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Les diagrammes suivants illustrent la mdthode. errata to I pelure, on d n 32X 1 2 3 1 2 3 4 5 6 F^LANR TRIGONOMETRY V"' l\k \- Mil SOLUTION OF TlilANGLKS. J. H. CHI{RKI.\[AN. MA. Snfnmlm.lent o/ Im„r,mcc /„•- tli,- /)o„„„w„ ,., ( ■„>„„/„ ■ l-HIlMSMPirv IN- INIVI'liSlTV ( ol.r.l:. iF, K.KONT.I FOUk'Ifl KDIIION. ^Vni. NWMKKO.K KXAMPUCS AXU Kor K-,. ,.;,• K K TA.U.KS .,K ...X^.KMMMs '>1 MMBKKS A.M. OF | IIK TKICO.Su.MI; | ,; ,rAl KAII.l^ KDri I II H\ AIJ'REI) BAKER, M.A. MATHI.M..TI.A, TL'T.IU. rNlviK-sriV .„,,,.,.■,.„., ,„„„s, I <» K <) \ I () <^OI'l'. CIA KK \- ( () 4 7 IK') N I S I- K K K T i-; a .s T. I S h o . LOGARITHMS. 1. The conunon logaritlim of a number is the index of the Jcmud^*'"" jHjwer to which ten must be raised in order to produce that numlier ; so that iu the equation 10' = N, X is tlie logarithm of the number X, and this is written z = log N'. In general the logarithm of a number to a given base is the index of the power to which the base must be raised in order to be equal to the given number. So that it a* = Ji, :• is said to be the logarithm of N to the base a. This relation is also thus expressed, x = log,^ 2f. Thus, since T = 49, 2 may be said to be the logarithm of 49 to base 7, or 2 = log. 49. Any }>o8itive number, except unity, might be taken as the base of a system of h>garithms ; in practice, however, only two bases are used, the common base 10, and the Napierian base, 2*7182818 In the following pages, unless the contrary is stated, the word logarithLi means common logarithm, 10 being tlie base. 2. The logarithms of numbers which are integral |>owei"8 of ten ure immediately known ; for example ; • W = 1000, log 1000 = 3, 10» = 100, log 100 = 2, 101 ^ 10, log 10 = 1, 10" = 1, log 1 = 0, 10-1 = 01, log 0-1 = -1, lO-!* = 001, log 00 1 = -2, 10-3 ^ 0-001, log 0001 = -3, For numbers greater than ten, the logarithms will be positive iutegei-a or mixed numbers ; for numbers between 10 and 1, Cliariicteris- tic «n to 10"'"" \ their logarithms will commence with — n and run up to — (n — 1), that is, will V)e — n increased by some decimal, and the characteristic for all such will therefore be n. Hence we have the following rule for finding the characteristic of the logarithm for any number : If the number be an integer or a mixed number, the characteristic is positive and is less by unity than the mimber of figures in the integral part; if the number be a decimal the characteristic is the number of the place of the first signi- ficant digit, counting from the decimal jmnt, and is negative. Thus for tlie following numbers 12345, 12346, 123, 054, 0000543, the characteristics are respectively 4, 1, 0, I, 4. fy. Tlio following fti-e tho rules on whioli arc f.,nn(le.I tliu inv..Mti^a us(>8 of logaritluns in porforniinj,' urithnH-tieal operations ; nil ',"'„['" (1 ) '''g (fi b) -- log a + log A. Lei X - log a, y ~ log b. so that Then, so th.tt or, (2).. r.et so that Then, so that or. (3) Let Then so that or, (4).. Let Then 10* = a, 10" = h. a h = 10' X 10" --. 10' + " X + i/'iH the logarithm of (a h), log (« b) ^ log a + log b. . . log -^ = log a - log b. X ^ log a,y = log b, a, W'' --: b. 10' a l W 10* = 10'-^ X — ;/ is the logarithm of ^*, 6 log J -=log« — loiji. log (f{") =r= n log «. X — log a, so that lO' = a. nx is the logarithm of «", log (a") = n log «. • log ("l/rt) = log a. X = log a, so that IC* = a. - ^ %/rt = a" = (lO')» = 10» IIHJIl),' Idj;- ahthiiiK III iiritliiiii'ti''iil <>p«Tatli)iiii. 'fT^- KO that X n or, is tl»' logarithm of *j/«, log ("|/rt) = - log a. 0. Any of tliese operatioiiH iimy be coin1)inelpH will l)o uviuliihlo only tor niunlx'rs oonsiHtinjjf of six figuroH or 1«'hh, that i» (diMre^Hnling tho ducinial point) for numtx;rH iimging from 1 to I ((00000 The inantissius, however, aro not entered for all those niiuiherH, but only for those terminating in the hiindr(>dH ; for the intermediate nuinl»er8, the niantiHsan must Ite calciilnted l>y aid of the principle that *he difference In'twcen the logarithmti of two numherH is proportional to the difference between the nntnhoi'8, when the numbei's are taken sutHciently close. Thus the difft'reneo between two consecutive mantiRsas in tho ttible corresponds to a difference of 100 between the numbei-s, and we obtain by a simple proportion the difference of man- tissa corresponding to any less difference than 100 in the numbers. E.ij. . . Required the mantissa for the logarithm uf 675347. From the tables, Number, 675400; mantissn, 829561 •' 675300; " 829197 Difference, 100 ; diflference, Then, by the principle, 64 regardii)g tli« (luciiiial |)oint, look in tlit! talili) for tlic first :{ tlgiircs of tlu; luiiiib'tr in tlio Ivft-hand coluntn, ami for tin- fourth tiguro in i]w top lino ; at the intoi-Hoction of tlio corrt'spoiiding lino and colnmn will ho found the inanti-sKa ; for th(^ tifth Hguro, look in tho tablo of proportional parts and takt' out tlu! nuinhor for that column ; and for tho sixth figuiv, also from tho tahlo of proporti., 1) , 5, diff. << 11 J) -7 327U<.)5, Mant. 515470 Til taki- out tlif iiiiiiilicr (• irii'S|M)iiti ill;; ti) a ^in'M lojj.ir- itlllri. and the uharacteristic is 2 ; therefore tlie logai'it))in of 327 0'.).') is 2 515470. 10. The reveree process of finding the number correspond- ing to a given logarithm is peiforined on the same principle. Disregarding the characteristic, look out in the tables for tho mantissa next below the given mantissa. In the correspond- ing line and column will be found the first throe and the fourth * TluM'i! are iiuiiierous tiibli's of logiirilliiiiK vn'il'thi'd. Four-dgiire lognrithnis of niimlwrs are given on a siiitjle eanl (witli luiti-logaritlniis on the reverse fuee) ]>ii1p- Iim1ii''1 liy Liiyloii.s, Lomloii, for tliruo llgiirea given ilirectly, iin hntwocn til)' iniiiitisNii thiiH t'oiiixl tinil tli(> ^ivcii one, niui iiiso tiiiit ln'tvvtM.'ii tlif foriiwr and tlu« next liij^licr in tlio tjiM«'n (wJiicli will Itc tin- ilitl'on'nce for 100 in tim nunilxT), l»y ii Hinqih! proportiou tlio tens and units in tho rcquinMl imndK-r ure found. The deciinul point must then Ix? insfitcd l»y consi dfi'atiiin of the cliamctoristic of the j^ivr-n logarithm. Kxiuniilu. Kiml tlio numluT c-orrcHiMimliiig to tlic given logarithm 'J. 7t»7 1 ".>S Tilt! iii.'HitiHsii next lu'low in 7<>7l.")er found in the table of proportional parts, nnd removing the decimal point in it one place to the right, look out again in the table of pro- portional parts for the number nearest to it, and the column in which this is found gives the sixth figure. The previous example would be thus worke7198 ; Mantissa next below, 707156, corresponding number, 5850 Difference In table of p. p., diff. next below is Residual difference 42, 37 0, This gives for the six figures, 585057, and the number rci^uirud is therefore •0585057. « limi.f We shall now exemplify the rules for jwrforniing arith- iiumiiti'i'm- nioticiil openitions by aii\ JHiiiii. Quotient, 4-2206 ; log, 0-62537 Ex. (2). Divide 2-7045 by 31279. Dividend, 27045 ; log, 0-43''00 Divisor, 312-79; log, 2.49525 Quotient, 0-0060465 ; log, 393684 Here the subtraction is 1"43. . , . — 0-49. . . —2 — 1. Ex. (3). Divide 465-94 by 0793. Dividend, 465-94; log, 2-60833 Divisor, 0-793; " 1-89927 Quotient, 687-57; log, 2-76906 Here the subtraction is 2-6 —0-8 —(—1). \ Ex. (4). Divide 0-0037095 by 0-00001605. Dividend, 0-0037095; log, 3-56932 Divisor, 0-00001005; log, 5-20548 Quotient, 23M2; log, 2-3G384 Here the subtraction is 0-5. . . , _ 0-2. . . . + (—3) _(-.'-,) w Use of 13. It is convenient to convert the process of subtraction .iritlinirtical _ _ ' _ . .piiipie- into one of addition by the use of what is called the arith- Illl'lltS. '' metical comjdement. Thus if h is to be subtmcted from a, instead of subtracting h, add 10 the result ; for h, and subtract 10 from a_i = a + (10 — 6) — 10. Tliis quantity (10 — h) is called the arithmetical complement of />, and is found by subtracting the first significant digit, beginning from the right hand, from 10, and each following digit from 9, including, in the case of a logarithm, the charac- teri.stic with its pro])er sign. For example, Number, 239-31 ; log, 2-37896 ; co-log, 7-62104 ; 0025177 ; log, 3-40100 ; co-log, 12.59900. The working of the previous examples would then stand thu.s. Ex. (1). Dividend, 32-495; log, 1-51182 Divisor, 7-6993; co-log, 9-11355 0-6253; Ex. (2). Dividend, 2-70-15 ; log, 0-43209 Divisor, 312-79; co-log, 7-50475 3-93684 Ex. (3). Dividend, DivLsor, 465-94; log, 2-66833 0-793; co-log, 1010073 2-76906 Ex. (4). Dividend, 0-0037095; log, 3-56932 Divisor, 0-00001605; co-log, 14-79452 2-36384 14. To raise a number to any power. Invcilutiofj. Rule. MultipJij the logarithm of the number hi/ the power, and take from the tables the number correspond iny to this product as a logarithm. Ex. (1). Find the sixtli power of 23-91. Number, 23-91 ; log, 1-37858 6 Required power, 18G8400(»0 ; log, ^<-27 1 4>S ^ Here the power lias 9 figures in its integral part, of which only 5 are lement is used, a ten must be subtracted from the result for each complement. Ex. (1). Find the value of (12-345) «5 670-59 X 50-323 Number, 12-345; log, 109149 « 5-45745 .... 5-45745 670-59 ; log, :J-82646 co-log, 7-17354 50-323; log, 1-70177 " S-29823 Required value, 8.4961 Ex. (2). Find ^ i log, 0-92922 Number, 5 ; log, 0-69897 " 6 ; co-log, 9-22185 3)1-92082 Required value, 0-94105 ; log, 197361 The operation here is this : logf? = ilogS = S(log5-log6) = i (log 5 4- co-log 6 • EXERCISE I. 10). 1. What are the characteristics of the logarithms of the following numbera to base 10: 3740, 33-492, 76495-9, -34781, -0000053 ? 144 2. Shew that log _= 5 log 2 + 2 log 3 — log 7 — 1. 3. Shew that lo^li^ m g^g =M-logl3-Iog2-]og3, 4. ijhew that log 8 -f- log 25 = 2 + log 2. 5. Prove log ^^|^ =log3-log2+ J(log 13-log 7-1). 6. Given log 6 = a, log 15 = 6, find log 8 and log 9. 7. Find tlie value of log J in terms of log 25. H. Multiply 1-372'1801 by 5. y. Divide 302135G9 by 6. 10. Find the value of T-4873051 ~ "s^Qooo^l _ k ^4721053). 11. From f of T--4214036 take J of 3-4729104. 1 -2. Given log 7-3335 == -8653113, log 7-333G = -86531 7'^ Hud log -07333572. ' ' 13. Find the logarithm of -06919583. log 6-9195 = -8400747, log 6-9196 = -8400810. 14. Find the logarithm of 56201-25. log -05G201 = 2-7497440, log 56-202 --= 1-7497518, 15. Find the logarithm of 2965845. log 2-9658 = -4721419, dif. = 146. 16. Having given log 2 and log 3, find the logarithui.s of the following numbers: 18, 25, 216, 1^, 6480 -0054 4 43-2, 720, i/I|, fThiH)'^. log 2 = -3010300, log 3 = -4771213. 17. Given log 2, find log -00016, log ( -00001 6) » 18. Find the logarithm of \ jl\'32xriS\ 21/27 lunifig given log 2 and log 3. 19. Given log 2-6201 = -4183179, lo,g26202 = 2-4183344. find the number whose logarithm is 34183253. ^ 20. Givenlog56248 = 4.7501071,log56249=:4-7.D01148, find the number whose logarithm is 2-7501113. : u 21. Given log 30-413 = 1 4830593, dif. = 142, find the uumbor whose logarithm is 4-4830G51. 22. Given log 49553 = 4G950700, dif. = 87, find the jiuniber whose logarithm is 3.0950741. 23. Given log 5-6043= -7485214, log 5-G044== -7485291, form the table of Proportional Parts, and employ it to find log 5G0-435G, and also the number corresponding to the logarithm 37485282. J/ . 24. Find the value of i (72489)'' X F 2-345G. log 72.489 =- 1-8G02721, log 2-345G = -3702540, log 18507 ^ 2-2G73380. 25. Find the value ^^i^ log 2 = -3010300, log 3 = -4771213, log 7= -8450980, log 100G5 = 4-0027894. ^3 26. Find the value of -pr= — j-— • |/4. I/O Given log 2 ^ 3010300, log 3 ^ -4771213, log 739148 = •8G87314. o- -p- 1 1 I -.1 .1 1 V ^»00 X -03 X -105 2i. Jut the tangents and cotangents ;nay liave any positive values. 19. As the angle A increases, retaining the same hypothe Tht>r niise, th(! perpendicular increases and the biuse diminishes vftlue. continually, and therefore the sine, tjingent, an') . = tan A. tan Ji. cot A + cot Ji 23. (cosec A — cot vl) = -j 24. cot^ + Hini4 1 + cos A 1 COH A -j- COS A cosec A. 25. cos cos y> — sin sin ^c ^= cos ' — sin " f . 2G. tan « tan /? — 1 = sin a — COS /9 cos ^ a cos " /3 27. sec -4 < 1 + cosec J (coH A — sin^J) ■ —cot.l. 28. sin ^ A tan A -t- cos A cot^ A =^ tan "' A f cot " A — 1. • 29. sin ^1 tan A + cos ^ cot A + 2 sin ^l cos ^4 = sec A cosec /I. 30 cot ' X — tan x = (cos ' x — sin x) sec " x cosec ' x. EXERCISE IV. Tlie formula! of § 21 enable us, having given the numeri- cal value of one of the trigonometrical ratios of an angle, to find the numerical values of the other trigonometrical ratios of the same angle. Thus, given sin A = h, to find cot A : cos ,s^ 1^'l-sin'^ V^l-I cox, /I = . — . = ; -: sin A sm A i = 1/3. 1. Given sin A = -, find cos A. 5 - 2. Given cos .4 = — , find sin A. 1 o 3. Given tan — -l, find sin 0. • a 4. If sin /I = .) find tan A. Q « • * 5. Oivon Bin x ^ -, find cot x. 5 t). (liven HOC x = -. find cot x. b t. (riv«'n 8ec x = ^/2, find oohpc x. 7 H. (Jiven sin + com = ■ , fin«l cos 0. 1». (iiv«'n cosec x + G Hin x" = T), find hui x. 1(». If (I + tun '^ A) cos A =: 2, find cos A. 22. The valnos of all thoso ratios am calculated for all T.iM.er fnictions, and so alt^o are some of the tangents and cotangents, their logarithms will have negative characttsristics, and to avoid the inconvenience of printing these, every logarithm of a trigonometrical ratio is increased \>y 10 before being entered in the table. To distinguish therefore the real logarithm ?"'"■ '.»'"''''' from that given in the tables, the latter will always be writ ''i>»'"kii>«1i " _ ' '' I'll flKIII tlll^ ten with an italic capital L, and it must always be borne in !^'"' '"«'" mind that 10 is to be taken from each such logarithm when used instead of the real logarithm, the operation being either expressed or understood. • For instance sin 30 Jog Bin ,30 L Bin .30 i = 5, log (05) = : 9 09897. 169S97, Also, tan 45 = 1, log tan 45 = 0, L tan 45° = 10 00000. 23. Again, since sin A X cosec vl = 1 , we have log sin A + log cosec A = 0, X sin ^ — 10 + L cosec A — 10 = 0, OH! 00 or, iiieiit of the tables of logiiiitliiiiic sines, &('. L sin A + L cosec A Antl similarly, Also, 20. L tan A + L cot yl = 20 L cos A + iy SOC ^ = 20. tan vl = - cos A A' log tan A = log sin A — log cos A, L tjin A — 10 = Z»8in A — 10 — {L cos A L tan A = L sin A -\-\^ — L cos J. 10), By aid of these formulas, if L sin A and L cos J be taVm- lated from to 45°, the values of the other logarithmic functions from to 90° can be formed. 24. In the ordinaiy tables, these logai-ithmic sines, cosines' ike, ai'e given for all angles from 0° to 90° at intervals of one minute, and it will be sufficient for most purposes to take out any required angle to the nearest minute ; but if greater accuracy be needed, recourse must be had to the principle of proi)ortional parts already explained in discussing the logar- ithms of numbers. The usual arrangement is that the angles from to 45° are placed at intervals of one degree at the head of the page, the it'inutes running down the left-hand column, while the angles from 45° to 90° are placed at the foot of the page, and the minutes run up the right-hand column. By this arrange- ment the same column is used for the sine of an angle and for the cosine of its complement ; and in the same way foi' the tangent and cotangent, and for seciint and cosecant. 25. Since sines and cosines are proper fractions, the tabular logarithms of them will always be less than 10 ; and since secants and cosecants are integers or improper fractions, their tabular logarithms will always be greater than 10. The logarithmic tangents will be less than 10 up to 45°, and after this will be greater than 10, and the reverse will bo 23 the case for the cotangents. The following table exhibits the changes as the angle pafises from to 90° : sine increases from to 1 ; L sin increases from — a to 10 cosine decreases " 1 " ; i/ cos decreases " 10 "— a tangent increases " "oc ; L tan increases " — a "-f. « cotangent decreases " a " 0; Z, cot decreases " -f a «_ « secant increases " 1 " a ; X sec increases " 10 "-fa cosecant decreases " a " 1; // cosec decreases " -f ex " 10. L tau and L cot are each 10 at 45°. 3 A be tabu- EXERCI8E V. 1. Find the tabular logarithms of the trigonometrical ratios of 30' and 45", having given log 2, and log 3. 2. Given L sin 22° 2G' = 9-581617/, L sin 22° 27' =r i)-58 19230, find Z sin 22° 26' 45". 3. Given L sin 38° 24' = 9-7931949, L sin 38° 25' = 97933543, find Z sin 38° 24' 27". 4. Having given Z cos 34° 18' = 9-9170317, Z cos 34° 19' = 9-9169455, find Z cos 34° 18' 25". 5. Given Z cos 57° 12' = 9-7337654, Z cos 57° 13' = 1) -7335693, find Z cos 57° 12' 24". 6. Given Z tan 32° 29' = 9-8039085, Z tan 32° 30'=- 9-8041873, find Z tan 32° 29' 27". 7. Given Z cot 16° 58' = 10-5155654, Z cot 16° 59' = 10-5151130, find Z cot 1G° 58' 18". 8. Given Z sec 73° 24' = 10-544ia74, Z sec 73° '>5' =. 10-5445314, find Z sec 73' 24' 36". 9. Given Z cosec 69° 34' = 10-0282238, Z cosec 69° 35' = 10-0281767, find Z cosec 69° 34' 54". U). Given Z sin 69° 7' = 99704902, Z sin 69° 8' =. y-9705383, find the angle whose Z sin is 99705261. 11. Given Z sin 16° 19' = 9-4486227, Z sin 16° 20' = 0-4490540, find the angle whose Z sin is 9-4488105. ^ 1 24 12. Given L cos 22° 28' = 9-9657199, L cos 22° 29' = 9-9656677, find the angle whose L cos is 996569 13. 13. Given L tan 51° 17' = 10-0960267, L tan 51° 18' = 10-0962856, find the angle whose L tan is 100962548. 14. Given L tan 30° 21' -= 9-7035329, L tan 30° 22' = 9-7037486, find A from the equation Z tan il = 9-7036421. 15. Given L cot 42° 12' = 100425 150, L cot 42° 13' = 10-0422611, find the angle whose L cot is 100423485. 16. Given L sec 47° 30' = 101703167, L sec 47° 31' = 101704546, find the angle whose L sec is 101703541. 17. Given L cosec 15° 21' = 105772220, L cosec 15° 22 = 10-5767620, find the angle whose L cosec is 10-5769821. 18. Having given L cosec 34° 31' = 102466882, L cosec 34° 32' = 10-2465046, find A from the equation L cosec A = 10-2466153. 19. Given L sin 28° 10' = 9-6739769, Z sin 28° 11' = 9-6742128, find Z cos 61° 49' 25". 20. Given Z cos 71° 45' = 9 -49577 16, Z cos 71" 46' = 9-4953883, find Z sin 18° 14' 10". 21. Given Z tan 52° 35' = 101 163279, Z tan 52° 36' = 10-1165897, find Z cot 37° 24' 50". 22. Given Z cot 36° 19' =- 101337003, Z cot 36° 20' =- 10-1334356, find Z tan 53° 40' 45". 23. Given Z cos 42° 26' = 9-8680934, Diff". Z sin 47° 34' 47". 1154, find 24. Given Z sin 20° 15' = 9-5392230, Z sin 20° 16' = 9-5395653, find the angle whose Z cos is 9-5394128. 25. Given Z cos 44° 20' = 98544799, Diff". -- 1234, find the angle whose Z sin is 9-8545671. 26. Given Z cot 57° 16' = 9 8080829, Z cot 57° 17' = 9-8078052, find the angle whose Z tan is 9-8080431. il 25 27. In the tables why tlo the same columns of differences answer for both sine and cosecant, tangent and cotangent, secant and cosine 1 28. In the tables for what angles will a column, which when read from the top is that for the sines of certain angles, answer for cosines when read from th© bottom I 29. Shew that the sum of the tabular logarithms of the sine and cosecant of any (the same) angle is 20, —also tliat the same is true of the cosine and secant, and of the tivngent and cotangent. 30. In increasing the true logarithm by 10 to fonn tho taV)ular logarithm, by what are we multiplying the trigono- metrical ratio ? SOLUTION OF RIGHT-ANGLED TRIANGLES. 26. Taking the triangle ABC, where C is 90°, and denot- Dhr;j'...,. small letter corresponding, the definitions of the trigonome- n'iSn',',i:le"" trical ratios give the following relations : of !>m. of'tiT. nnKlfS ill a . . a ri«lit-:inj,'l,(| Sin A = -, or, a = c sin A ; tii.nxi.. tan 4 = ^ a = b tan A; Bee A = -^ e = 6 sec A ; cos 4 — - 6 = c cos .4 ; cot J = - b = acot A; cosec A = - c = a cosec A. 27. From these relations, any two of the four qnantities T^i". part.. a, b, c, A being given, the other two could be fouud by aid (''".!" -u'lVa't of the tables of natural sines, cosines, ckc. ; and the i-emaining Ih 'uiluS;*' angle B, which is the complement of A, being thus found .soKvlI i 5-"iiur c'jiHes iitiiin. Ciiso 1. A sidi- a«(i iiii angle 26 also, the triiingie would be completely determined. Such a mode of solution would however be inconvenient, as involving long processes of multiplication, and we shall proceed to dis- cuss the iliffcrent cases of the solution of riyht-anyled triangles by means of the logarithmic tables. 28. Four distinct cases will arise, (1), an angle and a side; (2), an angle and the hypothenuse ; (3), the two sides; (4), a side and the hypothenuse. In cases (1) and (2), it is in- different which angle V)e given, as the other is at once known. The solution will be effected in each case by picking out from among the foregoing relations one which connects the quan- tity sought for -with two quantities which have been given or found, and it will be noticed that in each case there will Ix- two of these relations which would serve this purpose. If one involves a process of addition, and the other a process of subtraction, we shall always take the former. Case (I). Given a, A ; to fiud B, h, c. B^ 90°— A B found. h =z a cot A] Taking the logarithms of both sides. log b = log a + log cot A log b = log a -\- L cot A — 10 h found. c = a cosec A or log c == log a + Z cosec A — 10 .... c found. The li.V|Kitli- < iiiisi>anilan iiiiu'lc ''ivfii. Case ;i. TliH two tiiili's given. Case (II). Given c, A ; to find B, a, b. B=^0° —A B found. a =: c sin A, log a = log c + L sin A — 10 .... a found. b =z c cos A, log b = log c + L cos A — 10 ... . b found. Case (III). Given a, b; to find A, B, c. . a tan -4 = , , log tan A = log a — log b, L tan A — 10 = log a + colog 5 — 10 27 and therefore L tan .4 = log a + colog h A found. B=. 90°— J B founil. c 1=^ a cosec A, lo'' c = log « + Z cosec A — 10 . . c fouml. In this case it ia indifferent whether we determine A frt.ni tlie form.-.a tan yl =,• or from cot ^ = - Also there is not among h a our relations one connecting c with the given quantities (t, h, and although we know from Kuclid that c'^ --«'' + 6^ this formula is not convenient for logarithmic computation, and we therefore determine r by means of A, which though not (jlven has been already found U'e miglit also liave determined c by means of c = i sec A . Case (IV). Given a,c; to find A, B, h. sin ii = -, c log sin ^ = log a — log c Z sin ^ — 10 = log a + colog c — 10, C.isi' 4. A sill.' jiiiil till- li.vi"'t!i iliiisi' i;iii II. and therefore L sin A = log a + colog c A found. ii = 90° — J B found. b = a cot A lo" 6 — loa a + Z cot il — 10 .... b found. tound In this case it is indifferent whether we determine A from the formula sin A =-~, or from cosec A =:-. Also, there being none of C (I. the relations which connects h directly with the given quantities o,c, it is dot.3rmined by means of A which has previously been found ; it might also have been found from the formula b =^ c oos A. It is known from Euclid that h =c' — a , and h might have thus been found directly, but the formula is not convenient for logarithms. 29. The solution of iin isosceles triangle can be effected by aid of the preceding ; for such a triangle can be divided by a ))erj)endicular dropped from the vertex on the base into two right-angled triangles, equal in all respects, and by stdving the.se, the parts of the isosceles triangle also are deteraiined. ISilSOc'lf!* tliHn;;li-s S..lvr.|. tmmmmnmit I h 28 Kxaitii.i.s 30. Examples of right-angled triangles. Case (I). Given a = 129-5, ^ = 37° 07'. 90" 00' ^ = 37*07' B = f)2° 53' log b = log a + L cot A — 10. a= 129-5; log a, 211227 A = 37° 07'; Zcot^, 1012105 (i? found.) 6= 171 13; log 6, 223332 (6 found.) log c = log a + X cosec A — 10. log a, 2-11227 A = 37° 07'; L cosec A, 10-21937 c = 214-61 ; logc, 2-33164 (c found.) ■ Case (11). Given c = 31459, A = 40° 32'. £=dO° — A. 90° 00' " J = 46 32 5 = 43° 28' log a = log c + X sin ^ — 10. c = 31459; logc, 4-49774 A = 46°32 ; L sin A, 9-86080 (B found.) ft ^22832; logo, 435854 log b = log c + X cos .4 — 10. log c, 4-49774 A=iQ° 32'; X cos A, 9-83755 (a found.) b= 21642; log 6, 4-33529 (6 found.) Case (III). Given a = 2-7039, b = 3-4505. L tan A = log a + colog b. «.--= 2-7039; log a, 0-43199 6 = 3-4505; colog*, 9-46212 A = 38 ° 05^ ; LtanA, 9-89411 90° 00' ^ = 38 05' (^4 foiiml.) B = 5l'> 55' log c ~ log a + Z cosec A — 10. log«, 0-43199 A ^ 38" 05'; Z cosec A, 10 20985 (iifouml.) '^:JzA1^^L > ^^S<', 0.64184 Case (IV). Given a = 21, c = 21 -981. L sin A log a + colog c. ft = 21 ; log a, 1-32222 c=^ 21-981; colog c, 8-05795 (c found. ) A =72° 49'; X sin .4, 9-98017 /f_90° — A. 90" 00' A = 72° 49 (A fcinul.) i?=17° 11' log J = log a + Z cot A — 10. log«, 1-32222 A = 72° 49' ; L cot A, 9-49029 (B found.) b= 6-4940 ; log J, 0-81251 (6 found.) ■;: m ■ ^^^^H • 1 ^ 1 M^ 1 1 '^^ ■ ^Ul i i -i ■ SO The use of "traversc-taMea" may be briefly notetl in conncctiorr (vitli "plane sailing" an0° 4G' == 9747!)IL>5.' 11. «-3,c=:r), C^OO"; tind A and Ji. log3=-4771213,log5=.-G<)80700.L.sm3G=52'=9-778118G Lsin3G° 53'= 9-7782870. 12. «=^G00, c=UOO, C'=90°; find ^ and B log3=.477l213,]og7==-»4r,0980,Lsi„25»22'=^y.G31859r L8iu25°23'=^9-632125y. 13. «=l2,ft_ij,t' = 90°; findyfandi?. Iogl2 = l-0791812, log 19=1.278753G,Ltana2° 1G'=, 9-8002769, L tan 32° 17' = 9-8005567. 14. « = 21,i=27,C=90°;find4and^ lo.2=-3010300,log3=.4771213,Lcot3GO31'=l0-1305O6.> L cot 3G° 32' =10-1302028. 15. a = 200, c = 500, 6^= 90°; find b. log 3 = -4771213, log 45825 == 4-GG11025 log 7 = -8450980, log 4582G = 4-6G11 120 IG. c-=.3000,^ = 80°,C^-90^iind«andA. " L sin 80° = 9-9933515, L cos 80° = 9-2.39G70-^ log 2-9544 =. -4704093, log 5-2094 J: -7167877 v; H' 3-9045 = -4704840, log 5-2095 = -7167900, b. e = 4000, ,4 = 70°, C = 90°; find « and/. , '"' .^j: = 9-^-'^«517, log 3-7587 = -5750377, log 1 -3680 - -neosn L cos 20 =. 9-9729858 diff. = I15 ^ m~Sl7 18. «= 480, ^ = 70°, C^= 90°; find b and c ' ' '::ZZl'S.''''''-'''^''^'' 10. ^l^^O.. -7082509, log 1-7470 =-2422929. ^1'*^- -- 8;, tliff. =r= 245, 19. i = 3600, .4 = 75° r_on°. «. i 1 , ^^ — « ^ , o = ju ; hnd a and c. """"*-'**>«' 3S24 = -140(i3«. log 5!l877 = -7-72(i(J(). :i ( ;, 21. c = 294, A = 23° 30', C = 90"; find a and b, log 294 =.-4683473, Z sin 23°30'=9-G00r)997,logl 1-723=1 ■Ofi90388,diff.=37(). Zcos23°30'=9-9623978,log20-9Gl=l-4307360,dift.=lGl. 22. c = 328, « = 1 92, C = 90° ; find A and h. log 3-28 =-5158738, L sin 35° 49'= 9-7G7299G, diff.=1750, log 2G593 = -4247G73, diff. = 163, log 1-92 = -2833012, /. cas 35° 49' = 9 9089639, ditf. = 912. 23. c= 6-23, ^ = 64° 20', C = 90°; find b and c. log 6-23= -7944880, Ztan64°20'-10-3182r)04,log299-38=2-4762228,difr.=- 145 ZsinC4°20'-- 9-9548834,log691-20 = 2-839G037,diff".-- 63. 24. a =70-5, 6 = 96-5, (7 -=90°; find A and c. log 70-5 =1-8481891, 2/ tan 36° 9'= 9-8G36500, diff. - 2652. log 119-50 - 2-0773679, diff. = 3G3. log 9G-5 = 1-9845273, Z sin 36° 9'-^ 9-7707793, diff= 1729 25. 6=1218, c= 1282, C = 90°; find « and i?. log 1218=3-0856473, log 1282=3-1078880, L sin 71° 49' = {)()777r.23. 2/ sin 71° 50' = 9!)777938. Tables (seven-figure) will be required for the remainder of the Exercise. 26.i4 = 36° 21' 20", c = 74-8234, C -= 90° ; find b. 27. « = 784 -325, ^ — 60° 34', C - 90° ; find 6. 28. b = 29784, A -= 43° 24' 30", C = 90° ; find c. 29. 6 = 200, c = 249 C = 90° ; find a. 30. a = 416, c=740, = 90°; solve the triangle. 31. A - 37° 10', a = 124, C = 90°; find b and c. 32. a = 5, c = 13, C = 90°; solve the triangle. 33. a = 1100, c = 1 109, C = 90°; solve the triangle. 34. The base of an ioscelea triangle is 10, and the height ^0 ; find the vertical angle. 35. The side of a,n isosceles triangle is 30, and the base -10 ; find the vertical angle. 99 36. The sido of an iso.scclos triangle i.s^30, and tlio licight 20 ; find the vertical angle. 37. The sides of a triangle are 740-232, 740232, and 400; find the angles. 38. From the extremity of tlio diameter of a circle whose radius is 20, a chord is drawn, whose length is 12 ; what is the angle between the chord and the diameter ? 39. The sides of a rectangle are 10 and G ; what is the smaller angle contained by the diagonals'? 40. The hypothenuse of a right-angled triangle is 20, anres8io)i if the object be below the observer. 1. A i>erson wishing to ascertain tho height of a tower standing on a declivity, ascends to a point 80 feet below its base, and it then subtends an angle of 30" ; find the lieight of the tower, the inclination of tho side of the hill to the horizon being 30". 2. A person standing at a distance of 82 ft. 4 in. from the base of a tower, observes that the altitude of the tower is exactly 45^ ; find the height of the tower without I'efer- ring to the tables, the eye of the observer being 5 ft. 2 in. from the gi'ound. 3. A pei*son standing at the edge of r> rvec observes that the top of a tower on the edge of the opposite side subtends an angle of 60" with a line drawn from his eye pai-allel to the horizon ; receding 30 ft., he finds it to subtend an angle of 45°. Determine the bi-eadth of the river. 4. The angles of depression of the top and bottom of a cohmin observed from the top of a tower 108 ft. high are 30° and 60" respectively ; find the lieight of the column. lie a Ire :\') 5. Till' iiiii^'lcs of (li'iircssion ami flcviit ion ot' tlio top of ft coluiiiii oliscrvt'il from flic top mid lioltoin of ii fowcr 1(18 ft. Iiij,'li, aro i'J<» and ild n'.s|i»,'otiv(5ly. Find 'the lu'ij,'lit of the I'oliinin. ft. A and /^ aro two stations on ii hill sido ; tlm inclination of tlic hill to the horizon is 15'; the distance between A jtnd Ii is r»(l() yards. (' is the siininiit of another hill in tho same vertical ]>lane as A and Ii, on a level with A, hut at Ii its elevation altove the horizon is ."50 . Find the distanoo tVom A to ('. 7. A ship which is known to oe sailin;.^ due East at 12 miles an hour, was observed to lie .'}() to the East of So\ith ; ]/i. 'M)in. afterwards it was seen iu the Houth East. Find the distanoo of the ship when lirst seen. S. At the foot of a mountain tin; elevation of its summit is found to he •!;')'. After ascending fur two miles at a slope of ;50 towards its summit, its elevation is found to bo GO . Determine the height of the mountain. y. A person at a distance af L'O yards from the nearer of two towers in the same straight line with him, and 10 yards apart, observes them to subtend the same angle. Passing the nearer tower a cei-tain distance, he observes them again subtend the same angle, the oomitlement of the former. Find the heights of the towers. 10. A person standing on the hnnk of a river observes the elevation of the top of a tree on the opposite bank to be ■yV ; and when lie retires o() feet from the river's edge, he observes the elevation to bo 40'. JJ>etermine the breadth of the river. Given tan 51° - 1-2381, tan iG° = 10355. 11. A person on a tower whose height is 90 feet, observes the aroints distant 400 yards. B starts at riglit angles to the line joining the two points at the rate of 90 yards a minute. A starts in a direction to catch B as soon as possible at the i-ate of 1 50 yards a minute. Find how long lie will be before he catches him, and the direction in which he must walk. 23. A staff 1 foot long stands on the top of a tower 200 feet high. Find the angle it subtends at a place 100 feet from the foot of the tow^r. 24. The shadow cast at noon on the longest day of the year by a tower situated 51° 31' N. latitude was 124 feet ; find the height of the tower. 25. Find the height of a cloud whose elevation is 33° 10', and depression 45° when seen by reflection in a lake from a' station at a height of 150 feet above the surface of the lake. Kxtoiisioii of fill' (li'lliii- tiiiii iif tlic trij-'onomet- rical ratios til till- r,a.sc of MM angle Kivatcr tliaii 'MP. TRIGONOMETRICAL FORMULAS. 3L It is necessary now to extend oiu' definitions to the case of iin angle greater tlian one, bnt less than two, right angl(?s. Let CAB be such an angle, and be denoted by A. Produce CA through A and drop BC' perpendicularly upon it. The angle BAG' is called the supplement of A, and = 1S()° — A. We now define the trigonometrical ratios of th(! angle A to be the corrcH})onding ratios for the angle BAG' in the triangle BC'A, with the convention that AG' is to be considered a negative magnitude. Let p, b, h be the numerical values of the lengths of the perpendicular, base, and hypothcnuse in the triangle ; then [{( l.tlioiiw lii'twi'cii tlie ratios of an an^'li' and its Miipplcmcnt. sin A P. h IK," Ali tan A = «^C' ^ p_ AC -b AB _h AC -It AC ^-h AB h AC -b BC p AB h sin BAC sec A cos A = --' = cot .1 = - = — P b '"b h ' h :sin{180'' — ^ -ia.-aBAC = t.an(lSO" — ^1); = — sec BAC = — sec (180° —>1) ; =rT — COS BAC = — cos (180° —A) ; P = — cot BAC = - cot (180°— A) ; cosec A BC = cosec BAC = cosec (180° — A). Tiic ratios t.ir aiij,'li's j;riMtii' tlian >Mi^ I'ouml from llioseof angli's loss than yo'. 32. It will l)e seen on inspection that the ratios according to this extended definition will satisfy the same five funda- mental relations as before ; and although the complement of an angle (.1) which is greater than 90°, being— 90° — A, is a negative (piantity, and ceases at present to have any signification, we shall still say that the cosine, cotangent, cosecant of such an angh^ are the sine, tangent, and secant of its coinj)lement, and hereafter, if necessary, give a consis- tent interpretation to the quantity. 33. From the above it is seen that the trigonometrical ratio of any angle is the same in numerical vidue as the cor- responding ratio of its supplement, but bears a difierent sign except in the cases of sine and cosecant which bear the same sign. It is therefore unnecessary to construct additional 39 tables for angles greater than 90°, as the ratiof, for sucJi angles can be found from those of their aujiplements, which are less than 90°. Further, for such angles the tangents, secants, cosines, and cotangents being negative quantities, have no logarithms, and it is only for the sines and cosecants that the logarithms have real values, being the same as those given in the tables for the supplements of these angles. 34. We can now proceed to the discussion of triangles in general, to the angles of which, whether acute or obtuse, our definitions of the ratios will now aj)ply. # - The triangle being ABC, the lengths of the sides opposite Thno to the respective angles will be denoted by the small lettei-s I'eilKf '" corresponding. The triangle then is said to have six parts :— ^•^''Trts'Ili namely, the three angles. A, B, C, and the three sides a, b, c. i'" "'''"i"'- Ti. • 1 1 -n tnilllglc. It IS proved by Euclid that when three of these parts are given (one of them being a side), the other parts can be found There must therefore be three independent relations connecting these six quantities. One such relation is already established by Euclid, namely : i HI ^ + -B+C = 180° Two others we proceed to investigate. (1) O ne relation From C drop the perpendicular CD on AB (fig G) or on BA produced (fig. 7). Then in the righb angled-triangle CBD, GD^BC sin CBD = a sin B And in the right-angled triangle CAD, GD= AC sin CAD=h sin A, in fig. G, = h sin (180° — A) —6 sin A, in fig. 7 Hence a sin J? = i sin A Similarly, by dropping a perpendicular from A, we should obtain ft sin C ==c sin B, And hence sin yl sin 5 sin C a ^ ~lt "" ~n (-) "thrr. 40 Another 35, From these three relations (1 ), (2), all others can be relation ^ " ^ '' found inde- deduced, but for such as we require at present, it will some- pendently, ... . but actually times be easier to give proofs which do not directly depend from tti« on these. above. Resuming the figures and construction of the previous proposition, AB = DB + AD, in &g. 6. = BC cos CBD + AC cos CAD r=a cos B + b cos A, ^ Also, AB=-DB — AD,in&g.7. ^BC cos CBD— AC con CAD = acosB — bcos{180" — A) = a cos B + h cos A. * Hence, universally, c = a cos .8 + 6 cos A (3) I Deduction 36_ Multiplying the respective terms of this equation by general - . . sin C siu A sin B formulas, the equal quantities , , , cab sin (7= sin A cos .S + cos A sin B, but C is the supplement of (^A-{- B); therefore .Sin (A + BX sin {A + B) = sin A cos B + cos ^ sin ^ . . . . (4) 37. In the preceding, instead of A write 180° — A ; then sin {180°— (.4 — ^)} = sin (180°—^) cos B + cos (180°—^) sin B, *In this formula, writing it a b 1 = - cos B+- cos A c c suppose that C is a right angle. Then cos .fi= sin ^, a b -=sinvl,- = cos A, and, making these substitutions, it c c becomes l = (8in ^)3+(cos^)«. This is the proof alluded to on page 17, as not depending on Euclid, Bk. I., Prop. 47, but in fact being also a proof of that proposition. 41 be me- ?ncl 0U8 by m it or sill {A — B) = sin ^ cos J? — cos ^1 sin J5 (5) Again, in this for A \mte 90°— J ; then sin |yo°— (^+^)} = sin (90°— .1) cos B — cos (90°— A ) sin B, or cos {A + B) = cos A COB B = sin ^ sin ^ (G) The above proof of the last three formulas restricts the angles A and B to have their sum less than 180°. The formulas, however, are universal, but it is not necessary to extend them beyond this case, as it is the only case in which their use is at present reciuired. In the appendix a general proof will be found, applicable to angles of any magnitude. 38. In (4), and (6), putting B = A, we obtain sin 2 A = sin J cos ^ + cos ^ sin ^ = 2 sin ^ cos J. cos 2A = cos A COB A — sin J sin J = cos ^A — sin '^A, and therefore, (since cosM + sing ^ = 1), = 2cosM — 1 or = l— 2sin'^. Hin {A — /}) <'05(/« +/i>. (7) (8) Writing \A instead of ^, these become sin y^ = 2 sin hi co%\A, cos^ = 2cos2|.l — l = l_2sin2U . 39. Adding (4) and (5), we obtain sin (.4 + 5) + sin (.4 — 5) = 2 sin A cos B, And subtracting (5) from (4), sin (.4 + 5) — sin (^ — J?) = 2 cos A sin B. Dividing the terms of these two equalities, we obtiiiu sin (^+ g)+sii i (A—B) 2 sin A cos B 8in (A 4- ^)— sin {A — B)~2~^A'^hrB _ sin ^ sin B cos A tan^ tan B cos B sin /I & (03^1 in Urnis of sin i A , iDji M. 42 In tills formula, instead of {A-{-B) wiute A, and instead of (J — B) write B, and therefore also instead of A write ^ (-d-f ^), and instead of ^ write 1{A — B), and we obtain sin A -f- sin B tan ^ {A-\'B) sin .4 — sin jS tan ^ (^ — B) (50 EXERCISE VIII. (The Examples in this Exercise should be attentively studied, ) Prove the following relations : 1. Coa{A-B)=—coaJ,]80''-(A—B) [=— C08|( .) 2. tanM+^) = COS {(ISO" -A) + B'r = - cos ( 180° -yi) cos 5 + sin (180" -yl) sin fi =cos A cos B + ain A sin B. sin(^l + S) sin^cos5+co3^sinj5 tan^l+tani/ cos(..i 3. tan {A — B) = n\ cos^cosZf-sin^siuif 1-tan^tan/i tun A — tan B 1 -\- tan A tan B 4. tan 2 ^ = 5. cot (^ + 5) G. cot 2 A li I,. or tan A = 2 tan i ^ 1 — tan- A' 1 — tan'^ | A COS A cos i5 — sin ^ sin 5 cot A cotB — 1 sin A cos ^-|- cos ^ sin B cot A -j- cot B cot^yl— J. "2 cot J ■ 7. sin (A-{-B) = sin A cos B -^ coa A sin JS, sin (^ — B) = sin ,A cos J5 — cos ^ sin ^ ; .-. sin (A + B) + sin {A — B) = 2 sin yl cos i?. 8. sin (^ + B) — sin {A — B) = 2 cos ^ sin 5. 9. cos (A—B) + cos (A-{-B) = 2 cos ^ cos j5. 10. cos (^ — i?) — cos (il 4- -B) = 2 sin A sin B. 11. Since yl = ^ (i + B)-\-l{A — B), and B = ^ (A + B) -h{A-B); .: sin ^ + sin i? = sin {I {^1 + ^) -f- 1 (^ _ 7i)} + = , Ex. 7, 2 sin H^ + ^) cos J (.4 — 5). 12. sin ^ — sin 5 = 2 cos -^ (^ + B) sin ^^ — '^)- 13. cos 5 + cos ^ = 2 cos I (A + 5) cos |(i4 — B). 14. cos 5— cos ^ = 2 sin ^ (^ + 5) sin | (.1 — 5). I. ^ ; 43 ' ■ 1 + cos vl 1 + (2 cos'^ n:~^) - *'"' ^ "' • 10. sin (vl + Z? + C) = sin (.4 + B) cos C + cos(J + ij)sin C, =^ sill A cos 7/ cos C + sin Ji cus T cos A + sin C cos ^ cos 5 — cos A cos B cos C. 1 7. cos (^ + ^ + C) = cos J cos B cos 6'— cos ^i sin £ sin C — cos B sin C sin ^ — cos C sin A sin B. 1 8. tan(^l + iJ + C) = *5i^-"t*?.L^±*_'l"^r*''^" . Itan/i tanr 1 - tan J tan Ji ~ tan Zi tan C - tan C'tan yl * 19. sin 3 J :- sin (2.1 + .() = sin2.-lcos.'l +cos2ylsin/J. = 2 sin^ cos- J + sin^ - 2 sin' A, = 3sin.-l — 4 sin'' J. 20. cos3yl = cos(2yl+^) = cos2^cos.l -siu2^siii A, = 2cos''i'I -cosi - 2sin-yIfos J, - 4 cos'' A — 3 cos A. 21. tan3^=:tan(2.{4-.n^/'^-^^+A''^"^^ , 1 — tan 2 A tan A 2 tan A 1 — tan- A •f tan A -. 2 tan A ^ 1 — , — - — „ • tan A 1 - tan-vl 3 tan A — tan** A 1 — 3 tan^^ (Examples 19, 20, 21 may, of course, be obtained from Examples 16, 17, 18, by putting J .^ ^ = (7, 1 —sin'-M for cos- A, and 1 — cos^ A for sin- A.) 22. Let/1 = 18°. Then,since sin3G° =. cos(90^-3r)°) = cos54', • ■. sin 2 y1 =cos 3 J. 2 sin A cos A — 4 cos".4 — 3 cos A . Divide by cos A ; .-. 4 sin-^ + 2sin J = 1, or sin 1 8= ^ sin A = ?.' '^ "~ ^ , takiufr + sign, since sin 18° is positive. I l! •' ! .' M HXEHCISE IX. Prove the following relations : 1. aiu ir)°=sin (45°— 30°) ^^ sin 45° cos 30°— cos 45' sin 30° = 2. cos 15° t/3+1 2i/2" = sin 75" 3. cot 15'^ -=-- tan 75° = 2 + ^37 4. tan 15° = cot 75°= 2 — i/^\ 5. If sin a = I sin /S ^ ^, then sin (a + /9) = ^ + ^^•''^ 15 00s {a -\- jsy Vs — 6 15 6. If sin ^ = f , cos 5= I then sin (^ + B) = -fJ}^, 008 (4 + 5)= ^^-6^^ 12 7. If sin ^ = I, sin i? = ^, then sin ( A — B) = 8—31/5. ~15 4v' 5 8. If sin a = |, then sin 2 a = — _ , cos 2 a = 1. 9. If sin A = j%, sin if = |, then sin (45° + ^+i?) 79l/2~ 130 ■ _ 1/5 1 10 Given sin 18° = ^ , shew that cos 36° 4 ^^111 sin 36° = J s-V'o , sin 72° = ^^10 + 2V^5 i. ' li w ^ * EXERCISE X. Establish the following : 1. (sin + cos oy=^ 1 + sin 2 0. 2. sin 4 ^ = 4 cos 2 ^4 cos A sin id. 3. 1 — ^ sin 2 a tan a = cos^ a. 45 4. cosiA±HuxA=. VT^dr2~A. 5. cos ^ — sin ^ = i/2"sin (45° _ A). 0. sin A+cosA = 2 sin 45° cos (45° — A). 7. cos* a — sin* « = cos 2 a. 8. sin 4 A - sin 2 .1 = sin (3 A + A) - nln (3 J- .4) ' ■i cos 3 ^ sin A. ' 9. cos 2 ^ — cos 4 ^ = 2 sin A sin 3 J. 10. 2 sin 2 ^ cos A = sin 3 ^ + sin A, 11. cos 3 ^ + cos J = 2 cos A cos 2 ^. 12. cos ^y — cos 7 -f:/5)«ni (« — /?) f.r>u^ « „„. 2 a = tan' a — . cos^* a cos- /? taaV. 9. cos « + cos (a + 2 i3) ^ cos /(« + ,5) _ H ^ 10. sin (a + 2 ,J )= sin a + 2 sin ,9 cos (a + ;J). 11. sin («-2 ,?) = sin «_2 sin /3 cos («_/?). 1 2. cos (« + 2 /j) = 2 cos ^3 cos (a + ;3) - cos a. jasin(. + 2/3) = sin«-2sin«sin^^3 + 2cos«sin,^ U. Hiji80 = 8co840cos29cos0sin9. 15. cos 4 = 8 cos^ _ 8 cos- (^ -f- 1. 1 G. sin 3 J — sin .1 = 2 sin A cos 2 J. 17. cos 2 .1 + cos 3 .1 = 2 cos ~- cos ^' o ■ t> ■ 18. cos 3 vl — cos 4 J = 2 sin ^- ^^ sin -. 10. .sin y + sin 2.1 ==2 sin Ld cos ^' ^ 4 4 20. .sin 3^_sin ll=-2cos Ld sin ''J 244 21. 4cosJcos2ylcos3.I = (cos A + cos 3 A) 2' cos 3 .<. -2cos J cos 3 vl + 2co.s^3 J. = cos2 Jh-cos4J + U.cos0.4. 22. 4 sin ^ sin 2 ^ sin 3 A = sin 2 ^-f-sin 4 ^ - sin 6 A. 23. 4 cos A cos 2 A sin 3 J = sin 2 ^ + sin 4 A 4- sin G A . 24. 8 sin A sin 2 yl sin 3 .1 sin 4 ^ = 1 _ cos G ^ CO88 J+coslO.4. coso^- a 1 1, i ! 25. cos A -t- cos 2 A+coH^i A I cos cos A coh '— - - 1 '2 2 20. cos 9^ + 3 cos 7 ^1 + 3 cos f) .1 + cos 3 A --^ 8 cos' A cos G A. 27. cot' i4 — tan" A = CQ»*^— s'"*^ _ H cos 2/1 8 cos 2 X sin" i cos'-' yl 2 8in''2/| i-co84/l* - _ cos A COH ^ A . n . 28. . — _— , ; r - tnn 2 A. sin 3 yl — sin A 29. sin 2 a + cos « = 2 tan a + sec « 1 + tati* a „^ cosec 2

cos 2 = 1. I ;l 49 If A + A+ (/ = 180=, p„,v« tl.o following, 41-4r. : Oht^iined from tan (J 4. yj) = tan ( 1 80° - C). 42. cot J cot ^ + cot i5 cot C 4- cot C cot ^ = 1. 43. sin A H- «in ^ + sin C = 4 coh ^^ ^^ ^ * cos ^^-C0S-C08~. 2 ^^^ .> - = -«»«;, cos- .'^- sin r = 2 sin ?co8 ^ - .. ^^ A^B »> l-OS .. = J cos - COS _ 'J 'J 2 ' .-.sin^ +8ini^ + sin C= 2 co« |' | coh ^-Jl^-fcos = 4 cos - cos 1 cos - . '^ '2 '). 44. cos .1 ^cos ^.fcos C= U4 sin ;i' sin f ,i^ _ a») cot B+(a'~l cot (7 .-= 0. 25. 6cos*£ + c cos» :^ = * + * + c 2 2 9 • ..O 26. (6~c)cos£ = asm^Z=£ 2 2 8in> - sin^ 27. — .-i + 1^ ^ — « * c 6c * 28. 6 c cos» 2 + '^ <* ^^^\ + « * cos« ^ = ^'(a+b+cy. 29. l-tan4tan.?=— 1^. 2 2 a + 6 + c' 30. cot -- + cot X 2 2 cotd 2 a b + c — a 31. If a (s — c) = I a J, one angle is a right angle. 32. If c cos 5 = ft cos 0, shew that the triangle is isosceles. 33. If a sec if = 2 c, the triangle is isosceles. 34. If a = ft, c = 2 a sin -. 2 35. If c = 2 a sin _ then either the triangle is isosceles, or c' = a (a — ft). .'.{■■: in i ii Solution of oblique triangles, Four canes. Case I. Two angles and a side t-'iven. 64 SOLUTION OF OBLIQUE TRIANGLES. 42. Four distinct cases occur in the solution of oblique triangles, according to the way in which three parts out of the six which compose the triangle are selected, one at least of the given parts being a side. These are, (1), two angles and a side. (Euclid, B. I., Prop. 26.) (2), the three sides. (. . . . Prop. 8) (3), two sides and the included angle. ( . . . . Prop. 4) (4), two sides and an angle not included. ( The omitted case.) 43. Case I. Given A, B, a; to find C, b, c. To find C, C = im°—A — B C found. To find b, OJ? sin B sin A b a b a sin B sin A a sin 5cosec A, and taking logarithms, log b = log a-\- L sSxi B — 10 -|- Z cosec A — 10 = log aJ[.L«mB-\-L cosec A — 20 from which there is b found. To find c, sin C sin A c = a sin C cosec A log c = log a + X sin C + Z cosec A from which there is 20, c found. In this case it is indifferent which of the sides is given, as all three angles are at once 'mown. Case II. Tlie three silica given. 65 4*. Case II. Given a, 6, c ; to find ^, ^, C. To find A, we have, (where « = |(a + j ^ ,))^ tan i^ = ^/(« — *)(« — c) and taking logarithms, X tan ^ ^ __ 10 = ^ log (iri^Klnf) -iiIog(._6) + iog(,_,)^^^j and therefore '' ^ ZtanM = i{log(.-5) + log(,_.)+co]og(._a) + eoW.[ from which there is ^ ^ and therefore ^^ Jfo^nd whSi tlsT-Tl^-^t if h^^ '^ r^' -' ''- '' to fin^ ^ 1 u . ^^ however better in practice til! , ' ^*« analogous formula, and the sum of the three angles amounting to 180° will serve as verification We might also have used either of the formulag H 11 fn. • i . Pne of the qultitlT'-^^^Vil^^tH^^ *'? ''' '''''' logarithm imaginary. ' ' '" ^^ "^«**'^^' '*°d its 45. Case III. Given a, *, C/; to find ^, ,5. c. (a>M To find A, £. ^ ^' ^'""^ '"• ^A _ Bin i9 or Two sides and ttio in- cluded angle given. and therefore a . sin A ° + ^ = '*!^j1+ sin B a — b sin ^ — iii^ or _ tanj (yl + i9) tan l~Ji:ZJ^ from (9) ta„n^~^) = «_Z:|tan|(^,^). I! L^fTT.'. '.> 3 (* WaJ Now ^ (yl + J5) = ^ (180°-- C) = 90° — i C, and is known ; also tan ^ {A + B)^= cot ^ C, and therofore tan i (.1 — 5) = —, cot i C, •^ ^ a + 6 and taking logarithms, Z tan i (.4 — 5) — 10 = log (a — 6) + colog (a + 6) — 10 + i/Cot J C— 10, or Z tan A (^1 — i?) = !og(a— 6) + colog(a + 6) + Zcot^C — 10, from which ^ (-4 — B) is found ; also ^ (^ + 5) being known, we Iiave by addition and subtraction A and B found. A having thus been found, we obtain c from the formula sin sin A c a c = a sin C cosec A log c = log a + Z sin G + L cosec A — 20, (c found,) in which formula b, B might also be used in place of a, A. In this case c is known directly in terms of the given parts from c«=^a« + i« — 2a6cos C, but this formula is not adapted to logarithmic calculation, and it is preferable to find c by aid of one of the angles which have been previously found. Cise IV ^^- ^'^^^ ^^- ^iv®^ A, a,b ; to find B, C, c. Two sides ^" ^""^ ^^^^ there are sometimes two triangles which have the ana an angle given parts. For let A be acnie, and (fig. 8) drop the perpendicular liy tliem CD, which is equa. to 6 sin ^ ; then there can be drawn two lines, 'lmbl''iiir"' each = a, one on each side of CD, and if both these fall (as CB^^ (liscnsseil. CB,^) on the right of h, the two triangles ACB^, ACB^ will have the Fig. s. same three given parts. This requires a to be less than 6 and greater than CD ; if however a — CDf the two triangles coincide in a i ight- Fig. 9. angled triangle, and if a be less than CZ), no triangle exists having ' II 67 the given quantities for parts. Also if a = b, the triangle ACIi,^ vanishes, and only one is left, and if a be greater than h, the line Cli„ falls to the left of h, and the triangle so formed would not have the angle A, And in this case there is only one triangle. Again if A be obtuse (fig. 10), in order that a triangle may exist, Fiy n must be greater than b, and the other line equal to a will fall to the left of b, so that only one triangle exists. Collecting these results, we see that, when A is acute, if a<|6 sin A, there is no triangle ; if a = ft sin A, there is one only ; if a^b sin A and /), there is only one ; and when A is obtuse, if a <[ or=6, there is no triangle ; and if a > b, there is [one only. If ^ be a right angle, then a must be [> 6, and the two triangles on opposite sides of b are equal in every respect, and therefore only give the name triangle in different positions. The analytical solution which follows will of itself shew which of these varieties occurs in any particular case. To find B; 10. Solution. or sin B 6 sin A a and taking logarithms L sin B — 10 = log 6 + colog a — 10 -f L sin A — 10, whence Z sin B = log b + colog a -{- L sin A — 10. 13 I the liar les, p., Ithe kter mg This gives Z sin B, but as the Z sin of an angle is the same as the L sin of the supplement of that angle, there are two angles which have this value of Z sin, and both must be taken. Let -Sj, B2 be these two angles, the former being less than 90° and taken direct from the tables, the latter being its supplement. Let Ci, Cj be the corresponding values for C, so that A~B„ C,= 180° C',= 180° A- If both these values are positive, two triangles exist. \r^' ill I > i \ I h\ 58 Let Ci, Cj be the correspouding values of c. To find them ; sin Cj sin A Oj a Cj = a sin C\ cosec il. log a + X sin Cj + ii cosec A — 20. logci Similarly log c, log a -f Z sin Cj + -i!^ cosec A — 20. KxtllllplcH. A,U,n given tu llnd II'.] If the second value of C be or negative, the second solu- tion has no existence ; and if both values of G are negative, no soluticHi exists. Also if the value of Z sin J^ be greater than 10, there is no solution. 47. Examples. Case I. ' Given ^ = 120° 08', B = 24° 40', a -•= 981-23. C=1S0° — A — B. J = 120^08' ^= 24 40 [h.J 144 48 180 0= 35° 12' = log a +Z a = 981 -23 B= 24^40' J=120°08' sin J? + Z cosec A — ; log a, 2-99177 ; ZsinJ5, 9-62049 ; L cosec A, 10-06305 b = 473 -49 ; log 6, 2-67531 (C found.) (6 found.) [c] log c = log a -}- Z sin C + Z cosec A — 20. log a, 2-99177 = 35° 12' J ZsinC, 9-76075 Z cosec A, 10-06305 0-= 653 -99; logo, 2-81557 (c fornd.) 59 Case II. Given a = 75309, b a— 753-09 h=2 333-33 c= 666-66 =- 333-33, = 2* = 1753-08 log «= 876-54 » — o= 123-45 «— 6=543-21 «—c = 209-88 2-94277 2-09149 2-73497 2-32197 colog 7-05723 7-90850 7 '20503 7-67803 ^tanM = Hlo^(-^) + log(.-c) + co]og(.-a)^e.io..} log(8~b), 2-73497 log(s— c), 2-32197 colog(«_a), 7-90850 colog 8, 7-05723 2)2002267 ^^4 = 45'' 45'; ZtanJ^, l^^^Hii ^ = 91° 30' Ztan|^=^{log(,_e)+log(.-a)+coIog(.- log(»_c), 2-32197 log {8 — a), 2-09149 colog (s — 6), 7-26503 colog 8, 7-05723 2)18-73572 ^l:26o?6'" ^^''^^^^ ~^mQ {A found.) -6)+coIog»}. /. ten i C = Hlog(.-a) + log(._6) + coIog(.- log (*_«), 209149 log(*_6), 2-73497 colog («—c), 7-67803 7-05723 (^ found.) -c) + colog s|. (I. h. to /Imi Sivfn, colog *, 2)19-56172 ^ = 310 07'; Ztan^C, ~978086 C - 62" 14'. (C found.) mi [i-.i C] I i i lii r 1, ; I v«riticuti(in Verification. 60 A = 910 30' if =26 16 C = 62 U A+B + C = 180°. Hi; I a,h, C given, tu Una Case III. Given a = 209-88, b = 333-33, C = 112° 26'. Here, h being greater than a, we must interchange a, A with b, B in the formulas of solution. 90° 00 C = 122° 26'; ^ C = 61 13 J (5 + ^) = 90° — I C = 28° 47' VA »nd fl.j ^ <^'in i {J^—-^) = log (6— a) + colog (6 + a) + Z cot ^ C— 10. h = 333-33 a = 209-88; log, 2-32197 &>J /; — a = 123-45; log, 209149 /> + a = 543-21 ; log, 2-73497 ; colog, 7-26503 log (6 — a), 2-09149 colog (6 + a), 7-26503 J C = 61° 13'; Z cot i C, 9-73987 l(B — A) = 7° 07' ; Z tan K^ — ^). 909639 H^ + ^) = 28 47 ■> j5 il 35° 54' 21° 40' (B and A found.) log c = log a + Z sin C + Z cosec A — 20. ^ log a, 2-32197 * C=122°26';Z sin C, 9-92635 A= 21° 40'; Z cosec vl, 10-43273 c= 479-97; log c, 2-68105 (c found.) n V,me IV. Kx. (1). Given A = 57» 34', a =. 471)79, A == 54-324. Zmn^ = log6 + cologa + Z8invl-lO. ^=^^•321; log 6, 1-73497 AZiloli' *'''^*'^'«' 8-31896 yi_.)7 J4; Z sin J, 9-92635 [ /A= 72° 52'; L sin i?, 9.y8027 'A-=107°08'. ^1 = 180 -(^4./?^) ^1 = 57° 34' A= 72 52 (/?i and 7/. found.) t^=180 —(A + n„) A = 57° 34' A =107 08 C,= 180°— 1G4 42 = 15° 18'. C,==180°— 130° 26' = 49° 34' Hence there are two solutions. log e, = log a + z sin C,+ I cosec A ~ o() A~t%li' ^'"'^'" 9-88148 -1—0/34; Zcosecyi, 1007365 43269. logcr,, 1-63618 ^c, fbmul.) log c, * log a + Z sin C, + Z cosec vl - 20. r _Uo la, ^ !«8«' 1-68105 02_iD 18 ; ZsmC',, 9-42139 ^cosec,-!, 10-07365 , = 15. ^ogc« 1-17609 Ex.2. Given^ = 49° 41', « ==323-1, 6 = 21-808 Z sin£ = log b + colog a + Z sin J - 10 !Zol??^ ^" ^og^. 1-33862 ^— 49° 41'; Zsin.l, 9-88223 (c, found.) to lliiil tC.) Two Sllld- tiiiii.s. tv] rv] ft.ff.'/J.'iVCll, ti' liiul /;, j^i= 2° 57' ;Z sin 5, 8-71151 I ^2= 177° 03' CA and i^^ found.) r 62 to.) 'Jnc Holtitloii A —49" 41' Jii= 2° 57' C— 180 — (/( + /?,) ^ = 49° 41' 7/^— 177° 03' f\ 180° — 52° 38' 127° 22' C„=180°— 22G° 44' [\] A,(i,hgWen, III tliid U. Tho second solution does not oxUt. The value of c, can bo found as in the previous example. Ex. (3). Given A = 30°, a = 18-4, b — 38-9. Z sin ^ = log b + colog « + Z sin il — 10. 6 = 389; log 6, 1-58995 a =18-4; colog a, 8-73518 ^ = 30° ; I sin A, 9-G9897 Nil Mdllltioll. No solution exists. Zsin^, 1002410 /l.a./'xiveii, to Una K Ex. (4). Given A -= 128° 57', a = 21700, 6= 19342. Z sin .S = log b + colog « + Z sin ^ — 10. 6 = 19342 ; logi, 428650 a = 21700 ; colog «, 5-66354 vl = 128°57; Z sin J, 9-89081 m One solution ^ £^= 43° 53'; Z sin £, 984085 R^=z 136° 07'. Ci = 180 — (A+ £,) A =128° 57' -B,= 43 53 C, = 180 — (A + B,) A = 128° 57' .5^ = 136 07 C, = 180° — 172° 50' = 7° 10'. C,= 180°— 265^^04' -4, a, b given, to liuJ Ji, The second solution does not exist. Ex. (6). Given ^ = 163° 24', a = 42, 6 = 53-004. Z sin 5 =» log b + colog a + Z sin ^ — 10. b= 53-004; log 6, 1-72431 a = 42 ; colog a, 8-37675 A =163° 24'; Z sin A, 9-45589 B^= 21° 08'; Z sin B, 955695 ^2=168° 52' 9S -4 = 103° 24 -^1=*= 21" 08' C'^- 180°- 184= 3^ No solution exists. A = 163° 24' ^a*- 158° 52' C,-180°-3"22M6~ rr No solution, EXERCISE xrir 3- Given — l n «no ,. , , u - ^ ^ - 60 ; hncl the other angles. 4. .i = G0°,.==.v/6;i^2;findA •'■ '^ = ^'^^°'«-2,6 = v/2;findA «. ^ = 135°,« = 2,6=v^0;«olvethetHangle. '•' = 20,^==82',C=75°;find«. log 2 = -3010300, Z sin 23" = 9.591 «7sn lo;? 7-8914 — .«o7i K.I r . —''^ylO'oO, Diffl55"'^^°==^"^^^^^28. 8. a«36,i?=.44°,C=.l04°,findi iogi^^'ll'Soo'f^"^^^^-^-^^^^^^^^ Diffl92 '^'^"^^°=^8^i;713. ^^^• = 114 Z sin 75° 18'= 9 9855467 Diff. fori '=331. I ■f i 64 10. a^i/5676 = l, c=7; find J. log 2 = -3010300, L cos 57° 41' = 9-7280275, log 7 = -8450980, L cos 57° 42' = 9-7278277. 11. Given a = 18, 6 = 20, c = 22 ; find A. log 2 =: -3010300, L tan 25° 14' = 9.G732743, log 3 = -4771213, Diflr. for 1' = 3275. 12. Given a = 4, 6 = 5, e = 6 ; find B. log 2 = -3010300, L cos 27° 53'== 9-9464040, log 5 = -6989700, L cos 27° 54' = 9-9463371. 13. a = 1900, h = 100, C = 60°; find A and B. log 3 = -4771213, L tan 57° 19' =^ 10-1927506, /.tan 57° 20' = 10-1930286. 14. a = 18, 6 = 2, C -- 55° ; find A and B. L tan 56° 56' = 10-1863769, log 2 = -3010300, DifF. for 1' = 2763. L cot 27° 30' = 10-2835233, 15. a = 9, ft = 7, C = 64° 12' ; find A and B. log 2 =--3010300, Ztan 11° 16' =9-2993216, L cot 32° 6' = 10-2025255, Diff". for 1' = 6588. 16. 6 = 1590643, B^ 62° 6' 51", G = 53° 27' 20"; find a. log 159-06 = 2-2015610, L sin 64" 2.5' = 9-9551864, log 159-07 = 2-2015883, L sin 64° 26' ^ 9-9552469, log 162-33 = 2-2103988, L sin 62° 6' = 99463371, log 162-34 = 2-2104255, L sin 62° 7' = 9-9464040. 17. Given a = 222, I = 318, c ^ 406 ; find A. log 4-73 « •674''- ,1 X, log 2-51 = -3996737, log 4-06 = -6085260, L cos 16° 28'-- 9-D818117, log 3-18 = -5024271, L cos 16° 29'=- 9-9817744. 18. a = 85-63, b = 78-21, C = 48° 24'; solve the triangle. log 16384 = 4-2H4199, L cot 24° 12' = 10-3473497, log 742 = 2-8704039, L tan 5° 45' = 9-0030066, ' Diff. for 1' = 12655. log 67502 = 4-8293166, // sin 24° 12' = l>-61 27023, log 67501 =4-8293^02, Zycos5"45'15"= 9-9978062. log 3 vtsin 78° 19' ==9.9909077 Diff. fori' = 261. ' J-'in. for 1 = 141 J". 21. Given a ^25, 6 = 26 r.= 97. - i x, 1 o- , -o, c •= J< ; nnd the angles 4^ '1213, Z tau 28° r 40" = 9-7280074 ^ tan 3r56'50" = 9-7948986' ^ tan 31° 57' -9-7949455,' 22. a = 1^5, i ^ 13.5, C^ 650; finely and 5. 1 J58127, Z tan 51° 29' ^ 10-0991365, 23. . ^ 445, i=. 565, .1^44° 29' 53"; find i?. 0.C 445 = 2-6483600, Z sin 44° 29'= 9.84..,oo % 565 = 2-7520484, Diff. fo;' ^ L,t ' X sin 62° 51' =9-9492997 Diff. fori' =048. 24. A ^ 7Sr, B = 54° « = 97 < . « , . ' oi,a~-2a; find ^ and c. log 274 = 2-43775, X sin 48°^. 9.87107 og 226-63 =.2-35531, Zsin 54° = 9-' o ' . log 208-17 = 2-31842, Zsin 78° = 9 ■ J04J: 25. Given a - .i30, J = 3IO, c = I44 - find f1 log 392 - 2-59390 , ' ^""'^ *^^^ *^"sJ«s. 4 62.1:7923 ; i:^tz];T'' ;^3io=2-49i36 xcos ?o^r : ::^^' iog 144 = 2-15836, Zeo«340 4;'= .l^ll' 66 26. a == 30, b = 20,O= 78° ; find c. log 2 = -30103, L tail 13«52' = 9-39245, /. cot 39° = 1009103, L tan 13" 53' = 939299. L Hill 39° = 9-79887, Z cos 13° 53' =^ 9-98712, log 3-2412 = -51070, L cos 13° 52' - 9-98715. 27. a = 13, 6 = 37, y4 = 18° 55' 29" ; find P. log 13 = 11139434, L sin 18^ 55' = 9-5108031, log 37 = 1-5682017, Diff. for 1' - 3685, Lain 67° 22'= 9-9651953, Diff. for 1' = 527. 28. 6 == 149, ^ = 69° 59' 2", C = 70° 42' 30" ; find a. log 149 =^ 2 1731863, L sin 39° 18' = 9-8016649, log 22099 = 4-3443726, L sin 39° 19' = 9-8018192, log 221 = 2-3443923, L sin 69° 59' = 9 9729398, L sin 70° = 9-9729858. 29. If a = 22, i = 23, c = 25 ; find B. log 2 = -3010300, L sin 29° 5' =- 9-686708S, log 11 == 1-0413927, Diff. for 1' = 2271. log 13=1-1139434, h B i 'I 30. a = 75, i = 85, C ^ 75= ; find A and B. log 160 = 2-20412, Z tan 52° 30'= 10-11502, L tan 4° 40' = 8-9109. 31. a = 2820-9385,i = 1430-8485,^ = 14°59'49";fiii(l /?. log 2-8209 = -4503877, L sin 14° 59' = 9-412524.1, log 2-8210 = -4504031, Z sin 15° = 9-4129962, log 1-4308 = -1555789, Z sin 7° 32' = 9-1176125, log 1-4309 = -1556093, Z sin 7° 33' = 9-1185607. 32. c = 100, A = 50°, ^ = 70° ; find a and b. log 2 = -3010300, Z sin 50° = 98842540, log 3 = -4771213, Z sin 70° = 9-9729858, log 8-8455 = -9467224, leg 10850 = 4-0354297, Diff'. = 49, Diff = 401. 33. a = 230, b = 240, c = 12 ; find R log 11 = 1-0413927, L tan 72° 4S' = 10-5092GGS log 229 = 2-35983r)5, Difl'. for 1' = 4474. log 241 = 2-3820170, 34. a : 6 = 7 : 3, C = 6° 37' 24" ; find the other angles. 'log 2 = -3010300, L tan 8° 13' = 9159504(1, L tan 3° 18' 42" = 8-7624080, L tan 8° 14' = 9-1604569. 35. a = 21-217, 6 = 12-543, .4 =29° 51'; find j9 ami r. log 2-1217 = -3266840, L sin 17° 6' 40" = 9-4686806, log 1-2543 = -0934014, L sin 17° 6' 50" = 9-4687490, Z sin 29° 51' =9-6969947. 36. The ratio of two sides of a triangle is 9 : 7, and tlit; included angle is 47° 25' ; find the other angles. log 2 = -3010300, L tan 15° ?3' = 9-4541179, L tan 66° 17' 30" = 10-3573942, Ditf. for 1' = 4797. 37. a = 462, b = 220 5, ^ = 124° 34' ; find B and C. log 2-205 = -3434086, L sin 55° 26' = 9-9156460, log 4-62 = -6646420, L sin 23° 8' = 9-5942513, L sin 23° 9' = 9-5945469. il/?. 38. a = 95-372, b = 74-896, C = 59° ; find A, B and c. log 2-0476 = -31125, L cot 29° 30' = 10-24736, log 1-70268 = -23113, Z tan 12° -= 9-32748, log 9-5372 =-97942, Z sin 59° = 993307, log 8-5718 =-93307, i^ sin 72° > ' --^ 9-97942. 39. A = 41° 10', a = 145-3, h = 178-3 ; find B and C. L sin 41° 10' = 9-8183919, L sin 53^ 52' - 9-9072216, log 1453 = 3-1622656, L sin 53° 53' = 990731 38, log 1783 = 3-2511513. 40. a— '. = 4013-166,a+i=7906-72, (7-36°; findJand/i. log 4-0131 = -6034800, L cot 18' = 104882240, log 4-0132 = -6034908, /. tan 57' 22' = 10-1935848, log 7-9067 = -8979953, L tan 57' 23' = 10-1938630. log 7 -9068 = -8980008, 11 i. i' 68 EXERCISE XIV, Six or seven figure tables will be required for the following Exercise. ^ 1. ti = 31, J = 24, c= 11 ; find ^. 2. A = 23° 42' 43", B = 18°, a = 207 ; find h. 3. The sides of a triangle are 32, 40, 66 ; find the greatest angle. 4. a = 1-125, b = -875, C = 64° 12' j find A and B. 5. a = 70, 6 = 35, C = 36° 52' 12" ; find A and B. 6. a =14000, 6 = 15906-43, ^ = 45°; find the other angles. 7. c = 3727-593, A = 50°, B = 57° 53' 9" ; find a. 8. The sides of a triangle are as 4, 5, 6 ; find the largest angle. 9. 6 = 1-125, c = -875, ^ = 54° ; find B and C. 10. h = 1625, c = 13-75, ^ = 63° ; find B and (7. 11. ft = 1, c = 302943, i? = 19° ; find (7. 12. 0=100,^ = 45°, 5=10°; find a. 13. The sides of a triangle are 4, 5, 6 ; find the smallest angle. 14. b = 21, c = 9, ^ = 6° 37' 24" ; find B and C. 15. 11 6 = 14 c, ^ = 60° ; find i? and C. 16. a = 100, c = 125, C = 45° ; solve the triangle. 17. a = 500, B = 45°, (7=10°; find b. 18. a : 6 = 21 : 1 1, C = 34° 42' 30" ; find A and B. 19. The angles of a triangle are in A. P., and the greatest side is to the least as 5 to 4 ; find the angles. G9 20. One angle of a triangle ia 60°, and the ratio of the side opposite to it to the difference of the sides containing it, is 9l/3 : 2 ; find the other angles. . Note. — It will be well also to work the two foregoing Exercises and Exercise VI., using the tables at the end of the book, in which case seconds will be omitted and digits to right in values of the sides, when such values contain more than four digits. The results so obtained will be found sufficiently close to the answers given. it EXERCISE XV. In this Exercise the tables at the end of the book hav«5 been used. \. a= 74-5, B = 69° 59', C = 70° 43' ; find b and c. 2. « = 4730, b = 4016, J = 71° 4' ; find R 3. a = 420, B = 76° 42', C = 52° 29' ; find c. 4. a = 759, 6 = 1130, J[ = 40° 32' ; find B. o. a = -1063, b = -4182, ^ = 14° 24' ; find B. 6. a = 44-28, b = 14-76, C = 100° 30' ; find A and B. 7. a = 22J, b = 28J, A = 44° 30' ; find B. 8. a = 872-5, b = 632-7, C -= 80° ; find A and B. 9. a = 26, 6 = 74, ^ = 18° 55' ; find B. 10. Two angles of a triangle are 76° and 54°, and the side opposite the latter 809 ; find the other sides. 11. The sides of a triangle are 112, 88, 76; find the angles. 12. Two sides of a triangle are 1-732 and l-41i, and the included angle 75' ; find the remaining angles. 13. Two angles^ of a triangle are 65° and 85°, and tlw interjacent side is 12-5 ; find the other sides. 14. The sides of a triangle are 2376, 1782, 1188; find the angles. 'rf ■a mm 70 15. Two sides of a triangle are 908 and 640, and the included angle 62° ; find tlie other side. 16. The sides of a triangle are 1895, 188-5, 1236 ; find the angles. 17. Two sides of a triangle are 77-99 and 83-39, and the included angle is 72° 15' ; find the other angles. 18. The sides of a triangle are 102, 168, 128 ; find the angles. 19. Of three towns, the firat is 165 miles from tlie second, the second 155 miles fi-oni the third, and the third 72 miles from the fii-st ; find the difference in the bearing of the second and third from the first. 20. The angles of a triangle are in A. P., the greatest being twice the smallest, and the greatest side is 984-8 ; find the other sides. EXERCISE XVI. HEIGHTS AND DISTANCES. 1. Describe the observations and calculations necessary to determine the breadth of a river from stations on one of its banks. 2. A tree 51 feet high has a mark at the height of 25 feet from the ground ; find at what distance the two parts sub- tend equal angles to an eye at the height of 5 feet from the ground. 3. A pole is fixed on the top of a hill, and the angles of elevation of the top and bottom of the pole are 6(i° and 45° ; shew that the hill is | (v'^S +1) times as high as the pole. 4. The angular elevation of an object at a place A due south of it is 30° ; at a place B due w^est of A, and at a distance a from it, the elevation is 18°. Shew that the height of the object is 1/(2 '/S + 2) 71 f 5. An object 6 feet high, phxcetl at the top of a tower, oiibteiids an angle whose tiuigent is -015 at a place wlioso horizontal distance from the base of the tower is 100 feet ; shew tlmt the height of the tower is 170'23 feet nearly. 0. A person stationed on a promontory first observes a «hip at a point due north of him ; in a quarter of an hour it bears due east ; and after another quarter of an hour is seen to the south-east of him. t'md the coui'se the ship was steering, and shew that it was nearest to the observer 1 2 minutes after he first saw her. Ans. — An angle whose tangent is i, to east of south. 7. A person wishing to know the height of an inaccessible object, measures equal distances Ali, JiC in a horizontal straight line, and observes the angles of elevation at A, B, C to be 30°, 45° and 60° respectively. Shew that the .'"3" height of the object is AB^I - , and its distance from ABC m AB ~ . 1/2 8. The elevation of two clouds to a person in the same line with them is a. When vertically below the lower one, the elevation of the other is 2 a. Shew that the heights of the clouds are as 2 cos''' a : 1. 9. The elevation of a tower on a horizontal plane is observed ; on advancing a feet nearer its elevation is found to bo the complement of the former ; on again advancing its elevation is found to be double of its first elevation ; shew that the last station is - feet from the foot of the tower. 10. A pereon on the top of a mountain observes tlie * depression (45°) of an object on tlie plane below him : he then turns through an angle of 30°, and observes tlie depression of another object on the same plane to be 30°. On descend- ing the mountain he finds the distance between the objects is d. Shew tlmt the height of the taouutaiu is also d. n 11. At noon a column in the E.S.E. cast on the ground a shadow the extremity of whi(;h was in the direction N.E. ; the angle of elevation of the column l>eing a, and the dis- tance of the extremity of the shadow from the column c, shew that the length of the column is c tim a V 2 V'2. 12. The elevation of a tower standing on a Iwrizontal phvne is observed ; a feet nearer it is found to be 45° ; 6 feet neai-er still it is the complement of wliat it was at the fii-sc station. feet, Shew that the height of the tower is a a 13. From the summits of two rocks A, B at sea, the dij>s» a, ft, of the horizon are observed, and it is remarked that the summit of iJ is in a horizontal line through tlie summit of A ; shew that the rocks subtend at the earth's centre an angle wliose cosine is sec a cos /J. (Tlie dip of the horizon is the angle a line drawn to touch the eai'th makes with the horizontsil plane). 14. A pei-son wishing to determine the length of an in- accessible wall places himself due south of one end and due west of the other, at sucli distances that the angles the wall subtends at the two positions are each equal to a. If a be the distance between the two positions, the length of the wall is a tan «. 15. A ship, the summit of whose top mast is 90 feet from the water, is sailing towards an observer at the rate of 10 miles an hour, and takes 1 hour 12 minutes to reach him from the time c^ its first ap{)earance. Shew that the earth's i-adius is 4224 miles, the tangent from the mast head to the earth's surface being considei-eil equal to the arc beneath it. 16. A jjei-son walking along a sti-aight ix)ad obst^rves that the greatest angle that two objects make with each other Is* a ; fix>m the point where this hap\>eus he walks a yards, and the objects there appear in the same straight line making an angle /3 with the road. The distance between the objects is 2 a sin a sin /5 cos a + cos ^3 73 iat Is* id ill is 17. A balloon considered as a vertical object of given height floats at a constant height alK)ve the eartli, and sul)- tends angles a, /? at a place when the elevations of its lowest I)oint are A and B resi)ectively ; prove tan (A + a) cot A = tfoi (^B + jS) cot B. 18. AB is a tower at the foot of a hill of inclination ; C, D are two stations dii-ectly up the hill from B such that BC=CD; ACD = a,ADC = ft. Shew that sin a sin /3 coiO = 2 cos a sin /j 4- cos /3 sin a 19. From tho top of a tower the depressions a, ^ of two objects in the same horizontal plane with the foot of the tower are observed, and also the angle o> which thoy sulv tend ; the distance a between them is known. The height of tho tower is a sin a sin /3 V (sin'^ a + sin"'* (i — 2 sin a sin ^9 cos «>) 20. At each of three stations in the same horizontal plane, and at given distances a, b, c from each other, the elevation of a tower is observed to be a ; shew that the height of tho tower is, if a + i + c = 2 «, a b c tan a l/ s (a — o) (s~b) (« — c) 21. The angles of elevation A, B, C, of a balloon were taken at the same time by thi"ee observers placed respectively at the ends and middle point of a base a measured on the eartli's surface. Shew that height of balloon is a V^^ot* A +"^t'' C— 2 cot*^) " 22. To ascertain the height of a mountain, a ba.se of a feet was measured, and at either extremity of this l>i\se were taken the angles a, /5, formed by the summit and tho other extremity ; also at the extremity at which the latter was taken the elevation of the mountain was y ; shew tliat its - ... a sin a sin y height IS — : — r- . ® bin {a + li) 74 23. A column on a pedestal 20 feet high subtends un angle of 45° to a person on the gi-oiind ; on ap|)roaching 2(1 feet it again subtends an angle of 45°. The height of the column is 100 feet, 24. In the ambiguous case whei-e a, h and A are given to •lutermino the triangle, if c', c" be the two values found for the third side of tlio triangle, prove that c"' _ 2 c' c" cos 2 ^ + c"2 = 4 a* cos^ A. 25. In the case wliere the solution of the triajigle is aniViiguous, if k, k' be the areas of the two triangles which satisfy the given conditions, j)rove Ar^ + k'^ — 'IJc k' cos 2^ _ o^ {k + k'f ~~'IP' A, a and h being given. tie area i>\ a IriiiiiHli'. Ki«. (i, 7. KXPRESSIONS FOR THB AREA OF A TRUNOLE. 48. It is proved by Euclid (B.I., prop. 41 ) that the area of a triangle is half that of a rectangle having the same Vm.se and height. Now the number of stpiare units in the area of a rectangle is equal to the product of the numbers of linear units in the base and height respectively, which is briefly expressed by saying that the area of a rectangle is the product of the base and height. Hence the area of a triangle is half the pi'oduct of its base and height. In fig. 6, 7, area of triangle ABC = ^AJi.CD, = ^ c b sin A — I 6 c sin A. Again sin A = 9 2 sin ^ A cos h A from (7) (s — b) {s — c) I s (s — a) V 67\ 5 be be .from (11) 8 (s — a) (« — b) (s — c) j- Therefore the area "" \ {«(« — «) (« — *) (« — c) } 75 EXERCISE XVII. 1. Tho sides of a triangle are 3, 5, 7 ; find its nroa. 1 Tlie sides of a triangle ,vre 5, G, 7 ; find its area. ' 3. Find the area of a triangle whose sides are 60 70 and 110. ' 4. Find the area of a triangular field whose sides aro 4/1, 40Gand 635. 5. If p, ij, r bo the perpendiculars drawn from each of the angles of a triangle to the opposite sides, shcvv that tho area is equal to \v2>qr.ahc. 0. Shew that the area of a triangle is equal to »(« — «) tan ^- . 2 7. Shew that the ai-ea of a triangle is equal to 2abc B C 8. I{ 2s = a + b-i-c, prove that the area of tho tiian-Ie IS equal to ^ s"" tan ^ A tan i li tan | C. the 9. The sides a, i, c of a trianele area are in A. P., shew that 1/3 (3 6 10. The sides of a triande four-fifths the area of an equilateral same perimeter ; shew that the sid 2 a) (2« — i), re in A. P., and the area IS triangle having the 11. If the three sides of and 2 s ~ a ■}■ l> + c, shew that the es are as 7, 10, 13. a triangle he a + b, h + c, c + a. iT: area is equal to 6". a b c t :l I 70 12. If th(i throo Hides of a triiuiglo bo V'a + A, VL + v, vc -|- a, Hhew that its aroii is o(|\ial to \ Va L -\-b c^ c a . l.i. If /), q, r \w the reciprocals of tlio porpentliculars drawn fi'r)m each of the aiif^Ujs of a trianj^le to the opposite hides, shew that tlio area of tlie triaiight is e(pial to 1 V ip ^ q •¥ r) iq-irr—p) (p + r — q) (_p-\-q — r) 1 i. Fi'ove tliat the area of a triangle is equal to o' — 1« sin A sin li 2~' Mn{A — JJ)' 15. Prove that the area of a triangle is equal to if-rt^l/'c' (sin 2 ^ + sin 2 7? + sin 2 C). 1 0. Shew that the area of a triangle is equal to 2 fruin A sin Ji sin C (sin J -t-sinyy + sin C')" 17. Shew that the area of a triangle is equal to a" + b^ + <^ 4 (cot A + cot Ji + cot 6')* 18. Shew that the radius of the circle described about a triangle is equal to a ah c 2 sin A 4 X area of triangle 19. Pi'ovo that the area of the circle inscribed in a tri- angle is eipial to 2 X area a + b +c 20. Shew that the radius of the circle described to touch the side BC, and the sides AB, AC produced (called an escribed circle) is equal to ; area 8 — a EXAMINATION PAPERS OF THB UNIVERSITY OF TORONTO. Mi SENIOR MATRICULATION, 1874. 1. Define th« logaritl.m of u number to base 10, an.I deduce tbe propertios wbieh make logarithmB of value in tacihtatnig arithmetical operations. 2. Find the followinir • Zcosec85°10'33"; Z tan tan log ^31 362. 201' 3D-008^UiO-48 2010 3. Find the numbers and trigonometrical ratios corre- sponding to the following logarithms : T-\)045 ■ 4-501 • O-ni 7220 (sin); 9-998460 (cos). , ^ on , J 5V 4. Perform the following operations by logarithms : V: ^^^-01 . 100-1 ^008 85° 25' 362x200' '''''' ^7olT°"57p- 5. Define the trigonometrical ratios and co-ratios of an angle less than 90°. Express all the trigonometrical ratios in termsof the cosine. 6. Find sin {A + B) and cos {A — B). What are the values of sin 105°, tan 75°, cos IS*^? 7. Expre8ssinJandcos^(l)in termsof^, (2) of cos 2yl. 78 8. Prove that in any trianglo sin ^1 sin 11 sin O . h'^ + c^ — «* = — - — = ; cos A =r^ _ a c 2 bo .sm J ^^(^6^). ^.^^^ _ I ^ I ^^^_^^(^_j^ ^^_.^^ I •?«: 9. Obtain the logarithmic formuloe of solution in the fol- lowing cases : Having given (1) a, A, IJ ; (2) a, b, C; (3) a, b, c. 10. Having given a = 240, b = 3()2, c = 200, find .1 to seconds ; a= 1001, J = 85" 10', i5 = 90" 15', find b and c to three decimal places. 11. The peritneter of an equilateral triangle being 10, find its lieig'it and tlie radius of the circumscribed circle. 22. and D are two points lying directly south of A and Ji respectively, and such that 1 B subtends equal angles at each ; having given the distar ces A C, B D, and the area of A B C D, determine tlie disi ance A B and its bearing. No. Log. Diff. Angle. Log. L\ff. 362 558709 sin 19° 12' 9-517020 302 3900 591065 Ill cos 15' 9 999990 2010 3031 90 210 sin 85° 10' 9-998453 11 1001 000434 432 sin 4° 35' 8-902590 8027 904553 54 2000 301030 217 1(»04 001734 432 SENIOR MATRICULATION, 1875. 1. Write down the characteristics of the logarithms of 235, 2-308, -800, -00025. I 79 2. State the numerical limits between which tlie numbers lie whose logaritliraa have characteristics 6 ar.d 2^ 3. State the rules for finding the logarithms of products, quotients, powers, and roots. Find the logarithm of '^/•6723■t x (3.8826)'°, 4. Given log 2 = -30103, find log -00025. Calculate the values of '67-234 - /672-3 W8-26'"^^WriTo 34 X 388-26 67 X 462-75' 5. Explain how the size of an angle is expressed in Tri- gonometry. Find the complement 6f 66° 41' 4" and the supplement of 100° 5' 25". 6. Define the Trigonometrical ratios of an angle les« than two right angles. Find the ain, cos, and tan of 30° and 60°. 7. In a triangle {A > B) prove sin {A+B) = sin ^ cos ^ + cos A sin B, cot ^ 6', tan 4 {A~B)="l — ^ a-\-b 8. Prove the fonnulas tan^ ^ X = 1 — cos r tan I X = 1 + cos X 1 + sin X — cos X 1 + sin X + cos X 9. (1) Given f=. 672-34, /I = 35° 16' 25", C the triajigle. 90^ solve (2) Given A -. 50^ 38' 52", B .. 60" 7' 25", « - 412 the tr 67, solve lanjfje. 80 10. If s- the semi-perirneter of the triangle ABC, prove that tlie radii of the inscribed and circumscribed circles, are respectively. s tan i A tan ^ Ji tan J C, J s sec \ A sec | ^ sec J C. No. Log. Angle IjOg. 10473 02006 sin 35° 16' 25" 9-76154 11691 06788 cos 35° 16' 25" 9-91190 38820 58913 cosec 50° 38' 52" 10-11167 4121)7 61560 sin 60= 7' 25" 9-93807 46275 665.35 sin 69° 13' 43" 9-97081 49899 69809 54890 73949 67234 82759 FIRST YEAR, 1876. 1. State ard prove the rule for finding the characteristic of the logarithms of whole numbers. Given log -25 = — -60200, find how many digits there will be in the integnil part of (2-5)^". * a 2. Prove log a = x log a, log -j- — log a — log h, Evaluate the following by using logarithms : l>'80 X 1^2-7, t/^=ITx 18' Find the tabular logarithms of sin 45°, tan 60°, cos 30*^. 3. Shew that the logarithms of the trigonometrical ratios need not be entered for angles greater than 45° ; take sin A , tan A as examples, where A has any value from 0" to 180". Ifsecl20"c:; — , can a be found by logarithms'? 40 Adapt sin A — tan | A to logarithmic computation. 1 i .1., 81 stic ,ill tioa A, ?0". 4. Prove the following relations : sin'' ^1 =; 1 — cos' A tan' A - sec- A — 1. cosec A — sin A -- cos A cot A ■■ sec A + tan ^1 . sin -4 5. A person standing on one bank of a river observes that an object on the o])positc bank has an angle of eleva- tion of 45°, and going back 150 feet, the corresponding angle is 30°. Find the breadth of the river, 6. A vertical stick whoso height is 10 feet throws on u horizontal plane a shadow 7*74 feet long. Find the sun's altitude. Indicate l..)W the problem would bo solved if the shadow fell on a piano through the foot of the stick inclined at an angle d to the horizon, the line of intersection of the plane and hAv.>,)ii being perpendicular to the plane through the suji and -lick. 7. Prove cos (A + B) = cos A cos B — sin A sin /i, 2 sin" h A = 1 — cos A, (cos A — sin A)- - cos 2 .1 tan (45° — A), cos 6 + cos 3 1 — tan'- fl sin + sin '6 2 tan d 8. In any triangle establish the following : sin A _ sin li _ sin C b c' a cos i C ■■ jsis—c) ab 2 X area = b c sin A = cot A + cot b 9. In a triangle i? = 1 23° 40', 6 = 100, c =- 60, find A and C. .4 = 112° 40', 6 = 21 34, c= 2134, solve the triangle. a = 200, b = 77-4, C = 41° 50', find the area. 10. If (sin 0-i-cos e) =3 siu ^ + sin 20, find in de- grees, «kc. . i i 'I ■I , i m '; ' t ■I II 83 If 1 + sin (cos A — sin B)^. No. Log. Angle, Log. 20000 30103 tan 52° 15' 1011110 30000 47712 tan 52° 16' 1011136 41645 61956 sin 56° 20' 9-92027 77400 88874 sin 29° 57' 30" 9-69842 21340 32919 sin 41° 50' 6-82410 17761 24946 sin 19° 28' 9-52278 51623 71284 sin 19° 29' 9-52314 tan 26° 33' 9-69868 tan 26° 34' 9-G9900 FIRST YEAR, 1877. 1. Define the trigonometrical ratios of an angle, and write down the five relations connecting the six trigonometrical functions, — sine, cosine, tangent, cottmgent, si'caut. uud cosecant. 2. Explaiii the nature and use of logaritluns, and find the common logarithms of 2 J, 2 J, and ^(0162)''. 3. Perform tho following operation by logarithms : 1-28 (216)^^ 5 ^'f2 1-25 -81 4. Prove the following : sin vl + sin i? tan | (A + B) sill A — sin B A cot tan -^ {A—B)' - cot A + cosec A, sin A +si I .1 A +sin 5 A cos A + co^j .3 A + cos 5 4 A cot tan tan 3 A, A C08 A = coc -\- tan. 83 5. Find the values of sin 2 vl and cos 2 ^ in terms of the simple ant^lo A. If sm 2 ^ = cos 3 A , find A ; also its sine and cosine. 6. In any triangle, 0^ = 0^ + h" — 2 ai cos t'. From this single equation prove that any two sides of a triangle are together greater than the third. 7. In any triangle prove the following isolations : tan - = J(« — b) {8 — c)^ j^^^^ ^ i/s{s—a){s—b}(3—c), 2 ^ s (s — a) c = b cos A dz ya^ — b"^ sin- A. 8. The sides of a triangle are 20, 21, 29, find the area, and the angle opposite the greatest side. 9. If in any triangle = — ,then will .4 ^ 120°. b +c a — c 1 0. In deterniining an angle by its tangent, when the angle is near 90°, how would you proceed 1 Shew also how to find the sine of a very small angle accu- rately from the taoles. 11. A person undertakes to measvu'e the distance between two points, A and B, and proceeds 50 yards to G in a straight lino towards B ; and then meets an impassable barrier, and !is he has no instrument for measuring angles, he measures off a line in an unknown direction C J) = GO yards, and then measures A D 90 yards, and B D dO yards ; find A B. In the same question, if the person has neither instruments tor measuring angles, nor any trigonometrical tables, find A B, No. Log. Angle. Log. 20000 30103 tan .54^^ 44' 10-15048 30000 47712 tan r)4= 45' 1015075 40961 67174 sin 70' 31' 9-97439 sin 70' 32' 9-97444 sin 38" 56' 9-79825 sin 38° 57' 9-79840 84 I :i : i! JITNIOR MATRICULATION, 1877. 1, Define the logarithm of a number to a given base. Prove log m n = log m -\- log n. log,. N log, a -= log(, N log, h. 2. What are the advantages of employing 10 as a base] Shew how to find the characteristic of a numbei', part of which is integral. W. The mantissse of the logarithms of all numbers which differ only in the position of the decimal point, are the same. What is the object of always making the mantissje positive % 4. Given log 32953 = 4-r)1789r)0, log 3-2954 = •5179081, find log -003295345. Find this also by forming and employing a Table of Pro- portional Parts. By tliis table determine the number cor- respoiiding to the logarithm 3-5179025. Firid by logarithms the value of l/^0l28 X (12)^ X -J^ X 12-5. 5. Prove cos (.4 — B) = cos A cos B -\- sin A sin B. CCS Z^ — cos ^ = 2 sin h {A + B) sin I {A — B). cos'^ A — cos- 3 ^ — sin 4 A sin 2 A. 6. In any triangle shev/ that a = b cos G -^ c cos B. cof ^-^/ s{s — a) a — h C tan ,V {A — -6) = — ^-r cot a + 6 sill A - Hin B »iu {A — B) a-\-b ii-iif -|-|-agMg^g 86 7. a = 300, h = 400, c -= 500 : solve the triangln. a = 7, 6 == 5, c = 4- ; solve the triangle. a == 450, I = 250, C = 12° 3G' ; fm.l J and i?. 8. If an angle, the opposite side, and the sum of the other wo sides of a triangle bo given, sh( a' liow to solve the tuanglo. 9. If the sides of a triangle be in A. P., the tangents of half the angles are in //. P. 10. Three ciriles, two of which are equal, touch one another, and a fourth, lying between, touches them. Shew that the radius of this circle is r (/ + r) sin- } 8 ? COS' Id — r sin- 4 d' where is the angle between lines drawn from the centres of the equal circles to the centre of the other, and r' is the radius of each of tho equal circles, and r that of the other. No. Log. Angle. Log. 20000 3010300 cos 53° 7' 9-7782870 30000 4771213 cos 53° 8' 9-7781 18G 49353 69331 IG sin 50° 40' 9-889O044 70000 8450980 sin 50° 47' 9-8891675 Kin 22° 12' 9-5773088 sin 22° 13' 9-5776183 tan G° 18' 9-0429731 tan 08° 52' 10 -4 128096 tan G8° 53' 10-4131853 FIRST YEAR, 1878. 1. Define the trigonometrical ratios of an angle ; and shew which of them may have any magnitude w hatever, positive or negative, and which of them never can have a valu(> between •+ 1 and — 1. Shew that tho versed sine of an angle is equal to twice the square of tho sine of half the angle. ee 2. Prove tlio foiinula chord A = 2 sin i ^, and hence show that the chord of an angle w'.ll be jxjsitive while the angle increases from 0° to 3G0', and negative while it incixjases from 3G0° to 720». 3. Deduce formulas for expressing the sines and cosines of the sum and dift'ei-ence of two angles in terms of the sines and cosines of the angles themselves. Find sin in terms of sin 2 s'=i&csin^ = T/7(7zr^j^^TrM7j:r^ = — cos J ^ cos I i? cos J C (b) cos A = ^^jtfzz^' 2bc ' (c) c = (a + i)«^i^,^hentan^='iZZ^cot 4 C cosV' ^ ^TT ^ (c/) tan (^ + i?) = l±i tan ^. 88 12. Show how to find the radii of the inscribed and escribed circlea of a triangle in terms of its sides and angles. If R bo the radius of the circumscribing circle, prove tliat tiio product of tho perpendiculars from tho angles upon the opjiosito sides 8J{^ No. Log. Anglo. Log. 2.3154 304G3 4° 191' . tan . . 8-87859 07488 82923 12'25.V . tan . . 9-34319 15.3G 18G39 7° 32' . tan . . 9-12141 92178 9G4G1 73° 15' . tan . . 10-52143 20000 30103 9GO00 98227 42180 G2511 JUNIOR MATRICULATION, 1878. \. If be the circular measure of an angle between 0° and 90°, shew that sin y — ^ 0\ Show approximately what the dip of the horizon is for every mile of distance. 2. Shew that sin 18° = ^ (|/ 5 — 1), and hence shew liow to find the sines and cosines of all angles being nmltiples of 9° from 0° to 90°. 3. If A and JS are any angles, prove /i\ . / < , m tan J ±: tan B (1) tani.l ■±,B)=. ~ . 1 -f- tan A tan B (2) sec A + tan A = tan (45° + i A). /Q\ i. A. vers A (3) tan— = ^ ' 2 sin^ If cc = cos .4 cot i4, 2/ = sin A tan A, eliminate A. 89 4. Tn every trianglo prove tlio tnith of tlio following formulas : (1) Area = i a b sin C = Vs (h — «) (a — b) (s- (2) c = a cos li + b cos A. c). (3)sin^=^(liri)J»=^). (*) tan J{ tan (7 5. Define the terms logarithm, characteristic and mantissa- How are the logarithms of numbers less than xmity to be found from the tables, and how are they represented 1 Given log 4-9353 = -6933110, find the logs of -49353 and of (049353) . 6. Having ascertained the logarithm of four digits of a number from tlie tables, shew how to proceed to find the logarithm of the whole number. Given log 2 •30103, log 3 = -4771213, log 7 = -845098, i i i \ find the logarithms of 28 , 03 , 98 and 120 . 7. If a, b, c are the lengths of three straight lines drawn from a point making equal angles with one another, and straight lines be drawn respectively joining the extremities of a, h, c, the area of the whole triangle thus formed will be i-— (be + ca + ab). 4 8. Shew how to solve a triangle wlien two sides and the included angle are given. Ex. a = 705-432, ?»--= 1006-02, C= 70°. 9. Find the radius of the circumscribed circle of a triangle in terms of its sides and angles. If the centres of the escribed circles of a triangle be joined forming another triangle, shew that the circle circumscribing this latter triangle is four times the size of the circle circum- scribing the first triangle. rMAGE EVALUATION TEST TARGET (MT-3) WJ., (./ :<' y. v.. fJx 1.0 I.! 1.25 *ria IIM .1 i^ '""^ 12.0 1.8 U i 1.6 & '/r //, A "a ■c'l ^ ■> s^ m ^> w Photographic Sciences Corporation €3 % ^•V V :\ \ ^> ^ o^ .<' 9) ^i^ 73 WEST MAIN STREET WEBSTER, NY. 14580 (716) 872-4503 ^ ^"^V/ yA' <^%^ ' v^^ V ^ w< 90 10. A person at the top of a light-house descries a vessel just on the horizon; shew that he can ascertain the dis- tance of the vessel approximately by taking the square root of one and a-half times the height of the light-house in feet, and calling the result miles. •■■ No. Log. Angle. Log. 1772052 241188 100662 103543 2484765 3823555 0028656 0151212 sin 70° sin 66° tfl,n 35° tan 11° tan 22° 9-9729858 9-9607302 9-8452268 9-2886522 9-6064096 FIBST YEAR, 1879. 1. Define the logarithm of a number. Shew how the logarithm of a number to base e may be converted to the corresponding logarithm to base 10. 2. Prove the rule for finding the characteristics of loga- rithms. Why are the mantissas only inserted in the tables ? 3. Prove log — = log o -f- log 5 — log c. c log"l/a = - log a. n 4. Having given mantissa log 173300 = 238799 " " 173400 = 239049 construct a table of proportional parts for intermediate numbers. Find log 173*344 ; and write down the number whose log is 2^238854. 5. Given log 2 = 0-301030, log 3 « 0-477121, .3-»x^4 find the value of 21-6 Find the tabukr logarithms of cos 30°, sec 45° and tan 1 20°. :^ 6. Prove the formulas (l)tan^ ^'"^ cos A (2) C0Si4 = l/l— 8iir.4. (3) sin (A+Ii) = sin A coa B + coa A sin A (4) sin n -4 = 2 cos ^1 sin (n — 1) ^1 — sin (n — 2) A. 7. In any triangle prove the following: (n axniA=J ('-^) (^ — ^) ^ Ac (2) tan -2-=^-- cot 2- A—B.C . B—C . (3) sm — ~ sm - + sm — ^ - sin ^ + A 2 Bin Bin - = 0. 2 2 (4) tan il + tan J? + tan C = tan i4 tan B tan (7. 8. In a triangle (1) J = 53° 7' 48", B = 75°, a = 1056, find b and C. (2) o = 48, 6 = 30, and perpendicular from C upon opposite side =24. Find A, B, C, and c. 9. Obtain expressions for the area of a triangle. Ex. a = 70, 6 = 80, c = 90 ; find area. If I, m, nhe the bisectors of the opjiositc sides of a tri- angle, and 2azs=>l + m + n, shew that Area = | l/ff (a — I) {a — m) {a — n). 10. The altitude of a mountain from A is 30°, from B 36°, and from C 45°. A, B, and C are in the same straight line, and AB=BC= 1000 yards. Find the heightof the mountain. No. Log. Angle. Log. 30243 18944 30773 10560 12750 480629 277478 488168 023664 105518 sin 75° cot 36<' Bin 53° 7' 48' 9-984944 10138739 9-903090 1 t'l ,i; I ;■: ! 92 JUNIOR MATRICULATION, 1879. 1. Explain the terms cluiracteristic and mantissa, and state the rule for writing down the characteristic of the logarithm of any number. Write down the characteristic of -5, -0007 and 60050*3. What would be the characteristics of these numbers to base 100, and also to base — ? 2. Find the logarithms of y,^ -007 and (-5)-'. Find the index of the power to which 7 must be raised to produce 300. 3. Having given L cot 57° 30' = 9-804187 Difference = 279, find L cot 57° 30' 15", and find the angle, the Log of whose tangent is 9*80425 1. 4. Find the values of sin 30°, cos 30°, and sec 45°. Write down the tabular logarithms of these i-atios. 5. Px'ove the formulas, (1) sin A = sin (180° — ^) = cos (90° — A). (2) cos {A — li) ■= cos A cos B + sin A sin B. (3) sin 2 ^ = 2 sin A cos A. The angle BAG is bisected by AD. BC and BD are perpendicular to AC and AD. Pi'ove that BABC = 2 BDAD, and BA •AC = AD^ — BD\ 6. Shew that (1) sin 18° sin 54° = |. (2) 16 cos 20° cos 40° cos 60° cos 80° = 1. f 93 7. In any triangle, prove the formulas (0 <^==acoBB+bcosA. (2) tan I J =^ /5^2l>"^=^ M i>(a-S-^) • In the triant'le ^ /yr' fin: i "-eting AC inn. FiJ^n T" '* ''''''' "'^'^''« '^ ^^^^ t'iangle. ^ '" *^''"'s of the sides of the «• Solve the equations (1) sin'e + Hmi2d= 1. J 2 * ( cosec 2 + cos 2 ^' = J_ 9. Solve the triangles (l)^i = 2riO',C = 90°,a = 3141G (2) J == 74^ 53', ^ = 37° 5ry, = 300. wi^s::::::n«^:::;'^-'«^-riheai„^^^^ (- = 3. U16). ( 20000 30000 70000 3U1G 02323 30103 47712 84510 49715 9G531 tan 21° 10' f^in 74° 53' sin 37° 55' sin G7° 12' 0-G5205 9-98470 9-78858 y-9G4G7 ANSWERS. II :> I. (Pagb 12). 1. 3,1,4,-1,-6. 2. =log 2»X3» 4. =logl00x2 = . 7X10 C. log8 = i (a — 6 + i;, log9 = a4-6 — 1. 7. f log 25 - 3. 8. 4-8624005. 9. 1-4042714. 10. 1 8379346. 11. 2 -4506323. 12. 2-8653156. 13. 2 8400790. 14. 4 7497460. 15. 6 4721485. 16. 1-2552725, 1 3979400, 2 3344539, 0511526, 8115752, 3-7323939, 1 6478174 J 6354839^ 2 8573326, 0880456, 0263938, 1-6365006. 17. 4 20412, 1 4071244. 18. 1-9599947. 19. -002620145. 20. 562 4865. 21. 3041341. 22. -004965347. 23. 2-7485257, -00560^4388. 24. 18607. 26. 10065. 26. -739148. 27. 437499 9. 28. 124385. 29. 27-9866. 30. -000798595. II. (Page 14). 1. 8, I i, 3, I, 2. 2. 2. 3. 4. 6. 2.089. 7. (1) logc (3) m log a + n log b 4 log 3 + 3 log 2 y .V log c - d log b -69897. (2)3 5. 1-537. log 2 — 1 3 log 2 • log e ~ d log a log b - log a log 3 log a - log b 8. I, f . 10. 1 -5 =. i (2 + 1) = log (100 X 10)* = log 31 -02. . 11. 0103 = log ?£? = log ,?ij^,&c. 12. 13. 13. 27. 14. 1 log?£« = log?ll,&c. *260 '*1000' •67362. 15. 10^. 16. 5. 17. 100 or -J 001. 18. reqd. , log xv = 3, or y = }-2i-? = 1 09862. — logx logs (20)* = loga 2 + i loga 5 = log^ (4 X 5)*. 19. Ifybe 20. Tnie if III. (Pagb 19). Examples such as this exercise contains may easily be worked hy expressing the trigonometrical ratios in terms of a single one ; the identities being thus reduced to ordinary algebraic identities, may be verified as such. It will frequently be found convenient to express the ratios in terms of sine and cosine, simplify, and substitute the relation between sine and cosine. Thus Ex. 17, oof* + tan^ = sec^ cosec^ — 2, if cos" sin'' bin" cos'' cos''

". 18. 34° 31' 23-8" or 146"' 28' 36-2". 19. cos 61° 49' 25" = sin 28° 10' 36", 9-6741145. 20.9 4954522. 21.101103715. 22. 10-1330341. 23. 9 8081838. 24. 09M4' 27". 26. 46° 40' 42" or 134° 19' 18". 26. 32° 43' 51". 27. sin .r = L sin X = — L cosec x, L sin y = - L cosec y ; 28. Wlien 29. sin A = cosec X ' :. L sin x - L sin y — — {L cosec x — L cosec y), one angle is the complement of the other. — - — 5 ; .•. L sin .4 - 10 = - (L cosec .4 - 10) ; .-. L sin A + COB6C A L cosec A = 20. 30. By lO^", for log 10»o sin ^ = 10 -f- U>g sin .^ =3 X sin .^. 96 VI. (Page 30). 1. c -- 707, ^ = ^ - 46°. 2. a - 6 = 1G9-68, ^==46°. 3. ;, = 100, r. = 141-4, li « 45^ 4. a = VJ, b - 20-784, /i = «0°. ». 6 = 17•y2,c==34-04,lf = 30^ G. 6 = 831-3G,c=-9(K), ^ = 30'. 7. /> = 8;M3G, a = 30% 7i - G0°. 8. 44^ 29' 63". 9. 40^23' 37". 10. GO' 45' 30". 11. ^ = 3G'52'12", ii -53'7'48". 12. A - 25° 22' 37 " , fi = G4° 37' 23". 13. ^ = 32° IG' 32°, Z?= 57° 43'- 28". 14. ^ - 3G°31'44", iJ = 63°28'lG". 16.468-257. IG. n = 2954-42, 6 = 520-946. 17. (1 = 3758-77,6 = 1308-08. 18. b = 174-70G, c = 510-806. 19. a = 13436 4, c = 13iK)9-3. 20. h = 59-87G7, c = 138 24. 21. a = 117 232, b = 2G9-G1G. 22. ^-36° 49' 44", 6 = 206-932. 23. 6 = 299383, c = 091201. 24. A = 3G°9'3% c = 119-509. 26. a -^ ^(1282)2 -(1218)2 = V 64 X 2500 = 400, if = 7 1° 49' 10". 2G. GO 2593. 27. 442 540. 28. 40997-9. 29. 148 327. 30. 6 = G12, A = 34° 12' 20", B = 55° 47' 40". 31. 6 = 1G3-6G14, c = 206 2519. 32. 6 = 12, yl = 22° 37' 11 ", B - 07° 22' 49 ". 33. 6 - 141, ^ = 82° 41' 44", B = 7°18'1G'. 34. 28° 4' 20". Divide triangle into two right- iinglud triangles. 35. 19° 11' 18". 3G. 96° 22' 42". 37. Triangle i8 iBusceles, and may be divided into right-angled triangles. 31°5'32",74°27'14",74°27'14". 38. 72°32'33". 39. 61°55'39". 40. 8-988. 41 17-867; OD - 24 (coa 26°)3. 42. 152 62. 43. 905-98; 36 sin 64° = altitude. 44.6882. 45. 939 7 in. 46. 82-904 in. ; 8 sin. 56° = altitude. 47. By Geometry wo may shew that its area is eijual to that of a triangle whose sides arc the diagonals and included angle same as theii's, 92-72 in. 48. GG9-13in. 49. 44° or 136°. 50. 26° 19'. II !l i 1. 80 ft. 2, 6. 258 4- yds. VII. 87 ft. 6 in. 7. 49-2 mis. 11. 80-58 ft. 15. 49° 10'. (Page 34). 3. 40-98 ft. 4. 8. 2 -732 mis. 12. 193-50 ft. 16. 1097-77 ft. 72 ft. 6. 81 ft. 9. 4 1/'5, 6 ,/5. 13. 152-674 ft. 17. 421-99 yds. 10. 153 3 ft. 14. 81 -611 ft. 18. 89-9069 ft. 19. 124 4 ft. 20. 278 18 ft. 21. 93 97 ft. 22. 3 min. 20 seconds, 36° 52' 12". 23. 7' nearly. 24. 233 2 ft. 25. 715-93 ft. X. (Page 44). 4. If A lie between 0° and 135° cos ^ + sin ^ = + VI + sii 2^ ; between 0° and 45% cob A - sin ^ = + ^l-sin 2^4 ; between 45° and 135°, cos J. - sin ^ = - yi-si n 2 A ; between 135° and 180^, cos A±smA= - >^l±&in2A. 5. V2 sin (45° - A) 97 47. 0°or30° 48. eqiiivulent to sin 8 « = or 7i°. XI. (Page 46). 20° or 90°, 49. o° « + sin 2 « = . or 30°. 60. Equation «'" l('a- + 8in2x; thenco XII. (Page 51) f^ + ^j ""|=(«o»^ + l)8in| = 2co8^J? M-2?) sin A cos :^ = lo, cos 2 sin A f«°-|J 6. sin 2a B 2sin^ 8in^ + l-8in^-coV^VlZ _ 2 cot ^ ^ ,^t ^A~i = &c- 11. sin» hA:=l ft + c - «, ~ 4 sin A ^ c(w i A «in^ 2cos»^^J'28i„^i,„^^^^ 2(«4-6)8in'^C=a+6-(„ + cosC + ccosJ-(a + 6)eo8(7=.&c, sin Bsin C sin .4 - sin jBW^ 21. True if , 4- « cos B . ft) cos C=6cos C + (1 - cos A). Hi 20. True if cos j8 -j- « sin B cos B ~ ' sin (7 ~ * «os A sin i?' 2G. - sin B a sin^ ''^ ooa i (B + C) BinJJB - C) 2 cos h (B + C) sin ^ (^ -JTq = &o. 31. Roduce, sin C sin ,4 32. For then perp. bisects I !s to a--^h- to (.- - a^ + « 6) (a"/;;': 7^^*^ ""«- 35. Condtn. reduce.. 1 o^o ■^"'- <^^«K 63). 1. 30° or 160°. 2. 45°ornr.o q ^ „ 10. 116° 22' 30". il. 50. 28^' t "' ^'■»««'. 8 • 12. 55° 46' 16". 13. A - 117° 19' 11 ", B = 2« 40' 49". 14. A = 119"» 2G' 51 3 , B = 6° 33' 8 -7". 15. J = 09° 10' 10", B - 46° 37' 50". 16.102-3368. 17. 32° 57' 8". 18. J = 71° 33' 15", fi - C Bin n f ^ 60° 2' 45",c = 67-502 ; for latter part use — pT = — i / ^ "m" ' *^ a + b cosh{A-B) 19. 78°19'24''orl0r40'36". 20. 10.6816. 21. yl-5G°16'4', B = 59° 51' 10", C = 63° 53' 46". 22. ^ = 6° 1' 53' 9", li = 108° 68' 61". 23. 02° 61' 33" or 117° 8' 27". 24. 6 = 226-63, c = 208-17. 25. ^ = 84° 54', B = 69° 20', C = 25° 46'. 26. 32-412 ; in latter part use formula in ans. to 18. 27. 62' 22' 50, or 117° 37' 10". 28. 220-999. 29. 58° 10' 44". 30. ^ = 47° 50', B = 57° 10'. 31. 7° 32' 31". 32. a = 88 4552, /. - 108-500. 33. 146°37'30". 34. ^ -108"27'26-4",B=4°55'10-0'. 35. B = 17° 0' 46", C = 133° 2' 16". 30. 82° 10' 49", 60° 24' 11". 37. B ~ 23° 8' 33" G = 32° 17' 27". 38. ^ = 72° 30', B = 48° 30', c = 86 -718. 39. B = 53° 62' 30", or 120° 7' 24" ; C - 84° 57' 24, or 12° 42' 30 '. 40. A = 129° 22' 28", B - 14° 37' 32". XIV. (Paob 68), 1. 120°. 2, 159 0043. 3. 132° 34' 34". 40° 37' 50". 5. 110° 33' 54", 20° 33' 64 '. 0. or 120° 32' 40", C = 81° 32' 40" or 8' 27' 20". 8. 82° 49' 10". 9. 76° 47' 2 ',49° 12' 58". 50°45'22". 11. 80°30°or99°30'. 12.80-3218 14. 168° 27' 25", 4° 55' 11". 15. 71° 44' 30", 48° 15' 30". 10. ^ = 34° 27', B = 100° 33', b = 1737883. 17. 431 009. 18. 117° 38' 46", 27° 38' 45". 19. 70° 53' 30", 00°, 49° 0' 24". 20. G0°22'^46", 63°37'16". 4. 09° 10' 10", B = 63° 27' 20" 7. 3062-28. 10. 06° 14' 38", 13. 41°24'34". XV. (PageGSO- 1.110-5,111. 2. 63" 26'. 3.429-0. 4. 75° 23' or 104° 37' 5. 78° 7' or 101° 63'. 0. 02° 20', 17° 10'. 7. 02° 51' or 117" 9'. 8. 00° 45', .39° 15'. 9. 67° 19' or 112° 41'. 10. 97, 700. 11. 85° 50', 51° 30', 42° 34'. 12. 00°, 45\ 13. 22-60, 24 9. 14. 104° 30', 40° 34', 28° 50'. 16. 829 C 10. 71° 22', 70° 30', 38° 8'. 17. 60° 30', 61° 10. 18. 37°20', 93' 10', 49° 30'. 19. 09° 20'. 20. 642-9,806. 1. 0-496. XVII. (Page 75). 2. 14 09. 3. 1807. 4. 96523-5, or 95520 if four-figure tables be used. 6. For tan ^ -4 = */^l— ''^ ("' ''^- 3 (s-a) 12. v's(s-a)(a-?>)(«-c)-i v'(«» + 6^ +c»)a -2(a< +6* + r*), which use. 13. Fop », «, r, substituto „ » .t^ » ., • ' ' " ' '2 iirctt 2 art'ft '2 urea 14. Rendily obtained by using 22 of Ex. XII. 15. sin 2 .1 + »in2 B -{• am2 = i sin vl sin Ji sin C, - obtauicHl in fuinie way aa 43 of Ex. XI. 10. sin J -|- niii Ji + sin f = 4 cos - li C coa - ct« - , - 43 of Ex. XT 17. For cot A, Ac, Bubstitutc 2 2 h c cos A -i, , «fec. , and employ 12 of Ex. XII. 18. Let A li I' be 2 area ' ' * ^ the triangle, and O the centre of the circle ; then li C at \ ^ centre — 2 A; .". Bin w4 = -■ ,. . 19. If O bo centre of circle, radnia aro& A B C = A Ji+ O li C-\- C A = h r c -\- ^ ra -\- h r h 20. If bo centre of escribed circle totiching B V, area A B C = OAB-\-OAC-OBC=^rc + ^rb-ira = r($-a). EXAMINATION PAPERS. Senior Matncnhition : 1S74. 1. § § 1, 5. 2. 10001541 ; 10-255513 ; tmi "' 3!|? means the angle whose tangent is §Iif, so that tan tan "'ooi ' 2?>? I •0%G70. 3. -80270; 3900G-; sin 19° 12^ 33" ; cos 4° 40''22". 4. Si n 19° 12' 39 '^ ; 8 0274. 5. § 17. sin A=.^l- co8« A] tan A = 'yi"C('»''^ . &c. C. § 30 juid Ex. VIII. sin 105° = cos A sin (45° ■+■ 00°) = ; 3 ; ir— ^2^2 2^2 (2). §40. (3). §41. (4). §48. 9. (1). § 4.3. 7. §;}8. 8.(1) §34. (2;. §45. (3). §44. 10. (1). 38° 25' 18". (2). 100 "450, -802. H-^, 10 3 /J. 12. A circle may be described about J ii />('. Knowing the area and A C, B Dy \m may find the pcrp. distance bet. ^l C, BD. We may then proceed in dif. ways, c.(j. , A B'' — (i)erp.)^ + \ ^{BD -^ AC)\^ ; then find B V, ami thence angle CA li. Senior Matricnlation : 1S75, 1. 2, 0, 1, 4. 2. 100,000 to 9,999,999-9 . . . ; 3. §6. 6-8740G. 4. 439794. '10473, 1 1091. 79° 64' 35". C. § § 17, 31. § 20. -01 to 0999... 5. 23° 18' 50' , 7. §30. §45. jj 1 - cos X 2 sin ^ ^ x 1 + cos X 2 cos ^ ^ X " 1 4" sin .<■ - cos X 1 + sin X + cos X 100 2 sin ' J ar 4- 2 sin A r com .\ t, 2 cm ■■' i X -f. 2 Bill A j; cm I *; ~ *«• ^>- (!)• " ^ .'W820, 6 = 648 IK), 7i — 54° 4:f 35 ". " ^2). /> - 4«)2-75, c = 408-99, G^ 09° i:»' 4;r I 10. (1). r uruii I I area 6 «-c- --;,-• (2). /t = and tan J A tan i Zi tan ^ f ' = a 6e 4 X aroa , and sec ^ /I abe H s^n (■"• - a) (« - 6) (« - c) » X ar«a First Year: 1870. 1. § 4. 8. 2. § n. 4 ir>46, - -774. 9-84949, 10-23850, 9-93763. 3. Thus L sin 120' = L sin m' = 7. cm 30". a nmst of CMuniu bo nugativu ; wo may howover nugluct tlio sign, find ntnnoi'ical valuo of a, and profix -, =2 sin A A cm \ A - * ri =tan i ^ (2 cosa l^ A - 1) := tan i ^ cos A. 4. (4). COB B -/i ■* V 1 + HJn ^ _ |(l + sin Ayi 1 -f sin ^ •t A 4- cot ^ sin {A + £) -, if h Bin (' 9. (1). A ^ 20" 22' 30", C -= 29^ 57' 30 ". (2) Tri. angle isosceles a = 366-22, B = (7 = 33" 40', ('X). 510 23. 10. (1). 1 -= 3 sin tf ; 0-^ .19" 28' 17". (2). Eq. reduces to 1 + sin — OS ■{■ I - wn. 0, or tan Q — i; :. fl== 20" 33" 64'. 11. True if 2 - 2 sin ^ ccm 7i - 2 cos A sin B > cns^ A -\- sin" Jl - 2 c(« A sin ii, if 2 - 2 sin A cm'^ /i, if (sin A - cos'B)"^ > 0. cos B SHI'' .1 + 1- 1. 2217,21. r-232G5, last no. First' Year: 1877. 2. log 2h = log l^ = -39794 ; -35218 ; 3. 40901- ; only logs of 2 2 X 3* ,10* , and 3 are used until last operation. 4. (1) In left-hand side of identity, for A put its equivalent j^A + B + ^A-B, and for 101 (.3). sin A =- Bin (3 ^ - 2 ^) ; sin 5 A Ii,iA+Ii-l A - II. (2). cr.t A + cosec A = £2?._d+_l = 2 c<.s> } A "'" ^ 2 sin } A cfiH } .4 = sin (3^ + 2^), etc. ; .-. left-hand si«l» of identity -^ h\n 3 ^ + 2 sin 3 A v.m 2 A _ ,,, ,,. , . CC.8 3^ + 2 COS 3 A COS 2 .4 <*^- J^'«J't->"*"J »"1« = cos" d - sin" - cos" ^ + sin' i = cos w4. 5. J 38. cos 3 yl = 4 cos"" ^ - 2 2 3 cos ^ = 2 sin ^ cos -4 ; cos 4 = 0, or yl = ^ or any odd multiple of .J- ; also 4 cos" ^ - 3 = 2 sin v( ; .. sin yl = ±V5-1 , or sin i n TT +(-l)''vl| = ±V5-1 , giving other 4) "» >HI« J ^ ■ > ' - J 4- values of A, n beii- nr,y integer. cos wl = +- M ± v^f) .. , A/ - 8 ■"• *>. HO. (2), c» on AB. Then from Euclid, B. II., 12, CW— 20, .-. DN => 40 V2 ; .-. £JV- 70, and AB— 140. ♦ 102 Junior MatrictUation : 1877. From definition a log S log a = N,x * = a; log iV log a b X 1. §1. §5. log N log a x " = N ; similaxly x ' -■= N ; .'. &c. 2. Man- tissas same for same sequence of figures and characteristic determined by inspection. § 4. 3. § 7. By making man- tissas always positive, same sequence of figures always has same mantissa. 4. 3-6179009. 3295.357. 0049363. 5. Ex. VIII. Ex. VIII. = (oos ^ + coi 3 A) (cos ^ - cos 3 ^) = 2 cos 2 ^ cos A X 2 sin 2 ^ sin -4 = sin 4 ^ sin 2 A. 6. § 36. § 41. c sin C sin (A -f -B) §45. substitute and a -\-h sin A -{-Bm B sin J. + sin B ' reduce. 7. (1.) Triangle is right-angled, - J5 = 53° 7' 48", A = 36° 52' 12". (2). A = 101° 32' 14", iJ = 44° 24' 56", C= a -f , from which 34° 2' 50". (3). A = 152° 34' 24", B =- 14° 49' 3C", sin C cos h (A + B) sin ^ f7 sin A + sin fi cos ^ (^ - B) cos ^ (.1 - B) ' if a 4- ^> c> C be given, A — B may be found, and thence A and B. 9 a-b = b - c; :. sin J[ - sin B = sin B - sin G ; .-. sin i sin H^ - 5) = sin J ^ sin ^ (B - C) ; :. tan ^ C (tan i i - tani iJ) = tan i ^ (tan J £ - tan J C) ; &c. 10. If X be this radius, (r' + a;)2 = (r + r')2 -f (r + xy (r + O (r + «) cos J 9 ; «■= r(r-\- r-' ) (1 - cos ^ g) _ r' (1 + cos } e) - r (1 - cos i e) = &c. First Year: 1878. 1. tan and cot may have any value ; sec and cosec cannot lie bet. 1 and - 1. versin A = 1- cos A = 2 sin'* ^ A. 2 Radius is unity. 3. § 36. (2). 2 sin fl = ± V 1 + sin 20 + VI -sin 2 0, sin = ± V HI - cos 2 0). (3). True if 2 (1 - cos M) = 2-2 (cos -4 cos i -4 + sua A sin } -4), if &c. 4. From logj,, N = log^ N -r logg 10, where e is the base of the Napierian system. Let 6 = a*, c = 6^, . . . m = f ; .. m — a* ^ " ' * ^, log„ m = X 1/ . . . z = &c. 5. (1). § 48. (2). Area = J c sin £ (c cos B ±, ,Jh^ - c^ sin^ £). If x be side and its inclination to the line, Ag = ^ = ^^^^ (^^ _ y) ; .-. g^^^ = tan = a ; whence * is found and thence area. C. (2). a^ + V*' a' 103 h^ = c2. (3). sin C - 1 ; .-. sin^ C= sin (7 = sin (A + B) ^ am .4 cos £ + cos ^ sin J5 = sin2 .4 + sin^ £ 7 n^ A — 8° 39', B = 24° 51', C = 14G° 30'. (2). A --= 9° 13', B L 'zAt. 8. § 22. Equivalent to shewing that sec ^ sec M cosec J A = 4cosec2A 9.. = «-±|J2||. (g). -35730. 10.(1). F ' VT ^/r" .f ^ " "''' ^ ''"' ^ = *•'• (2;. § 39. (3). See Jix. XI. ,43. (4). sin ^ + sin 2 ^ - sin 3 ^ = 2 sin f J cos } A -am 3 A =2 sin f J. (cos, J A - cos M) = &c. (5). = co^vl^H-C(w£W-_siii^BjlJ. sin A) cos £ (1 + cos ^) - sin A (1 + dST^) '-±^-tan£.L+^»J _ co8.« cos A l±J?2i^ _ tan ^ . L±Z»Jf' cos A coaB which is at once convertible into an expression involving only tan J A, tan J B. 11. («). (3). For cos M cos i £ cos i C = ^^/3(. -a)(«-t)(s-c). (4). In(2)for«puti(a + fe + c). (6). § 40. sin C 2 sin i cos i (7 " sin A+sinB 'J sin H^ + i^) cos ^ (A - B) ~ '^*'- (^)- ^^ § 45, l^y tan - = cot i (C - B), and tan [^ + ^] = cotHO-£),ifcos(i^ + 5 + iC-iB)=o. 12. Radiusof inscribed circle = - ; radii of escribed circles = ■ ^ , &c. ; radius of circumscribed circle = 8^3 a ah c 8 — a perpendiculars = 2 sin A 4S 8 a^ 63 g3 Product of ab c ah c 04 /f' Junior Matriculation: 1878 1_ 8000 TT of 1. (1). See Todhunter's Trigonometry, (2). 360°. The dip of the Horizon is tlie angle a horiro'iL'line makes with the line dra^vn from the observer's eye tangent to the eartli's surface. (2). Find sin 9°, cos 9°, and then such formuLw may be used as sin {A + 9°) + sin (^ - 9°) = 2 sin ^ cos 9°, giving sin ( ^ + 9°), since ratios of other angles in formula would be already known 3. (1). See Ex. VIII. (2). sec ^ + tan J = 1 + Bin .4 cos.i COS i A -^ Bin i A cos i A — sin i ^ 2 sin 2 ^ ^ = &c. 104 (3). vers A sin A 1 — cos A sin^ = &C. (4). X sin A-\- y cos A ■ 1 y — * 1, tan 2 sin i A cos \ A tanM; .-. xf^\ J^y=Vl^ i^-)l' 4. (1). § 48. (2). § 35. (3). §41. (4). Obtained from fonnula of § 40. 5, (1). §U, 3, (2). §3. (3). r-G933116, 1-5644372. 6. (1). §8. (2). -7235790. -5997802, -4978065, 4200741. 7. Area= i ah sin 120° + ' - • -- &c. 8. 145- ^ = 44°, B = 66°, c = 1035'43. 9. 12 = a The radius of the circle may be shewn to be 2 sin J h cos } C cos i B + c -r-p 1 -i- 2 cos i -4, and this may be shewn 10. Let A be top, B bottom of lighthouse ; cos^B a equal to -• — 7- ^ sm A centre of earth ; AC a tangent to earth's surface. Then AE'^ hi h \ 8000 = ^B (2 BO + AB) = 5280 \^ + 5286) =" 5280 ^' "^''"•^^' = I /i nearly ; or AE = ,v^f Smiles. J'irai Fear; 1879. 1. 2 1. logjyiV = log^iV ^ log^lO. 2. I 4. 3. ?5. 4. Diff. for 1 is 25, for 2 is 50, &c. 2-239159 ; 0173322. 5. -00030243, 9-937531,10-150515 ; tan 120° is negative, and .-. has no log. 6. (1), (2). §21, (3). §36. (4). sinn^ + 8in(n- 2) ^ ■■^ sin {n-l A + A) ■\- sin {n-1 A- A) = 2 sin (h -1) A cos A. 7. (1),H1. (2). §45. (3). Expand and cancel. (4). Ex. XI., 41. 8. (1). 6=1275, 0= 61° 52' 12'. (2) ^-52° 7' 48", //=30, C= 96° 52' 12 ', c == 18 + 24 ^37 9. § 48. Area = 1200 ^5^ 62 + c2 — 2 P + }a 2 &c. a + 62 + c2 = J (^2 + ^2 4. n2). Also 2 i2 + I a2 =. a2 4. 62 + c2 = 2 m"^ + | 62 == 2 )|2 + f c2. Thence we may prove a* + 6* + c* =» ^/ (i* + *"-* + Area= J V(«'' + 6'' + c2)2-2(a* + 6* + c*) \ ^(P + m2 4- «2) i 2(J* + m* + n*) = &c. 10. If D be foot of mountain, and x represent height x"^ + 3 x2 = 2 BD 2+2 /1000\ 2 (1000)2; ..cot 2 36°— 2- It—) • Now numerical value X of cot 2 36° may be found from given table, and is 1 -8944. Thence x — 3077 '3 yards. 105 Junior Mairiculation : 1879. 1. (1). § § 3, 4. (2). T, 4, 4. (3). 1, 2, 2 ; 0, 3, h. 2. 1-66902, -90309. 2-9311 +• 3. 9-804117, 32° 30' 14 . 5. (1). §§31, : &c. 8. (1). sin2 9 = COS'' 2 ti, 4. i, — , V'2. 9-69897, 993763, 1015052. 17. (2). See Ex. VIII. (3). § 38. (4). True if sin J=2 sin } A COS J ^. True ii B A • A C = A B^ - 2 B D^, ii cos 4 - 1-2 8in2 i J. 6. (1). sin 72° ^ 2 sin 36° cos 36° = 4 sin 18° cos 18° sin 54°, and sin 72° =- cos 18°, .-. &c. (2). sin 20° • sin 160° = 2 sin 80° cos 80° = . . . = 8 sin 20° cos 20° cos 40° cos 80°, and cos 60° =- J. 7. (1). § 35. (2). § 41. (3j. B D= 4 o tan ^ A rtan^ = 2c^_^^,^^^ 1 - cos 2 = 2 cos2 2 e, &c. , = 90° or 30°. (2). First e(iiia- j^ 2 tion reduces to sin 2 <^ sin 2 t^ = — ; .-. from second ~7; si" 2 ,p 2 V " 2 + VI - sin''* 2>|'=— . e = 30°, »// = 45°; also sin 2 i// =- ' , V3 sin 2 « = I V3, impos. 9. (i;. 6 = 700, B = 68° 5 * (2). C = 67° 12', a = 314 16, 6 = 200. 10. Radius = area of triangle ^^ .irde = (* "^H* "-^li^---^^ X « « 3-1416 = 923-23. I' ; APPENDIX. SIN {A + B), &c. 1. To find the value of sin {A + B) and cos {A + B). Lot BAV (Fig. 11) represent the angle A, and CAD the angle '^^ " B ; then BAD represents A + B. From any point P in AD draw P3f, PQ perpendicular to AB AC respectively. From Q draw QR perpendicular to PM, and i^N perpendicular to Ali. Then L QPR ^ 90" - PQR = /J^^^ _ qj^^ ^ ^ Now sin {A + B)= sin P^ilf = iM AP ^ QN+PR ' AP ^QN- . ^PR ~'AP^~- = 9^ ^<2 , p^ AQ' AP'^ PQ AP PQ AP -= sin ^ cos ^ + cos ^ sin B. Also cos (A + B) = cos PAM = d^ AP AN QR AP AN QR AN ' AQ AP AQ AP QR PQ AP PQ AP = cos ^ cos ^ - sin A sin B. 2. To find the value of sin (^1 - B) and cos (A - B). Let BAC (Fig. 12) represent the angle A, and CAD the angle Fig. 12 B ; then BAD represents A - B. From any point P in AD draw PM, PQ, perpendicular to AB,AG respectively. From Q draw QR peqiendicular to MP produced, and QN perpendicular to AB. Then i QPR = 90° - PQR =.CQR = CAB=.A. 108 Now sin (A - B) — Fi{:. 13. Bin PAM = ^ AP + 93 AP "^ PQ AP PQ AP = sin -4 cos ^ — cos A sin P. Also cos (A - £) = cos PAM = -^ AP ^ AN+ QR AP ^AN ~ • ' AP ^AN AQ AQ' = cos ^ cos B-{- sin A sin P. 3. In Article 1 we have taken A and £ each less than a right angle, and m Article 2 their sum is less than a right angle The same results, however, are obtained whatever be the magnitudes or A and B. Thus let A and B (Fig. 13) have the magnitudes indicated by the hgure the lettering and construction being the same as in the preceding Article. Then LQPR = 90<> - PQB = RQA = QAN^18(y> - A. And cos (A -B)=cosPAB=- ^ AP ^_AN-QP AP~ AN AN ' AQ' AP'^ ■ AQ OR AP^PQ AP PQ AP = -{coaim°-A)co&B + 8in (180°-^) Bin.fi = cos A cos jB-f sin -4 sin B. And similarly in any other case. We may accordingly assume these formulas to hold whatever be the magnitudes of AwidiB. 109 LINE-DEFINITIONS OB THE TllIfJONOMETRICAL RATIOS. 4. The following definitioiia of the Trigonoinotrioal Ratios, formerly givon by most English writers, but now falling into disuse, are still sometimes referred to. Take any arc Ali (Fig. 14), subtending at the centre tlie angle ACB, and draw BP, AT at right angles to AC. Let .1 T meet VB produced in T. Draw OT, BP' at right angles to OC. Then BP is called the sine of the angle ACB to radius CB, A T is called the tangent, and CT the secant. Also BP" (or CP), OT' and VT', being corresponding lines for the angle BCD, which is the complement of ACB, are called resi)ectively the cosine, cotangent and cosecant of ACB. AP i& called the versed sine of ^. If we take the arc Ah, greater than one quadrant and less than two, then hp is called the sine, Cp the cosine, At the tangent. Of the cotangent, Ct the secant, Ct' the cosecant, and Aj) the versed sine. The radius is the whole sine, or sine of 90°. 5. If the radius of the circle be the unit of length, or, as it is expressed, to radius Hn'dij, it will bo seen that the above defini- tions exactly coincide with those already given in Article 17. For then each line in the figure will have for its numerical value the number of times it contains the radius, that is, the ratio it bears to the radius. Hence to radius unitij, the values of the . BP CP ^ . , , . ratios — — 7 , -——^ , &c., i.e., of the sine, cosme, &c., will be the same as the numerical values of the lines BP, CP, &c. ; in other words, the numerical values of the trigonometrical ratios from the definitions given of them in Ai-ticle 17, will be precisely the same as the numerical values obtained from the ^inc-definitions given above. »). The /*'nc-definitions explain the origin of the names sine, tangent, &c. The name sine, from sinus, bosom, is given to BP as being (half) the string of the arcus, or bow of which BA is half, which is brought up to the breast of the archer in discharging it. Tlie tangent ^Tis the touchinglinG. The secant TC is the cutting line. The cosine, cotangent and cosecant are so called as being the sine, tangent and secant of the complement. Fi«- !♦. in FORMULAS, &c. 7. log 10 = 1, log 1 = 0. l(.g = - oc . log (ah) = log a •{- log b, a log ^ = log n - log b. log a "= n log a. 1 log* Vtt=- log a. Any trigononiotrical ratio of an angle is the co-ratio of the coniplunicnt. 1 1 1 Bin ^ sin A = ~~r » tan A = z't—: » co8/l= t> tan A = v cosoc A — cot ^4 sec A cos A sin' A + cos" J = 1. sin 30° = i, cos 30° = -^- , tan 46° « 1. As the angle changes from to 90°. sin increases from . , to 1 ; L sin increases from — oc to 10 tan 0. . oc ; L tan , . — cx..-|-oc sec 1. . oc ; i sec lO-.-f-oc cos decreases ^ . . ; L cos decreases .... 10. . - . In the following four-figure tables of logarithms of nunibura, the first two figures of the number whoso logarithm is souglit will be found in the column marked N, and the third in the colunui at the top ; and oi)posite the first two figures and under the third will be found the mantissa corresponding to the tirHt three figures. The proportional parts are given in the cohunns to the right ; the part corresponding to the fourth figure will Ini found here beneath the fourth figure and opposite the first two figures. Thus to find logarithm of 83-47. Mantissa corresponding to 834 = 9212 Proportional Pai-t for 7 = 4 .-. log 83-47 = 1-9210 To find the number corresponding to the logarithm 2 -7648. 7648 7642 is mantissa corresponding to 581 6 is proportional part for 8 ; ;. 2-7648 is logarithm of 581-8. The tables will obtain numbers correct to four figures only. If, however, in any number a fifth figure be given, we may obtain approximate results by neglecting the fifth figiire or in- creasing the fourth by unity, according as such fifth figure be less or not less than 5. 10. In the tables of the logarithms of the trigonometrical ratios, the angles are given at intervals of 10' between 0° and 20° and between 70° and 90°, and at intervals of 1° between 20° and 70°. Tlire reason for this arrangement is that when the tables give angles between 20° and 70° at intervals of 1°, we can inter- polate as accurately for the minutes of any angle as when the tables give angles between 0° and 20°, and 70° and 90° at inter- vals of 10 minutes. In interpolating we must be careful to notice whether the number given in the colinnn of differences be a diflference for 10 minutes or for 60 minutes. ll.S Thus to find /. sin 33° 2'y'. L siti 33= -^ 9-7301 48 A sin 33° 25' =r !)-740f>. T.. find L cot 7!»^ 35'. Lot 7!> W = 9-2(jKO 3»J Difr. forJiO-rrrUo; 26' = ij^ of 115 = 48. Diti". for 10' = 71 ; 5' = j-;^ uf 71 = ;j(i. A cot 7!»' 35' ={) •2(544. Tl,,, Tal,l„, „.iU „,,tai„ ,.„j5|„, „,„„, J,. ,,_ .M,u,, „„,,„„„„„. „„„^ '"■ "»'."-«'« .i.„L„!;,i,,: :^ 9 lU mOARITHMS OF NUMHKUS. \ lo II 12 14 •5 16 17 18 «9 20 21 22 24 25 2b_ 27~ 28 29 3' 32 33 34 35 IMN 0(14 (»7!»: ll.-i!) I4(;i I7«l •2041 •-•:<(»4 •2788 MMM) 3010|3032 3054 3*2«3 .S4(!4 32'22 34-24 3()1 4314 447 4(>'24 477^1 4iU4 5I)'>I 5185 53]5 5441 j6^||5563 37 joGSiJ 3» 57!»8 39 5911 4» 42 43 44 45 47 48 49 S' 52 53 (5^ 61*28 t)'23 1 2 ' S 4 5 e «3 (NIH(! 0453,04it'2 0H'2h| 08(14 Ii73;r20(ii 14!»'2l i5'23| 17!»0 1818 •2(xi8 •2mr>\ '2330, -2355 •2577, '2H01 '2810 '2833 380'. 3!>75» 415(1 4160 3'243i 3444 ;{()3() 38'20 3!Kt7 01 '28 0531 080!) r23y 1553 1847 •2 1 '2' •2380 •2»V25 '2850 307: 3'284 3483 4330! H87 403!) 3055 3(574 3838 385(i 4014 4031 418^<420'2 450-2 4518 4()54 4()(>0 I 478«)48(M» 4!t'28 4042 50«>5 5070 "Uj>8;5'2n 53'28 5340 )4r»3 5405 5575 5587 5()!>4'5705 5809i 58-21 59'2'2 5933 '■>031 (»138 6243 633510345 !6435 !i653 6628 (>721 6812 6902 7076 7160 7243 6444 6542 6637 6730 6821 6911 6990 6998 7084 6042 6149 6253 6355 0454 6551 6646 (i739 6330 6920 7007 7093 716O|7108i7177 7'251 7259 4814 4<»55 5092 52'24 5;<53 5478 5599 5717 583 5944 6203 6365 6561 6656 6749 6839 6928 7016 7T0I 7185 7267 0170 0509 0934 1'271 1584 1M75 2148 '2405 •2648 •2878, 0253 064; l(M)4 1335 I(i44 1931 '2^201 2455 2()95 2923 0^294 0082 1038 1673 1959 2227 .'480 3096; 3304; 3502 3(592! 3874 4048; 4378 4533 4(583' 0212 0(507 09(59 1303 I6I4I 1903 2175 •2130 •2(572 2900; 3n8 33'24 35*22 3541 3711 1 3729 389*2 3909 40(55: 408'2|40!>9 42321 4: 4393 440!» 4548 4564 4698 4713 3J39 334f '249 4: 48'29 4843i 4W)4983| 5105 5119, 5'237 5'250 53(56' 5378 5490] 5502 56 n 1 5623 5729 5740 5843 5855 5955 5966 6053 6064 6075 4857 4997 513 5'2(]3| 53lii 5514 5635 5752 58(5(5 5977 616016170 0274 6375 6464 6474 6571 6665 6758 6848 6937 7024 7110 7193 7275 6180 (5284 6385 6085 iilOl (5484 45493 6580 6590 6(575 668416693 67(57 677616785 685716866 687 6946 6955 7033 7118 7202 7284,7292 7 8 9 »12 34 fi 6 0334 0719 1(»72 13(57 1 1399 1703 1987 2253 2504 '2718 '2742 •2945 2967 3160 3365 35(50 3747 3927 '265 4425 4579 4728 0374 0755 1I(M5 143( 173'. '2014 •2*279 25'29 27(55 '2989 3201 3404 3598 3784 3181 3385 3579 37(5(5 3945 4116 4*281 444()4456 4594 4(509 4742I4757 39(5' 4133 4298 )•> 4871 50Ti )145 5'2!76 )403 5527 5647 5763 5877 5988 4886 5024 5159 5289 541(5 5539 5(558 4900 I_3 5038 iT 5775 5888 5999 6096 6107 6201 '0212 6294 6304:6314 6395 6405J 6415 6503 (55T3 6599 6(509 i 6702 6794 5: 6884 696416972 7042 70507059 7126 7135 7143 7210 7218; 7*2*26 5172 5302 54*28 5551 567011 578(5 5899 6117 (5222 (5325 6425 6522 661 6712 6803 6893 6981 7067 7152 7' 235 1 7300 7308 7316 48 1 4 8 11 » 7 10 3 6 1(1 :i (5 3 4 5 3 4 5 7 8 9 •29 33 37 ' .'() 30 34 1 24 *28 31 ' 23 26 *29 21 '24 27 20 2'2 25 I 18 2? '24 j 17 20*22! 16 19 21 1 16 18*20 15 17 19 14 16 18 14 15 17 1;; 15 17 12 14 16 2 14 15 1 13 15i 11 13 14 11 12 14 10 12 13 6 7 9 "8 10 11 13 6 7 10 11 12 5 V 8 9 11 12 5 6 8 9 10 12 9 10 11 9 10 11 8 10 1 1 8 9 10 8 9 10 8 9 10 4 5 6 8 9 10 4 5 6 8 9 4 5 6 8 9 4 5 6 "6 8 9 4 5 8 9 4 5 6 8 9 4 5 6 '5 (T 7 8 4 5 7 8 4 4 5 6 7 8 4 4 5 6 7 8 3 4 5 6 7 8 6 7 8 6 7 7 6 6 7 115 LOGARITHMS OF NUMBKIIS.- (6'o/(/mHc./.) ^ 13 10 "91 91 9 9 8 8 8i 8, - i 1 8 "8 7 7 N 7.T24 7404 748'-' 54 I 55 I 57 I 7.V)!l 58 Ii7(l.*«4 59 |770!» 60 61 62 64 65 -^ "67" 68 69 7» 72 73 74 75 77 78 79 80^ 81 82 ^3_ 84 85 86 87 88 89 17782 |78r.:< !7!ti'4 I'Mf.i \h{){\ SI'-".* 1819:) 'h'_'(;i 882:) :8;w8 90 91 92 93 94 95 97 98 99 84M .S57.S 8()3;} 8751 SS08 88(55 8i»'_'i 8!»7(»| 9031 9085 9138 9191 9243 9294 9345 9395 9445 9542 9590 91)38 9685 973T 9777 9823 1 i 2 I 3 733'.' 7340 734s 7412 7419 7427 74iM) 7497 7505 75(i(» 757I 7582 7t>t2 7049 7057 7710 7723 7731 7789 HliO !I3I 8IM)0 80<;9 8130 8202 7790 7808 793H 8007 8075 8142 8209 8207 8274 8331 8395 8457 8519i 8579! 8039 8(ii»8 8750 8814 8871 8927 8982 8338 8401 1 8403 8525 8585 8045 8704 8702 8820 8870 8932 8987 78o;tf 78 -.'I 7945 8014 8082 8149 8215 8280 8344 8407 8882 8938 899; 9030 9042 904 9090 9143 9190 !)248 9299, 9350; 940(V >450 9494 9499 9090 9101 9149 9154 9201 920( 9253 9258 9304 9309 9355 9S 9405 9455 9504 9509 m) 9; 9547; 9595 9043 9089^ 9736 9782 9827 9808 9912 9950 9901 9872 9917 9552' 9557 9000 9005 9647 9052 9694 9099 9 9~7Tl 9745 9780 9791 9832:983(5 9877i988"l 992119920 99(551 9909 4:1 5 I 6 7 3 9 735(5 7304 737 7435 7443 7451 7513 7520 7528 7589 7597 7(504 7(5(54 7(572 7(579 7738 7745 7752 7810 7882 7952 802] 8089 8150 8351 8414 7818 7825 7889 789(5 7it59 79(5(5 8028 8035 809(5 8102 8162 81(59 S228 823. 8293 8299 8357 83(53 8420 8420 847018470 8482 8488 8531 8537 8543 '8549 85!»1 8597 8(503 8(509 8(551 8(557' 8(5(53 8(5(5! 87T0 87ir5,8722 8727 87t>H 8825 8537 8597 8(557' 87ir5, 8774 8831 8887, 8943 (98 :i8i 8779 87.S5 8837 8842 8893' 8899 8949 8954 9004:9009 9053 9058 • 100 9112 9159 91(55 9212 9217 92(53 9269 9315 9320 305 9370 941019415 9420 94(5019405 9409 9513:9518 iH)03 911^ 9170 92*'*' 9274 93; 9375 9425 9474 9523 9562 956(5 9609 9f) 14 9(557^9(501 7(>3|9708 9750' 9754 9795: 9800 9841 ! 9845 !»88T5: 9890 9930: 9934 9974; 9978 957j 9(519 90(50 9713 9759 9805 9850 9894 9939 9983 7380 7388 7390 7459 74(50 7474 753(5 7543 7551 7(512 7(5 lit 7(527 7(58(5 7(594 7~0i 77(50 77(57 777 57 7774 1 7832 7!t03 7i>73 8041 8109 817(5 8241 830(5 8370 8432 8494 8555 8015 8(575 8733 8791 8848 8904 89)50 •015 »009 (122 >I75 •227 •279 •33(^ •380 •4:«) •479 •528 •57(5 •(524 •(571 9717: •763' •809 •854' •89!^' •943 •987 7839 7910 7980 8048 8110 8182 8248 8312 837(5 8439 85(^0 85(51 8(521 8(581 8739 8797 8854 8910 8905 9020 9074 9128 9180 9232 928^ 9335 9385 9435 9484 9533 9581 9028 9(575 9722 97(58 9814 9859 99(^3 9948 9991 784(1 791 7 7987 8(^55 8122 HI 89 8254 8319 8382 8445 85(((1 85(5 8(52 8(58(5 8745 880; 8859 89)5 8!^7 1 9(^25| !K>7i 9133 9180 9238 9289 934(^ 939(^ (•44(I0 94S!) 9538 958(5 9(533 (• 90S0 9727fc> 977:*» 9818 I) 9863 9!MhS 9952 (• 9990 1 2 34 5 6.7 8 9 1 2 2 1 2 2 1 2 " 3 4 3 4 5 3 4 5 3 4 5 3 4 4 3 4 4 3 4 4 3 4 4 3 3 4 3 3 4 )5 ti 5 t; 5 t5 5 t; 5 <5 5 3 3 4 3 3 4 3 3 4 3 3 4 3 3 4 •> 3 4 •J 3 4 •> 3 4 3 4 •) 3 4 • 1 3 4 •> 3 3 •> 3 3 3 3 >> 3 3 •> 3 3 •> 3 3 '2 3 3 » 3 3 2 3 3 •> 3 3 •» 3 3 2 3 3 •) •> 3 •J >> 3 •) •> 3 •J 2 3 3 •) •> •J •> 3 •> 2 3 3 '2 •> 2 2 3 2 •) 3 2 •> 3 >j 2 3 2 2 3 it .) 5 5 5 5 4 5 4 5 4 5 4 4 4 4 4 4 4 4 4 4 4 4 4 4 3 44 3 I 4 3 3 4 .> (5 (> 5 (5 5 (5 5 5 5 5 (5 5 5 (5 5 5 5 5 (5 4 5 (5 4 5 4 5 (5 4 5 5 4 5 5 4 5 5 4 4 4 4 4 4 LOGARITHMS OF TRIGONOMETRICAL RATIOS. ANGLES AT INTERVALS OF 1(C. Sine. DtKF. roil 10'. Tano. Com. DiFF. FOB 10'. C0TAN(1. CuSINK. DlFF. KOH 10'. 0' 0' In. Ni'g. In. NcK. 1 Infinite. 10-(KK)0 1 90" (•' 10' 7-4037 3011 7-4637 3011 12-5363 10-0000 .'iO' 'JO' 7-7048 1760 7-7648 1761 r2-'2352 10-(MK)0 40' .•«)' 7-9408 1250 7-9409 1249 120591 10-0000 30' 40' 8-0<)58 969 8-0658 969 11-9342 10-0000 '20' 50' 816'27 792 8-1G27 792 11-8373 10-0000 89° 10' r 0' 8-2419 669 8-2419 670 11-7581 9-9999 ! 89" 0' 10' 8-3088 580 8-3089 580 11-6911 9-9999 1 50' '20' 8-3608 511 8-3669 512 11-6331 9-9999 : 40' 30' 8-4179 458 8-4181 457 11-5819 9-9999 i 30' 40' 8-4637 413 8-4638 415 11-5362 9-9998 1 '20' 50' 8-5050 378 8-5053 378 11-4947 9-9998 88° 10' 2° 0' 8-5428 348 8-5431 348 11-4569 9-9997 1 88° 0' 10 8*5776 321 8-5779 322 11-4-221 9-9997 50' 'JO' 8-6097 300 8-6101 300 1 1 -3899 9-9996 1 40' 30' 8-6397 280 8-6401 281 11-3599 9-9996 30' 40' 8-6677 263 8-6682 263 11-3318 9-9995 1 20' 60' 8 '6940 248 8-6945 249 11-3055 9-9995 87° 10' 3" 0' 8-7188 235 8-7194 235 11 '2806 9-9994 1 87" 0' 10' 8-74'J3 1)00 8 •74*29 2'23 11-2571 9-9993 1 50' 20' 8-7645 212 8-7652 213 11-2348 9-9993 40' 30' 8-7857 202 8-7865 '202 11-2135 9-9992 1 30' 40' 8-8059 192 8-8067 194 11-1933 9-9991 1 20' 50' 8-8251 185 8-8261 185 111739 9-9990 1 86° 10' 4' 0' 8 8430 177 8-8446 178 11-1554 9-9989 1 86" 0' 10' 8-8613 170 8-8624 171 11-1376 9-9989 50' 20' 8-8783 163 8-8795 165 11-1'205 9-9988 1 ' 40' 30' 8-8946 158 8-8960 158 11-1040 9-9987 1 1 30' 40' 8-9104 152 8-9118 154 11-0882 9-9986 1 20' 60' 8-9256 147 ' 8 9272 148 11-07-28 9-9985 1 85° 10' T,' Of 8-9403 142 8-9420 143 11-0580 9-9983 2 85° 0' 10' 8-9545 137 8-9563 138 11-0437 9-9982 1 50' Off 8-9082 134 8-9701 135 11-0-299 9-9981 1 1 40' 30' 8-9816 l'J9 8-9836 130 11-0164 9-9980 1 30' 40' 8-9945 125 8-9966 1'27 110034 9-9979 1 20' «)' 9-0070 122 9-0093 123 10-9907 9-9977 2 84° 10' COSIKE. DiFF. Fon 10'. COTAKO. Com. DiFI. for 10' Tano. Sine. DlFF. FOB 10' u; ANCJLES AT INTKllVALS OF W -(Continncl. ) Sish. PiKF. van 10', TAN4i. DlKK. FDIl 10. 120 1 ('iiTANIl. \ Cli.slNK. IIIFF, Foil 10'. f) 0' »'0I02 119 9 0210 10-9784 9-9970 1 1 84'- 0' 10' 1)0311 115 9-03.'«J 117 1O-9004 9-t)975 1 5(1' 'JO' «>042(; 113 j 9-0153 114 10-9547 9-9973 » 40' ;«)' 0()53i» 100 1 9-05(>7 ill 10!)433 9-9972 1 30' 40' !>•((( J4S 107 90(i7S lOS l't-!(322 9-9971 1 20' 60' 9 0755 104 9 0780 105 10 9214 9'!»909 •1 83' 10' 7 0' 0-0859 l(r2 9 0891 104 10-9109 9 '9908 1 83' 0' 10' »0!)()1 99 9-0<»95 101 10-90(15 9 9906 iJ 5(»' 'liV <)• 10150 97 9-ioix; 98 10-89(»1 9-99(W •> 40' 3(1' 1157 95 9 1194 97 lO-SWHi 9-9903 1 30' 40' 1 252 93 9 1291 94 10-8709 9-9901 •J 20' rio' 1345 91 9-1385 '.>3 10-8015 99959 o M2' Kt' 8 0- 0143r> 89 9-1478 91 10-8522 9-9958 I 82" 0' , !<►' 01525 87 9-1509 8!) l(>-8t31 9 9956 2 50' 20' 01012 85 9- 1058 87 l(t-8342 9-9954 '2 40' m' !)U)!»7 84 9-1745 8!> 1(»-.S255 9-9952 •y 30' 40' 0-1781 82 9-1831 84 10-8109 9 9950 2 20' 50' 91803 80 9-1915 82 10-8085 9-9948 2 81 10' !) df 9 1043 79 \ i 91 91)7 81 10-80(13 9 9940 2 81' 0' 10' 9'2(>22 7S ' 9-2078 80 10-79-J2 9-9944 2 50' •20' 9-2 KM) 7u 0-2158 78 10 -7842 9-9942 2 40' »)' 0-2170 75 , 9-22:il> 77 l(»-77()4 9-!)9-10 2 30' 40' 0-2251 73 ! ; 9-2313 7t» 10-7()S7 9 9938 2 •20' 50' 9-2324 73 1 9-2389 74 1(1-7011 9 '9936 •) 80 10' 10' 0^ 9-2397 71 9-2403 73 1(C7537 9-9934 2 80- 0' 10' 9-24('.8 70 9-2530 73 10 7401 9-9931 3 50' 20' 9-2538 (•>8 9-2()0!) 71 10-7391 9 •99-29 2 40' 30' 9 -2000 (>8 9-2080 7(> 10-7:v20 9-9927 2 30' 40' 9-2074 (50 9-2750 ()9 107250 ; 9-9924 3 20' W 9-2740 00 9-2819 08 107181 9-9922 2 79' 10' 11 0' 9-2800 r.4 9-2887 00 107113 9-9919 3 79' 10' 9-2870 04 9-2953 07 10-7047 9 '99 17 2 50' 20' 9-2934 ()3 9-3020 05 10(J!)SO 9-9914 3 40' 30' 9-2997 01 9-3085 Ot I0-()iM5 9-9912 2 30' 40' 9-3058 01 9-3149 03 10-0851 9 -9! 109 3 20' 50' 9-3119 00 9-3212 03 10-0788 9-94 9!t;»01 3 50' 20' 9-329G 57 9-3397 01 io(;i)03 9'9S99 2 40' m' 9-3353 57 9-3458 59 10-(>542 , 9-9890 3 30' 40' 9-3410 50 9-3517 59 10-0483 1 9-9S93 3 2(i' 50' 9 -3406 55 ! 9-3570 58 10-0424 9-9890 3 77' 10' Com. Cosine. DiFF. 'for 10'. 1 COTANO. 1)1 FK. Foil 10'. Taso Sine. DlFF. FOR 10'. 118 ANGLES AT INTERVALS OF W- (ContinuHl. ) 1 Sine. i DlFF. KOR 10'. Tano. Com. DlFF. FOR 10' CoTANO. Cosine. 1 1 DlFF, FOR 10 ■. i;r 0' 9-3521 54 9-3634 57 10-6366 ' 9-9887 3 77^ 0' 10' 9-3575 54 9-3691 57 10-6309 9-9884 3 50' •20' 9-3029 53 9-3748 56 10-6252 9-9881 3 40' 30' 9-3G82 52 9-3804 55 10-6196 9-9878 3 30' 40' 9-3734 52 9-3859 55 106141 8-9875 3 20' 50' 9-3786 51 9-3914 54 10-6086 , 9-9872 3 76° 10' 14" 0' 9-3837 50 9-3968 53 10-6032 9-9869 3 76° 0' 10' 9-3887 50 9-4021 53 10-5979 9-9866 3 50' 20' 9-3937 49 9-4074 53 10-5926 9-9863 3 40' 30' 9-3980 49 9-4127 51 10-5873 9-9859 4 30' 40' 9-4035 48 9-4178 62 10-5822 i 9-9856 3 20' 50' 9-4083 47 9-4230 61 10-5770 9-9853 3 75° 10' 15' 0' 9-4130 47 9-4281 50 10 5719 I 9-9849 4 75° 0' 10' 9-4177 46 9-4331 50 10-5669 9-9846 3 50' 2(y 9-4223 46 9-4381 49 10-5619 9-9843 3 4<»' 30' 9-4209 45 9-4430 49 10-5570 9-9839 4 30' 40' 9-4314 45 9-4479 48 10-5521 9-9836 3 20' 51/ 9-4359 44 9-4527 48 10-5473 1 9-9832 4 74° 10' IG" 0' 9-4403 44 9-4575 47 10-5425 i 9-9828 4 74° 0' 10' 9-4447 44 9-4622 47 10-5378 ; 9-9825 3 50' 20' 9-4491 42 9-4669 47 10-5331 : 9-9821 4 40' 30' 9-4533 43 9-4716 46 10-5284 ' 9-9817 4 30' 40' 9-4576 42 9-4762 46 10 5238 i 9-9814 3 20' 50' 9-4618 41 9-4808 45 10-5192 ; 1 9-9810 4 73° 10' 17' 0' 9-4659 41 9-4853 45 10-5147 ! 9-9806 4 73° 0' 10' 9-4700 41 9-4898 45 10-5102 I 9-9802 4 50' 20' 9-4741 40 9-4943 44 10-5057 ' 9-9798 4 40' 30' 9-4781 40 9-4987 44 10-5013 ; 9-9794 4 30' 40' 9-4821 40 9-5031 44 10-4969 i 9-9790 4 20' 50' 9-4801 39 9-5075 43 10-4925 9-9786 4 7-2° 10' 18" 0' 9-4900 39 9-5118 43 10-4882 9-9782 4 72° 0' 10' 9-4939 38 9-5161 42 10-4839 9-9778 4 50' 20' 9-4977 38 9-5203 42 10-4797 9-9774 4 40' 30' 9-5015 37 9-5245 42 10-4755 9-9770 4 30' 40' 9-5052 38 9-5287 42 10-4713 9-9765 5 20' 50' 9-5090 36 9-5329 41 10-4671 9 -9761 4 71° 10' 19° 0' 9-5126 37 9-5370 41 10-4630 9-9757 4 71° 0' 10' 9-5163 36 9-5411 40 10-4589 9-9752 5 50' 20' 9-5199 36 9-5451 40 10-4549 9-9748 4 40' 30' 9-5235 35 9-5491 40 10-4109 9-9743 5 30' 40' 9-5270 36 9-5531 40 10-4469 ' 9-9739 4 20' 50' 9-5306 35 9-5571 40 10-4429 9-9734 5 70° 10' 20° 0' 9-5341 34 9-5611 39 10-4389 9-9730 4 70° 0' Com. Cosine. DlFF. FOR lU'. COTANO. ! DiFF. FOR 10'. Tano. Sine. DiFF. FOR 10'. \ 119 ANGLES AT INTERVALS OF 1° i ! SiVR. Pirr. FOK F. Tang. Com. DiFF. FOB 1". COTANO. Cosine. DlFF. 1 FOK 1°. ! 1 20' 9-5341 202 i! 9-5611 231 10-4389 9-9730 27 70' 21" 9 5543 193 M 9-5842 222 10-4158 9-9702 28 69' 22' 9-5736 183 i 9-6064 215 10-3936 9-9672 30 68' 2;r 9-5919 174 ! 9-6279 207 10-3721 9-9640 32 67' 24' 9 6093 166 9-0486 201 10-3514 9-9607 33 t)6' 25' 9-6259 159 9-6687 195 10 3313 9-9573 34 65' 26' 9-6418 152 1 9-6882 190 10-3118 9-9537 36 64° 27' 9-6570 146 ' 9-7072 185 10-2928 9-9499 38 63' 28' 9-6716 140 9-7257 181 10-2743 9-9459 40 62' 29' 9-6856 134 9-7438 176 10-2562 9-9418 41 61° 30° 9-6990 128 9-7614 174 10-2386 9 9375 43 60' 31° 9-7118 124 j 9-7788 170 10-2212 9-9331 44 59' 32° 9-7242 119 ! 9-7958 167 10-2042 9-9284 47 58' 33' 9-7361 115 1 9-8125 165 10-1875 9-9236 48 57" 34' 9-7476 110 1 9-8290 162 10-1710 9-9186 50 56° 35' 9-7586 106 9-8452 161 10-1548 9-9134 52 55* 36' 9-7692 103 9-8613 158 101387 9-9080 54 54' 37' 9-7795 98 1 9-8771 157 10-12-29 9-9023 57 53' 38' 9-7893 96 9-8928 156 10-1072 9-8965 58 52' 39' 9-7989 92 9 9084 154 10-0916 9-8905 60 51' 40' 9-8081 88 9-9238 154 10-0762 9-8843 62 50' 41' 9-8169 86 9-9392 152 10-0608 9-8778 65 49' 42' 9-8255 83 9-9544 153 10-0456 98711 67 48" 43' 9-8338 80 9-9697 151 10 0303 9-8641 70 47' 44° 9-8418 77 9-9848 152 10-0152 9-8569 72 46' 45' 1 9 8495 74 10-0000 152 lo-oooo Tavo. 9-8495 74 45' Cosine. DfFF. FOR 1°. COTANO. Com. DlFF. FOB 1°. Sine. DlFF. FOR 1°. /snr H*' F.g9 (• f- , / /b \ ^^ HgJ ./ / / no". c — =r H- 4r / / A / 90^ H •* ys /3 90" ♦^ A rig.6 \ ij F.^.8 c r a , / b \a D .:\ Copp.Clarlc* i r^ ,.M ; i. -rs i. JWg,- ■ -- =^~ ■ ' -J.SX-. -j-r ■^ m ix am H * /60- 2/ / / W /3 90' \ \ \ ■^ V Kig Fig a r \ \ Fig. 10 C A ._\ Copp, ClArlc ft CoXifh. Toronto. xL\a ^■•s"v^ \ \\ v. Af N B Fi^l2. C A -R \ ><' D If M B -D Fi^'3 X M Fi6l4' 1 o T/ / / / / \ \ \ 1 X \ V 3 C \ > ■■■I *