhtch is Rule 1. KEMSnRATION OT BDITACKS. Id order to show «>e truth of Rule II., \et A.B=d, DEW', then Rule I., tho area of the ring ia d'x78S4— if'x'7BS*=(*— d") ^eSi=(d+d')x[d—dy7SSi, which ia Rule II. Example. What is the area of a ring incladed be- tnreen two concentric circles, the diameter of the greater beitig 12 inches, and that of the less 8 inches ? By Rule I. (12)' X 7854=1130976, area of the greater circle, and 8'x '7854^50'2656, area of the less circle; .-. 1 130976— 50 2656=62 GSS, area of the ring. Bj Rule II. (12+8)(12— 8)>C 7854=20x4x-7854 ■.62-832, the area of the ring as before. EXEBCISE9. 3. Required the area of the ring, the diameters of whose bounding circles are 5 and 4. Ans. 7'068fi. 2. The diameters of two coDcentrjc circles are 16 and 10; what is the area of the ring included between Iheir circumferences? Ans. 122-5221 3. What ia the area of the ring included between the circumferences of two concentric circles, whose diameters are 24 and 18? Ans. 197-9208. 4. If the diameters of two concentric circles be 15 bbiI 12, what ia the area of the space included between their circumferences? Ana. 63'61'ii Problem XIV. To find the area included between two arcs of eirelM Iiaving a common chord, when the chord and height oi each, of the segments are given. Note. If the segments be on the same side of the chord, the sp''^ included between the ares ia called a, loae. I , of each of the segments ACB, then their difference will be the the same aide of :hord, and their aum will be the area when the Kg- meols are on different sides oC Ihe chord. Rule. Find the aret ADB, by Problem XI., area required when the segments a Note. For the raetViod o^ (uiAin^&eTufims.^wnw* Note I,, Prob. VllL, tlio ni\e la ao tf Figures 1 and 2, that, it leiviirea ^^ "^ HaHSURATIOII OP SmtrACBS. 77 LK. Find the area included between the circular )g the common chord 12, the height of the one id the other 2. will be found, that the radius of the circle, the whose arc is 3, is 7*5. and that the radius of the 3, and that the aic of the first contains 106° 15' that of the second 73° 44' 24"; consequently b; SI., the area of the 6rat segment is -0087266 x 927289--48000O=447289 x (7 ■5)'= 25 160006; :ea of the second is -0087266 X 73 74 ='643499— :-163499x(10)*=zl63499. Hence if the arcs same side of the common chord, the area of the be the difference of these areas, or 8-8101; and ; on opposite sides, the jnclnded area will be the ese areas, or 41 -2099. EXERCISES. 1 the area of the space contained between two ircles having a common chord 36, the height of leing Q, and that of the other 6. Ans. 7929054, or 373-58856. 1 the area of the space contained between two cir- s having a common chord 30, the height of the [5, and that of the other 3. \aa. 41-7127651, or r I I 3 nearly the . straight Hn ea of a figure bounded by any curve md two other straight lines drawn : perpendicular to the > extremities of the c line. , Let the base divided into any limber of equal ' the perpend icu- 6 ; a', V ; a", b", :h meet the curve ints a, a', a", &c. sum of the first lerpendicularadd s sum of the remaining odd peipe'n<S.c\)\M*, voA.' ?s the sum of the even perpendVcvAana -, ftaa «>«&, / bf the third part of the common A.\ft\,auce o^ "Oa* liars, will be the area nearlj; aud, V\ie ■u.eas.e.T- "^^ Wz prapeadiculars are taken to one another, the more exaet wUl the approximation be. EsAiiPLE. Let Ba be 400 links, AB 110, ab 115, tt'6'119, a'^" 125, a"'h"' 128, a"h" 13], a'b'138, a-^'*J46, and PQ 145 ; what is the area of the figure ? Here the area of the figare will be {110+145 + ?(1I9+128 + 138) + 4(115 + 125+131 + 146)1 X 's" = 51550 links = 2 roods 2'48 poles. 1. If the base be 360, and it be divided into 6 eqnal parts, and the perpendiculars be in succession 20, 31, 43i 57, 08, 78, and 90 ; what ia the area of the figure ? Ans. 199W. 2. If the base be 196, and it be divided into 10 equal parts, and the perpendiculars be in succession 112, Jilt, 93, 80, 71, (JO, 52. 41, 30, 18, and 10 ; what is the area oftheiigure? Ans. ]208fif 3. If the base be 366, and it be divided into 6 eqnil parts, what is the area, the perpendiculars being bucmi- BiTcly 0, 20, 38, 50, (il, 70, and 78? Ana. imBl MENSURATION OF SOLIDS. In the mensuration of solids, the problem to be soW is, how often is a cube of a giren length of side containol in a solid whose rectilineal dimensions and form are ipwiil and the problem is solved in different vcaye, accoiding Ib the Tarioua forms of the bodies whose solidities areaoughb The number of times that a body contains a cubic indi, a cubic foot, or a cubic yard, &c., is called its solidi^. Table of Solid IMeasuke. 49li7g308SI)0a S874960a0 1DG4S000 /25435806105600o'iU7\Sn95'yiO(k4a\ll6m\i^'n68flOOl 613 MESSIRATION OF SOLIDS. i'J Problem I. To find the solidity of a priBm. Rule. Multiply the area of the base by the perpendi- cular height or length of the prism, and the product will be the solidity. DeuonstbatioK. If the BuperRcial asua of the boss be foDod, itod I height be then taken equal to the length of ihe measuring unit, iF will evidently conlain Jia many cubes of tbo required dimensiooB u Ihe base containa superficial units of thu same length of aide; and fiinoe the dimenHioDH of a prism &re tho same at all diatanoes from iW ends, the next uuit of length woald canlain the same number of eabes aa the fiiet, and so on throughout ; the number of nHmn'n^ eaita in llie wbalo will bo equal to the atiparfidal nnibin its eud mul' dplied into tbo Untar unite iu its length, whivh is the rule. Vote I. If the surface of u prism be required, it may he found by ilie following RcLE. To twice the area of its end, add the perimeter of the end multiplied into the height or length of the prism, and the sum iriil be its sur&ce. Note II. Tlie boat raethoii of giving boys a competent knowlcdgo of the mensuratjon of solids, is lo have the various iigui-as formeil of wood or pasteboard, which they may be mado to measure, and Ihenby leam both the theory aud practice at the same time. Example. Find the solidity and surface of a triangu- lar prism, the sides of the triangular base being 15, \% and 9 inches, and the length five feet? Here the area of the base =s/lQ%^Gx^= J'ifdi'd, (Prob. II-, Mens. Sur,)=54 inches =§ foot; which mul- tiplied by 5, gives I J of a foot for the solidity. AgaiD, to find the surface, we have twice the area of the md ^J foot, and the perimeter of the end is 3 feet, which multiplied by 5 feet, gives 15 feet for the surface of the «des, to which add the area of the two ends as found ahoTe, and we have the whole surface of the prism 15^ t^oaie feet mXEHCISES. I. "Wlint is the surface and solidity of a cuhe whose side 16 inches ? Ans. Surface lOj feet, and solidity 2-J-S cubic feet, 9. What is the surface and solidity of a cube whose side ia 6 feet? Ans. Each 216 feet. 3, What is the surface and solidity of a parallelopiped, *hose length ia 12 feet, breadth 2 feet, and ie^vV \. l«i^ 6 inches ? Ans. Saxface 90 feet, and ao^Sit^ ^ ieaV «aa y i » 10 Find the surfiice and solidity of a square ptism, the nde of its base being 1 foot 9 inches, and its length 16 feet? Ans. Surface 116^ feet, and solidity 49 cubic feet 5. Find the surface and solidity of a triangular prism, each side of the base being 2 feet, and its length 12} feetl Aub. Surface 7ii'4641, and the solidity 21-6506. t>. Find the surface and solidity of a pentagonal prisnii each aide of its base being 3j feet, and its length 15 feel 4 inches? Ans. Surface 310-485, and solidity 3231629 feet 7. Find the surface and solidity of a hexagonal prisnii each side of its base being J 1 inches, and its length 3 feet 9 inches ? Ans. Surfece 24>9912. and solidity 8-18664 feet 8. Find the surface and solidity of an octagonal priam, each side of the base being 6 inches, and the length H feet ? Ans. Surface 58'414'J, and solidity 16'89849 feet. PltOBLBM II. To find the surface and solidity of a cylinder. Rule I. Multiply the circumference of the base by titt; height, and the product Tvill be the convex snrfoce, tBt which add twice the area of the end, and the sum will ba the whole surfece. Rule II. Multiply the area of the base by the heigbfc. and the product will be the solidity. .,' Note. The demanstration of the rules are Uie Bams as thond Problem 1.; the area of the base U found by Problem IX, " ladon of Surfaces. Example. Find the surface and solidity of a cylinder ABDC, the diameter CD of its base being 16 inches, and its height AC 5 feet 10 inches. ^ince the diameter of the base is 10 inches, or 1^ feet, 31416xi(=41888is the circumference of the base, which being mul- tiplied by 5 feet 10 inches, the length, gives 2i-4346, the convex surface, whilst ■7854x (lj|)'x23 2-7925 feet, the area of the two ends; hence tttewbi' surface is 27'2271 feet. Again, the solidity is -7854 X V* X 5^=8-1448 cubic fe the solidity. I I. Find the snrfitce and solidity of a cylinder, the 3Sl meter of whose end is 3 feet, and its length 7i feet ¥ ■ ,. Surface 8\a^Sl ^eel, oa^i wVdit^ 53-01^ ITIOS OF SOLIDS. 81 3. Find the Bolidity of a cylindrical pillar of 5 feet dia- inet«r, its height being 30 feet, and find also its convex aur&ce. Ans. 58905 cubic feet, and its convex suifuee 471-24 feet. 3, The diameter of a circular well is 4 feet, and its depth 36 feet ; what did the digging of it cost, at 10a. 6d. per cnbic yard ^ Ans. L.8, 15b. lid. 4. If a cylindrical stone roller be 5 feet 3 inches long, and 1 foot 9 inches in diameter, nhat ta its solidity, and how often will it turn in rolling a field containing 4 acres, 3 roods, and 20poleB? Ans. Solidity 12'6278, and wUi turn 697993 times. Problem III. To find the surface and solidity of a pyramid. Rule I. Multiply the perimeter of the base by ka!f the ilatit height, to the product add the area of the base, and the anm will be the surface. RpLE II. Multiply the area of the base by one t!drd df the perpendicular height of the pyramid, and the pro- duct will be the solidity. DcMONHTKATiON. The B[dc9 of the pyramid are evident!)' triangles, ihc sum of whoBO bases is tlie perimeter of the base, and the coni- moii altitude of the triangles is the slant height at the pjruniid ; hence the Ham of all Uie bases, or the perimeter of the base of the ptnamid, multipliad by half the common aldtude, that is, /ui^Qie >^t height, wUl giro the smfaee of the eidea, to which add the area of the baee, and the smn will be tho nbole surface nf llic pyramiii, Kbieh ia Rule I. The second rule is demonstrated in Geometry, (prop. 1 03, cor. S-) Note L It is often neeasaary to find the slant height from the porpendicolar height, and the length of a side of tho base, the num- ber of sides in the base bemg also given. The slant height ia tlie hjpoteDDee of a right angled triangle, of whieh the perpendicular be^t. is one of (he sides about the light angle, and llie o^cr ia tlie perpendicular fmra the centre of the base upon the side; this side may be found by adding together tlio log. cotiui. -— (n being the lumber of Hides in the base) to the log. of Aa|/'the side; tlie sum di- mtiUHhed by 10 in the index will be Uie log, of the perjieiulicular; ^See Prob. VI. Mens. Sutf.) henoe it ia evident from Geometry [prop. 3S,) how the slant height can be found from the perpendicular beight being given, and conversely. KoTE ir. If (Area times the Bolidity of a pyramid be divided by the *rca of (be base, the quotient will be the perfendlcular altitude. Example. Find the surface and solidity of a square ppumid, the perpendicular height being 12 feet, aa4 ejw^w side of the base JO feet. I I I- 89 MKMITHATIOK OV Here, in order to find the perpen- dicular frgmthecentre upon the Bide cif the base, we divide 180" by 4, whioh gives 45°; then C Bis equal to cot.45°x^DE=4DE= 5;iiEainAC the Blant height ia =v'AB' + BC' = ^144+25 = v'l^=J3' h^^nce the slant surface is =40 X V = 20 X ]3=:2(iO: and the area of the base is evidently 10x10 = 100. The whole surface is therefore 260+ 100=360 feet, or 40 sq. yards. Again, the solidity is ^ the area of the base, 100x4, third of the perpendicular lieiglit, or 400 cubic feet. ' £SEBCISEB. 1. Required the surface and solidity of a square p mid, each side of whose base is 30, and the slant he 25. Ans. Surface 2400, and solidity 6 2. Required the surface and solidity of a hesag pyramid, each of the equal sides of its base being 40, the perpendicular height 60. Ans. Surface J2470765,and solidity 83138- 3. Find the surface and solidity of a trianguliirpyrai each side of the base being 20 inches, and the perpend lar height also 20 inches. Ans. Surface 797705 Inc and solidity 11547013 inches. 4. Find the surface and solidity of a pentagonal p mid, each side of the base being 2 feet, and the i height beiog 4 feet. Ans. Surface 268819 feet, and solidity 861 5. Find the solidity of an octagonal pyramid, «' perpendicular altitude is 15 feet, and each side of the 3 feet 3 inches. Ans. 2550 Pboblbm IV. To find the surface and solidity of a cone. Rule I. Multiply the circumference of the base by the slant height, to the product add the area of (he t and the sum will be the surface of the cone. Rule II. Multiply the area of the base by oite-thir the perpendicular altitude of the cone, and the pro will be the solidity. Deuonstrition. If tlie cone be couceiTed to be so oonBtitntec ilH surface may be cut by a. atraiglit line pBHHOg from the vertl the base, its EorCacQ lieiiig lVi:aU,^en off lutd eKteaded on • [ will be a aeotoi- of a circ\E,liie ateti.Ql'«^vitt'«a'\«V!iiuiHi^ v ■ HeiHtOBATIOK 0¥ BOUD^ 83 ■plying the length of its arc by half the ntdius of the circle; but its arc will be the cireamference of the base of the cone, ftnd the radiuB of the circle will be the slant height of the cone; to which, if tlie area of the base be added, the sum will be the surface of the cone. Hence Hula I. ia obvious. Kule II. ia dcmDDstrated in the treatise on Solid Geometry (Prop. 104.) Note L To find the slant height from the perpendicular height nnd the radios uf the base. To tbs square of the perpendicutor beight add the square of the radius of the base, and the square root cf the sum will tie the slant height. NoTB II. To Bud the perpendicular height from the slant height, mnd the radiaa of the base. From the squars of the slant height ■ubtract the square of the radius of the base, and the square root of the remainder will be the perpendicular height. Example. Find the surface and ^ solidity of a cone BCDA, the per- pendicular height AE being 24, and the diameter of the base BD 20, By Note I. A^=J{^W^<^ =26, and the circumference of the base wiU he 31416x20=62 832, ^ vhicb being multiplied by 13, half the slant height, gives the slant sur- q, &ce 816816. to (vbich add 31416 XlO» = 314-16, the area of the base, and the Bum 1130-976 will be the whole surface. Again, since the area ofthe base is 31416, if this be multiplied by 8, the third of the height, it giyes 2513*28 fot the solidity. XXERCISES. 1. Required the solidity of a cone, whose perpendicu- lar height is 10 feet, and the diameter of its base is 7 feet. Ans. 128-282. 2. Find the surface and solidity of a cone, whose slant height ia 25, and the radius of the base IS. Ana. Surface 1884-96, and the solidity 4712-4. 3. Find the slant surface and the solidity of a cone, ■whose perpendicular height is 5, and the diameter of whose base is 24. Ans. Slant surface ^90-0896, and the solidity 753-984. 4. Required the surface and solidity of a cone, its height heing 36, and the circumference of its base 94-248. Ans. Surface 2544-C57, and its sohdity 8482 32. Problem V. To find the surface and solidity of a frustum of a pyra- i; mid or cone. ' DEFjNiT/or.-. The irustom of a pyramid oi co-a^ Sa 'iaa i M part that remains after a part has been cut off by a plana parallel to its base. The part cut off will be a pyramid oi . cone similar to the origiual one. I Rule I. Multiply half the buu of the perimeters of tlie ^ jfeno.ends by the slant height, and the product will be th( surface of the sides; to this add the areas of the two endl, and the sum will be the whole surface of the frustum. Rule II. Add together Ihe square of a side of each enj, and the product of those sides, multiply the sum by thp tabular area belonging to the form of the base, and the product by ^ of the perpendicular height, the last product will be the solidity of the frustum. Note I. Tho tabnlar area belongiBg to the form of the bueviO . be gal a,t Problems VI. or VII. of Meusuraliuu of Surfaces. Note II. The solidity c&n also be eilcuLatod by tbe foUowiof Tulee: — 1st, To the ureu of each of the ends of tlic fruntura, uW a mean pmportioDal between them, then mukiptytfae Bum b^ ^oFlta height, and the product will be the solidity, ^d, Flod tbo solidity otillt whole pyramid or cone of which the EnistDiu is a, [lajt; tlteu (h^aillt of any line in the base, is to the difference between this cube ut the cube of the corresponding line in the top, as the soHdlty of ikl pyramid or cone to the solidity of the fVostum. DkNoKeiEauo^. Since each of the aides is a Irapoioid, ite anavil be fooad by multiplying half the sum of its piirallel aiflaa by A) perpendicular distance between them, and in each of th^ ud« thi perpendicular distance between tho pai'allel sides is the bshk^ lit. IheslanCheight of the frustum; also, since the sum of all the pMoM aides is evidentiy the aum of the perimelera of Ihe ends of the &»■ tum, the whole slant surface, or eurbco of the sides, is half the IHB of ^c perimeters, multiplied by the slant beigbt, lu which if A> areas of the two ends be added, the sum will be the BUlbm «Cl^ frnatUDi, which is Rule I. Let ABCD— P be a pyramid, nnd AO ft frustum of it conloiacd l>etweeQ the pbuies ABCD and KFGH, and Ice EFGH— P be the pyramid cut off by tha plane EF6H; it is evident tliaC the frus. tum AG ia the difference between tho wiiola pyramid ABCD — P, and liio pyra- mid cut off EFGH— P. Let now AB„fi and £F=&', also let the altitude PM be represented by a, then since the pyraiuids are similar, AB : EF : : PM : PL or 6 ; 6' ; : o -. PI,= ^iMiiKp 1 of the Iw , cor. S) ll* solidity of the pyramid ABCD — P a^pl^, and the pynouJ EFGH— Pis -pi^ — cs— x|, .'. the ftuBtum AG b^*'- fi—j _ — (6*— 6«). How \Xl 4«i iMaula t^ thfl Erattora ii •»- mi , BubstltDllng Cbis ID tile expresaion hand above for QiB aolidity of the frurtum, it beeorars p- ( , ) = p- (l^ + bb' f) = (4'+M'+6*) p- , Whicb is Rule II. The ralea given in tlio noKa ore easil]' derirabte from the above expression, or from the theorem that Gimilar Bolida ore ta one another Jis the cubes of Ihoir like parts. I D order to make the proof applicable to a cone, consider b and b' the radii Ot the bascB, and put 3'14I6 for;i. ExAsiFi.E. Find the surface and solidity of the frustum of a cone, the radius of the greater baee being 8 incheB, and that of the less 3 inchea, and the altitude of the fruB- tum 12 inchea. Since the altitude DG— 12 and GB=5 DB^ v'1^4 + 25= V'fi9 =13, and the sum of the circum- ferenccB of the bases =(AB+CD) X3-I416 = 22x31416=69 1152, which being multiplied by half the slant height, 6^, gives the slant sur- face 4492488 iachee; also the sum , of the areas of the circular ends is =(8'+3"}x 31416=73x3 1416 =229-3368, to which add 4492488, and we have the whole surface, 678-58o6 square inches. Again, the solidity i»= (8'+8x3+3^) x314I6x V ^97 X 3J416 x 4= 1218'9408 cubic inches. EXEItCISKB. 1. Find the surface and solidity of the frustum of a tri- angular pyramid, each side of the greater base being 15, and of the less 9, and the perpendicular heiglit of the frus- tum 12. An». Sor. 568 978794, and sohdity 763 834932. 2. Find the surface and solidity of the frnstum of a S({iiare pyramid, each side of the greater base being S feet, and of the leas 2, and the perpendicular height 12 feet. Ans. Surface 1983073, and solidity 156 feet 3. Each side of the greater end of a piece of squared timber is 28 inches, each side of the less end 14 inches, uid its length 18 feet 9 iuches; how many solid feet does it contain? Ans. 59'5486I1. 4. Required the surface and solidity of the frustum of a, bexagonal pyramid, the side of its greater end beio^ 4. feet, and the aide of its less 2 feet, and the "eex^ftiiitfsiiias 86 MSSBUSATIOV OF 80UDB. heJgtt of the fnisfum 9 feet. Ans. Surface 225-547-22' and the solidity 2I8'238384. 5. Fiod tbe surface and solidily of a frustum of a cone, the diameter of the greater end beinp; 5 feet, that of the less 3 feet, and the perpendicular height 4 feet. Ans. Surface 78'5]62. and the soUdity 5I'312a fe.t 6. What is the surface and solidity of the frustum of n cone, the di.imeter of the greater end being 20, that of tbe less end 10, and the slant height of the frustum 13 ! I Ans. Surface 1005-312, and the solidity 219912- ■ PSOBLEU YI. To find the solidity of a wedge. ItuLE. Add twice the length of the base to the length of the edge; then multiply this sum by the height of tbe wedge, and again by the breadth of the base, and onB'siilli of the last product will be the solidity. Demonstration. Wlien the length of the base BC is equal to that of the Ejp — |F edge EF, the wedge U eridently equal /)'. il to bair s prism of the biudo base and / i\ :\ altitude. / l\ M And (iMording as the edge is shorter A / J i T)! \ or longer than the baao, Ihe wadge ia V' f V" " ''■ 1 greater or leas than hnlf a prirnn of VJ— -tr ■. \ the Bame height and breadth with the B^ ^C wedge, and length equal to that of the edge, bjr a pyramid of the same height and breadth at the base *1Hi and the length of whose base ia equal to the difference of the lengUn of the edge and base of the wedge; putting now BC=L, AB=1. EF=/, and EG=y5, we will have t6/A^ziiAx(=tL=}=0=Jfta+JM (L— ')=iM(3/+2L— 20 = jM{2L+0, which ia the rule, Coa. If t=L, the rule will become ;fi/;(3L)=lML=i a prism o( Ihe same tiaae nnd height, as it evidently ought. Note. The surface of a wedge may be found bj the nilea for iho mensuration of Hurfnces, hy cal culating sepamlcly the arena of (h« bnae, sides, and ends, and addmg iheix surfaces logether for tfai whole surface. Example. How many solid inches are in a wedge, the length of whose base is 15 inches, its breadth 8 inchet. length of the edge 12 inches, and the perpendicular height 16 inches? Here twice tbe length of the base, (30 in.), added to the length of the edge, (12 in.), is 42, which multiply by the breadih of the base, (R in.), and by tbe perpendicular height, (16 in.), and we bave 5373 inches, which being divided by 6, gives 896 solid inches for tbe solidity. Or (15 X 2+ 12) X 8 X \6-^6=8Sft feii\\i.\wcVvw. HElel MMBORATies or ■ouiM. e length and breadth of the liase of a vreAge are SS and IB inches, and the length of the edge is 53 inches; ■what is the solidily, supposing the perpendicular height to be I7-I45inches? Ans. 3>]006 solid feet. 2. Required the solidity of n tredge, the length and breadth of the base being 9 and 4 iucliea, and the length of ihe edge 1 1 inches, and the perpendicular height 10 inches? Ans. 193^ cubic inches. 3. R'-quired the solidity of a wedge, the length and breadih of the base being 26 iind 18 inches, the length of the edge la inches, and the perpendicular height 28 inches? Ans. 3 feet 444 inch ei. | ^^B Problem ^^^H ^^^KBnd the solidity of a prismoid. ^^^^^M Bulb. To the sum of the areas of the two ends aM I fbnr times the area of a section, parallel to and equally distant from both ends ; multiply this sum by the per- pendicular height, and ^ of the product %Till be the so- lidity. (fasr For the ■ NoiE 2, For the demonstration of IUg aliove rule, aee ths Key. ExAMPLR. What is the solid content of a prismoid, whose greater end measures 12 inches by 8, and the less end 8 inches by 6, and the length or height 60 inches? Here 12x8+8x6=96-|-4a^I44= the sum of the areas of the two ends. Also ^-~ X ^- X 4 =10 X 7 X 4 = 280 = four times the area of a section, parallel to and equally distant from both ends. Therefore C144+2B0)x V=''24x 10=4240 is the solidity in solid inches, which is equal to 24537 cubic feet. 1. What is the solidity of a rectangular prismoid, the length and breadth of one end being 14 and 12 inches, and the corresponding sides of the other end 6 and 4 inches, and the perpendicular 30 feet? Ans. 17^ solid feet. 2. The length and breadth of a stone pillar at the greater end are 28 and 18 inches, and the leng\.H aai \«eaA'&i «^ the less end 16 and 10 inches, the perpendicular height being 9 feet ; ivhat ia the solidity of the pillar? Ana. 19-75 solid feet 3. How many solid feet of timber are contained in a rectangular beam, the length and breadth of the greater end being H'i and 20 inches, the length and breadth of the leBH end 16 and 10 inchea, and the perpendicular length 24 feet? Ans. 62| solid feet. 4. What ia the capacity of a coal waggon, the inside diniensionB of which are as follow: at the top the length DO, and breadth 56 inches; at the bottom the length 42, and the breadth 30 inches; and the perpendicular depth 48inchea? Ans. 7^^^ solid feet I Problem Till, I To find the curre surface of a sphere, or any segment or Kone of it. Rule. Multiply the circumference of the sphere by tb* height of the part required, and the product will be the curve surface, whether it be a segment, a zone, a hemi- Bphere, or the whole sphere. | I NoTK I. The height of the vhole sphere is its diameter j bout the whale surface of a sphere is the Bquore of the diameter mslli- ' plied into 3-14 1 6, which ia foiir times the area of a great cink tf the Bphere, or of a ciide having the Buae diameter as the sphsn. ^ Note 11. The eurfoce of a hemisphere is twice the area of ( [ circle, having the same radios as the hemisphere, j Note III. The deiaonstration of the above rule will be givoi i the Key. Ekahpls. Find the curve surface of a segment of ■ sphere, the height of the segment being 5 inches, and th« diameter of the sphere 12 inches ? Here 31416x12=37-6992, the circumference of the ■ sphere, which being multiplied by 5, the height of the palt . gives 188*496 square inches, the surface required. I 1. If the mean diameter of the earth he 7912 mile^l and the obliquity of the ecliptic 23° 28', find the surface ofJ the torrid zone, which therefore estends 23* 28', on es^ ' side of the equator? Ans. 78314300 square mita 2. The diameter and obliquity of the ecliptic being sbj*- posed the same as in the lust exercise, find the surface m the frigid zone! k.na. ftV^^AW «c>ywe inil»] MBReTTRATTOI* OT SOUDS. 89 he diameter and obliquity being the same as abore. 5 surface of the temperate zone 1 Ans, 51041500 square miles. 4. If the earth he supposed a perfect sphere, vrhnse dia- meter is "JQIS mileSj how many square miles are in its whole surface ? Ana." 196663000 square miles. FnoBLEM IX. To find the solidity of a sphere or globe. RtTLK I. Multiply the surface by ^ of the radioB, or by J of the diameter, aad the product will be the solidity. Rdlb II. Multiply the cube of the diameter by -5236, and the product will be the solidity. Note. The demonstratian of tbeae RuleB will be gJTcn in the Tiey. ExAMPLS. If the diameter of the moon be 2180 miles, and it be a perfect sphere, how many cubic miles of mat- ter does it contain? Here (Log. 2180)x3=3'336456!<3 =10015368 and Log. -5236 = 1-719000 Therefore the solidity is 5424600000 = 9-7U36S NoTB. Thia result ie given only to that degree of eisctnees that can bo obtained by using tbe Tables pven in tliia work; if the di«- taeter be uabed, and the cube be multiplied b; '5236, we will find the ■oliditr to be £4S46174T£-2 mUea. Tbia rceult differs from the last by less Iban a. three hundred thoosandtb part of the whole. The following answers are given as thc^ oan be derived by logarithms, ho also were the answers in the laM problem. BXBBCiaES. 1. How many cubic feet are in a globe, whose diameter is 3 feet? Ans. 14-1372 cubic feet. 9. How many cubic inches are in a ball of four inches diameter? Ans, 33-5104 cubic inches. 3. How many cubic miles are in the earth, its diameter being 7012 miles? Ans. 259333000000 miles. 4, How many cubic inches are in a globe, whose dia- ittetec is 16 inches ? Ans. 214466 inches. I Pboblbh X. LTo find the solidity of a segment of a sphere. RniiB I. To three times the square of the ladliw o^ \\» tt add the square of its height ; and thia aam m\^\A^\«& M l by the lieinht, and tlie product agaia by -5236, wlU give the aolidity. Rule IT. From three timps the diameter of the sphere mbtract twice the height of the segment ; multiply the re- mainder by the square of the height and by '5236. Note. For the demouslnitian of the abo\-e Rulea Bee the K»J. ExAMPLR. The radius A7t of r> tbe base of a segment CAB of a 'sphere is 5 iDches, and tts height Cn 4 iaches, nhat la the solidity o( j^i the segment ? By Rule I. 5'x3+4"=01x4 \ ■ =3(i4x-5236=l90-5904, the soli- \ dity. '-.^ _,y By Rule II. The diameter of the ' circle is (Geo. prop. 68) '^-^ = V = ^^h which beii^ multiplied by 3, and tntce the height subtracted, giTM 22| ; this again being multiplied by IG, (the si^uare of Ike height), gires 364. which again being multiplied by 5236. giTcB, as before, 190-5904 for the solidity. EXBRCIBBS. m 1. What is the solidity of the segment of a sfheKi I whose diameter is 20 and its height 9^ Ans. ndlWil r 2. What is the solidity of the segment of a sphere, tiw radios of its base being 10 and its height 9 ? Ans. 1790-4344, 3. What is the solidity of a spherical segment, the n- dius of whose base is 16 and its height 8? Ana. 3485-08IB- 4. What is the solid content of a spherical segment, tb( diameter of the sphere being 30, and the height of the wg- lent 24? Ans. 126669312. Pkoblbm XI. I > To find the solidity of a frustum, or zone of a sphere. RuLB. To three times the sum of the squares of ibe radii of the ends, add the square of the height of the zone ; multiply the sum by the height, and the product again by -5236, and the product will be the solidity. NoTK I. If the zone be in the middle nf the sphere, Ihenw ids will be equal, aril the solidity mnj' bo found by tlie M \le: — To six limes the B<\uR,re of Ihe rndiUB of Ihe eod sdi! foare of the height ot t^iicVuew ot Aib -inoe-, niuUk^ly the mud tf I MflmmATioir of solidb. Ihe beighl of the ione, and that prodnut by '5236, wl I the solidity. NoTB II. For the deroonstratioQ of those Ruira see the Key, , ExAMFLB, "What is the solid content of the zone ABCD, whose greater diameter, DC, is 24 inchps, ] the less diameter, AB, 20, and Ihe height mn. 16 inches? Here Dn=12, Am=10, and mn^ie. which being multiplied by 1 6, gives 15456; thia again being multiplied by '5236, giTes the solidity 8092-761 6. EXBRCISBS. 1. What is the solidity of the zone of a sphere, the dia- meter of whose greater end is 4 feet, the diameter of the less end 3 feet, and the height 2^ feet ? Ana. 32-725 solid feet. 2. What is the solid content of a zone, whose greater radins is 24 inches, the less radius 20 inches, and the dis- tance of the ends 8 inches ? Ans. I2532'88»6 solid inches. 3. What is the solid content of the middle zone of a sphere, whose top and bottom diameters are each 6 feet, and the height of the zone 4 feet ? Ans. 146-608 solid feet. 4. Required the solidity of the middle zone of a sphere, tlte diameter of each of the ends being 8 inches, and the hdriLt of the zone also 8 inchest Ans. 670-208 solid inches. Problbu XII- To find the solidity of a circular spindle. RpLB. Find the area of the revolring segment ACBE, which multiply by half the central distance OE. Subtract the product from i, of the cube of A13, half the length of the spindle; and multiply the remainder by ]2'56Q4, and the product will be the solidity. Note I. The Burfaee of a circular spiodla may be found aa fol- lows. From the product of the chord of the arc and radius of the ■arcle, subtract the [iroduct of tbo length of Ihc arc multiplied into the oentrol distance 0£, and the renmindor multiplied by 6-2S32 *ai be the surface. Note II- For the demonstratian of theee KoV^, vx "Odr Y^v.'j. 03 ME!tSOB4T»(WI OP 9m,W»i : 44728800 intral distance OE Example. The length AB of a drculnr spindle ADBC is 16, and the middle diameter CD 8 ; what A is tlie solidity and surface of the Here — — = — - — i= Y := 20, the diameter CF ; therefore the ra- dius CO=]a And AE^AO=8 -h10=-800000, the natural sine of F the angle AOC. Hence the angle AOCs^SS" 7' 48". anJ therefore the whole angle AOB=106'' 15' 36"=106-26''. 10&2G''x '0087266 = -9272885 i natural sin. KM)" 15' 36" = -4800005 difference of do. = -447288 multipljing by R'=100 area of Che segment This being multiplied by half the which is 3, gives 134-1864, this product being subtrscteo from |of the cube of AE=170'6666, gires 36'4802, whiA , being multiplied by 12-5664, gives ^8-42478, the soli- i dity. Ag^n, to find the surface, we have 16 X 10—2(106^ X -0087266 X 10x6) = 160— 1 1 1-27462= 48-72538 X 6'2832=306-1513. EXKBCIBE6. 1. What is the solidity and superficial content of a cif' nilar spindle, whose length is 48, and middle diameter 3B ^iaches. . (Solidity 17-31299 feet ■^"^- (Surface 327054 feet, 2. Find the solidity of a circular spindle, whose length IB 15 inches, and middle diameter 8 inches ? Aqb. 433-40932 inokei. ' 13. Required the solidity and surface of a circular epA-' He, whose length is 20, and its greatest diameter 15. i I ,„ 7 Solidity 2164m -*"^t Surface 817-636* Fboblgm xni. To find the solid content of the middle frustum or zoiK , of a circular spindle. RuLB. From the sijuare of half the length of the »?»• ■die, take ^ of ibe G^^u&ie o^ ^^\i 'Cat^\bQ'^\\x aC the luiMl), HXKSORATION OF SOLID S- 93 1 :tliofthe ' zone, miJtiply the remainder by the Baid half lengt zone ; from the product subtruct tbe product of ihe gene- rating aiea und central distance ; ihun the remainder mul- tiplied by 6-2832 trill be the content of the middle zone-. ExAMPLB. What is the eo- lidity of the middle frustum ABCD, of a circular spindle, ivhose middle diameter t/ih is 18, the diameter AD or BC of the end is 8, and its length 07-20? X jj» Join DC and Dn; produce \]. nm to 0, the centre of the arc EDCF, and join EO ; then DC is eTidently =or=20, and TM is =^^(m» — AD)=5. Again, Dm'^sDi'-J-jm*^ 100+25=125, and DK«-§-2rMi= the radius of the circle =125-^10^12-5, hence, since nv=:9, the remaining part of the radius or central distance ^=3*5, and therefore the square of Ei=(12-5)''— (3-5)2=16 j(9=144, Euisthere- &ne=12. Now, the generating area is the area of the segment CDn, together with the area of the paralleloOTam Dr; but the area of the segment CDm (Mens. Surf, Prob. 11,) = 69-88875, and the area of the parallelogram D*- is 20x4 =80. Hence the generating area =69-88875+80 = 149-88875. Now, by the rule 12 "-4(10) '=144— 33^ =110-6666, this multiplied by 10, half the length of the zoae, gives 1106'666; from this But)tract the generating area, multiplied into the central distance 3*5^524 6106, and there remains 582-053, which being multiplied by 6-2832, gires the solidity of the zone 3657-174+&C. 1. If a cask in the form of the middle frustum of a cir- cular spindle have its bead diameter 24 inches, bung dia- meter 32 inches, and length 40 inches, how many cubic inches does it contain? Ans. 27287-90178. 2, What is the solidity of the middle zone of a circular n'.le, whose length is 25, greatest diameter 20, and least eter 15 inches ? Ans. 385537 ft-et. Problem XIV- To 6nd the solidity and surface of a circular ring. Btn,B I. To the inner diameter add the thickness of the ring, multiply the sum by the square of Ike tbicVaftw, aa.\ 94 lassuKATttm ta souoe. that product again by 24674, the last product will be the solidity of the ring. RuLR II. To the inner diameter add the thickneSB of the ring, multiply the sum by the thickness of the ring, and the product again by 0-8696, the last product will be the surface of the ring. Note. The demoDstration of (he aboTo Rules will be found iii llw Key. Example. What is the solidity and surface of a circular ring, whose thickness AB or CD =4, and the inner diameter BC 8 inches? Here the inner diaraeter + the ^i thickness is 8+4=12. which being multiplied by the square of the thickness 16, and that product by 2-4674, gi»es the solidity 473-7408. Again, for the surface we have 8+4=12x4=48x 9-86i«3=473-7-108, so that in this example the surface and solidity are both expressed by the same number, but this can only be the case when the square of the thicfcnea is equal to the thickness, or when the thickness of the riijg is 4, as in the present example. 1. What ia the solidity and surface of a cylindrical ring, whose inner diameter is 14, and thickness 6 inches? .^^ J Solidity 1776 528 inchefc ■^''^- (Surface 11 84-352 indiM. 2. Reqnired the solid and superficial content of a cylin- drical ring, »hose thickness is 9, and whose inner diame- ter is 31 inches. , J Solidity 7994-376 inche*. ■^''^■\Superiicies 3553-056 inche*. 3. What is the solidity and Euperficial content of a tj- lindrical ring, whose inner diameter is 15 inches, aid whose thickness ia 5 inches ? , /Solidity 1233-7 incheB. P ADS. ^ s^pg^jjigg ggg^g j^y. Problem XV. To find the sorface and solidity of each of the r^nUr bodies. Dbp. a regular body is a solid bounded by plane free that are similar surfaces. The whole number of regular bodies that can be foroicJ in tbiB masner U &ie. MCMSUBATIOll OF BOLID8. 96 ] . The Tetraedron, or regalat pyramid, viid lia« fonr equal and limilar triangtUar faces. % Tbe Hexaedron or cube, irhich has six equal square hcea. 3. Tbe Ootaedrou, which has eight equal triangular faces. 4. The Dodecaedroo, which has fire equal jfenlagoiial 5. The Icosaedron, which has twenty equal and similar triaagnlat faces. If the following figures be drawn on |iastehoard, and the lines on the first and third row be cut half througb, and then folded together, they will form the five regular solids lepreseiited by the figure» in the second and fourth lowi. A Q ^ The following is a table containing tbe eurfaces and scdidides of each of the fiye regular bodies when the Uu%ti^ <tf id edge ia one. toKtoBJtxwK- OB aauaa. S0I.IIHTIE8 aSD Sl-RI'ACHS OF HEUIrt..K BoOtES. 1 1^^^ K.,™. Solidiiiei. Snrf.™. 6 12 20 HeiBBdrulL Octaedrob. Dodecttedron. ■117861 l-OOOOOO ■471+05 7-668119 2-181695 1-732051 6-000000 3-464103 20-645739 8-660254 Bole. Since Himilar surfaces are to one another as squarcB of their like sides, and since similar solids an one another as the cubes of their like sides, the saifaoe the table above, multiplied by the square of the lengi the edge of anj aimilar figure, will give its surface; the solidity in the above table, multiplied by Hie cub the length of the edge, will give the solidity. ExAMPLs. Find the surface and solidity of a regi octaedron, whose linear edge is 5 ? The tabular surface of an octaedron is 3-464102, wli being multiplied by the square of 5 or 2o, gives 86'609 the surface required. Again, the tabular solidity of an octaedron is '4714 which being multiplied by the cube of 5, or by 125, gi 58-925625, the solidity required. I!XEBCISEB. 1. What is the surface and solidity of a regular tetE dron, the length of its edge being 6 inches? , „ f Solidity. 25-455816 indi ■*°^\ Surface. 62-353836 inch . What is the surface and solid content of a kexi dron, the length of whose edge is 4 inches ? . „ J Solidity, 64 inch ■*''^\Surfiice, 98iDch 3. What is the superficial and solid content of a d« caedron, whose linear edge is 3 feet ? , (Surface, 185-811561 fs ■*"^- \ Solidity, 206-904213 fs 4. What is the solid content and superficies of a regu] jctaedron, whose linear edge is 10 inches ? .^, (Solidity, 471-405 inche) ^"^^ i Superficies, 346-4102 inchi 5. What is the surface and solidity of a regular icosi dron, whose linear edge is 8 inches? . f Surface, 554-256256 ioehi ■^^- V^Vi4A^A\\TQ2ria4 iachi UiftD 8tniT£7iira: LAND SURVEYING. Land Surreying is the art of fiading the extent of area of a field or an estate. Land is commonly measured hy a chain, invented bj Edmund Gunter, and hence called Gunter's chain. It is 66 feet, 22 yards, or 4 poles in length, and is dirided into 100 equal links, so that each link is 7-92 inches. An acre of land is one chain in breadth and ten chains in , length, or ten square chains^ and therefore contains 100,000 sqoare links. An acre of land is also 4840 square yards, and is divided into 4 roods, each rood is divided into 40 poles or perches, and each perch into 30| yards. The chain must always be accompanied by fen arrows, to the top of which is generally attached a small piece of red cloth, to make them visible ; and the surveyor will often find it necessary to have a few poles, with smaU flags attached, called station staves, to place at comers, or other particular points in the field or estate that he may be sur- veying, to render these points more visible from a distance. The surveyor will also find it very convenient to have a staff or rod, divided into 10 eqiial parts, corresponding to 10 links of the chain, with which to measure off-sets, or short distances, between the chain line and some bend or coiner of the field. A cross staff, which consists of two set of sights at right angles to each other, placed on the top of a Staff, to assist in finding the point on the main line, from which a per- pendicular to any bend or comer ought to be raised. The surveyor who has the above instruments is prepared to take such dimensions of an estate as will both enable him to lay down a correct plan of it on paper, and also to find its area, which are the two principal things required of him. FnoBLBu I. To measure a field of three sides. At each comer of the field place a station-staff; then if the sides be all straight lines, measure thcra, and the area j can be found hy(Mens.ofSurf.,Prob.n.); or, by the as- siitance of the cross-staff; find where a petpenditwlM. from one of the corners would fall on one o^ \,\ievL&^,v^>i^ 98 TUAim itvHvmttxm. measure that side, and the perpendicular upon it, from the opposite comer, and the area can still be found by the same problem, and a plan of the field can be drann Irom either dimensions. Note 1, Tbc dimeDBionB of the Egur« cau be laken in both v*js, Hnd if the area calculnted fnitii each be the same, the dimeaBtona have been carrectly laken ; if not, it should be measured again, t find where the mistake has been committed. NoTB U. Id order to measure a line, the assistant takes nid arrows in his left hand, and an end of the cbsin and one arrow i his right; (hen advandug In the place where he is directed, the fol- lower makes agaa to him wilh his hand, tUl he is exaetljr in a list witli the BtaUon to which he is advancing, and at the end of tbc chain he sticks into (he earlh the arrow which lie holds in hia ri^l hnnd; they then advance, the leader taking another arrow into his right hand, until the follower cume to the Hi^ arnin-, which he taka up with his chain Imnd, the leader at the samo time putting down second Brraw with his right hand ; they thus proceed till all II arrows are in the hauds of the follower. Tliis tslies place at the md uf the eleventh chain. The follower then advaucea to tlie leader, and counting the arro»s, delivers them again to the leader, places one down at the end of the cham, and then advances as be- fore, the follower at the samo time marking in his field book the n- etuDge tlial has token place, leal, if the line be long, an error of la Example. Required the lar field, ivhose three sides ar Here if we lay down the plan on a scale of 10 chains to the inch, and then on the plan measure the perpendi- cular on the side which is 1 4 chains, it will be found to be eiactly 12 chains; hence the area of the li eld Is !■«» X 600=840000 square links, which being reduced to acres, gives 8 acres, 1 rood, 24 poles. The same result will be ob- tained by finding the area fro ind plan of a trianga- 13, 14, and 15 chains. I (he three sides. EXERCISES. 1. What is the area of a triangular piece of land, iH* three aides of which are 720. 609, and 1044 links? Ans. 2acres,Oro.,2I4784r' 2. What is the area of n field in the form of an equil.i- teral triangle, each of its sides being 925 links? Aaa. ^ ■AM^s.'i TO.. la-iasa perch* -LAND SURVETII^Q. "96^ 3. What is the area of a triangular field, one of its sides heing 546 links, and the perpendicular upon it from the opposite angle 432 links, the perpendicular falling at the distance of 230 links^ from one extremity of the base, and 316 links from the other; required also a plan of the field ? Ans. 1 acre> ro,^ 28*6976 perches. Note I. When the field is triangular, hut not hounded hy straight lines. Fix, as hefore, station-staves in the three corners of the field; measore the three lines hetween these three comers, and the perpendicular distance of the hends in the sides of the field from these lines, at the same time carefully marking the point of the main line on which the perpendicular falls; the field will then consist of a triangle, and on its sides several small triangles and trapezoids, the areas of which must be added to the area of the principal triangle, to get the area of the whole field; but if the boundary of the field be on the outside of the main lines, such parts as may lie within the main line must be deducted. Note II. In measuring such a field, it is most convenient touse a field-book, consisting of three columns; in the middle column are marked the distances along the main line to any comer from which a perpendicular is to be measured, or to where the boundary of the fidd leaves the main line; then the perpendiculars from the various comers or bends in the boundary are to be marked in the other eoiunms, observing always to place the perpendicular in the right- hand column when it is measured to the right of the main line, and in the left-hand column when it is measured to the left of the main line. NoTB III. Begin at the bottom of the page and mark upwards, stating at the beginning what side of the field is first measured, and m what direction you proceed, and at each comer state whether you turn to the left or right. Mark also where the main line crosses any hedge, ditch^ or river, &c. The above directions being attended tu, it will be easy to lay down a plan of any field from Uie field-book, and also to calculate its contents, but it is sometimes convenient to draw a rough sketch of the field in the field-book, and mark .the va- rious lengths on this plan^ both of the main lines, and also of the perpendiculars upon them, from the various bends or comers in the boundary of the field. Example. Lay down the outlines, and find the area of a field, from the following field-book ? ^^B* AC ^^^H ■• 954 840 ^^^^H ^^V 154 560 ^^^^H ^^^ 208 ^^^^H ^^^^^m 000 RoffC 1 BC ^^^^^^t 694 ^^^^^^ 436 ^^^^1 ^^^^^t 000 RoffB ^ AB ^^V 785 ^^H Begin at A and go west. ^^B perp. on the left base line. perp. on the right. ^^H Area of the triangle ABC, n ^^V ftund from its three sides. /N-^ 1 ■ 78i //N?\ 1 ^B mi f 'i \N ^H \A H 2|24:« / \\ ^H tog. 1216o=3-085112 / \\ , ^H „ 431-5=:2-634981 ^ '^ ^H „ 522-5=2 718086 Offset A on BC. H ,. 262-5=2-419]29 347x69= 23943. ^M 2| f0857308 Offsets on AC. ^■Xce. 268320 =5-428654 ;0D „^ ^^W^^ — x74 = J6S)li H ^';^"X352_ 40128 H. ^^^;^^X280_ 29400 ■ ^H 56x57= »Id9 ^^^^ 1043S9 ^^^^^B AreaofAABC = 268330 ^^^^^V AVLole 372679= i ^^^^^^^^^^ 3 acres, 2 roods, 36-S8S«fM lAWB SUBVBTtXfl. ^ ESRRCISRS. Draw a plan of a field, and find its area, froit the ing field-book. CA 1252 37 lUOO 69 824 45 716 72 610 15 424 ^H < 55 212 ^H 000 EoffC 1 BO ■ 683 •* 636 40 ^H • 354 64 -* ^H ' 229 49 ; ^H 000 ^ KolTB ■ AB ■ 973 II 48 745 ^H 76 600 ^H 56 495 25 256 ^H 000 Begin at A, and go north-east. Ana. 3 acres, 3 ro., ]9'0648 perchea. Construct the field-book, draw a plan, and find the Df afield, from the following description: beginning at )Uth comer of a triangular field, where the station was : boundary, and proceeding along the west side lo- 9 the north, at the distance of )50 links, the ofTset on oandary to the left was 46 links ; at 210 links farther the perpendicular on the boundary towards the left tl links ; 1 60 links farther on the ol^act on the same iraa 34 links ; and the north-west corner also in the lary was 154 links farther on. Then turning to the , the main line continued in the boundary to the dis- of 209 links, and at a distance of 120 links farther in inset to the vertex of a triangulat \«ni Va W"ff iarf iras 48 links; 140 links fatlhet oiv V\it \a i 103 •LAtm euRTETniGi again coincided with the boundary, and continued 30 to the end, which was 113 liiilcs farther on. Turning again to the right, the boundary of the field lying towards the left of the main line, throughout its whole length, at 100 links, the offset was 20 links, at 200, 34 links, at 300, 50 links. at 400, 65 links, atiiOn, 30 links, at 600. 12 links, and _ the whole distance Co the first station was G64 links. ■ Ans. 2 acres, ro., 32-47 poles. PfiOBLBM II. To measure a field of four sides. KuLK. Measure a diitgonal and the perpend iculnn upon it from the other corners. Or, measure a diagonal and the font aides, then a plan of the field t ' stcucted, and the perpendiculars measured on the plan, or the area of each of the triangles may be found from the three aides. Note. Id nrdpr 1« prove tlie carrectnesa of the dimenioiii, mcBAurc the other diagoiukl; tben if its length on the plao corre^HHid with its length as measured on the field, the dimensions have been token correctly; if not, there baB been an error cooiniitCed, which most be dieeovered liefore proceeding fartlier. EXEItClBRB. I. What is the area of the field represented in the fol- lowing diagram, and whose dimensicws are as stated be* AE 238 BE 229 B AF 496 DF 388 AC 670 Ans. 2 acres, ro., 10-7I2 poles. 3. Bequired the area of a field, the di- mensions of which ta- ken as in the abore are AE 460, AF 742, and AC 984; the per- pendiculars being BE :W9, and PD 361 links, and a proof diagonal BD 72? links. Ans. 3 acres. 1 ro., 7-424 perchw 3. Required the area of a field, the dimensions of which are as follows, (the point F being in the diagtam nearer to A than the point E), AF 280, AE 594, AC JO20, and the pcrpendicuIaiB being DF 460, and BE 340. I.AND 8DRTETIN0. 103 ■ARK. When the four sides are not straight lines, it is neces- go round the houndary, measuring the four sides, and all the 1 and insets, from the angles of the boundary on the main lines; neasure a diagonal, whic& will enable you to draw a plan of the on which the perpendiculars can be measured from Ihe comers ) diagonal, and the area of the two main triangles calculated, idi add the spaces indicated by the offsetSi and deduct those ted by the insets, and the result will be the area of the figure. Draw a plan, and find the area of a field, from the wing field-book ; A being at the south-east comer, B e south-west comer, C at the north-west comer, and the north-east comer, and the four stations being all e boundary of the field. - BD 1420 Dia((onaI 1000 proof line. am toB AC 1500 Diagonal. 1000 • - Bo£PA DA 1200 1000 750 580 94 318 60 000 RoffD CD 838 144 490 86 300 000 RoffC BO 1156 40 1000 60 735 74 530 340 RoffB 1 IiAIIB SBSVXTfira.' 14 Begin ut AB 922 806 24 710 36 540 Cross 4S0 350 17t> 000 A, and go west. Ans. 10 acres, 3 ro., 36-4096 perehcJ. ^^H BuLS. Bj means of diagonals, divide the field into triangles and trapeziums, and find the areas of eacb of these figures separately, and add the results togcthei fui the whole area. Note. A five-sided tigure can be divided Into s trapezium and ) ttioDgle ; a Bix-sided figure cau be divided into Iwo traj>eziuma ; > seveu-aidad Rguro can bo divided intu two trspeziuniB and a triuglel and gencniU;, if r be the number of udes in a figure, it can be ^- vided into j — 1 trapeziums; and it the i^uotient contain a iaH, il representa a triangle, EXBRCIStS. 1. Required the plan and area of a field from llit following dimensions ? Pboblbh III. .LARD SUBTBTIFO- DA I Disg- 1756 i 1280 . *S4 B. 872 ■ R off D I along AD. C524 Prom Bonth-east to D. Ans. Ifi tWFM, I 10. 25 '376 perchei. Required the plan and area of a field from the follow- field-book and accompanying sketch ? B380 F({68 Begin at 1)80 600 E, and Diagonal 440 D. along AC- 120 D. go north-west to A. Ans. U Bvem 3 nods, 26'592 perches. SCRVETTIIO. 3. R«quired the pliin nnd area of a six-sided field, from the following field-book ? Phoblkm IV. To survey a piece of land in the form of an irregular belt. RcLB. Select tiro stations one at each end of the belt, such that A straight line joining these stations may be wholly within the belt, then measure this line and the off- sets from it upon the several bends or angles in the boun- dary, and calculate the area as before directed; if two sta- tions, one at each extremity of the belt, cannot be found, such that the straight line between them will fall within the belt, divide it into several parts, such (hat each part se* parately can be measured as directed above, then the sum of the areas will be the area required. KXBRCISKS, 1. Hequired the area of a belt from the following £dd'book and sketch. fTseta on the left of AB AB 230 1118 223 970 952 190 824 750 620 288 550 310 400 300 280 260 154 200 000 Begin at A. and offsets on the right of AB Ana. 5 acres, 3 rooda, 2195^ ^eiiTcLCT. Hbe LAKn B«trveit*4. B 2. Find the area of a field from the foUowing field-Look. V 210 1940 W 208 1800 12 ■l 202 1600 26 ^B^^-. 214 1400 30 ! ^^^^B- 1200 48 ^^^^B 1000 72 ^^^■l^ 800 117 ^^^B 110 600 145 ^^H 400 180 ^^K 200 206 ^^r 000 222 Ana. 4 mhm, 3 roods, 2-72 po'ea 1 3. Find the area and draw a plan of an irregular Geld from the folloivtng field- ook. 3425 3200 235 436 3150 3012 350 *- 576 2R24 2720 290 700 2400 2100 500 642 1830 1424 413 730 1200 ^ fil4 956 402 600 308 412 412 306 218 000 Ana. 29 acrws, 2 ro., 7 532 poles. PSOBLKM V. To measure a lake, mere, wood, or any field without entering it. RuLB. By the belp of a cross-staff or any other meant. fix three poles near to three corners of the figure, so tlat lines joining them may contain a right angle, and place a fourth pole in aucK a. ^oailitwi that it can be seen from eadi of the extreme poles &tsl \.\a.^X«4, mi w^ -oasa m, --^waible ^ 1 LAKP SUETSTlNe. Ii0 to another corner of the figure; then go round the four sides of the trape^nm thus formed and measure its sides and the insets on the sides of the figure. Construct the trapezium and find its area, from which subtract the space indicated by the insets, and the remainder wiU be the anea of the figure. Note I. The right angle in the figure is necessary to enable the surveyor to obtain the diagonal without any trigonometrical ealeula- tion; but if he has any instruments for measuriBg angles, he may place the poles in any oonvenient fMMition, and then measure one of the angles, and this with the data taken as above, will enable him, by the help of trigcmometry, to find the area of the figure, and to inw a phui of it NoTB II. The angle contained between any two lines may be measured by the chain or any graduated line in the following man- . ner. Measure 1 00 links or any other convenient distance along each of the lines containing the angle,, then measure very carefully the dis- tance between the extremities of these measured distanees, and half of this distance divided by the former, will give for quotient the na- tval sine of half i&e contained angle. Example. Find the area of a wood from the following field-book, the angle in the surrounding trapezium at A being a right angle. BA DC 1150 950 990 110 450 120 850 100 000 eoo RoffD 250 000 50 AD RoffB 1550 1000 CB 650 1340 350 50 940 60 IJO 160 640 50 000 290 Begin at A, and go east 000 , RoffC Ans* 12 acres, ^ > roods, 19 1 poles. PnOBLBM VI. To find the breadth of a riyer or other obstacle without crossing it. ^let She adjoining figure represent a river the breadth of L is required. w ^^^1 Bare LAVt> AFfiVdrt^ffi. ■From B to C mea- gure any distance _^_^ BC perpendicular to ^M AB. and make CD ^^^^^ ^^ ^J =BC, then measure ~ from D perpepdicu- lar to BD, till it meet the line AC produced in E,whict will be known by (he points A and C, at which ranrh must be placed, appearing in Ihe same line; then DE a equal to AB. For (lie triangles ABC and EDC have ihit angles at C eqaal, being vertically opposite and (he an^es at B and D right angles; also BC=CD, therefore (Geo prop. 6,) AB=DE. A'wOier method. Measure a line from B to C perpen- dicular to AB, and mark the point F, where a perpendicu- lar to AC meets AB produced, and measure BF; then bt (Geo. prop, (il ), FB : BC : : BC : BA, or BA ia equal to the square of BC, divided by BF. Note. If the impediment on the line were & house w taj other such obstacle, which only extended Tor a, short way; we can measun a line perpcndiculai' to tha line we want to mcaanre, till we »re bt jond the obetacla ; then measure perpendicular lo Iho but line, or parallel to Ibe original line, till we are past the obstncle ; if now we measure a perpendicular to this last line equal to the lirst perpendi- cular, and on the same side of Che BMond, we will have again ratorn- ed to the line we were firet mcasaring; and the diataaco of the point where we first left it and that at which we again returned to il Bill be equal to the second measured peiTcndicular, for Ihey wiJl be op- ]ioeiIe sides of a pamllelagrani. DESCRIPTION OF THE PLANK TABLK. The plane table is a plane rectagular board. bogany, and liaving a frame, by means of which a sheet of ilrawing paper i| may be fised on it. The ^^ frame is generally gradu- ated to as»>ist in laying olf distances. The table is placed on a tripod stand, and then made perfectly level, by n It has also a compnss B aUa.c\ii;4 IS of adjusting Bcrem. it, \o fttvable the sut- ■ UKs sokmiNG. Til tefor to place it in exactly the same direction at rarious' stations, where he may require to uae it. It ia also furnish- ed with a moveable index, marked I, I, on the diagram, commonly about 18 or 20 inches long, made of brass, and should be finely graduated, to assist in laying oif distances. It has upright sights at each end, also made of brass, and about five inches high, through which the poles at each station are seen, when the index is directed towards it. By means of this instrument, a. plan of a field or an estate can be laid down on the ground in such a manner, as Co determine the principal points, from which the plan can easily be filled up, and the contents calculated with conside- rable accuracy, by measuring the necessary perpendiculars ' and other lines with a scale of equal parts. Prdble.u "Vll, To survey with the plane table, by going round the field, and planting the table at each of the corners of the field. Let the table be placed , j-, ' as at A, having a sheet of ''. paper properly adjusted to , receive the plan. Place j the table level, and turn I it round till the compitss needle settle over the I fiewr-de-Ui, or north point; ;| then having placed poles at each of the corners, lay the chamfered edge of the index on the point A, and turn it round till the pole _ I at B be seen coinciding /' ] with the hair in the other sight ; then having first observed that the edge of the in- dex is still on A, draw a Hne from A to B along the edge ' of the index, and make it equal to 328, the measured dia- ' tance from A to B ; then remove the table to B, and plac- ing its centre over the hole where the pole was, adjust the table as before, and placing the index on the line AB, turn the table round till the pole at A coincide with the hair in the Bight of the index, when the needle of tlie compass will again settle over ^a fieiLr-de-lis as before, un- less acted upon by some local attraction. Apply now the index to the point B, and turn it round till the ijole at, C coincide with the hair ot the index ; measaxe We iiv4\a.t\'ijs 113 I^ND BtnmeTIKUIi from B to C 310, and drawisf; a line along the edge of the* index, make it equal to BC 3J0. Remove the Cable iiLtlie same manner to C, and adjuat it as before ; place the in- dex ott the point C, and turn the index till the point D coincide with the hair as before; draw the line CD 355; then remove the table to D, and hnving adjusted erery thing as before, make DE 318; lastlj-, remove the table to Ej and having adjusted it carefully as before, apply the index to the points E and A, then will the pole at A be seen coinciding with the hair in the sight of the indes, and the distance from £ to A, measured on the scale, will be equal to tlie distance measured on the ground, if no mis- take has been committed. Bj this means we obtain m exact plan of a field or an estate, and at the same time prove its correctness, by the two above-mentioned tests. If it were required to determine the position of any ob- ject, either within or without the plan, if not at too great a distance, this can easily be effected, if it be visible from two stations ; for we have only to apply the index at each of the stations where the object is visible, to the point of the plan representing that station, then turn the index till iJie required object coincide with the hair in the sight of the index, then draw a line along the edge of the index, and the object, will be situated somewhere in that line ; do tLe same at the other station from which the object is viable, and the intersection of the two lines will evidently be the position of the abject upon the plan, and its distance fiom any other point can then be measured on the plan. It J» evident that in this way any number of points can be de- termined, if they be visible from any two stations, so that the surveyor can determine as many points as he pleuM on the plan, and then measure their distances from bb; other points that are laid down on it. In the above plan we have then AB 328, BC 310, CD 355, DE 318. EA 450, and the lines CE and CA aa measured on the plan, are found to be 526 and 506 n- spectively, from which the area of the field can be deter- mined ; or the perpendiculars on CE and CA can be measured on the plan, and the area more simply detennin- ed from them ; and it will be found that the field contuu 2 acres 13-56 poles. The plan of a field or an estate can also be laid down by the assistance of the plane table with less labour than tbat described above, if two stations can be found either within or without it, whose distance can be ascertained, and from each of which all iW c(»m«xa «i *.V fisU. at estate can b« Ken; for we can place the table at one of tbe statiaos, and having made it level, and turned it round till ihe needle stand over the feur-de-lis, place ihe index on the point representing the station at which the table is placed, and tore it round till the pole at the other station appear to coincide with the kair in the sight of the index, then draw a line along the edge of the index, and, from a scale of equal parts, mnke it equal to the distance between the stations. The index being still kept over the point which represents the station where the table is placed, turn it round till it coincide with the direction of each of the cor- ners in succession, and dran- indefinite tines through that station in the direction of each of the corners ; then remove the tahle to the other station, and place its centre exactly over the hole in which the pole stood ; adjust the table as before, applying the index to the line representing the dis- tance between the stations, and, lookiiig back, make th& hMr in the sight to correspond exactly wiili the pole in the first station. Apply now the index to the point of the plan representing the second station ; turn it round till the poles at each of the comers appear to coincide with the hair cff the sight of the index, drawing at each coincid*nert an indefinite line in the direction ot the comer, through the station-point, the interaections of the lines drawn through each of the stations, towards the comers, will givn the proper position of the comers upon the plan, and these being joined, the outline of the plan will be obtained ; and the other lines which may be necessary for findingthe area, or tliose already laid down, may be measured on the plan with compasses, on a scale of equal parts. If any other points are to he constructed on the plan, their bearings should be laid down at each of the stations, in the siime manner as the comers. By this means the surveyor ob' tains A correct outline of the field or estate which he is tarveying, and also the position of any particular point reqnireil ; he can then by the chain measure any distances that maybe npcessnry for the more minute tilling up nf his plan, and then fiuish it off in any style that may be necessarj. A proof of the accuracy of the plan may be obtained, either by measuring certain lines on the field with the chain, and on the plan by compasses and a scale of equal parts ; if the plan be correct Ibese measurements will be the same ; if incorrect, they will be diffei-ent. Or Ihe table may be removed to a third station, ivhich is markrd on live plan, and the table being hero adjusted aa\)eto\e, v^ >^«i 114 1>AKS SinVSYISSi index be applied to the point which represents tbe Btatioa where the table is placpd, and at the same time to the point on the plan repreaenting any other station, then the pole placed in that station will be seen to coincide with tbe Imir in the sight of the index, if the plan be correct ; and te same will be the case with every other point laid dona itiieplan. pKon]:,EM VIII. To Burrey with a theodolite. The theodolite is an instrument used by BDrreyora fat measuring horizontal or veitical angles, and the most per- fect instrument for these purposes yet invented. Without it, it would be almost impossible to survey a large estate ot a county. It consists of two circular brass plates, finely adjusted to each other, the lower being graduated arooiid the edge, from 1" to 360°, and the upper has a vernier attached to it, so as to enable the surveyor to read the measured angle to minutes. The upper plate also carries a telescope with it, which is attached to a semicircular arc, by which the vertical angles are measured. The upper plate is furnished with levels, to enable the surveyor to place it perfectly level, and also with a compass, to enable him to find the bearings of various objects. In order to surrey with tbe theodolite, the surveyor must place up poles in the various points the position of which he wishes to determine, and having selected tvro convenient stations, he places the theodolite first in the one station, and having set the index of the vernier to 0°, Imm the instrument round till the pole in the other station hi seen to coincide with the cross hairs of the telescope; then fix tbe instrument by means of its screws, at the same time slackening the screw which fixes the horizontal plates to- gether, BO as to allow the upper plate to revolve freelj upon the lower, and the instrument is then fit (or use. Turn the upper plate, which carries the telescope along with it, till the pole in the nearest position to that station be seen to coincide with the cross wires in the telescope, and mark down the angle, as read from the horizonlnl fixed plate. Then turn tbe telescope round till tbe pole in the next station appear to coincide with the cross wires, marking the angle as before ; proceed thus tilt tbe ongalu bearings be obtained of all the points which you wiih to determine, and observe if the angular bearing of the second $tiition be ttie same as &»^ e«^, ti^'wik, {<n If it is not, the jnuro wwvOT iwc 1 Jo instTument must Lave been displftcod during ils reTolution, and it will be necessary to measure the angles again. HaTiDg got all tlie angular bearings of the points to be determined from the buse line at the first station, the theo- dolite must then be removed to the second station, and ad- justed in the same manner aa it naa at the first, making the cross wires in tlie telescope coincide with the pole in the first station ; then measure in succession the angular bearings of each of the points whose position is to be de- termined, as at the first station, observing, when the teles- cope is brought back to the first station, Uiat the beating of that station appear the same as formerly set down ; if it do not, it will be necessary to measure the angles again, until this be the case. When the distance from the station at which the theo- dolite is placed to the various comers, whose bearing it ia necessary to take, can be determined, it will not he neces- sary to take a second etation, as both the area and a plan of the field ot estate can be obtained from one set of angles. ExBBOisB. Find the area of .1 field, of which the adjoining figure is a sketch ; the theodolite havingbeenfirst placed a t A, the angles were found to measure as follows: ZBAC 94- 50', IC\D 44° 63', iDAE70''28', Z,EAF92''2',and/.FAB 57° 47'. while the distances were AB 504 links of the im- perial chain; AC 445, AD 325, AE 201, AF 648. The Uieodolite was then removed to B, and the following angles were measured : i^FBG 79° 5', iGBH ll-'SO', and ZHBL 68° 58'; while the distances were BF 572, BG 404, BlI 417, and BL 140. The above is the park in which Craigmillar Castle is Bituafed, the Castle being on the north side of BC, and east aide of BL. Ans. 5 acres, 2 roods, 7'6 poles. rHOBLBH IX. To find the surface, and draw a plan, of hilly or sloping The rules for finding the surface of hilly or bIq^\ii^ ground are the same as those for le^el gtoijjii ■, WvS.ii.Ns^s- 1 16 I.AWD aOBTSTntO. ing down a plan of hilly or sloping gjonnd, every line raeaanred on a surface inclined to the horizon, must be tt- duced in length, on the plan, by the folloning proportion ; R : eosinc of the inclination : : measured length : the length on the plan. Note. Some iand-BurreyorB ore of (i[iinioii, that in ineo^unn^ hilly ground, tbe horizontal surges of the base efaould be given for iha are& of the ground ; and in sapport of tb^ opinion tell ua, (hat 19 many treea or upright atalka of com will grow on the borizonUl surface aa on the sloping side of the bill ; but allhough this be per- fectly true, it appears to be the busineBs cS the land-aurveyoc to give the surface, uid tba business of the cultivator of Ihe soil, to coniider what the viktue of the surfaee is for &ny particular purpose. In lay- ing doKD a plan of hill; ground, bowerer, since all plans arc drawn OD plane aurfaees, it ia necoasary to reduce fhe length of the linas, 80 as to prevent diatortionH of Iho plan, which would be unavoidRbl» if any other method were adopltid. ExANFLK. A line of 960 links was measured np a hill, whose indination to the horizon was 17" 12'; what is the length of the base-tine ? r B Cos. \T 12* Surface line 950 k Base line 9075 log. : EXERCISES. 1. What is the length of the base-line correspondii^ to 1560 links, measured on an inclination of 21' 14'? Ans. 1454-1 Uak«. S, What is the length of the hase-line corresponding to SlOO links, the first 500 of which were measured on a slope of 9= 1 6', the next 900 on a slope of 18" 33', and the last, 700, on a slope of 23° 10' ? Ans. 1990-26 linki 3. What is the length of the base-line corresponding to 1800 links, one- half of which is inclined to the hoiizon, at an angle of 21°, and the other at an angle of 17° 12'? Ans. 1699^7 lint*. Problem X. To deduce from angles measured out of one of tlie tlk- ^ions, but near it, the true angles at the station. p.When the centre of the instrument cannot he placed in ^ vertical lino occu^\e4 V^ *.\ie wia nC % signal, the LAND B0RTETINO. la observed must undergo a "** , according to circum. '- KC be the centre of Ihe sla- P the place of the centre of _„_iMtruinent, or the summit of P'Ae obeeived an^le APB; it ia I nquired to find C, the measure of ACB, Bupposing there to be known APB=P, BPC= p, CP=d, BU=zL, AC=K. j Kbgc thff exterior angle of a triangle is equal to the sum ofthetwointerioroppositeangles, (Geo, prop, ]9),wehave, I with respect to the triangle lAP, Z.AIB:=iP+/.lAP ; aodwith regard to the triangle BIC, i^AIB=Z,C + iCBI. Making these two values of iAIB equal, and transposing iCBI, ive obtain LC=l.P+LlAP~LCBI, But the A» , CBP, giye sin. CAP = sin. IAP= ^sin, APC= B?^. riu. CBI= 8iu.CBP= f^. sin. BPC = ^■ "And as the angles CAP, CBP, are, by the hypothesis of W lite problem, altrays very small, their sines may be Bubsti- I tialed for their arcs or measures ; therefore _ (ifllnJP+p) dat^ R L ' Or to have the result in seconds, L /• The first term of the correction will be positive, if the Mgle (P+p) is comprised between 0° and 180°, and it will become negative if that angle exceeds 180°. The con- trary will evidently be the case in regard to p, since it is affected by the negative sign. The letter K denotes the distance of Ihe object A to ti>e right, L the distance of the object B, situated to the left, and p the angle at the place of observation, between the centre of the station and the object to the left. PnoHLEU XI. When a Hne is measured at an elevated level, to find its length, when reduced to the level of the sea. Let r= the radius of the earth, k= the height above Ihe level of the sea, at which the base ia mftaft\Kei.\ 'feevv since Ihe circumferences of circles are to oat a.n'i'Co.'a »* '^e 118 T.J their radii, r+A : r : : m', {the measured Icngtli), ; m (the tnielength= «j=W^l— -+^ — ^ + -_&c.j- ; bat the radius of the earth being very great in relation to the height of any mountain on its surface, all the terms of the series may be neglected after the second, which gives the reduced length :=m' — ■ — ■ hence Rule. Add the It^arilhma of the measured base, and of its height above the level uf the sea, both in feet, ami from the Euni subtract the constant logarithm '{■ZIQBQO, the remainder is the logarithm of the correction in feet, which is always subtraetive. ExAwpLB. If the length of a line be 22540 feet, at an eleTalion of 14600 feet above the level of the sea; what will be its length when reduced to the level of the sea? ^Log. 22540 =4-352954 Log. 14600 =4164353 85 17307 Subtract constant Log. 7319890 Correction =1575 =1'197417 .-. 22540— 1575 =22524-25 feet. Ana. EXERCISES. 1. If the length of a line be 31576 feet, at an elevation of 1800 feet above the level of the sea; what will be its length when reduced to the level of the sea? Ans. 3157a279fe<'t 2. At an elevation of 82Z feet above the level of the aea, a road was 5 miles ; what is ils length when rednceil ^^_.t9 the level of the sea ? ^^B Ans. 4 miles, 1759 yards, 1-963 feet. Problem XII. The Division op Land is one part of the land- measurer's profession, he being frequently reij^uired to divide a piece of land between suodry claimants, in proportion to their fespectiTe claims. In some Instauces, the division relates only to the quantities that each claimant is entitled to ; in others, the quality as well as the quantity must be considered, in doing whiah, the measure of the whole must be first accura«:\j astei\.a.\ne4, \i'3 aanu; d^ dw rules LAND SURVETIXG. 119 which have been fonnerlj given in this work, and then the division must be made according to the form of the figure, and the value of the claims. Cask I. To cut off a portion of a square or parallelo- gram, by a line drawn parallel to one of its sides. Rule. Divide either of the sides, adjacent to that to which the dividing line is to be parallel, in the ratio of the parts into which the field is to be divided, and through the point of division draw the required parallel, and it will divide the parallelogram in the ratio required, (Geo. prop. 58). ExAAiFLB. If a field in the form of a parallelogram contain 7 acres, and one of its sides be 200 links ; how must the other be divided, so that a line drawn through the point of division parallel to the given side may cut oif one-fifth part of the field ? Since the field contains 7 acres, its w^hole area in links is 700000, which being divided by 200, gives the adjacent side 3500, one-fifth of which is 700 links, the base of its fifth part. EXERCISES. 1. If a parallelogram be. 1200 links in length and 400 links in breadth ; what length must be taken, so that the area cut off by a parallel may be an acre ? Ans. 250 links. 2. From a square which contains 5 acres, 3 roods, and 10 poles, there is to be cut off 1 acre, 3 roods, and 5 poles, by a line drawn parallel to one of its sides ; at what dis- tance firom the corner must the parallel be drawn ? Ans. 233*6 links nearly. Case II. To cut off a part from a triangle, by a line drawn through the vertex. Rule. Divide the base in the same ratio as the parts into which the triangle is to be divided, and a straight line drawn through the vertex and the point of section will di- vide the triangle, as required. (Geo. prop. 58.) Example. A triangle whose base is 960 links, and area one acre, is to be divided into three parts, by lines drawn from the vertex to the base, so that the first part may be 24 poles, the second 1 rood 36 poles, and the third 1 rood 20 poles ; required the length of the segments of the base ? Since the parts into which the triangle is to be divided are 24 poles, ^6 poles, and 60 poles, and their sum is 1 60 poles, by the Rule of Distributive Pro]^ottvo\i, qx ?^\.\^^\- Aip, we have I I.Aia> BOBVIETIira. im : 24 : : OHO : 144 links, the first base. 160 :?<>:: 960 : 4afi links, the second bas«i] 100 : 60 : : yeO : 360 links, the third bane. 1 . A triangular field, whose base is 1380 links, is to le divided, by Tmes drawn from the vertex to points in (lie [riven base, so that the parts may be to one anoibcr in tlie proportion of L.3, 12b., L.5, loa., L.7, 5s., and 1..6, Bs.; required the segments of the base? Ans. 216, 345. 435, and 384 links, 2. A triangular field, ivjiose area is 2 acres 30 poles, is to bedividediiitobveparls, intheratioof 2, 5, 7, 9, andll, by straight lines drawn from a given angle to the opposite side, which is 1530 links; >vbat must be the length of ibe begments of the side ? Ans. 00, 225. 315, 405, and 495 links. Casb III. To cut off from a triangle any assigned part, by a line drawn parallel to its base, another side at leatt ituLB. From the area of the whole triangle subtract the part to be taken away; then state, as the whole triangle is to the remaining pnrt, so is the square of either of the sides to the square of the distance from the vertex to ihe point nbere the parallel cuts that side ; extract the square loot of this result, and the root will be the distance from the vertex required ; ihrougli this point draw a parallel to the base, and the thing required is done. £xAMPi.K. From a triaiigle which contains 4 acres, il is required to cut off I aere 3 roods, by a line drawn po- i-allel to one of its sides, the other two sides being 300 and :j(iO links respectively; at what distance from the veciex will the parallel cut each of these sides? The whole area is 640 poles, and the area to be cutoff is SKO p(>leB ; therefore the area of the remaining triuiigle ^vill be 3G0 poles ; hence fi40 : 360 : : (300)' : 50C25, the square root of which is 225, the distance from the vertex on tbe side which is 900. Again, 640 : 3fi0 : : (360)' : 72900, tbe square root of wUcb ii 270, tbe distance from the vertex on the side which iaSGO EXRIECISBS. 1. If tbe base of a right-angled triangle be 2100 liaki, and tbe perpendictt\at lOQUWaka-, a\,»W\. di&tauoe from LANK SUnVEYING. 1^^ tlie yertes must it line he drawn parailet to the buse, so that the area conbiined between the base ani] the parallel amy be an acre? Ans. 95! -2 Haks nearly. 2, la the same triangle, if a line be drawn parallel to ibe base, so tliat tbe area of ihe trapezoid may be 1 acre 2 roods; at what distance from tbe base must it be drawn 7 Ans. 74'2 links nearly. 3. If a trapezoid, containing 2^ aerea, be cut off from Ihe same triangle, by a line parallel to the hypotenuse, at what distance from the right angle will it cut the base and perpendicular? Ans. The perp. 872'9, and base 1833 links. Casb IV. When a given quantity is to be cut off from a field, by a line drawn from a point in one of its sides. RubB. Dmw a line by trial, cutting off as nearly as possible the given space; measure the space cut off, and lake the difference between it and the space required ; di- Fide tbe remainder by half the trial line, and the quotient will be the perpendicular altitude of a triangle, having the trial line for its baae, which triangle being added lo lh« space cut off if it be too little, or taken from it if it be too great, will give the space required. ExAMPLB. Suppose 3 acres are to be cut off from a field, by a line drawn from a given point in one of its sides, and that a trial line of 6'00 links has been drawn, and the space cut off is found to be 285000 square links, which is too little by liiOOO square links ; divide this remainder by 300, half the trial line, and the quotient is 50 links ; draw a perpendicular to the trial line, on the opposite side of the space, because it was too small, and make it equal to AO links; if its extremity be in the boundary, the inter* mediate part being a straight line, it is the point requir- ed ; if not, through its extremity draw a parallel to the trial line, and it will cut the boundary in the point requir- ed ; join this point and the given one, and the thing re- 'is done, as is evident from (Geo. prop. 2S). MISCFLLANKOUa EXBRCISKS. (ill a ridge of grass cost, at 10 guineas per 49*7 its length being 1 7^ yards, and its breadth (!^ yards ? Ans. r..2,9s. 0|d. 2. What is the area of 18 drills of potatoes, the length being 454 links, and the breadth 62 hnks; and what will they cost at L.24 per acre V Ana, 1 ro, 5 0368 poles; cost US, \Jrt,.\i^*>, 122 LAso auKvrriNG. :). ^Yba^ is the value of 3<> drills of potatoes, at 3s. fid. per 100 yuidti, ilie niean leiigtii being 2/2 yurds? Ans. L.14, 5b. 7j<l. 4. What will 60 drills of turnips cost, at L.21. ISs. per acre, the menu Irn^tli being 11)6 j'urds, and the diiit<iiii.'e between each drill -J Itol ? Ant. L.35, 49. 7^(1. 5. If the breadth of a ridge be 5^ yards, what leujiib muBt lie taten to aiate half an aeref Ans. 440 yank tt. To preveat iujuriug the uropn, a land-surveyor mea- sured one side uf a iriaiigulnr field, which be found to be (i54 links, and the angles at its extremities 5(1° 20' and 64" 15'; required the area of the field in acres? Ans. 1 acre, 3 roods, 313056 pol«. 7. The annua) rent of a triaogular field is L.43, I 5b., its base ineasurea 2500 links, and perpendicular 1400 liaks; what is it let for per acre ? Ails. L.2, 10s. 6. A field in the form of a right-angled triangle is to be divided between two persons, by a fence made from the right angle meeting the hypotenuse perpendicularly, at tie distance of 880 link-s from one end ; required the area of each person's share, the len^rth of the divisiou fence being 660 links? Ans. 2 acres, 3 ro., 2464 poles, and 1 nav, 2 ro., 21-36 poles. Q. A genileinan has a rectangular garden 100 feet long and 80 feet broad ; ivhat must the breadth of a walk routul it be, so as to occupy half the garden ? Ans. 12-9844 feet nearlj. 10. The side AB of a triangular field is 40 chains, lit' RO, and C.\ '25; required the sides of a triangle, parted oft' by a division -fence made parallel to AB, and proceed- ing from a point in CA, 9 chains from the angle A? Ans. 16, 19-2, and 256 chaint 1 1. The side .^B of a triangle is 050 links, and the side AC 560 links ; it is required lo bisect it by a line drawn from a point in the side AB, 4U0 links from A; find the distance from A, where the line will cut AC? Ans. 455 links. 12. What is the annual value of a pentagonal field, it L.2, 5s. per acre, the side AB being 926, BC 536, CD 835, DE 828. EA 587, and the diagonals AC and CE 1194 and 1223 links respectively? Ans. L.18, lOg. 7jd. SPECIFIC GRATITT. 123 SPECIFIC GRAVITY. ie gravity of a body is its weighty aa compiured t of another body of the same size, whose weight ered as the unit of measure. The body used as iard or unit of measure is a cubic foot of rain or .water^ which at a temperature of 40° Fahrenheit's leter, weighs exactly 1000 avoirdupois ounces. n&c gravities of bodies may be expressed either by against them the quotient arising from dividing [ght by that of an equal bulk of water, or by writ- ist them the weight of a cubic foot of the body^ in )ois ounces; the first can be derived from the y dividing by 1000, and consequently the second first by multiplying by 1000. In the following will give the weight of a cubic foot of the sub^ I avoirdupois ounces as its specific gravity. SFECIFIC GaAVITIES OP BODIES. if EARTHS, METALS. /ental . . 2590 nyx . . 2638 loudy . . 2625 f oriental white 2730 , of Piedmont , of Malta , Spanish saline , of Valencia . , of Malaga Of . m • - B, long . f short . ripe , starry 9 crude . , glass of . ^lass of natural lolten Giant's Causeway 2864 >f Judea, . 1104 2000 M, . . 3549 mtal, . 2723 2693 2699 2713 2638 2876 1078 926 909 2313 2578 3073 4064 4946 3594 5763 Bismuth, molten, . . 9823 Brass, cast, not hammered, 8396 Brass, wire-drawn . 8544 , cast, common , 7824 Cohalt, molten . . 7812 hlue, glass of . 2441 Copper not hammered, 7788 , the same wire drawn 8878 Chrysolite of Brazil, . 2692 Crystal, pure rock of Ma- dagascar, . • 2653 Crystal of Brazil, . . 2653 Calcedony, common . 2616 — ' , transparent . 2664 i veined . 2606 Camerlian, pale . . 2630 Chalk, British . . 2784 , simple . , 2613 Diamond, white oriental 3521 , Brazilian . 3444 Emerald of PerU| . 2775 Flint, white . . • *15i^V '-'■ — > Eg^ptaaxL • ^ti^^ Gk>ld, pure, csiftt, \iviX> t^\> ^ 19369 Pktina, crude, in grains. 15602 , guinea of Geo. 11., 17150 , fuiDeaofGeo. III., 1763.1 mered, 19500 , trinket standard, 13709 20337 Gamet of Bohemia, 4189 21042 Giniwl, . . . 4000 B2069 Gypsum, opaque . 2168 Pearl, virgin oriental 2684 230C Pebble, Englidi 2609 ., rhomboidal . 2311 Prasinni, . 25S1 — — .oimeifonn cry Blallized -2306 Peat, hard 1SS9 Glufls, green . 2643 Porcelam, china, . 2385 , white . . 28!)2 Porphyry, red . 2;6i ^~~, bottle . 2733 — .green . 2fi7« , Leith crjBtal , 3169 Quartz, cij-alallized . 26S5 — ; fluid . . . 3329 Ruby, oriental 42!1 Grsnile, red Egyptian 2654 -.Brazilian . 35 Jl Hone, while razor 287<i Hyacinth, conunon . 3637 hakened . . 7B31 Iron, cast . 7207 , bar 7788 , do. not hardened TBlf Jade, white . . . 29.50 10471 2966 lOSIl ~;^e". ■ . ■ . 2983 . Paris gtandard . lOlJi Jaaper, clear green . 2359 , Bhiliing of Geo. n. 10000 , red . . 2661 , shilhng of Geo. Ill 10S34 , yeUow . . E7I0 , French Poiu 10408 — ^ — , veined . 2696 Sapphire, oriental 3991 .bloody 2628 . . Brazilian 3121 iaxgon of Ceylon, 4416 Spar, white sparkling 24M Lead, molten . 11352 ■ . red do. 3111 , ore of black lead 4059 , green do. 2r« , do. do. vitreous 655S , blue do. . SCSI , ore red lead . . 6027 SSH Limestone, 3179 SBIJ ■ , white fluor . 3158 Sardonyn. pnro 25M . green 3182 Stone, paving . . SIK 4756 , caller'B Sill Molyhdona, 4738 , grind . . 9u: Mercury,soUdoreongea!ed 15632 ,niill . 24S1 , fluid . . 13G68 SIS , natural calj of 9230 1981 _,pm:ipitatered . 8309 Sulphur, native 2»31 , bro«n cinabar 10218 -, molten . im 6902 Tin, pure Comisli melted, Nickel, molten . 7807 and not hardened . 7251 Marble, green . . 2743 ,red . 2724 Talc of Muscovy mi 2B3B , black crayon 30» ■ , Pyreuean . , do, Germui . »4( 26 95 . yeflo* — , white Grenadan 2705 ,bh«:k . . aS , violet Italian . 2858 , white . , green Egjptiaa t&66 VlviB'^". ■ , Op»l, '\ . . . ^■V\t\-«l«Ol'DO, . I r BTECIFIC GRAVITT. 135 Wolfram, . 7113 Wine, Bourdeaos . 994 ZinCj molten 7191 , Madeira . . 1038 , Port 997 LIQCORS, OILS, ETC, Acid, aulphuric . - 1841 KESIM, QUMS, iND ANIMAL -, nitric . 1271 , inuriatio 1191 Aloes, socotrine . , 13B0 1033 1359 , white aoetooB lOU . 1398 1558 Bcea wax, yellow B66 , formic 994 , white , 969 837 Bono of an ox, . 16fiG , highly rertified 629 Butter, 942 , mixed with water Camphor, S89 7-BlbB alcohol . 867 Copal, opaiiue . . 1149 , 6-BthB do. , Chinese . . 1063 , 5-BlhB do. 920 Fat, beef . . 823 , 1-aiha do. 943 e2« . 3-Utl« do. . 9(i0 , hog's . 937 , 2-athH do. 973 ,y^tl . . 934 — , l-Blh do. 9BS Gamboge, . . . 1232 897 , Arabic 1207 . 1453 Ammontac uqum Beer, pale . !(I33 , brown . ID3-I — -, Eapborbia H24 Cjder, .... 101 B , seraphio . . 1201 Ether, gnlphuric 739 , Ditric . 909 -, do. of A oppo 1235 , nmristic . 730 933 , aoetic Milk, woman's . .cow's . in a loon heap, 836 . 1745 1450 1820 1032 Honey, . . .asses' . 1036 Indigo, 769 ^..e-we'a . 1041 Ivory, 1B2S , goat'a . 1035 Lard, . 948 .Siare'a . . 1031 Myrrh, . . 1368 , cow's clarified . 1019 Opium, . . . 1336 Oi), essential of torpentin 870 Spermaceti, 943 ^, do. of lavender 894 Tallow, 942 — , de. oTcloTeB . 1036 Shoemaker's wax, 897 1044 — , do. of olives . 915 — , do. of almonds 917 — , do, of lintsatd 940 Alder, 800 — , do. of whale . S93 Apple-tree, 793 — , do. of hempseed 926 Ash, the trunk, . 84.'. — , d*. of poppies 994 Bay-tree, 823 Beach, . . . 853 Tnipentine, liquid 991 Bon, French . 912 Urine, human . ]01 -, Dutch , 1328 Water, ram or distilled 1000 ■ , Brazilian red 1021 ,9m (average) . 1026 ' Campeach)- op log wood, 9\* of Dead Sea, 1340. Cedar, wild Wine, Biirgundy 992 , Paleatino ■■ ^^^J SPBCmC GHATTITT. ^ Cedar, Indian . 1315 MMtic-tree, 349 , American , 6BI Mahoganj-, 1063 Citron, . 726 Map4 . . 761 Cherry-tree, 715 Medlar, , .. 944 Cork. . . 240 Mulberry. Spanish B9T Cypress, Spanish . Sit Oak, eo years old . im Ebony, Americaa 1331 Olive-treo, 9W . .Indian . . 1209 Omnge-tree, . 7a5 Elder-tree, 695 Pesr-lree, , 6S1 Elm, trunk of . 671 PomeBTanate-tree, . 1351 eno Poplar, . . 311 Fa, male . . 630 , while Spaniah S29 ■ , female . 493 Plum-u™, . 783 Hazel, . . fiOO Vino, '. . . U27 lacmin, SpnniBh 770 Walnut, 671 Juniper-tree, . BS6 Willow, . . . iBi 703 Yew, Dutch 781 Lignum vite, . . 1333 .Spanish . HW 6M ^K WEIGHT AND BPE nnc GKAviTirs of diffekbnt gases. ^H FallKolleit'B Themoqi.5 =. Barometer 30 mches. PSptc tn^vil;. Weight cublcfML Atmospheric air, . 1.3 535'0 gn. HydrogeD gas, . , 0-1 43.75 Oxygengaa, . . 1-435 627'813 Mtrousgoa, . . 1'4544 63fi'333 Ammoniac gas, . . '731 1 319'B33 SnlphureouB acid gas, , 2-7611 1207-978 H Problem I. I Given the specific gravity and solidit; of a body, to find its -weight. RuLR. Multiply the solidity in feet by the tabular spe- cific gravity, and the product will be the weight in avoir- dupois ounces. Example. Wlat is the weight of a block of Pyrenesn mntble, containing 7"25 cubic feet? Here we find the tabular specific gravity 2726, tberefoK 272Gx7-25=:197tS3-5 avoirdupois ounces, or 11 cwt,31l» 3^oz. EXERCISES. 1. IFhat is the weight of a log of mahogany containing 35| cubic feet? Ans. 1 ton 3 qrs. 9 lb. 9J oi 2. What is the weight of a piece of French box, mee- auring 3| cubic feet? Ans. 1 cwt 3 qrs. 3 lb. 8 □£ 3. What is the weight of a piece of cast iron containing 450 cubic inches? Acs. 1 ewt. 5 lb. 5 oz. nearly 4. What is the weight of a Ic^ of beech which ia 20f«l long, and 1 foot 6 incVica at^uiwe? Atia- \ ton \ c-jA,. \ cj. \Q ft. 4 m ^ ^ SPECIFIC GKATITT. 127 5. What is the weight of a cjlindrical pillar of Egyptian CTanite, its diameter being 3 feet 3 inches, and its height 15 feet 5 inches? Ans. 9 tons 9 cwt. ] qr. 18 lb. 5 oz. 6. What ia the weight of a hollow cast iron cylinder, whose length ia 5 feet, outer diameter 30 inches, and the thickness of the metal 1^ inches? Ans. 15 cwt 3 qra. 1 Ih. 13 oz. Problem II, GiTen the specific grayity and tveight of a body, to find its solidity. KuLB. Rednce the given weight to ounces, and divide by the specific gravity of the body, and the C[U0tient will be the solidity ia cubic feet. Example. How many cubic feet of male fir ore in a log which weighs one ton? Here 1 ton reduced to ounces is 1 x20>: 112x 16— 35840, which being divided liy 550, the specific gravity of male fir, gives the quotient 65^"^ feet, ESERCISES. 1. An irregular block of red marble was found to weigh 2 Ions 7 cwt. 13 lb. 12 oz., how many cubic feet did the block contain? Ans. 31 cubic feet. 2. If a mass of molten sulphur weigh 15 cwt. 2 qrs. lb. 2 oz., how many cubic feet are in the mass? Ana. 14 cubic feet. 3. How many cubic feet of green porphyry are in an irregular block which weighs 7 cwt. 1 qr. 24 lb. 4 oz.? Ans. 5 cubic feet. 4. How many cubic feet must a vessel contain to be capable of holding a ton of rain water? Ans. 35'84 cubic feet. 5. How many cubic inches are in an irregularly formed Teasel, which contains 5 lb. 7 oz. avoirdupois weight of run Tvafer? Ans. 150'33d cubic inches. Pboblebi III. To find the specific gravity of a body, its weight and solid content being given. Rule. Divide the weight in avoirdupois ounces by the solid content in feet, and the quotient will be the specific grarily. Note. If the eolidity he in inoheB, multiply the weight in avoirdu- paiftoauceB by 1728, and divide llie product by the solidity in cubic »FLE. If a mass of 9 solid inches weigh 15 oi, 1 it, » the specific ^ javity of the body^ Here hi,', ounces, multiplied by 1728, (see note) gires 26028, whitli bemg divided by 9, gires 2892, the specific gravity; hence the body ia white glass. EXERCI3K?. 1. If a mass of 4^ cubic feet of stone weigh 7 ewt 3c|T8< 261b. 15 oz,, what is the specific gravity of the stone; and find from the table what kind of stone it ib. i„, J Spec- g"'- 3182. \Oreen limestone. 2. If a mass of Sfeet of wood weigh 1 cwt. 8 lb. 12oz., what 18 its specific gravity? Ana. 644. 3. If apiece of silver of 4^ cubic inthes weigh 1 lb. lOo*. 41 dr., what is the specific gravity of silver? Ans. 10474neariy. 4. If a cubical piece of iron, whose side is 5 inches, ireigh 35 lb. 3 oz. 6 dr., what is the speci'fic gravity of iron! Ans. 7788. 5. A bar of melal 6 feet long', 4 inches broad, and 2 inches thick, weighed 1 cwt I qr. 22 lb. 4 oz., what wai the specific gravity of the metal, and find from the tahh m yibat kind of metal it was? ■ Ans I ^P^^'^" e™''*y 7788. ■ ' I Hammered copper or bar iron' F Pbobi-km IV. To find the specific gravity of a body without knowing its solidity. Cash I. When the body is a solid heavier than water. Rule. Weigh the body in air, and also in water, then stale ; The weight lost in water, Is to its weight in air. As the specific gravity of water 1000, Is to the specific gravity of the body. Note. A body may be weighed in water by suspendiag it bea i fine chord attached la the ami of a tiaUnce, eo tUnt the body na} deBcend inio the water while the other scale in which the weights aW rBmaios oat of tha water. The body descending into the water "ill dJsjilacG a quantity of water equal to its own bulk, and will bo buoyed or preasod upward with a force equal to tha weight of llm water displaced, and hetica will lose a part of its weight equ&l la llw same bulk of water; and from this the truth ot the rule is obvioiA ExAMPLB. If the weight of a body in air be 40 ob., anil in water 10 oz.; what is its specific gravity, and what hodj is it? Here the weight lost in water is 30 oz,, and hence 30: 40 : : 1000 : 1333, the body's Byecifio gravity, and by in- specting the table, we ?iu4 \l lo \ie ligTvu-m. ■uVI'b, vnanc GXATRfr. 1. If die weight oFa body in air be HO ot., and in rain fit distilled water 42 oi. ; what is the specific gravity of tfiebody? Abb. 6250. 2. If the weight of a piece of metal be 1 5 or. in air, and 14 OS. in water; what ia its specific gravity? Ans. 15009, 3. The weight of a stone in air is 26 oe., and in water 16 ox, ; what is its specific graTity? Ans. 2600. 4. Find from the table what substance that ia, which weighs 4 lb. in air, and 3 lb. in water. Ans. GirasoL 5. What is the specific gravity of » body which weighs 15 ox. in air, and 14 oz. ] dr. in water. Ans. 16000 Cask II. When the body ia lighter than water. Bulb. Attach to it a piece of another body heavier than water, so that the mass compounded of the two may sink together. Weigh the denser body and the compound body separately, both out of the water and in it, and find how much each loses ia water, by subtracting its weight IB water firom its weight in air; and subtract the less of these remainders from the greater j theu use this proportion : Ab the last renainder Is to the weight of the light body in air. So is the specific gravity of water, 1000 OBuces, To the specific gravity of the body. Example. Suppose a piece of filbert tree weighs 15 lb. in air, and that a piece of copper which weighs 18 lb. in air and 16 lb> in water is af&ied to it, and that the com- pound weighs (f lb, in water ; required the specific gia- Tily of the filbert 1^ Here the weight lost by the copper was 2 lb,, whilst the mass, which weighed 33 lb. in air only weighed 6 lb. in water, it therefore lost 27 lb ; hence the difference of the weight lost is 25 lb. ; and therefore by the rule we have the following proportion: 25 : 15 ; : 1000 : 600, the answer. EXBRCISKS. 1. If a body lighter than water weigh 20 lb. in air, and a body heavier than water weigh 12 lb. in air and 10^ lb. in water, and the two together weigh 4^ lb. in water; what is die specific gravity of the lighter body ? Ans. 769/b. 2. If a piece of wood weighing 14 lb. in air be attached to a piece of metal which weighs 11 lb. in air and 10 lb. in water, while the compound mass weighs 2 lb. in water; £ad the specific gravity of the wood ? Ai«, SJft-^^. 3. A piece of metal whicli weigha \6 \\i, vo. ^lv^ %&^- k SPECIFIC GRAVITT. 14 lb. in water, is atdichcd to a piece of wood weighing 20 lb. in air, and the compound mass is found to weigh only one pound in water; what was the specific graTitj of the' wood ? Ans. 606 nearly, j 4. If a piece of metal which weighs 14 lb. in wr and 12^ lb. in water, be attached to a piece of cork weighing 3 lb> in air, and it be found that the compound mast weighs also 3 lb. in water; what is the specie c giavitj of cork? Ans. 240. Case III. To find the specific gravity of a fluid. BoLG. Take some body of known specific gravity, weigh it both in and out of the fluid, and find the loss or weight in the fluid ; thea say, As the weight in air of the body Is to the loss of weight in the fluid, So is the specific gravity of the body To the specific gravity of the fluid. ' ExAUPLE. If a piece -of metal whose specific gravity ii 19500, weigh 9| lb. in air and 9^ lb. in a fluid ; what k the specific gravity of the fluid ? Here the weight tost in the fluid is half a pound ; and therefore by the rule we have 9| 4 : : 19500 : 1000, the specific gravity of the bodj, it was therefore rain or distilled water. EXSRCISKS. 1. If a piece of bar iron weigh 7 lb. 9 oz. II dr. _ ^ and in a fluid 6 lb. 11 oz. ; what is the specific gravity «( the fluid? Ans. 940, lintseed «L 2. If a piece of green Egyptian marble weigh 2 lb. 9 oii [' 11 dr. in air, and 1 lb. 6 oz, 5 dr. in a fluid ; what is tlw. pecific gravity of that fluid? Ans. 12411 3. If a piece of unhammered copper, weighing 2 lb. 8 ol 9 dr. in air, lose 5 oz. of its weight in a fluid ; what is ibt specific gravity of tho fluid? Arts. 9SX' F110BI.BM Y. To find the quantity of each ingredient in a givett com- pound of two ingredients. KuLE. Find the specific gravity of the compound, and of each ingredient ; then multiply the weight of the nu by the specific gravity of the body, the quantity of whit you wish to find, and by the difference between the fie gravity of the mass and the other body; divide this nw duct by the difierence of the specific gravities of the booii^ multiplied into the specific gravity of the compound and the quotient wiVl ^ve &e (^aa.^<;i.*.^ qC that body. 131 ExAUPLB. A mixture of gold and copper weighs 18 lb., and its specific gracity is 14520, Ihe specific gravities of gold and copper being reapectively J9258 and 77"f ; how much of each was id the mixture? ix(U530— 77Bt (19-250— 77Ba)itU, lexTTBOxClHaSB— 146'J0) (19258— 7788)xl4S2o' = 14-012, the gold, = 3-988 the copper. 1. Amislare of gold and silver weighed fij lb., and its specific graTity i*aB 15800; what quanlitj of each metal did the mixture coatain^ Ans. 62-8177 of gold, 221823 of ailver. 2. A mixture of silver and copper weighed 100 lb., and Its specific gravity was 8530; what quantity of each meial did the composilion contain ? Arts. 33-92 lb. of silver, 6608 lb. of copper. 3. A mixture of pure platina and silver weighed JiO lb., and its specific gravity was 16420; how much of each metal did the composition contain ? . I 62-5865 of platian. ^^^- I 17-4135 ofBilvcr. 1. Gauging is the art of finding the quantity any vessel can Or does uontain, and conipnaea the methods of ineasuring the dimeDsioDs, and therefron) calculating the cnntenta. * 2. All diniensioDfl ave taken in inches and tentlis of incht-h, sod the conti^nts are foimd in the same, by the rules for men- nntion of Bolids, whose furma are sitnilar to those of tlie Tea- ttta gauged. 3. These contents are converted into pounds weight, ga!- lone, and bushels, by the use of ei-rtam known nnmbeni, called divisors, multipliers, and gauge-points, and which arr contained in the following Table. In Lhis table, the divieors for cii'des for flint and plate glass, and the divisors for arjuarus for the di^ereat kinds uf saap,ui(l far the imperial gallon and bushel, ai'e those legalized, as the standard of ineuaurernent, by act of Parliament. All the other divisors for circles are obtained by dividing their res- pective diviBora for squares by '7S5396, and the other divbors for squares by imiltiplying their respective divisors for circiea by -7B5308. The divisors are the measures of capacity, The multipliers are the refiprocals of the divisors, and are obtained by dividing nnity by the relative divisors for sqaaies or circles. Gauge-points are the square roots of the divisors; they aN chiefly used for calculations by the sliding rule. 4. Tabular divisors, multipliers, and gauge-points may be found for any form of regalar superjicies, and for any sUod- ard of capacity. In general they are given in treatises of gaog- ing; but as they are seldom or never used in practice, they have been omitted from this. Sncb, hov>ever^ as may be de- sirous of knowing them, and having thetn at command, may easily compute them by the following rules. For Conical Figures, — Multiply the number of cubic inches, in the measure of capacity, by 3, for divisors; thebc- tors and gauge-points may be obtained from these as before. NoTB. In using this t&ble or rule, the whole depth mast bo For Regular Polj/gons. — Divide the measure of capacity by the tabular multiplier given in the table of polygona for di- visors. (See page 6!l.) Proceed as before for multiplien and gauge-points. When the middle area of aroj regular frustuia is taim, — For the gauge-points. — Take the square root of dx times the divisors, for polygons, for squares; and for circles, the gtpisw _ nota of six times the circnlar measure of capacity. fi. Many calculations in gaaging are made by the sUding- P THE Sl:IDIN0-RaLE. The sliding-rules generally used are of two forms. One is 1'2 inch broail, and '6 loch thick; tlie other is 1-7 inch brMiI and -2 inch thick; both are 9 or 12 inches long. Tho former lias four sides, in each of which is a groo'e and a slider. One side is marked A on the upper edn, snJ MD on the lower. The opposite side is marked D. Of the other two sides, one is marked SS, tha other SL. On the first side, the mark A is placed at the end of alinr, called the Une of numbers, which is divided into 100 parts. ia such a manner as to constitute a scale of graduated loe*- rithms, numbered from 1 to 10, and ninningfrom left to rignt. In this line is a brass pin, marked 1MB, at 2218']92, Si"! wnoiher IMG, at 277"27i- The former is the divisor (m imperial bushels, the \a\,let ioi wtv^rLni. ^Utnis. The line marked MO on the same aide, is for computing ualt-floon ; its numbers stand In inverted order to titoee un tKe line A, snd ore aa placed that 10 ia oi>poBitc to 1MB n A. On the opposite side, which la marked D, there is hut one ine, numbered I to 32 on the upper edge, and on the lower ■2 to 10. Thb line ix the same m construction as the line of nnmhera A, but has a double radiua. In the upper portion of the line, at 18-7)192, is a brass piu, marked IM.U; this is the circuhir gauge-point for an imperial gallon. In tlie lower part, at 49'0977, ia another pin, marked MS, wliich is the gauge-point for squares, and one at -531441, marked MB, the gauge<point tor circlei^for an imperial biiahel. The sides marked SS ami SL are used for ullaging, or finding; the quantity in ca!<k9 that arc neither full nor empty. SS signify segment standing, and SL Begment lying. These linea occupy both the upper and lower edges of the mde, and their numbers are placed from left to right. Of tlie sliders, two are marked B and two C. All tlie lines on the bees so marked are similar, and eciiial to the line of nuniberg A. On the under face of one slider are marked the gaajre -points, divisors, and factors for squares and circles, for a, gallon and huahel. Another slider baa on ita under face three lines; the first or uppermiist is divided into inches and tenths, the second is marked splid., and the third 2d va- rietj'. These lines are for reducing the 1st and 2d variety of casks to cTlinders, hut in practice are very seldom used, Aiders having the same letter, are, in use, joined together, ao that ttie letters are the extremities, and thus tiiey form sntt are rend as one slider having two radii. Either pair uf diders mav be used in all computations. The other form of sliding rule in common use has hut two sliders, which being of the same thickness as the stock, are grooved, to slide on tongues on the stock. Both the faces of these sliders have on them the lines of numbers. On one edge of the stock are two lines of numliers, so placed, that 1 on tho upper cuts 1'63 on the lower. On the other edge are also two aimilar lines, but placed so that 1 on the upper cuts 1'227 on the lower. In all other respects both rules are alike. G. To estimate the value of the numbers on the sliding-mls. On all sides the value assigned to 1 determines the value of the rest of the numbers on the same line. If the assigned value be ■!, 1, 10, 100, or 1000, then 2 will be -2, 2, 20, 200, or 3000, and the intermediate divisions will have correspond- ing values. On the line marked MI), 10 may be read 1, and then the others must be proportionately decreased. 7- To multiply by the sliding-rule. Rule. To 1 on A set the lesser factor tn\ ^B, a^vi t^iwisS. the greater faclor on A will be the product c KxiKPLE 1. Multiply Sii by 7. cipiHi^ite to I iin A set 7 on B, (tnJ against 56 on A will In 31t2 on B. 2. Multiply !WM hy 48. Ana. 17472 3. Multijilv 75-0 by 6-S. Ans. *09* 4. Multiply sag by -2. Aus. 164-6. Note, a umple way lo read the product is, after Bxiog the rule, to give the nuDiericBl value of tlie Jint figure at the greater bclar In Qie vnil fi){ure of the lesser factor, and valuing tlie prodnet le- cordiugly. Thus, in the 2d enunple, 6, the uni'f ligure o[ the lesser Tactor, after taking the Taluo of ;!, tlio first figure of the greater bc- lor, and whiuh in hiadrrdf, bemniea SQ^, imd the tesser fiiclDr tboi reads lUOO, wbidi caaily leads tu the correct value of the pradad, 17*72. 6, To divide by the a!Ming--riile. Agninst tba div'dfind on B eet the divisor on A, flJui lite to 1 on A will be the quoticDt on B. 1. Divide .'WO by 4JS. 1 B to 45 on A, and beneath 1 on A is 8, the SB- Divide 85'5 bv 9. AnB. 9-5. Diviile 1470 by 42. Ans. 35. 4. Divide 1728 by 14-4. Ans. ISO, KoTE. WlioQ the divisor ie greater than the same nnmbei <t figyiras of the dividend, the quotient coniiats of as many Agnret u the dividend hiia more tliwi Uie diviaor; but fthen Icea, the quonsiLl lias another figure in addition. 9. To work Proportion or the Rule of Three on the slidinj- Rur,£. Set the first terra on A to the second on B, then Bgainat the third on A will be the fourth on B. BxAuPLE I. If 14 books cost 35s., how mui-h will 44 bookE To 14 on A set SJSs. on B, and 44 on A cub llOs. on B. 2. Wbatisthefourthproportioniil toSl,27,nuii8l? Ans. 24a a, Kequiied the third proportional to 48 and G4'^ Ana, 8d^ Note. The third proportional is found by using the aecond M for bothia second and third term. 10. To square a given number. RubB. Set I on B to 1 on D; ogainat the given number D will he its square on B. Example 1. What ia the square of 9! Fliice 1 on fi tu 1 on U, und against 9 on tlte lower line •£ D will be 81, the answer, on B. 2, What is the square of 731 Ans. ilB*- The s*|narB of s.viy number may he a!?o found by (Art, 7) ^ Another methud ts to 'w>(iivt Ute '£\&«n^ wdv^t (be gjna GAUGINO. 135^ number on B to the same number on A, and against 1 on A wUl be the square required on B. Note. If the given number be on the lower part of the line on the side D^ then the square thereof will consist of twice the number of figures that are in the given number ; but if it be on the upper part, then of one less than twice that number. 11. To extract the square root by the sliding-rule. Rttle. Set 1 on B to 1 on D, and against the given num- ber on B will be the required root on D» ^Example 1. What is the square root of 49? Place 1 on B to 1 on D, and opposite to 49 on B will be 7 oh D, the root required. 2. What is the square root of 13*69? Ans. 3*7. 3. What is the square root of 420-25? Ans. 20-6. 4. What is the square root of 22500? Ans. 150. Note. If the given number have 1, 3, 5, 7, &a, figures, the root will be on the upper part of the line D; but if of 2, 4, 6, 8, &e., it will be on the lower part. 12. To find a mean proportional between two given num- bers. Rule. Set one of the numbers on C to the same number on D, and opposite to the other number on C will be the re- quired proportional on D. ' Example 1» Required the mean proportional between 4 and 16. To 4 on D set 4 on C; against 16 on C is the answer, 8 on D. . 2. What is the mean proportional between *20 and 301 Ans. 24-49. 3. What is the mean proportional between 72 and 288? Ans. 144. 4. Required the mean proportional between 2-2 and 250? Ans. 23-45. 13. To cube a given quantity. Rule. Set the given number on C to 1 on D, and against the given number on D is the cube required on C. Example 1. Wh^lis the cube of 3? To 1 on D place 3 on C, and beneath 3 on D is 27, the answer on C. 2. What is the cube of 9? Ans. 729. 3. What is the cube of 7*5? Ans. 421-875. 4. What is cube of -4? Ans. -064. 14. To extract the cube root of a given number. Rule. Move the slider either way till 1 on the line D and the given number on C are opposite the same number^ wvd this number is the required root« J 3d oacging. ExAMPLB 1. What is the cube root of 343T Find 343 on C, then move the slider till 1 on D k against T 1*11 C, and 34.1 on C is against 7 on D, and 7 is the answer, ■ S. Required the cwbe root of 17S8? Ana. IZ. ■ 8. What 13 the cube root of -SI Ana. -926. I 15, To find the areas of squares, circles, paralledogiams, &qi} ellipses, by the pen and the aliding-rnle. Note. All nreaa are as soUik, having a depth of ene inch. Rdle. 1, Find the surface in inches and tenths, by the role* for the " mensuration of sarfacec," then divide or multiply by the proper diyiaor oi mnltiplier in the table, (Art. 3.) Rdlg 2. Set 1 on C to the proper gauge-point on D, anj against the given side or diameter of a square or circle on D, will he the answer required on C. RuLB 3. To the proper square or circnlar divisor on A, set one aide or diameter on B, and opposite the other side m diameter on A will be the answer on B. Note. In computing the areas of pamllelograma of ellipses \j Rule II., the mcaa proportionBl between their «des or diimetsnl muBt ba firat found by (Art. 12.) ' ExiHTLB 1. What is the area, in bushels, of a eircnlff room, whose mean diameter b 200 inches? By Rule 1. Now 2824-2003, is the tabular circular divisor, and -000354 the tabular circular multiplier for bushels. Therefore 260x280-=-2824-2903=23'93 bushels. Am. And 2e0x2«0x-000'3fi4 =23-93 do. km. By Rale II. The tabular circular gaose-poiot for bBsheli • is fi3-1441 ; then set 1 on C to 63-1441 on D, and agtunet 260 ' on D is 23-93 on C, the answer in bushels. By Rule III. The tabular circular divisor for boaheb \t ■, 2824-2903. 1 Then to 2824-2903 on A, set 2(50 on B, and opposite to 260 " on A will be 23-93 bushels on A, as required. EkavpubZ, Required the area, in gallons, cd a sqnai^, whose side is 120-7 inches. Bv Rule I, The tabular dirisor for ^usres i> 277-27^ and the multiplier is -003606, for gallons. Then 130-7x1 20-7H-277- 2 74^ 62-5 gallona. Ani. And 120-7xl2O-7X*0036O6=62-fi gallons, Ans. By Rule II. The tabular gauge-point for squares b 1R-6S1S for a gallon. Tberel'ore to 16-6515 on D sot 1 on C, and againat 120-7 on U wiU be 62-5 on C, the answer in gatlonk | By Rule III. To 277-274 on A set 120-7 on B, and ow^-- site to 120-7 on A, will be on B 52-5 gallons. Am< ExAJBPLE 3, There \a b.q e\^^^aa ■«'b«* S.raaT>-<«nie i" "**■ oAVonro. 137 is 154 inches, and its conjugate diameter 55 inches; what is its snpeifieifis in poands of plate glass? Ans. 688-68 lb. ExAitFLB 4. Giren a frame 45 inches long and 15 broad, to fmd its area in ponnds weight of hard soap, hot and cold? A«. f 24107 lb. hot. , -'^^t 24-87 lb. cold, ExAMFLB 5. How many pounds weight of flint glass are there in the area of a flint glass pot, whose mean diameter is 27*2 inches? Ans. 67*25 lb. 16. To flhd the contents of solids of greater depth than one inch. Bulb L find the content in inches by the rules for the mensuration of similmr solids, tben divide or multiply by the tabular divisor or multiplier proper to the required answer. Rule II. By the sliding-rule. When the vessel is square or circular. To the proper gauge-point on D, set the depth on C, and against the given side or diameter on D will be the answer on C. Bnt if the vessel be a parallelogram or an ^pse, find a mean proportional between the sides or the dia- meters given, and with this, as a side or diameter, proceed as before. Example 1. A brewer's cylindrical mash-tun is 144 inches in diameter and 82 inches deep; how many quarters will it hold wbenl^U! Bulb I. By the divis(»r for circles for a gallon. 144xl44xd2.^d5d-036da.l879 gallons=29 qrs., 2 bush., 7 galls. By the multiplier for circles for & gallon. 144xl44x32x*0028d2=sl879 gallons =29 qrs., 2 bush., 7 galls. Rule II. By the gauge-point for circles for a gallon. To 18*7892 on D set 82 on C, and against 144 on D is 1879 on C, which being gallons, and reduced, gives 29 qrs., 2 bush., 7 galls. Example 2. There is a square guile tun whose side is 48*7, and depth 27*4 inches; what number of gallons would fill it? Rule I. By the divisor for squares for a gallon. 48r7x48*7x27-4-j-277 274x=234-37 gallons. By the multiplier for squares for a gallon. 48-7x48-7x27'4x-003606=234*37 gallons. Rule II. By the gauge-point for squares for a gallon. To 16*6515 on D set 27*4 on C, then against 48*7 on D will be 234*37 galls, on C, the answer. Example 3. A soap franco, 45 inches long and 15 broad, is fiUed to the depth of 51*4 inches with soap 8ilicated\ h.^^ many pomid^ hot and cold, does it contaixi\ 138 GAVomo. , RULB I. n ti. 1- ■ „ i 45X1 6x51 ■4-f-24-n4.=l 443-33 lb. hot. By the divBOTB, | 45,^1 5^51 -i-i-ai' 30=1 489-05 lb. cold. m.l,».„„I.Uli.,. S45xl5x51-4x04169=1442-fBlbhol. Bj themultipliMS, ^ 4.,xl5xar4x-04292-<1489-10 lb. cold. Rdlb II. Find the mean proportional (Art, 12), which a 26 nearly. Then to 4-9031 on D, set fil-4 on C, and against 20 on D tj 14431b. hot on C. Ans. To 4-827 on B, set 51-4 on C, and a«ainat 26 on D is 1489 lb. cold on C. Ans. Example 4. Grain to the depth of 14-8 inchea ia pnt ml* an elliptical couch frame, whose transverse diameter is IIS-Z, and conjugate 111 inches; required the number of bnaheltit uontains? RiTLE.T. By the divisor for circular bushels. 140-2x111 x14-Q-h28Z4-2903=84-44 bushels. Ana. < By the multiplier for circular buahels. 145-2xl]lxU'8x'000354=84-43 bushels. Ans. RuLB II. The mean proportional between 145-2 andltl, »126'96, (Art. 12). Tlien to G3'1441 on B, set 14'8 on C, and against 126-95 oa D. U 84-43 bushels 0: " ' Example 5, IIow many pounds weight of plate glass metal will fill the frustum of a cone, whose dinmeler at the top is 2(1, and at tho base 27 inches, and whose height is •'W incheit Rule I, By the proper tabular divisor. 27'+2^+(27x20)xYH-14-387S=1160 lb. Ans. KuLB II. The mean proportional between 27 and 20 is ■" (Art. 12.) 1 to ■," on C set 3-703 on D, and opposite 20, 23-2, 27, is 278, 375, 607, on C. Total, 1160 lb. Ans. ExiucPLE 6. Required the content, in imperial bushels, of malt hopper, in the form of a truHtum of a square pyrjmtJ, and whose depth is 3(5, aide of the lower haw 14, and of the upper 34 inches! Rule I. By the proper tabular multiplier. 34+14=48; and 48*^ (34^04) xy'-WWifi 1=98-93 bushels. K II. The mean proportional between 14 and 31 ii (Art. 12.) 1 Then place y on C to 47-0977 ■ — D ia, lO'O, 26-8, (!2-5, on tliegj-eater bo'mg 75 \D\igMi4&ft\viii\^^rtQ»i OAUGINO. 139 long and 35 broad, and the depth of the solid 30 ; to find how many pounds of hard soap cold it can contain. RuLs I. By the proper tabular divisor. 75x60 » 3750 45x35 » 1575 iisx!^x4»1020q 15525 5=Y 77625-^27-14=2860 lb. Ans. Rule II. The mean proportional between 75 and 45 is 58*^9 50 35 41-8. Then to 27*14 on A set 75 on B, and at 50 on A is 138-2 on B. ^7-14 27-14 99 W 581 45 » And286xV=2866'lb." Ans'! >9 41-8 35 » 99 89-6 58-2 286 17. Gauging open vessels. All vessels which admit of having their dimensions taken from within are considered open vessels. Some of them are regular or nearlv so, others are irregular in form, but all can be reduced to the forms of solids. For the purpose of facili- tating the taking of account of quantities of goods contained, and securing correctness in calculations, many are fixed and tabulated; while the ascertained lengths, breadths, or diamcr- ters^ and the computed superficial areas of others, are record- ed in a dimension book. To fix a vessel is to determine its poeition and dipping place, to tabulate one, is to compute* and form into a table, the area at every inch and tenth of an inch of the depth. 18. To gauge a malt- cistern or couch-frame, soap-frame, ot other rectangular vessel. Note. In rectangles the shortest distance from end to end, and side to side, is the true length and breadth. Jhtle^ Measure with a proper instrument or tape, the in- ternal length, breadth, and depth of the vessel, and calculate the area in the denomination required. The areas of malt- kilns are computed according to their forms. 19. Precedent of a dimension-book for Maltster's utensils, and a Soaper's soap- frame. Mr G. H. Maltster. Ko. of Malt fitoiuo. 9 Ctotamt. I 41 HI 37-Oj ! i4 i08*5 1910 6 I a U>8'2 5-2<> 110-8 16'57 Couches. I a 27*4 leao 240 1452 5 1 111-6 111 i •<i|7-26 KUns. No. of KUn Length. Breadth. Area. ,l»t. \ \ 2nd. m. OAtroiso. Messrs J. T. & Co.'s Hard Soap-frame. -SS' :^ BrodUi. „-.^ c»,..™. 1 4fi-0 150 21-107 28-47 aeUn. Hms IM ■; large oi 20. To gauge a cylindrical vessel. Note. In circlea the greatest disCaoaa between uiy two poiiUs of flie circumfproiico is the true diaroeier. Rl'ls. Measure the diameter and perpendicular depth with- in, and caluulatu the area iu tlte capacity rerj^iiired. 21. To gauge & vessel ia the form of a &uatuin of a pyn- RuLS. Measure the sides or diameter at the top ontl bot- torn of the vessel, and the perpendicular depth; find then Nota. In measuring circnlar veasela il each end cross diametei-s, und Sod the mi fimajl vessels ore gauged in the above man and tabnlaled. 22, When the side or diameter 8,t any aswgned depth of • frustum of a pyramid or cone b required; Bulb. Divide the difference between the top and botton sides or diameters, hy the depth of the frustum, and the quo- tient will he a common factor, hy which multiply the yn- posed depth; and, if the given depth be reckoned from tlx smaller end, tht prodoct must be added to the less side o; " meter; but, if the depth be reckoned from the greater end,tt>i product must be subtracted from the greater side or diame- ter; and the sum or ditfereuce will be the diameter o: quired. KxiupLB. Given a, frustum of a cone, whose top and bot- tom diameters are 28 and 41, and height or depth 62 inches,U find tlm diameters at 7, 12, 16, 23, 31, and 44 inchea from tiK First *1—2B=1 3, di8erenca of the given diameltOT. Then 13-^-5^=-25, the cooinwn factor. F«tpr. DqWhu. Prnducli. DIamelcri. (■ 7= I'75"| (29-75\ r 7' 13= 3-00 31-00 .13 Now2ix- llZ ',Z .-.2Btbetopdiam.= f^lJU^Ul 31= 7-75 • 35-76 31 U4=ll-00 US'OOJ L4i. Were the products subtracted from the greater diameter thi remainders would he \,\ie d\e.mete'nU,\.\^«R£ae distances bun tils bottom d[ ilie fruaWm.. CAuaina. 141 gauge, fis, anil tabulate a Tessel n-lticli is nearly cir- Rtnai. Find the top diameter, which multiply by -7, and the product will be the side of the inscribed aquare nearly. By triab find the exact eides, mark the angular points, from each let fall a plumb line, mark where the plummet touches the bottom, and then, withaehalked line, strike a line each way across the bottom, and pussjng tlirough the oiipMite marka. Strilce lines up thui^dea, joining the top and bottom points, and the veaw! will be quartered. Determine the dipping place, from which let Ckll a plumb line; mark where Uie plummet touches the bottom, and from this mark take the Teasel's perpendicular depth. Divide the depth into frustums of 10 inches deep, beginning at the top, if tlie vesspl is to be gnuged for dry mches, that is, when the distance between the top of the liquor contained and the dipping place, is to he af- terwords taken to determine the quantity in the yessel, but if gauged for wet inches, which is when the actual depth of the liquor is taken, beginning at the bottom. At the middle of each frustum take cross diameters on the lines on the sides. Find the mean diameter and area of each fhtstnm. Multiply the area by the depth of the frualum for its content, and these contents added together will be the conUnt of tile whole vessel. If the bottom be nnlevet, measure water into the vessel till the bottom be just covered. Find the deepest depth, and o«er it fix the dipping place. Should there be any incumbran-ces, calculate their solid con- tent, and deduct the proper proportion from the frustum, or pan of a frustum, in which situate. To tabulate a vessel. Calculate the areas at every inch and tenth of an inch, as may be required, of each frastum, which areas, if the gauging be for dry inches, must be suceeaNvely deducted from the full content, if for wet inches, be similarly added to the quantity to cover the bottom. When the aides ol a vessel are much curved, the frustums ahouid be only 4, S, or 8 inches deep. Large vessels, and when great exactness is required, are divided into 8 equal parts at the top and bottom, by quMter- ing as above, and then dividing each quarter into two equnl parts, and striking lines between the ends of each correspond- ing diameter. In these cases there are four cross diameters, whose sum must be divided by 4 for a moon diameter, by which the area must be found. ExiUPt^ Of the gauging and tabulating to wet laches of a vessel, nearly circular, by the above rule. i Mr J. P.'s Spirit Receiver, gaiiced 12tli January 1814, by J. R. and W. B. 1 Cr.»l>l^..c~ H^n =. — ■SSSL- 1 1 1 ! ^ 1 4 ..r. n^tb i>.c. U1-* lD.ip-perinea.ure l^T.Drplh. El w^we? 74S7 1 S _ c..,„,. I«IU!« INCUMBRINCE. A beam across the top, whose length is 67'i breudth 03'), and de])th 05-2 inches, and whose ares b •7417 gallons, which mufit be deducted from the area of the first Wi inches from the top. Precedent of tabulating to the incli. WetincHei o.«.. i"hl Gallon!. lnTh"«. GMllam. Drip ,% of lowest area iowEst area 3 5 B 10 -2d ureii 11 12 21-0001)0 14 IS 16 17 18 19 20 Sdarea 21 22 23 34 25 25 ar 28 332-74488 356-86758 380-59028 404-51298 428-43668 452-36838 476-28108 22.53770 29 30 4thar«a 31 32 32-3 ,', Sth area 32-i 33 34 35 36 37 37-5 57M3H 70H6M ai-ssm 24-SSI60 65-19388 83-74448 U4-2!)G08 138-84768 163-3!)928 187-95088 212-S0248 237-05408 23-92270 723-3148 745-OSIB rsi-sMT 7SS-7S«3 766-1981 TtTiat aoe-OBW 829-0308 849-97M1 860-44784 (98-81878 521-35648 543-S94I8 566-43188 588-90958 611-50728 634-04498 656-58268 284-89948 308-8-2218 NoTJi — For the sake of cnncipenpsa tho first addition only of tif J area of each fmBtuni Le shown. Had the tabulation been to tb* tenth of BQ inch, each area niuat have beeo divided by 10, beliiK being added. From this table another is conBtrueted in whj^ MEMl pilous only are given, and in the conBtruction of whicli all bdsw- Hii-teiilhs IB rejected, ani att b,\™-^b VaV™ tn he another gallnk Thus, the content in the above «i.Uea,\,iimft \B'Siw*-«wiA,\a *iieiii- c-ond table be 2\2 gallona, tni &« co'n\oiiV »*. \^ m*** «>#^>a 35? gallons. OAUOINO. 143 24. To gauge a vessel nearly rectangular. Rtn:.E. Find the perpendicular depth; divide it into sec- tions of 10 inches or less each. Take two lengths and two breadths at the middle of each section; add the two lengths and two breadths respectively together, and divide by two for a mean length and breadth, which multiply together, and re- duce the product into the capacity given ^r the area. Mul- tiply the area by the depth of its section for the content there- of; add all these contents together, and the sum will be the content of the whole vessel. Example of a Brewer^s mash tun gauged by the above rule^ Mr R. H.'s Mash Tun, gauged 10th Feb. 1844, by T. R. and J. S. IMllS Men Length. Mem Breadth. Arru in Oalloni. Areas in Contents in 1 illOIU. .I^enguu. BraaatDs. Qri. B. Galloni. Qrs. 4 3 3 4 4 B. 4 5 7 1 2 Gallons. 13 10 10 10 30 98-6 100*4 1027 I0AH> 107*3 99*2 100-8 1031 105*(» 107*2 969 100*6 102-9 1060 107*2 636 66-1 68*0 69.6 71-8 62-9 66-0 68*4 70*1 72.0 63*2 660 68-2 698 71-9 22-5392 23*9423 25*3061 26-42a3 27*8038 • •• • •• ••• • •• 2 2 3 3 3 6 5392 79423 l*3(>ni 24283 3-8038 5*0096 7-423«l 5-0610 0-2830 6-0380 *>. Total depth. i :;ont( mt. 20 5 7*81'46 To tabulate the above, proceed as in (Art. 23), but to qrs., (ushels, gallons. • 25. To gauge and fix a rectangular back or cooler. * Rule. Measure the mean length and breadth, and find the area. Cover the bottom with fluid; take the depth in various p&ces, add the dips together ; divide the sum by the number of dips, and the quotient will give the mean depth. Try till a depth of the fluid is found, as near the side of the cooler as poedble, equal to the mean dip. Fix this as the dipping place^ snd opposite to it, on the edge* of the vessel, make a mark, and specify also the distance the dipping place is from the point marked. Measure the whole depth of the cooler; cal- culate the area to every tenth of the depth, after deducting therefrom the mean dip, and tabulate, as in (Art. 28.) Should the place of the mean dip be inconvenient, any other place may be taken, by marking its position, and showing on the vessel how much deeper or ebber it is than the mean dip, which dif- ference must be subtracted from or added to the mean dip, as the case may be. A mean dip answers only when the bottom is covered with fluid ; under other circumstances it would be better to fix the dipping place at the deepest depth. Example of the Gidng and gauging ot a Teclaxk^vjA^ax \i^'c?^ c»^ cooler. * c £ , £ 3 * i. * 144 OADOINC. Let the fignre ABCD represent m rec- tangular back or cooler, whoso mean leogth, AB, is 17s inciies, and mean breadth, CD, 124 inebes; let the deptha b« at <i (t3-6, at b Oa-3, at c 08-8, at rf 03-^ at e 033, at /"oaA at g 03'0, at A 03-2, - at ■ 02-S, and at A 03-0; to find the area and meGui depth. Now 175xl24-217'XI, which, if gallons be the denomination reqaired, being divid- ed bv 277-274, gives 78'2619 galk., or 2 barrels and 6'2GI0 rbUs. = area. Then 03-6, +O;i-3+O3'8+03'2+O3'3+03'0>O3'O +03-2+02'9+03'0=31-2 and 31-2-1-10, (the number of dipi ta- ken), =312«tnean depth. By trial a dip =3-1 ia found at 0, opposite to it is made the mark^', and it being i inchM from the side, a note to that efidct ie also marked ob Um edg» of the cooler. 26. To gauge a copper with a riung ci Rdie. With a straight edge or cord, hetd tightly acrori the mouth of the copper, aa a diameter, and with a Une * plummet, find the perpendicular distance between the tc, the crown, which should be ita centre, and the straight edge or cord; measure this distance, and it will be the depth of tha body of the copper. In the same manner measure the perpendicular distance be- tween the straight edge or cord and the lowest point of the bottom, where the crown rises from the sides. Twice the di»-' tanco that one perpendicular ia from the other, will be lb*' diameter of the splierical seement forming the crown, anil their difference in length will be the altitude of the s^men^ or rise of the crown. * Quarter the top and bottom, so s,a to touch, but not cut tiia- crowTi, and join tne corresponding points by lines, drawn with. a straight edge and lung pescil, up the sides. Divide thai depth of the body of the copper into frustums of 4, G, 8, &c»} inches deep ; take cross diameters in the middle of each fru^j tum; find the mean, and comnute its area, which raultipl|l| by the proper depth, and add the products together. Calci^i late the content of the frustum containing the crown and tbsi content of the spherical segment or crown, by the rule fiJt computing the solidity of St segment of a sphere, page 9()>i From the former subtract the latter content, add the remain-t der to the sum of the products or contents of the other ftiw, turns, and the total will be the content of the whole copperi The most expeditious and satisfactory way of finding thH content of the frustum containing the crown, ia to ^i^^g the crown with liquor, and then draw it off into ft propelj I'he tabulation. m\i.a\.^)e\a'\iaxit\'Q,&[<£ixa, Bud galloni) a 0At7QING. 145 for dry inches; hence the areas xnttst be successirely deducted from tile fall content. Example 6f a copper With a rising crown. Let the figure ABCD repreeent a copper with a rini^ crown, to he ^ ganged and in<diedy and whose dimensions, ta- ken by the above rale, are depths BF 48 and GC 54 inches; cross diameters, at 6 inches from the top, OO'O and 90*1; at 17 inches, 85*3 and 85*5; at 26 inches, 89-2 and dOl ; at 83 inches, 76^4 and 75'8; and 39 inches, 70-8 and 71*4; and also at 45 inches, 67*4 imd 67*4 inches; and let the cross diameters at the top of the crown bead's and 64*3; and at the bottom of the crown 59*8 and B0*2 inches. j^QYT <»'8H-g*'3 «64*0» mean diameter at the top of the crown; and ^'^^^'^ «a60*0=mean diameter at the base of thecrown. Then 642+602+(64x60)x 5^, the height of the crown, X •002832=65*3399 gallons = the content of frustum CFDH. Again, Y=30= radius of base of the crown, and 64 — 48=6 ssheight of the crown; whence (3x30^+62)6x*6236x-003606= 30*9950 gallons = content of the crown, page 90. Therefore 65*8399— 30*9950 Bd4'3449 gallons required to cover the crown. Form of Table for Dimension-book. A. B.'s copper, gauged 16th March 1844. f 19 10 8 6 6 6 48 lOCllM fifom tha top. 6 17 86 33 39 45 Cron Obunetera. Mean Dlaine> ten. ft^'OWl 85-3 85*5 80S dOl 76«4 75-8 70-8 71 '4 e7'4l67-4 To cover th« Crown. Deeth and Content 90- 85 80< 76 71 67 Area* In OaUoos. 22 9438 20-6583 18'1738 16-4(i40 14-3192 12*8676 34-3449 Areas in B. 2 2 2 1 1 1 Gallons. 4*9438 26583 01738 7-4040 5-3192 3-8676 Content in B. F. Gallons. 7 5 4 2 2 2 25 2 2 2 1 3 2 5-3256 8*5830 1*3904 8-4240 4-9152 5-2056 7-3449 6-1887 27. To gauge a copper with a falling crown. Rule. Find, as in (Art. 26), the perpendicular depth, from the straight edge to the angle where the body and crown of the copper are joined. Quarter, and take cross diameters ; find the area and content of each frustum ; the content of the crown^ by measurement or calculation; add tYi^-wVc^a <wcl- tents together, and tabulate, as in the piecedvn^ ^tVa!(^<^« 1* lo gan^ a Teasel \rith a fall or drip, 'hen a vessel U in an indiaed poaition, tlie quantity d oesMry to cover the bottom is aaid to be the fall or drip. RuLB. Measure into the veasel water sufficient to cov the bottom, or ascertain the quantity required, by the rules for finding the solid contents of ungulas. Find the greatnt depth of thu fall or drip, to which let fall a plumb line from a point level with the lower side of the top of the vessel; deduct the depth of the drip, and the remainder will be th« depth of the vesael, as positioued. Divide this depth inti proper segments, of which find the areas and contents, ae cording to (Art. 23), and the sum of the contents, added to the quantity for the drip, will be the whole content of the vessel in its inclined state. Tabulate as before. Should the fixing of the dipping place over the greatrrt depth of the drip or iall be inconvenient, any other point ma]' be taken, if the difference in height between the two paints be added to or subtracted from the depth, as the case may U. 29, To gauge and tabulate an irregular oral vessel, bj means of ordinates. Let the two following figures represent the hases,OT battoiu ' 'wo irregular oval V' Fig. 1. GAUOIKO. 147 To find the centre, and to quarter the base, of an irregu- lar elliptical vessel. Draw any lines, as ab and cd^ Figure 1, parallei io one another; bisect them in e and/; through e and /draw the line gh^ the bisection of which is O, the centre of the vesseL With as a centre, and any distance, cut the periphery in h and /, which join, and bisect kl in m. Through the point m and the centre draw AB, which is the trans- verse diameter. Through the centre O, and perpendicular to AB, draw CD, the conjugate diameter ; then the distances AD, DB, BC, and CA, win be equal to one another, and the figure is quartered. To draw the ordinates. Parallel to AB, the transverse diameter, Figure 2, draw MN, cutting CD, the conjugate diameter in H. On the lines AB and MN, and on either side of H and 0, the centre of the vessel, set off any even number of equidistant points, as Jl, 1'; 2, 2'; 3, 3'; 4, 4'; 6, 6'; 6, 6'; 7, 7'; 8, 8'; 9, 9'; 10, 10'; the sum of all the distances of which shall be less than the transverse diameter at the top of the vessel. Through these points draw lines terminating both ways in the periphery, and they will be the ordinates. To draw lines from the ends of the ordinates up the sides of .the vessel* Fix a string or straight edge across the top of the vessel, so that it shall be exactly over the respective ordinate on the bottom, which will be the case when the plummet of a plumb- line, ffidling from any two distant points in the string or Btraight edge, shall touch the ordinate. Then with a chalked cord strike a line up the sides between the ends of the ordi- nates and the points on the top cut by the string or straight edge. Draw such lines from the ends of all the ordinates, and of the transverse and conjugate diameters. Fix and mark the proper dipping place, (Art. 23,) and ascer- tun the perpendicular depth of the vessel at that point. Divide the depth into proper frustums of 2, 4, 6, 8, or 10 inches, beginning from the top and leaving the odd inches in the bottom frustum, if the tabulation be for dry inches, but if for wet, beginning at the bottom and leaving the odd inches in the top frustum. Take the ordinates in the middle of each frustum, except there be a curb or cover through which the gauges must be taken, and if the tabling be for dry inches, then deduct the thickness from the top frustum, and take the or- dinate in the middle of the difference. Should the sides of the vessel be other than perpendicular, the points for the ordinates must be marked, so that a line drawn between any two correlative points, will cross the per- pendicular depth at right angles, and at proper distances from the top. To find ihegroBB area and contant of eacViitvx^Wxsu^ Its Rule, To the snm of the first an J Inst ordinateB, adil fiinr tiniea the aum of the even ordinatei, and doable the a um of i the W9t; then to the product nf the total -multipUad by the eqnidiatance of the ordinates, add the prodnct or the aiiiti of the extreme ordinatea, multiplied by the sum of the segments, that is, the difference between the Ivanaveifle diameter and tlie part intiTcepted by the cxlrame ordinates. One-third nt tliis Bum divided or multiplied hy the proper divisor or factor for B(|iiare8, will be the area in gallons, which multiplied by llu height of the frustum will give the content thereof. Measure the quantity required to cover the bottom, to whitli add the contents of the several frustums, and the si the content of the wliole vessel. Find the depth of the drip, subtract it from the gross deptli. the remainder, deducting the curb as before directed, will be the net depth to be tflhulated. When there are incumhTsnces, compute their areas and va- tents, which deduct from the areas and cuntenta of the r ~ tire frustum, KxAHFi.B of the ganging- of an irregular vessel. Suppose Figure 2, page 141, to repreafat the base of ai guiar ovul vessel, on whieli 11 euuidistant urdinates have bcA taken, and also in the middle of every frustum, according t< rule, and that the dinieunlons, areas, and incumbrances, an I in the following scheme; — ate|?.J|:!|«.. --'blfP 77-" '»■»«■» m-smtI Incumbrances. Two iieanis inside the lop, each 80 i lonit, 04 broad, and Ofi'2 deep. Area 2'4809 gallona, deducted from the first 06'2 inches from the top. 1i "*f GAUGING. 149 leraiionsfor finding ihe areas for the several frustums. FnutuoM. 1 1 8 3 4 5 6 '»«'»[£« 24-7 SI '3 253 84*0 25*6 86*5 27*2 88-6 3(H} 3*7 317 827 eme ordinate*. 46-0 49*3 52-1 55*8 1 007 64*4 . Second V Fourth jcs, < Sixth 4 Eighth ^ Tenth 47-2 896 760 7«'-5 490 48-6 7«»*4 76-7 71*1 49-5 ■ m 71*0 77-1 71*5 50*9 51-4 71-6 775 72-2 51*6 52*5 ■ 72-8 780 73-0 53*2 72*9 78*3 73-5 54*1 ordlnates. 318-3 4 316*5 4 320*4 4 324-3 4 328*9 4 302*4 4 ordin. x 4 =s 1249.2 1266*0 1281*6 1297-2 13756 1329*6 e Third ) Fifth * 1 Seventh C Ninth (Ji-4 74-8 751 62-1 0^*2 75*3 75-7 62*9 62-8 75-6 76 63-5 63« 76-2 !76*6 64*4 64 3 767 770 66*3 65*4 77-2 77*3 66*0 irdinates 27S4 2 2761 2 2779 2 880*8 8 283-3 8 285-9 2 wd. X2s=» do. x4s do. 546*8 1249*2 46-0 552-2 12660 49-3 555*8 1281-6 52*1 561-6 1297*2 55*8 566*6 1315-6 607 571-8 1329*6 64*4 1842*0 1867*5 1889-5 1914*6 1948-9 1965-8 lie equidistance. 18420* 18676. 18895- 19145- 19499- 19658- nneordnates egments 460 3-4 49*3 4-8 5-8 i^*l^ 6-8 6)7 8*1 64-4 9-2 rd.X8umofseg. 156*40 18420- 236*64 18675. 30218 18895- 379-44 19146- 491-67 19429- 592-48 19658- 18576-40 18911*64 19197*18 19525*44 19920^ 20250-48 c inches 6192-1333 6303*8800 6399-0600 6508*4800 6640-2233 675a-l6'J(J 77^41 ans 3 nbrances 22-a321 2-48U9 227362 230784 23-4730 23-9482 24-3447 cum. deducted 19*8512 227352 23-0784 23*4730 23-9482 [24-3447 5»r wet inches tabulate as in (Art. 23); if for dry, as in 28), only commence with 1*9 the depth of the curb instead ill." To gauge a still. Is are of various forms. Those in common use consist lead and body. The head is generally a conical frus- the body a copper with a rising crown, and covered I spherical segment. le stills are too small to admit of the internal dimen- being taken easily and accurately. Such may be gauged dng the external dimensions, and dedxiclvsY^ \X\«tAxw«v ickuess of the copper or metal of wMch. TCia^"fc\ \s>qX ^^^^ \d moat satisfactory method would \)e to ft\V ^^^^^^ Vv^^^ nd draw it off into a given measure. ISO I A usma. To gniige those which admit of their dimensions being ia- ktii interiiully. Role. From the point where the glohular part of ths hody heRiaa, extend a, line na a diameter; measure the perpeo- liicular dlBlance between the centre of the crown nnd the top of the glohaler part, which will be the whole height otlhe atill ; from this deduct the depth from the diameter tfrtht trrown, and the remainder will he the hei^lit, and the diame- ter the huge of the g-lobular portion. The part below tha diameter will be a copper with a rising crown, and ita dimen- gione mnst be taken and content found by (Art. 26). The glo- bular part wilt be n spherical zone, and its content must ba found liy the " Rule for meaanring the aolidity of a zone of a aphfre, pai^'e 81>. The Bum of the contents thua found will he the whole content of the body of the still. To gauge a still head, Rvi.B. If it be of an irre^lar form, divide it into patti, whose foriiiB are regular. Take the internal dimenaions, if possible, of each ; if not, the external dimensions, and allow for the thickness of the melal, Compute the contents by tlie " Rule for the Mensuration of Solids" whose figures are simi- lar. Add the contents obtained to the content of the whnle of the body, and the sum will be the full content of the »tiU. EicAHrLB, Suppose the annexed figure to representa^til!; Athe head, a frustum of n I'one; B the breast, a ifone of s sphere; and C the body, a copper with a rising crown; and the diuensiona to be as follows: of the head ab, the height 2ti inches, diameters at 1C\ tapJfflV& and 12-1, ' at bottom lij 23-8 and 24'2; depth of breast br 13'0j diameters at A; 238 and 24-2, at A/ 6!)-« and 60-4; depth ' of body 27-0, diame- tt-rs at 4'6 from top, c, 5(i-8 and 6r'2, at 13-5 from c, 50-0 and 60-4, and at 22-8 from c, 45 '3 and 45'1; depth of rrown, de, 4, diameters at top, m". '" " and 43-2, and at bottom, op, 39-7 and 40-4 incbea, to find hole content. ^^the whi =12-0, ■ ^!i?:r-*JL =niVQ,>C&stt GAUGIKO. 151 12«+24^+.(12x24)VX*002832-21*885e96 gaUs., the content of the head. 2d, ^^ -12-0, and ^^^ -30-0, then 12a+308+ *^ xl3x2x-002832=81-019744 gaUs., the content of the hreast, 3d, 55f±*?:?=,57.0; then 572x-002832x9«82;81051 2= con- tent of Ist frustum. ^^°*^ - «60-2; then 50-22x-002832x9ifc64-230779=: 2d, and^^^^*^«46-2; then 45-22x-002832x9«52W3003= 3d, 199*114294 galls, the content of the hody. 4th, ^!l±i51=43-0, and ¥1±^=400; then432+402+(43x40)| x*002832sil9'518144 gallons, the content of the frustum con* taming the crown; and »'7+^-^ ^20; whence (202x3+42)4 x*5236x*003606» 9-183726 galls., the content of the crown; therefore 19*518144— 9-183726=10*334418 galls, to cover the crown * conseouentlv 21-885696 + 81-019744+199114292+10-334418=312-354160 galls., whole content. These dimensions, areas, and contents, should he formed into a tahle, as in (Art. 26), and inserted in the dimension-hook. When stills are gauged with a view to tahulating, the height or depth from the top of the hreast to the top of the crown is divided into frustums of three or four inches deep, and cross diameters are taken in the middle of each, and the areas and contents are calculated as for cylinders. Malt Gauoing. Malt is made from harley, grain, or pulse, which must he covered with water for at least 40 hours in the cistern, and then removed to a couch-frame, where it must remain for 26 hours; afterwards it is put and wrought upon the floor until ready for drying upon the kiln. During the manufacture, gauges must be taken of it in cistern, couch, on floor, and oc- casionally on kiln ; and that gauge which shows the greatest net quantity is set forward, and charged with duty. » While in cistern or couch, one hundred bushels by gauge are estimated as 81-5, and while on floor or kiln, as 50 bushels net. Cisterns and couch-frames must be constructed with their sides at right angles to one another and to the bottom. Cis* terns are not to exceed 40, and couclviiaixi^^ ^ vclOcv^ vol depth. GJlVGISG. To gauge malt in ciatetn, coucii-frame, or kiln. ItDi.B. Take 5 or 10 dips of the malt, odd them together, and divide the suiu by tlie numtipr of dipa, for a meaii depth, by which multiply the arsa of the vessel, preTionsly comput- ed, for the quajitity contained. ExAUPLB 1. How many bnahels of malt are there in a cl»> tern, the area of which is 6'67 bushels, and the dips 26-4, 2S'3, a4'9, 28-0, 26'8 inches. Now 26-4+25'3+24-9+'28-0+26-8=]31-4, which -^3, tha number of dips, gives 36'2. This multiplied by 6' 57, produces 172'13 bushels. Ana. By the sliding-nile. Place 6-67 on B to 1 on A, then beneath 20-2 on A is 17S-I3 onB. Ana. ExiMPLE 2. The aren of b couch-frame is 8"11 bushels, and the dips of the malt therein 22-0, 21'3, 21-9, 22-4, 23-0, 21-^ 22'0, 22'2, 23'1, and 22'5 inches; how many bushels does it hold? Now (22'O+81-8+2]-9+22-4+23'0+21-7+22'O+2E-2+23'l +22'fi)H-10, the number of dips is 22-2; fhia multiplied by 8-11, gives 180'04 bushels, Ans. By the sliding-rule. To 8-11 on B set 1 on A, and against 22-2 on A is 180'04 on B. Ana. 32. To gauge malt on floor. Rule. Find the depths in various places, as in (31), and compute the mean depth. Measure nith a tape the length and breadth of the malt on floor. Should the sides or ends be slant, take the dimensions from the upper edge of one side or cud to the lower edge of the other. When tile fonn of the floor is irregular, divide into regular forms, measure tlie diniensiona, and calcnlale tiie content of each separately; add the contents together for the whole con- tent. E.'CAMPLB 1. There is a Sooe of malt whose length is 450, breadth 372, and mean depth (U-3 inches; what number of bushels is there (herein? (460x372x04-2)h-2218-1&2, the divisor for bushels, t»31Q-9 bushels. Answer. By the rule. To 4-2 on D set 4S0 on C, and against S72 on A is 316'9on C. Ans. Example 2. An irregular- formed floor of malt is divided into two pieces, of wh'ic^ owe laftX-^itW mean depth, 278 tliB fcreadfh, and 304 inchea Uic \ew¥jX^', ^^\^n.■0l\'6■^■\axBwsa^^!^ OSl, tcngth 553, aMbteaAV^ iflaw^eaatt-, t<<Piiv-H&,'^'-^-^- eonteut of the floor inbusWVat GAUGING. 153 Noiir (8©4x2r^08^2)-5-2218«192t« 121-9 And («6dxl08x05-l)-^2218-192s 137*3 Tota], 259*2 bushelB. Ans. By the hiding rule, let, To 3-2 on D set 304 on C, and against 278 on A is 1 21*9 on C 2d, 6-1 „ „ 553 •„ „ 108 „ 137*3 Total, 259*2 Ans. 33. To reduce the quantity of malt injone stage of operation to the quantity in another stage. Ist, To reduce cistern or couch bushels to net. Rui.E. l^fultiply the gauged quantity by 81*5^ and divided the product by 100. %, To reduce floor or kiln bushels to net. Rttijb. Multiply the gauged quantity by 50, and divide the prodact by 100 ; or divide the gauged quantity by 2, the quotient will be in either case the anstver. 3. To reduce dry barley to floor or kiln bushels. RuuL Multiply the given quantity by 2. 4. To reduce dry barley to cistern or couch bushels. Rule. Multiply the given quantity by 1*227. 5. To reduce floor or kiln bushels to cistern or couch bushels. BuitS. Multiply tlie given quantity by -CI 35. 6. To reduce cistern or couch bushels to floor or kiln bushels. Rule. Multiply the given quantity by 1*63. ExAUPLB 1. Given 250 bushels of malt in couch-frame, to determine the con'esponding quantity in floor and net bushels. Now 250x1 •63=407*5, the floor bushels required. And 250x81•5-^100=203•75, the net bushels required. ExAMP&B 2. A circular kiln, whose area is 23*93 bushels, is covered with malt, of which the mean depth is 7*5 inches ; required the quantity net, and corresponding cistern bushels? Now 23-93x7*5«179*4 bushels on the kiln. Then 179.4-i-2=89*7, the net bushels, and 179*4x-6135«110-0 bushels in cister% as required. Cask Gauging. % Casks, from the difl^erences in their foTm^,\\«bNfe,^«t 'Ccv'Sk mrpoae of ganging, been divided into four Nat\^\.\ft^, «Jgt'fc^^^^5^ /rith the solids to which their curvatures Yiov^ W\^ <:i\ft^^^"^ ^^' 1S4 eemTilance. Those which are much curred between Ibe middle and ends, are considered as the middle frustum of « spheroid, and are called the first Tariety; those which are cor- respondingly less curved, aa the middle zone of a parabolio spindle, the second variety; those whose similar parts tn very slightly curved, as two equal frustums of a parabulic conoid joined togetlier at the greater ends, the tliird va- riety; and those which are straight betweeu the middle and the ends, as two equal fjiistume of a cone joiued together St the greHter ends, tlie fourth variety. 35. Of instruments used in cask gauging. These are the diagonal rod, head-rod, bnng-rod, and OOH and long callipers; they are adapted to casks of the 1st varirtj. The diogonul rod is i feet long and '4 of an inch square, nod folds by joints to 12 inches in length. On one aide is a diago- nal line for from 1 to 240 gallons; on the upper edee of U» aide, from left to riglit, is a line for ullaging a half-hogsheld lying; and inversely, from right to left is a line for nllaeiDg a half-hogshead standing. The second side is divided into inchot and tenths. On the third side are two lines for nllaging a barrel lying and standing, two for a hogshead lying and ataod- ing, and reversely, two for a kilderkin lying and standing, and one for a firkin standing. The fourth aide ismarked with tw* lines for ullsging a puncheon lying and standing, two for a bolt lying and atunding, and reversely, one for a firkin lying ""■' principal line is the diagonal line, and is formed thus: I _ periment, a cask containing 144 gallons has a diagonal of 40 inches or very nearly, hence 144 is marked on the rod againsl 40 inches; then since the contents of casks are as the cubes of their diagonals, as 144 : 40^ : : any other diagonal : cnbeof the quantity. Tlie head-rod oonsista of two parallel parts, joined at tk ends, and a slider which moves between them and has t*a f^ces. On the end of one face is tixed a piece of brass crooked in the form of _|~ which projects one inch; the lower edge of this face is divided into inches and tcntha, reading from tb* right, on the upper edge is the line of numbers com men cio^at 25 on the left end; the slider, on this face, has attached anin- dex of brass, fixed perpendicnlarly, from which to the right end Is continued the line of inches from the rod; on the Wl of the brass are two scales, one marked spheroid, the other a line of inches; on the upper edge of the slide is the line i>f numbers, having a double radiua, and beginning at IS'TSKV tlie gnuge point for circles. On the upper part of the rod ani on the slide, of the other fuce, are lines of numbers, and on the lower part of the rod the line for uliaging, marked The bang-rod is hnlf-an-inch square. Two sides are diTidcJ | into inches and tenths; another marked, imperial area, ebo** I the area in gallons wHea B^^^i^>l \ii ^-Va &\B.uu;t«r of a circN; GAUGIRO. 155 ind the fourth expresses, when measuring the diagonal from lie lowest point of the head to the middle of the bung, the sontent of the cask. The cross callipers are two rods, sliding on one another, and laving at one end of each a perpendicular leg; the sliding parts lie divided into inches so as to show the distance the 1^ are ipart. The long callipers are similar to the cross callipers, but with horter legs and returned at right angles. 36. In finding the content of a cask by the diagonal rod; — »lace the rod so that its bevelled end shall touch the lowest loint of one head of the cask, and observe the number on the liagonal line cut hv the centre of the bung; turn the rod to he other head and repeat the same; the mean of these two [oantities or numbers will he the content sougl^. 37* When the content is to be found by the callipers, the Messary dimensions are the head diameter, the bung diame- er, and the length. To find the head diameter. Put the crooked brass at one nd of the head rod into the chimb of the cask, move the Uder, till the piece of brass or index on it touches a point f p the opposite chimb, and against the index will be the dia- leter in the lower part of the rod, except the index is with- at the rod, when the end of the rod will be against the dia- leter on the slider. Cross diameters must be thus taken at Oth ends of the cask, a mean be found between each pair, and x>m these a third mean Ji>e taken, which will be the head dia- leter* To find the bung diameter. Place the bun^ rod perpendi- nlarly in the bung hole, observe the dimension cut thereon Y the inner edge of the hole or stave, for the vertical diame- t; lay the cross callipers with the legs downwards, over the liddle of the cask, which is generally the centre of the bung- ole, expand or contract the callipers till the legs touch easily )th sides of the cask, deduct two inches, or more or less as le thickness of the stEives may warrant, &om the dimension town on the callipers, the remainder will be the horizontal ameter, and the mean, between it and the vertical diameter, ill be the bung diameter. To find the length. Apply the long callipers along the cask • as to touch both ends; do this in several places, the mean f the dimensions shown by the callipers will be the length. Should the cask be of a variety other than the first, it is istomary to make an allowance in the length, according to LB form. 38. To compute the content of a cask by the head rod. Set the index on the slide, to the head diameter on the wer part of the rod, on which find the bung diameter, and ark the number it cuts* on the scale marked ^\|\ie\ orot. \^^ isa dtde. Applj this number to tlie lower scale on the slidr, and beneath it on the rod is the mpari diameter. Fix the leli end of the sUde to the lensth of the cask on the upper part of tlie tod; find the mean diamelci- on the upper Ime on the elide, and over it, on the up|>er pArt of the rod, is the content iu gallons. Note. Wlien nnj dimension for finding the content of n out bj the head rod is tliroira off tlie role, work with hslt thnt dimenaon, Uid doubli! the answer. If the Begtnent or inean dLainet«r lie throivn off, double titc mean disnieter, and tahe \ the answer. If the mean diameter and length be thrown off, donUl both,' and take ^ the answer. 39. To compute the cojitent of a cask by the sliding-nil«. Subtract tlie head from the bung diameter, and inuttipl; the difference, when it is Add the product t-o the head diameter, the sum will bellu mean diameter. Then act thelen^h on B tn the gauge-p<nnl UK D, and against the mean -diami^r on D will be the conteiil onB. 40. To compute the content of a cask by the pen. Find the luean diameter by (Art. 3!)), multiply its aqiure by the proper tabular mulliplier, and this product multiplitd hy the length will give the content. Example I. It is required to find the content of a caalt of the first variety, whose dimenaiona are, head diameter W'Z, bung diameter S3-0, and length 35'G incliea. By the head rod, "Set the index on the slide to 282 on the lower part of thft rod, and over 83-0 on tl>e same line, on the scale marked spliF- roid ia S'3; this applied to the lower scale on the slide, cutt on the lower part of the rod 31-6, which is the mean dian»t«r. Place the left end of the slide to 3fl-6 on the upper part of U« rod, and over 31 '5 on the upper line of the slider, atsodien tlie rod 100 gallons, the content. By the sliding-mle. First, Sg-O—Sa-Z^J-O, and4-8x-68=3-a; hence 283+ 8^3- 31-fi, mean diameter. Then set 35'6 on B to IS-'USS on D, and at 31'& on D cute on B 100, the content in galloua. By the pen. The mean diameter is 31'5; then »J-Jx31-5x-O02832x36G— ACfe^SiieeSiwoft^'OQtwBUnt GAUGIKO. 157 ExAMPLR 2. Let the head diameter of a cask 1)e 22*7, the mng diameter 81*3, and the length 50 inches; required its intent? By the head rod. Set the index to 22*79 ^^^ against 31*3 will he 6 on the scale ipd.) which found on the lower scale, b against 28*7 on the od. Then place the end of the slide to 50, on the upper part I the rod, and against 28*7 on the slide will be 116^ gallons^ he content. By the sliding-rule. Suppose the cask to be of the second variety; then 31.3—22*7 «-6, and 8*6 x'64— 5-5. Again, 22*7x5*5 «»28-2, he mean diameter. Now set 50 .on B to 18*7892 on D, and beneath 28*2 is 112*6 ^Is., content. By the pen, The wiean diameter is 28*2 ; then 28*2x28*2x*002832x50*0— 112*6 gallons, content. Ullagtno. 41. A cask is on ullage when the liquor it contains does not 1^ ity^ and to compute the quantity contained is uUaging. The d^pth of the liquor is called wet inches, the remainder »f the bung diameter or length dry inches. A cask on its jde is said to be lying, one on its end standing. The dimensions necessary for finding the ullage of a cask ire^ the bung diameter, if the cask be lying, the length, if it be landing, and the wet inches, together with the content. Should the content not be given, it must be found hy the »receding rules. If it be fifiven or ascertained, and the cask be standings neasure with a bung rod, or some other proper instrument, hrough a hole in one end, the length, and the inches on the od, wet with the liquor, will be the wet inches. If the cask « lyiuff, measure the vertical bung diameter, and the inches ret wiu be the wet inches, for all practical purposes. 42. To find the ullage of a lying cask by the head-rod, slidr ng-mle, or ullage-rule. Set the bung diameter on C to 100 on the line marked SL, nd against the wet inches on C will be the segment or mean rea on the line marked SL. Then set the content on B to 1 n A, and beneath the mean diameter or segment on A will be he ullage quantity on B. Note. When the wet inches are less than ^'^ of the bong diame- )r, the segment or mean area is found on the upper part of the line L, on the first form of Rule (5); and on a line marked Seg. Ly, omediately beneath the line marked SL, on the second form of Ittle (5). II qnot 43. To ull^e 8 lying cask by the pen. Divide the wet inches by the bung diametev ; if th Snotient be lees than '500, gulitract ^ of the difference frn le quotient; if the quntient exceed '500, add 4 of the diSt laotient be less than '500, gulitract ^ of the difference Ment; if the quotient exceed '500, add + of the ^i the quotient; the remainder or aum nmrtiplied by content will give the ullage quantity. ExAUPLB. The content of a lying cask is 124 gallon^ bnng diameter 3<!'0, and the wet 24'8 inches; how gaUona does it holdl Cy the Blidiug-rule, -} Set m-i) on C to 100 on SL, and against 24-8 on C i . I SL, the segment; tlieii set 124 on B to 1 on A, and below on A is 03 galls. Ana. on B. By the pen, 2i-fi-^m=-SdO; from '630— '500— '190, which -^4='04I 1. ,.,„..., ^^^P The rule by tite pen m ubvioualy only an approxlmatioa ^^^B 44. To 6ud the ullage of a standing cask by the sliding ^^^^ Ullu^e rule. ^^^■^ Use the side or line marked SS., instead of the aide or ' ^^^B , marked SL., and proceed aa in ullaging a lying caek. (Art. NotG. When the wet inches are Icfs tlion ,', of the lengtb, , segmeat or mean area is roaud on the uiiper pai-t af the line SS| the first fol-m of Rule [5); and on a line marked Seg. St., in diately below the lice mai'ked SS., on (he second form of Rule (j 45. To find the ullage of a standing cask by the pen. HnLK, Divide tile wet inehea by the length of the cA) the quotient be under '500, subtract ,'„ of the diHerence . the quotient ; if above 'SOO, add ,'o of the difference ta quotient. The remainder, or sum multiplied by the ' of the cask, will give the ullage quantity, E^jtHFLB, Find the ullage quantity in a standing t ^ whose content is 120 gellons, length 43-0, and wet By the sliding-rule. Set 430 on C to 100 on SS, and beneath 17'R on C i« an SSi then place 120 on B to 1 on A, and beneath 4«>Di I will be on B 48 gallons ullage. I ^^^^ Now 17-6-f-43'a=-401), and -500— ■409=-O9t. Tht* ^ ^^■'^lO^'OOQl, and-409— '00t)='400, ttnd-400»cl20= ^^^V the ullage quantity. GAUGING. 159 Miscellaneous questions. 1. Bequired the divisor, multiplier, and gauge point for a Done of nint glass. Ans. Divisor 33, multiplier '030303, gauge point, 6*7445. 2. What are the divisor, multiplier, and gauge point for a sentagon in bushelsl Ans. Divisor 1289-2889, multiplier '0007756, gauge point 35-9067. 3. What is the circular gauge-point for gallons, when the niddle area of a regular frustum is taken? Ans. 46*024. • 4. Multiply 74 by '027, by the sliding rule? Ans. 1-998. 5. A malt couch is 7*44 bushels in area, and mean depth 18*3 inches; required the content in bushels, and in net malt I . „ f 136-15 bush. ^^•\ 110*96 net. 6. What is the quantity of malt on a floor whose length is f66, breadth 164, and depth 03*1 inches. Ans. 175*5 bush. 7. A soap-frame 45 long and 15 inches broad, has in it hot ilieated soap to the depth of 61*4 inches, — how many pounds loes it contain? Ans. 1443*2 lb. 8* Given a plate glass pot whose depth is 21 inches, and MBS diameters at 3*5 from the mouth, are 24*2 and 23*9; at 10*5 ore 25-3 and 25*5; and at 17-5 are 27*4 and 28*1; to de- mnine the pounds weight of metal it will holdl Ans. 967-44 lb. Note. Plate glass pots are tabled for drj inches. 9. A floor of malt is found to be 250 bushels, and its cor- ^ponding couch was 158*8 bushels; from which would the Inty charge arise, and upon how many bushels net? Ans. Couch; 129*42 bush. net. 10. The length of a cask of the first variety is 43*0, bung 6*5, and head 28*2 inches; how many gallons is its content! Ans. by the rule, 141* I ,, by the pen, 140*7 J ^*^^* 11. Given a cask whose content is 108 gallons, bung diame- er 33*5, and wet 24-1 inches, to find the ullage quantityi Ans. By the rule, 84*5 ) ,, By the pen, 83-6 J ^^^^^ .12. Divide 228 by 67 on the sliding rule? Ans. 4. 13. There is a mash tun whose top diameter is 70, bottom liameter 50*4, and depth 40*0 inches; required the content, ad tabulation at each inch, in qrs., bushels, and gallons! Ans. Content, 6 qrs. 3 bush. 6 galls. 14. What is the area in gallons, of an ellipse whose diame- ers are 72 and 60 inches? Ans. 10*1952 galls. 15. Required the content in bushels, of a frustum of a one, of which the top diameter is 24^ the bottom d\ax£ke\>^x^) nd the depth 53 Inches? Aus. VI 'W WsJto^ ■ . depti GAVaiKG. IS. The length of a cooler is 212, the breadth 148, and the . deptba at different places, 3-8, 3-6, 4'4, 4-4, 3-9, i% 3% i% 4'3, and 3'9 inches; how many gsllons does it contain? Ans. 452-63 giil, . 17. A copper with a rising crown has to be inched and ta- bulated, for dry inulies, in barrels, firkins, and gallons, the dimensions of which are — depths 48, frotn the centre of th« crown, and S4 froTa tile rising of tlie crown to the top of the vessel, the cross diaincters at 6 trom the top 7B'0 and 74'6, at 17, 74-8 and 74-4, at 26, 69'6 and 69-0, at 33, 66-6 and 65-S, it 39, 61-6 and 61-2, and at 45, 57-1 and 57-2 inchea, and Ui« quantity to cover the crown. 22 gallons: wliat is the conlent of the copper? Ana, IS bar, 2 firk, 5'04 galli. 18. Of a standing cask the length is 40-0, wet 19-0 invlu^ and the content 207 gallons; how much doea it conlaiii? Ans. 87 i galls. 19. The depth of a flint glass pot is lfl-5 and its mean dia- meter 2^3 inches; find tlie area and content, Ans. Area 8i-273G Ih,, conti-nt 1503-5G1S1.. 20. Suppose the dimensions of a guile tun, in the form of Uie fhistum of a square pyramid, to be depth 30 inches; the length of a side B inches from the bottom, 32'5, at IS inchee, 9-J, and at 26 inches, 42'5 inches; it is required to find the ai in gallons of each irnstom. Ans. 1. 3'8094 galbni 2. B-34S8 do. 3. 6-5143 do. 21. Required tlie content of a cask whose head diameter ii 17-4, bnng diameter 19-6, aad kngth 23-7 mches? Ans. 24 veariy. 22. Ilowmany bushels of malt are there in a floor, of whieh the length is 863, breadth 28S,aud depth 1-7 inches? Ans. 188-5 biuli. 23. What quantity of hard soap, hnt, 19 coiitsined in a cy linder, whose diameter is 3G'5 and depth 71'4 inches? Ans, 2667-81 lb. 24. Let the altitude of tile globular part oFa still be S inchei, and the cross- dial nete is at the top and bottom be 27'2 and 26'8, snd 55'2, and M'8; aUo the cross diameters oFthe b«df of the still at 4-5 inchea down be 59-8 and GO-2; at IS-fiincbH fl3'8 and 64-4; at 22-5 inches 64-0 and G4'G, and at 32-5 inclw 62-0 and 02-4 inches; and let the quantity required to cover the crown be 3fi gallons; required the content in callonsl Ans. 5iM)-S6Sgall>. GENERAL EXERCISES. 1 the difference between tlie areas of two rect- ■npnlar fields, the one 560 links bj 426, and the other 280 Jinks by 213? Ana. I nc. 3 ro. 7 272 poles. 2. The aides of three squares are 12, J5, and 16 feet tespecllTely; find the side of a sq^uare that is equal in area to all the three? Ans. 25 feet. 3. If the side of an equilateral triangle be 99 feet; re- l]Uired the side of another, whose area aha!] be one-ninth of the former? Ans. 33 feet. 4. Find the sum of the areas of the two equilateral tri- mgles mentioned in the last question ? Ans. 523-94573 square yardi, 5. How many yards of paper Tvill be required to line a ihest that is 5 feet 3 inches lon|T, 3 feet 2 inches wide, and I feet 6 iiicbeB deep, the paper being Iti inches broad ? Ans. IBi yaids, 6. What quantity of canvass, yard wide, will be required ] cover the convex surface of a conical tent, the diameter f the base being 18 feet, and the perpendicular height in lie centre 12 feel? Ans. 47-l24yard8. 7. What are the three sides of a right-angled triangle, hose aides about the right angle are to one another as 4 1 3, and whose area cost as much to pave it at one shiU og per square yard, as the pallisading of its three sides I8t at half-a-crown per lineal yard '? Ads. Hyp. 25 yards, and the sides 20 and 15 yards. 8. A roof, which is to be covered with lead, weifjhing lb. per square foot, is 30 feet 3 inches long, and 15 feet incbes broad ; how much lead will be required to cover ? Ans. 37 cwt. 2 qrs. 19 lb. 14 oz. 9. Suppose the plate of a looking-glass is 30 incbes by I, and it is to he fraraed with a frame of such width, that I aniface shall be three-fourths of the surface of the iBs; required the width of the frame ? Ana. 4'3I 15 iuches nearly, 10. How many bricks, each 9 inches long, 4i inchei'^ nad, and 3 inches thick, roust he taken to huiid a feet long, 20 feet high, and one foot thick ? Ans. 28444J.' 11. If the circumference of a circle, the peiiraetM ot & lare, and of an eguiJateral triangle, be eac\i ^ ^eW. what IB the area of each of the figuree in square jards, and which has the greatest area 9 Ans. Circle II '45916; squared; and triangle 69282 square yards; and hence the area of the circle is the greatest. 12. If the diameter of a circle, the aide of a eqnare, and the side of an equilateral triangle, be each 100 feet; nhat are their areas, and which is the greatest! Ans. Circle 7854, square 10000, and triangle 4330-13 sqnare feet; and hence the area of the square is the greatest. 13. The rent of a farm is paid in a certain fixed nam- her of quarters of wheat and barley; when wheat is at 55*. per quarter, and barley at 339., the value of the poriioni of rent paid hy wheat and barley are equal to one anothei; hut when wheat is at fios., and barley 4l8. per quarter, the money rent is increasc^d by L.7- What is the rawn- rent ? Ans. 6 quarters of wheat, 10 quarters of barley. 14. There is a wa^on with a mechanical contrivaiicr, hy which the diSTerence of the number of revolntion* of the fore and hind wheels on a journey is noted. The cir- cumference of the fore-wheel is a (lOi) feet, and of the hind-wheel b (12) feet; what is the distance gone over, when the fore- wheel has made n (1000) revolutions more than the hind-wheel 1 Ans. — — -. or 15 miles, 7 furlongs, 60 yarfd 1,1. Tliere are four numbers, such that if each he roulli- plied by their sum, the products are 352, 504, 396, and 144; find the numbers r Ans. 7, 14, 11, and 4. 16. A man being asked what money be possessed, re- plied, that he bad three sorts of coins, namely, half- crowns, shillings, and sixpences; the shillings and sii- pences together amounted to 409 pieces ; tlie shillinss and half-crowns 12,54 pieces; but if 42 was subtracted from the sum of the half-crowns and sixpences, there would le- main 1103 pieces; what did the man possess in all' Ans. 995 half-crowns, 259 shillings, 150 sixpences; in all L.14l,ls. 6d. 17- Find two fractions whose sum is I, and whose pro- duct is .,3,? Ans, -jJ, and.*,. 18. Suppose a ladder 100 feet long placed ai^ainst a per- pendicular wall 100 feet high ; how far would the top of the ladder move down the wall, by pulling out the boliont thereof 10 feet! Ans, 50126 feet, or 601512 laehe^ 19. A may-pole having been broken hy a blast of wind, it was found that the pcirt broken off measured 63 Jni, and by falling, the W^'^iiBwiAe Mn'stWatfaa ^oad, at the distance of 30 feet from ihe foot of the pole ; it is rc- (jnireil to detennine what wns the height of the pnle when slandinsr upright? Ana. 118-3085559 feet. 20. Find four numbers in arithmetical progreasion, whose common diflerenee is 4, and I heir continued product ]76985? Ana. 15. 19, 23, and 2". 21. Find the distance from Eddystone Light-house to Plymouth, Start Point, and Lizard, respectively, frora the follomng data: the dislance from Plymouth to Liiiard being 60 miles, from Lizard to Start Point 70 miles, and from Start Point to Plymouth 20 miles ; also the bearing of Plymouth from Eddystone Light-house is north, that of the Lizard W.S.W., and that of Start Point E. by N. Ans. From Eddystone Light-house to Lisiard 5312 miles, to Plymouth 14-19 miles, and to Start Point 1713 miles. 2'2. At an election foul candidates offered themselves, and the whole number of votes was 5219 ; the number for the first candidate exceeded those for the second, third, lud fourth, by 22, 73, and 130 respeclirely; how many roted for each? Ans. For the first 1361, for the second 1339, for the third ]2!I8, and for the fourth 1231. 23. What ia the area of an isosceles triangle inscribed in k circle, vrhose diameter is 24, the angle included by the mual sides of the triangle being 30 degrees? Ans. 134-354. 24. The sides of a triangle are respectively 40, (30, and W feet ; find the radius of the vnscribed, and nlso of the iicumscribing circle? Ans. Inscribed S^^IS, and the ircamacrjhing 10|^lg. 25. Required the aide of an equilateral triangle, whose rcB is just tiro acres ^ Ana. 079(319 links. 26. A field in the form of an equilateral triangle con- lins juat half an acre ; what must be the length of tether, xed at one of its angles, and to a horse's nose, to enable im to graze exactly half of it ? Ans. 48*072 yards. 27- The area of a right-angled triiingle ia 60 yards, one f the sides is 8 yards ; reijuired the other side, and the ypotenuse? Ans. 15 and 17 }'urds, 28. The area of an isosceles triangle inscribed In a rcle is half an acre, and the angle contained by the equal des is three times the angle at the base ; required the igles of the triangle and the diameter of the circle? ns. Yectical angle 108°, anglea at the base 3<i°, diametei 'the circle 121'367 yards. 164 CSXBftAI. SXKSCISE8. 29. Fiad the expense of gilding a spbeie at Hd. pn ijuare inch, its diameter being 2 feet? Ana. L.Il,6s. 2j<i. 30. Find the expense of inclosing a piece of grDunil in )f a Bector of a circle, whose arc conUina Sl)°, i area is a quarter of nn acre, at the rate of 'i lilliags per lineal yard? Ans. L.37, }Qs. 4^d. , 31. It 13 required to find the area of a circle vrhoae ra- Ib equal to an arc of 70 degrees of another ciitlf « diameter is 20 feeta Ans. 468-922 feel . If a circle he described Tcith radius one, and a Recond nth a radius eijual to the circumference of the first, )ind f third with a radius equal to the circumference of Uie Kaacond ; what will he the area of the third circle ? Ans. ^BflfrSa. 33. If the vertical angle of a triangle be 100°. the dift ftxence of tlie sides 4^, and the dilference of (he segnienU of the base made h_T a perpendicular upon it from the tvi- tes 5 feet; what are the sides and angles of Ihe triangle, and what is itsarea? Ana. y60387, J-IUIS?, and 1»h 6-03699 i the angles are CI" 69' 7", and 10' (T 53", and the area 2-94l.'>9. 34. What will the diameter of a globe be, when its snt- fftce and solidity are both expressed by the Bame nambert .All*, a 35. If a round cistern be 26'3 inches in diameter, and L$6 inches deep ; bow many inches in diameter must a hold three limes as much, the depth being the ' RBine as before? Ans. 45553 inchn. 36. How many spheres 5 inches in diameter will hr equivalent to a sphere 20 inches in diameter? An*. 61- 37- How high above the earth's surface must apenon be raised to see a third part of its surface ? Ans. To the height of the earth's dinmeht. 38. Sapposing the ball at the top of St Paul's'Chatth to be 6 feet in diameter; what would the gilding of it oome to, at 3^d. per square inch ? Ans. L 237. It's. H- 39. Three persons having bought a sugar loaf, want to divide it equally among them, by sections parallel to tlic base ; what must be the altitude of each person's shan^. supposing the loaf to be a cone, whose height is 18 inchtt! Ann. 12-480U the upper part, 32430 the middle part, and 2-2756 the lower part. 40. If a cubic foot of metal be dmwn into n wire of Vn of an inch diameter ; what will be the length of the wite. sJJoiring no loss in the metal i hvA. \&;ift\1 feet, or 31 i mile«- CENBBAL EXBIlCieBS- 165 -41. If a sphere of copper, of Due foot in diameter, was In be beat out intn a circular plate of ^'^^ of an inoh thick ; what woold be iis diameter? Aue. 113)37feet. 42. If a round pillar 7 inches in diameter have 4 feet of stone in. it ; of what dianieler is the column, of equal length, that conlains 10 limes as much ? Ans. 22136 inches. 43. What is the area of a triaagiikr field, whoae three Ndes are reepectivelj 1717 links, 1515 links, and 808 liskar Ans. 6 acres, 19 poks. 8954 yards. 44. The perpendicular, from the verteZi on the base of an tquilatexal triangle, is 10 feei<; find (he sides of the triangle, and the diameter of the circumscribing ciccle^ Aos. Each aide is 61^3) and the diameter of the circle ifl 13^. 45i Giren the base 160, the vertical angle 100°, and the difierence of the sides 8'7365, to find the other sides and angles? Ans. The sides 100 and 1087365, and the angles 37° 59' 19" and 42° 0' 41". 46. Given the vertical angle of a triangle 120°, and the three sides to one another as the numbers 7. 5, 3 ; also the numerical expression of the area equal to ten times the perimeter, to find the rest. Ans. The sides are 93J ^3, %|V3. and 40^3"; the area is 2000^3 ; and the other angles are 38° 12' 48", and 21° 47' 12". 47. Given the vertical angle of a triangle 80", the base 100, and the sum of the other two sides 150, to find the Bides and angles. Ans. The sides are 92 313, and 57fi8ti8, md the angles are 65° 22' 55", and 34° 37 5 '. 48. The sum of two sides of a triangle is 300, and their difference is 100, and the angle contained by these two ttdes is 60°; find the other angles, the tliird side, and the uca ? Ans. The angles are 90° and 30°, and the third lide ia 100^3=173-205, and the urea is 5000^3= 1B60254. 49. What is the solid content of a cone, the diameter if its base being SO inches, and its perpendicular altitude ©inches? Ans. I 81805 feet nearly. 50. What is the solidify of a segment of a sphere, irhose diameter is 10 inches, and the height of the seg- aent4incheBi Ans. 184-3072 inches. 51. A gentleman wanted to know the contents of a iqilare field, but had forgotten the dimensions, only he -emembered that the distance between a large oak which ;rew within the field and three of its cornets, in & «m<lc(a- ive order, were 116, 156, and 166 ja.ida; ie(^\tei iA»a f 186 GENEBAL EXEIICISG3. eontentsof the Belli and theside of the tiquurn? Ans, Cm tent 813585 ncies, and the side 198-437 yards. 52. How often will the fill of a conical glasg, 2;^ inch, ide at ihe top, and 2^ icches deep, be contained in an< ther in the form of a t'ruslum of a cone 3^ inches wide - the top, 2 inches wide at the bottom, and 4^ inches deej Ans. 8,Yj tima 53. How many apheriotl lenden halls, of half an incl dinnicter, could be obtained from a cubic foot of lead, saf^ posing no waste of metal in casting? Ans. 2()401ft 54. How menj cones, of one inch diameter of base, anli one inch perpendicular altitude, would be equal in solidity to a sphere of 6 inches diameter? Am. 433.' 55. If the diameter of a lub at the top be 30 inches. uC its depth 2 feet, what must the bottom diameter be, fio M it may contain 8 cubic feet of water? Ana. 2'0044fut> 56. If a ctstem. 4 feet long, 3 feet wide, and 2^ M deep, were proportionally enlarged in all its dimeniioiu,N as to hold 4 times as much ; what would then be its iu< mensioDS ? Ass. 6-3406 feet long. 4'7622 feet wide, avt 3-9685 feet deep. 57. If the base of a triangle be 30, and the other liilH 25 and 20 ; find the segments of the base made by a pf pendicular upon it from the Tertex, the segmenU msJo by a line bisecting the vertical angle, and the leoglb (^ a lise drawn from the vertex to the middle of the bw- Ans. The segnuents of the base by the perpendicular b« 1875 and 11-25; the segments by the line bisecting t\it vertical angle are 16^ and 12^, andjhe line from the TW-i tei to the middle of the base is ^287-5=17 nearly. I 58. Find the dimensions of a room, in the form of », rectangular parallel Dpi ped, the solid content of which iBj cubic yards is expressed by the same number as the supe^ ficial yards in the walls, ceiling, and floor together, asn such that its length may be double of its breadth, and il^ breadth double of its height ; find also how many peOJ I the room would contain, each having the fourth part ' Bquare yard to stand on? Ans. 14 yards long, 7 ) broad, and 3^ yards high, and would contain 392 peopli 39. The area of a triangle is 59(11-16 square yards, perimeter 370 yards, and an angle 85° 27' 34"; reqiiii the side opposite to it? Ans. TABLE I. rHMIC SINES, TANGENTS, AND 8ECANT9. TO KVERt POINT AND QUAkTEK POINT OF TriK C0MPA8S. " Co-inev ThubbiiI. CoUng. Secant. Coscc, A I S03 Itt 000000 0. 000000 9.909477 aeomg 9.89790* 8.9933B8 9.99S274 ai71S*T 11.808681 la000623 11.006603 10,002096 1ft 828763 1ft 001726 11,309204 11,008898 lft833480 lft700T64 10614429 la 637176 loiiT'ieo 10 369008 lo'28«960 ia-i34973 10197641 ]ft 172916 i! 9" W ii 571 934 9.991fl74 8,298863 0.086786 9.308788 S.US0886 9 4S1939 0.073641 9.SBSm ia701838 Ift60I31« 1(1618061 ]a446853 ia008*26 10.013314 ia019116 ia03SI69 ib.03B86- Ift043e37 :o.*o66e5o 0.»0fl«15 9.0S6163 9.946430 9.933350 a617XH 0.674820 B. 737067 9.777700 10.383776 ia326171 ia272043 10.233300 B7 06* 9.904S38 9.888185 9.860790 aS7OI09 9.014173 9.967296 lftl7S107 1ft 139801 1ft 086837 lft(M3706 1ft 0061 72 limi8]6 miiMBlO 9.849185 10.000000 '1'HiiKene. Conee. B TABLE n. ^T LOGARITHMS OP NLMDERB. 130 160 ITO 23 33 34 ai 1361728 1.380211 1897940 44 46 L 623240 1.033468 L643463 1,663213 (a (34 65 1.7S63aO 1.792il93 L799M1 Leoei80 L812913 83 84 86 1,1108486 1JJ19678 1,924279 1,029410 m Da m 2T S8 39 1.431 1.44; 364 68 98 C3" 14 479 <m 47 48 49 *>_ 63 64 66 1.672098 1,681241 1.690196 _1.B98»70 liTieoos 1,724276 1.732394 1.740a63 67 68 7U 7i 7a 74 75 ■76 77 78 79 L 826076 1.8t!2A09 1838849 1,846098 1.861268 L863323 ],ee!U.f3 1.876061 1.^0814 1,886491 1.HI2O06 1,897637 87 88 89 90 ^91 92 03 94 06 96 07 98 99 1.939619 1,944483 1,949390 1,934343 1960041 L9637S8 1.068483 1.973128 1,07772* 1,082371 1,086772 1991226 1,996635 HS 128 m 32 33 U4 36 W 37 .19 .511 631 tbbo .M8 l.fiTff 1. 69: MS 178 Mi 202 f4 56 57 69 1.J481B8 1.766876 1.76342S l.'77S15l ■ A ;. 4 1 6 * 1 M ,1 1... 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S 1 8 18 3~ r~5- « 1 * 1 6 TT 1731--) M7468,4478a3|44T778 147933 448397 t&m IM 8861 [lOIS 9170 9478 mx 9787 91M1 160095 154 iOiCa 4505.-17 ]lfi0711 lS086r 151018 151172 161336 151479 1633 164 IJMO 3093 240O 3563 3700 3859 3012 3165 163 3*71 3624 3777 3930 40B3 4235 4387 1640 489: 163 4097 5160 5302 5454 6606 W58 SOK 8063 153 6B18 6870 6821 6973 7136 7276 7670 7731 163 803S BlBl 8336 8638 8789 8940 8091 9243 151 96*3 9BH 9845 0996 460146:460296 «0*47 460697 1B0748151 ssoia^mr 463847 462097 4U314B 463296:463445 463694.463744 150 41H2 uai 4340 4490 luay 4788 1936 6085| 5231149 e533 B6S0 68^9 6977 6126 6271 64231 6571 8719140 70W 7164 7312 7460 7608 7766 7001 8053 8300148 84SS 8613 0968 «0U6 87»0 6D38 470263470110 908; 93331 170701 0380 170861 9627| 9675:148 tn«3S 15S5 1733 1878 3171 "m4|*'2610;146 2903 3049 3195 3M1 36331 sni 3836^ 40711*6 4363 4S0S 4853 4799 5090 6236 6381 653^146 B816 6063 6107 0363 8397 6543 0687 0S33| 8976 145 145 47726^ 4T7411 17;55S 17770(1 477844 4T7989 478133 178278 478133 Wll 8856 8899 9143 9387 9431 9576 97191 Baa 144 480151180394 180725 18080! 181013 iaU56 481209 16B6 1739 1873 301* 3159 2445 268i 3018 3159 3303 3445 3687 873( 3872 4443 4585 4737 iaeo soil 5679 0863^ 6005 6147 63S< 8430 8714 6855 6897 113 8269 8410 141 9S37 8677 9818 141 490099490239 iTKX\-.. ■. .■ ■■... ! -^'11 i:)0911 19I08i;491222 140 149150'2,40ir^,41)17- '1 !'ja34149a4ai492S3iil«) 3900 3040 3737 3876 4O15130 43!M 4133 6128 5287 6406,130 MS3 5323 6616 6653 e79lll30 9068 Szoe 7621 7759 7897 8036 6173 138 8448 8586 8863 8899 0137 0550,138 e824 KW3 500090 600336 600374600611 S00785 600923137 soiioe 501333 1170 1607 17441 1880 3017 3161 329i:i37 E664 2700 3837 3973 310ft 3346 447l[ 4807 3618 3666136 S927 4063 4100 4336 505G93 4743 4878 6011136 606380505431 606567 606828 505964 606099:506234 6B40 6776 6911 7046 J 181 7310 7451 7586 799J B12f. 8395 8530 8664 8799 8934 9337 9471 9710 9871 5100M 610113 610377 610411 B10879 S10S13 1182 1616 son S151 3284 3551 2818 2961 3484 3617 3883 1016 1149 4383 133 mi 4813 1946 6311 5476 5609 5711 133 600(1 6139 6371 6536 8800 ffi)33 70e4;132 7328 7460 7592 7866 7^ 8119 8251 8382133 518040 618777 618909,5I004il619171i619303 6I9t)4'519.'.6tl 519697X31 9D59 52OO0( 52022] 52B(63 5201&1,620«15 520746 620876 621007 1 131 oma 1661 1922 BIKJ 2576 2066 3006 3226 3356 3486 3618130 3876 4006 426B 4306 1626 4666 4915 6174 5301 6434 636.3 6^3 5822 6061 62!0 129 64G9 6398 6727 6856 6985 7111 7343 7501 129 7759 7388 8016 as 8102 8531 8660 8789 129 9045 9174 9303 9687 9815 9943'630072 128 S3032Sfi304565305B4 530712630840530068:5310965312331 1351 128 1 1 1 3~7~3~T"0~5"~P6~ri~T'ft \ ^"^^^ 3 92ifl 9401 5371088 6 2913 3U06 3 SWIfi 5765 * 1aao\ 756S b\ Biee 9:)43 e 330935391113 3912) 7 8607 3873 - 8 M63 4627 9l l)l!)9l 6ST4^ I 6481 T30ffl 7483 91241 93001 703ffl^3T(W! . _ S17A 3360 40ld 4198 G84H 6039 767ffl 7853 94871 966e 1|381U« 2737 29171 309! *7ia 4891 6321 6409 667; 11381476 38n ■ 33771 at . 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M. 1 1 3 3 1 4 1 6 1 d 1 7 1 8 1 mmi^ BTsiHWiJO 973266'973313( 973405 973461 97340719 1 3630 3083 3774 a 40fil 4097 4143 4189 4335 4281 4327 4374 4420^ 3 «12 4668 4604 4696 4742 4788 4834 «88M tBIi 5018 6064 6110 6166 6348 6394 EaM 6524 S670 5610 6707 6763 6708 6801 6937 6983 6075 6131 6167 6313 6368 7 0300 0396 6442 6633 6S79 6626 6S71 6717 8 6808 6854 6900 6992 7037 7083 7129 7176 9 726( 7312 7368 7403 7449 7495 7541 7686 763i SfiO 077724 977769 977815 OTTSaiSmM 97799f 978043 ff60899 aiBi 8226 8272 8317 8363 8454 8500 8M 3 8637 e6»i 8723 8774 8819 8911 8956 9003 3 9093 9138 8184 9230 9275 9366 9412 9467 4 9Mf 9694 0639 977( 9821 9867 99U £ 980049 880094 8801+0880186980231 980376 SSO323Ba0367A ( 0603 0549 0694 0640 0730 0776 Offll Hli T 0957 1003 1184 1239 1371 Hf ^ 1466 IWl 1647 1637 I68G 1738 8 1864 1909 19541 3000 3041 309( 213f 31S1 ^ 0^271 983316 9^^ 982407 982*62 9S2497i88a543:9S368SiS^8^ i ^M\ 1 3723 2769 2814 38Se 3904 2949 3994 8040 BfiH ^H 3 3220 3266 3310 3356 8401 3446 3491 aSW ^1 8 3626 3671 3710 3762 3807 3863 3897 S842 aeer ^H * 4077 4133 «67 4312 4267 4309 4347 4389 4487 H » 4827 4672 4617 4662 4707 4753 4797 4842 4887 ■^ ( 4977 6033 60B7 5112 6167 6203 6247 6393 63S7 G43Q 6471, 6516 5561 e606 6661 6696 6741 B7Bfl, 8 6S7Si 59201 69651 BOlOl 605S 6100 6144 6189 6234 ( 633*1 6360' 6«al 64S8I 6503 0648 86031 66371 66831 fl7( 986772 B86817 986861 086996 B87040|9S70e5]987 13019 7319 7364 7309 7363 7398 7443 : 7066 7711 7756 780d 7846 7890 7934 7979 8024^ 8113 81B7 8203 8247 8291 8830 8381 8425 847« 8559 8604 8648 8683^ 8737 8783 8826 8871 eSlfl 9005 9049 0094 913^ 8183 9337 9372 93iq 3aG^ ' 9460 B494 9539 95Bd 9628 9673 S717 9761 980a ^' 9895 9939 0983 Hi 8990339990383 990438 04721 05161 0661 0006 0660^ 06M ^^ 9' 0T33I 08ZT 0871 0816) 090ol 10041 10491 1093| llWl ■ [&0:g91226 J9I370I >0t31fl l9913S9 99140i( 991448 B01492 991636 981680* . 1 1689 1713 1768 1803 1846 1890 1936 1979 3 2111 3168 3300 3244 3288 2333 2377 2431 ; 26*4 2698 2S42 2686 2730 2774 3819 2863 3995 3039 308! 3172 3216 3360 3304 3436 3480 3669 3613 3657 3701 3746 ; 3877 3921 396) 4009 4097 4141 4186 4317 4361 4404 4449 4403 4637 4681 4626 4767 4801 4845 4889 4933 4977 6031 6066 9 6196 S240 6384 6338 6372 6410 dim 6504 56471 MioiBsaas 596679985733 996767 B968ir 6074 6117 6161 6306 6249 6393 6337 6380 ; 6512 6566 669! 6643 6687 G731 6774 6818 & 1 3 6S4[i{ ima 7080 7124 7168 72l3 7356 Hi « 7386, 7430 7661 7606 7048 7693 ^H " 7823 7867 7910 7064 7998 8041 8085 8129 ^n « B369 B303 8347 8391 8477 866* ^fl T 8696 8739 8783 8869 8913 8956 9000 // a 9131 9174 9218 926: 8306 9348 8393 9436 / 9/ a^fls 960H 9W4 MSt^ w* wea 9S26; 9870 wm iz s:} 1 i\a\4y6\'i\%\fcl ■ 1 (O DEGREB.) 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LSI 315 15530 990393 121 009G01 1457C721I0' 8542;«;tS 990651 121 009319 ]l.«9i;200 85tlOii:37 HI 315 oaoso- 009097 110014200 853980 ne 2ts 1510S2 9^11150 0OS811 14613S'2n6, 85381 W .« 31S 16«»>3 9;JI40U 116202'20a' 8Si7*' « !T8 3U 1617^4 9U1002 853014 38 «5 3» 15153; 991914 131 116510)207 863491 iH tU 3U 992107 121 llfle34 207 B533fl( 1. '>**.:« ayjiao 007590 1407581207 86334L 3" 29 llLloUlff a 992072 421 10.U0732S 10140882 207 U-BSJIIB no ail 16-HBl 421 007075 1470M20T 862991 28 M7 au IMMKJ 9^3178 1471311207 8.53861 27 ,7fl ail KfJAiS 20 101 ail mmr 993ia3! 421 H,;;.2620'2fi ■32 ai3 15»()S U9;i9:ifl, 421 OOOiiOl «a 313 15W10 994 189 421 UlJi-'ll les 313 15t(13 99«41 421 005551 U77h-i'li. llO 313 1531S1 984094 421 IW53(l6 m aia 155058 994947 421 00.^.1 kVJ 851997.20 mrm^ iai52t«fl 9.99519! IS9 313 16280 99545L 421 004543 '851747 18 (a: 318 995706 851022 17 UM ai3 9959.'i7 851497 16 »2 213 iSUh 090210 003790 14aa28|209 85137i, rem 312 15229 990463 0135.37 14876l'209 aM24<i m 213 152164 990715 121 148879209 la ia« 313 163030 990008] 121 149001209 85U9!W 13 lawos 097221 421 003779 119130209 8.50870 11 as 212 151782 002527 149255 209 850715110 ml 213 iaisiess "TaiiTHB" 121 10.0<tfl!7i 107149331209 9.830019 (^ 3U 161528 997979 421 14960T|210: B50493 IBQ 311 998231 001761 1496.32:210 86O30S m 311 151374 151148 993431 098737 121 0015J6 0012^3 14B7582U 860242 85011 ITS 311 161031 121 B49990 \M 311 160894, B90342 000758 649864, 15U:«8 121 000505 B19738 ISS: 211 150641 999747 421 000263 1503892lS 849611 1.50516^210l 849445 10.<H«)I>W 421 s«.„i l->u^. \ *(i\ww™- ^ 1 iS^^TABLE IV. A TABLB Oe NATOEAI, SIMM. Ul «■ 1 l> A> . *■ ITl* I *' 1 4' > *' > *'-™. iUM.m,»ln^J.iK»m 0523aB iMifrae 1 087i5ti 1WS38 131869 13917360 0361!H), 063030 104818 133158 13MUISI vinM5i» 03JM81 063917 087735 106107 133447 1397«,M ttWMMTS msssa oaa77i OSXWT U70837 088035 106396 1I3T39 I4O037J6T 400UM umu OWMJ O0S198 070917 088315 lOMHK 1S9IU4 1 14Utf6W At»l45t 018(H)7 03031)3 0S3T88 071307 088606 IU5976 1SI313I 140613 w «'W171S 0IH1»7 OMdU 064079 071497 otsaau loewM 133(»l' 140901 M KS oiiHsa OUOUM OMaau 071788 U891B4 106S63 12389U , 141188a UIU7TS uuUik) i>«»4?4 1U68U 134179 1414775* KWWW OS4U60 ofaaaa 089763 107133 U446r 1 141T<G'a| loioujm OJuasi 037806 038007 056341 0736W 0000531107431 121766' 43UiaM U'OOSWO (M0S5^ 073948 09034a, 10771O U604ff usaiM ■ IHWMUIIO'JOW 038380 073338 090633 10799i) 1363ilil, tMHMq ^■m>t«' mixa aWfi78 06«U3 073538 090933 106389 i:iS6^ fiVUffl! ^■Ruii irjl534 0389B9 066403 073818 091313 108S78 11t6910 , U33WM| 13fU9i» 1 ItUttMH HK*^^ (WlSlfl 03y3CO 066693 074108 09ISU3 108867 BBIhsh O33106 03!IS»I 036963 091791 109166 07468U 093081 109445 iMTTtf i4«anm 074979 093371 lumat 137066 ]448Mi1 Vims oiin33 057*64 076:i6M 09366(1 1 100311 i37afiti 1 imeSa amoMiSKwaiHlU 040713 068145 U76569 093050' 103ia 13T6»'l44Mii| ui|l«X>llW 1 iKUfiW 076849 093339 10603 137990 lluHafl sjootBfflB uaaasi IMI3tU 068730 076139 093639 10891 ll383U>!l4SM« i»W»UIO (UU*1 (Hi685 069016 076439 093819 lllWO ! 138SU7 «4187fl 076719 094103 1U4H9 138796 013168 069697 077009 094398 111768 139084 1^^^ 059^7 0773BU <I94«87 1 113047 13WT3 a;[o»T8M l«5.fU5 0B4977 113336 139661 aoW727 0*8177 in:\fm IHIOKW 077879 w>^m "«"7.^i iirhioM 096367 095556 113<Ei5 112914 139949 laU338 04;Wiy FNiLIH!) 07^459 096846 113303 lau&M ■ai|«»B0i7|oa«4fw IM^Ii)i0:00l339il':w;49 096135 713483" laws' »j'UlltKIIM (MHTO) 044-01 l>Ullii"J,U7SHWit U9U435 113781 laiioa ayuuUciouliKJTim 07U329 096714 1140T0 laiisi 3*0.KWW)H)a73i0 044783 063310 079619 097004 114339 131IISD MOIUISI omoiu 079909 097393 U4I148 laUMS aU 010473 037U3a 045363 063791 080199 0B7583 114937 193368 y7 ui07«a 0Z831a 0456M 063081 oao489 097873 US336 133645 38U11UM 02SS03 046W4 063871 080779 098163 U5516 liuiiiw ^U1IM4 033794 046335 063661 OS1069 laaiai laoawul 40011U»5 030085 ' oiossa 063993 081369 iiwws ia»4io 15068U3II 41011IW6 039375 o*B8ie 064343 081649 099U3U 133898 43012317 omm 047106 064633 081939 099^30 11667 I 13U98U omsj 047397 064833 0»»3S 069009 116960 134274 M;oia7i» 047688 066113 099899 117249 134663 «l'013090 46013380 030639 047878 066403 08380S 100188 U7637 134861 030839 W83e0 086693 088098 100477 117836 135139 47013671 031130 048659 066984 100767 USllA 136m 4S'|)1396'J 031411 048860 0U63I4 0B367S 101056 11B4U4 4U0113KI 019140 066664 083968 0348 86UJ 36004 Sll!ui4fi44 mm. 049431 060854 084368 084547 0836 9ii 8»&i 1303U3 51,014835 033363 oSfsT "067145 119270 aoE&rteii^ a3,oism 033674 060012 0S7Ua 0S483 11965J 'S'^ S3 U1S4I6 OUl'864 067735 *I3 64 0157OT Od£4 AS 015998 0«SJ<I6 0B5 66 omasa 033737 00859(1 0Su9n 57 0165S0 034037 051464 0U888G 066 •* .« 016871 034318 051756 0691 6| I8>f f> m'l\3 fi'J (117163 034009 0530« A,m 60U17i63 034S99 U533SII 1 IMiQ7fi6 OS ob 04538 U1389 Ml .1 ■MU.ul S J "'«"" «»" "*''' ""' 5* ?i — ^^ » i ^,MS A TARI.B OF NATURAL BINES. 631 ».f 0,156*34 TmisT 'wfesn 2419M 268810 276B37 292972B0 1 156;a:J i7;ro:ts imu95!-^l!(li 2691 Oil 276917 B 107009 i74aii iBiijM ; lioawi 26**81 270197 «1«;2US ITi-WS laiuuGMosTrt.-. 2-mti9 2»;M)li,,W 4 ixisa* 174794 lUiy.'il,2U!>i.5'l MliftISS ■ifmvi 271.768 29;!4W.W 6 157471 17601*) IIW237 ! »>itt3+ KllkilW a4a3.'« 2eiW:'4 1 277«M 39:lT{Uft-. 8158158 17.5,J67 iWiWiwratjiy uiimii 2*iei5 2W15U5 277^10 201iHO;e4 71.Wt45 19:M)7 ! aWiWS ^J6ftU 34:eS7 aeoTss a77oi»4 2U43IH|63 P1587y3 imsmt 1 310187 337;il8 344179 361066 I377S71 294696163 » H33 (W 4Ma 26d47 3- * .»48 6 iWH W JO 0^' 3784.12 2flfifiSlfl0 93SMJ 20 9U8 1 ^»&(-J0|4i) J4 "M 9i » iJ W 4^604. liB (6 « ID 771M4 •9 29 51.46 ( d W «>H 2b. U •9 bJJ JflUW* 29 09 43 fi 18 [.28006 » » M 2N 6464 28094( 9U- B4434I 2) OJOkOB i- *^ m 2- 994 '^ TI6 \ 30 wuea Uwaisa Lffliffja aootsB^i ■M 2603«olafi jm'js4<iis aooToeao ^lir-« \'Mm jmB63si m. 1 30 36138 M loesafl M 36 201 W 34^ r30B33S til- J0237 a* m 302fl4 38 w 1 4H. 1 "B 46 30293 JS «+ >9U LftJH IttHnSU J86IU6 803303 1 »n a7i04nlwwn 3'»4mio ^ U« J8 UtU dU. 6U (H4 T8 363798 27flAn0 SST36 3040331 ja ZZI 3MU39 0880 287039 804. 9304 Sfl* MM 80 2879 804687 6 1 aoaoiJ sum 337880 8.W«W 27 440 288 Dfi 304864 »W9a7 23096 337908 BIHftS. 37 M 368*6 306 «H1 M J3L«S WLfl AS 66 ^jono 388763 3064 imiw 6644 3723fO 68 2- 27^660 289002 2893 TOS6B6 1069 2 en 80189 ■iOfi -uti-ca 27 •mn KHL6|8 UU 2K» 273400 "90 46 aoosoa 8 27167 ■W4 30-oeol w ■>7T8flfl 39- •» W 6 IMBfl "7 » "vm 30 T3 ■^ o9 W 168 ■^ 98 2B 637 ■/■J*«4 2*1 :«7 27.W78 20181.1 308464 2 winaiie ->?4fl68 36S6:n 270368 1 29-2094 308740 1 eonam inowo aoTQis r}*m "41 fl-" j^Mio * "OMW v^wm!L2«*^Sl^ U. BO - H i 64 A TABLE Of NATURAL BINES, »»• ■3JHI73T 40075*^619 4383FPS 374ffl?r l|lO"'M ja4 1 -J 3.MAI i748-rlj9-P90') 4. 8S 438833^ sLn-i^ (jS It 140 «1 OJ 4( ^ 4)i<m7 Sm. H Ml 4JB40Q 43n6s3 1 4'UOI 3 37l(^rtfll30l8(^ iJami 1394111 I mi WOO 6|jlOC-6 feOoS 30i070 4II80U5 439936 MOffiCT 1 i7JIS AJOfl ( 24 aspur wAm 4341)19 iHoeatw TJlUBoJ JJ 49rf 34.«.).l 36U"W 3 0404 31C200O 408jOO 434483 tiimm 8^311229 327 68 344 Ob 36064. j-e76j 30-8^ 4W801 4247.26 «4«l4an O^UIBOO 3280t. 34449 360811 39J140 40' 127 4.4900 4ttBM ujiaoae )2a»7 144 5 34.K)25 361082 377571 mu7_ 40939- 43,rfSir 44IHS| ^ai" 3930 6 4(mfiS8 3Z8W. 345 '08 3fll6B5 JT7841 303942 40- "I 9^141 340671 36IS96 3 8in 304300 41 -i ^ 4 ff l2siJ8B8 3-WI6 345844 362167 378379 J944T7|4I w 19S13I6* 3:iOU)l 34611 362438 37S649 304744141 1 4 U ■iam^ J4a.]90 J62700 8-8018 306011 410's. 4 >n, 44 « *t ITJlSTll SJO'K 346663 8B2US0 3~Jia7 305r8 41 4 imta-r fc)O.H 3469ai 303 11 J-J45b 300640,411614 427J58 <430lJ|| 1 tl" « ll>W 7 J3jt*13 4I177J 1 IJ Hi i 1 J094 (06(180 1 4IJ04fl 1 i. m A7 ^ lKoa<^ ilM H 41 31 4J»14 '44^^ 1 JJ 1 J4* 3UJJ6 JSUOJ- MJ 4 4t 7 4 t4\ 4441 14 ip J.J1H87 34a«« 161600 jaoa " 4 4 4 n B 94J15049 (S-lBl 148573 364977 1 issi -oau J3243> MStt5 30j14S 381'BJ jeiifj 1 33!riO 340117 366418 WWi 37 a« dSaOSi J49390 366689 381H 1 1 jsiatBTw 333258 3JWi' 31.5900 3S 40 3 a 15 414104 429UMI 44f(.7T3l| 39317020 JSaBJ,) J4»Jj jnfl 31 W +lu 308482 414429 43U->4Ui4MS9V^| 3iajJc« 333iW7 i> 1 7 J6.501 JM6S.J 398^40 4146CU 4»)6ulMunH l«l»l W^S. JJl (HH, iOW-lb 414S68 4.307-4 1 44W)MH J.l.J.t.'i. t- V 1 iK, 2 1 Kf g 4 J ^3 36 " * J 87 3a3ij tdW«fl « M , JBMUW 3H48,f 4 11 *«■ 1 4 39 )]U7iJ' 336i 4 di^tb(td6lil!kH> Ideal 1 4I114U|4I « J'.W* 33im7 3S^931 1 3U9-J0G 1 3^»3U( 41 14IS 4 ', 1 l«a ikj- XXSl J6MH JUU4 6 4 1681 1 4J320fn 33-095 3534 360 47 iHjOOt 4 1948 4 td^Mm J3730!) 3h 174 MdJUtH 33 J4J |45'fil*.W JJUl WHJJl J7».7 JHG 11 4U'2747 4 4631!1716 3381 to 3j4)ft 3 JS 1 weiffl 4030U 4 47 321000 3;ie4Si dfi48J5 371 m 3K 47 4a279 4 48333^66 3.38738 366107 J71J68 jy 10 4S322M1 339(1 366J79 ^11)38 3877H4 403SI1 4 S03Z381S 339^ Sfisari * 3395^9, JUiO^J 37J1 8 388320 404344 aaoeai 300194 37244B 0SS5M 40401 4 is 340100 3S(Hbe 34U38U MB 38 37^-18 37 01*8 3WW, 4 jsei 4 34«6Bd(i?i WO 381 34uri\367ia\,irjB-B 3fJ 5T'a24 43 341200l357BWl\Kli jn WJlJi-s . a83''aH8 34147 iB78S 31\(« ■w J ->A u:A«. 3*1 47iar)&*Wi"r4aJl\*JW - - ,-^iv .™,a, ^ \^^^^\i ^^"'-' «wi^^ w«i^ ^^>'y■!%>^*Ti NATIJBAI. BINES. b5 ai^aa-Xoa "lain ***'"■ MUS2 .46 «S ra|*G2366 i8 54 S3. ^ 9fil *W ^ 59 sn 1)9. 6433 IS 'V SK 43 Ml « m4i SSflR •u 47S6» 011904 6. U oaj H64 J06 86 «M fl( U3 50-^ 49J4 50 6JS 46 jl. (■94 { lib BBMmU J7l56ii4nfl M 'K (HI 226 I <ai5i titl*SO 1 654 i 611073 [ 6M9S 641896 6Sfi 80 C t WiV ft\<!MAW\ fl . ..^^[ -w'-^ •>."'-\ ^t'-C S P NATCBAL BINea. fHP 7ilSE0 731780 7919^ 733163 733766 721967 783108 ffil ViftMi ' V7n< 1 75M73 r TU0S3 1 756853 ] 7W0U ) Tfieaa* T»3S32 733730 733997 73«m 73IA30 73*717 1 7349U 7 73511a 7 73fl309 7 74M78 745a7U HBBM 7*6087 7S7375 7S7565 7B7J6S 7579*6 76ISe»6 7«6793 76a6a(s 7a88U 709038 lOftll* 76)HiXf 777329 ; 777fiia ] 777685 7 777878 7 77S(Wi> 7 779520 J 779702 J 77UBei 7 780067 7 7910*6 791924 791*01 5 800731*1 i 800006*; 1 801060 *l a 801951 *« " 801*284. 80160?*; 801776*: 801IM9'4: ► 73*773 t 791974 ' 730174 i 79WT4 t 730975 I 736175 1 7a(375 1 736575 1 736U97 1 736394 <i 736*91 " 736S87 786884 7 737081 7 737277 ] 787174 737870 737867 738063 738359 738*55 738S51 747806 747798 747991 748134 758703 7 761«93 7 7fiO0H9 7 7*9148 749341 749534 749726 749918 750111 750303 771810 VriBSA 773179 T7338* 779MS ) 7«)«4 7 ) 740609 7 } 740806 7 i 741000 7 r 741195 ■ 7 I 729IM13 J 730169 ) 730961 I T30MD •3iamh* 7418H6 741781 M197e 743171 742366 742561 '59989 7A'i372 ) 7GOa»|i7 i 760317 17 J 760406 1 7«0flW T6D784 760973 761 IGl 761350 761737 7 761915 , 7 769104 1 7 76 2292 7 T6:!*80 . 17 JUSG 77*677 174781 T74B44 T7519S 776ffl2 775498 7B2789 7 782970 7 7B!J161 7 B 791757 " 791935 809*70 38 792112 81.1964437 799990'80:;8173S! 709*07 802991'3ai T92644 8031S*;34 J 803337,33 ) a035n'32 i 80368431 i aoas.vTiBO 794061 794338 79*415 79*591 794768 79*944 7951 U) TS4921 765109 769296 765^3 776679 775863 77804S 7762;lO 776413 8IV.75(i 10 8l>rill:;8 18 ; 'ikwso I mam n '951196 ! 80G273 10 , '9(l002leiH)44i;i5l 796178 ' 806617,14 1 '96351 1 801(788 13 'g6.V)0 1 806960119 '96706 !S07139lll! BO73l»(10' \ ^k^] <tu ^TTT , 797057 I S0747SI 9i 786S7S 797233 1 807617 8i 786750 797408 1 807818 7 1 786836 1 7975»4 80799(K 6' 787114 T97769 806161 6 787294 797935 1 SOOM*, <' .«™M...>*n™iiwma\n'i»Yis>\^»»iWi ^ - 7ft5&57 \ n«i^V«i'«A"'^£^\'^^-i i ^o - I a» - ■, - ^ ^ -^'* '^-^^ i . ^ 68 B N R Ea iUhjiss k II i' MH W^ -TIT" WlfiT ^■^ir V"''"H.a m 84^ ti»8Ue 4S^ eS4B7 tew UU&f -W. tA 6 1181 82MJ38 3li!li04 848WM 65 W> 8B986 a^dOO SJaMi S4t» 8S a2U£J SdOU (MUtUU 843U *wuua rkioaia 8:0 830 83U 34J 2B 868- SllM 8:M>48a 830J;)7 8JU0.JO H4ir 8D8JI14 8<l 868513 W WAHSO an uH 840^ 84lt08(i tUMWl 8i> e 80-63 MKM 88I8HIS ni810891 S^^UWd 83082' 34M0e~a)J »a 8S88U ft7fiU laBiiow Si w 83W84 840ati 84Ucflt: BaSMU «IU7 13 8UKM 82 83 4B 840 >^»m BJJOU 8b U Irti? 14Bll*wla. « aj jwj S4utM2 BiM jU as 2fi8 sasuM lGBlla7i,!UtM 8a 84 03tf 8j(WS BjIMUM aoB uu a 16^L7U » 84 U Mioou Boasso U8^ 3J 1M rt 331 M 89UI \mmi iB'Siaafla ftwa 84(188 sjtww sbooo MOlSlUMa Sffli esa- 84 aj 80 I8a04tt aiisiMffi 8ia6*i 8II343B 8411182 as '691 29 a:«8ja7ftl R,»U6 SSmi 84 W 8fl KlSimi i9 833 UO 84 UO wsiaioi 8ja JG saaaa 84. !«HU82 812 W;SliM39 -J4U 3J3. U a8'8l3778 SAff SAKfaS W aiBiaw: saatti rsj (<4j ss 85 BM BO 48 austuie e- »hHt> 85 04 HbbJ9 aiiaiias* 12a i4046 (H 51 er w, XJlSHUa S:,445b Mi A^ >1 JU44 8b tSlSl^fi^a 1 MJ20 SUM!? 1 8439 HSJUDb 86 34 8Ur«l 8i tfO 8J4&f7 8«0 H5. jalS, 3aS149o9 8:( IMU 8^4088 8M jJJ. ObANMi J080 a' S3**) I844.JM B*Hk. 1 J5 ' J7H N, AtiOUS W a)8 « ■M 8 *1H 4J816.JW W 1J M 1 ■^ 8H&Slill| -M81(M74 U4 » 84 8o4 | B(M«<J ■mj W iMlSeti W JO (* -t, ^ 8.>4! 8fai8Jb IN. U8ur» 4rHiufiua JU 63644(1 M,«8. 8&£0li3 H0!ij»82 u^ (« aa 4 81UU 7 8:/69 1 JbtiU) 84bOJ8 Slrj 8t>4128 4881,146 Bros 8J6 G4 H4 8WJ64 »6t- 4U817aia aa we.m >mM sasa 8M4 I5OS174S0 82741 OSJ 84U503 8561 rfi 8MflU7 5iaiTW« ^ «a 8466.58 .58 8W 52 817815 ■» 83 8468 8a5S«l 8b4*l« .13817082 b3 WJ 84ai 666 54818150 84 1 68 818818 1 Af 818118a -88 v(. 84 8J4 m. a US t ■'\ o'Tl tp-aa- 1 8«~^ iTBl 'JUBKUTU SMiHlt : 90C6H ' 911178:2 wwuos au!f i7<j I wears i visuoo MBM6»1 8DUd(M lloe7W lows TISO 686 aJHHi I i»(i9- 3 9 38 «eu 98 a» "■ -■ " ■ ■■ 92741^ il 1197^10 e ll!l7fl 9 S 107 IS S 1127836 H 9398111 It 9:W9D1 7 MOOBO 940 89 a 91ST0eSB »4Se07M 04038 946 mSS tH04B6 !H(U 5 IHOeSfi 94<i!i6S» IM06W ft464aj60 HU S3 »4U»A,4tt IMOflB '■- J 9 JWH |] SHSoaS " ■" 9 903709 a 9 03831 1 9 Ii4 S U(I3B58;9 hi: It 904083:9 ua 9 W B 004207 19 5 9 S5> 1 MMOai 9 043 9 «f S 90445S 9 (j I 9 S-y I 90467919 ftS 8 9")< :i 904TU3I913U0 S 8 1 904827 912 20 9 7 904961 B 2"39 S 90 5075 9 ^3fl8 3 !>05198|9 1 905322 1 U 596 906«fl|y 2" 5 9 S Mn.'iG'.l I 9 ..S34 9 Jx i S i)05C93|9 Jj tiO 3 90.W10 9 31 9 MO I 1 'JOdSiSa \0 « 04 , \) 9*1608219 109 l»tf> OitO I 5 jaoaa 93" 84 Is aCi'Aa'- j'ia -la*' Tia' rtJk^ t a>^ ^ io 94907'' 1 1M44 950063 U W4«M 9fil Silf ' nsos 6 6 tl 50606 fi{H| Mm im >4 0SD786 3U^ 3 96K9T:M9801 9 B66473 1 960BT3 OGAMS 9%H3 9 90,5024 OTOnU 1 9en7oo 97on84 aHftWlM SUllOl 9e5776l97IH5S / ' eo Safi3(W ' 9612fi3 1 B65S»a6 iWWMlXaHWU / /'itri7--TT>" 981 J-i* o tewa, 1 1 ■"V -ti.o ' A TABLK OF NAT0nAL ilNEfl. 7l| ' i^' 11987731 Ui987779 ssnwfl 99^^(110346^2 ft«.lte96-S64|8e8&» 989a4^fiO|| 9P03U* IWJoHJ 9941 I'.JI, 1j9 »i J9«64d 999401 9903*0 991!6 994 obn 99941] U9985aS8 IWUSSS 99-^ J 9'H 99942J 99986»eT ffitOU^ flaosog J9 5 991 Jf 41 199877 5*11 [wufiia 992 92 994 J. tmsisa B&yjea 9B«ei7 9947h .JUOJ <mBsaB3 99oejii 99-2862 994792 99«4 J Mril«8!38 l>92S9a 900TW 990-48 99-296fl iai\)S»iT.I t*lU885n 9908^7 9a.*m is'Msaes 9ffll86G 99306S lti!»S40U BldBOa iO.103 17 9*«-T") DHWM 911 J7|4 J0 938j«. JM JO 90J ■>« 4.lMMg583 n 4U 91J8JJ 991 7B3 J"!.!!*. W1S2I)^ P91S57 61(9'^ JO Jta'9l4J94J ej989l)8.t 9^2234 eeo^ 1 fl7 99 u wnsii- VU30 11)2511 114491 99'2Mfa 1 994622 Biuei 994<r2fi 994(1* 9940)48 gwi^o Q9«ai 994183 994.14 09jjU7 997008 99j6B 1 99v5H9' 19,'Hil (97063 99 7fi •)■) )r wriiu 998"74 IWgi 99a»l 8991 99a'iU8 9« 9 9996^98 99904 29 J99672a , 1W«) 49998937 I 990ae~ 999971 S« .990694 99eg71ifi 1 999701 99ft97S3« > 999709 99997828 ■ 9TO716 9<t99S03» M 999722 9999B121 l9ff)7'«) 99998330 rf9jS8* 407314 I 998441 '999'^ 99% 99 Jd6 99845 999274 J959-r D47a '99W 3 1 9992SS 999 K 99990619 999743 99998G18 999749 99998817 699T66 99998916 999703 999S901S 999-S 999996 994.'-( 1E3=I I n TABLE VI L ] Amount Dt L.l laid M compound intcrut roi u; number of fouL ' 1 i!w.Ww , i,iK>iV^ hkMk^ lo^KKXJ i.mm. LoAuouo: iMm a' l.(l6<^2S 1.06U900 1.0712^ 1.061600 1.002035 3 1.0768BI i.i»a7a7 1.108718 1.121861 1.141166 1.10U813 1.136SI» 1^88869 1.193619 1. 131406 1.160371 1.187686 1.316663 1.948183 1.278282 ]gM e 1.1S»6»3 1.I94U52 1.329265 1.265319 1.S03380 1.4O7100 umOB J i.iaaaso 1J172279 1.315939 1.360689 8 1.3]MI)3 1.916800 1.368669 1.423101 1.4n4S6 i.«9 8 1^48883 1.304773 1.486095 1.M13W IMM 10 1.380K6 1.313916 i;il0699 1.480344 1.562960 1.6388M i.mM 11 1.31 «)87 1.384331 1.499970 1JB9164 1.6338D3 l.Tlim J^jH 1.344889 1.435781 IJHIOBB 1,601032 1-696891 1.378611 1.468531 1.58:»66 1«H1074 1.772196 ijBsM ^^m 14 1-113974 1,613590 1.618695 1.731876 1,801916 ijirnn kS9 IS M4K!08 1.557067 1.876349 3.07«M HBS 16 1.18t6<>a 1.604708 1.733086 l[87298i 2;02«ffU ajffisn itSM 17 IJUIGIS 1.663848 1.794876 1,947000 3.ll8Bn sxim ^H 18 Lsseese 1.702403 l.a6748e 3.035817 2.208479 a.tomil IS 1.598690 1.763508 1.033501 9,106849 3.307880 s-Bjewi i.saMils 1.806111 1.960780 2191133 3 411714 26taiM g| 1.679^83 1.80029S 3.059131 JJ7876S i7Bg*aa a* 1.731671 1.916108 a.i3iHia J 360019 3633669 sjciun 33 1.761G11 14B3587 3.i08114 i464 16 3 763160 a071Bl<l 1 31 I,e0871i6 3.039791 9.je3J3S 3563304 .i 876014 asasiw 1.863944 2.003778 3.368945 9 66.1636 aoosiBi 3.3a68Si 38 1.900eB3 2.156591 3.116959 2 7724701 8140679 a555e» 1.947800 2.831289 2.6^1567 9883388 3.282010 a7a3» 38 1.996496 2^87938 3.BJW17i .^998 03 atwioo asm 99 3.010407 2.366566 3.7118-8 3118861 3.5840^8 4iims> ao rrniMS 2.427382 S.806794 3.^3398 a *MiB i.mni a: >.lflO007 2.500080 2.905031 a913857 tmxa 83 2-3KITfi7 3.008-08 3 508069 4 068961 4.;«w 33 2.368851 2.052336 3.111913 364S3S1 4 37«M0 6,ilWlB» St 3.31fiaa3 2,781908 3 794318 4.ieej(, , w ss 3.37330S 2.813862 3.916089 3.432635 3.898978 8.450386 4.103933 2.493349 2.086327 3.571039 4.^68000 a074783 3.608011 39 3.618674 aiS7027 a826a-2 4.816368 BJibS 1 1 40 3.685064 a363038 S*»2C0 4 801021 6.810.*. J S iOWM 3.752190 3,859899 4.0978J4 499J0S1 6.0-8101 42 3.890996 a4606a6 4.341.f58 5109781 6351615 3.801620 asAifii? 1.380703 £400196 6 637138 8j*«g3™l 44 3.963S08 3.671152 1.G13312 5.618615 e»3<iiJ3 4S 3.037903 3,781698 4.-na359 5.811176 7 248- ) ' «'""■]?; 46 B.11386I 3.S9*W4 1.868011 8.071833 7Ji4-l 1 1 47 3.191687 4.0118!J6 6.0J7284 8.317816 48 3.371490 4.133252 5.213^ 6 870528 46 4,S.^ZI9 5.39e06u 6833349 3^437 109 4.383906 5.58192; SI 3ja3036 5,780a90 63 a.6U113 ijxmfis fijiaarid 7 (Ml 63 3.701390 4.790112 0.193108 1J4 1 1 a7nap2jj 3.888773 1.9.tll25 6.**83J 5.08^149 1 B.fi*!lll KM II li ^■M 3^86993 6.334613 1 H.R66J01 mmLit.06Bm 6391861. llAOBSSl I li lit ^^^V 1 .1.1877 H3 6.KI340V\T364aea U ^^Bl *.393m r,.-ioim' I6\i6«t 1 A ^^^L«.8997» 1 &.t>ai60a' IXiVKSt^ \ ^^^3 *■ TABLE VIIL 76 The prwenl nine of L.l due at the end of amj Number of Yesri. &Oirpircen- [ajprcWt. ,.p««nu T^ -ftifiHlO .9fi6184 IMm* .ysuasi .9i>39« 3 .951SU .9«.'i96 J);i35!l ,fi34B.«) .015730 .907039 3 .!) 161*2 .901943 .883996 .876397 .8G!M38 .839811 4 isofissi .88S*a7 JI71443 .854804 .838581 .793094 S .HSSBM 341973 .831937 .803451 >a3636 .747368 a .882297 .767996 .746216 .704961 » .8tl2fl5 .813093 .785991 .769918 .734838 .710681 .666067 8 .820747 .789409 .769413 .730690 .703185 .876839 .637413 ^ .800739 .786417 .733731 .703687 .673904 .644609 .601898 .T811S8 .744094 .708919 .676664 .64.3938 .6i:i913 .658395 .T6ai4« .723431 .684946 .640681 .616199 .636788 |2 .743556 .701380 .661783 .634597 Ji89U64 ;5S6837 .496969 ,72Si30 .680951 .839404 .600574 JS64273 .630331 .468838 .707737 .661118 .577475 .639973 .443301 i** .890166 .641863 J»6891 J»6364 J116730 :*81017 .417366 i« .e7.%25 .633187 *7870« .M390e .4»*4e9 .mm ^^ .857196 .606018 .657304 .613873 .473176 !436397 .S713fl« i« .841166 .68739S .638361 .»361S .453800 .416531 .360344 is .SiSBlS .570386 .«2015S .474643 .433303 .395734 .330613 ^ .619371 ■663676 .SOSAGS .456387 .41464.3 .376889 .311806 ai .535386 Ji376«e .485671 .4BS834 .306787 .;I68943 .394166 29 .880865 .631893 .409161 .421966 .370701 .377605 3a .666697 J(06S93 *S3a88 .405736 .363360 !336S7l .361797 94 A53876 .4919D4 .437967 .390131 .347703 .310008 .346979 35 .477606 .433147 .376117 .833731 .395303 .293999 28 .536335 .463896 .408838 .360689 .318403 .381341 .319810 27 .613*01) .450189 .396013 .340817 .304691 .267848 .307368 38 .600B78 .437077 .S816M .833477 .391671 .196630 39 .188661 .434348 .869748 .320661 .»79016 .343946 .184657 SO .476743 .411987 .366378 .308319 .367000 .331377 .174110 31 .399987 544330 .396460 .366503 .320360 .164366 sa .388337 .333690 .344600 .309866 .164967 33 .443703 .377036 .331343 .374094 .J83B71 .199873 -146186 3* .4.51905 .366045 .310476 .363663 .333896 .190356 ■137913 as .431.571 .366383 .399977 .353415 .214364 .181390 .130106 36 .411094 .346083 .389833 .343669 JW5D38 .173667 133741 97 .401087 -334683 .3S00S3 J«14397 .196199 .1644.36 -1167S3 38 .391385 .336236 .270563 .335385 .187760 .186606 -109339 89 J8I741 .361413 .316631 .179665 .149148 -103066 *0 .373431 .306667 .3626rJ .308289 .171929 .142046 «97333 41 .363347 .297838 .344031 J»0027a J64636 .136383 .091719 43 .354W5 .3S8S69 .335779 .193676 .167440 .128840 -086637 48 .345889 .380543 .237806 .186168 .160681 .133704 -061630 .337404 .379373 .330103 .178046 .144173 .116861 ■077009 .339174 .364439 .313639 .371198 .137964 ■073060 46 .331148 .356737 .305468 .164614 .imm ! 105997 -068638 47 .313S13 .349369 .193630 .168383 .100949 -0646r>8 48 .806871 .241999 .191806 .163196 !l30S9S .096143 -060998 49 .398316 .334960 .186330 .146341 .116693 .091564 457546 .300943 .338107 .179063 .110710 .087304 ■064388 61 .383848 .331463 .173998 -13S301 .106043 .083051 4161316 82 .378833 .180097 .101380 .079006 •04831S S3 .870189 .161496 .136093 .097014 .0^6330 .0(6581 «4 .368679 .303670 .168035 .130383 mssn «]>»««» » JW7J51 .196767 .15»»768 .115ftM \ J8S«M6\ J»KSA\»«* S8 1 .350879 .191036 .146660 .111201 \ joatft\%\ .'*««\'}.\'«« Br / .3*4700 .140784 .106930 \ JO»\SEa\ s«v^iAT ^ / 338790 180070 .136076 1 .102811 \ iynwa\K*»ga\J iS 1 .232986 / 174836 .131377 \ .osasA i \ S3\VSS\ \ .«*»».S 0J^ 337384 /. 109733 .126934 \ .09506 O \ jtfl\'«S\-'= ^ I 1 ^^^=^= — - ■ 76 TABLE IX. 1 The aamunl of L.l per ttnnuai in any number of Vea™. 1 l.UWHMJO l.umiiMJi l.UIMHiu, l.UKJiiUO i.uwxmu 2 OSOOoS 2 0^ a.035000 9.03500UI S.04UU00. 2.046U0I 3.975025 3.106225 a. 1216001 3.137035 4.1SMW 4.31«64, 4.279181 s.3Masg 6JW9138 5.862466 e.4i8aK e.47071< ftSSTTW 8.468110 8.S5015S 6.633976 6.718892 T.MT4ao 7.770408' iism* 8.01U15i a73mi6 9.061687 9-311336 0.854310 10.150106 10.36S49U 10-583795 ii.uimsa 11.731393 12-006107 19.a88a« 13.483480 12,8077IM 13.1*1992 13.486861 13.8U178 13.795553 I4.6(J19B2 15.025805 le.inMia 16.6177(« le.ll.'fOSO 10.626838 17.'l699i; 16.618053 17.670988 18.9331^;^ ]■.;■:-.- .. .-;« lT.Oai027 1&5989H 10.ai5USl aci;023688 19.380226 30-160881 2o.97in;)o 21.821631 2271!)-;^!: j- ao,8e4;3o 21.761588 33,705018 2a 697512 33.386349 23.414436 24.499691 35.046413 26^S66<Vl .-,.--■ ..VI 33.946007 26.116868 36.357181 37.671339 29.0636«ia ■l•).i^i^J^^•nt M.in-jmk 30 2G.544658 26,370374 38,379683 39-778079 27.183274 V8. 876486 30.269471 81.869209 33-783137 29 2S.86286(( 30.530780 33.338902 3*.a47970 36-303378 23 30.584427 ai.4528S4 34,460111 36-617889 88.937030 24 33.349038 31.426470 36.666328 39-083604 41.689196 34.157764 36.459364 38.949857 41.646908 44.565210 *t 16 4 .5 0015 jn 3 00 40 09634 1. 9060 081il 50 1331 H.660 Si Jn 3985980 13 U933 41, -W "7 90 688 5. 9^3333 68 4(2683 58290 46 H8C0 !•■ 13 6 3 1390 03 «l 46 >» 50J02e7S * J ■i 48 a a S. .502759 J 60d54O34 66 07 j S3IIU885 1 M-oasw 60 1t> R 69 33948 6 3^m 60 64 S3979 w 6 4f 554 08701 2S39S0B B- 02.196 93 i. 66«080i 86 483899 80 i 8 6SJ333 8904 1 ,« 81.610131 9^ 71 .1 48 81.55403* 90 i 87.607885 IK .[( U 04 jS 49 8l!l310J 97.184349 .13 ".t Bl 100.921458 11-1807 J 13t, S 53 101.444491 m 0931971112 S3 oao556oe id'mim 1» i li6i 111.756996 131 V6S /lI5.5i5O021 the /119.439G91 W /133.425667 PS /127.611329 » lai-flgsij" FO (135.991590 163.0634J7\l9GB^^«»* ■ ! 5= TABLE X. 77 The preacBt i-flliia af L.l per Bnnum for any Nuinbei of YeBrB. ^ '«i«c""- aMrc*t.|at.-r™^ 4p^c.nMl(p„.™tl,M,.,.™l h1p„r™i rr .(175010 .970874 .96618* .kifito, kmx-i, k,i.^ lU'Liyiil !i a 1.91742* 1.913470 1.893004 1.880095 1.872UUh : 3 3.856034 2B386I1 2.S01637 3.775091 2.748964 \. * a073079 3.629895 3.587520 3.466100 a 4,6*6828 4.679707 4,389977 4.329477 4.312364 ! e 6.B08135 B.4I7191 5.»iAo--ii 6-157872 5.075693 4,917334 T G.31S391 6,33038.3 iMHAi u.iwms'i 6.893701 6.780373 6.682381 B T.17013T 7.019692 0,873058 0.73^1745 0,696386 6.463213 6.209794 9 7.970866 7.78Gli« 7.607»S7 r.43S,i;i2 7.288790 7.107823 6,801693 ID B.7fi3064 8.316006 auoawi 7.912718 7.721736 7.360087 11 9,614209 9.:^ h 4 8^28917 1106414 imsas 19 10.2677((fi 9,y540-M 9 66Ui4 938.10 4 91186^ 8 803253 8.383844 13 10.983186 10,634B'J6 1OJ0273S 9 t&M 1 > U 11.090913 11.296071 1J*306J0 If 1 ^ 1' u 13.381373 11^07930 11JU7411 IS 13.066003 13 094117 17 13.713iae 13.166118 U 651321 IB 14.36S184 13.753513 13189083 19 14.878881 14.323799 13 70(H37 IJIkI 1 » 15,589162 14,877475 14 12403 la-wj 1 1 91 16.184649 14 697974 14.0"911,0 114 1 SS 16.T6S413 16,936917 ]6 167 1J5 L4 4.11ir 11 1 33 17,333110 16,443006 16 620410 34 17.884986 18.936543 I6 05S.WS 25 18,434378 17.418148 16 4S1 1 11 96 18.960611 17.876843 IbH^ 'l BT 114e40U 18.33 031 17 -s I 1 aa ia,964fi89 18.704108 171.1 U 1 89 90.453660 19,188466 18 0k 1 1 80 30.930293 19.60O44I 18 jgjii 17 ) i. 10 Sh 4 1 i 4 IIJ W83l| 31 81.396407 18 73037H 17 6884D4 16 544391 83 21.849178 20..'!88766 19O0I4S66 17 87355^ 16 788891 16 802677114 084043 Ha 38.291881 20,763783 3l,l3fc37 19 3B02OS 18 H7B4t. 1? ^Sfl' 16 00364914 230230 34 aa.73378B 19 7l)08« 1S41119B 17 246758 10192904114.368141 SA 23.1*5157 21.487220 -OOOOOUl H 604613 17 46101 lflJ7419414 4J824b 36 33.S6636I 18908 8- 1 666041 1654686 14.620987 ar 23.967318 19 14.n 117 %3l4n 16 71128 14 736781 38 34.348603 111.^ 8C4|]8 04.f990 Jb^89.il4 846019 ae 84,730344 22.908 15 lJ6S44S6|lti22'J(>,% 17 01 04lll4 949075 M 35.103775 33.U4773 21.,J6a073 19.. 92,, 4 18,401584 17.10908615,046297 1*1 35.466122 23.413400 31.599101 19.993052 18.566109 17,294366 15, l;i80l6 .43 36,820607 33.7<U36» 31.a«6fl;i aul85627 18.723550 17.423208:i6.23464S 13 36.166446 83S81903 32.063689 20,370795 17.545912.15.306178 M 36.603849 94.35^4 33.382791 19^018983 17.66277316,383188 46 36.833034 a4ju87ia 33,496460 .■0,720040 19.16(a47 17.77407015.455833 ,'« S71M1 24.776449 32 700918 ^884054 19388371 179aOI*7 6o4 37 487483 36034708 2^899438 1M2B36 19414709 981016 6 6>Wn 8 iS J7nai54 35,^66707 23 091"^ ^1 195131 19,535807 ,49 280 'm "6 50 667 19 661398 816872- f 2 fiO 7") A J4M613 214SJ185 19 762008 1825692jl (,8(1 'fil ■« 6 21 617485 19867950 I83J8! 158 T 6 53 65 1 475N lH4 8f7 1 S.1J93 es 8726 o 2i 0()6345 8 41140, I6<>0C974 6* 99 m ■,■66 «^ W1 ■i&^y.W,! •*«»« ■0S 1 JS A- W.W.^M>^. ^ '^ SIS T. ;:i^^ Wji P 78 TABLE XI. ^ L«nSthB of Circulu Area to Rwliiu 1. J -rr JS 1 .(1174533 16^ ^^7936^7 1 m>ti S .08«»8« 17 .2W17060 «j .8736046 3 5818 3 8 .odiatm 18 .3141693 60 1.047197B 3 « .0698132 19 .3316126 70 11636 S .0873065 .3490668 1.3963634 6 T .I0471BB .1W1730 23 JW6S101 .383973* BO 100 1,5707963 1.7463293 6 7 17453 8 .1398263 .4014267 11» 1.9198623 a 33271 a .1070796 at .4188790 ■120 2.1Ht*30fil 9 36180 10 .1T4S3Z9 .436.3333 3,2689380 sme .IBlSSfla 36 .4637856 £.4434610 SO .aoii*3»fi r*7iaa80 S,(J179939 30 8736G 30 .asmfis .48809^2 160 40 116365 *o; 14 .a4*MGl 39 .6061ifi5 170 2'9fl70597 60 1454M 60 IS .■jauw* 30 .6-23S98fl IBO 3,1415937 60 1 TABLE All. • 1 1^; rHittT -KStT — TnT |(Ut..| Amt-T oiSn T ■nm- ooi?e4§li3 W rssT olSee^ST *+ ITOIM { 006 0021660618 '4 1,096 0107338664 ^ 0191 1 b| 4 l!o476 0201 01 OM.'!ai373a 013flS79347 1.0B Oil 11 OlSfi 00B395031B ,0385 0138B00003 1,0626 4 OIB ooeieooiaa ,035 aaa l| 00TS3i4J78 .0376 0169881064 l!0575 rm ooseoonia 017033S393 6 1,06 TABLE XIII. ^^H UKrulNumbem.wicbtheii'J'aguithini. ^^^H „ ,. . . «* 1 AreBorcircle (do. do.) .7gM T« Conlont of sphere (do. do.) Jme T.TI Surface of a sphere (do. do.) 11416 .tf Numb^ofeecondsiiiSflD'' USeuOO CU Numbor □rarcn of 1° in tlie radius fiT.39fiT8 l.TI 'BaBeofNnperinnLoBarithms 3.71823 .43 /jJI!odaliiiolcomramho^^t\'Caai -VHUdM T.6S 1 Inches inn meUe 'ft^W* TJ« //common tropical year in mton «.\m ia'!* m^ffii ^ 1 Inchea in a pendulum, flhictv VOoTUte* «s<^™4'.-v g,jj(|a