GIFT or t!rs. Gordon Lillibrldge Engineering- l,^']^!.,^^^ yuv^. -''fj3 SOLUTIONS OF THE EXAMPLES IN HALL AND KNIGHT'S ELEMENTARY TRIGONOMETRY. ^" MACMILLAN AND CO., Limited LONDON • BOMBAY • CALCUTTA MELBOUENE THE MACMILLAN COMPANY NEW YOEK • BOSTON • CHICAGO DALLAS • SAN FRANCISCO THE MACMILLAN CO. OF CANADA, Ltd. TORONTO SOLUTIONS OF THE EXAMPLES IN HALL AND KNIGHT'S ELEMENTARY TRIGONOMETRY BT H. S. HALL, MA., // FORMERLY SCHOLAR OF CHRIST'S COLLEGE, CAMBRIDGE; LATi: MASTER OF THE MILITARY SIDE, CLIFTON COLLEGS, MACMILLAN AND CO., LIMITED ST MARTIN'S STREET, LONDON 1914 a^ 05 ^ o ^Xo COPYRIGHT First Edition 1895 Reprinted 1898 W^/VA additions 1899 Reprinted 1902, 1905 iV^w Edition 1905. Reprinted 1909, iQif, 1914' G,v\ .\- -Vo u ENGINEERING LIBRARY l^i^ PREFACE. In preparing this Key two objects have been kept in view. It is intended first to save the time and lighten the work of teachers, and secondly to afford help to those who study Mathematics without the guidance of a teacher. Accordingly the solutions have generally been given in the most simple and natural manner, with frequent reference to the text and examples in the Elementary Trigonoimtry. In particular, the solutions which involve logarithmic work have been presented in the fullest detail, so that with the help of the Key, a teacher will be able very readily to discover and correct mistakes in the work of his pupils. For very many of the solutions I am indebted to Mr H. C. Playne of Clifton College, and my thanks are due to him for valuable help all through the book. H. S. HALL. January, 1895. The present Edition contains solutions of all the examples introduced into the Fourth Edition of the Ele- mentary Trigonometry. For many of these I am indebted to Mr H. C. Beaven of Clifton College, whose valuable help I gratefully acknowledge. H. S. HALL. October, 1905. 85. 06 > » ^ » > > , > » > > ELEMENTAEY TRIGONOMETRY. EXAMPLES. I. Page 4. 7. 69^ 13' 30" = -76916 of a right angle = 76^ 9r 6f;-7". 8. 19° 0' 15"= •21125 of a right angle = 21? 12' 50^ \ 9. 50° 37' 5-7" = -562425 of a right angle = 56s 24^ 25". ^0. 43° 52' 38-1"= -487525 of a right angle = 48? 75' 25". 11. 11° 0' 38-4" = -1223407 of a right angle = 12« 23' 40-7". 12. 142° 15' 45"= 1-5806944 of a right angle = 158^ 6' 94-4'\ 13. 12' 9"= -00-22"^ of a right angle = 22' 50". 14. 3' 26-3" = -000636 of a right angle = 6' 36-7". 15. 56^ 87' 50'* = -56875 of a right angle = 51° 11' 15". 16. 39e 6' 25" = -390625 of a right angle = 35° 9' 22-.5". 17. 408 r 25-4" = -4001254 of a right angle = 36^ 0' 40-6". 18. 1« 2^ 3^' = -010203 of a right angle = 55' 5-8". 19. 3B 2' 55" = -030205 of a right angle = 2° 4.3' 6-4". 20. 8« 10' 6 -.5" = -0810065 of a right angle = 7° 17' 26-1". 21. 6' 25" = -000625 of a right angle = 3' 22-5". 22. 37' 5" = -003705 of a right angle = 20' 0-4". 23. Let the angles expressed in degrees be A and /> ; then .-t + i.^ = ^ X 80° = 72°, and A-]i = 18°. Hence A = 45°, B = 27°. 24. If u in the number of degrees in the angle, n + — ^'' = 1-'^'^ i whence n = 72. H. E. T. K. ^ 1 MEASUREMKNT OF ANGLES. [CHAPS. X y 2^. dere - =. numb'^r of degrees, and • - = nuinber of grades in the angle. X 9 7/ Therefore ,,. — ,-- . ' ^ ; whence we obtain 50x = 277/. ()0 10 100 "^ s t 26. lloic = number of degrees, and —~^ — ,—- = number of grades in oO X (50 100 X 100 .<; 9 f the angle. Therefore tt^ = ,-7; x , -- ; that is, 2o0s = 81/, on 10 100 EXAMPLES. II. Page il. [The following five solutions will sufficiently illustrate this exercise.] 3. From fig. of Art. 17 we have a- = b- + c^ = 400 + 22o = 62o ; whence 4 4 3 5 a = 25, and sin C = -_ , cosIi = -, cotC=-, sec (7=-. o o 4 6. Let rt = 15, h = d; then c^^a--b'^={a + b){a-b) = 2-ix&; whence 4 3 4 c = 12, and sinC=-, cosG— , tanC= . o o o 8. In the third fig. of Art. 17, let yiC = 41, JJ>' = 9; then £C2 = 412-9- = (41 + 9)(41-9); 40 9 whence BC = 40, and sin ^ = — , cot -4 = — . 10. Here CD = 2])E. Hence if ED = a, DC = 2a, KC = a^^D. The required ratios may now be written down. 11. From the right-angled a ^/>C we have IiC='6[); also from the right- angled a J C7> we have DC = 11. The required ratios may now be written down. EXAMPLES. Ilia. Page 17. Examples 1-25 are too easy to require full solution ; the following eight fiolutions will nufficc. 10. ( 1 - cos* e) sec- 6 = sin- ^ x = tan"'' 0. cos- 6 12. cosec a V 1 - sin- a = . — xcosa = cota sm a 15. (1 - co8= (?) (1 -h tan'^ d) = sin= mc-0 = - / ^ tau^ 0. 19. (1 cos-A){\+(-ot"A) = Hh\-Aco&cc^A = l. I — Til.] EASY IDENTITIES. 20. siu a sec a ^./cosec'* a - 1 = x cot a = 1 . cos a 22. sin- d cot= d + sin2 5 = sin^ ^ (1 + cot= 6) = sin^ e cosec- 1?= 1. 23. (1 -f tan2 ^) (1 - sin2 6) = sec^ 6 cos^ (? = 1. 1 sin" d 25. cosec- - x sec- a ' cot- a cos^ a sec- a cos- a ' = sin- a sec- a. sin^ a cos- a ,^ -,- X 1 1 + sina + 1 - sina 2 15. 1 irst side =: r-^ = ., =2 sec- a. l-sm^a cos- a ^. ., tan a (sec a + 1) +tan a (seca- 1) 2 tan a sec a 16. First side= ^ — — ^ ' = , .. sec-a-1 tan- a 2 sec a 2 sec a cos a 2 _ = = . = — r — =2 cosec a. tan a sin a sm a ^. ., 1 1 1 sin^a 17. lMrstside = , . ., + ■, = , ■ ., + t"^ — • -r ■^'' 1 + sm-a 1 1 + sm-a 1 + sin^a sin- a _ l + sin^ a.. 1 + sm^ a 18. First side = ( — - + . — ) (sin (^ + cos — -. ' = cot 0. 2sm-(?-sm^ sin ^ (2 sin ^- 1) 29 First side = ''''" ' ^''' ' " ^^ - '^'" ^ ^^ " '^ '^ 1 + sin 1 + sec ^ _ cot2 (6ec2 0-1)- sec2 ^ (1 - sin2 0) (1 + sin^) (1 + sec^) cot2^tan2^-sec2(?cos2^ 0. (l + sin^)(l + sec^) 30. tan- a + sec- /3=: (sec- a - 1) + (tan^ ,8 + 1) = sec- a + tan'-* p. , , - tan a ~ t an g + cot / 3 _ tan.,'3 _ tan a tan /3 + 1 tan a _ tan a cota + tan/:{~ 1 ~ tan /3 ' tanatanjS + i~ tan^ ~ T" tftn 3 tan a 33. First side = cot a tan a tan /3 + tan /3 cot /3 cot a = tan ^ + cot o. G EASY IDENTITIES. [CHAP. 34. First side = sin^ a ( 1 - sin- /3) - ( 1 - sin^ a) sin- (i = sin- a - sin- a siu= /3 - sin^ ^ + sin- a sin- /!i = sin-a - sin-/3. 35. First 8iae = (l + tan-a)tan2^-tan2a{l + tan2j3) = tan- ^ + tan- a tan'^ /3 - tan- a - tan^ a tan- ^ — tan- /3 - tan- a. 36. First side = sin- a cos- /? + cos- a sin- /3 + cos- a cos- /3 + sin- o sin- /3, the other terras cancelling ; this expression = (sin- a + cos- a) cos- ^ + (cos- a + sin- a) sin- ^ = cos2^ + sin-/S = l. EXAMPLES. III. c. Page 23. 1 A 1 1. sec^ = cos^ ^l-s\n^A L 1 \^^ y-i , , cos A J\ - sin^ A J'6 \ ,„ ^"* ^ = sin ^ = --sIKl— = -2--^ 2 = ^^• 2. sin^=— i^!l:i__ [Art. 32, Ex. 1] Vl + tan- A ^^ , 1() 4 3 4 9 3 o o cos A = cot ^ . sm ^ = -.-, = „ 4 5 o 5. sin e=sj\- cos--' ^ = y/^l - 1 = ^^1 = ^--. ^^ cos^ 1 7^8 1 — / 7^ 24 25 6. cos /I = /^l - 8in2 -•! ■= a/ 1 - :y?li = ^ • Therefore sec .-J = "-^ , sin^ 7 25 7 tan A = - = -- X — = -- . cos -4 25 24 24 8. coseca=^l + cot2a. [Art. 27.] cot a cot a cos a = cot a sin o= = . — coseco ^1+cot-a III.] KELATIONS RKTWEEN THE TRIGONOMETRICAL RATIOS. 7 10. cosec A = '.- —, ; cos A = ^1 - sin^ A ; sec A = =-. . sin A ^ ^i_ sin-^ .^ , sin A sin ^ ^ , Jl - sin- A tan A= , = . — ; cot A = - - — ■ — . cos J ^,/l-sin2.-l sin J 11. Here sin A =cos A, so that tan J = 1. .-. cosec ^ = /^/l + cot- ^= >/l + 1 = ^2. 12. tan A = -- : = --^j^/^- — .Jl-sin2.4 « V "- m n VI ■■— X 13. iy-cot2^ = ?2-2j3; .-. /■i(cot-^4-l) = rr. ;>- cosec- ^ = (/'^ so that sin <^= - . 14. In the diagram of Ex. 2, Art. 33, let PQ = 2m, rR=zni- + l; then IlQ-^={m^ + lf - {2)1,)- = {),>■' - 1)\ .: RQ = ufi-l. tan A = — ^^ — , sm ^4 = — ; — - . 2m in- + 1 1 « rni • '' tan a - 3 . , ^ /lH9 12 16. i he expression = , - ; but tan a = / __i — 4 tan a - 1) ' V 25 5 ' 2.12 •j> ^1 • 5 9 ., the expression = = _ =3. o 17. The expression ='' — ^ = 4; — = .-\ — + (7 EXAMPLES. IV. a. Pack 2G. Let E stand for the expression to be evaluated in each case ; then 6. ;■;=-- (i)=x^^?.j^k/3)^ = ^. 8 TKIGONOMETRICAL KATIOS OF CERTAIN ANGLES. [CHAP. . IT 14. We have 15. We have ^' V2; ~ \/2 V2 • ^ ' 2 • U/2 j ~ U/2F2~ ' 4""^- 2 ' 4 = 2' _x/3 .•. x — 6. 2 EXAMPLES. IV. b. Page 28. For Examples 9 — 14, see Example 1, page 28. 20. Second side = 1 + sin^ A sec- A = l-\ tan^ A = sec^ A = cosec- (00^ - J ). 21. Fh-st side = sin A cot A tan .-I cosec A — 1. 22. First side = cosec .-i - cot J sin A cot A = cosec J (1 - cos-\t() = sin^. 23. First side = tan^ A cosec- A - sin- A sec- A = sec-.-t -tan''*^ = l. 24. First side , . , . cos^^ + sin-^ ^ . , ,„^o = cot^ + tan^ = — ; — . ; = cosec ^ sec ^ = cosec J cosec (90 - A). sm^icos^ T^- X -1 cos ^ cot A 25. First side = -. -=co8^. cosec A cos A T^- , . T cosec^Jtan^^ cot .4 .„ . 26. First side= . — ^, =cot2^. tan A sec- A = cosec-' A -1 = sec- (90" - ^) - 1. IV.] COMPLEMENTARY ANCLES. 9 ^. , .. tau A sec A cot-^ A , _ 27. First side— — ^ _._, , . -„^^.. 7 ^ =sec A = ^Jta.n'^ A+l. cosec- .1 cos- A 28. First side: sin- A 1 - cos- A 1 - cos ^4 1 — cos - = 1 + cos J = 1 + sin (90° - .4). 29. First side- - ^ , ^^ ^.^^ =cot^ (1 - sin.4) = tau (90^' -^) -cos J. cot .-1 (l + siuyf) 30. •'■ cos A tan A = sin A ; .-. .1=1. 31. sec^ A ~ X tan .4 = 1; tan- A =jrtiin A ; X = tan A . EXAMPLES. IV. d. Page 3L 9. l + tan--i^ = 2tan2^; tan2^ = l; .-. tan^= ±1; .-. ^ = 45°. 11. l + tan-^ = 3tan-^-l; 2 tan- ^ = 2; .-. tan^=±l; .-. ^ = 45°. . 13. cot-^ + l+cot2^ = :^; cot2^=l; •. cot^=±l; .-. ^ = 45°. 15. 2cos2^ + 4-4co3'-^t? = 3; 2 eos^ 6 = 1; .-. ^ = 45°. 17. 12sin-^-4sin^-l = 0; (6Bin^ + l)(2sin^~l)=0; .*. sin^ = - or - - : 2 6 •. ^ = 30°. 10. l+cot2 = 4cot-^; 3cot2^ = l; •. ^ = 60°. cot^=± 12. l + tan2 + tan2^ = 7; 2tan2^ = 6; .-. tan^==:±/y3; .-. ^ = 60. 14. 2(eos'-^-l + cos2^)=:l; 4cos2^ = 3; .*. cos ^= i J-^ .-. ^ = 30°. 16. 6cos-^-cos^-l=0; {3cos^+l)(2cos^-l) = 0; , 1 1 .-. COS^=:.^ ^^ ~S' .: d = m°. 18. 2-2cos2^ = 3cos^; 2 COS- ^ + 3 cos ^-2 = 0; /. (2cos^-l)(cos^ + 2) = 0; .-. 008^ = .-,, BO that ^ = G0". 10 19. tan ^/3 sin ^ - 12 = ; .-. (2 sin - ^/3) (3 sin -3tan^ + l = 0; .-. (2tan^-l)(tan^-l)=0; .•. tan ^ = 1, or - . .-. ^ = 45°, or2G°34'. IV.] MISCELLANEOUS EXAMPLES. 11 MISCELLANEOUS EXAMPLES. A. l^A.iE ?,± 21 29 3. If ^ be the angle, we have sin 6=~- , so that cosec ^ = -— . Ai ^ /r 721? >/(29 + 21H20-2l] 20 Also cos^=^l-(^-j = ~ = ,^. 1 15 15 4. tan A ~ — — = — - — _- - , = -Q ; Vcosec--' ^ - 1 J'2x 32 8 sec i4 =^1 + tan'^ A = - — ^ = — . 5. First side = cosec" A -cot^ ^ - 1 = 0. cot^ = -=TK; sec^=- = — ; secC = - = — . a 40 c 9 a 40 8. See Article IG. 1 4- COS 6 9. First side = cos^(l- cos ^) — ^-~ =l-cos2^ = s:n2^. ' cos^ ^. , „ ^ o 1 + cot-a Jl+coi-a 10 We have sec- a = l + tan- a = -:; — ; .-. seca = ^ — . *" cot-a cot a Also cosec' a = 1 + cot' a ; so that cosec a = >^/l + cot- a. 11. First side = 3(^3)%|.2 + 5.1-|.(fy=l + | + 5-.^ = 6. -.« • 1 "^ I'a — I — 5- Jinr + n^ 12. sma=: , ■ - ■ : - - = , ^; seca = v/l+tan''a=:- . Jl + cot^a Jm^ + ii' « 13. m sexagesimal minutes =-- — -- right angles, n n centesimal minutes = :r— r — , _ . right angles. 100 X 100 WW, ••• 60^96 = 100-^100' whence m = -o4n. 14. cos A = .yi - sin^ A= ./ 1-^= +-, since A is acute ; 4 5 .-. tan yl + sec ^ =- + ;- = 3. o o 15. First side = tan ^ cot A sin A cot .-1= cos A. 16. -R(^ =7 ■'^02 + 212 =^841 = 29: , i?P 20 , QR 29 ^"'^^ = i'^ = 21' ^°^^^^ = iep = 26 12 MISCELLANEOUS EXAMPLES. [CHAP. siu'-^ a - cos^ a 17. Firstside= — . . sm acosa = l -cos-a - cos-a = l 2cos-'a. sin a cos a 18. See Art. 39, Ex. 1. 19. Second side = (V3)2- 2 . QY - '^ (^2)2 = 3-^ - ^ = 3 - 2 = tan3 60^ -2tan-15°. 20. (1) 3sine = 2cos2^; (2) 5tan^-seca^ = 3; 2sin3(? + 3Bm^-2 = 0; otan ^- 1 - tan2^ = 3; .. (2sin^- l)(siii' = 60°. (I 2 a 2 2. -; .-. C'=60°, Z? = 30^ a 2 v.] SOLUTION OF RIGHT-ANGLED TRIANGLES. 13 6. c= Ja^+^= .v/75 + 3x75 = 10 ^3. sin^ = - = *{-; .-. JB = 60°, ^=30°. C a 7. Z> = c = 2; .-. JJ = C = 45°. 8. a = x/fo^T^i^ ^3^36^727 = 9. a=jW+? = 2j1. ^:,^c=\ = \', .-. C = 30°, Ji = 60°. 9. i? = 90°-yl = 60°. 10. C = 90° -25° = 65°. ^ X T, , ^ /o ,o ^rr & = acosC = 4x-4226', - = tanB; .-. & = 9V3.^3 = 27. . ^ . ,_,' a > V c = rt sin (7 = 4 X -9063. - = sec£; .-.0 = 9^3.2 = 18^3. 11. B = 90°-^ = 36°. 12. £ = 180°-C-yf = 90°. a = ccos£=8x-8090; a = 6 cos (7 = 6 x -4540; 6 = csin£ = 8x-5878. • c = & sin63° = 6 x -8910. 13. ^ = 90°-C = 53°. 14. C=180°-^-J5 = 90°. rt-6cosC=100x-7986; a = 6tanJ = 20; c = &sec^ = 40. c = & sin C = 100 X -6018. 15. ^ = 180°-75-C = 90°. 16. .4 = 180° -5-0 = 90°. 6 = c = 4; a=sJW+c^^4^2. b = a cos (7=4; c = a sin C = 4^3. ,, , 49 __„ 18. a=csin^=50x-62 = 31. 17. a = Z>tan^= — - = 700. 19. c = a tan (7= 100 X •8647 = 86-47. 20. a = 6 sec = 200 x 4-89 = 978. 21. C = 90°-.4 = 54°. 22. sin (7 = - = -37; .-. C = 21°43'. a = c tan ^ =100 X -73 = 73; 7?_qno r AQoir. t = csec.4=100xl-24^124. ^ r ^J L o^ & = acos C=100x -93 = 93. 24. c = jTi^b^ = 7124609 = 353. 23. (7=90°-i^ = 50°36'. tanB-^ - ^'-1-200- c = &cotIJ = 25x 1-2174 = 30-435; ^<^^-^-^ 225 ' a=&coseciJ = 25x 1-5755 = 34-3875. .-. 7? = 50° 24'. ^ = 90°-i? = 39°36'. 22-75 25. cos^ = ^^_' =-91; whence^=24°30'. Hence 5 = 65° 30'. a^c sin J =25 x •4147 = 10-37, approx. 14 SOLUTION OF RIGHT-ANGLED TRIANGLES. [CHAP. EXAMPLES. V. b. Page 30. 1. P=180^-30°-120° = 30° = .(; 2. c = /?D cosec 30" = 20 ; .-. CB=CA =20; n = BDcosec'i5° = 10^2. :. BD^BG Bin ^0° = 10 JZ. 3. Since5+C = 90°; .•- J=90''. .45 = L'7) sec 30° = 10^3 ft. ^C = JjBtan30°=10ft. AD = AB sin Z0° = 5 J^h. 4. Let QS be the perpendicular from Q on PR. ThenP7i: = 8sec60==16. ,Si? = 8 cos 60° = 4. .-. ,S'P=16-4 = 12 5. -S(2 = 36 tan 53°.:= 47-77. i?Q = 36 tan 35° = 25-21. .. RS = SQ -RQ = 22-5Q. 6. We have z PPQ = 180°- 135° = 45°; .-. z gPiJ = 45°; .-. QR = QP =20, and Z gPS-90°- 25° = 65°, .-. S:Q = 20 tan 65° = 42-89, .-. P,S = 42-89 - 20 = 22-89. 7. Let AD be the perpendicular and let AD = x. Then z P^D = 90°-45° = 45°= z >4PD, .: DB = DA=x. DA X ^'>'^^ = ^^^^^°i ••• ^340=^^=^5 ••• x{^3-l)=40V3. •'■ «^ = ^-^ = 20v'3(vA3 + l) = 20(3 + V3). .-. perpeudicular = 20(3 + ;^S) = 94-64. 8. Ijet DC = x. Since z DCP = 90° - 45° = 45° = z CBD ; .-. DB = DC = x. A A ^"^ *o-oio/ x + 41-24 , .,,, And v^y-, = cot3o°18 ; .-. = 1-41'24. DC X :. a; + 41-24 = l-4124.r; .-. a: = 100; that is, DC = !;/; = 100. 9. The perp. A J) = 20 sin 42° = 20 x -6691 = 13-382, , ^ AD 13-382 „,^„ *^"^=Ci) = iM3-8 = '^'^^^' whence C = 36°2d'. EXAMPLES. VI. a. Page 42. For Exaini)les 1 — 5 see figure on page 40. 1. LetPC = height of chimney, JC = 300ft., then elevation = z BJC=30°, .-. PC= JC tan 30°= 100^/3 = 173-2 ft. VI.] EASY PROBLEMS. 15 2. I^et B be the top of the mast, 50 = 160 feet, A the boat observed. Theu iBAC=30°, iABC = m°. .: distance required = JC = 5C tan 60^ -160^/3 = 277- 12 ft. 3. Let BC represent the pole, and AC its shadoNv. Then tan A =---, = ^--75 = JS ; .-. angle of elevation = 60^. ALf ^sjo 4. Let BC represent the tower; A the position of the observer. Then AC = S^'&ii. and z/UC = 30^ .-. height of tower =AC tan 30° = -7- = 50 ft. Distance ^jB = 5Ccosec 30° = 2BC = 100 ft. 5. Let AB represent the ladder, and BC the wail. Then AB = 45 ft. z ABC = m°. :. height of wall :^BC = AB cos 60° ^ 22-5 ft. 45 lo Distance .4C = .-1B sin 60° = — J- = 38-97 ft. 6. See figure on page 9. Let DE, BC represent the mast?, and OGE tlie horizon. ThenZi'OC = 33°41'. BC=40ft. BE^mH. And O(7=^Ccot33°41' = 60ft. OA'=Dii: cot 33 41' = 90 ft. .-. distance required = OE - OC — 30 ft. 7. See figure on page 40. Let BC represent the cliff and A the observer. Then iBAC = ir 18'. i5C=132yds. 132 /. distance required = .4jK = 7?Ccosec 41° 18'= ^- =200 yds. •ub 8. See figure on page 9. Let BC, DE represent the chimneys, and O the observer. Then 00= 100 yds. lBOC^IT'I'. Now BC= OC tan 27° 2' = 51 yds. .-. Di; = 7?O+30yds. = 81yds. 9. See figure on page 41. Let FT be the tower, and Q, R the two points of observation. Then zPQi? = 30°, z7^i?r = 60°, (?i? = 100yds. .-. z iipg= 60° -30°= 30° = zpgi?, .-. i;p=J?g = 100 yds. .-. height of tower = i2P sin 60° = 50^3 = 80-6 yds. 16 EASY PROBLEMS. [CHAP. 10. Let AB be the flagstaff, BC the building, /> the point of observation. Then 7)C = 40ft. l ADC = 60°, lBDC = ^0°; 80 .-. I DAB = 30''= A ADB; .: BA = BD = -iO sec 30°= - = i9-VJ it. 11. See figure on page 41. Let FT be the spire, R, Q the two points of observation. Then QR=200ft. /.PRT = 45°, iPQR = 30°. lRPT = i5°; .. TP=TR. Let X ft. = height of spire. .-. X = -^^^ = 100 (1 + V3) = 273 -2 feet. 12. Hee figure on page 43. Let CD represent the post, and AB the steeple. Then CD = 30 ft. i AC K = 30°, i A DB = 45°; .-. iDAB = 4o°= aADB, .-. BA = BD = x feet, say. A V r — RO .-. tan 30°= -— = ; .-. x (^/3 - 1) = 30^3. .-. x = 70-98 ft. That is heights distance = 70 -98 ft. 13. Let B be the top of the hill, and C the point on the horizontal plane vertically below B. Let D be the position of the balloon wlien the obser- vation is made. Draw DE perpendicular to BC. Then I?C = 3300ft. iBDE = 30°, iBAC = 60°; :. AC = BC cot 60° = 1100 J3ieet. And BE = DE tan 30° = AC tan 30° = 1100 feet. .-. DA =EC = 3300 - 1 100 = 2200 feet. .*. the balloon rises 2200 feet in 5 minutes, ^^ . 2200 X (50 ., , _ ., , that IS, frr/TT^ — n — e miles per hour, or o miles per hour. JL / oU X o X O 14. See figure on page 44. Let OA represent the monument, B, C the two objects, OP the horizontal line through ; Then zPOC = 30°; .-. lOCB = 30°. L BOB = 45° : .-. L BOA = z OB A = 45° ; .: AO = AB = 100 feet. Let X feet = CiJ = required distance. "^•"^ r+"l-0b = ""'='»° = J3= .■..r=lO0U/3-l), .'. distance required = 73*2 feet. VT.] EASY PROBLEMS. 17 15. See figure on page 43. Let AB represent the monument, CD the tower. Then J 7? = 90 feet, and the angles are as in the figure; .-. DB = AB cot G0° = 32^3 feet; .: A E = GE tan 30° -. BB tan 30° ^ 32 feet. .-. height of tower = CD = EB = 96 - 32 = 64 feet. 16. See figure on page 44. Let OA represent the cliff, B, G the two boats. Then O.-f = 150ft., zOJ5^ = 30°, zOCi^ = 15°; .. zi?0C=15°, and BG = BO. :. required distance = J5C' = 7? = . 4 Ocoaec 30^ = 300 ft. 17. See figure on page 44. Let represent the top of the hill, /»', C the milestones. Then zOB.4=45°, zOCJB = 22°; .-. z £0/1 =45°, so that JO = /(/?. Let .10 = J2? = x yards. Then 4^ = "^-^1^ = cot 22° =2-475; .-. l-475x = 1700. AG X .-. height of hill = x = 1193 yds. nearly. 18. See figure on page 44. Let OA represent the lighthouse, B, C the two rocks. Then OA = 80 yds., z OBA = 75°, Z OCB = 15°, z CO A = 75° ; .-. AB = OA cot 75° = 80 X -268 yds. Let CB — x yds. Then ar + 80 x -268= 0.4 cot 15° = 80 x 3-732, .-. j- = 80 x 3-464 = 277-12 yds. .-. required distance = 277 -12 yds. EXAMPLES. VI b. Page 47. 1. Let A, B be the two positions of the observer, P, Q the two objects. Then AB = 800 yds., and PQA is a straight line making Z PAB equal to 45°. Also zPi^ J =90°, lQBA = -i5°. .: QA = QB=QP. And QJ = /IB cos 45° = *^^^ = 565-6 yds. P.4 =2Q.4 = 1131-2 yds. Thus the required distances are 565-6 yds., 1131-2 yds. 2. Let A, B be the two positions of the observer, P, Q the two ships ; then APQ is a straight line at right angles to AB. And AB = 3 miles, iABP='60°, zJ7^Q = 60°. .-. BP = AB sec 30°= ^ =3-464 miles, BQ = AB sec 60° = miles. Thus the required distances are 3 "464 miles, C miles. H. E. T. K. 2 18 THE mariner's COMPASS. [CHAP. 3. litt O represent the harbour and OX, oE, OS, OW tlie directions of North, South, East, West. Let P, Q be the positions of the two ships at 2 p.m. Then iP0W=2S\ iQOE = 62''; :. z POQ = 90^ Also 0P = 2x 10 = 20 miles. 0Q = 2x 10^ = 21 miles. .-. distance = PV=>sy 202 + 21^= 29 miles. 4. Let be the position of the lighthouse, and P, Q the points at which the steamer enters and leaves the light. Then PQ lies East and West, and OP, OQ are the directions of N.E., N.W. .-. zPO(3 = 90°, zP(?0= z(?PO = 45°, 0P=0Q = 5miles. .•. PQ = j2o + 25 = 5,^2 miles. .-. steamer sails 5^/2 miles in 30^/2 minutes, that is, the speed of steamer is 10 miles per hour. 5. Let 0, P, Q be the first positions of the ship and lighthouses. OA the direction in which the ship is sailing, A its second position. Then 0.i= 10 miles, Z 0.4P = 45°, Z.4OP=90°, lPAQ = m°. :. zPg-rl=45°-22i^ = 22^° = P.-ig; :.PA=Pq, And OP = OA = 10 miles, PA = OA sec 45° = 10^/2 miles. .-. OQ = OP + PQ= OP + P^ = 10 (^/2 + 1) = 24 -14 miles. .•. distances are 10 miles, 24*14 miles. 6. As before let be the port, and ON, OE, OS, OW the directions of the cardinal points of the compass. Let P, Q be the positions of the ships at the end of an hour. Then 0P = 8 miles, 0^ = 8^3 miles. And zPOiV=35°, zQOS = 55°; .-. zPOQ = 90°. .-. distances apart = PQ = J&i + 3 x 64 = 16 miles. Also tanQP0 = -^-=V3; ■- Z gPO = 60°. o .-. iQPO- lPON = Q0°-So° = 2o°; ■. bearing of the second vessel as observed from the first is S. 25° W. 7. Let A be the liglithouse, 0, P the two positions of the vessel. Then AP the direction of S., AO the direction of E.S.E., OP the direction of S.S.W. .-. z A OP = 90° ; z PA O = 90° - 22^° = 67A° ; and AO = A miles. .-. P0 = ^0 tan 07A° = 4 X 2-414 = 9-656 miles. .-. the vessel sails at the rate of 9-656 miles per hour. VI.] THE MAIUXER's COMPASS. 19 8. We have L CA U = 10° + 50° - 60°, Z A li C = 180° - 50° - 40° = W, ii(7 = 10 miles. .-. J 7? = ^C cot 60° = -.J =: 5-77 mis.; .4C' = /iCcosec60=: "=11-54 mis. 9. Let be the lighthouse, and A, B the two positions of tlie ship. Then Z GAB = Z OB A = 45° ; OA^OB = lo miles. .•. .-(J) = lov/2 miles = — -^^tt; knots. .•. the ship in 1.^ hours sails — ^- knots. bl) , .^ ., 15^2x60 24x2, , ,, , . „^. ^, , , .•. in a day it sails — ~~ x — ;;-- knots, that is, 29o-09 knots. 69 o 10. Let be the lighthouse, A, B the two positions of the coaster. Then AB is in direction S.E., and OA is in direction N.E. ; /. z OAB = ^0°. Also Z ^ OB = 45° + 15° = 60°, and 0A = ^ miles ; .-. AB = OA tan 60° = 9v/3 miles; .-. coaster sails 9>y3 miles in 3 hours. .-. rate of the coaster's sailings 5-196 miles per hour. Also OB = AO sec 60° = 18 miles. That is, the distance of the coaster from the lighthouse at time of second observation = 18 miles. 11. Let P be the position of the vessel when it is N.E. of A and N.W. oiB. Then z.4PB = 90°. Also zP.-iB = 45°-15°:=30°. .-. P.4 = .47? cos 30° = 6^/3 miles. Now the direction S. 15° E. is at right angles to the direction E. 15° N. .-. the ship crosses AB at right angles. Draw PA'' perpendicu ar to AB. Then P.V=JP sin 30° = 3^/3 miles; therefore the ship will reach >V in 3 /3 YTT- hours, that is in 31-176 minutes. .-. the ship will cross the line at about 31' past midnight. 12. Let P, Q be the two spires. Then zPJP = 90°, and z P/?(3 = 37^- 7i° = 30° ; .-. Z QBA = 90° - 37i° - 22r = 30^ ; .-. lBPQ = SO°= iPBQ; so that gP = QP- 1-5 miles. .-. .4P = PQcos30° = ?^ miles. 3/3 .-. the train travels -- ■ miles in 2 minutes, 4 3 /3 that ia, ^ x 30 miles per hour, or 38*97 miles per hour. 2—2 19 a kasy problems. [chap. EXAMPLES. VI. c. Page 48 a. 1. ;/ = 83tau23^44' = 83x-43i)7yards = 109 ft., approximately. 2. /t = 173tauG3° = 173x 1-9626 ft. = 339-53 ft. 3. h = 200 sin 04"^ = 200 x -8090 metres = 101-8 metres. 4. (Z = 500sin23^ = 500x3x-3907ft. = 586-05 ft. , 1760 ^^ , 5. The distance between two consecutive posts = -^^^ = 80 yds. Then required distance = 80 tan 16^42' = 80 x -3000 yds. = 24 yds. 6. Let ABCD be the square, and let the line be drawn from B to E, the middle point oi AD. Then tan ABE = ^= -5, whence z ABE = 26°34' ; .-. /Ei?(7 = 90°-26°34' = 63°26'. 7. Lot D be the middle point of the base BG of the isosceles a ABC, in whicii J7; = 3^a. Then cos DBA = ^ = ;^. = -1666, BA (> whence /.B = 80°2o'= iC; .-..-l = 19°10'. 8. With the figure on p. 41, QJ? = 160ft., lPRT = i5°, z i'Qr=21°48'; also BT=liT. If li is the required height in feet -A^ = tan 21° 48' = -4000 ; ^t+lbO .-. /t = -4/i + 64; .-. -6/i = 64, and;i = 107. 9. With the same figure as in Ex. 8, Q/i = 100yds., LBliT=b4°2^', lBQT^2TV2'. Let h be the height in feet; then t - = cot 27^2' = tan 62° 48' = 1 -9458 ; h and iZT = /t cot 54° 24' = ;i tan 35° 36' = /ix -7159; .-. 300 + /»x-7159 = ;tx 1-9458; that is, l-2299/f = 300; whence ;t = 244, nearly. Or thus: Since lRFQ = 2TV1', .. Pii = gi^ = 100 yds ; .-. h = 300 sin 54°24' = 300 x -8131 - 244. VI.] EASY PROBLEMS. 19 B 10. With the same figure and notation as in Ex. 9, 1760 + i^r^ ^^^ ^go ^^, ^ 2 ggg2 ; n and J2r=;itan3o°=//x-7002; .-. 1760 + hx -7002 - /t X 3-3332 ; that is, 2-6330^ = 1760; whence /t = 668, nearly. 11. Let AB represent the top, and DE the bottom of the trench. Draw AC and BF perp. to ED and DE produced. Then CD = 8 tan 12° = 8 x -2126 = 1-7008 ft.; whence (7^ = 10-7008 ft. .-. £F=4-2992 ft. 4 -2992 In A EBF, tan B = — g— = -5374 ; whence 5 = 28° 15'. 12 Let C and 7) be the first and second positions of the observer; then £ ACS = 90\ andiADB^US°24'. .: z^DC= 36^ 36'. Now ^ C = 630 tan 36= 36' = 630 x -7427 = 467 "9 m. 63 63 „^,„ AD = ^^,^^, = — ^- = 784-7 m. cos 36 36' -8028 13. Let .4 be the point of observation, B the top, and C the bottom of the tower. Draw AE horizontally to meet BC in E. Then z.EAC~n°, and EG = AD = 30 ft. AE = EG cot 17° = 30 tan 73° = (30 x 3-2709) ft. = 98-127 ft. Again I>*£ = ^£ tan 42° = (98-127 x -9004) ft. = 88-3506 ft.; .-. required height= (30 + 88*3506) ft. = 118-35 ft. 14. See fig. on page 44. Let OA=x, AB = y; then ?/ = xcot35° = a;tan55° = a;x 1-4281, TOO + y ^ ^^^ ^^o ^ ^^^ ^go ^ ^,Q jQQ . X .-. 700 + X X 1-4281 = a; X 4-0108 ; thatis, 700 = xx2-5827; whence j- = 271. 15. Let O be the lighthouse, and ^, i? the two positions of the ship. Then z 50^ = 90°, zOBJ = 32^ ^i5=15mi. Now 0.-(=^Bsin^J50 = 15 8in32° = 15 X -5299 = 7-9485 mi. 20 RADIAX MEASURE. [CHAP. 16. Tjet O be the house, and A , B the two positions. Then / BOA = 00^, / on A = 52°, AB = 2 km. Let OC he perp. to AB ; then 07? = 2000 cos 52" = 2000 X -0157 = 1231-4m. 00=07? sin 52° = 1231-4 X -7880 = 970-3 m. 17 Let L he the lifzhthousc, and S^, S^ the two positions of the ship. Then z SiOSr, = Q0°, L .S2.S\L = 56°, LSi = 12 mi. rr Now (S',.*?o= = — ^mi. ; and since the ship has been sailing _ ' " cos 56 5592 " 12 6 of an hout, the number of miles per day=7^— ^ x -x 24 = 441-5. 18. Let B be the battery, and .Sj, .So the two positions of the ship. Draw 'f^fi perp. to i^.^o; then z 0ec fig. on page 47. Here lBAE = W, z£.-1C = 4P; .-. zLMr = 90°. Also z^ON' = 90°-41° = 49°, and Z 7?C.V' = 15°; .-. z. 4 07; = 49° -15° = 34°. Kow JU = ^0 tan 34° = 20 x -6745 = 13-49 mi. J50= =- — ;r7r = 24-12mi. cos 34° -8290 EXAMPLES. VII. a. Pa(;k 54. For Examples 1 — 22, see Art. 64; the following solutions will sutiice as illustrations. 57.V 237r 6. Kadian measure of 57^ degrees =." tt = y . 14| 27r 7. Radian measure of 14V degrees =.-- tt = ,- . 37i 10. Radian measure of 37^ degrees= ^^^ x 3-1416 = -6545. VII.] RADIAN MEASURE. 21 68? 11. Radian measure of 68| degrees= - ^ x 3-1416 =^1-^ ,on X •^•111^> = 7^ -o X -2618:- 1-1090. 4 X 180 4x3 ^n 77r ,. 7x180, ..^ 16. T- radians = — -- — degrees = 28 . 4o 45 1 sn° 19. •3927radians = -3927x-— — [Art. 63] 31410 8 22. 2-8798 radians = 2-8798 x = "^ ',! x ^^^"^ tt. x 180"= lOo''. IT 31416 12 ^_ 36-54 _ 2-03 ^3. ^ere--^^^---— ; 00)24 (30 ) 8-2-4 .-. ^=-203 X— = -638. -54 __ „ ^ 116-046 ^,,^ 25. Here - = —g^= -6447; eo)45 6^ 99 «0 >^76^ .-. ^ = -6447 x^„- = 2-0262. 'O^e 7 180 , 27. A radian = — degrees; . , . ,. 180x60x60 .'. no. of seconds in a radian = — = 180 X 60 X 60 X -31831 = 206265 nearly. 28. Since 1° = ^^ radians, the radian measure of 1" '■"^^ -0000018. 180 X 60 X 60 EXAMPLES. VII. b. Page 56. 5. cot2^ + 4cos2| + 3sec2^=(V3)- + 4(-l)'+3 (iy = 'i + 2 + A^JS + 1) {JS - 1) = 3-^/3-^3 = tan=^ 60° - 2 sin 60°. 6. In the figure on p. 14, if BC represents the chimney and AG the shadow we have 7. (1) Firstside=:2tan2^ + 5tan^ + 2 = 2 (1 + tan2 ^) + 5 tan ^ = 2 sec- ^ + 5 tan d. cot" a cosec^ a - 1 (2) ,— = T^=coseca-l. 1 + cosec a cosec a + 1 8. Expressed in radians the third angle = 7r "(t + VI^u* "^^^ sexa- gesimal equivalent is 22^°. 9. Let X be the number of degrees in the angle; then a:=14 ( , — - X ) +51, or x- j;=:51; whenoex — 67A. \180 / 45 10. With the figure of Art. 45, we have a=6tau60° = 6^3; c = Z> sec 60° = x 2 = 12. Also the perpendicular from C on AB-b sin 60'^ — 3^3. VII.] MISCELLANEOUS EXAMPLES. B. 25 /^v T-,- 1 -3 i« i « sin ^ co8-. 17. (1) a^ + &^>2a6, since (a- fe)- is positive. Therefore ^ , > 1. Henoe cosec ^ = — -- — is possible. 2a 6 2a 6 ^ (2) a2 + l>2a: so that a + ->2. Hence 2sin^ = a + - is impossible [Art. 16J. 26 MISCELLANEOUS EXAMPLES. B. [CHAP. 18. The height of the balloon = GOO tan G0° ft. = 060 ^3 feet. .-. the balloon rises GGO x ^/3 feet in 1-5 minutes. .^ . 660x^3x60 r,rr ■^ ^ .■. it rises ^> - ?r -.„v,,^ > or 8GG miles per hour. 1-5x3x1760 ^ 19. Let x^ be the common difference between the angles, then they are 36°, 36° + x°, 36° + 2a;°, .-. 3x + 3x 36 = 180; whence .r = 24. .•. the angles are 36°, 60°, 84°, or -, -, --radians. 5 3 lo 20. First side = sin- a ( 1 + tan^ /3) + tan- /S (1 - sin- a) = sin- a + tan- /3. 21. Let CZ) = ar = the perpendicular. Then z C7i7) = 180°- 116°33' = 63°27'; .'. 5: = 7:>i5 tan 63° 27' = 2Z)i3. And .T = DA tan 42° = ( ^ + 55 j X -9 ; whence a; = 4-5 x 20 = 90. rtrt /iv -I-.- i. -T cos a (1 + cos a) + sin- a cosa + 1 22. (1) First side = — — r^ jz , = ^- — yz , = cosec a. sin a (1 + cos a) sin a ( 1 + cos a) /n\ -n- i. -J 1-cosa/ 1 \ /I sin a (2) First side = — ^ cos a ) = (1 - cos a) = tana-sina. ^ sin a \C08 a / cos a 23. First side = 1 + -^ l4-./3\- 2 + v/3 2 1 + cos 30' \' iO 1-^3/ 2-^3 1 s^3 l-cos30°* 2 24. Let the man start from A and walk to B, and let C denote the position of the windmill. Then we have lAGB = 90°, iCAB = 30°, BC = l mile. .-. ^i? = i?Cco8ec30° = 2 miles. .4 (7 = 7? C tan 60° = 1-732 miles. And rate of walking is 2 miles per half hour, or 4 miles an hour. 25. The complement oi -— = ~ -'-— = - radians. 8 2 8 8 26. (1) 3sin^ + 4-4sin2^ = ^-, (2) tan ^ + -^ = ^^^-^ , 8 8in2(?-6sin^4-l=0; ^/3 tan2^ + 2 tan ^- v/3 = 0; •. (4sin^-l)(2 8in(?-l) = 0; .-. (v/3 tan (? - 1) (tan ^ + ^/3) = 0; .-. sin^=-T, or o ' - tan(?= -, or -»^/3. .-. ^ = 30°, orl4°29'. .-. ^ = 30°. Ylli.J TRIGONOMETRICAL RATIOS OF ANY ANGLE. 5 sin a - 3 cos a 5 tan a - 3 20 - 15 5 i^V 27. Here sin a + 2 cos a tana + 2 4 + 10 14 28. First side =-i.:±-i^2id_-x sin J- cos ^ , sin J- cos yl sin2 y( - sin ^ cos ^ + cos2 ^ cos .4 . —^ ; — sin A cos A sin ^ (1 -sin J cos J) = z r— j — = sm A. 1 - sin A cos A lOK.O 29. The distance = 195-2 cosec 77° 26' yds. =— =200 yds. •976 -^ 30. We have 70° = ^ radians. lo Itt ^1 2*? .'. distance required = 27 x ^- feet =^ x -^ = 33 feet. lo 2 7 EXAMPLES. VIII. a. Page 70. 18. sin 420° = sin (360° + 60°) = sin 60° = **^- , I 2 20. tan ( - 315°) = tan ( - 360° + 45°) = tan 45° = 1 . 22. cosec ( - 330°) = cosec ( - 360° + 30°) = cosec 30° = 2. 24. cot -r^ = COt(47r + j J=cot^=l. 26. tan ( - '^) - tail ( - 27r + ^') = tan ^ = x/3. EXAMPLES. VIII. b. Page 72. 1. Tlie boundary line of 120° lies in the second quadrant, "3_ 1 4~~2' .-. cos 120° = - ^1 - sin'-^ 120° = - ^1 - ? = ^ ^ ,„^„ sin 120° .-. tan 120°= = -./3. cos 120° ^ 2. The boundary line of 135° lies in the second quadrant; .-. sec 135°= - Vi+lan^l35°= -^2; and sin 135° sec 135°= -1; .-. sinl35°= — ^/2 28 TRIGONOMETRICAL RATIOS OF ANY ANGLE. [CHAP 3. The LouBdary line of 240° lies in the third quadrant; .-. sec 240°= - Vl + tan2 240°= - 2 ; .-. cos 240°= - .- . 4. The boundary line of 202° 37' lies in the third quadrant; cos A _ 12 sin .1 5 •. cos A=- Vl-sin-^J= - a/1- 169= - ri- Also cot.i 5. The boundary line of 143° 8' lies in the second quadrant ; and 3 4 cosec A = 1| ; /. sin A = y', .". cos A= - >/l - sin- ^ = - - ; .'. sec A= --T . Hence tan A = ' — -, = - -- . 4 cos A 4 6. Tlie boundary of 210° o2' lies in the third (quadrant; cos A / -I c q .-. smA=-fJl-cos^A=-./l-rp,= --. Alsocot^ = sin A 3 27r 7. The boundary line of — - lies in the second quadrant; o 27r ^ 27r 1 . 27r /, ^^ir ^/3 and sec — = - 2 ; .-. cos — - = - .^ . .-. sin — - = + . / 1 - cos- -— = - . 27r 2. ^°^y 1 sm- 8. The boundary line of -.- lies in the third quadrant; 4 '^V' COS - = - ^ / 1 - sin- --= - -^^ ; Stt „ ■ sin — sec -T-= - ^'2, and tan - - = — '_ =1. 4 4 OTT cos . 4 9. We have sin A = i ;^1 - cos'-* yl ^ i ^ 1 144 _ ^ 109~'^13 sin y( 5 tan A ^. = ± -— - . COSyl 12 IX.] MISCELLANEOUS EXAMPLES. C. 29 EXAMPLES. IX. PA(iK 79. 6. Expression = l.(- 1)2-2 (-1). 1 = 1 + 2 = 3. 7. Expression = 3.0. ( - l) + 2 . 1 - 1 = 2 - 1 = 1. 8. Expression = 2.{-l)-.l+3(-l)3-l = 2-3-l= -2. 9. Expression = 0x0 + l-(-l) = l + l = 2. MISCELLANEOUS EXAMPLES. C. Tage 80. 1. If tan -4= - J , the boundary of A will lie either in second or fourth quadrant. [See figure on page 72.] In either position the radius vector = ^3- + 4- = 5. 4 4 Hence cosA'OP=--; cosA'OP'=-. o o 2. First side = (2 + sin A) (1-2 sin A) sec A = (2-3sin^ -2 sin^^) sec^ = (2 cos- A -3 sin A) sec A = 2 cos A -S tan A. 3. a = Vc^^^= v/PlMlO^Sp =. 21 ^1 - ^ = ^^ . sin ^ = - = ''i^ ; whence A = 60°, B = 30°. c 2 4. A lies between 180° and 270°; / /95\ 2 94 7 .-. tan A = + a/scc* ^ - 1 = / f y j - 1 = ^ ; and cot A=^-. 197r 5. We have 19° = r-^ radians. 180 Also the radius of the earth = 3900 miles. 197r .*. required distance =—^ x 3960 = 19 x 22 x 7r = 1313 miles nearly. IbO 6. Let AB represent the cliff and P, Q the positions of the two boatK. Then JJ5 = 200 ft., z^PB = 34°30', i AQB==18°40' ; .: (^JS = , ^7)' cot 18° 40' = 200x2-96 = 592 ft. PB^AB cot 34° 30' = 200x1-455 = 201 ft. ,-. required distance = (?i? - P/i = 301 ft. 29 a miscellaneous examples, c. [chap. 7. Since the boundary line of A is in the third quadrant, A=- Jl + t^u^A= - y^l + ^= -|, sec .'. cos A= - -, and sin A = - -. o o 3 / 3\ 4 3 4 .-. 2 cat .4 - 5 cos ^ + sin .-1 = 2 . ^ - 5 ( ~ ^) - -= = K + 'i-^ = S/a 4 \ 5/ 5 2 o 8. We have 71°36'30" = 71001° = - ..^ '^ radians. . ^ ,. ^. TI-GOItt 15x180 ,.,^..^. , .-. required radius = lo-: TacT ~ 7i-fi01 =l-^'003 inches. ^ ^. , ., tan3^ cot=J^ 9. First side = o'u + 27i sec^ cosec^ 6 _ sin^ cos^ 6 _ sin-* ^ + cos^ 6 cos ^ sin 6 sin (!^ cos ^ (sin2 d + cos2 ^)2 - 2 sin2 ^ gogS 1 _ 2 ginS ^ cos^ sin ^ cos ^ sin cos f^ 10. Let AB represent the flagstaff, BC the tower, and let D be the position of the observer. Then /7?DC = 68°11', iADB = 2°10'; .: i ADC = '70°21'. Let 2?C = x ft., then a: + 24 = DO tan 70° 21', and DC = .r cotGB^ll'. .-. a; + 24 = a; cot 68° ir tan 70° 21'; .-. a: + 24 = x x 2-8 x •4 = 1-12j-; whence x = 200 ; that is, the height of the tower is 200 ft. 11. If tan A = -5, and tan I? = -3333, from the Tables we have ^ =2G°34', 7? = 18°2G'; .-. ^ + ^ = 45°. 12. We have (4 tan ^ - 3) (3tan S\So = 20 mi. .-. FSr, = 20 tan 43° = 20 x -9325 = 18-6o mi. ^^1 = -^4-o = -^ = 27 -346 mi. ^ cos 43° -7314 EXAMPLES. X. a. Page 87. 1. cos 135° = cos (180° - 45°) = - cos 45° = - 2. sin 150° = sin ( 180° - 30°) = sin 30° = i . 3. tan 240° = tan (180° + 60°) = tan 60° = ^^3. 4. cosec 225° = cosec (180° + 45°) = - cosec 45° = - ^2. 5. sin ( - 120°)= - sin 120°= - sin (180° - 60°) = - sin60°= - "^ . 6. cot (-135°)= -cot 135°= - cot (180°- 45°) = cot 45°= 1. 7. cot315° = cot(180° + 135°) = cotl35°= -1. 8. cos ( - 240°) = cos 240° = cos (180° + 60°) = - cos 60° = - ;3 . H. E. T. K. ; no CIRCULAR FUNCTIONS OF [CHAP. 9. sec ( - 300 ) = sec 300 = sec (180^ + 120'-) = - sec 120° = - sec (180^ - 60°) = sec 60° = 2. 10. tan— - = tan ( TT-- J= -tan-= -1. . 47r . / 7r\ . TT -^3 11. sin — = sm(^7r + - j= -sin-= --^. 27r / 7r\ TT o 12. sec— =sec ( TT-- j= - sec -= -2. / 7r\ TT 13. cosec I - ,7 ) = - cosec - = - cosec — = - 2. '6w\ Sir f Tr\ w 14. cos - — Ucos — =cos TT-- ^-cos- = 15. cot( --^,!^ )= - cot ^= -cot f7r-^J=cot^ = v'3. 16. cos (270° + ^)= cos (180° +90° + ^)= -cos {90° + y() = sin ^. 17. cot (270^ - .-I) = cot (180° + 90°-^) = cot (90° - .4) =tan A. 18. sin {A - 90°) = - sin (90° - ^) = - cos .4. 19. sec {A - 180°) = sec (180° - ^ ) = - sec J . 20. sin (270^ -A) = sin (180° + 90° -A)= -sin (90°-^)= -cos^. 21. cot {A - 90°) = - cot (90° -A-)= - tan A . 22. sin('^-0=-sin (^|-^)=-cosf^. 23. tan (^ - tt) = - tan (tt - ^) = tan d. 24_ sec(^-6'^=sec(7r + ^-^W - sec(^-^j= -cosee^. 25. Expression = tan .4 cos A cosec -4 =1. 26. Expression = - sin .4 + sin ^ -(-sin^i) - (- sin.4) = 2 sin^. 27. Expiession = sec^ A - tan'M = 1. EXAMPLES. X. b. Page 91. 1. cos 480° = cos (300° + 120°) = cos 120° = - ^ • v/3 2. sin 900° = sin (3 x 360° - 120°) = - sin 1 20° = - ^g ' 3. cos 780° = cos (2 X 360° + 00°) = cos 60° = .^ . X.] CERTAIN ALLIED ANGLES. 81 4. sin ( - 870°) = sin ( - 2 X 360° - loO°) = - sin 150° = - ^ . 5. sec 900° = sec (2 X 360° + 180°) = sec 180° = -1. 6. tan ( - 855°) = - tan 855° = - tan (2 x 360° + 135°) = -tan 135°= - tan (180°- 45°)= tan 45° = 1. 7. cosec( - 660°)= - cosec 660°= - cosec (2 x 360° - 60 ) = cosec60° = -- . v'3 8. cot 840- = cot (2 x 360^ + 120 j = cot 120° = cot (180° - 00=) = - cot 60-^ = - 4o • 9. cosec ( - 765°) = - cosec 765° = - cosec (2 x 360° + 45°) = - cosec 45" = - ^2. 10. cos 1125°= cos (3 x 360° + 45°) = cos 45° = -^ . 11. cot 990° = cot (3x360° -90°)= -cot90° = 0. 12. sin 855° = sin (2 x 360° + 135°) = sin 135° = sin (180° -45°) = sin45° = ~- 13. sec 1305° = sec (4 x 360° - 135°) = sec 135° = - sec 45° = - ^2. 14. cos 960° = cos (3 x 360° - 120°) = cos 120° = - cos 60° = - - . 15. sec (-1575°) = sec 1575° = sec (4x360° + 135°) = sec 135° = - sec 45° = - ^2. _ „ . 157r . / , 7r\ . TT 1 16. sin-^-=sin(^47r--j = -sin-=--^. 237r f r. Tr\ , TT 17. cot-— - = cot ( Ott -" j = -cot -= -1. -.r^ 77r /ri ''■X "f n 18. sec— = sec ( 27r +— j=sec — = 2. 19. cot-3 =cot(^67r--j=-cot- = cot-=^-. fiir 7r\ /, TT 7r\ /« tX tt 2 20. «ec(^- + -j=sec(^2. + ---j=sec(^2:r--j=sec- = ^-^^-. /3 21. cos 6 — ^ = cos 30° ; .-. ^ = 30° satisfies the equation. And cos 30° = cos (360= - 30°) = cos 330°. There are no angles whose boundary hnes are in the second and third 3—2 32 CIRCULAR FUNCTIONS OF [CHAr. quadrauts which satisfy the equation since the cosine in those quadrants ie negative. ,'. the positive angles are 30^, 330^^. And the negative angles are - (360° - 30°), - (300° - 330°). That is, the angles are ±30'', ±330°. 22. 8in^= -J = sin(180° + 30°) = sin210°; .-. ^'^ 2 10 Ms a solution. Also sin (360° - 30°) = - sin 30° "-= - 5 ; -■. ^ = 330° is another solution. Thus 210°, 330° are the positive angles. The negative angles are - (360'^ - 210°), - (360° - 330°) ; .-. the required angles are 210°, 330°, - 150°, -30°. 23. tan ^ = - ;^3 = tan (180° - 60°) = tan 120° ; .-. ^ = 120" is a solution. Also tan (360° - 60°) = - tan 60°= - ^/3 : .-. ^ = 300° is another solution. Thus 120°, 300° are the positive angles. The negative angles are - (360° - 120°), - (360° - 300°) ; .-. the required angles are 120°, 300°, - 240°, - 00\ 24. cot ^= -1= -cot 45° = cot 135°. Also cot 135' = cot (180° + 135°) = cot 315°. .-. the positive angles which satisfy the equation are 135°, 315°. The negative angles are - (360° - 135°), - (360° - 315°). .'. the required angles are 135°, 315°, -45°, -225°. 25. Let the radius vector OP start from the position OX and revolve in the positive direction till it reaches the position Ol\ such that lFOX — A. Then let it revolve in the negative direction throu;j;h an angle of IHU', reaching the position OV . Then FOV is a straight line, and z XOr = A - 180°. Draw TM, P'M' perpendicular to A'A". Then the a' OPM, OP'M' are geometrically equal. OP' OP Then sec {A - 180°) = ^^^^, = - ^^^^= -sec A. 26. Proceed as in Art. U7. Let the radius vector first revolve from OX through tlie angle A to the position OP. Again, let it revolve from oX through 270° and then further through an angle A to the position OP'; draw PM, P'M' perpendiculars to A'A''. Then from the equal A ' OPM, OP'M', we have P'M'= - OM, 0'M' = PM ; P'M' M X.] CERTAIN ALLIED ANGLES. 33 27. Let OP be determined as before, and then let the radius vector turn back in the negative direction through an angle 90° to the position OP'. Draw perpendiculars as before. Then cos {A - 90°) = ^' - f^p = sin A. 28. First side = tan .-1 - tan .4 - tan ^ = - tan A = tan (360° -A). „. ^ ., sinJt tan.-l cos ^4 29. Fn-st side= - - - . — — — - . - ; — - = 6mA. tan A - cot A - sin A _- „ . -sin^ -cot^ cos.-l , , . o 30. Expression = — -. — ^ r ,- + --1 + 1 + 1 = 3. - sin A cot -4 cos A „- „ . cosec vl cos .4 cos^^i „ , 31. Expression^ ~ , — ; — r= . „-— =:cot'J. - sec A - sin A sin-* A nn -r. • - sln ^ . scc -4 . ( - tan .4) 32. Expression = ~- — -'—X-l r = - 1. sec^ ( - sin^) tan .4 33 . First side = sin f tt - ^ - (9 j sin I^ + tt-^J cot (tt + ^ + ^J n4 ■ . IItt . /^ .^7r\ . Stt . t 1 cJ4, sm a — sin , = sin ztt + — - ) = sin -r- = sin - = -— ; •4 V 4 / 4 4 ^2 37r TT 1 , , , COS a = cos — - = - cos 7- = — r=i • Also tan a = - 1. 4 4 ^/2 .-. Expression = - - - 2 - 2 = - 4. EXAMPLES. XI. a. Page 97. 1. sin {A + 45°) = sin A cos 45° + cos A sin 45° = —3 (sin ^ + cos ^ ). 2. cos (4 + 45°) =cos^ cos 45°- sin ^ sin 45°= (cos^-sin^). 3. 2 sin (30° - .4 ) = 2 (sin 30° cos .4 - cos 30° sin J ) = cos .4 - v'3 sin A . 4 Q Q I 4. cos^=-; .-. sin^=v. cosZJ=-; .-. sin 2? = - . 5 o 5 o •. sin (^ + B) = sin ^ cos 2? + cos ^ sin B = 1 ; 24 cos {A- B) = cos A cos £ + sin -<4 sin B-~-. 34 FUNCTIONS OF COMPOUND ANGLES. [CHAP. 6. sec^=— ; .-. cosJ-^^^, sin^ = ^^. cosec B = -:; .". sin I> = ^ , cosB=^. 4 1 85 .-. sec (A + B) = zfz • — r' -"v— ~ oc • cos A cos B - sin A sin B do 7. sin 75<^ = sin (90"^ - 15°) = cos 15° = cos (45° - 30°) = cos 45° cos 30° + sin 45° sin 30° = '^ Ig ' 8. sin 15° = cos (90° - 15°) = cos 75° = cos (45° + 30°) /3- 1 = cos 45° cos 30° - sin 45° sin 30° = \-,^ . sin(a + ^) ^ sjnajcos^cos asin^^^^^ ^^^^^ cos a cos § cos a cos /3 siMa - ^) ^ sin^cos ^ - cosa^in ^ ^ ^^ ^^^^ ^ sin a sin /3 sm a sin /3 cosjo-^ ^ c_o_s acos /3 + sin a sin ^ ^ ^^^ . ^ ^^^ ^_ cos a sin ^ cos a sm j3 12. cos(^+iJ)cos(^-I>') = (cos A cos if - sin A sin ii) (cos A cos ^ + sin A sin Z^) = cos^ A cos- 1> - siu^ A sin^ i) = cos2^ (1 - sin^i?) - (1 - cos2 .-1) sin^L' = cos''' A - sin- B. 13. sin(yl+i5)sin(J-7>') = (sin A cos i? + cos A sin 7^) (sin A cos L' - cos A sin ^) = sin- A cos*'^ B - cos- A sin- J5 = (1 - cos- ^) C0S2 B - C0S2 A (1 - C082 B) = cos^'B - cos^ A . 14. cos (45°-^) -sin (45° + .•!)= — r{cos^ + sinJ - sin.-f - cos^}=0. 15. cos(45° + ^)-Hsin (.1 - 45°) =-y^(cosyl -sin^ + sin.-l -cosyl) = 0. 16. First side = cos A cos B + sin ^ sin B - sin ^ cos B - cos -«4 sin B - (cos .1 - sin A) cos B - (cos ^ - sin ^) sin B — (cos .1 - sin A ) (cos U - sin B). XI.] FUNCTIONS OF COMPOUND ANGLES. 35 17. First side = cos A cos B - sin A sin B + sin ,1 cos li - cds A sin B = (cos .-1 + sin A ) cos B - (cos A + sin A ) sin B = (coB A + shi A) (cosB - sin B). 18. First side = 2 (sin A cos 45° + cos A sin 45°) (sin A cos 45° - cos A sin 45°) - 2 X ^ (sin --1 + cos A) x -j (sin ^ - cos A) = sin-^ -cos^^. 19. First side = 2 . -y^ . (cos a - sin a) x -y^ (cos a + sin ct) = cos^ a - sin- a. 20. First side = 2 . -— . (cos a + sin a) x - (cos ^ - sin ^) V'S ^2 = cos a cos /3 + sin a cos yS - cos a sin j3 - sin a sin ^ = cos (a + /3) + sin (a - /3). 21. As in Ex. 10, it is easily shewn that the first term of the expression = tau ^- tan 7. Thus the first side = tan ^ - tan 7 + tan 7- tan a + tan a- tan^ = 0. EXAMPLES. XI. b. Page 100. 1. We have tan ^ = - , tan^ = -; 1 1 6 5 7 3, We have cot ^ = ;^ » cot B = -\ ... „, cot A cot J5 - 1 ^ .-. cot [A +B)= ——- -— = 0, ^ ' cot^ + cotJ5 ' 7_ 5 tan( - )- 2^ ^ tajj .^ t^jj 2? ~ 1 4 1 " 35 ' - , ,,^- ,, tan 45° + tan ^ 1 + tan^ 6. tan (4o + .J ) = i-_-^^-;j^5rs5rj = i-tanT^ " 36 CONVERSE USE OF THE ADIHTION FORMULAE. [CHAP. tan 45° - tan A 1 - tan A 6. tan{4o -.^)^ r+t^rr45--ta^ = 1 + tan^ ' 7. cot(^-^):^ 77-^t-^— 1- ^ ^ cot^-cot- 4 cot , cot ^ - 1 ^ , , cot ^ - 1 8. cot(^+^)= . TT cot ^ + 1 cot a +cot - ■i ,.r.^ .^n. tan 60° - tan 45^ J'6-l , ,.^ 9. tan 15" = tan (GO- - 4o») = r + tan 60' tan 45 ° =1^5 = ' ' ^"■ 10. cot 15°=cot (45° - 30°) = m^J = 2 + ^/3. 11 . cos (.4 + bTc ) =r cos ^ cos (i>* + C) - sin .1 sin (B + C) = cos yl cos 7> cos C - cos J sin J5 sin (7 - sin A sin B cos C - sin A cos />' sin T. sin ( ZC^B + C) :^ sin {A - B) cos C + cos {A - B) sin C = sin ^-1 cos B cos C - cos A sin J3 cos C + cos yi cos B sin C + sin A sin i> sin C. — tan M -- ii) - tan O 12. tan .4 - 2? - C = -:i — / /. m. — r^ ^^' ^ ' 1 + tan (^ - i^) tan C tan A - tan B 1 + tan A tan JS -tan C (tan u4 - tan /)) tan (7 1 + tan A tan B tan .4 - tan B - tan O - tan A tan J? tan C 13. cot(^+i>* + C) = 1 + tan A tan i) - tan B tan C/ + tan C tan .1 cot (yl+iJ) c ot C'-l "corC + cot'iA+B) (cot .-1 cot fi - 1) cot C ^ cot 7>' + cot A ^ cot A cot 1! - 1 cotC+ -- r-TT- — T cot i^ + COt /I cot A cot B cot C - cot .4 - cot B - cot C cot ii^t C + cot (foot A + cot .4 cot .B - 1 ' XI.] CONVERSE USE OF THE ADDITION FORMULAE. 37 EXAMPLES. XI. c. Page iui. 1 . First side = cos (A+B -B) = cos A . 2. First side = sin (3.1 - A) = sin 2A . 3. First side = cos (2a- a) = cos a. 4. First side = cos (30° + A+ 30° - A ) = cos 60° = J . 5. First side = sin (60°-^ + 30° + .-!)= sin 90°= 1. 6. First side = cos 2a cos a - sin 2a sin a — cos (2a + a) = cos 3a. 7. First side = tan (a - /3 + /3) = tan a. 8. First side = cot {a + ^ -a) = cot /3. 9. First side = tan {iA - 3.4) = tan A. 10 cot ^- cot '>^ — *'*^^^ cos2^ _ sin20cos^-cos25sin^ sin 6 sin 20 ~ sin 6 sin 2d _ sin (2(? - 9) sin ^ :C0sec2^. 11. l + tan2^tan ^ = l4 sin 6 sin 2^ sin d sin 2^ sin 2d sin ^ cos ^ sin 26 + sin ^ sin 20 cos 2 f^ cos ^ cos ^ cos 2^ cos {20 - 6) . = sec2^ cos cos '^6 oi'ti 0/3 oil-, a I 12. l+cot2^cot^ = cos cos 2^ sin 20 sin ^ + cos 20 cos ^ sin ^ sin 20 cos (2^ - 0) sin ^ sin 2^ = cosec2^ cot^. 13. First side = sin (2^ + ^) = sin 3^ = sin(4^-^) = sin 4^ cos (9— cos 45 sin (?. 14. First side = cos (4a + a) = cos 5a = cos (3a + 2a) = COB 3a cos 2a - sin 3a sin 2a. EXAMPLES. XI. d. Page 104. 1. Hereco8 2^ = 2cos2^-l= --. «7 38 FUNCTIONS OF 2 A. [chap. 3 4 3. We have sin A = - , and cosA=-; o o 24 .-. sin2^=2sin^ cos^ = — :. 2o ., 5. By Art. 124 , sin 2d=^^ *^" ^- = ~ , •^ 1 + tan^^ 25' _^ l-tan2<9 24 cos 26 = r = — , 1 + tan- d 25 7. See Example, Art. 122. sin 2A _ 2 sin ^ cos .4 _ l + cos2^ 2cos2^ - sin 2A 2 sin A cos ^ 1 - cos 2A 2 sin^ A 1 < 2 sin2 ^ 1-cosyl 2, A 10- sin:4- = -— 4 ^-*^"2- 2sii>-cos- ^4 , , 2 cos- — 1 + cos^ 2 A 11- -sin:4-^--^5 ■^^^^*2- 2 sin — cos — 2 2 2 1 12. 2cosec2arr - = . =secacoseca. sin 2a sin a cos a T o X , i. ^in^ a + cos^ a 1 ^ ^ 13. tan a + cot a= — ; — = -. = 2 cosec 2a. sin a cos a sin a cos a 14. cos-* a - sin^ a = (cos- a + sin- a) (cos= a - sin- a) = cos 2a. 1 c ^ + + co^" <* - sin2 a 2 cos 2a ^ . ^ 15. cot a -tan a = — : =^r-. = 2 cot 2a. sin a cos a 2 sin a cos a 16. By Art. 116, cot 2A ^^t^cot^-1 ^ cot^^ - 1 cot A + cot yl 2 cot .4 1 cot ^ - tan A _ tan .4 1 - tan- A ^ '• c^t A + tan ^ 1 ' = 1 + tan- A ^ ^°^ ^"^^ ,- — -. +tan A tan ,4 iQ 1 + cot- A _Bvo?A+ cosi^4 _ _1 ^°' 2cot^ ~ 2cot^Bin2T~2'sin'7c'o8 2~^°®^*^^'^' XI.] FUNCTIONS OF 2^. 39 cot=^ .:! -r- 1 _ 1 + tan- A _ ^^' c^t2T3i-i"tan2j[-^e°2J. «^ 1 + sec^ 1 , , . „^ 20. — . = -+l=cosd?+l_2cos2--. sec ^ sec ^ 2 Eec(9-1 , ^ ^ . nO 21. — ^ =l-cos^ = 2sin2-. sec ^ 2 „„ 2-sec2--= :i — = tan2a + sec2a. cos a - sin a cos-^ a - sm^ a cos 2a __ cota-1 cos a -sin a 1 - sin 2a ro -n i ^ ■./^^, -, 10. r i = -. — = ^ . [See Example 2, p. 107.1 cot a + 1 cos a + sin a cos 2a 11. 12. l + sin^_ (co^ + sin^y cos^ + sin^ 1 + tan^ cos^ .,e . ,,d ~ 9 ■ e~ ^ ^ d cos- - - sin-^ - cos - - sin - 1 - tan - ^ ^ Jit Jt dt J.J d . 6 6 , cos---sni-- cos - + sin- cot+l cos^ 2 2 2 2 2 1-sin^ { d . $\' e . e J ^ cos - - sm - cot - - 1 2 2 2 (^cos--sin-j 42 FUNCTIONS OF 2 A AND SA. [CHAP. A 1 - tan — .X . l-sin.4 *^ . /,-o ^\ 13. sec .4 - tan A = = —•^-~ = tan 4o° - - . cos A , A \ 2 J 1 + tan - ^ ^ , , 1 + tan- , i>i X . , sin .4 + 1 2 ^f.rro^ 14. tan ^ + sec ^ rr ----- = ^^ cot 45° - - cos A ^ ^ A \ 2 1 - tan — (I-) 1 2 sin^ (t+o) 1-cos.^.., . . ^ -ic c ^ -1 V^ 2/ V2 / 1 + sin^ 15. Second side = ———=^ —— = .^-—- 2C0S-^--r-j 1+C0S^-+^ 16. (2 cos ^ + 1) (2 cos ^ - 1) = 4 cos2 .4 - 1 = 2 cos 2.4 + 1. 2sin— cos — ^. , ., 2sm^ + - 2(p) + sin (26* + - ^ + 20) = sin (3^ - 0) + sin (0 + 30). 20. 2 cos (3^ + 0) sin (^ - 20) = sin (3(9 + + ^ - 20) - sin (36* + - ^ + 20) = sin (4^ - 0) - sin (2^ + 30) . 21. cos (60° + a) sin (60° -a)=^ {sin 120° - sin2a} = ^ (^"^ ~ ^"^ ^") ' EXAMPLES. XII. b. Page 114. For Examples 1—12, see Examples on page 113. cos a - cos 3a 2 sin 2a sin a , ^ "iq = ^ . = tan 2a. •^ sin 3a - sin a 2 cos 2a sm a ^ . 5a a 2 sm -— cos - Usm 2a + sm 3a 2 2 a =: , = cot --: . ■ cos 2a -COS 3a ^ . oa . a 'i 2 sm — sm ^ . 5^ . 3^ - 2 sin -TT- sm -^p „ _ co s4g-cos^ tan ^^- sin ^- sin 4^ ~ .^ o^j" 36^ 2* -2cos— sm ^^- 44 TRANSFORMATTOX OV l»ROJ)UCTS AXD SUMS. [CHAP. cos 2^ - cos 12^ _ 2 sin 7^ sin 5^ _ •*■"• "sin 12^ + sin 26 ~ 2 sin 76 cos 58 ~ ^" ' * 17. First side = 2 cos 60° sin A = 2x- sin A = sin ^ . 18. First side =: 2 cos 30° cos .4 — ^3 cos ^4. 19. First side = 2 sin - sin {- a) — -2 x -j-^ sin a = - ^^2 sin a. ^. , ., 2 cos a cos (a - 3i3) 20. First side =,^-^ — ^^=cota. 2 sin a cos (a - 3^) -r.- . •-, 2 sin (2^ - 0) sin (^ + 2A + 2 sin 3.4 cos 2.4 =:2sin3J (l + cos2i4) = 4 sin '6 A cos*'' A . I XII.] TRANSFORMATION OF PRODUCTS AND SUMS. 45 7 First side = ^^" ^iL"tA°°^ 3asin 2a _ siii2a (1 + 2 cos 3a) _ cos 2a + 2 cos 3a cos 2a ~ cos2a (1 + 2'^06 3a) ~ 8 First side = (^^" ° + sin oa) + (sin 2a + sin 4a) (cos a + cos oa) + (cos 2a + cos 4a) _ 2 sin 3a^cos 2a + 2 sin 3a cos a sin 3a ~ 2 cos 3a cos~2a + 2 cos 3a~cos~a "^ cos~3a~ **° ^'** 17- X •;, 2 cos 5(9 cos 2^ - 2 cos 3^ cos 2^ 9. r list side =- — ^— 1 — cot 2^ 2 cos 5^ sin 26 - 2 cos 3^ sin 26 ~ 10. First side = - (sin 5 A - sin .4) - ^ (sin 5 A - sin 3.4 ) = 2 (sin 3.-1 - sin .4) = cos 2A sin A. 11. First side =- (cosTA + eosSA) - - (cos 7^ + cos ^) = 2 (cos 3.-1 - cos ^) = - sin 2^ sin ^. 12. First side =- (sin 5(9 + sin 3^) --(sin 50+ sin ^) 1 , . = 2 (sin 3(9 - sin 0) = cos 20 sin 0. 13. cos 5^ -sin 25° = cos 5° -cos 65° = 2 sin 35° sin 30° = sin 35°. 14. sin 65° + cos 65° = sin 65° + sin 25° = 2 sin 45^ cos 20° = ^2 cos 20°. 1 5 . First side = 2 cos 60° cos 20° - cos 20° = cos 20° -cos 20° = 0. 16. First side = 2 cos 48° sin 30° + cos (180° - 48°) = cos48' -cos48° = 0. 1^. sin= 5.-1 - sin2 2 A = sin {5 A + 2^ ) sin {5A -2A) = sin 7.-1 sin 3.4 . 18. cos 2.-1 cos o.-l = 2 (cos 7.4 + cos 3.4) = ^ (2 cos^ "^ - 1 + 1 _ 2 sin^ --\ o7A . „SA = cos- 2 -sin2-^- 19. First side = 2 sin a cos (/3 + 7) + 2 sin a cos (^ - 7) = 2sina{cos(/S + 7) + cos(/3-7)} = 4 sin a cos /3 cos 7. H. E. T. K. 46 TRANSFORMATION OF PRODUCTS AND SUMS. [CHAP. 20 First side = 2 ein 7 sin (a - j9) + 2 sin (a + ^) sin 7 = 2 Bin 7 {sin (a-/3) + sin (a 4-/3)} = 4 sin a cos /9 sin 7. 21. First side = 2 sin (a + /3) cos (a-/3)-2 sin (a + /3)cos(a + /3-f27) = 2 sin (a + /3) { cos {a- ft)- cos (a + jS -f 27) = 4 sin (a -f /3) sin {ft + 7) sin (7 + a). 22. Fn-st side = 2 cos — r-^ cos -_- + 2 cos ^r~~ ^^^ ~o^ a+ft/ a- ft = 2 cos — o"^ ( C08--~ + C0S a + /3+27 \ 2 j . ^ + 7 7 + a a + /3 — 4 cos ----' cos - cos - . 23. First side = 2 sin A {cos 2^ - cos 120=} = 2 sin A jcos 2^ +-1 = 2 sin .i cos 2^ + sin ^ = sin SA -smA+ sin A = sin 3.4. 24. First side= 2 cos 6 Icob J^ + cos 2^1 = 2 cos ^ ( - ^ + cos 26 ) = - cos + 2 cos 6 cos 2^ = - cos d + cos Sd + cos ^ = C08 3^. 27r 25. First side = cos ^ + 2 cos — - cos 6 = cos ^-cos ^ = 0. 26. t'irst sidc=- {1 + cos 2.4 + 1 + cos 2 (60^ + ^)+ 1 + cos2 (fiO° - .<)} = 2 {3 -f cos 2.4 + 2 cos 120"* cos 2A } 1 3 = - {3 + cos 2.4 - cos 2A\ = '-. £1 £1 27. First side= - {3 - cos 2.4 - cos2 (120° + .4) - cos2 (120= - A)\ = - {3 - cos 2.4 - 2 cos 240° cos 2A } It 1 3 = - {3 - cos 2^ -f cos 2.4 } = r . XII.] TRANSFORMATION OF PRODUCTS AND SUMS. 47 28. cos 20^ cos 40=^ cos 80^ = ,^ cos 20° (cos 120° + cos 40^) = 2 cos 20° ^ - - 4- 2 cos2 20°") = T (4 cos'' 20° - 3 cos 20°) =^ cos (3a^ = - . * 4 8 29. sin 20° sin 40^ sin 80°= ^ ^in 20= {cos 40° - cos 120°} = -sin20= -i^-2sin2 20=l -^sin60==^^. 4 8 EXAMPLES. XII. d. Page 119. 1. First sider=2cos(.l + Z?)sin{^-i5) + 2sinCcosC = - 2 cos C sin (A -B) + 2 sin {A + B) cos G = 2 cos C { sin {A +B)- sin {A - B) } —- 4 cos A sin B cos C. 2. First side ^ 2 cos (.1 + B) sin {A -B)-2 sin C cos C = - 2 cos C {sin {A -B) + sin (A + B) } = - 4 sin .1 cos B cos C. 3. First side := 2 sin '-^— cos '-^— + 2 sin - cos - -^ J 2 2 o (? ( ^1-i? A+B) = 2 cos - jcos — ^ + cos -^ [ , .4 £ C = 4 cos -r cos — cos — . 2 2 2 4. First side = 2 sin "^J^ cos ^^^ - 2 sin - cos - 2 2 2 2 o CI A-B A+B) = 2 cos - - cos -^— - cos -^ - . . A . B C = 4 sni - sm -^coq-. 5. First side = 2 sin -t? gin ?^ + 2 cos^ - - 1 o C I . B-A . B + A] = 2 cos - |sm —2- + sin -^| - 1 = 4 cos ^ sin ^ cos - - 1. 4—2 48 RELATIONS WHEN A -\- B + G = 1S0\ [CHAP. A . B . C 4 cos 2 sin - sm 2 ^ c 6. First side = ^:= tan -tan-. 4 cos — cos — cos — 7, We have tan tan — =1; (tan^ + tan|)tan^^ A B 1 - tan — tan — Multiply up and transpose and we obtain the required result. „ ^A . . B + C . B-C 2co8^— -f 2 sin — - — sin — ^^ 8. First side = — . ,^ ^ . B+C . B-C 2cos2---2 8in — - — sin- ^ £i Zi . B+G . B-G ^ . B G s,„ .__ + sin -2- 2sm-co8- ^ ^. - ■ B + C . B-C - ^ B . C -^^2^^^2- sm sin — - — 2 cos — sin — ^ ^ II £i 9. First side = 2 cos (.4 + B) cos {A-B) + '2 cos^ C 4- 4 cos A cos B cos G = - 2 cos G [cos (J - -B) + cos (.4 + i?)} + 4 cos .4 cos B cos C = - 4 cos A cos 5 cos C + 4cos,-l cosB cos C = 0. 10. We have cot {A+B)= cot (180° - C) = - cot C ; cot A cot B-\ ^ ^ .'. =: - cot C ; cot A + cot B whence by multiplying up and transposing we obtain the required result. _,. ^ ., 8in{i? + C) sin{C + .4) sin (.4+ J?) 11. First side = . V, . - ' . —. — \— — • -■ — ^.- r— /> sin B sin G sin C sin .-1 sin A sin B sin A sin B sin C . -., ^ — -T-TT-.— . o v^- • 0-.% = cosec .4 cosec B cosec (7. sm-^ sin-i?8in- C 12. First Bide = ir {3 + cos 2/( +cos 2i? + cos 2C + 4 cos.-l cos ii cos G] It = ^ (3 - 1), by Example 9, a = 1. XII.] RELATIONS WHEN A + B -[- C = 180^ 13. First side = -^ (3 - cos A - cos B - cos C) 49 = 2\^^-l-'*sin-sm-sm-j [See Ex. 3, p. 119.] = 1 - J sin — sin — sin — . ^ 2 Z 14. First side = -{2 + 2cos2 2^+cos47i + eos4C} = 2{2 + 2cos2 2.-{+2cos2(l? + C)co8 2(ii-C)} = 1 + cos 2A { cos 2{B + C)+ cos 2 (ii - C) \ = 1 + 2 cos 2A cos 2B cos 2C. 15. T^--^--^-:^^ cot/J + cotC . cotC + cot.^ , cot.-l + cotJJ ^ , 1 ■ 1 1 1_ 1^ cot B cot C cot C cot J cot A "^ cot B = coti?cot C + cot(7cot^+cot.4 cotB = 1. [See Ex. 10.] 16. ^^--^^^— tan .-t tan i? tan C IGcos^^cos-- cos2 sin ^ sin B sin C "cosJco8J5cosC • A ^B Tc lb cos-^ — cos^ — cos- — ^ Jt ^ o-^ A ^ . B B . C C 2 sin 2 co^ 2^ • "-^ sin - cos - . 2 sin - cos - cos A cos BcoaC. 16 cos^ ~ cos'-* - cos^ - 2 2 2 , A^ B ^ C tan - tan - tan - "2 cos.yl cos iicosC* EXAMPLES. XII. e. Page 121. 1. First side = ,^ {sin 2a - sin 2/S + sin 2p - sin 27 + sin 2-y - sin 25 + sin 25 - sin 2a } = 0. 2. First side - cot 7 - cot /9 + cot a - cot 7 + cot /S - cot a = 0. [See Ex. XI. a, 10.] 50 MISCELLANEOUS IDENTITIES. [CHAP. ^ . a + 8 a-B . . a + /S a + /3 2 sin — — ^ cos -,^ ^ + 2 sm ^ "^ cos - — 3. First side = ^^ a-rf „ . a + /i TT^ 2 sin -^^ cos — 2^ - 2 sin — ^ '^ cos -^ a-B a + f3 cos /^ + cos „ A> 2 2 a 8 — ^ = cot - cot •- . a-d a+B 2 2 cos — -^ - cos —7^ 2 2 4. Fh-st side = sin a (cos/3 cos 7 - sin /3 sin 7) - sin /3 (cos a cos 7 - sin a sin y) =:C0S7 (sin a cos /3 - cos a sin /3) = cos 7 sin (a-/3). 5. First side = co8a (cos ^3 cos 7 - sin /3 sin 7) - cos )3 (cos a 00s 7 -sin a sin 7) = sin 7 (sin a cos ^8 - cos a sin /3) = sin 7 sin (a - p). 6. First side = cos .-I cos 2.-1 + sin A sin 2A - (sin A cos 2^ + cos A sin 2.4 ) = cos(2,-I -.4) -sin (2.4+^) = cos^ -sin 3^. . ^. a(l-tan2e) 2btan^ ., , ,.,,-, 7. acos2«+tsm2()=-L___i + ^_^-_^, [A,t. 124.] — ^ ' _( == a . =a. ~ a" + b'' a- + 6- a' + b- 2 tan A 1 - tan- A (1 + tan .4 )- - 2 tan- A 8. sin 2.4 + cos 2.4 = ^^ ^^-^.^4 + 1 ^.Ta^tQ " iTtan-^^":! ' ^, 4tan^(l-tan-^) 9. sin 4^ = 2 sm 2^ cos 2A = — r-r-— . — s- , ... • ** (l + tan-j4)- 10. We have /t 4- -R = 45°; ^ , ^ „. tan .4 + tan J7 /. tan (.4 + 1^) = , — r- — ^ = 1; ' l-tan.4tani> .-. tan A + tan B + tan A tan B — 1 . .-. (1 + tan -4) (1 + tan B) = 1 + tan .4 + tan B + tan A tan B = 2. ., cos ( 15^ - .4 ) cos ( 15° + .4 ) -r sin (15° - .4 ) sin (15° + A) 11. First side = -. — -r-p- tt — -^ ,1 ^o , — n ■'•■'■' sin (15 -^) cos(lu +.4) 2 COS 2A 4 cos 2.4 sin 30° - sin 2.4 1-2 sin 2 A ' ^^ ^. , ., cos'-' (15° + .4) + sin-^ (15° + A) 2 12. Fa«tside = — ^^--^.-^— ^y^^^^^o^^) -sin(3d° + 2.4) 2 4 ~ 1 J'i . ^ . ~ cos 2.4 + J'i sin 2.4 ' cos 2/1 +^^ sin 2.4 ^ 2 2 XII.] MISCELLANEOUS IDENTITIES. 51 ,^ ,.. ^ ., sin M + 30^) siu U - 30°) 13. t irst side = , . , .,-c jt rj— l>Ao-^ cos {A + 30") cos {A - 30°) _ cos 60° - cos 2.4 _ 1 - 2 cos 2.1 ~ cos 60° + cos 24 ~ 1 + 2 cos 2i ' 14. First side = (4 cos^ A - 1) (2 cos '2A - 1) = (2 cos 2,4 + 1) (2 cos 2J 1) = 4 cos"^ 2,4 -1 = 2 cos 4,4 + 1. „, , , , , tan ^ + tan + tan li' - tan ^ tan tan i/* 15. We have tan {d + (^ + \p)= —— — --^ — , — , ^ ^ , , ^ 1 - tan d ta,n

^ ,—- sin 10° cos 10° sin 20" 4 sin (30° -10°) sin 20° :4. / 5-1 1 v/5 + 1 (ii) Second side = sin 18° + sin 30° = ^—, — + - = ^ — = sin 54°. » ' J. V 4. 4 2 Xir.] EASY MISCELLANEOUS EXAMPLES. 52 C 21. Here sin = sin 5, cos C=- cos 7?. .-. Second side= 2cos=2? = 2 (l-sin-i?)=2(l - sin7?sinC). o^. l + co82i? l + cos2(7 22. cos- 7? + cos2 C= ^ + z = 1 + - (cos 2B + cos 2C) - 1 + cos (i?+ C) cos [B - C) — l-cosAcos{B-C), for cos {B + C)= -cos A, = 1 + cos^^ - cos A {cos A + cos {B — C)} = 1 + cos- A - cos A {cos {B-C)- cos {B + C) } = 1 + cos-^ - 2 sin B sin C cos ^. 23. As in Ex. 22, cos"B + cos2 (7 = 1 + cos (i5 + C) cos (5 - C) = l + cos^ cos(B- C), for co3{B+G) = cosA, = 1 + COS' A + cos A {cos (i? - (7) - cos {B + C) } = 1 + COS" A + 2 sin B sin C cos ^. 24. cos2^ cos 2B - cos'B cos 2 A = cos- A {cos"B - sin-B) -cos^B{cos-A-sin-A) = cos-B sin- A - cos- A sin-B == (1 - sin275) sin2^ - (1 - sin2^) sin27? = sin^A-sin^B; whence, by transposition, we have the required result. oc + -no . .no sin 50° sin 40° 25. tano0°-tan40°= — _ — - cos 50° cos -40° _ sin 50° cos 40° - cos 50° sin 40° "~ cosT0° cos 40° _ sin (50°- 40°) _ sin 10° "cos 50° cos 40° ~" cos 50'^ sin50° _2 sin 10° _ 2 sin 10° -sin 100"^- "cos 10° -2*^^1<^- 52 D SOLUTION OF TRIANGLES. [CHAP. 26. Here tan<> = 2; .-. sin2. = j^= A. =^= -8 „, l-tan2^ 1-4 3 and cos 2^=, —7 — rr^ = q — r=-r=-*o- 1 + tan-^ 1 + 4: 5 Again, from the Tables, cot^= -5 gives ^=63°26'. .-. 2^= 126° 52'. A9in2^ = sin(180°-126°52') = sin53°8' = '8, from the Tables. cos 2^= -cos 53° 8'= --6, from the Tables. 27. If tan a= -362, the equation may be ^^Titten tanacos^ + sin^ = l, or sin a cos ^ + cos a sin ^ = cos a. .'. sin(a + 0) = cosa = sin (90°-a). Now from the Tables, a = 19°54'. .-. sin (19°54' + d) = sin (90° - 19° 54') = sin 70° 6'. .-. 19°54' + ^ = 70°6', or 180° -70° 6'; whence ^ = 50° 12', or 90°. EXAMPLES. XIII. a. Page 128. a2 + &2_c2 225 + 49-169 105 1 ^ ^_ 1- ^°^^=-2air-=— 210— =2l0 = 2' •'• ^==^^ ' 62 + c2-a2 9 + 25-49 15 1 . ,.„ 2- ^^^^= 2bc = 30 ==-30=^-2> •-^"^^^- 3- ^°'^- 267^- 5a2V3 - 2 ' •• ^-^"• Also a = c; .-. C=^ = 30°; hence .8=180°-^ - = 120°. , h'^ + c^'-a'^ 961+98-625 1 ' ^^^ ^- ^"^^ = -W- = -- 434 V2— = 'J2 ' •■• ^ =^^ • XIII.] SOLUTION OF TRIANGLES. 53 5. Leta = 2, 6 = 2f,c = 3J. , 64 100 4 .^ Then cosC = l^-J- = ; /. C = 90°. "3 jr- + c"-a"' 4 + 6-(4 + 2^3)_3-^/3_V3-l_ . .n.o. ^ c2 + a2-6-_6 + 4-t2V3-4_ _^+v/3^ _J. . rj^.^o. i °°"^= 2ac -"276l73Tir"v''-2(3 + v/3rs/2' ^ .-. (7 = 180° -75° -45° = 60°. &2 + e2-a2 4 + 4-2V3-2_ 3-^3 _V3. . ._ooo. 7. cosJ = — 267"= 4(^3-1)'^ -2 (73-1)- 2 ' •• ^"'^ ' c2 + a2-&2_4-2V3 + 2-4_ 1-V3^^_JL^ ^°'^ = ~^^^ T72 (^/3"'- 1) - ^/2 (V3 - 1) v/2 ' .-. i? = 135°; .-. a = 180° -30° -135° = 15°. 42 + 52-72 16 + 25-49 8 1_ ^^,70008'. 9. cos C = _ — . — ^- = j^ = ~ 77^ = ~ c = - cos * ^ ^^ ' ^' 2x4x5 40 40 5 .-. C=180°-78°28'=101°32'. 10. c2 = a2+ 6" - 2a6 cos C=4 + 4 + 2 V3 -2. 2 (v^3 + l)- = 6. 11. ;;2 = c- + a2-2cacos5 = 9 + 25-2. 3.5^ - -j =9 + 25 + 15 = 49; * 12. a-^b'^ + c"- 2bc cos .4 = 49 + 36 - 2 . 7 . 6 X -2501 = 49 + 36-21-0042 = 64, approx. ; whence a = 8. 13. a2^6- + c2-2&cco8^=64 + r21-2.8.1l( - ^^j = 64 + 121 + 11 = 196; .-. a = 14. 54 SOLUTION OF TRIANGLES. [chap. 14. Zy2 = c2 + a=-2cacos^ = 9 + 49-2.3.7{--5476) = 81, approx.; A i = 9. 15. &2^c= + a2-2cacosi5 = 48- 2-4^/3 + 24 -2.2 ^6. 2^/3(^/3-1).^^-^ = 48-24^3 + 24-4.3.(4 -2^3) = 24; .-. & = 2 ;^/6 ; whence A^B^ 75°, and C = 30^ 16. a2 = i2 + c2-26ccos^=4 + 6 + 2^5-2.2(V5 + l)^^^^ = 4 + 6 + 2;^o-4 = 6 + 2^/5; .-. a=:V5 + l; whence C = .4 = 72°, andP = 36^ 17. C=:75°; whence a=c. 6 sin .4 V8(V3 + 1)2 Also a = sin jB 2^2 = 2j3 + 2. ._ ftsinC ,^ ^3-1 2 ,- . Also A = 105° ; whence a = ^^^ = ^6 . ^ sin 5 ^ v^3 + l 2 _ 2./2 •.73-^^ + -^' 19. C=30°; whence a = ^4^= ^^ = 2; sin G J2 a sin B sin^ -^''^^-J^ = ^'^^^' on Hcrc" = "^^^-"^"^^°-^^^ + ^/^ _V3 + 1 '^"" a sin^ sin 45° ~ 2^/2 /v/2~ 2 * 21. sin A = a sin C v/3_l c 2^ * 2 2' •. i = aco8C + ccos^ = 2 (^ - -") + 2^/3 . ^- = 3- 1 = 2. J XIII.] SOLUTION OF TRIANGLES. 55 _- . . UBinB 3 v/3 1 22. sm.4= ^- = 3-^.V = 2; c = a cos B + b cos ^ = 3 . - + 3 ^/B . -^ = 6. 1 . . /o V-r 2 23. We have {h + c)"^ - «- = 36c ; 24. cos .-1 = 6- + c--a- 1 .•. — jr^ = X ; whence -4 = bO . 2bc 2 h'^ + c'-a^ 12 + G-(12 + 6V3) 26c 2 . 2 v^3 . ^6 6(1-^/3) V3-1 „.o = -127^-- 2^2^=-^°^^'' .-. .1 = 105°; similarly C= 30°. •• '^°^ = -^- = V6 = 72' ^-l^ence5 = 45°. 25. The sides are proportional to ^^3 + 1, 4^3 - 1, ^JQ ; .•. cos .4 = = ^ ^- -^ — - 2hc 2 V6 (s/3 - 1) 6-4^3 _ 2^/3U/3-2)^ J^ + l 2V6(v/3-l) 2V3.v*<2(^/3-l) ^/3 + l : 2^= -cos 75°; -•. ^ = 105°. . ^ c . . V6 v/3 + 1 ^3 sin G= - sin ^ = , , — - . -^^r— ,^ = V ; a ^/3 + l 2^2 2 ' .-. C=:60°, andB = 15°. 26. Here6 = ^^^^ c = ^^^, ^=60°; • ,,o_ (v^6 + V2)" + (V6-V2 )2 2.4 1_12_3 "16 16 •2~16~4' . ^ c . ^ ^/6-./2 2 V3 ^3-1 smC=-8m^ = V_N__.^ = '^^^. whence C = 15°, and B = 105°. 56 THE AMBIGUOUS CASE. [CHAP. EXAMPLES. XIII. b. Page 132. .'. 7? = 60° or 120"^; and since a<^h, both these vahies are admissible. Hence C ^ 00'= or 30^. rtsinC7 c = — . — - =2 or 1, on reduction, sin A .-. i( = 60° or 120°; and since c, or 3-v/3, on reduction. 7. sin J - ^ sin C _ --^^^.^ . .^-^2 " 272" " .*. .4=75° or 105°; and since c<.a, both values are admissible. Hence iJ=^y0°or60°. c sin B b = —. — - =2J6, or 3v'2, on reduction, sm (7 ^ ^ XIII.] RELATIONS BETWEEN THE SIDES AND ANGLES. 57 8. sin li =- sm A = --^ • — -3 — = 1- a 4 4 .-. ^ = 90°, and there is no ambiguity. t- = b--(r = {b + a) [b -a) = [8 + ^'80) ^/8U = 4 (2 + Jo) . 4V5 = 16^/5 (2 + ^/5 1 .-. c = 4 Jo + 2^5. .•. ^-1 = 60° or 120°; but neither of these values is admissible as in each case the sum of the angles would be greater than 180°. Thus the triangle is impossible. EXAMPLES. XIII. c. Page 134. 1. Follows at once from Art. 137. 2. The first side = />- -f c- - a^ + a^ + c'- - b'^ + a- + b"'-c^ = ir + b^ + c^. 3. The first side= =b^-c^. 4. The first side = (6 cos A-\-a cos B) + (c cos B^b cos C) + (c cos A-\-a cos C) = c-\-a-^b. 5. The first side = a (1 - cos C) + c (1 - cos A) = a + c- (a cos G-\-c cos^) = a-vc-b. « mi 1-1 acos£ cosJ5 6. The second side = ^ -,= -. a cos G cos G 7. The second side = 7= - ^=tan^, c cos A c cos A 8. Put^-=-; — -= . — ~= . „; then sm A sm B sm G the first side = k (sin J5 + sin G) sin — = 2k sm — - — . cos — - — . sin — ^, . A A B-G = 2k sin - cos — . cos — ^r— 2 2 2 B-G = k sin A . cos B-C = a cos — ~ — H. E, T, K, 58 KELATKJNS BETWEEN THE SIDES AND ANGLES. [('HAP. , , k (sin ^ 4- sin 5) . „ (7 9. The first side = — ^ ,- . - „ ' Bin2 -- ^ . A+B A-B 2sm^2-cos-2- ^ = ,. C C ""-2 2 sin - cos — o a A-B . G = 008^2- sin - A-B A+B = C08— g- cos ^— cos A + cos B = 2 * 10. Thefirstside=A;{sin^ sin{2?-C) + + } = h {sin {B + C) sin {B-C} + + } = fc{sin-^^-sin2C+ + }=0. , ., sin2^-sin2J5 11 The second siae= . ., ^ ^^' sm- C _ sin (J + B) sin (A - B) _ sin { A - B) ~ sin (A +B) sin [A +B)'~ sin [A + B) ' sin^ A - sin2 B sin {A + B) sin {A - B) 12. The second side = ^.^, ^, _ ^.^,3^ = sin (C + .J) sinic" J) sin C sin (^ - B) _ c sin (^ - i?) "" imFsin (C^:4) "" ftsiM^" - -"^ ) * EXAMPLES. XIII. d. Page 13(5. 1. Let ABC be the triangle, in which i?= C = 2.1, and a = 2 ; then 5 + (; + ^ = 5.-l=:180'^; .-. .4=30°, B = C = 1T- a . . a sin 2.4 ^ , , J^ -H 1 ,- . 1 and b = c= . — , sin i^ = -^ — — = 2a cos J =4. =^.) + l. sin A sin ^ -t 2. A=lSO°-B-C = m'': If .4L> be the perpendicular, then c~A 1) cosec 45" = 3 ^/2, ^ = ^l>cosec75° = J^ = 3(V0-v/2) a = Bi> + DC = .4i>cot45° + /iI>cot7o^ = 3 4-3(2-V:i) = 9-3x/3. XIII. ] RELATIONS IJKTWEEN THE SIDES AND AXCI.KS. 50 _ ^^+ c- - a- ^ 28-16^3 ; 24 - 12 y/S - 4 _ 48-28^3 ^- ''''^- ~ "26c" ~ '" 4^2(9^5^3) ~ A'^'-o'^JS) = ^1^ 7\/''^)( 9 + 'V3 ) _ 3 - 3 ^/3 _ jSj-l ' J2.Q ""6^2 ~ 2^2 ' .-. .4 = 180^ -75°= 105"^. .ir^r-'^inA-hJlz^/^ v/3 + 1_n/3 (v /3-1)(V3 + 1)_V3. and iJ = 180° -105° -60°:= 15". 4. We have 6-a = 2, ab = 4; .-. h + a = 2 ^5, rejecting the negative sign, .-. a = ^5-1, 6 = ^5 + 1; . ^, 6sin^ J5 + 1 J5-1 J5 + 1 ■'■ ^^^^--^7-=V5^r 4- = -^ ' .-. B= 54° or 126°; .-. C = 108°or 36°. _ . csini^ -__ 1 1 V^ 5. Bmc=-^ = 150..^.^^-^ = -2-' .'. C' = 60° or 120°; both valuea being admissible since &' 2siu f7;+^-jcos- ■*•^• a sin J sin^ , . -1 A " siii.^ 2sin — cos — 2 2 .A . /A ^\ (?> + c)sin- = asin I 2 +^ ) B+ — ] cos — _ _ ^/ '^ 12- i;:^^ - sin ii + sin C ~ "Sn^'+'si^ " ., . B + C B - C 2 2 in(j? + g)cosg cos - cos — -— XIII.] RELATKXNS BETWEEN THE SIDES AND ANGLES. 61 13 First side = ^ " cos (-^ - -^) cos M + i J) ^ 1 - ( cos^-t - sin ^ /J) 1 - cos (A - C) cos {A + C) 1 - (cos2 A - sin-^ C) - si_^^_£+sin^ B _ a2 + b- sm-J + sm-C fH + c-^ 14. ^Ve have c* - 2 (r/2 + 6=) ^2 ^. „4 + a"b- + ft-' = ; .-. (c2 - a2 + flV; + ft2) (c--! - a- -ab + b-) = 0; .: c' = a^ + ab + b^, or u'-ab + b-'. But c2=a2 + 62 + 2«6cosC; .-. 2 cos (7=1, or - 1 ; (7 = 60°, or 120^ 15. See figure of the Ambiguous Case on page 131. (1) Ci-c._^ = B^B.2 = 2B^D = 2acosB^. /ON (7j - C -r, ^^ CD bsinA (2) cos -1— _- = cos B^CD=~~j = . (3) Cj, To are the roots of the quadratic c- - 2b cos .-1.0 + 6- - a- = ; [Art. 150.] .•. Ci + Co = 26cos.l; CjCo=//--C = DJ =^-i±^: also DB.^'-'^ ; 2 ^2 .■,cos^B.CD = ^''': = ^h±^^' . ' (75o2 2(Ci2 + C22)' .-. COS B.CB.. = 2 cos2 B.^CD -1= ^^-i±^-^' _ i ^ .^^^2 62 RELATIONS BETWEEN THE SIDES AND ANGLES. [CHAP. 17. From the given condition we have sin C cos A+Q, cos C sin C = sin B cos J + 2 cos B sin B, or cos J (sin C- sin £) = sin 25 -sin 2(7. This easily reduces to , . B-C B+C ^ ^ . B-C B-C cos A sm — -- cos — ^r — = 2 cos A sin — ^r — cos — ^ — . Now cos — — cannot equal 2 cos -— ^r — ; hence we must have cos ^ = 0, which gives A = 90° ; B — C or sin — ^r — = 0, in which case JB = C. 18. Since a, &, r are in A. P. ; we have a -& = & -c; .-. sin ^ - sin B = sin B - sin G ; „ . A-B A+B ^ . B-C B + C .: 2 8m — - — cos — ^r — = 2sm — — — cos - _ — ; 2 2 2 2 . A-B . B-C sm - — sm — - — ■? » ■ . A . B . B . C sm - sm - sm - sm - .•. cot— -cot — = cot — -cot -^ ; A Z ^ ^ That is cot — , cot — , cot — are in A. P. 19. Let k = - — r = -^-p = -^-^ ; then smA smB sm C ^ ^ . , ;c2 gin ^ sin ^ + (7 sin />' - C first side = -. — =: : — f- + sm jD + sin C 2 rsin^Msiu^B-sinS 0) n " L sin7?4-sin(7 "^ '^ J = k^[sinA (sin Z? - sin C) + -f ]=0. MISCELLANEOUS EXAMPLES. D. Page 138. 1. (1) tan2^cot^-l = ^ ^ ^^^l + tan^^^^^^^^^ 1 - tan^ 1 - tan^ 6 ,_, . .^ sin a sin ^ - cos a cos ^ (2) sm a- cot ^cos a = ; — = -cosec^cos(a + ^). XIII.] MISCELLANEOUS EXAMPLES. D. 63 2. c- = a'^ + b-- 2ab cos C = 48^ + S5^ - 48 x 35 = 132 + 48x35 = 1849; /. c=43. -y(¥ 3. Here tan a= ^ / ( :^- ) - 1 =y ; //17\" 8 also ^^''^=\/\i5) -l = i5 = cota; .-. a + ^ = 90°; :. tan (a + /3) = 00 , and cosec (a + jS) = l. , ™, .2 sin 8a cos 15a cos loa 4. The expression = „ . ^ „- = ^ = - 1, 2 sin 8a cos 6a cos ba since 15a = tt - 6a. 5. Ffrst side = - (sin 3^ - sin 6 + sin 59 - sin '60 + siu 76 - sin 5^) = 5 (sin Id - sin (?) = sin 3^ cos 4^. 6. a2^2 + 4 + 2V3 -2 (^3 + l) = 4; whence a = 2. . . . ^ b sin A J2 1 Again ^'»* = -^r- = 2^/2 = 2 = .". B = 30', or 150*^; but the latter value is inadmissible since c is the greatest side. Therefore G = 180° - (45° + 30°) = 105°. 7. (1) 2sin2 36° = l-cos72° = I-sml8= 10 = 1 - 'J^ = ^-'^■ = Jo sin 18^ 4 4 (2) 4 sin 36° cos 18° = 2 (sin 54° + sin 18°) _^5 + l J5-l_ sin 3a cos 3a sin 3a cos a + cos 3a sin a 2 sin 4a 8. — . i = -. = — .- „ =4 cos 2a. sin a cos a sin a cos a sin 2a y. cos^- 2^^ - - _ — , .-. ^ = 30°. .-. B = C = ]: (180° -30°) = 75°. 64 MISCELLANEOUS EXAMPLES. D. [CHAP. 1 sin a 10. (1) Each expression easily reduces to -. — - . sin ocL „ 3a CL -. „ 3a , (2) cos a + cos 2a + cos 3a = 2 cos -- cos - + 2 cos^ iT' ^ 3a / a 3a\ = 2 cos — ( cos - + cos j - 1 a 3a , = 4: cos a cos ^ cos—- - 1. 2 2 11. (1) First side = 26= sin C cos C + 2c2 sin ii cos L' = 26 sin G (6 cos C-\-c cos B) = 2a6sin C = 26csin^. (2) First side = h (sin A sin B'^ + + ) = fc(sinJ5 + CsinB-6' + + ) = fc(sin2B-8in2C+ + ) = 0. ( 12. tan(^+i3) = tan(360°-"C + I)); tan A + tan B tan C + tan T> \ 1 - tan A tan B~ 1 - tan G tan 7) ' .•. tan A + tan JB + tan G + tan D = tan G tan D (tan J + tan B) + tan ^ tan B (tan C + tan D) ; or tan A + tan 2? + tan G + tan I> \ — tan .^ tan B tan {7 tan D (cot A + cot 7) + cot C + cot D). EXAMPLES. XIV. a. Page 145. For Examples 1 — 3 see Arts. 151, 152. For Examples 4—7 see Arts. 162, 163. 8. log 768 = log (28 X 3) rr 8 log 2 + log 3 = 2 -8803613. 9. log 2352 = log (24 X 3 X T^j =4 log 2 + log 3 + 2 log 7 = 3-3714373. 10. log 35-28 = log (^^^^^'j = 3 log 2 + 2 (log 3 + log 7-1) = -90309 + 2 (-4771213 + -845098 - 1) = -90309 + -6444386 = 1 -5475286. 11. log x/6804 = ^ log (22 X 35 X 7) = I (2 log 2 + 5 log 8 + log 7) = ^ (-60206 + 2-3856065 + -845098) At = 1-9163822. XIV.] LOGARITHMS. 65 12. lug i/-00lG2 = I log (^') = I (log 2 + 4 log 3 - 5) = 1 (3*2095152) = 1 •411903. no 1 r^c.',n ^ 217-21 , 196 , V 13. log -0217 = log- ^^^^^ =log..— =log 9000 ~ "9000 "152x10 = 2 (log 7 -log 3 -log 5) -1 = 2 (-8450980 - 1-1760913) - 1 = 2-3380134. 14. log cos 60^ - log ( - j = - log 2 =. - -30103 = I 15. log sin3 60'--3 log h^ = |log3 - 31og 2 = -7156819 - -90309 = I '8125919. 16. log L sin a cos a 2 sin a cos o 2 cos 2a = . ^ = 2 cot 2a. sin 2a 8. c^ = a-' -i- Z)2 _ 2a6 cos C = a^ -f (4 - 2 ^3) a' - a^ (^/3 _ l ) ^/3 ^ (2 _ ^/3) ^2, .-. a== (2 + ^3) C-, and sin^ A = (2 + ^/3) sin^ C ; . ., , 2 + V3 4 + 2^3 4 s • . V3+1 .. sin ^=^5^ ^rr- . 2^2 Hence ^ = 75^, or 105°, and the latter value must be taken, as J =75° would make the triangle isosceles. Hence also 2^ = 45^. 2 tan 2a . x-^ ^ ^ « 2 tan a 9. tan 4a = ^ — - — ~^- ; substitute tan 2a = = .r- . 1 - tan- 2a 1 - tan2 a MISCELLANEOUS EXAMPLES. E. [chap. 10. (1) First side = a2(l-2sin2 2J) + 6-^(l-2sin2^) = a2 + ^2 _ 2(i2 sin2 B - 2b^ 8in2 ^ = a2 + 62 _ 4^2 gin'j j> = a2 + i;2 - 4a sin B . b sin A=a^ + b^- 4ab sin A sin B. (2) First side = 2;^c (1+ cos ^) + + = (26c + &2 + c2-a2) + + = {a + b + c)-. 11. From the equation c* - 2c2 (a^ + b-) + {a* + b*)=^0 we have {c2 - (a2 + ab V2 + &2) } {c2 - (a2 - ai ^2 + '>') } = 0- Equating the two factors separately to zero, we get 1 whence 12. We have a^ + b^-c ^ _ _1_ "2^6 ~x/2°' v/2' C = 45° or 135°. S(A+B) S{A-B) ^^ , 2 cos — ^^^^ — - cos ^ ^ ■' + cos 3C = 1 ; 2 cos 2 3(^-1?) or /'270° - -^ j cos -^"^-^' + cos 3C= 1 ; ^ . 3C 3(^-5) , ^ . ,3C ^ - 2 sin — cos — ^— ^r + 1-2 sin^ tt "= ^ 5 - 2 sm -- [^c 2 3 (^ - !>•) + sm- UO; 3(^-Ii) ,, . 3Cr '6{A- 2 sin .^ l^cos cos 2 J 3(^ + B) ^]-^ , . 3^ . 3ii . 3C ,^ 4 sm — sin --- sin -— = 0. 2 2 2 Since A, B, G are the angles of a triangle we must have one of the angles — - , -^ , or - - equal to 180'^. That is, one of the angles of the triangle must be 120°. EXAMPLES. XV. a. Page 155. 1. log 49517 = 4-0947543 log 49510 = 40947450 diff. for 1 = 87 j34 "3 48 20 1_ diff. for -34= 29:58 log 49510 rr:4-694756| log 49510-34 = 4094780 .-. log 4951034 = 0-094780. 2. log 3-4714 log 3-4713 diff. for -0001 •5405047 ^404921 120 •026 1756 2 52 J diff. for -0000020= 3 270 log 3-4713 = -5404921 log 3-4713020 = ^540492T \ XV.] 3. THE USE OF LOGARITHMIC TABLES. log 28497 = log 28490 = diff. for 1 = 4-4547991 4-454783 9 152 6l08 15 2_ 21128 diff. for -14 = log 28496 =4-4547839 log 28496- 14 = 4-4547860 . log 2849614 = 6 -4547860. logGOS'U =4 -7840036 1 6 432 5 360 log 6081465 = 6 -7840083 7. log x = 2-8283676 log 67354 = 2-8283634 diff. = 42, and diff. for 1 = 64 ; 42 21 .-. prop', increase =-- = —-= -66 ; ^ ^ 64 32 .. .r = 673-5466. 9. log J = 3-9184377 log -0082877 = 3-9184340 dift".= 37, and diff. for 1 = 52 ; .-. prop', increase = ;r7c= '71 ; o2 .-. a: = -008287771. 11. log x = ^ log 142-71 = ^(2-1544544) = -3077792 log 2 -0313 = -3077741 diff.= 51, and diff. for 1 = 213; 51^ 213 a; = 2-0313-24 . prop'. increase= -— -=-24 ; ^ J. O log 57-634 = log 57-633 = diff. for -001 = 1-7606788 1-7606712 76 •25 3 80 15 diff. for -00025 = 19 00 log 57-633 =j.-76067^| log 57-63325=1-7606731 6. log x = 4-7461735 log 55740 = 4-7461670 diff.=~~ 65, and diff. for 1 = 78; , , 65 5 prop'. Increase = — = . : -83; .-. j: = 55740-83. 8. log a: = 2-0288435 log -010686 = 2- 02881 52 diff. = 283, and diff. for 1=406; .-. prop', increase = "—="7; .-. .r = -0106867. 10. log a; = 1-4034508 log -25319 = 1-4034465 aiff.= 43, prop', increase = -_^ : .-. x = -2531925. 9 1 4 - = -2o 12. log 13-894 = 1-1428*273 281 6 811-1428561 7 .26 log a: = log 1-3894 = diff. -1428570 -1428273 297 and difi". for 1 = 313; , . 297 prop', increase =- .-. a- = 1-389495. = -95; 72 THE USE OF LOGARITHMIC TABLES. 14. log20G91 = 13 log 24244 = 5 -3840043 7 2^5 14|5-384'6168 log a; = -3846105 log 2-4244 = -3846043 diff . = m; .-. prop'. mcrease= --=-u3; .-. X = 2-424463. 8 [chap 6-3157815, 63 J^y^ 20 16^3157895^' log X = log 2-0691 = diff.= •3157895 •3157815 80, I • 80 8 .,^ .-. prop', increase = — = — = -38 ; .-. o-rr 2-069138. EXAMPLES. XV. b. Page 159. 1. sin 38° 3' = -6163489; diff. for 60" = 2291. 35 prop', increase = ^ x 2291 = 1336 •^64825 3. cosec 55° 21' = 1-2155978 ; diff. for 60" = 2443. 28 prop'. decrease = ^^ x 2443: 60 1140 1-2154838 4. sec nC TABLES. 73 10. L sin (9 -L sin 44° 17'=: 176; H. L cos 55=30' -7. cos ^=1205; Jiff, for 60" =1295; diff. for 60" =1838; and i!;Jf5x60" = 8"; ^ = 44*^17' 8 .-. ^ = 55° 30' 39". 12. prop', increase = x 3313 L tan 24° 50' = 9 '6653662 ; diff . for 60" = 3313. 52-5 : 2899 9¥656o6l 13. The required angle is 42-5" less than 40^5'; 42-5 1. .•. prop', increase- 60 X 1502 1064 L cosec 40= 5' = 10 1911808 L cosec 40° 4' 17-5"= 10-1912872 EXAMPLES. XV. c. Page 161. log 300-26 = 2-4774975 1 15 8 11 6 log -0078915 = 3-8971596 1 6 9 5 4 22 •3746609 log 2-3695 = -3746567 42 2 37 Thus the product = 2-36952. 2. log 235-67 =2-3723043 8 148 3 5 6 log 357-84 =2-5536889 3 36 8 _ 9^ 4^9260131 log 84336 = 4-92601 30 Thus the product is 84386. H. E. T. K. 74 THE USE OF LOGARITHMIC TABLES. [chap. 3. log 153-J4 = 2-1853721 1 28 9 25 6 log 2-8632 = -4568517 5 76 3 46 log -075836 = 2-8798754 4 23 6 3 4 1-5221148 log 33-274 = 1-5221050 98 7 91 70 5 65 Thus the product is 33-27475. 4 log 1-0304 = -0130059 5 21 1 42 subtract -0130081 1-4328656 2-5801425 log -038031 =2-5801377 48 4 46 20 3 23 log 27-093 =1-4328571 5 81 2 32 4 64 1-4328656 Thus the quotient is -03803142. 5. log 357-83 6 4 = 2-5536767 73 4 2-5536845 3-5037539 log 11218 5-0499306 = 4 -0499154 4 152 155 8 log -0031897 =3-5037498 3 41 3-5037539 Thus the quotient is 112184. XV.] 6. THE USE OF LOGARITHMIC TABLES. /O log 21-856 =1-38'J5707 3 60, 2 4J0 l;339577l" subtract 2-2512610 logx =3^83161^ log 1225-5 -3-0883133 280 8 28£ .-. j;= 1225-508. log -017831 =2-2512488 5 122 2-2512610 7. log 3-7895 = -5785819 6 69 log -053687 =2-7298691 2 16 1-3084595 log -0072916 = 3-8628228 1-4456367 log 27-902 =1-4456353 140 9 140 Thus the required value is 27-90209 8. log -83410 = 1-9212181 3 1 9 6 47 log -58030 3 Thus the cube is -580303. 9. log 15063 1-9212183 3 1-7636549 = 1-7636526 23 22 4-1779115 2 8 2 30 log 6-8482 5 I 4-1779120 ^355824 = -8355764 ~20 19 Thus the fifth root is 6-848293. G— 2 76 THE USE OF LOGATllTHMIC TABLES. 10 log 384-73 :^ 2-5851561 1 11 [chap. 5 I 2-585157 2 "" -5170314 log 3-2887 = - 517024 3 71 5 66 ~5'0 4 _53 Thus ^/384^ = 3-288754. log 15*732 4 — : 1-1967839 111 13 1-1967950 log 1-2301 2 9 •0920612 -0920536 76 70 60 70 Thus v/l5-7324- 1-236122. log 1034-3 9 r= 3-0146465 379 25 1 3 26 log 35324 :r^ 5-5480690 6 74 6 3 5-5480773 3 1-8493591 add log 2273-0 4 2 1 3-0146871 1-5073435 1-8493591 3-3567026 = 3-3566950 76 76 Thus the product is 2273-54. 12. Let a-: 1-0356270 and 6 = -7503269; then a^ -h- = {a + b){a- b), and a + b = l -7859539, a-b= -2853001. log 1-7859 = -2518571 5 122 8 7 3 9 2 19 log -28530 =1-4553018 1 15 1-7071722 log -50953 1-7071698 24 2 17 70 8 69 Thus the difference is -5095328. xv.] THE USE OF L(XiARITHMIC TABLES. t i 13. log a; = - log 34-732fi + ,. log 2-o3894 - I log 4-3yG82. log 2-5389 = -4046450 4 6_8 G 1^46524 ^74421 add -7958146 •8632567 log 7-2988 = -8632515 52 9 54 Thusx = 7-ii9889. 14. log -0037258 =3-5712195 1 12 6 7 9 1 05 add 3-5712215 1-7505167 log 4-3968 =r -6431367 2 20 •6431887 log 34-732 =1-5407298 6 75 ^5407373 3 4-6222119 subtract -6431387 5 I 3-9790732 •7958146 log -56301 =1-7505161 7 5 4 62 1-7505167 3-3217382 2-6608691 log ^045800 =2-6608655 36 3 29 To 7 67 Thus the mean proportional is -04580037. 15. If X be the required number, we have x= — ^- log -037517 =2-5742281 8 93 6 70 2-5742381 I^2791230 i-2951151 log -19729 1-2951051 100 4 _88 120 6 110 Thus x= -1972945. (-43607528)- ' log -43607 =1-63955621 5 501 2 2'0 8 79 1-6395615 2 i-2791-23U 78 THE USE OF LOGARITHMIC TABLES. [chap. 16. If X be the required number, we have 29-302564 x -33025107 56712-43 log 29-302 5 6 4 = 1-4668973 74 8 9 59 log -33025 1 7 = 1-5188428 13 92 subtract -9857498 4-7536783 4-2320715 log -00017063 = 4-2320554 6 161 152 3 90 76 log 56712 Thus tlie fourth proportional is -0001706363. 4-7536750, 31 23 4-7536783 17. Let X be the required number, then X = \/(-035689r^ X (2-879432) log 2-8794 o -4593020 45 30 -4593068 3 log -64406 14 1 1- 3779204 j098422y 1-7105069 1-8089298 1-8089263 35 34 . log •035689 = -^ (2-5525344) •J 5 = 1-7105069 Thus the geometric mean is -644065. XV.] 18. Here THE USE OF L()(;AKITHMIC TAHLES. _(7836-43)ix(357^814)^ 79 .r = (32-7812)^ log 7836-4 r=3^8941160 3 17 4 3^8941183 •9735295 add •5107315 1^4842610 subti-act •5052083 •9790527 log 9^5291 = : -9790519 8 2 9 Thus the fourth proportional is 9-52912. 19. og sin 27° 13' = 1-6602550 GO ^ 2*°« 491 1-6603041 1-8414768 1-5017809 log -31752 =1-5017711 7 1 98 90 20 14 Thus the required value is -3175271. log 32-781 =1-5156222 2 26 3 1 1-5156248 •5052083 log 357-81 =2-5536525 4 49 5 j 25536574 ~^073l5 log cos 46= 2' = 1 •8415095 iiihtract ^~ X 1310= 327 60 1-8414768 20. cot 97° 14' 16" = - cot 82° 45' 44", sec 112° 13' 5" = - sec 67° 40' 55". log sec 67'^ 46'= -4220725 — x3092= 2834 -4223559 1-1038011 1-5-261570 log -33585 r:z 1-5-261454 116 9 116 Thus the required value is -335859. log cot 82° 45^= 1-1045420 px 10103= 7409 I- 10380 11 80 THE USE OF LOGARlTHiSlIC TABLES. [chap. 21. log sin 20° 13' = 1-5885375 1^3429 1148 log sec 42° 15' -r 1306408 i-- 574 1-0693495 Thus the required value is -4221836. log cot 47° 53' 15 subtract -— x 2540 oO add = 1-9562154 = 635 1^9561519 1-6693495 log -4-2218 1-6255014 1-6254977 37 31 60 62 22. log 324-13 = 2-5107192 log sill 113° 14' 16" 6 8 log 417-24 3 1 80 10 = 2-0203859 31 1 5-1811174 1-9632566 7 logs = logsin66°45'44", sin 66° 45' =1-9632168 ^ X 543 = 398 1-9632566 log 12427 2 5-0943740 = 5-0943663 77 70 2 7 7 Thus the required value is 12427-2. 23 Here a = ^^}^^ , and sin B = sin 60° 45' 42". sin B log Bin 35° 15' log 378 -25 log 250-23 = 1-7612851 983 i-7613834 1-9408130 i-8205704 = 2-5777789 2-3983493 = 2-3983394 "~ 99 87 120 121 log sin 60° 45' = 42 so''™'' = 1-9407634 496 1-9408130 Thus a = 250-2357. XV.] THE USE OF LOGARITHMIC TABLES. 81 24. (1) log tan ^ = -y (log 5 -log 12) log 5= -6989700 log 12= 1-0791812 1-6197888 log tan ^= 1-8732629 log tan 36^ 4,7 rr 18731668 961 961 x60"=22' 2634 .-. ^ = 36" 45' 22". (2) (3sin(9-l)(sin<^ + l)--0. .-. sin^= - or - 1. log sin ^ = - log 3 = 1-5228787 log sin 19° 28' = 1-5227811 976 976 .ri7»x60" = 16". 3o72 .-. ^ = 19° 28' 16". 25. X = sin 23° 18' 5" x cot 38° 15' 13" x cos 28° 17' 25". log cot 38° 15'= -1032884 subtract — x 2598 = 562 bO -10323-22 1-9447579 log sin 23° 18' = 1-5971965 J. X 2932= 244 bO ~ 1-6452110 log -44178 =1:6452061 49 5 49 log cos 28° 17' = 1-0447862 25 subtract ^^ x 680 = bU 283 1-9447579 Thus x = -441785. 26. log cos 32° 47' = 1-9246535 2 1-8493070 log cot 41° 19'= -05 59928 3 i 1~905 2998 1-9684333 log sin 68° 25' = 1-968428 6 47 |9X60" = 6". Thus ^ = 68° 25' 6". 81 A THE USE OF LOGARITH.MIC TABLES. [chap. EXAMPLES. XV. d. Page 163 c. 1. 3. 5. 7. 9. 11. 13. 14. log 2834 = 3-4524 log 17-62 = 1-2460 log a: = 4-6984; whence a; = 49940. log -00567 = 3-7536 log •0297 = 2-4728 log a; = 4-2264; whence :r = -0001685, log 31-9 = 1-5038 log 1-51= -1790 log 9-7= -9868 log a; = 2-6696; whence a: = 467-3. log 17-3 = 1-2380 log 294-8 = 2^4695 log a: = 2-7685; whence x = -05868. 2. log -2179: log -08973 = logo;: whence x - log 2-38: log 3-901: log 4-83 = logo-: whence x - : 1-3383 : 2-9529 : -3854; : 2-429. : -3766 : -5912 -9678 : -6839 : -2839; 1-923. log 925-9 = 2-9665 log 1-597= -2034 log 74-03: log a;: log 15-38: log -0137: 3-1699 : 1-8694 4. 6. 8. 10. 12. 1-3005; whence a: = 19 -97. : 1-1809 : 2-1367 1-3236 -0207 log x = 1-3029; whence .t= -2008. log 8-034= -9049 log 1893 = 3-2772 logx = 4-1821; whence x = 15210. log 3-7: log 8-9: log -023: log a;: whence a;: log 43: log 8-07: log -0392: logic: whence x ■- log 2-035: log 837-6: log^: whence x •■ log 487: log 6398: : -5682 •9494 : 2-3617 1-8793; : -7573. : 1-6335 :_-9069 : 2-5933 1-1337; : 13-60. : -3086 : 2-9230 : 3-3856; : -00243. : 2-6875 : 3-8060 log ar = 2-8815; whence a; = -07612. log 14-72: log 38-05: log 387-9: logo:: : 1-1679 : 1-5804 2-7483 : 2-5887 •1596; whence a; = 1-444. log 276 = 2-4409 log •0038 = 3-5798 -0207 XV.] THE USE OF LOGARITHMIC TABLES. 81 H 15. log 2-31 = j363G log -0561 = 2-7490 log -037 = 2-5682 log 3-87= -5877 log 1--13 = -1553 lo^, .QQCf-^ ^ g.(j5gQ 1-0871 ^:^7 3-2957 loga- = l-7914; whence a: = 61-80. 16. loga:=-log5-l=^(-7076) = -3538; whence a; = 2-258. 17. log a- = - log 11 = -(1-0414) = -3471; whence a: = 2-224. 18. log x = ^ log 82-56 = - (1-9168) = -6839 ; whence x = 4-354. 19. loga; = -log 10-15 = - (1-0064)= -2516; whence a- = 1-784. 20. log X = 4 log -097 = 4 (2-9868) = 5-9472 ; whence x = -00008855. 21. 5 log 2-301 = -3619 x 5 = 1-8095 ; whence x = 64-49. «« 2, ^, o^ 1-7103x2 ,,,^^ 22. ^log51-32= ^— = 1-1402; whence .r = 13-81. 4 2-9494 X 4 23. p log -089 = ^ = 1-3997 ; whence x = -2510. 24. log -0137 = 2-1367 log -0296 = 2-4713 = 4-6080 log 873-5 = 2-9412 2 )7-6668 4-8334 ; whence x = -0006814. 25. log83 = l-9191 logl27 = 2-1038 -log 92= -6546 I log 246= -4782 2-5737 2-5820 2-5820 loga;=l-9917; whence a; =-9811. 26. log -678= 1-8312 log 9-01= -9547 •7859 log -0234 = 2-3692 2)2-4167 logx = 1-2084 = log 10-15; .". X = 16, to nearest integer. 81 C THE USE OF LOGARITHMIC TABLES. [CHAP. 27. (i) If X is the mean proportional between 2-87 and 30-08, x=j2-S7x30-6S log 2-87= -4579 log 30-08 = 1-4782 2) T93(il log a: ="•9080; whence a- = 9-29. (ii) If X is the third proportional to -0238 and 7-805, XX -0238 = (7-805)^; •■ ^ = ~^^ - 2 log 7-805 = 1-7848 log -0238 = 2-3766 log X = 3-4082 ; whence x = 2560. 28. Here .1;= {4/347-3 x >>y256-4}^ = (347-3)^ X (256-4)1^. hog 347-3 = -4234 D j^ log 256-4 = -2409 log x = -0643; whence j: = 4-616. 29. 5 log a; + 3 log ?/ = log 5, 2 log a; + 7 log y = log 11. These equations give , 71og5-31ogll , o log 11 -2 log 5 iog-= — 29-' ''^y= 29 • 7 log 5 = 4-8930 5 log 11 = 5-2070 3 log 11 = 3-1242 2 log 5 = 1-3980 29) l-7'688( -06099 29)3^8090 ( -1313 288 90 27 39 100 .-. loga;= -0610; .-. log7/= -1313; whence a; = 1-151. whence ?/ = 1-353. 30. log Z = log 2-863= -4509 log ^ = log 32-19 = 1-5077 2)2-9492 ^- Vr ^ 1-4746 log 2= -3010 log7r = _-4972 ~2728; whence the required value = 1-874. I XV.] THE USE OF LOGARITHMIC TABLES. 81 D 31. logm= log 18-34 = 1-2634 log r2 = 2 log 35-28 = 3^0950 4-3584 log2= -3010 log ^mv2 = 4-0574; whence -mi;'^— 11410. 32. (i) log J) = log 93-75 = 1-9719 log r" = 4 log 1 -03 = -0512 log _pr" = 2-0231; whence 2)r''= 105-4. (ii) log r^ = 3 log 5-875 = 2-3070 log 355 = 2-5502 log7r= -4971 log 113 = 2-0531 log 4= -6021 log7r= -4971 3-4062 log 3= -4771 4 4 log - 7rr3 = 2-9291 ; whence - 7rr3=849-4. o o 33. log m = log 33-47 = 1-5246 log ^7 = log 32-19 =1-5077 logf-= log 3600 = 3-5563 logr= log 9-6= -9823 5^09 2-4900 2-4900 logF=2-5909-, whence F= 889 -8. 8x537-6 2^- 4x8-1416' log 3= -4771 log 4= -6021 log 537 -6 = 2-7305 log 8 -1416= -4971 3-2076 1-0992 1-0992 3)2-1084 logr= '7028; whence r = 5-044 2s 578-6x 8=^ log 578-6 = 2-7624 log 64 = 1-8062 4-5686 2 log 31 = 2-9828 log /= 1-5858; whence /=38-53. 36. n log 3- + log?/ = 8 + log 8-7 71 log 73-96 + log27-2o = 8 + log8-7 = 8-9395 log 27-25 = l-43o4 .-. n log 78-96 = 7-504L • « _ _Z^2i^ - 7 '5041 18^6.9 ) 7504-1 ( 1015 'log 73 -96 "1-8690 2S1 = 4-015. r 81 E THE USE OF LOGARITHMIC TABLES. [CHAP. Sr 3x33-87 37. ,-3: •iir 4x3-1416* log 3= -4771 1ok4= -6021 log 33-87 = 1-52U8 log 3-1410^ -4971 2-0069 1-0992 1-0992 3) -9077 logr= -3026; whence r = 2-007. 38. Let d be the diameter; then 4 /dy lo>;fi= -7782 qTT - =(36-4)3; alog3;j-4=4-6833 "* V-^/ 51615 ^„ 6 X (36-4)3 loKTT ^ -4971 «^= ^ -; 3 )4-9644 : '^ 1-6548 .-. 31og(Z = log6 + 31og36-4-log7r, antiloKl -6548 =45-16. Thus ^ = 45-16 cm. 39. 21og?; = logr + logi7-log289 = log 4000 + log 32-2 - log 5280 - log 289. log 4000 = 3-6021 log 5280 = 3-7226 log 32-2 = 1-5079 log 289 = 2-4609 5-1100 6-1835 6-1835 3) 4- 9644 =3 log rf 2)2-9265 log V = 1-4632 ; whence v = -2905. 27rr Let E = -— , ; then log £ = log 27rr- log?;- 2 log 60 V X bO- log2= -3010 log v = 1-4632 log7r= -4971 2 log 60 = 3-5564 log r = 3-6021 30196 4-4002 3-0196 log jE= 1-3806; whence £ = 24, approximately. EXAMPLES. XV. e. Page 163 h. In Examples 16 — 19 let the expression be denoted by x. 16. log sin 27°13' = 1-6602 17. log sin 47°13' = 1-8656 log cos 46°16' = 1-8397 log tan 22°27'= 1-6162 log x= 1-4999; log j- := "^2494 ; whence a: = -3161. whence a- = 1-776. 18. log sin 34n7' = 1-7507 19. x = cos 28°14' x cos 37^26'. lug tan 82° 6-= -8577 log cos 28^4' =1-9450 _'^084 log cos 37° 26' = 1-8998 logcosl2°37'= 1-9894 f^rrr logx = -:6l90; ^'^^^^'' whence x = 4-159. whence x = -6995. XV.] THE USE OF LOGARITHMIC TABLES. 81 F 20. 7 log tan x = log 11 -log 13. 21. log 32-73 = 1-5149 log 11 = 1-0414 log 27-86 = 1-4449 log 13 = 1-1139 log sin 30° 16' = 1-7025 7)1-9275 log a6 sin C= 2-6623; log tan j; = 1-9896; whence a; = 44^ 19'. whence aftsin C=4o9'5. 22. (i) wa2cot- = 32cot22i° (ii) ^" sin— =5 x (3*3)2 sin 36°. = 32 tan 67^° log 5= -6990 = 32x19-3136 2 log 3 -3 = 1-0370 = 77-2544. log sin 36°= 1-7692 1-5052; .-. required value = 32-00. 23. tan = ^-r^^ sin 56° 14' ''^ sin 56° 14'. 1-35X-65 log -7 = 1-8451 log 1-35= -1303 log sin 56° 14'= 1-9198 log -65 = 1-8129 1-7649 1-9432 1-9432 1-8217; whence = 33° 33'. _. - 2x32-78x19-23 ^^. ,^, 24. 1 = --^^^ cos 57^47'. log 32-78 = 1-5156 log 38-46 =1-5850 log cos 57°47'= 1-7268 2-8274 log 52-01 = 1-7161 25. Expression logZ=l-1113; whence Z= 12-92. 2 X (48)2 sin 23° 32-19 x( -63)2 0082 23°' log 2= -3010 log 32-19 = 1-5077 2 log 48 = 3-3624 2 log •63 = 1-5986 log sin 23° =1^5919 2 log cos 23° = 1-9280 3-2553 r0343 10343 2-2210; whence value of expression = 166 -3. 82 SOLUTION OF TRIANGLES AVlTll LOGARITHMS. [CHAP, EXAMPLES. XVI. a. Page 166. -„. ., s(s-a) sls-b) s(2s-a-h) 1. First Bide = -^ ' + '- = — -' ■- .". s-b) 2. First side ^. ^/^^^^^ • ./^ V s (s -b) V « (s - o • "-^ ^. , 1-cos^ "" 2 (s-b)ls-c) ca 3. First side = ^ = ^ =-^ r^— — >^ 1 + coSjB ^ . o^ ^c (s-c)(s-a) 2 sin-^ - o a{s-b) _a (a + c - b) c{s — a) b{b + c — a)' „. , b(s-b)(s-c) a{s-c)(s-a) 4 First side = -^ j^ '- + -^ '- - be ca {s — c){s-b + s- a] _c (s-c) _ ■ s - c. „ , , , . -1 /(s-a) is -b){s-c) 5. Each of the expressions reduces to ^ / -^^ . 6, tan - _ .^ -,-(7-r&p - V 24^ " 2 * C _ / s{s- c) _ /21 X 6 _ 3 7, ^^^2-V is-a){s-b)-\ ~8l^7-2- _ _. ^ ., s (s-a) + s(s-b) + s(s-c) 8. First side=^ ^^ '— — ^, ^^ abc _ s{3s-(a + b + c)} _ s^ abc ~ abc ' « TT ^ J h-c s(6-a) ^ ... 9 rirst Bide = . -^, — ^ + two similar terms a be (b-c)(s^-as) . . „ = i +two similar terms abc __ s^{{b -c) + (c-a) + (a-b)\- s{a(b -c) i-b{c - a)-tc(a- b)} ~ abc = 0. XVI.] SOLUTION OF TRIANGLES WITH LOOAIUTIIMS. 88 EXAMPLES. XVI. b. Vmuc 109. i. ^"'.2~V ab "V5x8~VlO" C 1 log sin - = - (log 7-1) = 1-9225490 log sin 56° 47' = 1-9225205 diff. 285 285 prop', increases x 60" = 20*6"; o27 ^= 56° 47' 20-6", and C'= 118° 34' 41". 285 60 827 ) 17100 ( 20-6 lHo4 5GU0 49U2 Z. tan^-^ ^^^_^^^ -V 67x27" V log tan -=1-6634464 = logtan24°44'13"; .-. ^ = 49° 28' 26". '128 603 log 128 = 2-1072100 log 603 = 2-7803173 2)1-3268927 1-6634464 3- ^^4 = \/^- - \/2^ 15x5 X 4 X 6 4 ^^'2 B log cos — = log 5 - - log 2 = 1-9463950 log cos 27° 53' = 1-9464040 diff. 90 90 prop', increases ^^^ x 60" = 8-07" ; bo9 B log 5 = -6989700 I log 2= -7525750 1-9463950 30 (50 2-23 j 180O ( 8-07 1781. lUOO .-. -=27° 53' 8-07", and 7? = 55° 46' 16". H. E. T. K. 84 SOLUTION UF Till ANGLES WITH LOGAPvITH^MS. [CHAP. | 4. cos - = C_ A- (s-c ) _ /9x2_ /^ 2 ~~ V """oft"' "v^xG^Vk 6 lU' C 1 logcos-=^(logG-l) = 1-8890757 log cos 39° 14' = 1-8890644 diff. 113 113 prop^. dec7'ease=-ir — x 60" = G-G"; c J =39° 13' 53-4", and C = 78° 27'' 47". 113 60 1082 ) 67800 ( 6-6 619-2 5880 6192 G /J^-a){x-b) /-75x2 /¥ log tan I = i {2 log 2 - 1} =^ (1-0020000) = 1-8010300 log tan 32" 18' = 1-8008305 diff. 1935 , . 1935 prop', increase = 279g C '• 2 X 60" = 41-5"; = 32° 18' 41-5", and C = 64°3r23". 1935 60 2796)116100(41-5 11184 4260 2796 14640 ^ ^ C /(s - a) {s - b) /3x4 /I 6. tan - = ,^ -^ ^^ -^- ^ V 15^ ^ V 10 log tan ~= - ^ = 1-500000 log tan 17° 33' = 1-500042 diff. 42 42 prop*, decrease = -^ x 60" = 5 '7" ; .-. ^ = 17° 32' 54-3", and C = 35°5'49' 60^ 439) 2520(57 219.') 3250 XVI.] SOLITTTOX OF TRIANGLES WITH LOGARITHMS. 85 7. Let«^4, 6 = 10, c = ll. 2 V ah ~ V 2 • 2 • 40 ~ V 25" ~ V cos 30 ¥ c 1 log cos - = - (log 3 + 1 - 6 log 2) = 1-8354707 log cos 46^ 47' = 1-8355378 diff. 671 671 prop', increase = -^- ^ x 60" =30" 1345 C l + log3 = 1-4771213 6 log 2 = 1-806180 2 ) i-6709413 i-8354707 -s- = 46°47'30", and C = 93°35'. 7? — log tan — = -log 2 = 1-6989700 log tan 26^ 33'= 1-6986847 dift". 2853 2853 „ prop', increase = ,, _, ^ 3159 -. ^ = 26° 33' 54-2", and J5 = 53^7'48 63x7 1 21 X 8 ~ 2 • 2853 60 = 54-2". 3159 ) 171180 ( 54-2 15795 13280 12G3G P 7' 48". 5940 Again tan^= V^^^-^^^= / ^ =i 2 V s{s-c) V21x7 7 log tan — = 2 log 2 - log 7 = 1-7569620 log tan 29^ 44' =£7567587 difif. 2033 2033 prop', increase = —- x 60" = 41 -5. C .'. - = 29° 44' 41-5", and C = 59° 29' 23". .-. ^ = 67° 22^49". 2 log 2 = -6020600 log 7 = -8450 980 1-7569620 2033 60 2933 ) 121980 ( 41-5 11732 4660 2933 17-270 7—2 86 SOLUTION OF TRIANGLES WITH LOGARITHMS. [CHAR 9. tc ^''"2-V ' s{s-b) -V'-^ 2''9''"'-V3' s (s J) 1 log tan - = -(logo-log3) = •1109244 log tan 52° 14' = -1108395 diflf. 849 849 prop', increase = —— x 10" = 19 '5": * 43o logo = -0989700 log 3 = ^771213 2 p2218487 -1109244 B = 52° 14' 19-5", and B = 104° 28' 39' , C /{s -b){s- a) /I 3 2 2 Again tan-=^-_^l_^^=^/^-x-x-x^ = s/3x5* C 1 log tan - = - ^ (log 3 + log 5) = l-4119o44 log tan 14° 28'= 1-4116146 diff. 3398 3398 prop', increase = — — - x 10" = 39 ^ : 870 .-. -=14° 28' 39", and = 28^57' 18". .-. ^=46° 34' 3". EXAMPLES. XVI. c. Page 173. 1. tan A-B a-h . C 1 ^ , cot-- = ^cot30°= — ^/3. a + b 2 o 10^ A — B 1 log tan — -- = log 2 + - log 3 - 1 = 1-5395907 log tan 19° 0' = 1-5394287 diff. 1620 , . 1620 prop', increase = —-- x 60" = 24". 4U84 log 2 = ilog3 = 3010300 2385607 -5395907 XVI.] SOLUTION OF TKIAXGLES WITH LOGARITHMS. 87 .-. — 2^^' = 19° 6' 24", and^i±^=60=; /. A = 79° 6' 24" ; B = 40° 53' 36". 2. , G-A c-a ^B 8 tau — — - = -— cot ^ = — cot 32° 30'. 2 c + a 2 10 C — A log tan -^— = 3 log 2 - 1 + log cot 32° 30' = -0989027 log tan 51° 28' =^988763 diflf. 264 264 prop', increase =-^, x 60"= 6"; 2592 C — A :. — jr^ = 51° 28' 6", G + A = 57° 33'; 3 log 2= -9030900 log cot 32° 30'= - 1958127 1^989027 C=108°58'6"; ^ = 6°1'54". 3. , B-A b-a G 5 , ^ J. B-A , , 1 logtan ^ =l-31og2--log3 = 1-8583494 log tan 35° 49' = l-8583357 diff. 137 ^1 ; 137 prop'. increase=— — x60" = •3"• 2bb2 ' B-A :. —2— = 35° 49' 3", B-^A = 60^; iIog3= -2385606 3 log 2= -9030900 1-1416506 .-. B = 95° 49' 3" ; A = 24^ 10' 57". 88 SOLUTION OF TRIANGLES AVITH LOGARITHMS. [CHAP. 4- *- ^-T = i^c o"' 'I = ^ ""^ ''° ''' = m '=°' '''° '''■ logtan?-,-- = l-39124Ul log tan 11° 3' = 1-2906713 diff. 5778 prop', increase = ^y,m" = ry2"; 6711 3 log 2 -2 = 2-9030900 logcot22^15'= -3881591 1-2912491 ^— ^ = 11°3'52", and ^±^ = 67° 45'; 2 a .: jB = 78°48'52"; C = 56°41'8". _ ^ G-A c-a ,B 10 , ^„ 5. tan --=,:p7, '=0*^=32 '=°"''^^=- log tan ?-^ = 1 + log cot 17° 21' Id" - 5 log 2 = 1-5051500-1-5051500 = 0. .-. -"— = 45°, and ^— = 72° 38' 45" ; -. C= 117° 38' 45"; .4 = 27° 38' 45 6. tan— „— = "— ^ cot^= ,cot30°15'. 2 a+6 2 4 log tan ^-J^ = - 2 log 2 + log cot 30° 15' = 1-63214 log tan 23° 13' = 1-63240 diff. 26 log cot 30° 15'= -23420 2 log 2= -60206 1-63214 26 prop', decrease = ^ x 60" = 45" ; ^-^ = 23° 12' 15", and ^-^ = 59° 45' ; :. y4=82°57'15"; i^ = 36° 32' 45". XVI ] 7. 8. SOLUTION OF TRIANGLES WITH LOGARITHMS. 89 *-^=::^-*f=^-^^«^i^' I — c log tan '- - - = 1-3556602 log tau 12° 46' = 1-3552267 difif. 4835 prop', increase = --.-r, x 60" = 41'"; A-C J + C = 12° 46' 44", and ti±^^61°46'; log 71 = 1-8512583 log cot -- = -2700705 21213288 log 583 = 2-7656686 1-3550602 2 .-. A = 74° 32' 44" ; C = 48° 59' 16". tan ^—^ = f— ^ cot i = l cot 32° 30'. 2 b+c 2 5 log tan^^— ^ = i-9739640 log tan 43° 18' = 1-9742133 diff. 2493 2493 prop', decrease = ^— - x 60" = 59' 2odl ^ - ^ = 43° 17' 1", and^ = log 3= -4771213 log 5= -6989700 1-7781513 log cot 32° 30' = -1 958127 1-9739640 2 57° 30'; .-. B = 100° 47' 1" ; C = 14° 12' 59". 9. Here ^A-B a + b^ C ,, G cot — ^^— = tan TT = 2 tan — . 2 a-0 2 2 A — B .'. log cot --r — = log 2 + log tan 15° 5' 2-5" = 1-7316286 log cot 61° 41' = 1-7314436 diff. 1800 prop', decrease = — -- x 10" =35 -7": o04 log 2= -3010300 log tan 15° 5' =1-4305727 2^5 10 x838 = 209 l-731623fi •. ^i^ = 61°40'24-3", and ^i^' = 74^54'51-5"; ^ = 136° 35' 21-8"; iy = 13^14'33-2". 90 SOLUTION OF TRIANGLES WITH LOGARITHMS. [CHAP. EXAMPLES. XVI. d. Page 174. 1. Here ^ = 180°- 114° 45' = 65° 15'. _ asmG _ 100 sin 54° 30^ ~ sin ^ ~ sin 65° 15' logc = l-0525317 = log 89-646162 ; .. c = 89-646162. log sin 54° 30' = 1-9106860 log 100 =2 1-9106860 log sin 65° 15' = 1-9581543 F9525317 csin^ 270 sin 55° _. . .^. 2 2. a = . ^ = -.— ^n^- = 270 sin 55° x -.^ . sm C sm 60° ^3 .-. log a = 1 + 3 log 3 + log sin 55° - - log 3 + log 2 = 2-4071977 log 255-38 = 2-4071869 diff. 108 108 prop', increase = — r^ x -01 = -0064 ; .-. fl. = 255-3864. log 270 = 2-4313630 log sin 55 =1-9133645 log 2= -3010300 2-6457584 I log 3= -23856 07 2-4071977 3. h sin C 100 sin 62° 5' sin I? sin 72° 14' log c = 1-96749 = log 92-788; .-. c = 92-788. log sin 6-2° 5'= 1-94627 log 100 = 2 1-94627 log sin 72° 14' = 1-97878 1-96749 4. Here A = 180° - 148° 40' = 31° 20'. a sin n 102 sin 70° 30' h = - logZ> = 2-267 = log 185; sin^ sin 31° 20' log 102 = 2-000 log sin 70° 30'= 1-974 1-983 log sin 31° 20'= 1-716 2-267 XVI.] SOLUTION OF Till ANGLES WITH LOGARITHMS. IJl Again c = a sinC sin J log c = 2-283 = log 102; .-. c = 192. log 102 = 2-009 log8in78°10'=L990 1-999 log sin 31° 20'= rne T283 5. Here c = a sin G 123 sin A J2 sin 15° 43' ' log = 2-5066124 log 321-10 = 2-5066403 diff. 279 279 prop'. decrease = ——:X -01 = -02066. loo Thus = 321-0793. -log 2= -1505150 log sin 15° 43'= 1-4327777 log 123 1-5S32927 2-0899051 2-5066124 a — bsinA 1006-62 sin 44° sin B sin 66° log 1006-62 = 3-00-28656 log sin 44° = 1-841 7713 2-8446369 log sin 66° = 1-9607302 log a =2-8839067 .-. rt = 765-4321. 7. Here A = supplement of 75° 45' ; _ 1652 sin 26° 30' ■■ sin 75° 45' log 6 =2-8852436 log 767-80 = 2-8852481 diff. 45 45 prop^ decrease = ^ x -01 = -008 ; c = - b sin G 1006-62 sin 70° sin B sin 66° log 1006-62 = 3-0028656 log sin 70° = 1-9729858 2-9758514 log sin 66° = 1-9607302 logc =3-0151212 .-. c = 1035-43. log 1652 =3-2180100 log sin 20° 30' = 1-6495274 2-867^374 log sin 73° 45' = 1-9822938 2-8852436 Z> = 767 -79-2. Again c = 1652 sin 47° 15' sin 73° 45' logc =3-1016030 log 1263-6 = 3- 101609 6 diff. 66 66 prop'. decreai>e=-.^x •1 = -019; o44 log 1652 =3-2180100 log sin 47° 15' = 1-8658868 3-0838968 log sin 73° 45'= 1-9822938 'saoieo^ c = 1263-58. k 92 SOLUTION OF TRIANGLES WITH LOGARITHMS. [CHAP. EXAMPLES. XVI. e. Page 176. 1. sm A = — = j_^ sm 41° 10'. log sin ^ = 1-7293399, .-. ^ = 32° 25' 35". log 145 = 2-1613680 log sin 41° 10^= 1-8183919 1-9797599 log 178 = 2-25 0420,0 P7293399 2. & 127 ein B=- sin A = ~ sin 26° 26'. a 8o log sin jB = 1-8228972, .-. i5 = 41°41'28"; and since a is c there is only one zoo Ji JLO solution. From (ii) we have log sin C=log 5 - log 8 = 1-79588. .-. C=38°41', or 141° 19'-, and .4 = 111° 19', or 8° 41'. Now in the obtuse-angled triangle we have , a sin B 200 sin 8° 41' sin^ sin 30= log 6 = 1-7809601 log 60-389 = 1-780957 8 diff. 23 23 prop', mcrease = ;^ x '001 = -0003. .-. 6 = 60-3893. log 200:= 2-3010300 log sin 8° 41' = ri789001 1-4799301 log sin 30° = i-69 89700 P78U9G01 94 SOLUTION OF TRIANGLES WITH LOGARITHMS. [CHAP. EXAMPLES. XVI. f. Page 180. ^ B_ /{s-c){s-a)_ /3^4_ /I, 1. ^''°2" V "^Ts^^^ ~ V 12.5" V 5' PI 1 .-. log tan -^ = - 2 log 5 = - - (1 - log 2) = I-65051oO log tan 2-1° 5' =1-6502809 diff. 2341 2341 '3390 2341 prop^ increase = .,y„^ x 60" = 41-4" ; - = 24° 5' 41-4", and 1^ = 48^ 11' 28". 2 ^^''2 "" V '^s{8-c) " V 12:3"" V 3-^' .-. log tan 1 = ^(1 -log 2- 2 log 3) = 1-8723637 log tan 36° 41' = i-8721123 diff. 2514 2514 prop', increase = -„^^ x 60" = 57-2" ; .-. ^ = 36° 41' 57-2", and C = 73° 23' 54"; a :. ^ = 58° 24' 43". A 6 + c^ B-G 2. ^^* 2 = ^-0*^^-^ = S2^-^^°=W*-^^°- \ :. log cot '- = 8 log 2 - 4 log 3 + log tan 12° = i-8272293 = log cot 56° 6' 27" ; .-. - = 56°6'27", and J =112^2' 54". 2 ' .-. B + C= 67° 47' 6", and Ji - C= 24° ; .-. jB = 45°53'33", and C=21°53'33". XVI.] SOLUTION OF TRIANGLES WITH LOGARITHMS. 05 3. sin ^ := - , if ^ is the less of the two acute angles. log sin A = log 2 - log 7 = I-4o5932 logsinl4°ir = l-455921 diif. 11 11 prop*, increase = — — x 60" = 6". .'. 4 = 14° 11' 6"; .-. ^=90° -14° 11' 6" = 75° 48' 54". 4, Here a = 2183, A = 30° 22', B = 78° 14', C = 71° 24'. , a sin 5 2183 sin 78° 14' sin^ sin 30° 22' log 6 = 8-6260817 227 diff. log 4227=4 = 3-6200733 84 84 prop', increase = — _, x -1 = -0815 ; /. 5 = 4227-4815. log 2183 =3-3390537 log sin .5 = 1-9907766 3-3-298303 log sin ^ = 1-7037486 3-62608i7 5. Now B-C b-c .A 1 ,,,oin. tan — zr— = - — cot TT = -: cot 11° 10'. 2 h+c 2 9 log tan ^~^^- = log cot 11° 10' - 2 log 3 = -70465 - -95424 = 1-75041. .-. ^^ = 29° 22' 26", and ^^ = 78° 50'; :. B = 108° 12' 26"; C = 49° 27' 34". c sin^ """"Tin^a ' .-. log a = log 2 + log sin 22° 20' - log sin 49° 27' 34" = -30103 + 1-57977 - 1-88079 = -UOOOl ; /. a = l, approximately. 96 SOLUTION OF TIUANOLES WITH LOGARITH^VIS. [CHAP II 6. tan ^-^ = ^ cot -^ = -56234 cot 29° 21' 3 2 a + 6 2 Nov,- log cot 29° 21' r= -250015 log cot 29° 22" = -249715 cliff, for GO" 3T)0 /. prop' . decrease for 3" = — x 300 = 15 ; /. log cot 29° 21' 3"= -250000. A — B - .-. logtan—-r= 1-75 + '25 = 0. ^ A-B ^ ,. A-B . .'. tan — ^r — = 1, so that — o" — ^ 5°. Also ^— = 60° 38' 57"; whence ^ = 105° 38' 57", 75 = 15° 38' 57". . ^ bBinA 12 sin 30° 7. sm7?=--_= ^-_; .-. log sin B = 1-07918 + 1-69897 - -95424 = 1-82391 ; .*. ^ = 41° 48' 39" or 138° 11' 21", both values being admissible since a<6. .-. C = 108° 11' 21" or 11° 48' 39". _ & sin Ci _ 12 si n 108° 11' 21" -^^^'^ ^~^^hrB^~ sin 41° 48' 39" ' .-. log c = 1-07918 + 1-97774 - 1-82391 = 1-23301 ; .-. c = 17-l. Similarly from c = -^--^r" . we easily obtain c = 3-68. ^ G a-b .A-B I .._„ 1 8- ^^"2=;7T6"^*~2- = 2""*'^-^ =2' .-. log tan ^ = log 1 - log 2 = 1-6989700 log tan 26° 33' = 1-6986 847 diff. " 2853 .'. prop', mcrease = -^^ '^ "0 = o4'2 ; .. ^ = 26° 33' 54-2", and C = 53° 7' 48". < I XVl.J SOLUTION OF TRIANGLES WITH LOGARITHMS. 97 Hence ^^ = q-^- 2{V 6" , and''^~^ = i5°; a a :. ^ = 108°2G'G", i} = 18°26'6". 9. (1) Let « = 1404, Z> = 960, .4 =32" 15'; ,, . ^ h sin A 80 then sin i> = = , ^- ..,,„ ., ^, : a 117cosec32°15" . . log sin B = 1 + 3 log 2 - (2 log 3 + log 13 + log cosec 32° 15') = 1-5621316, on reduction. .-. I? = 21° 23'; .-. C = 126°22'. (2) Let a = 1404, b = 960, B = 32° 15' ; then '^^=S0^^^o-YE''^ .: log sin yl = 21og 3 + log 13 - (1 + 3 log 2 + log cosec 32° 15') = 1-8923236, on reduction. .-. .4 = 51° 18', or 128° 42' since the solution is ambiguous. .-. C=96°27', or 19° 3'. 1-^ 10. ^\ e have tan —^ = ^-^ cot - = cot - 1 - cos (b ^A ^ „(b ^A = cot — = tan- \ cot tt , 1 + cos 2 2 2 ' c 10 where cos = - = — - . 6 11 Hence log cos = 1 -log 11 = 1-958607; .-. = 24° 37' 12". fi — C (h A Again log tan — - — = 2 log tan ^ + log cot - = 2-677782 + -495800 = 1-173582. .-. ^^' = 8°28'56•5'^and ^^ = 72° 17' 30"; .-. i5 = 80°46'2(>5", C = 03° 48' 33-5". 98 SOLUTION OF TRIANGLES WITH LOGARITHMS. [OHAP. 11. . ^ 6 . . 1071 . _.o sin ji5 = - sin A = -t^ttt- sin oO° ; a o7o log sin B = 3-029789 + 1-884254 - 2-941014 = 1-973029 log sin 70° = 1-972986 diff. 43 log sin 70° 1'= 1-973032 log sin 70° = i-972986 diff. for 1' 46 43 prop', increase = t^ x ^^" = ^^" 5 .-. i3 = 70°0'56", or 109^ 59' 4", both values being admissible since a 8 100 bULUTlON OF TRIANGLES WITH LOGARITHMS. [CHAP. c 28-58 16. HeresmC=-^ = ^,^:321 log 28-58 =1-4560622 log 57-321 = r7583LS8 T-6977484 log siu 2r- 54' = 1-697654 5 diff. 939 Diff. for 60" = 21' 16; 039 .-. prop', increase = ^^j^ x60" = 26". -. (7 = 29° 54' 26"; whence yl =60° 5' 34 17. Let G be the right angle, and A = 18° 37' 29" ; then 284 _ a _ ^ "" ^n 2 ~ sin 18° 37' 29""' log 284 =2-4533183 log sin 18° 37' 29"=r5042917 logc =2-9490266 log 889-25 = 2-9490239 27 5 25 20 4 20 .-. c = 889-2554 feet. log sin 18°37' = 1-5041105 29 60' x 3748 = 1812 log sin 18° 37' 29"= 1-5042917 B-C b-c .A 1 ,„. 18. tan-2--^— cot2=8.v/3, B — C 1 log tan — 2 — = - log 3 - 3 log 2 = 1-3354706 log tan 12° 12' = 1-384871 1 difif. 5995 B-C Diff. for 60" = 6112; .-. prop', mcrease = — - x bU = o\) . = 12° 12' 59", and — |^' = 60° ; -. U = 72° 12' 59", C = 47° 47' I". XVI.] SOLUTION OF TRIANGLES WITH LOGARITHMS. 101 19. Let AC be the ladder, C the window, and B the foot of the wall; then from the right-angled triangle ABC, Sin 72° 15' ' \ogb = 1-6270585 -1-0788175 ,',!', V ^' = 1-6482410 '- . .' log 44-487 = 1-6482331 ' ' ; 79 •.'•.,, 8 78 .-. length of ladder = 44-4878 feet. on * -^^-^ a-& .C 9-99 , , , 20. »-"^- = „-T6'=°'2=53-9l'='"l'='<'- log 9-99= -9995655 log cot 17° 30' = -5012777 1-5008432 Diff. for qq" = 2892 ; log 5-3-91 = 1-7316693 ^^^„ 1-7691739 •■• P^°P'- '''''''®*^^ = 2892 ^ ^^" = 38"; log tan 30° 26' = 1-7689922 diff. 1817 .-. ^^-^ = 30° 26' 38", and ^-^ = 72° 30' ; .-. .4 = 102° 56' 38", 5 = 42° 3' 22". 21. tan ^ = ^ cot ^ = ^ cot 23° 37' 30". 2 b + c 2 38-9o log cot 23° 37'= -3592844 30 Subtract -.y- x 3441 1721 bO -3591123 log 11-29 = 10526939 iTJ^J^^ Uiff. for60" = 2743; log 38-95 = 1-5905075 , . 2413 _^„ ^^„ ^ .-. prop', increase = x 60 = 53 . Ti — C — log tan — -— = 1-8212987 log tan 33° 31' = 1-8210574 diff. 2413 2743 ^^-^ = 33°31'53", and ^±^ = 66° 22' 30"; .-. £ = 99° 54' 23", (7 = 32° 50' 37". 8—2 102 SOLUTION OF TRIANGLES WITH LOGARITHMS. [CHAP. 6 sin .4 25-12 sin 47° 15' 25-12 sin 47° 15' Agam a - ^.^ ^^ - ^.^ ^^^o 54. 33' log sin 47° 15' = 1-8658868 log 25-12 = 1-4000196 • - 1-2659064 1-9934760 . . loga = 1-2724304 iog 18-725 :^ 1-2724218 86 4 93 cos 9° 54' 23" ' log cos 9° 54' = 1-9934844 Subtract ^^ x 220 -: 84 00 _: 1-9934760 .-. (1 = 18-7254. 99 sin-- A-")(^-^^ - /136M2> log 1361-1 = X 1024-48 log 1024-4 3-1338900 64 3-0104696 8 339 6-1443999 6-6013840 2 ) i-5430159 log sm - 1-7715079 log sin 36° 13' = 1-7714702 diff. 377 B 2173-84 * log 1837-2 =3-2641564 log 2173-8 =3-3372196 4 80 6-6013840 Diff. for 60" =1724; prop', increase = :. _ , . x 60" = 13". = 38° 13' 13", and B = 72° 26' 26". 23. A _ /(s- b){s^) _ / 6-4405x14-911 *^^ 2 ~ V "s (s-a) "V 52 -1248 x 30-772 log 6-4405 = -8089196 log 14-911 = 1-1735068 4 116 1-9824380 3-2052113 2 ) 2-7772267 log tan '- log tan 13° 44' 1-3886134 1-3880 837 5297 A 9114 7728* log 52-124 =1-7170377 8 66 log 30-772 =1-4881557 8 n3 3^2052113 Diff. for 60" = 5475; 5297 prop', increase =.^^g x60" = 58". = 13° 44' 58", and A = 27° 29' 56 XVI.] SOLUTION OF 'J'KIANGLES WITH LOGAllITHMS. 103 Again tan — B_ / 14-91 2 " V 52" 12 9114 X 30-7728 248 X 6-4405* log 14-9114 log 30-7728 log tan B log tan 49° 27' diff. 1-1735184 1^4881670 2^616854 2-5259639 2) -13572 15 -0678608 -0677338 1270 log 52-1248 = 1-7170443 log 6-4405 rn -8089196 2-5259639 Diff. for 60" = 2558; 1270 prop', increase = ^r— - x 60" = 30 2558 B = 49° 27' 30", and £=98° 55'. = 53° 35' 4". 24. ^ C-B c-b A 202-949 ...oo^.o.,/. tan — -— = — - cot - = z-77,-— -- cot 51° 36' 27". 2 c + 6 2 1497-597 log cot 51° 36' = 1-8990487 27 Subtract -~ x 2595 = bO 1168 1-8989319 log 202-94 =2-3073677 9 193 2-2063189 3-1753949 log tan ^^ =1-0309240 log tan 6° r = 1-0300464 diff. 8776 log 1497-5 =3-1753668 9 261 7 203 "3-1753949 Diff. for 60" = 11909; .-. prop', increase = :^^^ x 60" = 44" ; —^ = 6° 7' 44", and — "J ^ = 38° 23' 33" ; .•.C = 44°31'17", 2? = 32°15'49". 104 SOLUTION OF TRIANGLES WITH LOGARITHMS. [CHAP. To find a, we have a = h sin A G47'324 sin 103° 12' 54" sin B sin 32 15' 49' Now log sin 103° 12' 54" = log sin 76° 47' (3". log sin 76° 47' = 1-9883415 6 60 log 647-32 X 297 = 30 log a log 1180-5 i-9883445 = 2-8111190 27 2-7994662 i-7273911 = 3^20751 = 3-0720660 91 73 180 185 log sin 32° 15' = 1-7272276 49 60 X 2002 1635 1^7273911 1 25. a = 1180-525. &sin^ 23-2783 sin 37° 57 a = BinB sin 43" 13' log 23-278 = 1-3669457 3 56 log sin 37° 57' = 1-7888565 1-1558078 log sin 43° 13' = 1-8355378 log a = 1-3202700 log 20-905 = 1-3202502 198 187 a = 20-9059. XVI.] SOLUTION OF TRIANGLES WITH LOGARITHMS. b siu C 23-2783 sin 81° 10' 105 Again sinB sin 43° 13' log c = 1-3GG9503 + 1-9948181 - 1-8355378 = 1-52G231G log 33-591 = 1-5262229 87 7 90 .'. c = 33-5917. 26. c sin B 2484-3 sin 72° 43' 25" sinC log sin 72° 43' := 1-9799339 25 60 log 2484-3 X 393 = 164 = 3-3952040 sin 47° 12' 17" log sin 47' 12' = 1-8655362 ^xll69= 331 00 log& log 3232-8 3-3751543 1-8655693 = 3-5095850 = 3-5095788 62 54 8 1-8655693 Z^ = 3232-846. Again a = c sin yl _ 2484-3 sin 60° 4' 18" smCT ~ sin~47°l5rir' = 1-9378220 x727= 218 = 3-3952040 log sin 60° 4' 18 60 log 2484-3 3-3330478 log sin 47° 12' 17" = 1-8655693 3-4674785 = 3-4674749 36 30 6 log 2934-1 .-. a = 2934-1-24. 10(3 SOLUTION OF TRIANGLES WITH LOGARITHMS. [CHAP. «r, X C'-J3 c-b A 4367 ...o.r, 27. tan-^-^— cot- = ^gg-^cotl5°45. log tan -^-^ = 3-6401832 + -5497060 - 3-6690378 = -5208514 log tan 73° 13'= -520568 1 diff. 2833 Diff. for 60" = 4568; 2833 .-. prop', increase = -=777. x 60" = 37". ^ 4568 ^-^ = 73° 13' 37", and ^±5 = 74° 15'; 2 A .-. (7=147° 28' 37", B = r V 23". a = sin C : 3-6548501 1-7180851 Again log 4517 log sin 31° 30' = 3-3729352 log sin 32° 31' 23" = 1-730490 7 log a =3-6424445 log 4389-8 =3-6424447 csinyl 4517 sin 31° 30' sin 32° 31' 23" ' log sin 32° 31' = 1-7304148 =^xl981= 759 bO log sin 32° 31' 23" = 1-7304907 a = 4389-8 nearly. 28. sin A = a sin G 324-68 sin 35° 17' 12' 421-73 S-o ><"»*= log sin 35° 17' = 1-7616424 357 log 324-68 = 2-5114555 2-2731336 log 421-73 = 2'62o0345 log sin ^ =1-6480991 log sin 26° 24' = 1-648003 8 diff. 953 Diff. for 60" = 2544; 953 .-. prop', increase = ^^Tr>^ 60" = 23"; ^ 2544 ^=26° 24' 23", and .-. J5 = 118° 18' 25". XVl] SOLUTION OF TRIANGLES WITH LOGARITHMS. . . , csinB 421-73 sin 61° 41' 35" Again 107 Bin G i 5in35°17'12" log sin 61° 41' = 1-9446501 35 60^ 680= 397 log 421-73 log sin 35° 17' log 642-75 6 = 2-6250345 2-5697243 12" = 1-7616781 2-8080462 = 2-8080421 41 41 .-. b= 642-756. 29. Bin C: c sin A 435 -6 sin 36° 18' 27' a 321-7 log sin 36° 18' = 1-7723314 27 60 log 435*6 X 1719= 774 = 2-6390879 log 321-7 2-4114967 = 2-5074511 log sin C =1-9040456 log sin 53° 17' = 1-903958 7 diff. 869 Diff. for 60" = 943; prop', increase = — - x 60" = 55" ; (7 = 53° 17' 55", or 126° 42' 5", both values being admissible since a' = 37°37'-5. Again. ^^^sin_C;^16;8sin45^r sinB sin37°37'-5 ' log 16-8 = 1-2253 log sin 45° 7'= 1-8503 1-0756 log sin 37° 37'-5 = 1-7857 log c = 1-2899; whence c = 19-49. TK . B-G b-c ^A 57-8 15. tan -2- = — cot - = ^^ cot 18° 30'. log 57-8 = 1-7619 log cot 18° 30'= -4755 2^2374 B+G_ log 108 = 2-0334 —g— -71°30', Ti — G ' Ti c log tan — -— = -2040 ; whence — ^ = 57° 59'. :. 5 = 129° 29', and C=13°3r. Again. a = "-'i° ^- = 252^'° 37° sinC sin 13° 31' ' log 25-1 = 1-3997 log Bin 37° = 1-7795 1-1792 log sin 13° 31' = 1-3687 log a = 1-8105 ; whence a = 64-65. 16. tan -^ = _ cot - = ^,^ cot 30°. log 6-48= -8116 log cot 30°= -2386 rO502 ^ + ^_rno log 60 -48 = 1-7816 ~2~~ ' log tan -^ = 1-2686 ; whence ^^=10° 31'. .-. C = 70° 31', and B = 49° 29'. 9—2 112 F SOLUTIONS OF TRIANGLES WITH LOGARITHMS. [cHAP. 17. tan — - = ^^ ^°* 2 = 22 ^^^ ^^ ^ = U "°* '^^ ^ ' log 4= -6021 log cot 41° 7'= -^90 log 11 = 1-0414 ~2~-'^'^ ^'*' log tan ^i^ = 1-6197; whence^- =22° 37'. .-. ^ = 71°30', andi? = 26°16'. 18. Let a = 2b, and C=52°47', then tan ^1^ = ^ cot 5 = i cot 26° 23' -5 2 a+b 2 3 = 5x2 -0152 = -6717 = tan 33° 53'. .'. ^^^ = 63°36'-5, and ^-^ = 33° 53'. .-. ^=97°29'-5, andL' = 29°43'-5. log 44 = 1-6435 log cot 9° 8'= -7938 2-4373 B-\-G log 218 ^ 2-3385 2 ~ ' log tan ^^= -0988; whence 5-^ = 51° 28'. £1 2 .-. 7? = 132° 20', and C = 29°24'. ^ li-A b-a a 6-43 ,^^0.0, 20. tan -2- == ^^ cot - = --3 cot 60° 49'. log 6-43= -8082 log cot 60° 40' = 1-7470 •5552 ^±^-29° 11' log 41-63 = 1-6194 2 ~ ' log tan ^- = 2-9358 ; whence ^— = 4° 56'. Again, = .-. i5 = 34°7', and ^ = 24° 15'. asinO 17 -6 sin 121° 38' 17-6 sin 58° 22' sin A sin 24° 15' sin 24° 15' log 17-6 = 1-2455 logsin58°22' = i-9301 1-1755 log sin 24° 15' = 1-6135 log c = 1-5620; whence c = 36-48. SOLUTIONS OF TIUANGLKS WITH LOGARITHMS. 112 G b sin A _ 100 sin 40° -^ A t ~ sin B ~ siiTTO'^ log 100 = 2 log sin 40°= 1-8081 rsosi log sin 70° = 1-9730 22. h = log a = 1-8351 ; whence a = 08-41. asini? 85-2 sin 42° 23. a sin A sin 31° log 85-2 = 1-9304 log sin 42°= 1-8255 1-7559 log sin 31° = 1-7118 log 6 = 2-0441; whence Z> = 110 -7. csin^ 5-23 sin 49° 11' 5-23 sin 49° 11' sin C " sin 109° 34' ~ sin 70' 26' log 5-23= -7185 log sin 49° 11' = 1-8789 •5974 log sin 70° 26' = 1-9742 loga= -6232; \Yhence a = 4-200. ftsinO 873 sin 71° 35' ^ ^^ sinX' ~ sin 42° 58' * log 873 = 2-9410 log sin 71° 35' = 1-9772 2-9182 log sin 42° 58' = 1-8336 logc = 3-0846; whence c = 1215. _csin^ 60 sin 60° ^^' " ~ siiTc' ~ sinr40°l0' * log 00 = 1-7782 log sin 60° = 1-9375 1-7157 log sin 40° 40' =1-8140 loga=l-90i7; whence a = 79 -75. 112 H SOLUTIONS OF TRIANGLES ^yIT^ LOGARITHMS. [CHAP. 26. a = cs'mA 3-57 sin 51° 51' sin G sin 40° 26 log 3 -57= -5527 log sin 51° or n. 1-8056 •4483 Iogsin40°26' = r8120 loga= '6363 .-. a = 4-328. h = csinB 3-57 sin 87° 43 sinC sin 40° 26' log 3-57= -5527 log sin 87° 43' ^ 1-9996 •5523 log sin 40° 26' = 1-8120 log 6= -7403 .-. Z^ = 5-499. 27. ashiE 125-7 sin 65° 47' ^^y[T~ sin 61° 34' log 125-7 = 2-0993 ^ log sin 65° 47' = 1-9600 2-0593 log sin 61° 34' = 1-9442 log& = 2-1151; whence /> = 130-3. 28. a c sin A sin C 92-93 sin 72° 19' ~ sin 24° 24' * log 92 -93 = 1-9081 logsin72°19' = i-9789 1-9470 log sin 24° 24'= 1-6161 log «= 2-3309 .-. ^1 = 214-2. h = c sinB sin C 92-93 sin 83° 17' ~ sin 24° 24' " log 92-93 = 1-9681 logsin83°17' = l;9970 1-9651 log sin 24°24' = 1-6161 log & = 2-3490 .-. /> = 223-4. 29. h = a sin B sin 4 4-375 sin 49° 30' sin 60° log 4-375 = log sin 49° 30' = I log sin 00° =1 logfc = .-. Z> = 3-841 0410 •8810 ^220 ■9375 •5845 a sin C sin A 4-375 sin 70° 30' ~ • sin 60° log 4-375= -6410 log sin 70° 30' = 1-9743 •6153 logsin60° = i-9375 log c = -6778 .-. c = 4-702. XVI.] SOLUTIONS OF TRIANGLES WITH LOGAltlTHMS. 112 I 30. ABC A^B+C 180^ 14 7 12 ~ 12 .-. .1 = 15^ I? = 60°, C=105'= = 15°. a = 31. b sin A _ 89-36 sin 15° sin B ~ "s'irTeO^ log 89 -36 = 1-9512 log sin 15° = 1-4130 1-3642 logsin60° = i-9375 log a = 1-4267 .-. a = 26-71. sin B = c = hsinC _ 89-36 sin 75° sin B srn"60° log 89-36=1-9512 log sin 75° =1-9849 r936l log sin 60° = 1-9375 log c = 1-9986 .-. c = 99-68. 6 sin .4 62 sin 82° 14' a 73 log 62 = 1-7924 log sin 82° 14' = 1-9960 1-7884 log 73 = 1-8633 log sin i? = 1-9251; whence ^ = 57° 18'. 32. sin C= c&xnB _ 63-45 sin 27° 15' ~"& ~ ^r62 ' log 63-45 = 1-8024 log sin 27° 15' = 1-6607 1-4631 log 41-62 = 1-6193 log sin C = 1-8438; whence C = 44=16', or 135° 44', both values being admissible since h - log sin o = - o ^°o ^ ~ 1'5485. = 20° 42', and /I =41° 24'. X -^-C h-c ,A 11-29 ,^.3oo^, ^ 43. tan-^3— = -— -cot- = — — cot23°37'-o. 2 h^-c 2 38-9o logcot23°30'= -3596 subtract for l'-5= 5 (the mean of dilTs. 3 and 7) •3591 log 11-29 = 1-0527 1-4118 log 38-95 = 1-5905 B-C B + C = 66°22'-5, log tan '-^—^ = i-8213 ; whence ^^-^ = 33° 32'. '. i? = 99°54'-5, and C = .32°50'-5. XVI.] SOLUTIONS OF TRIANGLES WITH LOGARITHMS. 112 M . . _b sin A _ 25 -12 sin 47° ID' _ 25 -12 sin 47 ' 15' gam, a_ ^.^^, _ ^j^^ g^^d 54/. - sin 80" 5'- 5~' log 25-12 = 1-4000 log sin 47° 15' = 1-8658 r2658 log sin 80° 5' -5 = 1-9935 log a = 1-2723 ; whence a = 18-72. 44. tan-^r— =— — eot- = — — - cotlo°4o'. 2 C + & 2 4667 log 4367 = 3-6402 log cot 15° 45'= -5498 ^^"^^ ^tl^-74°15' log 4667 = 3-6691 2 " '* ^"^ ' log tan -^= -5209; whence ^^ = 73° 14'. .-. (7 = 147° 29', and7? = l°l'. . _ c sin yf _ 4517 sin 31° 30' _ 4517 sin 31° 30^ gam, a- ^.^ ^ _ gj^ ^470 29' - ^~sin 32°~3r"" " log 4517 = 3-6549 log sin 31° 30' = 1-7181 3-3730 log sin 32° 31' = 1-7304 log a = 3-6426; whence a = 4391. csin^ 435-6 sin 36° 18' 45. sinC = a 321-7 log 435 -6 = 2-6391 log 36° 18' = 1-7723 2-4114 log 321-7 = 2-5074 log sin 0=1-9040; whence C=53°17', or 126° 43', both values being admissible since a'I> = 192 ft., i>^A' = 216ft. Let AB = X ft., Z EAD ^ iCAB^d, l DAB ^ a. 216 ^ 192 ^ ^, 6 Then tan(a + ^) = — ; tana = ; tau(?=-; X X X X' x2^6 X 192 216 192 6 — + - .T X 198 X , 6 192 1 X X X X From this equation we obtain a: = 48*,y6; .-. breadth of river = 48 ^6 = 117-6 feet nearly. 13. We have at once from a figure, tan a = - , tan /3 = - , tan 7 = - X ^ X X 116 Now MEASUREMENTS IX ONE PLANE. a + ^ + 7 = 180°; . tan a + tan /3 + tan 7 = tan a tan /3 tan 7 ; [chap that is, a b c ahc .-.- + -+ = a// iLi JU JL {a + b + c) x^ = ahc. ■A > I 14. See figure on page 188. Let P be the top of the hill, A and B the points of observation ; then aPAC = 45°, iB AC = 15'', iPDC^ld''; :. iBPA=15°-A5° = S0°= IPAB; .: PA = 2AB cos 30'^ = 500 ^^3 yards ; .-. height of hiU = PC = PA sin 45° = 250 ^6 yards = 750^/6 feet. 15. We have Z CBA = 30°, aBCA = 135° ; .-. Z BA C = 15° ; i5Csiul35° „ 2 I AB = sin 15° 1760x3x /3 - 1 ft. = 1760x3(^3 + 1) feet; .-. height of mountain ^AB sin 60° = 880 x 3 (3 + ^/3) = 12492 ft. 16. See figure on page 188. Let A, B he the two points of observation and P the top of the hill; then in the figure iPAC^a, iPDC^y, AB = C ft.; .'. LAPB = y-a, I ABP = Tr - {y - §), sm (7 - a) .•. height of hill =:^P sin a = c sin a sin (7 - /3) cosec (7 - a) feet. 17. In the figure let P be the top of the mountain, and A, B the two points of observa- tion. Then ^£ = 2?^ = 800 ft.; Z BAE = 15°, Z BET) = 30°. Also ZPX> = 75°, ZP.-1C'=60°; .-. Z JPD = 15°. From A A BE we have ^P = 2.J A' cos 15° = 1600 cos 15° ft. ; J f <75T, ID ^i? sin 120° and from aAPB, AP = -. — z-^— sm lo = 800 V3 cot 15° = 800 ^3 (2 + ^3) ft. ; .•. height of mountain = AP sin 60° = 400x3 (2 + ^3) = 4478 ft. ai)proximately. 4 XVII.] PROBLEMS DEPENDENT C)\ GEOMETRY. 117 EXAMPLES. XVII. b. Page 190. 1. In the figure of page 18(> let L'P = 25 ft., PA = 15 ft., CA ^x feet. Then -- = ?^ = ^ [Euc.vi. 3]; X lo o .•. ., = TT ; whence .T = 80; .T^ 9 thus the width of the road is 30 ft. 2. Let C be the position of the observer, B the top of the statue, 4 the foot of the cohimn. Then if CB — x feet, we have X a I CA=ra = S [Euc.vi. 3], X 1 that is, — r^T-r- = - ; whence x = aj2. Jx' + Ua^ ^ 3. See figure on page 189. Let BL be the flagstaff, LA the tower, C the observer ; then BL = a, LA = b, DA = CE = d, EA = CD = h ; also BC-= CE-2 + EB^ = cP + {a + h- hf, CA^= CE^ + EA- = d^ + /i2. IW_BL , (a + 6 -;?)- + (?- _a". CA~l2'' " W+d^ ~b-^' {a + by-- 2h {a4-h) _a-^-h'\ h^ + d^ ~~lF'' a+b-2h a-b Now or h^ + d^ h- ' whence (a -b)d'^ = {a + b) b- - 2bVi -{a-b) h"-. 4. See figure on page 190. From 0, the centre of the circle, draw OL, OM perpendicular to AB and DC respectively; then L, M bisect AB, DC. Let DC = 2x feet. Then z/3 = ^zDO(7at centre = z COM. Now CD = 2x^2 CM = 2 OM tan ^ = 2EL tan ^ = {a + b) tan ^, since 2EL = EB + EA=a + b. 5. With the same figure and notation as in the last Example, we have ED = Ar, -= 20 ft. , and (i = 45°. .'. 2x = {EB + EA ) tan 45° = 2EB + 20 ; .-. x = EB + 10. II. E. T. K. 10 118 Again proble:ms dependent on geometry. [chap. ED. EC = EB.EA = EP{EB + 20). .-. 20(20 + 2a;) = (a:-10)(x + 10); whence a: = 50. Thus the height of tlie cohimn is 100 ft. 6. Take the figure on page lUO, interclianging the letters xi and !>. Then •we have lBDA= iBCA = a-^; lABD= /.ACE = 90°-a; .: iBBC^^- iABD=:a + ^-90°. Now from a CBD, BD sin DBC _BD sin (a + /3 - 90^) ^ I i CD = BD sin BCE AB sin BAD sin B~DA~ sin BCE a sin BCE ' sin (a-^)' .-. CD = a sin (a + /:i- 90°) cosec (a -/3). 7. See figure on page 191. Let CB be the j)illar, BA the pedestal, E the point where the pillar 'sub- tends its maximum angle 30°. Then using the same construction as in Ex. iii. page 191, we have a = 30°, ^J = 60ft. Now / AEB = L ECB = ^ / EDB = \ (90° - 30°) = 30°. CB = 2CF= 2DF tan 30° = 2 x 60 x ^^ = 40 ^3 ft. 8. Let 0, P be the two posi- tions of the observer ; let AAPO = d. Then lABP, in alternate segment, =d. „ . _ AP sm 3 Now AB = — . — -^ sin 6 _ c sin a sin |3 sin 6 sin (a + 6) __ 2c sin a sin /3 C cos a -cos (a + 2^) ' But, from A OPB, a + 2d + ^ = 180° ; .•. AB = 2c sin a sin /3/ (cos a + coR/3). 9. Let OA be the tower, ^17? the flagstaff, P, Q the points at which the flagstaff subtends equal angles, ]l the point at which it subtends the greatest possible angle ; then since z APB= L AQB ; .'. B, A, P, Q are coucyclic and OP . OQ = OA . OB. XYTI.] PROBLEMS DEPENDENT ON GEOMETRY. 110 A^'ain since ABU is the greatest angle subtended at a point in fX^ by the str. line AB; .-. a circle can be drawn to pass through A, Li and touch OU at R; .: OA.OB=OIi"; .: OP.OQ = OR-; that is OP, OR, OQ are in geometrical progression. 10. Here ABCD is a cj'clic quadrilateral; AB BD BD DC sin ADB sin BAD sin BCD sin CBD' .: AB sin CBD= CD sin ADB. 11. Since ABED is a cyclic ciuadrilateral, we have IADC= zEBC = y, and iBDC^^. Also lBDA=y-p, and l ACE = Tr - (a + ^ + y). j^^_ BD sin ^ _ AB sin (a + ^) sin 13 "~ sin (a + j3 + 7) ~ sin (7 - j3) sin (a+^a + 7) * 12. Here the points P, Q, R, S, A are concyclic, and lRAS= aPAQ, since AR, AS are perpendicular to AP, AQ; .: PQ = RS= ^400 + 100 - 2 x 200 cos 30° = V^OO^OO^ = 12-4 ft. nearly. 13. Let A, B he the two beacons, P, Q the positions of the ship at the end of 3 min. and 21 min. respectively. Let iABP = a, iPBQ-O. Then it is easily seen that Z0.4P = 90° + a, /.OAQ = ^0° + a + e. Also from the a OBQ v;e have a+ ^+90° + a = 13.j°, so that a -f ^ = 45° - a. If PO=x, OQ = lx; AP hence x = sin 45 II 5 —sin (90° + a) = AB ^/2 . sin a cos a o«in2a (1). Again 7-r = V2 Bm 45° ^ ' = ABJ2 sin (a + d) cos (a + 6) K K If = — - sin 2 (a + 0) = — sin (90 - 2a) = - cos 2a. V"^ v^ >/2 10—2 .(2). 120 PROBT.E^rS rH>:PEXI)ENT ON GEOMETllY. [CHAP Squaring and adding (1) and (2), we have 1 50.1- = 25 .r = - mile. :. the ship sails a mile in G min., or at the rate of 10 miles an hour. Again if ?/ miles be the distance from at which the beacons subtend the greatest angle, we have And the ship will travel tbis distance in ^^ x — , or 3 ^/7 minutes. 14. Let AB be the flag- staff, BC the tower, D and E the first and second positions of the observer respectively. Then since AB subtends the maximum angle at £, the circle round ABE touches DC at E, so that z BEC= L EAB = d, suppose. .'. a + d^ lEBC = dO''-d; .'. 2^ = 90° -a; also iEAD = a + d~p = 90°- ^-/3. AE sin a Now AB = sin (5 + a) ' AE = -: ED si n/3 _ AB=-.-";:^ a sin a sin /3 sin (90^ - > - /3) cos{d + li)' 2a sin a sin j8 '' sin {d + a) cos (^ + /S) sin {2d + a + /3) + sin (a - ^3) 2a sin a sin ^ ~ cos)3+siu(a-j3J * EXAMPLES. XVII. c. Page 195. 1. Let A be the top of the hill and B its projection on the horizontal plane tli rough P, Q. Let .42? = .T yards ; then BP = BA = x, BQ = BA cot ^0° = xj'i; .: 3x2 = a;- + 500-; .-. 0- = 250^/2; tbat is, height of the hill = 250 ^2 yards = 1000-5 feet. XVII.] MEASUREMENTS IN MORE THAN ONE PLANE. 2. Let P be the top, and Q the bottom of the spire; then 121 A Q = 250 cot 60° = ^^.^ feet, BQ = 250 cot 30° = 250 ^/3 feet ; .-. AB" = BQ''-AQ^-= 250- ( 3 - ^) ; .-. AB = 2o0.—^.^ = — ;^ feet. 3. Let P be the top and Q the bottom of the tower ; then lPAQ = QO°; .: Q A = '660 cot 60°= -j- ieet smd QB=QP = SGO feci ; .-. breadth of nveY = AB= ^/BQ^-- QA- =360 ^l - \ = V20^, 4. See figure on page 194. Let CZ> be the steeple, then Z C4Z) = 45°; .-. CA = CD = x; and iCBD = \b°', :. CB = .c cot 15° = a; (2 + ^3) ; .-. a:-(7 + 4V3) = a;2 + 4a2; 2a2 a2(4-2j3)_ /6 feet. .•. x- = 2^3+3 ^3 .-. height of steeples -HV"=«(^^ -3-4) 4/3 5. Let AB be the lighthouse, and let CP, T)Q represent the two positions of the observer, 21, N the extremities of his shadow at each place. Jjr'M Let AB = x feet, then x : BM=PC : CM =6 : 24; .-. BC = BM - CM= 4x - 24. 122 MEASUREMENTS IX MORE THAN ONE PLANE. [CHAP. Also x: nX=QD :T)X^C): SO; .: BD - BX -DN=5x-30', :. 25 {x - 6)- = 300- + 16 (x - G)- ; .-. 9 (.T- 6)2=: 3002; whence .t = 106 or -94; .-. the hght is 106 feet from the ground. 6. Let A be the balloon, and C, B, D the points of observation at Avhich the angles of elevation of the balloon are 60°, 30^, 45^ respectively. Let X yards be the height of the balloon ; 2x then AB = x cosec 30° = 2.r, AG = x cosec 60° = -" , AD = x cosec 45° = .r ^/2. Now 2AD- + 2BD'- = AB^ + AC^', :. 4x2 + 2 X 8802 = 4j;2 + -^-; o whence x = 880 . ^ ^ = 440 ^^6 ; .•. height of balloon rr 440,^/6 yards. 7. Let A be the top of the mountain, BC the base of length 2a, D its middle point. Let X be the height of the mountain ; then AB = AC = x cosec 9, AD = x cosec C= 19-54 yards. Again, BC tan 10° 45'' log J;C= 1-2911 log tan 10° 45'= 1-2780 log7;C = 2-012(J .-. />'C= 102-9 yards. XVII.] PROBLEMS REC^UIRING THE USE OF TABLES. 125 3. If the required height is x feet, we have 500 sin 26° 3-i' x sin 12° 32' ^~ ^il4° 2^ * log 500 = 2-6990 log sin 26° 34' = 1-6505 log sin 12° 32' = 1-3364 1-6859 log sin 14° 2' = 1-3847 log a; = 2-3012 .-. a; = 200-1. 4. Let B be the position of the boat, FG the flagstaff, CD the cliff, then, if CD = x feet, we have 30 sin 43° 46' . ..,., sm 2° 6 log 30 = 1-4771 log sin 43° 46'= 1-8399 log sin 44° 8' = 1-8429 1-1599 log sin 2° 6' = 2-5640 log a; = 2-5959 .-. height = 394 -4 ft. X Again, BD = tan 44° 8'' log .T = 2-5959 log tan 44° 8' = 1-9869 log 5D = 2-6090 .-. distance = 406 '4 ft. 5. With the figure on page 184, we have I PBC =10°, iPAC = 5°; PB = AB = 52S0 it. Now PC = PB sin 10° = 5280 sin 10°. log 5280 = 3-7226 logsinl0°=i-2397 2-9623 .-. PC = 9l6-8ft. Again, IiC=I?P cos 10° = (1 X cos 10°) miles = -9848 milee. 126 PROBLEMS REQUIRING THE USE OF TABLES. [CHAP. 6. Let B be the point in the road which is vertically below the observer A. Let DC be the telegraph post, and let a horizontal line through A meet DCinE. Then /D^£ = 17°19', iACB = 8°SG'. Again, AE=BC=- 15 "tan 8° 36' log 15 = 1-1701 logtan8°36':-i-1797 log 4E = 1-9964 .-. J£ = 99-17 ft. DE = EA tan n° 19'. log ^^ = 1-9964 log tan 17° 19' = 1-4938 log DE = 1-4902 .-. D£ = 30-91 ft., and DC= 45-91 ft. 7, Here we may take the third figure on page 131. Then ^(7 = 60 miles, CjB.= CJBi = 30 miles, z C^ Bo = 20° 16'. 60 Now sin B 30 sin 20° 16' = 2 X -3464 = -6928 ; ■. J>' = 43°51', or 136°9' (sincea<6). lACBo = 23°do', z^CBi = 180°-64°7'. _ 30 sin 23° 35' ^^2- sin 20° 16' • log 30 = 1-4771 log sin 23° 35' = 1-6022 1-079:5 log sin 20n6' =-1-5396 log .4 5. = 1-5397 Ji>'., = 34-65 miles. XVII.] PROBLEMS REQUIRING THE USE OF TABLES. 127 ,„ 30 sin 64° 7' ^^«^^°' '"^^= sin 20^6'-- log 30 = 1-4771 log sin 64° 7' = 1-9541 1-4312 log sin 20° 16' = 1-5396 log .4i?i = 1-8916 .-. yll?i = 77-91 miles. Thus the train must travel at the rate of 11*55 miles or 25-97 miles per hour. 8. Let C be the doorstep, E the point of observation on the roof ; then CE = h. Let AB be the spire and let ED drawn horizontally meet AB in D ; then /LACB = 5a, /.AED = 4:a. Also ^EAC = a. Let AB = x. m jri • - /isin(90° + 4a) . ^ Ihen x = AC sin oa = ^, .sin 5a sin a = h cosec a cos 4a sin 5a. GB = X cot 5a = h cosec a cos 4a cos 5a. In the particular case a=7°19', 4a = 29° 16', 5a = 36° 35'. 39 cos 29° 16' X sin 36° 35' Then ^^ sin 7° 19' • log 39 = 1-5911 log cos 29° 16'= 1-9407 log sin 36° 35'= 1-7753 1-3071 log sin 7° 19'= 1-1050 log a;= 2-2021 .-. a; = 159 -2; thus the height is 159-2 ft. Again , CB = ; ^ = — ^L^^-, • tan 5a tan 36° 3o log a: = 2-2021 log tan 36° 35' = 1-8705 log CJ? = 2-3316 Thus the distance is 214-5 ft. 128 RELATIONS OF A TRIANGLE AND ITS CIRCLES. [CHAP. EXAMPLES. XVIII. a. Page 206. 1. Area = J, X 300 X 120 sin 150'^ = 300 X 00 x= 9000 sq. feet. 2. 26—171 + 204 4-195 = 570; .-. area = ^/285 xTi478r>r90 = ^/lO'^x 572x27^ = 15390. 3. Let a = 10, h = 147, c = 119 ; then s = 168. 2 / . -.^— ^-, — -^ri 2x12x7x49 84 4. Let a = 39, h = AO, c = 25, and denote tlie perpendiculars hyp^, p.^, 2h- Then area = ^52 x 13 x 12 x 27 = 12 x 13 x 3. 2A ^, 2A 117 2A 936 ••• i>i = — = 24, P2=-r=-E > 2^s a b ~ 5 ' ■^^~ c 25 30- sin 22i° sin 1121° 30^ sin 22^008 22^° 30- _. „^ 5. Area = - ^ ^.^- ^3^, = ^^^50 = ^ = ^^^ ^^l" ^'- 6. The diagonal bisects the parallelogram; .-. area = 42 x 32 sin 30° = 672 sq. feet. a2 7. Let a yds. be the length of a side. Then area = a^ sin 150° = -^ ; .-. |- = 648; .-. a = 36 yards. :. length of a side is 36 yds. 8. 13 + 14+15 = 42; .-. A = 2:= 4^2 (sin2 A - sin2 1^) = 42^^ gj^ {A + i?) sin {A - B) = 4i?2 sin C sin (.4 - B)=2Rc sin (^ - i^). «« Tn- . . -, . ^o sin2 A - sin2 B sin ^ sin B 30. Fu-st ..de = 4B^ ^ . ^5^^j3^,y 4i?2 sin (^ 4- 7J) sin A sin 2> 2 22? sin A .2RsmB . sin C «?> . ^ 2 =28^^^ = ^- 31. (1) '\Yeh&\e 2A = apj^ = bp^ = cp.^; 1 1 1 _a + b + c _ s _1 Pi p- Pi ^^ ^ ^ ,^,1 1 1 a + b- c s-c 1 (2) — + = =-— = -. i^i Pz P-i 2A A 7-3 32. ('1 - r) (r, - /•) {r., ~ i) := G47i^ sin^ ^ sin^ | sin^ ^ [Ex. 21] = 4i?r2. [Art. 212.] XVIir.] EELATIONS UF A TPJAXGLE AND T'lN CIRCLES. 131 33- Vr rj\r rj [r rj - ^^^ "7^ t^^^-H] 34. 4 A (cot ^ + cot B + cot C) = 2bc cos .1 + 2ca cos B + 2ab cos C' = ft2 + c'^ - a2 + c2 + a2 - t^ + r,'-i + b-^ - c2 = a- + t2 + c--. QR ^ - f c - a a - 6 _ (6 - c) (s - a) + (c - «) (5 - 6) + (a - M (s - c) ^1 r^ '3 A _ s{b -c + c-a + a-b)- [a {b - c) + b jcj-a) + c{a-b)} _ ^ _0. 36. a-bV- (siu 2A + sin 2B + sin 2C) = 4a^b^c^ sin A sin i5 sin C = 4:. be sin ^ . ca sin 7? . rti sin C = ^2^}\ 37. «cosyH-&cos5 + ccos(7 = 2i^ (sin^cos^ -f sin i? cos B + sin C cos C) = R (sin 2^ + sin 2B + sin 2 C) = 4R sin A sin ^ sin C. 38. rt cot ^ + & cot J5 + c cot C=2R (cos A+cosB + cos C) gsm-sin- = 27i ( 1 + 4 sin — sin — sin — = 2{R + r). . A 39. (i + c)tan-=2J2(sini? + sinC) L=.47^cos^^'cos:^^ cos — 2 = 2iJ(cosJ5 + cos(7); A .'. {b + c) tan — + two similar terms = 4i2 (cos ^ + cos i? + cos C). ABC 40. t' (sin ^ + sin 2i + sin C) = ir cos - cos - cos - At A ^ ...p . A . B . C A B C = IbR sm - sni - sm — cos — cos — cos - 2 2 2 2 2 2 = 2R sin A sin B sin C. l.']2 INSCRIBED AND CIRCUMSCRIBED POLYGONS. [CHAP. 41. Goa- ^ + cos2 - + cos2 - = - (3 + cos ^ + cos B + cos Cf) 2k 2t A A =: - ( -4 + 4 Sin -- sin B . C\ EXAMPLES. XVIII. b. Page 210. 1. By Art. 214, the area = - . 10 . 9 . sin 3G° = 45 X -588 = 20 -46 sq.ft. 3 2. By Art. 215, the permieter = 2 . 15 . ^ . tan 12° = 45 X -213 = 9-585 yds. q 135 X ''^13 The area = 15 . ? . tan 12° = j-^^ = 7-18875 sq. yds. 4 4 3. In fig. of Art. 215 let ^B = 2.'r = side of regular hexagon. Then OD = a; cot 30° = x ^3. .-. area of inscribed circle = Sirx'^. Again, in fig. of Art. 214, if ABhe the side of the hexagon, 'OA=AD cosec 30° = 2.T. .-. area of circumscribed circle = 47r.r2. .-. ratio of areas is 3 to 4. 4. With fig. and notation of Art. 217, area of pentagon = 5^ J) . OZ) = 5?-2 tan 36°; 22 .-. 250 = 57-2 tan 36° ; 7r/--^= x 50 cot 36" = 216-23 sq. ft. 5. Area of circle = 7rr-= 1386; whence r = 21. By Art. 214, perimeter = 16r sin 22° 30' = 16 x 21 x -382 = 128-352 in. 6. Area of circle = 7rr- = 616; whence r=14. By Art. 215, perimeter of pentagon = 2 . 5 . 14 tan 36° = 140 x -727 = 101-78 ft. 7. Area of circle = 7rr2= 2464 ; whence r = 28. Now in Art. 214, if AB is a side of the quindccagon, 07) = 28, and diameter of required circle = 2,10 = 202) sec 12° = 2 x 28 x 1-022 = 57-232 ft. XVIII.] INSCRIBED AND CIRCUMSCRIBED POLYGONS. 133 8. Let r be the radius of the circle ; then 5 27r for the pentagon, 50 = - r- sin — , 2i o 12 TT for the dodecagon, area = -— r- sin - ; f Art. 2141 2 o area of dodecagon _ 12 sin 30^ ' • 50 ^ T ■ sin72° ' .-. area of dodecagon = 60 cosec 72° = 60 x 1-0-515 = 68-09 sq. ft. 9. Let the perimeters of pentagon and decagon' be denoted by 10^/ and 106 respectively. Then as in Example 2, page 209, TT area of pentagon =oa^ cot ^ , 5 area of decagon = — - - cot -^ ; .-. 2«2 cot 36° = 62 cot 18°. Now cotn8»=i±i^:= r^ = |i-f; 1 - COS 36° V5 + 1 3 - ^/5 ' i + ^i-i and cot=36-= ^ = ■ ±^5 ; 4 cotn8° _20 ■* cot2 36°~T~ «^ _ v/o « _ 4^5 -■ b^~ 2 ' ""^ 6~72' 10. Let 2na be the common perimeter, so that 2a, a are sides respec- tively of the two polygons. Area of polygon of ii sides = - . 4a2 cot — . 4 11 271 TT Area of polygon of 2n sides = —- . a^cot — ; 4 271 .-. ratio of areas = ( 2 cos — sin ,-^ ) : ( sin - cos ~ ) \ 11 2nj \ n 2n) 2sm|.co8=.jJ = (2cos^):(l+, 2 cos - sin ^ « 2» TT cos — n H. E. T. K. 11 134 INSCRIBED AND CIRCUMSCKIHP:D POLYGONS. [CHAP. 11. In the fig. of Art. 211, if ^7) = ^, OA=E, we have 7? = rtCosec n In the fig. of Art. 215, if OD = r, we have r — a cot-. . TT . TT sm - sin a f 1 + cos — j 2(1 cos- — 2» ^ TT •—(I cot. ^ . TT TT „ . TT TT 2)1 2 sm - - . cos r— 2 sm ^r- . cos ^r- 2?i 27i 271 2w 12. Let p, /?, rf represent a side of the pentagon, hexagon, and decagon respectively inscribed in a circle of radius r; then p = 2r sin 36°, h = 2r sin 30^ (l = 2r sin 18°. ^4,.2 /'l^T^N^'j ^4,-^ sin236°=i;-. [See Ex. Ait. 126.] 13. A, = l nr^ sin — , B,^ nr^~ tan - , [Arts. 214, 215] .4 , = - 2n . r- sin - , jB., = 2n?-2 tan — . 2 2 ;t •* 2n .-. yti7?i== 7i-/-*. 2 sin — cos — tan— = /I V sin- — — .i 2^. 1^2 n n n n -^ Thus v4„ is the geom. mean between A^ and B^. 1 + cos — Again --- + --= — ^— — -( — — ^— — — — ^ 7i/--sin— 7??- tan— 7//-sin — 71 n n '^""''"k 1 2 1 ^ „ . TT TT ., , TT Bn 2)in Sin - cos - )ir- tan ^, - 2n 2;i 2/i Thus l^o is the harm, mean between A.^ and B-^. XVIU.] THE EX-CEXTRAL AXD PEDAL TllIAXGLES. 135 EXAMPLES. XVIII. c. Page 218. 1. Required distance A ,„ . A . B . C A . B C = r cosec - = Ui sin - sin .^ sin - cosec — = -iR sin — sin - . 2. A--i = ''i cosec -^ = -iiv sin -^ cos - cos - cosec - = 4A' cos - cos - , IjZ? = r^ cosec i 90"^ - ^J j = r^ sec ^ = -i/i sin ^ cos - , I^C = j-j cosec MJO^" - - J = i\ sec - = U{ sin ^ cos — . 3. (1) From the fig. of Art. 219, we have 1 ICG area = ^^ I^I.^ . I.^C=--iR cos ^ . r.j cosec ^ = 2i;'r3C0t ^ = 2i?s. A 1 -i B C (2) Ai-ea = 2 A's = 2 E . - = , A cosec ^ cosec — cosec -^ . [Art. 212,] 4. We have J/^ = IC cosec 11^ C = IC cosec — . ■i- B C i B C .-. rl/j . /J. . 7/3 = 4i? sin — sin — sin - . IJ . IB . IC . cosec ^ cosec -jr cosec - £1 i Jj 2 2 2 = 4Ji. J.4 .IB .IC. 5. Perimeter of pedal triangle = B. (sin 2^ + sin IB + sin 2C) = -liJ sin J[ sin B sin C In-radius of pedal triangle = 4 . ^ sin (90= - A) sin (90"' - B) sin (90^ - G) L^rts. 225, 212J = 2ii cos J cos 5 cos C. a /i\ ^ ^^ , ^ cos^ cos 5 cos C a- b- c- a b c _ '2hc cos ^ + 2cg cos ^ + 2a6 cos C 2ahc 2a6c ~ 2^<6^ * 11—2 136 THE EX-CENTRAL AND PEDAL TRIANGLES. [CHAP. (2) —2- 0+ + r=-— -COSyl + + 1 A 2abc = 0. 7. Let Pj , P2) P3 be the radii, then 72 C JT IC Pj = 4 — . sin — cos — cos -^ = 27? cos A sin jB sin C. [Arts. 225, 212. ] 8. (1) rtcosyi, &C0S7?, ocosC, 180^-2^, 180°-2J>\ 180'' -2C are the sides and angles of the pedal triangle of the triangle ABC. Hence in any formula connecting a, b, c. A, B, C, we may replace the sides by «cos^, b cos B, c cos G respectively, and the angles by 180'^ - 2A, 180^^ - 2B, 180^ - 2C respectively. 4 R C 4 P C (2) a cosec - , b cosec - , c cosec , 90° - ^ , 90° - - , 90° - — are Z 2i 2i 2i L 2i the sides and angles of the ex-central triangle of the triangle ABC. Hence the proposition follows. 9. We have SV- =^B? - 2Rr [Art. 228]. . . Jv- - 2Iir must be positive ; tliat is J? - 2r must be positive. Hence jR can never exceed 2r. 10. If R = 2r; I we have SP = R^ - 2Rr = 0; ' .•. the centres of the circumcircle and in-circle coincide, and hence the tri- angle is equilateral. 11. 672 ^ f^];^2 ^ SL/ + SLJ^ = R~- 2Er + R- + 2Ri\ -f R- + 2Br.. + i?= -f 2Br^ = ^B'-\-2R{r^ + r„ + rz-r) = 12R' [XVHL a. Ex. 25]. 12. (1) a. AI^ + b. BI^ + c.CP=r'^a cosec^'^+ + = abcr^ \ + 4- {{s - b) (s - c) abcrh abcf's- = abc. (s -a)(s- b) {$ - c) 6? (2) a. Ali^ - b . BI^- - c . CIj^ := i\'^ ( a cosec^ ^ - 6 scc^ ^-c sec2 ^^ \ ^ , o f 1 1 1_) ^^ |(s-6)(s-c) s{s-b) s{s-c)f abcr." -Is - (s-c)- Is-b) ] abcr.^ Is - a)^ = *— ^— , .-^ - - V ' = ^-- \. '- =abc. s{s-b){s- c) A^ XVlll.] THE EX-CENTRAL AND PEDAL TRIANGLES. lo7 OG A OBC 13. (1) We have — = ^-^^.. OG OH OK _ sum of areas of a" OBC, OAC, O AIi _ ^ '' A^'^ BH'^ (JK~ aABG (2) Since A, H, 0, K are concyclic, HK=AO sin A ; ^ ^ a cos A ^ , thus A0=—. — ,- =acot^. sin^ .-. OG + aeotA = OG + AO = AG, and the resiilt required is reduced to that already proved in (1). KH Rsm2A „ 14 Circum-radius of a AHK = _-^ — -. = ^.^ , - =RcosA\ ^^' 2 sm A 2 sin A :. sum of circum-radii of a' AHK, BKG, CGH=R {cos A + cosB + cosC) ( , . A . B . G = E ( 1 + 4 sm - sin - sm - = E + r. 15. We have A^ = ^- ^, _7r A^ _'K 1 /tt A\ 2=2~¥~ 2~2 V2~ 2/ ' _7r Jo_7r IJTT l/^_ i^M •'^~2~"2"~ 2~2 (2 ~2 \2~¥yj ' 1 ^^ ''2" 2 138 THE EX-CENTRAL AND PEDAL TRL\NGLES. [CHAP. similarly ^- = l + ( " ^^^J"- (^^ " t) ' IT When n is indefinitely increased, then ^„= ii„= (7„ = - . 16. (1) 06'-= E2-8i?2cos^ cosBcosC [Art. 230] = i?2^2E2(i + cos2.4+cos2Z? + cos2C) = 9ii2 - 2i22 (1 - cos 1A + 1- cos 275 -!- 1 - cos 26') = 9i22 - 42?2 (sin2^ + sin^ B + sin2 C) A B C (2) We have A0 = "IR cos A ; 1.4 = r cosec ^ = 4i2 sin — sin - ; Zl^O = (90°-J?)-^=^; B C B C C - B :. or- = 4R- cos2 4 + kj/js ginS si,v2 _ i6iJ2 cos A sin - sin - cos — —- = 4 72- ( cos^ ^1 + 4 sin- - sin^ ^ - cos ^ sin B sin C - 4 cos A sin^ - sin- - j = Mi- ( cos2 .4 + 8 sin- - sin- - sin"- ^^ - cos A sin 2? sin O j =r 2r- - 47?2 cos A (sin I? sin C - cos A) - 2;.2 _ 47^2 cos A cos £ cos C. (3) We have ^ B G C — B AO = 2RcosA; I j^A=i\co&ec- = 4:R cos — cos-; Al^AO= ^^ ; B G B G G — B :. Oli^ = 47J2 cos2 A + IQR- cos^ — cos^ - - 167?2 cos ^ cos - cos ^ cos — ^ / Tl C Ti G \ - U{- ( cos2 A +4: cos- — cos2 - - 4 cos A cos^ — cos^ — - cos J sin B sin G J = 47J2 ( cos2 ^ + 8 sin2 - cos^ ^ oos^ - - cos ^ sin B sin G j = 2ri2- 4/22 cos ^ (sin j] gin C-cos J) = 2rj'- - 472- cos A cos B cos C. < < J XVII I.] QUADillLATERALS. 139 17. Let W, p, pj, p.,, Ps represent the circum-radins, the iu-radius, and the ex-radii of the pedal trianj^'lo; then, since A, J!, C are the ex-centres of the pedal triangle, we have, as in Art. 229, r- = R'-^ + 2R'p^ ; also J2' = ^' . .-. fi + g"- + ]fi = 3 JJ'2 + 2R' (pi + p.:, + P3) = 3E'2 + 2R' {4:R' + p) [XVIII. a. Ex. 25] = llR'~ + 2R'p =z UR'^ + SR'-^ sin ^ sin ^ sin f- ^ ^ ^ = IIR"^ + 8i2'2 cos .4 cos i? cos C ; [Art. 221] .-. 4 (/-' + ^2+ 7,2) = 11222 ^ 87^2 cos A cos B cos C. [Art. 225. ] EXAMPLES. XVIII. d. Paoe 223. 1. Let r be the radius, and a, h, c, d the sides of the quadrilateral. Then we have 2.5 = ra + rh + re -f rd ; .•. r=-. (T 2. Let ABCD be the quadrilateral having sides AB = BC=S; CD=DA=4. Then z 7^.4 C = Z BC-t ; lDAC= lT)CA; I BAD + A BCD = 180"; .: i BAD = A BCD = 90°, and the a" BAD, BCD are identically equal. Thus it ea.sily follows that the bisectors of the Z =* BAD, BCD meet on BD which bisects the angles ABC, ADC. .'. a circle can be inscribed in the quadrilateral. If r be its radius we have r(3 + 3-f4-i-4) = 2 (area of quadrilateral) = 2.4.3. 1 1 5 Also radius of circumscribed circle = ZfJL> = y/3-+ 42 = ^^ = 2^. 2 2 ^ ~ 140 QUADRILATERALS. [CHAP. 3. Let ABCD he the quadrilateral, and let AB = 1, BC = 2, CD = i, DA = 3. Then AG^ = 5 - 4cos ABC = 5 + 4 cos ADC ; also AC^ = 25 -24 cos ADC; .-. 5 + 4 cosylD(7= 25 -24 cos ^DC. ^r.^ 20 5 •*• ^^^^^^ = 28^7' and area of quadrilateral = ^4 x 3 x 1 x 2 = ^24 ; J24: .-. radius of inscribed circle = - ^ = '98 nearly. 5 4. Let ABCD be the quadrilateral, and let AB = QO, BC ==25, CD = 52, DA=S9. Then ^^2^602 + 25^ - 2 . 60 . 25 cos ^BC = 52 x 13^ - 2 x GO x 25 cos ABC; also ^(7- = 522 + 392-2x52x39cos^i)C' = 52xl32 + 2x52x39cos^2JC; .-. cosJJ5C = 0, that is lABC = 90°; and hence the i ADC = 90°; .: ^02 = 602 + 252 = 52x132; .. AC=65; d5 1 Also the area= ^28 x 63 x 36 x 49 = 6^ x 72= 1764. 5. Let ABCD be the quadrilateral, and let AB = 4, BC = 5, CD = 8, DA=9. Then since AB + BC = 9, AC must be less than 9 ; .-. diagonal BD = 9 ; 1 ^ 2 . ^ ^/77 and cos ^ = ; .*. sm A= ; cosC7 = ^-^; .-. smO=-^--; .-. area = | {ad sin A + be sin C) = 18 ^J^ + 20 . ''-j^^ '^ J^^ + ^ v/H- XVIII.] QUADRILATERALS. 1 41 6. Since the quadrilateral is such that one circle can be inscribed in it and another circle circumscribed about it, therefore ^ ad -be , . ^ 2 J abed r » * .vj < i^i i cos A = —, — ,- , and sin A = — ^ — , — ; [Art. 2.^4, Ex.] ad + bc ad + bc .-. area = -{ad sin A + be sin C') = ijabcd. If ?• be radius of inscribed circle, - (a + 6 + c + d) = area = ^abcd. 2 J abed a+b+c+d 7. We have S- = {cd - 2abcd cos ^ cos 6\ . „A + C 4S2 = 2rt&cr? (1 - cos 4 + C) = 4abcd sin^ -— - A + C S = sjabcd sin — ^r — . 13. If /3 be the angle between the diagonals, we have S = -fg sin p. ... /V-'-4,S2=/2^2cos2/3 = |{(a2 + c2)-(Z>2 + rf2)}2 [Ex.10] = -T {2bd - 2rtc)2, since a + c = h + d; that is, 4:S^=f-9'' - {ac - bd)"-. 14. (1) By Euc. VI. I), we have {ac + hd) sin /3 =/;5r sin /3 = 2.S' = (arf + be) sin yl . ^ (a2 + c2)~(&2 4,tZ 2) " 2(ac + 6(/) * XVIII.] QUADRI LATERALS. 143 ^6) ^^'^\2~ l + cosfi~\c + ay' - {b- d)'' (b+dy'-{c-a)- {b + d + c-a)(b + d-c + a) [c + a-b + d) (c + a + b-d) , or (c + a -b + d) (c + a + b-d)' (b + d + c-a) {b + d-c + a) (L- = — - — = 7071 sq . yds. 2 144 MISCELLANEOUS EXAMPLES ON TRIANGLES. [CHAP. A 2A 2A 3. We have _ , — , s - a s - b s -c .-. h = c, and s-h = 2 {s-a); c+a-h , .'. 2~ =h + c-a; that is, - = 26 -a; that is, 3a = 46. 4. If «, h, c are in a.p., we have a + c = 2b; 1^ 1 _s-a + s-c _2{s-h) _2 " r^ r.^~ A ~ A "^3' .•. i\, r„, r^ are in h. p. 5. We have s = - + - + ^ . X y z 'V \^ y zj y z X \J y z 6. We have ^^JI.^J ^ sirt^ ^ Bin^^ +_^) . sin {B - G) sm C sin (^ + -B) ' .-. sin {A - B} sin {A +B) = sin (5 - C) sin (7? + C) ; or siu2 A - sin2 B = sin^ i? - sin^ C ; tliat is, «-, h-, c- are in a. p. 7 First id _ " ^^^^ + ^ sini? + c sin (7_ a'^ + 6- + c '^_ fl2 + &2.|.c2 sin ^ + sin i? + sin (7 ~ a + b + c ~ 2s * 8. First side = — ((( + b + c) sin - sin - sin — sin A ^ 2 2 2 _abc {s-a){s-b){s-c) A^ - 2A • ^' ^6^ - -A=^- 9. By Ex. 20, XVIII. a. we have ABC first side = abc cot — cot — cot — w ^ ^ = abc /T(^Z3I ^(g-6) s{s-c) V (s - 6) (s - c) ■ (s - c) (s - a) * (s - a) {s - b) dbcs = ~~r~ = 4i?s2 ^ 4JJ (,.^^,.^ + ^^^^ + ^^^ ) [XVIII. a. Ex. 24.] XVIIl.] MISCELLANEOUS EXAMPLES ON TRIANGLES. 1 i5 A B ^ G i\ u r, n+ro + r-j 10. tan-4-tan.^+tan-=^ + --f-= ^- 11. First side= be = !Vt!2±l3_ . [XVIIL a. Ex. 24.] (s -c){s- a) ^ 1 / "^ V (s-«] c) - a) (.s - h) -^ { he (s - a) + ca {s -b) + ab{s- c) } — {(&c + crt+a&) s -Sabc} (- + ^---) \a 6 c s / c s/ A abcs^ = 4Es2| = + -" + 1113' a b 12. First side=f ^^ ) " V A J - a + s - 6 + s - c\ _ 4 /'I 13. We have - = 6, 2s = 70; .-. A = 6x35. 4s f s-a (- + - + -) Let rt, 6 be the sides containing the right angle; then also -a6r=A=:6x35; .-. a6 = 12x35; 2 a2+&2^c2={70-(a + 6)}2; .-. = 702 - 140 (a + h) + 2a& = 702 _ 149 (« + &) + 12 x 70 ; .-. a + & = 41 = 20 + 21, and a& = 12x 35 = 20x21; .'. the two sides containing the right angle are 20, 21, and the hypotenuse = 70-41 = 29. t 14. It is easily seen from a figure that ^ = tan A ; .-. --\ h r = 2 (tan A + tan B + tan C) f h = 2 tan A tan B tan G= abc Ifyh' 14H MISCELLANEOUS EXAMPLES ON TRIANGLES. [CHAP. 15. Let r, ?•' be the radii of the inscribed circles; then since the peri- meters of the triangle and hexagon are ecjual we have 6r tan - = 12/ tan ^ ; whence Sr = 2r'; wr- _ 4 ■'• 71^2" 9' . . aucas of inscribed circles are as 4 to 0. 16. Terimeter = R (sin 2A + sin 2B + sin 2C) = 4J? sin .-1 sin B sin C _ a 6 c _abc ^ 2R' 2R' 2R'^2R^' 17. Area = s ^-^ cos ^ 4J2cos - sin (90° - '^ j 2 2 2 = 8if 2 cos - COS — cos — Ji ^ ^ = 2R- (sin A + sin L' + sin C) -n, 7 ^ ahc{a + b + c) 4A 18. Let Oj, O2 be the two circumcentres ; then O^^On is at right angles to -4 C at its middle point. Draw O^N^, OoN^ perpendicular to AB^; then O. O., = N-, N„ cosec A = ^^ T— -, . ^ - ^ - 2 sm yl 19. Let/, (7 be the diagonals; then by Euc. vi. i>., {ac + bd) sin/3=/£7 sin/3 = 2>S. [Ai't. 231.] . ^ 25- ac + bd 20. We have r.II^ = r. IG cosec Z/jC = ICr cosec ~ = IG. IB. :. r^II^ . II n . IIi = Lr- . I/?- . IC-\ 21. Sura of squares of sides of ex-central a = UR^( cos- 4 + cos2 ^ + cos2 ^ \ 2 2 2 = 16122 ('2 + 2 Bin ^ sin ^ sin ^"j [XII. d. Ex. 13] = 327^2 4. 32JJ2 sin 4 sin ^ sin ^ 2 2 2 XVlll.] MISCELLANEOUS EXAMPLES OX TRIANGLES. 147 22. Since a circle can be inscribed in the qnadrilateral, .'. a + c — b + d; and since a circle can be circumscribed about the quadrilateral, .-. cos ^ = ^"'i'^^^f ,^^"^ [XVIIL cl Ex. 14] ^ 2 (ac + bd) *■ _ {a + cf^b + d)" + 2 {ac ~ bd ) ~ 2 {ac + bd) (tc ~ bd ac + bd ' 23. (1) We have 21- + 2 .^-^ ^ U^ + c^ ; •i 2wi- + 2.^ = c2 + a-; 4 2n2 + 2.^ = o-- + 62; .-. by addition, 4 (Z^ + »i2 + n^) ^ 3 (a2 + b- + c^) . (2) First side = j { [b- - c~) (26'- + 2c" - a'-) + two similar terms} 1 "^4 = 0. — [{2 (64-c4)- a- (62- c-)} + ... + ,..] (3) 4.P = 2b" + 2c"-a"-; .-. 16 (Z^ + »i^ + 7i-i) = (2//- + 2c2 - a^y^ +... + ... = 9 {a^ + b^ + c*), on reduction. 24. We have ri + ro + r3=4E + r [XVIIL a. Ex. 25], 7Vo + r„r3 + 7-./i = s- [XVIII. a. Ex. 24], r^nr3 = rs2; [XVIII. a. Ex. 15] .'. the equation whose roots are r^ , ro , i'^ is x^ - {AR + r) a;2 + s-x - rs^=0. 25. Let PQ be the tangent to inscribed circle parallel to BC, and draw AHD at right angles to PQ and BC ; Ai ^P.^Q ^Jf Jfl" /AD-2r\- then ^ — — — ' ' A AB.AC AD AD ( AD-2r y- _ . 2A aY_{s- a)- s 2AJ s'-* A, A (s - a)' s- Ao A, r, = -,—" v2 > ^y symmetry. \s-by^ (s-c)2 148 MISCELLANEOUS EXAMPLES OX TRL\XGLES. [CHAP. Otherwise. Let the sides and perimeter of triangle APQ be denoted by a b C S A, A rT7 im But - = - = - = -;.-. -i = 7,. [Eiic. VI. 10]. Ai A That IS, (53^2 = 72- 26. ^^^ is perpendicular to the bisector of the angle .-1 ; .•. MN is parallel to IJ-^ ; thus the sides of the A LMN are parallel to the sides of the excentral A and the triangles are similar. A Also iV/iY=2(.s-«)sin-; A B 2 (s - a) sin — . 2 (s - h) sin - aLMN MN.nl ^ ' 2 ^ 2 •• AliVs hh'hh 4i2cos^.4iicos| 2 ^ _ (s-a)(g-fc)(g-jc )__A^__r^ 4E% 4iiV 4J2-' 27. Wehave lPBG= lPCB = A', .: I QPR = 1S0° -2A. Similarly i PQR = 180'' -2B, i QRP = 1S0° -2G. ACsinB _ bsinB ^ 6_ Again ^^ = ^i^iQC -■iiE2£"2cos5' similarly ^^ = 2^' 6 c ft cos (7 + c cos i? _ a ■ ^^~2'cos^"^2cos(7~ 2cos£cosC 2cos£cosC* 28. (1) '^Ve have xjc sin - +;?;& sin ^ = ^^ ^ ^^^ ^^^ ^ ' .-. _27(6 + c) = 26ccos-; j 1 il_^'+f _1/1 , 1 •■• p"""^ 2~ 2bc ~'2\b^ c 1 J? 1/1 1\ 1 c'_i/i ^\ Bimilarly - cos -^ = ^ ^ + «j ' r ^°' 2 " 2 V^i "^ b) 1 .4 1 J? 1 (^' 1 1^1 .-. - cos - + - cos — + - cos 73 = - + r + - . J) 2 q 2 r 2 a o c J XVIII.] MISCELLANEOUS EXAMPLES OX TRLVXGLES. 149 S^r/j-f-cos - cos — cos - ('2) "We liave vqr= ,, . , " , " . , _ 2^-7)2^2 (sin ^+ sin 5 + sin G) ~ {b + c)(c + a)\a + b) _ 4A ahc {a + h + c) ~ (?> + c)~(c' + a) {a-vh) ' pqr _ abc{a + b+c) ■'• 4A ~ '{b + cy{c^a)Ja:+b) ' 29. Let O be the orthocentre; then G, IF, K are the middle points of OL, DM, ON respectively; therefore the sides of the aLMN are double the sides of the pedal triangle and parallel to them. Therefore also the angles are equal to the angles of the pedal triangle. (1) Area of ALil/^'=- . 2acos^ .2&cos5 . sin (180° - 2C) = 4:ab sin C . cos ^i cos B cos G = 8A cos A cos B cos C. (2) ^L = ^G + OG= csin7?+ ccos7?cotC _ ccos(.B-C) __ g cos (J^ - C) , sin G ~ sin A ' .-. AL sin A + BM sin B + GN sin G = acos{B - C) + b cos {G -A) + c cos (A - B) = 2iJ {sin^B + Ccos ^^^+ ... + ...» = i? [(sin 2J5 + sin 2C) + ... + ...] = 2R [sin 2 A + sin 2B + sin 26'] = 8i2 sin A sin B sin G. 30. Let P be the centre of the circle inscribed between the in-circle and the sides AB, AG ; then (1) ^_Z^^sin4; ' ~ ' a " 1 • ^ 1 - sm — •'■ ^a = ^ • ~4 ~^ ^^^" — A~ ' [Compare XL f. Ex. 15.] 1 + sin- (1) Also '^Z+^^ + '^JZ^^:,!. ^ ' 4 4^42' •■• Jw^ + sJr^Ta + s/r^ri = r ( tan = r. H. E. T. K. TT-B^ TT-G ~—r— tan — — - + . . 4 4 ,.„.) 12 150 MISCELLANEOUS EXAMPLES ON TRL\NGLES. [CHAP. 31. It is easily seen that the triangle XYZ is the same as the triangle PQIl in Ex. 27. •. Perimeter; a n + + 2 COS B cos C 2 cos C cos A 2 cos A cos B R (sin 2 A + sin 2B + sin 2C) 2 cos A cos jK cos C = 2 Ji tan ^ tan B tan C I Area a 2*2 cos I? cos C ' 2 cos G cos ^ a& sin G . sin 2(7 4 cos A cos JS cos C R- tan ^ tan B tan (7. I 32. Let A'FZ be the triangle formed by the tangents, and let the perpendiculars from A^ Y, Z to the chord be represented by x,y, z respectively. Then the area of aXYZ= aBXG- AAGY+ aAZB ^\x.BG-\y.AC + ^~z.AB. X DG Now -— = -— , since OGX is a right angle. G ^ v. G C/ ,, GX DB DB Also --r = -— =: GO 01) I) are concyclic. since 0, By G, X X — - BG.DB :. aBXG = ^.DG .DB(DG-DB)=^^-^^—^\ if a, h, c denote the dis- 2p ' 2p tances oi A, B, G from D. .•. area of aXYZ = cb (f -b) -ca {c -a) + ba {b - a) _cb (c-b) + ba{b-a) + ac {a-c) _ {c -b)(b - a) {a- c) _ BG . CA . AB - 2p ~ 2^ ' for GA =c -a= - {a -c). XVIII.] MISCELLANEOUS EXAMPLES. V. 151 MISCELLANEOUS EXAMPLES. F. Page 228. 1. We have cos (a + /3) = cos (180° - 7 + 5) = - cos (7 + 5) ; .•. cos a cos /S - sin a sin j3= sin 7 sin 5 - cos 7 cos ; .•. cos a cos /3 + cos 7 cos 5 = sin a sin /3 + sin 7 sin 5. „ -c- , ., cos (15°-^) sin 15° -sin (15°-^) cos 15° 2. Fn-st side = . ., . ^ ^_-5 ^ sm lo° cos 15° 2sin/( = 4 sin A. sin 30° cos A sin /i sin C + 1 - cos- A 3. cot ^ + sin ^ cosec Z? cosec C = sin A sin i? sin cos^ {sin J5 sin C + cos(B4- C')} + 1 siu A siu B sin C cos A cos i? cos C+1 sin ^ sin^ sin G ' which is symmetrical with respect to A, B, C. . ,,, , . ^. &sin^ ^8. sin 30° 1 4. We have sm ii = = ^ = -.- ; a 2 V^ .-. B= 45°, or 135°; [Art. 148. (iii)] ; .-. C = 105°, or 15°; asinC a ■ -.eo — - = 4 cos lo°, or 4 sm 15° ; sm A .: c = V6 + ^2, or ^6-^2. c M\ ATT I iioo^ 000 2 cos 18° sin 36° 5, (1) We have cot 18° tan 36° = 2 sin 18° cos 36° sin 54° + sin 18° sin 54° - sin 18° V5 + l + (v/5-l) = s,/o. v/5 + l-(V5-l) (2) sin 36° = sin 144°, and sin 72' = sin 108° ; .-. first side = 4 (2 sin 72° sin 36°)2 = 4 (cos 36° - cos 108°)2 - 4 (cos 36° + sin 18°)^ = 4 (^^^ + ^Y"~^)' =5. 19 9 152 6. MISCELLANEOUS EXAMPLES. F. log 2 = -30103; .-. log 4 = -60206 log 3 = -47712; .-. log 9 = '95424 log 11 = 1-04139 [chap. 2-59769 .-. log -0396 = 2-59769; .-. log (-0396)90= -180 + 53-7921 = 127-7921; .-. number of ciphers before the first significant digit in {•0396)9*^ is 126. 7. Let P, Q be the two positions of the observer; then zgPi? = 30°, z Q/>»P=45°, Pg = 50 yards; 8. First side = 2 + - {cos 2a + cos 2^ + cos 27 + cos 2 (a + /3 + 7)} = 2 + cos (a + i3) cos(a-^) + cos(a+/3 + 27) cos (a + /3) = 2 + 2cos(a + /3) cos (/3 + 7) cos (7 + a). sin2 40° + cos2 40° 2 9. (1) tan 40° + cot 40^ = cos 40° sin 40° , = 2 sec 10°. 2 sin 40° cos 40° sin 80° (2) tan 70° + tan 20° = tan 20° + cot 20° : = 2 cosec40°. sin 40 -, as in (1), 10. (1) First side = 2 sin 4a- 2 cos 6a sin 4a = 2 sin 4a (1 - cos 6a) = 4 sin 4a sin- 3a = 16 sin a cos a cos 2a sin- 3a. (2) r u'st side = sin i=- - 2 cos — sin - = 2 sin - cos - - cos -— 7 V 7 7 . . TT . 2ir . 37r = 4 sin - sin — - sin -— - 7 7 7 ..7r.57r.37r = 4 sin- sin - sm — . XVIIl.] MISCELLANEOUS EXAMPLES. F. 153 11. We have sin C = '-^ = ^-~~^^q^ = -j^ ; .-. C = 45° or 135°; [Art. 148, (iii)]; .-. .t( = 105°or 15°; _ ^sin A _ ( 6-2V3);^/2(V3 + l) (6-2 ^3) ^2 (^3 - 1 ) *• """ sine " 2^2 '""^ 2^2 ' that is, a=2;^/3, or 4^/3-6. 12. Let C be the rock and A, B the two positions of the ship. Then we have z BAC^ L BCA = 671° ; .'. £C' = -B^ = 10 miles, and AG = 2AB sin22i° = 10 ^2 - x/2 miles. [Art. 251.] sin2 B - sin2 G 13. First side — 1 . +... cos jB + cos G ' " cos^C-cos^^ ~ cosjB + cos G . + ... = (cos C - cos £) + . . . + ... = 0. 14. We have 1 1 2 4 cos 6 cos a 2 cos 20 + cos 2a ~ cos 9 ' :. 2 cos2 (9 cos a = 2 cos^ ^ - 1 + 2 cos- a - 1 ; .*. cos^ e (cos a-l) = cob2 a - 1 ; .*. cos"^=cos a + l = 2cos-^ ; whence cos = >/ 2 cos - . 1 5 . We have sin a cos o = sin^ /3 = ^(l-cos2|3); .*. cos 2/3=1 - 2 sinaCosa=(cosa-sin a)^ = 2cos2( j + a) . 16. See figure of Art. 223. OG = BG cot BOG = BG cot G = ccosBcotG = 2R cos B cos C; similarly 0/f=2JJcos Ccos ^, Oi:= 2ii cos /( cos B. 154 MISCELLANEOUS EXAMPLES. F. [CHAP. „T , cosu-e 17. >>e have cos^ 1-e cos M 1 - cos 6 _l-e cos w - cos w + e _ (1 + c) (1 - cos u) 1 + cos^ ~ 1-e cosu + co&u-e (1 - c) (1 + cosu) ' ^ /l + e . 71 18. The sum of the squares on the sides containing the right angle = 4{l + sin(9)- + 'i(l + cos^)- + cos2^ + sin2^ + 4cos0(l + sin^) + 4sin^(l + cos0) = 8 + 8 (sin ^ + cos ^) + 4 + 1 + 4 (sin 6 + cos ^) + 8 sin 6 cos 6 = 9+12(sin0 + cos^) + 4(l + 2sin ^cos0) = 9 + 12 (sin ^ + cos ^) + 4 (sin + cos ^)- = {3 + 2(sin^ + cos^)}2. .-. hypotenuse = 3 + 2 (sin^ + cos ^). 19. In the figure of Art. 227 let be the centre of in-circle of 1x^2^3' and o'l the centre of the ex-circle opposite to 7j. Let R' be the circum- radius of IiIJ^s- Then, as in Art. 220, 00i = 4E'sin ^^2^1-^3 but Z I0I1I3 = ^ - 4 ; and E' = 2R ; [Arts. 221, 222] ; .-. OOi = 8Esin('^-^^ = 8Esin--±^. 20. The sides of the ex-central triangle of the triangle I1IJ3 are 4J^'cos'^^, 422' cos ^^, 4E' cos ^^ , [Art. 221], 4 4 4 that IS, 8iico8- , , 8i2 cos — ^— , 872 cos — . . 4 4 4 21. We have (1 + cosa) (l + cos/3) (l + cos7) = (l-cosa) (l-cos/3) (I-C0S7); a i3 7 , . a . iS . 7 .-. COS - COS ^^ cos ^ = ± sin r sni ^ sni ^ ; CL 3 ^ .•. each expression = 8 cos* - cos* ^ cos- ^^ a /ii 7 . a . /3 . 7 = it 8 cos - cos -, cos ^ . sui ,^ sin - sin - 2 2 2 2 2 2 = ± sin a sin (i sin 7. ( XVI 1 1.] MISCELLANEOUS EXAMPLES. F. 155 22. Let a, /3, 7, 5 be four angles such that a + ^ + y + 8 = 180° ; then cos (a + jS) = cos (180° - 7+^5) = - cos (7 + 5) ; .•. cos a cos j3 + cos 7 cos 5 = sin a sin /3 + sin 7 sin 5 ; similarly cos )3 cos 7 + cos 5 cos a = sin /3 sin 7 + sin 5 sin a ; cos 7 cos a + cos /3 cos 5 = sin 7 sin a + sin /3 sin 5 ; .•. by addition we have the sura of the products of the cosines taken two together equal to the sum of the products of the sines taken two together. 23. (1) III . IL. . II3 = 47? sin - . -iR sin ^ . 4iJ sin - [Art. 220] — 16i2- . AR sin — sin — sin — ^ Ii u = l&R"-r. (2) 7Zi=+ 70/3-= 16iJ2 sin2 ^ + 16i?2 cos^ "^ [Arts. 220, 221] = 16^2. 24. (1) Let a, ;S, 7 be the angles ; „5 ,C ^A , B+G B-C ,A cos- - + cos- - - cos^ — 1 + cos cos —^r- - cos- - 2 2 2 Z Z £ then cosa = ^ ^ ^ 2 cos — cos — 2 cos ^ cos -- . ^ £-C . „^ ^ . ^ B C sin — cos — -— + sin^ — 2 sm - cos - cos -^ " I ^ G "^ ' ~ ~ 2 cos — cos — 2 cos 2 cos - = sin — = cos ( 90° - — Thus the angles are 90° - ^ , 90° - f , 90° - ^ . Z z z (2) Here _ sin2 2B + sin2 2(7 - sin^ 2A _ 1 - cos 2{B + C) co s 2{B- C) - sin- 2A cos a- 2^m 25 sin 2C ~ 2 sin 2i? sin 2C __ - cos2>(co8 2(.B-C) + cos2/t ~~~ 2~sin 2B sin 2C _ - cos 2A {co s 2 (Ji - C) - cos 2 (J3 + C ) } _ _ ^ 4 ~ ^2 sin 2jB'sirr2(7 ~ Thus the angles are 180° - 2 A, 180° - 21^, 180° - 2(7. 156 MISCELLANEOUS EXAMPLES. F. [CHAP. 25. The expression = {sin {e + a) + sin (6 + (3)}" - 2 sin (6 + a) sin (^ + /3) - 2 cos (a -(3) sin {9 + a) sin {6 + /3) = <2sin ^ — ^cos^— ^( -2 sin (^ + a) sin (^ + /3) (1 + cosa^) = 4 cos2 -^ jsin2 — \ - sm (O + a) sin (^ + /3) [ = 2 cos2 '^~ ^ { 1 - cos (2^ + a + /3) - cos (a - /3) + cos (2^ + a + /3)} = 2 cos2 *^^ (1 _ cos"^^), which is independent of 6. 26. See figure on page 220. Since the quadrilateral is described about a circle, .•. a + c = b + d; that is, a-d = b-c. Now a^ + d^-2ad cos A = BD" ^h^' + c^ - 2bc cos C ; .'. (a - d)2 + 2ad{l- cos .4) = (& - c)- + 2bc (1 - cos C) ; ,•. a'iM =43° 10' 24". Also PC= 1566. Let AB = x ft. PB sin 43° 10' 24" X = sin 13° 14' 12 log sin 43° 10' =1-8351341 1^x1347= 539 logP^ =3-2741376 3-1093256 log sin 13° 14' 12" = 1-3597858 log X =3-7495398 log 5617-4 = 3-7495353 45 6 46 PB = 1566 cosoc 56° 24' 36". logcosec56°24'= -0793961 subtract — x 839 = 503 oO __^^_^ •0793458 log 1566 = 3-1947918 log PB =3-2741376 log sin 13° 14' = 1-3596785 12 ^^x5369= 1073 bO 1-3597858 Thus a; = 5617-46 ft., whence it easily follows that the speed of the train is 21-3 miles per hour. 29. Let A represent the harbour, C the fort, B the position of the ship when 20 miles from G. Then JC=27-23 miles, (75 = 20 miles, z (7.45 = 46° 8' 8-6". 27-23 sin 46° 8' 8-6" sin B: log 27-23 = 1-4350476 logsin46'^S' =1-8579078 174 ^xl2l5 log 20 1-2929728 = 1-3010300 log sin/; =1-9919428 log sin 78° 59'= 1-99192-20 ^208 20 Diff. for 60" = 246; 208 prop", increase = —j-r X 60" = 50-7". a^ = 32° 51' 42-1". 158 GENERAL SOLUTION OF EQUATIONS. [chap. In aACB^, In aACB^, 20 siu 54° 52' 0-7" i~ sin 46° 8' 8-6" ' 20 sin 32° 51' 42-1" -~ Bin 46° 8' 8-6" ' log sin 54° 52' =1-9126551 log sin 32° 51' =1-7343529 4 O 1 J^x889= 10 g„„xl956= 1372 log 20 =1-3010300 log 20 =1-3010300 1-2136861 1-0355201 log sin 46° 8' 8-6" = 1-8579252 log sin 46° 8' 8-6" = 1-8579252 logAB^ =1-3557609 logJ/?., =1-1775949 log 22-686 =1-3557579 log 15 -052 =1-1775942 30 2 38 ^^1 = 22-6862 miles. .-. ^^2 = 15 -052 miles. Thus the time taken is approximately 2*27 hours or 1-5 hours; that is the ship will be 20 miles from the fort in 2 hrs. 16 miu. or in 1 hr. 30 min. _ . 1 . TT 1. sm^ = - = smg; EXAMPLES. XIX. a. Page 235 2 TT .-. e=7i7r + {-iy'-. 3. cos^ = - = cos|; e = 2mr± IT 3 5. cot^= -V3 = cot f - ^j ; 7r .-. d = n7r- — b 7. cos2a = ^; cos ^ = ± . „ TT ^ I IT = 2/t7r ± - , or 2/i7r dL I tt - - . . 1 . TT sm0 = -^ = sm-; -. = n7r + (-l)« 7r TT 4. tan ^=^3 = tan ,y ; o .-. ^ = 7l7r + -. 6. sec ^ = - ,^/2 = sec — .-. ^ = 2;i7r±— . 4 8. tan2^ = |; •' . tan 5 = ± - = tan Jo \ Both of these are included in //tt ± TT XIX.] GENERAL SOLUTION OF EQUATIONS. 159 9. cosec2^ = -; o .-. cot3^ = J; o .-. cot ^ = ± -y,3 = cot ( ± -- v/3 V -^ ^ = ;t7r± 10. cos 6^ = cos a ; 11. tan'-^ ^ = tan- a ; .'. tan ^= ±taua = tau(±a); .•. ^ = «7r±a. 3* 12. sec- ^ = sec- a ; .•. tan- ^ = tan- a; 13. tan 2(? r= tan <^ ; 14. cosec3^ = cosec3a; .'. 3^ = 7J7r+(-l)'*3a; 15. cos3^ = cos2(9; .-. 3^ = 27i7r±2^; . « 2n7r /. 6 — 2mr, or 16. sin 0^ + sin = sin 3^; .-. 2 sin 3^ cos 2^ = sin 3^; .-. sin 30 = 0, 1 or whence cos 20: 2' = -^- , or 7l7r±^, o o 17. COS - cos 70 = sin 40; .-. 2 sin 40 sin 30 = sin 40; .-. sin 40 = 0, or sin 30 = - ; whence = -j- , or o" + ( ~ ^) Tg • 18. sin 40 + sin 20 - (sin 30 + sin 0) = ; .-. 2 sin 30 cos 0-2 sin 20COS = 0; .: 2 cos . 2 cos — sin - = ; T0 .'. cos = 0, or cos— = 0, or sin- = 0. ^ (2;i + l)7r (2?i+l)7r . .-. 0= ^ — , or ^ j~ — , or 2jt7r. 19. As in Example 18, we obtain 4 cos cos 40 cos 20 = ; whence 0: (2n+l)7r (2n + l)7r (2/t + l)7r or , or 8 160 20. whence 21. whence 22. 23. whence 24. 25. GENERAL SOLUTION OF EQUATIONS. sin 56 cos d= sin 66 cos 20 ; .: sin 6^ + sin i6 = sin 8^ + sin id ; .'. sin 8(9 -sin 6^ = 0; .-. 2 sin ^ cos 7^ = 0; .-. siu^ = 0, or cos 7^ = 0; a (2;i + l)7r 6=mr, or — — -^ . sin 11^ sin 4^ + sin 5^ sin 26' = .-. cos 7^ -cos 15^ + cos 3^ -cos 7^ = .-. 2 sin 9^ sin 6^ = - 7nr nir sJ2 cos 3^ - cos ^ = cos 56 ; ^ .-. ^2 cos 3^ = 2 cos hd cos 26 ; .-. cos 3^ = 0, or cos 2^ = 6 = (2u + l)7r TT "~ -n-^ > or 7i7r±_. o 8 sin76'-;^3cos4^=sin^; .-. 2 cos 4^ sin 3^ r= ^3 cos 4^; .-. cos 4^ = 0, or sin 36* = ^; l + cos^ = 2sin2^; .-. l + cos^ = 2-2cos-^; /. 2cos-^ + cos^-l = 0; 1 .-. cos^= -1, or -; .-. ^ = (27i+l)7r, or 2«7r± 7r tan2 ^ + sec ^ = 1 ; .'. sec2(9 + sec^-2 = 0; .-. sec^= -2, or 1; 27r .-. ^ = 2n7r±— , or 2n7r. [chap. • XIX.] GENERAL tSOLUTIOX OF EQUATIONS. IGl 26. cot^ ^ - 1 = coseo ; .-. cosec- 6 - cosec ^-2 = 0; .-. cosec 6 = 2, or — 1 27. cot ^- tan ^ = 2; .-. cot2^-l = 2cot^; .-. cot 2^ = 1; „ nir IT ••■«=2- + 8- 28. 2eos^=-l; .: d = 2jnr:L— (1), 27r 2sin^=V3; .-. ^ = n7r + ( - l^y (2). From (1) we see that the multiple of tt must be even, and from (2) that the sign before the second term must be positive when the multiple of tt is even; 27r .-. = 2n7r + -g-. 29. sec^=V2; .-. ^ = 2n7r±| (1), tan^=-l; /. d= nir -'^ (2). From (1) we see that the multiple of tt must be even, and from (2), that the sign before the second term must be negative ; 4 EXAMPLES. XIX. b. Page 237. :. tanjj^ = tan ( -- - cos 6' o C = 2?j7r+^, or 2//7r+g. XIX.] 6. GENERAL SOLUTION OF EQUATIONS. cos - ; .■.e-l = 2n..l; = 2inr + z.~ , or 2mr 12' 12 1G3 7. 8. 9. cos ^ - sin ^ = -.- ; .'. COS (-^)^ COS TT 3' TT ^ TT TT 77r .-. = 2mr + T^ , or 2?f7r - y^ . cos & + sin ^ + ^2=0; .-. COs(^-|)=.-l: .-. -7 = 2?i7r±7r; 4 .'. = 2w7r+-r» OJ^ ""^r j-. 4 4 cosec + cot = /^'S ; .-. l + cos^ = ^/3 sin ^; - cos - -V- sin 0= — 2' cos TT ^ 27r whence which may be written 10. TT = 2mr + - , or (2/t-l)7r; o TT ^ = 2«7r + -, or (2u + l)7r. cot ^- cot 2^ = 2; .•. cosec 26^ = 2 ; ''=2^ + (-l)"S' 104 GENERAL SOLUTION OF EQUATIONS. [CHAP. 11. 2 sin ^ sin 3^ = 1; .-. cos 2^ -cos 4^ = 1; .-. cos2^ = 2cos2 2d^; .•. cos 2^ = 0, or - ; . (2«+l)7r TT • • ^ — -, , or ?<7r±-. 4 b 12. sin 3^ = 8 sin^^; .-, 3 sin d-\ sin3 ^ = 8 sin^* 6 ; .'. sin ^ = 4 sin^^; .-. sin^=:0, or ±-: 2' TV :. d = mr, or mr±-i . b 13. tan ^ + tan 36' = 2 tan 2^; sin Ae _ 2 sin 26 cos 3^ cos cos 20 ' .'. sin 2^ = 0; whence ^r= — , 2 ' or cos2 2^ = cos^cos3^; .-. 2 cos2 20 = cos A0 + cos 2^ = 2 cos^ 20 - 1 t- cos 20 ; .'. cos 2^ = 1; whence 6=mr; :. all the values are included in = — . 14. cos ^ - sin ^ = cos 26 ; :. 2 sin — sin - = sm 6 ; a :. sin-=0, whence 6 = 2mr^ . 30 6 or sm — = cos - ; ••■2=2«T*(-2-2); TT 7r .'. d=mr+~, or -^ = 2/i7r--; 4 2 TT TT .'. the values of 6 may be written 2mr, nir-i- , 2mr+ -^. I I XIX.] GENERAL SOLUTION OF EQUATIONS. 165 15. cosec^ + sec ^ = 2y^2; .-. sin d + cos d = 'l J2 sin d cos d \ :. cos [ ^ - — J = 2 sin ^ cos d = sin 26 ; , ^ 2mr TT _ TT whence 6= — -- + -: , or 2inr+ — . 6 % 4 16. sec^-cosec^ = 2;y/2; .-. 1 - sin 2/? = 8 sin2 d cos^ ^ = 2 sin- 2^ ; .-. sin 2^= - 1, or ^ ; IT HIT , ^,„ TT .-. ^=n7r--, or _ + (-l)n^^. The equation may also be solved in the same way as Ex. 15. 17 _i L_ = 2. ■^^' cos 4^ cos2(? cos 2^ - cos 4^ = 2 cos 4^ cos 20 = cos 6^ + cos 2^; .". cos 6^ + cos 4^ = 0, 2 cos 5^ cos ^ = 0; .-. 6' = (2n + l)|, or 5^ = (2n + l)|. 18. cos 30 + 8 cos30 = O; .-. 4cos3 = cos0; .•. cos — 0, or ±- ; , (271 + 1) TT ^ TT ^ / 7r\ .-. = ^ 2 ' 2"^^3' 2«7ri/7r--j; ,-. . .1 1 n ^ (2n + 1) TT TT that is, the values of 6 are ^ -— ^ — , ?i7r ± - . 19. 1 -t- V^ tan-^ = (1 + ^3) tan 0; .-. (^3tan(9-l) {tan0-l) = O; .-. tan = 1, or -— ; - TT TT .'. = 7i7r + -, or ?i7r + 7r. 4 fa H. E. T. K. 13 166 OENEIJAL SOLUTION OF EQUATIONS. [cHAP. 20. tiui^ e + cot^ <9 = 8 cosec^ 20 + 12 ; .-. sin^ 6 + cos' ^ = 1 + 12 sin=* cos^ 6 :. sin-* - sin- d cos^ d + cos* ^ = 1 + 12 sin=5 d cos^' d :. 1-3 sin- cos'- ^ = 1 + 12 sin=* ^ cos^ .: sin2 ^ cos- (4 sin cos 5 + 1) = whence sin 2^ = 0, or - -; •• ^-T' T"^^ ^ 12' 21. sin0 = ^/2sm(p, ^/3 cos ^ = ^/2 cos 0; .-. by squaring and adding we have sin^ ^ + 3 cos- ^ = 2, that is, 1+2 cos- = 2, whence cos ^ = i -^ ; TT / IT .-. = 2mr ± — , or 2inr * ( "f - j TT both of which are included in = mr^- . v/3 v/3 Again, we have cos 0= -r^ cos ^=t ±— ^ ; ^ "■ ^ / TT .-. (f) — 2mr^-,r, or 2/(7r-t ( tt - -; which are both inckided in (f> — mr^- . 22. cosec^ = ,y^/3 cosec0, cot ^ = 3 cot 0. By squaring and subtracting we have 1=3 (cosec- 0-3 cot2 0) r= 3 (1 - 2 cot- 0) ; 1 , TT .-. cot = ± -— ; whence = ;<7r ± - . Also cot ^ = ± ^3 ; whence = mr^- . 23. sec \\ whether n be odd or even. Hence the whole formula comprises angles formed by starting again from TT TT OY and turning through an angle --.-{- 1)" a ; that is, - 7 ± a. Hence tlie second formula also comprises angles whose boundary lines are OPj or OP^. EXAMPLES. XIX. c. Page 242. 1. Tjet ^ = sin-i-^; then cosec^ = — . .... 169 , 25 .-. ^ = cot-^^. 2. Let ^ — cosec"^ —-; then cosec ^ = —, and cot ^ = —- . o 80 .-. ; then sin^ = rt, cos^=>yi-a2, cos0:==?^ sin0=^l-&2; .•. cos [6 - 0) = 008^ cos + sin ^sin0=:b sjl -a^-^-a sjl- 6^; .'. sin-ia-cos-iZ> = ^-0 = cos-i {h Jl-a^ + a jY^Jr). [In some of the examples which follow diagrams may be used with advan- tage as in Examijles 2 and 3 of Art. 249.] 4 2 3 2 2 19. sin~iv + cos-i-7--=cot~i7 + cot-i2 = cot-iT; = cot~i -7. 5 ^/5 4 6 11 «« 1 63 ^ , ,1 . 63 , ,5 20. cos-i gg + 2 tan-i ^ = cos-i g- + tan-i -^ -tan-^^^^?^-+tan-i^ -tan - gg +tau ^^ = tan-ij|4-tan-i^| ^ , 192 + 3L5 ^ ,507 , ,3 = '"" -756380--*^" 676=^^^" 4 = sin ^ -= . o XIX.] INVERSE CIRCULAR FUNCTIONS. 171 vi + n 21. tau~^ m -h tan~^ n = tan~^ 1 - mil __„„-! 1-mn 1-wm = cos ' = cos~^ — , = cos ^ s/(;/i + 71)'^ + (I - 7?m)- ;^1 + wt2 + n- + wi-'„2 1 - nvi 20 1 fi 22. Let cos-i— = (9, tan-^--=0; J.' bo -, 1 . ^ 21 . 16 63 then we have sin ^ = ^,t , sin = - - , cos = — ; thatis, ™-'i-*--'i = ''-* = --'^5- 23. Letcos-i ^^:^e, then cos ^ = VV ^^" ^ =^ /q- Now cos f ^ - ^ J =cos ^ cos ^ + sin $ sin ^ ^ /2~3 ^ /n: V 3 * 4 ' V 3 * 4 ^ v/6 + 1 2^3 • thatis. cos-^^-cos-^.^. 24. Second side = 2 1 - a:-* 1 - .t2 - tan (^tan-i ^^ ) = tan (2 tan-^.T). 25. Second side = tan-'a - tan-^/y + tan-^/; - tan-^r + tan^'c = tan~'a. 26. Let tan-'.c = a, tan-^y = /5, tan-'2 = 7; then a + /3 + 7 = TT ; .-. tan a + tan /3 + tan 7 = tan a tan /9 tan 7; [Art. 13/5, Ex. 2.] that is, X + ?/ + 2 = X//Z. 172 INVERSE CIRCULAR FUNCTIONS. [CHAP. 27. We have » = cot ^ >J^cos a - cot"i ^/ ^ 'V cos a = cot-^ r-^±i ^ cot-> A^^^^ : 1 - cos a /I / A. / — ,v/cosa V cos a ^„ 4 cos a (1 - COS a)'-^ cosec'' 1 , +" /l + cosa\- ^u = l + cot-u=[ , ; \ 1 - cos a/ 1-cosa , „a ,'. sinM = — - =tan--, 1 + cos a 2 EXAMPLES. XIX. d. J^age 244 1, siii~^ar = cos ^ x = sm~^ ^1 X-, , 1 •. x= sjl-x-; whence x— ^ —i^ 2. taii~^a; = cot ^a; = tan~^-; x 3. tan-i (.r + 1) - tan-i (.r - 1) = cot"! 2 ; 2 1 .-. tan- ^ — , = tan~i - : whence .t = ± 2. x^ 2 4. cot~^ x + cot~^ 2a; = -;- ; 4 ^ , 2j;2 - 1 Stt •■• ^"* -3.r =T' 2a:2 - 1 Stt .•. 2x- + ;-}.T -1 = 0; wlience x — j^- — 5. sin-^ x - cos-^ X = sin"^ (3.c - 2) ; .•. sin (sin~i x - cos~^ x) = Sx ~ 2 ; .-. X- - jr^^ . >,/rT^= 3.T - 2 ; that is, 2.t2-1 = 3.t-2; .•. 2.r2-3.r+ 1 =0; whence .r = l, or ^. XIX.] INVERSE ClllCULAU FUNCTIONS. 173 6. cos"' a:- siu~^x = cos""^ j:;^/3 ; .•. cos(cos~i X- sin~^x) = X;^/3; .". .r ^1 -x'^ + x /Jl - X' = X Jii ; that is, 2.1; Jl-x-=xJS; whence x = 0, or i-. 7. tan-i + tan-i ^ = j ; x-2 x+2 4 a: - 1 .r + 1 , x-2 x + 2 IT x-—\ 4 a;2-4 .'. 2a;- = 1, or a;=i— — . 8. 2cot~i2 + cos-i- = cosec~i.r; 5 .•. cot~i - + cot~^ - = cosec"^ a: ; 4 4 1-1 .-. cot ^ . — cosec~^j;; 2 .". cot""^ - — y = cosec"^ X = eot~^ y/.r- - 1 ; 49 ^ , ^ 25 .'. 2rP)=a;^-l; whence a;=±-r-r. 24*^ 24 9. tau-i X + tan-i (1 - a:) = 2 tan-^ Jx^^ ; .-. tan-i,--i ,=tan-i ^^^~^' ; 1-x + x^ 1-x + x^' .•. 1 = 4 (.c - .T-) ; whence ^ = o • 10. cos 1 - -— ^ - cos-i :j — — =2tau-i.-r: 1+a- 1 + 6- ' . 2a , 2/> .'. tan-i „-tan-i , -- = 2tan-i.r; I- (r 1 -b- .: 2 tan-i a - 2 tan-i b = 2 tan"! x ; .". tan~^ a - tan~^ b = tau~^ a; ; whence x = — ^^ . 1 + ab 17i 11. 12. 13. where 14. whence f-.hat is, INVERSE CIRCULAR FUNCTIONS. sm-i — „ + tan-i --5 = cos t , ,0 ; 1 + a- 1 - .r- 1 + 0- .-. tan-i _ _ + tau-i - — .,=tan-i ; 1 - a^ 1 - a:- 1-6- 2 tau~^ rt + 2 tan~i.r = 2 tan~i 6; whence x — ^ , X--1 . . 2x 47r . 2tan-i-|:^, +^=0; b - a l + ah' tan -1 .T--1 3 2a; 27r l-«2 3 ' TT tan~i ic = ^ ; whence x = /^/3. 26 26 sin" -^'-=sin-i - tan-i • a2+62~" ,6- ' -, ^'' 1 + — 1- — a- a^ 2a 6 2cd 4^., X x- + y y = bc + ad, x = ac - bd. sin[2cos-i {cot(2tan-ia:)}] = 0; .-. sm pcos-i -cot (tan-i j^„j - Uq. .'. sin 2cos~^ — — - =0; x= ±1, or 1 x-^^ ±2.r; a:=±l, or i(li^2). [chap. I I h XIX.] INVERSE CIRCULAR FUNCTIONS. 175 15. 2 tan-1 (cos 6) = tan-^ (2 cosec 6) ; 2 cos is a right angle ; .-. PQ^ = PP^ + QP- = 200- (12 + 4) = 200- x 4- ; .-. PQ = 800. Again, the distance of P from A P> = PB sin 45° = 200 (^3 - 1) = 140-4 yds., and the distance of Q from AIi = QB sin 75° = 200 (^3 + 1) = 540-4 yds. XX.] FUNCTIONS OF srJLMULTiri-K ANGLES. 179 EXAMPLES. XX. a. Page 255. i A 1. Wheu — lies between - 135° and - 180", sin — is negative, therefore 2 ^ in the first formula of Art. 254, the negative sign must be taken. 2. 5" lies between 135° and 180°; and therefore A A I sin - is positive, cos — is negative. 2 ^ . A /l-cos4 ^ ) 119 109 5 2 ~ 13 A 11 ^ 1 11? + cos^ , / "^109 12 2 ~ ~ 13 A A . 3. Here sin — is negative, and cos — is positive : 2 -^ 161 A /I -cos .4 . / "^289 15 .', sm- /l - cos A 2~ \ 2 > 2 17' 1 1" A /l + coaA I ' 289 8 ^°'2^ V 2 = V — 2~ = 17 A A A 4. Wlien — lies between 135° and 225°, cos - > sin - and is negative ; :. sin - + cos -^ ~ - sj\ + sin ^ , and gin cos— = + ^/l + sin^. A A A 7. ^\lien - lies between 45° and 90°, sin -> cos - and is positive ; .-. sm- + cos-=: v/l + sin^=^ ^"^25^5' , A A ,- . . L 24 1 and sm--cos--=Vl-sin^ = ^/ 1-25^5' . A 4 A 6 180 FUNCTIONS OF SUBMULTIPLE ANGLES. [CHAP. A A A 8. When — lies between 135° and 180°, cos ,5->sin ^ and is negative; ■ ^ ^ n — ^-. /-. 240 7 ••• ^^"2+^°^2=-^^ + '"'^==-V 289=-17 and «^"2-^"'2 = ^^"'^°^=\/ "^289 = 17' . ^ 8 A 15 9. (1) We have sin A + cos A= ,y 1 + sin 2/1 (i) , sin .4 - cos .4 = - »Jl- sin 2^4 (ii). From (i) we see that of sin A and cos ^4 the numerically greater is positive. From (ii) we see that cos A is the greater. Now cos A is greater than sin A and positive between the Hmits 2;/7r - V aiid 2mr + — . 4 4 (3) We have sin^ +cos.4= - v/l + sin 2A (i), sin A - cos A— -{- >Jl - sin 2.4 (ii). From (i) we see that of sin A and cos .4 the numerically greater is negative. From (ii) we see that cos A is the greater. Now cos A is greater than sin A and negative between the limits 2mr + —r and 2mr + -r- . 4 4 A A A 10. 1^0= 120°, sin — > cos — and is positive , .-. sin^+cos-=,yi + sin^, a a . A A I . sin — -cos —= ^1 -sin -4. .-, 2sin— = ^1 + sin /I + ^/l-sin^. XX.] FUNCTIONS OF SUBMULTIPLE ANGLES. 181 -. X r,,o 1- COS 15° 11. tan7*°=- . ,.„- =cosecl5°-cotl5° sin 1d° cot 1-42^° = ^o- o = cosec 285° + cot 285'^ sin Joo = - cosec 75^ - cot TS^-^^ - ^—^ x "^—^ - (2 - ^/3) - -^2 (V.-3 - 1) - 2 + ^3=^2+^3-2-^6. 12. sin 9° = ^ { v/s'+V^ - V5 ^^^5 } = ^ { v/5^236080 - V2-'7639320 } = ] {2-288... -1-662...} 4 _-626... ,_. = j =-loD... . 4 13. (1) As in Art. 251, 2 cos ^= ^^/2+^2; 8 but 4 sin2 ^'^.^ 2 - 2 cos ^ = 2 - J2+J2 ; .-. 2sin~=:72-x/2 + ^/2. (2) tan ir^ 1-5' = ^-i^lll^— cosec 22i° - cot 22^ = _-J 1 _2V2T:,/2 v/2--V2 v/2-1- s/2 ^^^^ + ^^ [Ai-t. 260.J = v'4 + 2j2-(^2 + l). 14. (1) cos^ + sin^''2. As 6 increases from — to - - , the expression is negative and decreases numerically from - ,^f2 to 0. As d increases from ,- to 27r, the expression is positive and increases from Oto 1. (2) ainO-^f^cosO^-lfsmO-'^^ cos <^ j ^2 sin ( <^ - ^ j . From 6 — Oio r , the expression is negative and decreases numerically o from - ^/3 to 0. From ^ = - to ^ + - , or ,. , the expression is positive and increases from 3 2 3 b From ^ = ^ to - H — r, or . , the expression is positive and decreases ^ O t-v ri Oto 2 b 2b' 3 from 2 to 0. 47r TT 47r IItt From ^ = -^ to — + — , or -^ , the expression is negative and increases o ^ o o numerically from to - 2. IItt From 6= — to 27r, the expression is negative and decreases numerically b from - 2 to - J'6. , MM • sin-6> + cos2 1 15. (1) i he expression = -..r- .,.= — =-sec2^ ^ ^ sin-^^-cos-^ cos 26^ = - sec 0, where = 26. We liavc therefore to trace the changes in - sec 0, from 0^0 to 2 tt. From to -- , the expression is negative and increases numerically from - 1 to - oc . From _ tu TT, the expression is positive and decreases from cc to 1. From TT to , the expression is jjositive and increases from 1 to go . XX.] FUNCTIONS OF SUBMULTIPLE AN(JI.ES. 18,3 From -— - to 27r, the expression is negative and decreases numerically from - X to - 1. ._. „, . 2sin^(l-cos the 2 tan A numerator and denominator must have different signs. But when ^ = 320^, tan A is negative ; therefore we must take the sign which will make the numerator positive, A- 1+ yi + tan"^ 2 ~ tan A + have the positive sign prefixed /l - 008 2,4 3. tan .4 =: ± A / f- , and since tan A is positive the radical must ced. ^ , /13 - 12 1 *""^^=Vi3Tr2 = 5- 1 / 1 _i_ (JOS A A A 4. cot--= ± A / , — t; and when — lies between 90° and 135°, cot — 2 V 1 - cos A 2 2 is negative ; A cot /o-4_ 1 V5 + 4~ 3' 5. The general solution of cot 2^ = cot 2a is 2^ = 7t7r + 2a; therefore ^ = ^ (W7r + 2a), (1) when 74 = 2m, cot (9=:cot (w(7r4-a) = cota, (2) when 7i = 2//i+ 1, cot (^=: cot ( //itt + ^^ + a |^ -tana. 6. sin^=sina; .-. ^=:7i7r + ( - l)'*a ; hence in finding sin - we have to o consider all the values of sin ( — + ( - l)** - j . Give to n in succession the values 0, 1, 2, 3,... . Proceeding as in Art. 261 we shall find that the first six angles are a TT a 27r a a -Itt a ott a 3' 3~3' ¥"^3' "^"3' 3~'^3' 3~"3' and that the other angles are coterminal with one of these. 14—2 184 FUNCTIONS OF SUBMULTIPLE ANGLES. [CHAP. 27r a . ( /tt a\| „• ""-a Now «i^-3- + r^'°f -U"3;f = '^" 3 a\ . a sin I TT - - 1 — am - ; sin( ^ + - 1= sin ^tt + | ^ + - II- = - sm'^ 3 ' TT + a . 6 .a . TT - a .TT + a Thus the values of sin - are sin - , sin —— , - sin — ^ . 7. Here d = mr + a, and we have to find all the values of tan ( "o" + ^ j • Give to 71 in succession the values 0, 1, 2, 3,... ; then we shall find that all the angles are coterminal with one of the following : a TT a 27r a a 47r a ott a 3' S'^S' T'^3'''"^3' ¥"^3' y^3' and as in Ex. 6 it may be shewn that the tangents of these angles assume one of the three forms a ^ TT + a TT -a tan - , tan — - — , - tan . do o 2inr 8. Here 30 = 2n7r±3a, or ^="„- =^a, and we have to find all the values o r ■ /2u7r \ of Bin I - - ± a 1 . Now n must be of the form 3m, or 3m + 1, or 3m- 1. If 71 = 3m, sin ( ^^ ± a ) = sin (2mT i a) = ± sin a ; /2mr \ . /. 27r \ . /27r \ if n = 3m + 1, sin ( - ± a | = sin ( 2m7r + — i a 1 = sin I -^- ± a 1 ; if 11^:^111-1, sin f-lp'ia j = 8in (2m7r-^±aj= -sin f ^ia ). 9. Here 3' whence x = 50; that is, the height of the Hagstaff is oO feet. XXI.] LIMITS AND AlTllUXIMATlOXS. 187 11- ^^^ "'^-180x60^-"^^""^' Lt. l^'''"'\= ''''— x-= "" n=0\ n J 180 X GO n 108O0 „=0V n J 180x60x60 n 648000 , „ 1 .. . 27r , n . 'iTT ., / . 27r 27r\ 12 . nr- sin — = tt?- . -- . sm — — tt?- sin ; ; 2 n 27r // \ n n J 2ir 27r 27r . but when » = x , — =0, therefore the Ihnit of sin = is unity. Thus the required limit is tt/-. 13 r...(^z^\^L,.(^.rJ^::::j\.^ ^^' e=o\dBine 0=0^ d ' 03=0 \ 2 d I 2 ' 0^.o\ tanmO + tannd J 9^o\ md + nO ' ^ ^ in" - n- m - n VI + n. 15. cos ^| + ^')=^-cos^-^sin0 = ^-|^ = -491 nearly, 16- '''■''''- ^.^^^ A— ^O'-'^^''^^^^^ •.sin 30^10' .30".siu(^H-3jM 1 11 V'-^ . 11 = 2"°^%00+ 2 ^^"iToOO 1 llv/3 .^^ 188 LIMITS AND APPROXIMATIONS. [CHAP. 17. cos (^ + ^ j = -49; and cos- = -5. .-. 6 ii a. very small angle, so that approximately .•.^-^ = •01; = 39-7' nearly. EXAMPLES. XXI. b. Page 271. 1. Let the distance be .t miles; then by the rule on page 269, we have .: .r = 12; that is, the distance is 12 miles. 2. Let a feet be the height of lighthouse above the sea level ; then 152 = -^, or a = 150; that is, the height of lighthouse = 150 feet. 3. Let the distances in miles of the horizon visible from the masts of the ships be x^, x.^; then ' x,2=— |'^-^ = 49; .-. x,=7, .-. the greatest distance at which one mast can be seen from the other = x^ f .f ._, = 15 miles. ,t' I XXI.] DISTANCE AND DIP OF THE HORIZON. 189 4. Let the distances in miles of the horizon seen from the two masts be X, 1/ respectively, then -^ '^' "^ ' •■■ ^ = ^' also x + ij = 20; whence ij = 11. Height of mast of second ship=: -|- feet = fZr feet = 80ft. Sin. o 5. Let .T, y be the distances in miles of the horizon visible from the mast and the lamp respectively ; , 3x73i 21 then x^= — 2~^ ■' •'• ^~~2' 35 and x + y = 2S; whence ?/ = y . 2?/2 352 1225 „^ .-. height of lamp =-^ = -77-= ^ it- = 204 ft. 2 in. 6. From the formula on page 271, we have 10 /2x2G40 10 number of degrees in dip of horizon = — ^ 3x 1760 ~ II ' .-. dip of the horizon = 54' 33", nearly. 7. The greatest distance at which the light must be visible is the distance of a point exactly opposite a point on the shore midway between two lighthouses, and 3^ miles from it. This distance = / 12^ + ( - j = \/ 4 ^'2 ^ 1 • w X-, 2 625 625 - , ^_. ., ^. .-. height of lamp = - x ,- = -7. feet = 104 ft. 2 in. o 4 o 8. Let the height of the hill be h miles, then we have 1 -hI = -- ^2h ; .-. - — — Aj2h; whence h=2; that is, the height of the hill = 2 miles =10560 feet. 100 DISTANCE AND DIP OF THE liOKIZON. [CHAP. 9. Height of hill .= ^--^^5^'^''^' feet = 610 ft. nearly. The chp of the honzon = --- ^ -— -^ x ^-^^ degrees = ~ijll6 minutes = 20' 13", nearly. 10. We have ^ = j7 ^'^h, where /; -height in miles, 11 V 3x1700 -10 /: 11 V 9 — £ /^ "OC) V 11 7 , where «= height in feet, X 1700 sin 40 cot ^ _ 2 sin 2^ cos 20 cot ^ ^■'•' vers 2^ cot- 2^" ~2 sin^y^ot-W" sin^ 20 cos ^ 8 cos-* sin=* (9 cos 20 cos 2^ 0<3 ' 12. Lt /_^i£i^ot^ ^ - // /8cos^\ sin ^ - cos- 2 2/ XXL] DISTAXCE AXD DIP OF THE IIOUIZOX. 191 13. (1) Lt.(^'''L^^L^^\=Lt. e-a\ e-a ) = a . ti -a 6 + as sin -^;— cos — - 2 2 2 ^^' /:'• -TTT- =1' [Art.2r,(>.] .-. required limit = Lt. ( cos — ^ ) =eoso. = a\ 2 J y . a-d . a + ^v /a \ /sin ^r- sin — ^2) LL /cos g-cosa \ ^ ^^ / 2 2 = J.? a + d ^^ -sm-^-|=-sma. 14. Let .47? = ,32, yl = 31; then tanC = ^ = l + — ; 31 31' .•. C is a little greater than 45° ; .*. G =--{-6, where d is small ; •1 1 + 6> _ 32 ■ l-6'~3l' . /, 1 T 1 7x180 , 10^ . . ^ = - radians = ^^ x — .^ degrees = ^ = ;-;4' 33" : .-. t'=45°54'33", 2? = 44° 5' 27". 15. We have Z 7?P.4 =a= iJiAP- .. An = IiP; AB BP sin 3a ^ . . • • T^n — r>V. = ^ = — =3-4 sm- a = 3, 7jC' BC sin a since the object is distant and therefore a is small; that is, JB = 3/?C, nearly. 192 DISTANCE AND DIP OF THE UORIZON. [CUAP. 16. We shall first shew that - — — is positive, h being the + h radian measure of a small positive angle. ^, . , ,. dtRn(0 + h)- + htiin0 (tAn d + h -t&n d) -hta.n0 This fraction = ^^ — —-^ — — = — ^ -— — d{0 + h) d{0 + h) sin h h sin _Gos cos {0 + It) cos(^ _ ^ sin /< - /i sin^ COS (^ + //) 0{0 + }7) ^ Ylf+h) cos cos {0 + h) ' sin h sin ^ . „ , ,.».-, Now . —C'^-a ^^h<0 [Art. 27'2.] n and cos (^ + /<)<: 1; .•. sin h > h sin cos (0 + h) ; that is, the fraction is positive, and /. — -^ - - ' > . :. — - — continually increases as increases. TTT, ^ rv tan ^ . „ TT tan When ^ = 0, - -- = 1; when ^ = , , -=qo. '2 Thus the proposition is established. MISCELLANEOUS EXAMPLES. H. Page 283. 1. cos '2a + cos 2^ 4- 2 cos (a + ^) = 2cos (a + ^) {cos (a - /3) + 1' = 4 cos (a + /3) cos2 ^^^ , sin 2a + sin 2/3 + 2 sin (a + j3) = 2 sin (a + ^) {cos (a - /3) + 1 } = 4 sin (a + j8) cos^^— ^; .-. hypotenuse = 4 cos- — ^ >/cos- (a + /3) + sin^ (a + /3) = 4 cos^ ^^- . 2. If the in-centre and circum-centre are at equal distances from BC, we have jRco8^=r = 4ii sin -, sin sin—; I) .> () f ^ -^ aj .•. cos A = cos A + cos B + cos G -1; .'. cos/i + cos C=l. XXII.] MISCELLANEOUS EXAMPLES. H. 193 3. Let 6 be the required angle ; then we have tan (45° + ^) = 2; .-. log tan (45° + 6)= -3010300 log tan 63° 26' = -3009994 difif. 306 .: prop'. increase = ,,Y— X 60" = 5-8" = 6", nearly; .-. 45° + ^ = 63°26'6"; .-. required angle 18^ 26' 6". 4. Denote each expression by E ; then E-={1 - sin- a) (1 - sin-j8) (1 - sin27) = cos2a cos^/Scos^y; that is, £ = ± cos a cos 13 cos y. 5. Let ;?! , p.. be the distances of the chords from the centre, and let r be the radius; then Pj = r cos 36° ; similarly i>2 = '* ^^^ 72° ; .-. distance between the chords =p^ -P2 = ^ (cos 36° - cos 72°) _ / V5 + 1 V5-l\ r Also the sum of the squares of the chords = 4r-sin2 36° + 4r2Rin2 72° = br'^. 6. Let .1 be the point at which the railways meet, then we have to solve a triangle in which ^=60"^, a = 43, ft = 48. It is easy to see that the solution is ambiguous ; hence from the third figure of Art. 148 we have CD = 48 sin 60° = 24 ^3, .4D = 48 cos 60° = 24. Also B-iT) = ^/432-{24^/:3J2 = J\2l =11. .-. ^^1 = 24 + 11=35 miles, ^i5., = 24- 11 = 13 miles. 194 MISCELLANEOUS EXAMPLES. H. [CHAP. 7. a = cos~i - + cos-^ f- ; a .". cos .'. 1 - cos- a = — . i- cos a + Vr, ; a- ab b- .r" 1x11 1/' that i K, sin- a - '— , cos a + '. ., . a- ab b- A R C 8. We have p = 47v cos — , q^-iR cos — , r = -l/v cos — ; ^ ^ Z a 2Ri^mA . A .'. - = , = sin — ; ^ 4Kcos4 2 rt2 /^2 c2 2abc . .,A . .Ji . .,C , . A . B . G .-. —„ + — + — -h -= sin- — + sin- — + sin- - + 2 sin — sin — sin - p" q^ 7- pqr 2 2 2 2 2 2 = 1. • [Ex. 13, p. 120.] 9, Let P be the top of the tower, and let .c be its height ; then FA = -^ , PB = J^-; sin a sin /S and RA -+OA'^ = FO" = FB- + 0B~ ; x^ „ x^ hence the height of the tower sm-'a sin-/3 2 _ (g- - fc°) sin- a sin^ ^ _ (a^'-i'^sin^ a sin= /3 ^ ~~ sin'-^ a - sin- (3 ~ sin (a + /3J sin (a - ^) ' ^a^ - b- sin a sin ^ /^/sin {a + /3) sin (a - 8) 10. This example follows readily from the results proved in Exami)les 18, 24, 25 of XVIIL a. r- + r^" + r.r +,;•/= (r^ + r., + r^ - r)- + 2r {i\ + r.^j + n,) - 2 ( r.,r.^ + ror^ + rjro) = IfJfi^ + 2 { (r,ro + rr,) + (r.,?-., + rr^) + {r.-p\ + rr.,) \ -4 {rjr., + ror.^ + r,rj) = 16/i- + 2 (a& + ?;c + ca) - 4.s-2 = 1Q,R- - a- - b- - c\ > p I XX 11.] MISCELLANEOUS EXAMPLES. H. 195 11. (1) Let iliAD^O; then iCAD^^A-O; sin (A -6) _ CD _ BD _ sin 6 _ ■ ■ sin (J Ti') ~AD~ AD ~ sinB ' sin (J -^)_sin {A + Ji) sin 6 ~ sin B ' .-. cot 6 - cot A = cot A -r cot B ; that is, cot BAD = 2 cot .1 + cot B. (2) Draw AM perpendicular to BC, then 211.1/ 2DG-2MC 2 cot ADC :^ AM AM {B G - MC) - MC _ BM MC AM ~AM~ AM ■■ cot B - cot C. 12. See fig. of Art. 223. a BG + GG BG GC ^ ^, ^ ^, Then p=—0G- = OG-^ 0G = '^'' C + ianB; ,/l + ^--'-^2t^nC, P c a\ (c a h + - -) + - - q r V Vr p <1 EXAMPLES. XXIII. a. Page 291. 1. Here the common difference is 2a ; ^, sin »a . a + 2/1 -la sin- / '' . 7r-2/3 sm — -—^ 13. -5 = ]^ { (cos d - cos 3^) + (cos d - cos 56} + (cos ^ - cos 7^) + . . . } 2 = -{ncos^-(cos3^ + cos5^ + cos7^+...+cos27i+l^)] = — ?i TT-- « cos (h + 2) ^. 2 2sin^ ^ ^ 14, ,S = - {(sin 4a - sin 2a) + (sin 8a - sin 2a) + (sin 12a - sin 2a) + . . . } it = - {(sin 4a + sin 8a + sin 12a + . . . + sin 4ua) - n sin 2a} sin 2?ia . „ , ^ , n sin 2a = ^ . ^ sin 2 (n + 1) a — • 2sm2a ' 2 1 sin (2a - a) 1 *i sec a sec 2a = -— = coseca. <^ 10. st,cahBi.^a j,Qgjjgoy2a cos a cos 2a = C03ec a (tan 2a - tan a). Similarly, sec 2a sec 3a = cosec a (tan 3a - tan 2a), sec na sec (u + 1) a = cosec a { tan (/i + 1) a - tan na} . By addition, (S = cosec a {tan (u+l)a - tan /ta}. XXIII.] SUMMATION OF FINITE SERIES. 199 16 . cosec 6 cosec Zd = cosec 26 {cot 6 -cot 36), cosec 3^ cosec 56 = cosec 26 (cot 36 - cot 5^), cosec (2n -1)6 cosec (2/t + 1) ^ = cosec 26 {cot {2)i - 1 ) ^ -- cot (2/i + 1) ^ j By addition, S = cosec 2^ {cot ^- cot (2u+ 1)^}. 17. tau - sec a = tan a - tan , .a a , a , a tan -2 sec - = tan ^^ - tan -^ , tan ~ sec 2^, = tan ^^ - tan ^^^ By addition, 5f = tan a - tan 2n K .,n r. o cos 2a sin a 1 sin 3a - sin a r 18. cos 2a cosec 3a =-. — ;r — -. = - , — . — - — -. sin 3a sni a 2 sin 3a sin a = ^ (cosec a - cosec 3a), I cos 6a cosec Da = - (cosec 3a - cosec 9a), It cos 3**~i . 2a cosec 3** a = - (cosec 3"'~^ a - cosec 3" a). Zt By addition, -^ = s cosec a - cosec 3" a. 19. sin a sec 3a = 2 sin a 2 sin a cos a cos 3a 2 cos 3a cos a _ sin (3a - a) 2 cos 3a cos a = - (tan 3a- tana), sin 3a sec Oa = . (tan 9a - tan 3a), sin 3"- la sec 3** a = - (tau 3''a - tau 3«-ia). 1 By addition , ,S' = - (tan 3" a - tan a) . £1 15—2 200 SUMMATION OF FINITE SERIES. [CHAP. \ 20. Let AB be the diameter, Pj, P^, P.i...P,,^i the pomts of division of the arc of the semicircle starting from the end B ; then IT 2ir ^ (n-l)7r .4Pi = 2a cos - , AP^ = 2a cos — , . . . AP^_^ = 2a cos ; TT 27r Sir {n-l)ir) 2u .-. sum of distances = 2a ( \2 An) All) TT An = rt I cot -. — 1 ) . V 4/i / 21. Let Pj, P2, P3... be the angular points of the polygon beginning with that nearest to XX' in the quadrant XOY. Letjpj,|J2» Vz-- ^^ perpendiculars from Pj, ?„, P3... on XA", and (/i, ^2? 93--- ^^ perpendiculars from Pj, Pj, P3... on IT . Let Z P-^OX = d, and let r be the radius of the circle; then p-^ = rsmd,p^ = r^\.n{d-\-—\, ^;3 = rsin(^H ),•••; .-. Sp-r jsin (? + sin ( 6> + — )+ sin( ^ + — ) + ... to n termsS sin TT . ^i n = r sin =:0. . TT iS Sin - n Similarly Sq = 0. EXAMPLES. XXIII. b. Page 294. 1, 2S = 1 + cos 2^ + 1 4- COS 6^ + 1 + cos 10^ + . . . 8in2n^ 2^ + 2^ + (n - 1) 4^ =v?i + — .— cos ; sin 2d 2 . ' • „ n sin And • 2 4 sin 20 XXIII.] SUMMATION- OF FINITE SERIES. 201 2. 2S = l-cos2a + l-co82('a + ^^ + l-co8 2('a + ^-^ + 4a + (u-l) — Sin TT n = n - ■ cos ^—^^-^^.^ = n .IT 2 sm - n 3. 2S = 1 + coe 2a + 1 + cos 2 /'a - -") + 1 + cos 2 (a - — ^ + . .. sin TT ' n = n + — — cos ^— i^__^ =: n . IT 2 sin- 11 4. 4S'=3 sin d - sin 3^ + 3 sin 2$ - sin 6^+ 3 sin 3^ - sin 9^+ ... = 3 { sin ^ + sin 2^ + sin 3^+ ...} - (sin 3^ + sin 6^+ sin9^+ ...) 3 sin ^ sin ^-^ sin ^ 2 . 3^ ''° 2 " ' sin- sin — S = 3 m'n *^^ sin (" + 1) ^ • 3n^ . 3 (n + 1) ^ o sin — - sin — - — sin — — sin — ^ - •^ 2 2 2 4 sin - sin - 5. 4,S=3 -jsina + sin fa + — j + sin(a+ '^]h-...1 - |8in3a + 8in3 L + |!:\ + sin 3 /'a + — ) + ...[ sm ^ ^.- ^ n sm 37r . ^ '^ w ~3 sin -in • TT 2 Htt sin - sin — 71 n =0. 202 SUMMATION OF FINITE SERIES. [CHAP. 6. 4Sr=3 jcosa + cos( a j + cos ( a- — j + ...- „ / 27r\ ., / 47r\ + cos 3a + cos 3 1 a j + cos 3 ( a j + . . 2a-(7i-l — . „ 6a-(;i-l)-- sm 7r " , sm Btt n 3 — cos . \ =— cos IT Sin - n sin 37r n = 0. 7. We have tan(9 = cot^-2cot2^, 2 tan 2^ = 2 cot 20 - 22 cot 2^ 6, 23 tan 23 ^ = 2^ cot 2^ ^ - 23 cot 2» 6, [XI. d. Ex. 18.] 2"-^ tan 2«-i ^ = 2''-' cot 2"-^ ^ - 2** cot 2» ^ ; by addition, S = cot ^ - 2" cot 2" 0. 1 1 1 8. '^^ + 77- + -^ + ' 2 cos a cos 2a 2 cos 2a cos 3a 2 cos 3a cos 4a : - \ sec a sec 2a + sec 2a sec 3a + sec 3a sec 4a + . . . } 2 ^ = -coseca {tan (n + l)a - tan a). [XXlll. a. Ex. 15.] 9. oi-n 9/? sinS d sin 2^ = ^=^ (1 - cos 2^) ; sin2^Bin2^ = sin 26 sin 46 Replacing 6 by 2^, we obtain Similarly, 1 . o^. . .n sin 4^ sin 8^ - 8in2 26 sm 4^ = — . 1 . o,. . a. sin 8^ sinlG^ - sin^ 40 sm Sd = _ :r^ — ; 4 o io 1 „ , . . . sin 2»» sin 2"+! 6 sin2 2«-i0siu2«0 = 2"-! 2" 2«+i bv addition. _ sin 20 sin2"^i0 XXIII.] SUMMATION OF FINITE SERIES. 203 10. 2 cos e sin2 ^. = cos ^ (1 - cos ^) = 1 - cos^ - {1 ~ cos 0) \ 6 6 :. 2 cos d sin2 ^ = sin- d-'2 sin" ,^ . a Replacing 6 by , we obtain 22 cos ^ sin2 ,^2 = 2 sin2 ^ - 2"^ sin^ |, . Similarly, 2^ cos —^ sin^ — = 2^ sin^ -^ - 2^ sin^ ; Q a a a 2« cos , , sin2 - = 2"-! sin^ ^— - 2" sin^ - ; /a .-. by addition, S = sin^ d - 2" sin^ - . 11. We have a; X tan-^ — ; r^r „= tan~^ . ■ , = tan <■ — tan ^ ^ , n(n + l) + x2 I x^ n n + l and hence tan~^ ; — r, = tan~^ x - tan~i - ; 1 . 2 + x- 2 *^^^"' 2:1^^= *"°~i- *"""'!' tan~^ — -. — „= tan~^ — tan""^ n(n + l} + x- n « + l' by addition, iS = tan~^ x - tan~^ X n + l 12. tan-'— ^,= tan->i^l-" = tan-^ (» + !)- tan~^ ?r .-. tan-i ,— , — ,., = tan-^ 2 - tan-i 1 ; 1 + 1 + 1- tan~^ , i, = tan~^ (n + 1 ) - tan~^ ?t ; 1 + n + 7r .-. -S' = tan-i(n + l)-tan-i l = tan-i (u + 1) - |, 204 SUMMATION OF FINITE SERIES. [CHAP. J 2n _ _ _^ {l + n + n^ ) -{l-n + n'') ^^' ^^"^ 2 + n'' + n'~^'^ l + {l+n + 7i"-){l-n + n^) ' = tan-i {1 + 11 + n-) - tau-i (i _ ^ + n^) . 2 .-. tan-^ ^ , ., - . = tan~^ 3 - tau-^ 1 ; 2 + 1- + 1* 2n tan-i -^ s i = tan-^ il + ii + n^ - tan-^ (1 - n + n'-) ; :. S = tan-i {l + n + n-) - tan-^ 1 = tan^^ {l + n + ir) - j . 14. tan*i , ry i =tan~i - — ~ „— Vr .> ~^ ■^ 1 - 11' + 11'^ 1 + {n2 + n ) (7i2 _ n) — tan~^ (?i2 + n) - tan- (/j^ - ji). 2 .-. tan~i - — -5 — ,, = tan-^2- tan~^0; 1-1^+1* 4 tan-^ — ^^ — ^^ = tan~^ 6 - tan ^ 2 ; 4 2« tan""^ , — ., — , = tan~^ (n~ + n) - tan^ (n- - n) : 1-7}' + 11* ^ ^ .: S = tan-V^ + '0- 15. Let be the point on the circumference of tlie circle, and P, Q, R... the vertices of the polygon beginning with that nearest to O. Let L be the other extremity of the diameter tlirongh O, and let z ()LP = d; then 0P = 2r sin d, 0Q = 2r sin ( <9 + '^ V OR = 2r sin ld+~\, ... :. sum of the squares of the chords = 4r2 jsin^*^ +sinM ^ +'!! J 4-sin-( ^ + -'^ J + ... to 7i terms I yt - uus 2^ - cos 2 ( ' IT 2 V 2n + 1 2n + 1 sin > 2« + l . (n + l)7r sin^ 2w + l . / nir = 2r sin + - ^ IT \ 2n + l sm ^ 2n + 1 .-. PA, + PA^+...+PA.,^^, = PA., + PA^+...+PA.,^. 20G SUMMATION OF FINITE SERIES. [CHAP. 18. Letpi, po, Pzy-Pn^^ *^^6 perpendiculars; then as in Ex. IH, we have /. 27r\ /, 47r\ Pj^ = r-rcoad, p.^ = r -r cos id -\ 1 , p.^=r -r cos Id-] 1 , ... (i) :^p-={r-rcosd]-+ \r-rcos(e + —][ +... to 7t terms = 7ir- - 2r2 I cos ^ + cos ( ^ + — ) + . . . I + r2 jcos2^ + cos2(^ + — )+...[ . Now cos ^ + cos ( ^+-^ )+•... =0; [Art. 297.] ,. ■Zp^- = nr^ + - cos^ + cos ( ^+ — ) +•••• =0; ^ 'n + cos2(? + cos2(<^ + — j + ...( = ^' +-2' = ~2- (ii) I.p^ = nr'-^ - Sr^ icos ^ + cos ( ^ + — | + . . . ton termsl 1^ + -^ hi + cos 2^ + cos 2 0.3 f / 2ir (-V + .., , ,3 cos ^+3 cos ^H + 4 1 V w = nr^ ■ 3nr^ 5/17-' 4- cos 3^ + cos 3 f ^ + — )+... [ 2 2 ' since all the series of cosines vanish by Art. 297. EXAMPLES. XXIV. a. Page 301. 1. We have 1 1 • 1 1 ^ 1 • o 1 - cos a + r sin a = -, - cos /3-t- 7 sin /3 = - ; a b c a be and the required result follows at once by cross multiplication as in Ex. 1, p. 297. 4. cos (a + /3) = cos a cos /3 - sin a sin § 9 V*> *> 1 C^ -D- C- - CI'' u' + h^ a^ + h'^ as in Exuiiiple 2, page 297. XXIV.] SYMMETRICAL EXPRKSSIONS. 207 c „a- (i _1 + cos (a - /3) _ 1 + cos a cos /3 + sin a sin /3 ^' cos-—--- 2 ~ " 2 2 (a- + 62) 6. sin 2a + sin 2/3 = 2 sin (a + ^S) cos(a- j3). ^F^om Example 4, a^ find that sin (a + B) = — ; a^ + 6- and cos(a-^) = ^-^., + ^,-— =.--,^^. 7. sin2a + sin'-/J=(sin a + sin /S)^ - 2 sin a sin j8 = \^^^) -^W- [See p. 298.] 8. Here a and /3 are solntions of a cos 5 + & sin ^ = c ; hence as in Example 4, we find cos(a + /3) = — ;j — 5-:,, and therefore sin(a + /3)= „ ,„. A • ^ . ^ sin (a + S) Agam, cot a + cot S = -; — ^^ sin a sin ^ 2ah c2 - a2 — 2 , i o -^ 2 , 1,0 • [See p. 298.1 9. By squaring and adding, we have 2 + 2cos(^-0) = a2+62. Again, cos {9 + (f>) = ^^^-^ . [Ex. 4, p. 299.] And 2cos ^cos0 = cos (^ + 0) + cos (6^- 0); .-. 4 cos ecos) 11. tan e + ta.nd> = ^——^ 2ab ^{0^ + 1^)^41/^ [See Ex. 4, p. 299 and Ex. 9 above.] . d +

2 2 12. tan - + tan ^ = , = , . 2 2 d

cos C), that is, if S sin^ A = 2 (1 + cos A cos B cos C). Now sin2 A + sin^ i>' + sin^ C= . (3 - cos 2A - cos 2B - cos 2C) = (4 + 4 cos J cos J5 cos C). [See XII. d. Ex. 9.] 20. Expressing a, b, c in terms of R, this identity will be proved if we shew that S sin A cos^ J = n sin .4(1-4 cos A cos B cos C). Now 8S sin A cos^A = 42 cos-^ sin 2A = 2S(l + cos2^)8in2.4 = 22sin2.4 + Ssin4J. Now 2 sin 2A = 411 sin A , and 2 sin 4^ = - 4n sin 2.1 ; [Ex. 7, p. 301] ; .•. S sin .4 cos'yi = IT sin .4-411 sin A cos A = II sin A (1- i cos A cos Z/ cob C). XXIV.] SYMMETRICAL EXPRESSIONS. 211 21. :^a^ COS {B - C) = 2iJ2«- sin A cos {B - C) = 2RZa- sin {B + C) cos {B - C) = i?:2ft2 (sin 2B + sin 2C). Now a- sin 27> + b- sin 2.-1 =2a sin B . a cos B + 2/; sin .1 . 6 cos A = 2a sin B [a cos B + h cos .^ ) = 2ac sin 5 = 4 A. Hence Sa^ cos {B-C) = 12RA = Sabc. 22. (1) We have, from the given equation, (6 sin ^ - c)- = a2 cos^ d^a'^{l - sm-O), or {a^ + b-)sm^d-2bc&me + {c--a^) = 0; and this is the required equation since by hypothesis it is satisfied by sin a and sin /S. (2) The required equation is x^ - (cos 2a + cos 2^) x + cos 2a cos 2/S = 0. Now cos 2a + cos 2,3 = 2 cos (a + /S) cos {a - j3j _2(a-2-62) 2c^ - a^ - b- ~~^+W' a^ + b"^ ' [See solutions to examples i and 6 above.] Again, cos 2a cos 2/3 — cos- (a - /3) - sin- (a + /3) (2c^-a2_^2^2 _4a2fc2 Hence, by substitution the required equation is obtained. Otherwise. Let cos a = y, then the given equation may be written {ay-c)^^bHl-y^), or (a2+&2)y2_^2 + c2=2ac7/ (1). If a; = cos2a = 2 cos'^a- l = 2?/2- 1, 1 2 ^ + 1 we have y'= — ^ . Substituting in (1) we obtain the equation which reduces to k (a2 + 62)2 ^2 _ 2 (a3 - b^) (2c2 - a- - b-) x + {a* + b* + 4c-' - 2a262 - 4a2c2 - ii'V) = 0. 212 IDENTITIES. [chap. EXAMPLES. XXIV. b. Page 307. 1. S sin (a - 8) sin (/3 - 7) = S (sin a cos - cos a sin 6) sin (/3 - 7) = cos ^2 sin a sin (^ - 7) - sin 62 cos a sin (/9 - 7) = 0. 2. 2 (cos /3 COS 7 - sin /3 sin 7) sin (^ - 7) = 2 cos (/3 + 7) sin (/3 - 7) = ^S(8in2/S-sin27) = 0. .-. S cos /3 cos 7 sin (/3 - 7) = S sin (S sin 7 sin (^ - 7). 3. Ssin(/3-7)cos(j3 + 74-^) = S sin (/8 - 7) {cos 6 cos (/S + 7) - sin sin (/3 + 7) } = cos 02 sin (|3 - 7) cos (/S + 7) - sin ^2 sin (^ - 7) sin {^ + y) = - cos (9 S (sin 2/3 - sin 27) - - sin (9 2 (cos 27 - cos 2/3) 2 A 4. cos2(/3-7) + oos2(7-a) + cos2(a-j8) = 2 cos ((3 - a) cos (a + /S - 27) + 2 oos^ (a - ^) - 1 = 2 cos (a - |8) {cos (a + /3 - 27) + cos (a - ^) } - 1 — 4 cos (a - /3) cos (a - 7) cos (/3 - 7) - 1. 5. 2 sin /3 sin 7 sin (/3 - 7) = - 2 {cos (/3 - 7) - cos (/3 + 7)} sin (/3 - 7) - -T 2 sin 2 (/3 - 7) - . 2 (sin 2/3 - sin 27) 4 1 = - II sin (/3 - 7). [Art. 306, Ex. 2.] 6. cot(a-/3) = cot{(a-7)-(/3-7)} _ cot (a - 7) cot (p - 7) + 1 ^ ~~~cot"03r^7)^cot'(a'-^) ' by multiplying up and transposing we obtain the recpiircd result. XXIV.] IDENTITIES. 213 7. 22 sin 3a sin (/li - 7) = cos (3a - /3 + 7) - cos (3a + /3 - 7) + cos (3/3 - 7 + a) -cos(3j34-7-a) + cos(37-a + /3) -003(37 + a ~ (i). Combining the first and fourth terms, the second and fifth terms, the third and sixth terms, and dividing by 2, we obtain 2 sin 3a sin {^ - 7) = sin (a + /3 + 7) {sin 2 (/S - a) + sin 2 (a - 7) + sin 2 (7 - j8)} = 4sin(a + /3 + 7)nsin(^-7). [Art. 30G, Ex, 2.] 8. 42 cos^ a sin (^ - 7) — 2 (cos 3a + 3 cos a) sin (/3 - 7) = 2 cos 3a sin il^-y) = 4 cos(a + /3 + 7)nsin(/3-7j. [Art. 30G, Ex. 4.] 9. 42 cos (^ + a) sin [6 - a) cos (/3 + 7) sin (/3 - 7) = 2 (sin 2d - sin 2a) (sin 2/3 - sin 27) = sin 202 (sin 2/3 - sin 27) - 2 sin 2a (sin 2/3 - sin 27) = 0. 10. In the identity Zbc {h~c)= -n(b- c), put a = sin- a, b = sin^ /3, c = sin- 7 ; then 6-c = sin-/3-sin-7 = sin {^ + y) sin(/3-7); .-. 2sin2/3sin2 7sin(/3 + 7) sin(/3-7)= -nsin(/3 + 7) . II sin (p-y). 11. In the identity Zbc {b-c)= -Jl{b-c), put a = cos 2a, t = cos 2/3, c = cos27; then Z>-c = cos2/3-cos27=: - 2 sin (/3 + 7) sin (/3-7) ; .-. - 22 cos 2/3 cos 27 sin (/3 + 7) sin (/3 - 7) = SH sin (^ + 7) . n sin (/3 - 7). 12. Ill the identity Za^ {b- c)= -U (b-c), put a = cos 2a, iy=:cos2/3, c = cos27; then 6-c = cos2/3-cos27= ~^ sin (;3 + 7) sin (/3 - 7) ; .-. 2 2 cos22a sin (/3 + 7) sin (/3-7)= -8n sin (p + y) . II sin (/3-7). Also 2sin(/3 + 7) sin(/3-7) = 0; whence bv subtraction, we have 2 (2 cos2 2a - 1) sin ijS + y) sin(p -y)= - Hll sin (;S + 7) . II sin (/3 - 7). H. E. T. K. 16 214 IDENTITIES. [chap. 13. In the identity Sr?^ {h - r) = - {a + b + c) TL {h - c), put a = sin a, 6 = sin/3, c = sin7; % .■ Z siu^a (sin/3- sin 7)= - (sin a -f sin ^8 + sin 7) 11 (sin/3- sin7)...(l). But 2 sin a (sin /S - sin 7) = (2); multiply (2) by 3, and (1) by 4 ; then by subtraction we obtain :S (3 sin a - -i sin^ a) (sin ^ - sin 7) = 4 (sin a + sin ,3 + sin 7) 11 (sin /3 - sin 7). 14. If a + h + c = 0, then a^ + b^ + c^ = Sabc. The condition a + b + c = is satisfied, if rt=:sin (/3 + 7) sin (/3-7), and b and c are equal to corresponding quantities. 15. The condition a + b + c = is satisfied, if a = cos((3 + y + d) sin (^3-7), and b and c are equal to corresponding quantities. 16. ^^^e proceed exactly as in Art. 309, and shew that a-f ^ + 7 = «7r; .-. 3a + 3/3-7-37 = 3/(7r. From this relation it is easy to shew that 2 tan 3a = n tan 3a ; ^ 3.r - .ir 3.r - x^ 1 - 3x- 1 - 3x2 17. Put .T = cot a, ?/ = cot/3, 2 = cot 7; then cot ^ cot 7 + cot 7 cot a + cot a cot ^ = 1 ; cot /Scot 7- 1 •. cot a = - cot (/3 + 7) ; cot 7 + cot /3 .-. a = mr- (p + y), or a + ^ + y = inr ; :. 2a + 2^ + 2y = 2mr. From this relation it is easy to shew that cot 2,3 cot 27 + cot 27 cot 2a -f- cot 2a cot 2/3=1; iijz 4zx ijcy .-. Zx{l-i/){l-z'^) = 4xyz. XXrV.] IDENTITIES DERIVED BY STTRSTTTTTTTON. 215 EXAMPLES. XXIV. c. Page 311. 1. If A+B + C=0, then cotC=-cot(A+B)= ; ; ' cot 7/ + cot .4 that is, cot B cot G + cot G cot A + cot A cot B = l. The given condition is satisfied if A = 2^ + y-Sa, B = 2y + a-S^, C = 2a + ^-'dy. 2. (1) In Example 4, Art. 133, we have proved that 4 sin a sin/3 sin 7 = 2 sin (/S + y- a) - sin {a + ^ + y). In this identity first replace a, /3, 7 by ^ + 7, 7 + a, a + /3 respectively, and secondly replace a, /3, 7 by 2a, 2^3, 27 respectively. Thus 8nsin(/S + 7) = 2sin2a-2sin2(a + /3 + 7), and 4nsin 2a = S sin2 (/3 + 7- a) - sin 2 (a + /3 + 7). .-. 8nsin(;8 + 7)+4n sin2a = 22 sin 2a + S sin 2 (/3 + 7-a) - 3 sin 2 (a + /3 + 7) = 22 sin 2a + 2 {sin 2 (^ + 7 -a) -sin 2 (a + ^ + 7)} = 22 sui 2a - 22 cos 2 (/3 + 7) sin 2a = 22 sin2a {1 -cos2 (^ + 7)} = 42sin2asin2(/3+7). (2) In the first part of this Example, we have seen that 8n sin a = 22 sin (^ + 7 - a) - 2 sin (a + ^ + y). By replacing a, ^, 7 by /3 + 7-a, 7 + a-/8, a + /S-7 respectively, we have 4n sin (/3 + 7 - a ^ = 2 sin (3a - /3 - 7) - sin (a -!- /3 + 7) . .-. 4n sin (/3 + 7 - a) + BIT sin a = 22sin(/S + 7-a) + 2sin(3a-/3-7)-3sin(a + /3 + 7) = 22 sin (:/3 + 7 - a) + 2 { sin (3a - /3 - 7) - sin (a + /3 + 7) } = 22 sin (/3 + 7 - a) - 2 2 cos 2a sin (/3 + 7-a) = 22sin(/3 + 7-a){l -cos2a} = 42 sin-a bin (/3 + 7 -a). 16—2 216 IDENTITIES DERIVED BY SUBSTITUTION. [CHAP. 3. (1) This is equivalent to proving that in the pedal triangle a'^ - ir- = 2R'c' sin (A' - B'). Now in any triangle ir~ -h'^ = 4:R'^ {sin- A -siiv^B) = \P? sin {A + B) sin [A - B) = 2R . 211 sin C . sin {A - B) = 2Rcsin{A-B). (2) This is equivalent to proving that in the ex-central triangle a^2 _ 6^2_2jjj^ci sin (A^-B^). This identity has been proved in (1). (3) This is equivalent to proving that in the pedal triangle A' 2 (&' + c') tan — = -li^'S cos ^ '. Now in a)iy triangle A A S (& + c) tan — = 27v2 (sin B + sin C) tan — ,^^ . B + C B-C . A A = 47?2 sm — ^- - cos — - — sin — -^ cos — B-C B-i-C = 4ixScos — — cos — — = 2i?S(cosi^ + cosC) = 47? (cos^ +cosB + cos C). 4. We have sin^ 2^ = 4 sin- a sin* 7 = (1 - cos 2a) (1 - cos 27) ; /, cos2^\ /. cos20^ •. l-cos-2^= 1- cos2^7 \ cos 25/ ' 1 \ „„-/. 1 cos 20 f-^ + — ^J=cos2 20 ( 1 \cos .-. COS 20 2j8'^cos257 ''"" "" V"^cos2/3cos2o/' COS 2^8 + COS 25 1 + cos 2p cos 25 ' 1 - COS 20 _ (1 - COS 2/3) ( 1 - cos 2 5) ^ ■ ■ 1 +cos20 ~ (i + cos 2^1 + cos 25) ' ,•. tan- 6 = tan- (5 tan- 5. XXIV.] IDENTITIES DERIVED BY SUBSTITUTION. 217 5 . We have tan^ ^ = tan^ ^ tan^ ^ ; 1 - COS 7 1 - cos Q 1 - cos " 1 + cos 7 "" 1 + cos ^ ' 1 + cos 1 - cos a cos 7 1 - cos j3 cos 7 ~ 1 + cos a cos 7*1 + COB /3 cos 7 ' Componendo and Dividendo, 1 _ l + cosacos/3cos27^ cos 7 ~ (cos a + cos /J) cos 7 cosa + cos/S-1 .-. cos- 7 = — ; ' cos a cos /3 . „ (l-cosa)(l-cos^) .■ sm2-y= -^ ' = (6eca- 1) (sec yS- 1). ^ cosacos^ ^ ;v r / 6. By solving for cos Q, we have sin^ 8 cos^ a — sin^ a cos^ /3 cos ^ = . ., ^ —^ X- . sm- /3 cos a - sm-' a cos /3 By substituting l-cos^/S for sin^^, and 1-cos-a for sin'^ a, we have eos^ a - cos" /3 cos^ = (1 - cos- /3) cos a - (1 - cos^ a) cos /3 cos a + cos ^ 1 + cos acos/3' 1 - cos 6 _ (1 - cos a) (1 - cos ^) _ ■■ 1 + cos ^ ~ (1 + cos a) (1 + cos /3) ' .-. tan2 - = tan- .- tan^ ^ . 7, From the two given equations, we have tan 8 , , tan 7 sec a = ; — ^ , and tan a= .— ^ ; tan d sm ^ tan^ iS tan= "V _ 1 . tan- ^ sin^ B tan- /3 cos^ B tan^ 7 = 1; 1-C0s2fl 1-C0S2^ ,•. tan^ ^ COB- B - tan^ 7 = 1- cos^ B \ 1 + tan= 7 sec- 7 .-. cos2^= — - — s-3 = — ^ . l-htan-/3 sec-/3 218 IDENTITIES derivp:d by substitution. [chap. 8. We have be cos a cos 2) gin a sin /3 = a2 + Z>2 - c-. 9. From the given equation, we have sin^ a (cos 6 - cos a)2 = cos2 a sin^ O = cos- a (1 - cos^ Q) \ :. cos^ Q -1 cos a sin^ a cos ^ - cos^ a = ; which is a quadratic in cos B with roots cos /3 and cos 7. .•. cos^ cos7 = cos'*a. Similarly, from the equation cos^a (sin^- sin a)2=sin2a cos^ ^ = sin2a (1 - sin^^), we may shew that sin /3 sin 7 = sin'* a. cos j8 cos 7 sin S sin 7 ., . „ .'. ^„ ^ H -f-., — = cos- a + sm2 a - 1 . cos"^ a Bin- a 10. We have fc^cos j3cosa + fe sina + (fcsin/3 + l) = (1), and A-^ cos 7 cos a + fc sin a + (A; sin 7+ 1) = (2); whence by cross multiplication, cos a _ sin a . 1 ^ ^2 (sin /3 - sin 7) " fc^ sin (^ - 7) + ft2 ^^qs 7 - cos /S) "~ k^ (cos /3 - cos 7) ' cos a. sin a 1 /y + 7~ , /^- 7 . /3 + 7~ , . /3 + 7* cos —A— « cos "^ -- + sin ^ - k sm '^ it 2i £1 a .-. cos2^-— ' + ^/ccos '— -'+sin -- 'V =A;2sm- -^-' ; .-. /r ( cos2 ^'^ - sin2 ~^^ + k (sin (3 + sin 7) + 1 =. ; .-. /c2 cos /i COS 7 -h /i (sin /3 + sin 7) + 1 = 0. XXIV.] IDENTITIES DERIVED BY SUBSTITUTION. 219 Otherwise. Form a quadratic equation in sin d\ thus k^ cos2 a (1 - sin2 ^) = (1 + A: sin a + A; sin 6)- ; sin /3 + sin 7 — sum of roots of this equation _ coefficient of sin 6 coefficient of sin- d _ 2 (l + ksma)k ~ ~ k'^l + k'^coa'^a)' , , . ^ . , 2(1 + A; sin a) .^. .: k(.mfl + . my) =-~^^^^,-^^^ (1). Again, form a quadratic equation in cos^; thus k- (1 - cos2 ^) = (1 + ^ sin a + k^ cos a cos 0)^ ; cos /3 cos 7 = product of roots of this equation _ A-2 - (1 + A- sin g)^ - A;2 - A-"* cos'^ a 1 + 2A; sin a- k^ cos^ a ■• '■■^"^'^^^^^=— TT^^-cos^^ (^^' By adding (1) and (2) we obtain the required result. 11. ^^e have sin- a cos ^ cos + cos^ a sin /3 sin + cos- a sin^ a = 0, and sin'^ a cos y cos + cos^ a sin y sin

2 tan /3 tan 7, and two corresponding in- equalities. [See Art. 316.] 6. Since (1 - sin a)^ is positive, 1 + sin-a> 2 sin a. Similarly, l + 8in2/3>2 sin/3. j .-. 2 + Bin2a + sin2/3>2sina + 2sin/3. ; \ 7. sin ^ + 008^=^/2 sin f (9 + ^ j . '. '1 8. 008^ + ^3 sin^ = 2 sin (^ + ^j . 9. a cos (a + ^) + 6 sin ^ = a cos a cos ^ + (ft - a sin a) sin 6. .-. maximum value = ^/rt- eos'-^a+^ft - (i sin a)-. [Art. 317.] 10. P cos 6 + q sin {a + d) = {p + q sin a) cos ^ + g cos a sin 6. .'. maximum value = \/(2> + q sin a)- + (/- cos- a. 11. sin a + sm /3 = 2 sin — — ^ cos — — = 2 sin - cos — r— . Z Z £1 £1 \ maximum value — 2 sin - . 12. sin a sin /3 =- {cos (a - ^) - cos (a + /3)} =5 {cos (a - ^) - cos cr}. , 1/1 \ • 2*'' .". maximum value = - (1 - cos a) = sin-' ^ 2 n — nns 2 J ^ sin(a-|-i3) sin (r 13. tana + tan/3= ^ ^-'= --• ^ cos a cos j3 cos a cos ^ M By Art. 319, the denominator is a maximum when a = /9, and in this case tan a + tan ^ is a minimum, its value being 2 tan - . XXV.] MAXIMA AXD MINIMA. 221 I , sin a + sin B 14. cosec a + cosec yS = — -. . — , '^ sin a sin /i 4 sin — - - cos —~ 2 sin —^ cos — jr-^- Cos(a-/:i) - cos (a + /:i) o^ - iS oa + /3 cos^ — -^ - cos- -7,-- 2 2 I _ . a+^/ 1 1 ~^^^ 2 f a-8 a + B'^ 7~^ TTft \ cos -^ + cos -,y^- cos — ^ - cos --J- P Since a + ^ is constant, this expression is least when the denominators are greatest, that is, when a = ^ — ^. Thus the minimum value is 2 cosec - . 15. C0S.-1 cos L' cos C = --cosC {cos {A -i^) + cos(^+B)} = - cos C{cos (^ -5) -cos C}. i . Supposing G constant, this expression is not a maximum unless A — B. Similarly, we may shew that the given expression is not a maximum unless A—B = C= 60°. In this case its value is cos^ 60° or - . o 16. cot.4 + cotJB + cotC = ^!''^;^t''^l + cotC= . ^""^ ^ + cotC. sin^sini> Bin^smi? Supposing C constant, sin^ sinJ5 is a maximum when A = B, and in this case the given expression is a minimum. Thus the given expression is a minimum when A=B=C, and its value is 3 cot 60° or ,^3. 17. sin2- + sin2- + sin2- = ^ (1 - cos^) + ... + ... 3 1 = - - - (cos A + cos B + cos C). As in Example 1, p. 315, it is easy to shew that cos A + cos B + cos C is a maximum when A=B = G, and in this case the given expression is a mini- 60° 3 mum, its value being 3 sin- — or - . 18. If G is constant, A+B iB constant, and therefore sec A + sec B is a minimum when A =B. [See Ex. 2, p. 316.] Thus the given expression is a minimum when A=B=G, its value being 3 sec 60° or 6. 222 MAXIMA AND MINIMA. [CHAP. 19. tau=:| + tan2| + tan2| = ( tan — + tan -^ + tan — j - 22 tan — tan — = ( tan - + tan 2 + tan - j - 2. ABC As in Example 16, it is easy to shew that tan -- + tan — + tan — is a minimum when A—B=G, and in this case the given expression is a minimum, its value bemg 3 tan- -^ or 1, 20. cof^A + cot'B + cot"G= (cot ^ + cot 7? + cot C)^ - 2S cot B cot C = (cot A + cotB + cot C')2 - 2. But it has been shewn in Example 16 that the right side is a minimum when A=B = C, and in this case the given expression is a minimum, its value being 3 cot^CO^ or 1. 21. 2 (a sin2 e + b sin Ocosd + c cos- 6) = a{l- cos 26) + ft sin 2^ + c (1 + cos 20) = a + c + bsm29-{a-c)co82d (1). The greatest value of ft sin 2^ - (a- c) cos 2^ is Jb- + {a-c)-, which is less than /s^/4ac + (a-c)- or a + c, since b'^<4ac. Hence as in Art. 317, the maximum and minimum values of (1) are a + c±Jb^ + (a^c)^. 22. sina + sin/3 + sin7-sin(a + /3 + 7) = 2 sm — — ^ cos „ - - 2 cos — sin —~- , . a-\-B . y + a . 3 + y = 4 sin — - sin '-^r— sm -^^ • A Z ^ The expression on the right is positive, since each of its component factors is positive. .-. sina + 8in^ + sin7>sin (a + /3 + 7). 23. Let x = a cosec 6 -b cot 6, and put cot 6=t; then x = a ^/l + t^-bt. Ah in Ex. 2, page 317, we may shew that x> sja^- b'^; •. a cosec 6 -b cot 6 > >/a"-^ - ft^. I XXV.] MAXIMA AND MINIMA. 223 sec*^-tan^ 1-tan ^ + taii2(9 24. Let a = sec2^ + tan^ 1 + tau ^ + tan- ^ ' .-. tan2(?(x-l)-f-tan^(x + l) + x-l = 0. i In order that the values of tan 6 found from this quadratic may be real, we must have {x + l)2>4{a;-l)2; that is, - 3.r^ + lOx - 3 must be positive; that is, {3x -l){x- 3) must be negative. Hence x must lie between 3 and - . o 25. Denote the expression by x; then _tan-*(9 + tan2^-l |: ^~ t&n^J^tB.n'd + 1 ' .-. tan*^(j;-l)-tan2^(.i; + l) + a; + l = 0. In order that the values of tan^ 6 found from this equation may be real, we must have (^ + l)2>4(a; + l)(a;-l); .-. (.r + l)(5-3a;)>0. Thus the greatest value of a; is - . 26. We have {a^ + b' + C-) {cos^ a + cos" j3 + cos- y) - (a cosa + 6 cos/3 + c 0087)2 = (b cos 7 - c cos j9)- + (c cos a- a cos 7)^ + (a cos /S - 6 cos a)-, the minimum value of which is zero. .-. minimum value of (rt- + 62 + c2)(cos^a + cos2/S + cos27) - ^2_o. Again, {a + b + c){a coii^a + b 008^/3 + 0008-7) - (acosa + 6 cos^ + c cos 7)- = be (cos p - cos 7)2 + ca (cos 7 - cos a)- + (ib (cos a - cos /3)^, the minimum value of which is zero. .-. minimum value of {a + b + c){a cos- a+ b coa^ (i + c cos-y) - k' = 0. 224 ELIMINATION. [chap. EXAMPLES. XXV. b. Page 324. 1. By squaring each of the given equations and adding, we obtain — + ^ =2. 2. By transposition, we have asec6:^xta.nd + y, b secd = x-y tand; whence by squaring and adding, (a2 + 62) sec2 d = {x^- + 1/) (1 + tan^^), or x'' + y-' = a^ + b". 3. We have cos ^ + sin ^ = (7, and cos2^- sin2^ = 6; h :. cos^-sm^=:- ; a 62 a~ 4. We have a;2=(sin ^ + cos ^)2=l + sin 2^, siu 6 cos 6 _ 2 ^^^ y = '^e ^ shT^ ~ sin 2^ ' .-. y(x2-l) = 2. 5. By addition and subtraction, we have cot d = — ^ , and cos ^= -s" » . a — 6 whence by division, ^^^ a^h ' /a-6\2 fa-h\' , /«-^' 4^6 that is, (a2-62)2 = 16a6. 1 cot2^ + l cosec-^ 6. Here ^ = ^^^^-^^^0= '^^otd' = "coTf ' 1 cosec2^-l_ cot^g and y = cosec d - ^^^^ = -^^sec'^ " " c"^^^ ' .-. x-y = cosec^8, and xy'^ = cot'^d. But cosec^ d - cot2 = 1; 4 2 e 4 , XXV.] ELIMINATION. 225 1 . ^ cos^tf 7. Here a'^ = ^— --sin(?=: . , ' sin ^ sin 6 ,,1 ^ sin2^ and t'= --cos^ = .; cos 6 cosd :. a%'^ = cos^ 6, or a^Z> = cos ^ ; and ' a%^ = ?'nv' d, ox ah- = sin 6; 8. By substituting for cos 3^ and sin 3^, we have 4a; = 4a cos^ 6, or x-^=a^ cos 6, and 4?/ = 4a sin-' 9, or ?/ - = a^ sin ^ ; .•. x^ -\-y^ = a^. 9. By transposition, we have ^ a tan^ 6 X (1 + tan^ 6) = a tan*' ^, or a: = s-„- , sec^ ^ tt SGC^ B and ?/(sec2^-l) = asec^^, or ?/ = — — ^; .-. x^y^ = a^ tan^ ^, and x^y^ = a^ sec^ ^. .-. (.T2?/3)t - (a:3^2)| ^ ^^2 (sec2 ^ - tan^ ^) = a^, 10. Here x = a cos ^ (2cos2^- l) = a cos ^ (4cos2^- 3) = a cos 3^. Similarly y = b sin 36, a;- w'' , a- O'' 11. Here sin ^ sin a + cos ^ cos a = a, and sin 6 cos /3 - cos sin /S = 6 ; .-. sin 6 cos (a - j3) = a sin /3 + 6 cos o, and cos d cos (a - /3) = a cos fi-h sin a ; .-. cos- {a-^) = a'^ + b^-2ab sin (a - /3). 12. Here .T-r?/ = 3 - (1 -- 2 sin' 2^) = 2 + 2 sin^ 2^, and X - ij = 4i sin 2^ ; .-. a:= 1 +2 sin 26 + sin2 2^ = (1 + sin 26 f, or x^ = 1 + sin 26. Similarly, - + 2cf/>cos^. Again, a: + b = acos d + 2hco^- d = coa d {a + 2h cosd), and y =: sin d {a + 2i cos d) ; .-. (a- + i)'- + 7/2= (a + 2^ cos 61)2 = j,(:r2 + j/.,^2)2. XXV.] ELTMIXATION. 227 I 18. Componendo and dividendo, we have a _ tan (^ -f a) + tan (^ - a) _ sin 20 b ~ tan (^ + a) - tan {d-a) ~ ain^a ' .-. 6 sin 2^=3 a sin 2a. Also hco&2d = c - aCQ&2a; .-. h'^ = c^ -2ac cos, 2a -\- a-. 19. By squaring and adding, we have x'^ + i/ = ar [2-2 (cos Zd cos 6* + sin 36^ sin 6)] = a^ [2 -2 cos 26) And 2a- - x- ~y^ = 2a- cos 2d. But a; = 2a cos 2^ sin ^; .-. ^a*x-= {2a- cos 26)- {Aa- s.in- 6) = {2a'^-x^-f-f{x^ + ,f-). 20. Solving the given equations for x and y, we have x = a (cos 6 cos 2^ + 2 sin 6 sin 26) ; .-. 2.r = a { (cos 3^ -f cos 6) + 2 (cos ^ - cos 3(9) [• = a(3 Gos^-cos3^) = a(6cos^-4cos=^ 6). And y = a{^ sin 2(9 cos (9 - cos 26 sin ^) ; .-. 2y = a{2 (sin 3^ + sin 6) - (sin 3^ - sin 6) } = a (3 sin ^ + sin 3^) = a (6 sin ^ - 4 sin^ 6). .-. 2{x + y)=a (cos^ + sin 6) {0 -4 (cos^^-cos^sin ^ + sin^^)} = 2a (cos 6 + sin 6) (1 + sin 26) ; .'. x + y = a (cos ^ + sin 6)^. Similarly, x-y = a (cos 6 - sin (9)3 ; .-. {x + y)^ + {x-y)^^ = 2a^^. 21. From the first equation, (.r sin 6* - y cos 6)- = (x^ + y-) (sin^ 6 + cos2 6) ; whence .r cos ^ + ?/ sin ^ = ; cos 6 _ .sin ^ 1 y ~ - x ~ ^x'^ + y- ' By substituting in the second equation, we have 1 ff :^\ _ 1 or "^' + ^■'=1 228 ELIMINATION. [CHAP. 22. From the first equation, we have hx cos 6 + ay sin 6 = ah ijain' 6 + cos^ d ; .-. h^ (.r2 - a-) cos2 e + 2rtZ;xy cos 6 sin ^ + a^ (j/-' - //■:) sin- (9 = 0. From the second equation, we have (?/2 _ 62) cos'- ^ - 2xy cos ^ sin 6* + (x^ - a^) sin^ ^ = 0. Multiplying this equation by ah and adding to the preceding equation, [IP [x^ - a-) + ah (2/2-t2)} coB^ d + {a"^ {y"- - h"^) + ah {x^ - a'^)] sin2 6' = 0; .-. {/).r2 + ay2_a&(a + 6)} (tcos^^ + a sin2^)=0. .'. &.r2 + a?/2-a6 (a + 6) = 0, or h cos2 ^ + a sin^ ^ = 0. cos2^ sin2^ From the last result, = 5— ; a -0 but { (y2 _ &2) cos2 ^ + (a;2 _ ^2) siu^ ^ }2 ^ 4^2j^2 cQgS Q gi^^-i ^ . .-. {a {y^ -b^)-b (.r2 - a2) } 2 z= - 4:ahx'y-- 23. We have 4 cos (a - 3^) = m (3 cos ^ + cos 3^), and 4 sin (a - 3^) = ?« (3 sin 6 - sin 3^) ; whence by squaring and adding, we have 16 = m2 (10 + 6 cos 4 24. We have = v,^ =:tan^tan0, ^^' y cot ^ + cot ^ „ . , /^ N tan ^ + tan0 But tana = tan (^ + 0) = , , ^,, ; ^ ^' 1 - tan ^ tan (p ' .'. tan a — y-x 25. We have a2 + 62 = 2 + 2cos ((?- 0); .-. a2+62^2 + 2cosa. 26. We have a sinS ^ + 6 cos2 (9 = 1 = sin2 d? + cos'' ^ ; .-. (a-l)tan2 + sin- (p ; .*. [h - 1) tau- = 1 - a. But a- inn- d = h'^ iQ.n- (f> ; a° (1 - h) _ ?>2(i_a) ■ a - 1~ ~ 6-1 "' .-. (/(/;- 1) = ± /> (a -1). Rejecting the upper sign, we have a-\-h = 2ab. 27. From the first two equations, we have a b I cos-^^ ^^"-2^ ^^^~2 j;2 v^ _ _ 2 ^-^ _ . o a o + 7^ — sec TT — sec- ■ . a^ b^ 2 2 oo \\r 1 rt tan (9 + tan <6 , 28. \v e have - = — -^ = tan ^ tan 0. 6 cot ^ + cot

- { (tau 6 + tan 0)- - -1 tan ^ tan 0} = fe2a2-4fl6. 29. We have a cos^ ^ + fo sin- ^ vi cos^ a sin^ d + b cos^ ^ 7i sin- ' a + & tan^ dm 1 a tan- ^ + 6 7i tan^ tan- ^ ' .-. b tan^ ^ =^ fe, or tan^ 0-^1. .'. u tan^ 0= ±7?i. By adding together the first two equations, we obtain a + b = VI cos^

»t)- = Z;- (a + 6)- - f- (c- + 2am)2 ; .-. (a- - Z^2) (a + Z;)--^ = 6-2 (a2 - fcS) _ iabcm (« + &); ,-. 4a&cm := (a - ^) [c- - (a + b)'^]. EXAMPLES. XXV. c. Page 334. 1. ^y putting x = y cos 6, the given equation becomes 3 1 cos^ - — , cos 6 ;. = 0. y'- r 3 . COS 3^ ,, But COS'* - V COS 6 . — = ; 4 4 .-. ., = ; whence y = 2. cos3— = •< = o ' whence cos 60 --\ % ij fi ^ :. 3^ = «.360°±60°; .-. ^ = 20°, 100°, or 140°. But x — y COS 6^2 cos 6 ; and therefore the roots are 2 oos 20°, - 2 cos 40°. - 2 cos 80°. k XXV.] APPLICATIOX OF THE THEORY OF EQUATIONS. 231 2. By putting x = y sin d, the given equation becomes 3 1 sin^^ - — sin ^ + -^ = 0. But sin3^-?sin^ + ^''',^^ = 0; .". y^ = 4; whence y = 2. ., sin 3^ 1 1 , • o/, 1 Also — , — =-„- = - ; whence sin 3^= - . 4 y* 8 2 Hence as in Art. 328, the roots of the equation are 2 sin 10°, 2 sin 50°, - 2 sin 70°. 3 '3 3. Put .r=ycos^, then cos"^ :rcos^-^ = 0. y' y' T. X ,^ 3 ^ 0033^ ^ But cos3^--cos^ i — = 0; 4 4 .•. y^ = 4; whence y = 2. .. cos 3^ JS ^S . „^ V3 Also - -_ = 2L^ — :^ ; whence cos Sd = -— : 4 7/-^ 8 2 .-. 3^ = n. 360° ±30°; .-. ^=10°, 110°, or 130°. But x = y cos ^ = 2 cos 0, and therefore the roots are 2 cos 10°, - 2 cos 50°, - 2 cos 70^. 3 /2 4. Put a: = vsin^; then sin^^ - -p-; sin ^ + -— ^ = 0. ■n . • ^^ 3 . ^ sin3^ ^ But sm3^---sin^ + — - — = 0; 4 4 .'. 4 —IT it ~ — 4 — ' 4 Aa-^y^ 4 .-. 3^ = n.l80^ + (-l)"3^; .. e^A, GO°-A, 120° + A, 180°- J, 2iO'' + A,... .-. sin ^ = sin .4, sin (60°-^), sin (240° + ^). But .r~)/ sin 6 — ~ sin d, and therefore the roots are •^ a -sin A, -sin (60° -J), - i sin (60° + .4). a a 'a 3a2 ^ 2a3cos3.4 „ 6. Put x = ycos^; then CO s^^ l^cos^ 3 = 0. ^ , ^ 3 ^ COS 3d 2a3 cos 3^ cos 3^ l2 = J ; whence 2/ = 2a. A-lso . — . — ^ o — . , 4 1/"* 4 .-. 3^ = /I. 360° ±3/1; .-. d = A, 120° ±^. But x = y cos ^ = 2a cos 6, and therefore the roots are 2a cos J, 2acos(120°i.4). 7, (1) From the theory of quadratic equations, we have sin a + sin /3 = — (1). a By supposition, sin a + 2 sin /3— 1, .-. sm/3 = l + -. But asin2/3 + fosin/S + c = (2), .-. {a + h)^ + b{a + b)+ac = 0. (2) Substituting from the equation csina = asin/3 in (1), wo have a sm 8 + c sm 8= ; •^ a .-. a (a + c) sin/3— - he. c But sin a sin /3= ; a .-. a-sin-/3 = c-; whence a + c= J^b. XXV.] APPLICATION OF THE THEORY OF EQUATIONS. 233 8. We have tan a + tan B = ~ . '^ a Also, by hj'pothesis, a tan a + h tan /3 = 26 ; whence (/>- a) tan^ = 6. But atan-^-6tan/S + c = 0; .-. oU--b-^{h-a) + c{b-u)- = 0. 9. We have tan a + tan ^ + tan 7 = 0, 2a — X and tan a tan /J + tan a tan 7 + tan /3 tan 7 = 2a — X .: tan a tan 8 - tan- 7 = . "^ a Now tan-a + tan-j8=(tan a + tan/3)2-2 tana tan/S 2 (2a - a;) a ' = tan^7-2 tan-7- a .'. a (tan'^a-f tan-;3)=r - a tan27- 4a + 2x. But, by hypothesis, it (tan'-^ a -r tan-/8') = 2a; - 5a ; .•. 2x - oa = -a tan- 7 - ia + 2x ; .-. tan-7=l. But a tan37 + (2a-a') tan7 + ?/ = 0; .-. a tan 7 + (2a - x) tan y + y — 0; .'. {'da - x) tan y + y = 0; .'. Ba-x= ±?/. 10. We have cos a cos /3 + cos a cos 7 + cos ,3 cos 7 = &. Also, by supposition, cosacos^ + co8acos7=26; .•. cos ^ cos 7= -b. Again, cos a cos /3 cos y= - c; .'. icosa = c. Now cos^ a + a cos- a + b cos a + c = : .-. c^ + abc- + b\- + b^c = 0. 234; APPLICATION OF THE THEORY OF EQUATIONS. [CHAP. 11. As on page 329 we may shew that 27r COS a, cos ( -^- + a I , cos (I-) are the roots of the cubic , ^ 3 ^ cos 3a „ cos^ 6 - -7 cos 6 - — — — = 0. 4 4 Write cos 6 = -- - ; then — .— sec^^ + - scc^d -1 = 0; sec ^ 4 4 /2ir \ /27r \ ! a + sec I --^ + a 1 + sec I — - a 1 = 3 cos 3a sec a + sec I -rr + a 1 + sec [ -rr - a )= - ;-. r — • 12. The values of sin 6 found from the equation sin 3^ = sin 3a are 2ir \ . / 47r sma, Pin(— + aj, sml-— + a); that is, these three quantities are the roots of the cubic equation sin3^-^sin^ + — "^'' = (1). 4 4 .-. ; .-. cot-asin'*^-2sin-^cos2^ + tan-acos4^ = 0; .•. cot asin^^- tanacos2^=:0; sin-^_cos2^ sin- a cos^a sin2'*^_cos2«^ sin-" a cos-" a sin2«+2^ cos=«+2 6' sin-" a -f cos-" a 1; sin2 = sin w 2a y sin w 45i a2 + ^2 _ _j2 ^ ^2 ^ 2^ - X- _ 02 + x2^J/2 a- + h'' + C' a'+h-^c' I 4(51 + 52 + ^3) 4A = cot ^ + cot 2? + cote. [XVIII. a., Ex. 34.J (2) By squaring the result just obtained, we have cot^ w = cof- A + cot^ B + cot2 C + 22 cot B cot C = cot^^ -t- cot^iy + cot2 C + 2, since 2 cot i? cot C = 1 . .-. l + cot5w = (l + cot2.4) + (l + cot2 7?) + (l + cot"C); .'. cosec^ w = cosec* ^ + cosec- B + coaec^ C 242 MISCELLANEOUS EXAMPLES. 1. [XXV, I 8. On page 195 of the FAcmentary Trigonometry , suppose that ABFD is a vertical plane running' N. and S., and that CG is drawn in a S.E. direction. Also suppose that AB = a, and AC = HJ^a\ - then 2JC2=(169a)2-fl2 = 28560a2; " .. CG2 = 2i5 02 = 57120a-'; .-. CH2 = 57121a2, .-. CH=ima. I 9. Let ABC be a horizontal section of the two walls, and let (p be the inclination of the wall AB to the meridian, so that 7 - is the inclination of BC to the meridian. The length of the shadow of the wall AB measured along the meridian is (I cot 6 ; hence the breadth of the shadow (which is measured at right angles to AB) is a cot 6 sin 0, .-. h = a cot^ sin0. Similarly, c = a cot ^ sin (7 - 4>)\ :. c = a cot d (sin 7 cos

^~ %' '^~~h' « mi ^ X • 1 / cos A sin A \ /cos A sin A \ 3. The first side= -. — ^ . H" "^ — 7 + 1 \sinJ cosyiy \sin.4 cos A/ = (cos^ A - sin- A ) -^ (cos- A + sin- A ) = cos2 A - sin" A = {1- sin^ A ) - sin^ A = l-2Bin2^. MISCELLANEOUS EXAMPLES. K. 243 , 5 13 3 5. Let .475^15, AC=30, then Bince cos60° = i, it ia easy to see that CB is at right angles to i? .4. .-. CB = CA sin eO° = 30x^^1^ =25-08. 6. Let AB be the tower, BD the clilT, and C the point of observation; then if BD = a; f t. , CD = yit., we have 50 + .v = y tan a, x=>j tan /3. By division 50 ^U^ ^Um^Si X tan/S 1185 79" 50 5 ••• y = y^,anda; = 790. 7. If X ft. be the length of the arc, — = radian measure of 10° ; whence :c = 30x-^x 10 = 5-236. 8. Here tan a = — ; .-. sin a = — , cos a = — . Id 17 17 9. Here 4 sin^ ^- (2 + 2v'3) sin ^ + ^3 = 0; whence (2sin ^ ^-^t: = 77 • ^^' 180 7 180 4o op 1 n The arc traversed in 30" = — x -— x 1760 yards = 176 yards. oO bU .-. \[ d be the number of yards in the diameter, 176 22 . -d- = 45' •■•'^ = ^^0- i r » MISCELLANEOUS EXAMPLES. K. 24j5 19. See Art. 35. 20. (1) Fir.tside^lx(^^-l) 'sin- A , \ sin^ A - cos'-^ A cos- A = sec- A (sin- A - cos^ A ) = second side. , „ ^ .- sin- a sin- S sin-asin^iS (2) Second side = — . ., ^ r-;; ^ ' sin-/S sin- a = sin-a - sin-jS = First side. [Examples III. b, 34.] 21. ^Ye have cos jB =- = -405; anda + c = 281. c .-. c(l + -405) = 281; whence c = 200, a = 81. Also 6 = V^^-^ = 7^3439 = 183 nearly. «^ rr,, T arc 495 . 3 ^^«„_ 22. The radian measure = — ^. — = 7, — --=-— — :r^= -09370. radius 3 x 1760 32 Q With this as unit a right angle would be - -^ — =16 '7552. ^^ ,,,_,.,., . „ „ {cos 6 sin ^1 „^ . „ „ , 23. 1) First side = sin 6> cos ^ -^^ — --+ ^^ =cos2^ + sin2^ = l. ^ (sin d cos 6\ ,^.T^- . • -. cot cot- 6 cos 6 (2) First side= ^ x x . „ ■ sec 6 cosec d sin-* ^ ,„ „ cos ^ cos ^ = cot3^x-^ — - x -^ — - = cot^^. sm ^ sin 24. If O be the jDoint of observation, l BOA =45°, and the line drawn from B perpendicular to OA bisects it at a point A'. Then OA : OB = 20 A' : 0Jy = 2 cos45° = v/2 : 1. I 25. (1) First side = (sin^ A + cosec^ ^ - 2) + (cos^ A + sec- A - 2) = {nin-A + cos2 A ) + (cosec^ A-l) + (sec^ ^ - 1) - 2 = cot- vl + tan- A - 1. I (2) First side = 3cof-^- 10cot^ + 3 = 3 cosec^^-lOcot^. I 2-? 2 - cot /I 2 5 27. The expression = ^^^^^— ="-27= "31 * H. E. T. K. 18 24G MISCELLANEOUS EXAMPLES. K. 28. 2cos^cot^ + l-cot^-2cos^ = 0; .-. 2cos^(cot^-l)-(cot^-l)=0; .-. (2cos<^-l)(cot^-l) = 0; .-. ^ = 60°, or 45°. X IT 1217 29. If ;c iuches be the length of the arc, - = -— x — .^ . In the second circle ^ = — .,. x —^^- x - ; o 180 10 1217 .-. sexagesimal measure of ^ = ^r.— ji = 15° 12' 45" 30. Take the figure of the Example on p. 41, and let PT = x yards, IiT = y yards. Then ;c = ?/ + 110, since zPgT=: 45°. X Also y=x cotQO°=-j: . A a; = 55(3 + V3) = 2G0-2G. ,-,. , • -> 1 + cos^ 1-cos^ . .„ . 31. First side = - , - , , - 4 cot^^ ^^' 1-COS^ i + COS^ _ (l + cos^P-(l-cos.4)2 ^^^^^^ (1-cos^-l) (1 + cosyl) 2. 2 COS ^ 4cos2^ _4cos^ (1-cos^) sin^'-i 4 cos^ sin- A 4 1 - COS- A 1 + cos .J 1 + sec A ' 32. (1) 8 (1 - cos- ^)- 2 cos ^=5 8cos2 + 2cos0-3 = O (2cos^-l)(4cos 2rt?> ; .-. cos ^ < 1, which is possible. 36. Let ACB be the hill, A being the summit and C a point halfway down. Draw AD, CE perpendicular to the horizontal line through the object 0. Then AD = 2CE, and BD = 2BE. Now OB + BE = GEcot^, and OB + 2BE = 2CE cot a; ,^. r u u, .• 0B + 2BE OB + BE ^ ^ therefore, by subtraction, — -— =2 cot a - cot ^ ; BE Ti'F that is, -^=:2cota-cot j8; also ^^;^ = cot ^. 37. From the figure in Ex. 2, Art. 46, we have ».-. & = csin45° = 50x-- =2ov'2. If p be the perpendicular from C, ^ = a sin 45° = 25. 40. Since tan ^= ± 1, the angles will be those coterminal with 45°, 135°. 225°, 315°. 41. Take the figure of Example II. on p. 43. Jjet AE^x, CE:=y; then I I X „„. .T + 42 -=-96o: =1-0. y y a; + 42_lG0O_ 635^ ■■ ~^~ 'Mib ~ "^905' 905 whence i; = 42 x — -^ = 63 , approximately ; DOl) ^5 = 63 + 42 = 105. 18—2 248 MISCELLANEOUS EXAMPLES. K. 42. We have tan 15» = ff^|^= i / (l + f ) = ^^, = 2-^/3. Similarly tan 75° = 2 + ^3. Again , 1 + tan^ 6 = 4 tan 6, tan2^-4tan ^ + 1 = 0; 4.4- /To .-. tan^ = ^^|^-=2db^3; .-. ^ = 75°, or 15°. 43. The first side = 1 + sec 6 + tan 6 + cosec 6 + sec 6 cosec d + cosec 6 tan 6 + cot 6 + sec 6 cot ^ + 1 = 2 + sec ^ + tan 6 + cosec ^ + sec ^ cosec ^ + sec ^ + cot 6 + cosec 6 ^,-. ^ ^^ sin 6 cos ^ 1 = 2(l + sec6' + cosec^)+ +-. — 4- cos 6 sin ^ sin 6 cos 6^ .V 2(sin2^ + cos2^) = 2 1 + sec 6» + cosec 6) + ^ . ^ -— ■ ^ sm ^ cos y = 2 (1 + sec ^ + cosec + tan ^ + cot 6). 44. I^ee figure and notation of Art. 25. 45. We have 2 (1 - sin^ ^) = 1 + sin ^ ; .-. (l + sin^)(l-2sin^) = 0; .-. sin^=:-l, or sin^ = -; whence ^ = 30°, 150°, 270°. 46 sin (270° + /!)= -sin (90° + ^)= -cosy(. But cos A= ^ x/l - sin^ A= ±'S; .-. sin(270° + zl)=i-8. 47. See Art. 113. ^ r. mi • 2 sin 2ii sin ./4 » . on 48. The expression = —^^ — : — - = tan2.4. See Art. 89. '■ 2 sm A cos 2^ 49. tan A = Vsec2"^ - I =±^--l = rfc_. The angle is coterminal witli 150° or 210°, and the tangents of these angles are ecjual hut opposite in sign. MISCELLANEOUS EXAMPLES. K. 249 50. Take the figure of Example on p. 41, and let PT=x yards, iPQT=i5°, lPRT=&0°. Also (?i?=1760, and QT=PT; ■'- ^^^='^^'''' = -''-^ .-. a;(v/3-l) = 1760v'3; .-. a: = 880 (3 + ^3) = 4104-10. 52. (1) First side = sin-a(sin-a + 2cos2a) = (1 - cos2 a) (1 + cos2 a) = 1 - cos-* a. (2) First side = sec 2 ( ^ - ^ j [Art. 124] = cosec 20. (3) cos 10° + sin 40° = cos 10° + cos 50° = 2 cos 30° cos 20° = ^3 sin 70°. a tan ^ - 6 a^- b'^ 53. The expression = atan^ + & a^ + b^ 54. Multiply all through by cos 18°, then we have to prove that 4cos2 18°-3 = 2sinl8°. First side = 4 {l-(^~~J} -3 = l^i^-3 = ^^ = 2 sin 18°. re -U n ^^C 20x10 1 1 56. Here ^= ,. = -~ — -^- =.-• radius 00 x 00 2 9 1 7 70 • D = - X 180 X — — — - — fi * ° -^ 9 22~11~^T • cr? rri • 2-3 cot a , , 12 57. 1 he expression =- — — , where tan a = - - , 4-9 tan a 5 2 -3x 5 12 3 4 -9x 15 12 5 ~4 • ~ 352* _88\ 250 MISCELLANEOUS EXAMPLES. K. w 2 sin J Bin 6 58. (2) First side-. +^2 2 cos — sin 6 ^/^li-lh'j'-'- __ ,,, m • C08.4 COS (7+(-cos C)(-cos^) 59. (1) The expression = --: -^ — . - ^ '^ —.-— ^^- ^ ' ^ cos.4smG'-sinC(-cos.t() 2 cos A cos C . ^ 2 cos A sm o sin u4 cos A +two similar terms sin A sin B sin C 1 /sin 2^ + sin 2B + sin 2C 2 (2) Tlie expression = sin ^ sin B sin C = 2. [Art. 135, Ex. 1.] ) 60. Let ABC be the horizontal equilateral triangle, and let FQ be the flagstaff. Then since each side subtends an angle of 60° at P, the top of the flagstaff, the triangles PCB, FBA, PAG are equilateral. Let a; be a side of a ABG\ then AQ=i Then from a BA Q, X 2 sec 30° ^3 * PA^=^AQ''+QP-, X* whence x2 = — +10000; a;2 = 15000, or x =50^6. 61. The first expression = sin ^ + cos + sin2^ + cos2^ = sin^ + cos^ + l. Similarly, the second expression = sin 9 + cos 6 -1\ :. the product = (sin 6 + cos d)'^-l = 2 sin 6 cos 6 — sin 26. 62. Second side^ sin''' o "^ ^°^^ "o ^ ^^^ ~ o ^^^ ~o = - { 1 - cos (^ 4- 0) + 1 + cos {6 - 0) 1 - sin d - sin 0. = l-sin ^-sin + .-r {cos (^ - 0) -cos(^ + 0)} — 1 - sin ^ - sin + sin ^ sin = (l-8in0)(l-sin0). MISCELLANEOUS EXAMPLES. K. 251 2 cos a sin d cos a , • i • • j i ^ <• /. 63. The expression = ^ . ^ . ^ = -^-^ , which is independent of 6. "^' ^ 2 sin /S sin ^ siu /3 H + G H — C 64. Fust side = sin — ^ — cos — ^ H ... + .. . = - {(sini5 + sin C)+ ... + ...] = sin A + siu B + siu C. . 65. ^^ e have 2 1 - SIU- 6 cos- ^ ^ sin ^ ^ . 6 e' cos- 2 sin- cos ^ ^(^ 2sin2;;-( 2cos2^-l ] ; l! /. 4 cos^ o - 2 COS --1 = 0; „d 2±V'-^0 1±V5 .*. cos" - = COS 36°, the other value being impossible. 66. Draw ZW perpendicular to XY and let ZW=x. Then aXZY is right-angled at Z. And ZZ = Zr cos 30° = 100 ^/3 yds. Again, from aWXZ, irZ=Xi; sin 30° = 50^3 = 86-6 yds. 5585 67. We have 327r x 1000 = -— - x 3 x 12; whence tt =3-141, approxi- mately. /,o • " ' '> r, • o 1- cos 2a 68. sm-a + sm2^ + 8in27= +... + ... 3 1 = - - - (cos 2a + cos 2^ -f cos 27) 3 1 = - - - { 2 cos (a + /3) COS (a - /3) + 1 - 2 8iu-7} = 1 - {sin 7 COS (a - /3) - sin- 7} = l-sin 7 {cos{a-^)-cos(a-i-/3)} = 1-2 sin a sin )3 sin 7. 252 MISCELLANEOUS EXAMPLES. K. 69 (1 First side =( -+ -~- , ) (2 cos 2^ cos J) ' \coSxi COS 2.4/ = 2 (sin 1A cos A + cos 2.1 sin ,4 ) = 2 sin '61 . (2) Multiply all through by 32 ; then Second side = 2 + cos 2.-1 - 2 cos iA - cos 6A = 2(1- cos 4^ ) + 2 sin iA sin 2A = 4 sin2 2.1 + 4 sin2 2.J cos 2 A = 4sin2 2.-l (1 + cos 2.4) = 16 8in2.4cos2.4.2cos-.4 = 32sin2yl co8*^. 2 cos 13a sin 10a cos 13a 70. The expression = — . — ^— = r^— ' ^ 2 sin 10a cos 6a cos 6a cos 13a _ cos (tt - 13a) 71. We have cot {A+B) = 1. Therefore cot A cot B -1 = cot A + cot B ; .-. 2 cot A cot B = l + cot A + cot B + cot A cot B = (l + cot^)(l+cot^); cot A cot B _ 1 •* I'+cot^ * iT+cot^ "" 2 • 72. For the first part see XI. d. Ex. 15. Then tan ^ = cot ^ - 2cot 2d, 2 tan 2l9 = 2 cot 2^-4 cot 46, 4 tan 4^ = 4 cot 4^-8 cot 8^ ; by addition tan ^ + 2 tan 2d + 4 tan 4^ = cot } = 1 - cos2 {A+B) = sin2 (A + B). (2) First side = 2 (sin 5A - sin A) - (sin SA + sin ^) — 4 cos 3.4 sin 2.4 - 2 sin 2A cos A = 2 sin 2^ (2 cos 3.4 - cos .4) = 4 sin A cos A (8 cos^ A -7 cos .4) = 4sin/I cos^vl {8(l-8in2j)-7} = 4 sin 4 cos- ^ (1 -Ssin^.^). 254 MISCELLANEOUS EXAMPLES. K. 77. If r is the radius of the circle, and AB one side of the square, we have 27rr = 3, and Ztt =r I X 1-4142 X -3183 - -6752 feet » = 8*10 inches. 78. Here ^fi = 2rBin54°, £C = 2rsin30^ CD = 2r sin 18°. And it remains to prove that sin 54°= sin 30° + sin 18°. [See Examples XI. c. 9.] 79. First side = (2 + ^3) + (2 - ^3) - 1 - 2 = 1. 80. We have, by addition, cot ^ = 2 {m + n). Also, by subtraction, cos = 2 {m- n). cos- d ' sm 6 : 16 (m^ - n')^ = —^ = cot^ x cos2 ^ ' sin2 e = cot2^(l-sin2^) — ICmn. 81. (1) First side = - [sin (2/3 + a) + sin a- sin (27 + a) -sin a] = -[2cos(a + /3 + 7)sin(^-7)]. i ,^., ^. , ., /sin2^ sin.4\/ 1 1 \ I (2 Fn-st side= ^- - ■ + ~, ' \cos24 cos^y \cos^ co&dAJ sin A cos 3.1 + cos A cos 2A cos A ' cos A cos 3-4 2 sin yl cos 2/1 cos A ^ . . = ^ ,, , ^ . = 2 sm A sec ^ sec oA . cos 2^ cos-* A cos o.-i 82. 2 cos 6° cos 66° = cos 72° + cos 60° = sin 18° + ^ _ J5-1 . 1_^5 + 1 -~T~'^2 4~* MISCELLANEOUS EXAMPLES. K. 255 2 cos 42° cos 78° = cos 120° + cos 36° 2"^ 4 ~ 4~ • .-. 4 cos 6° cos 66° cos 42° cos 78° = - . 4 83. Put 2.4 =45°. 84. The distance required is evidently equal to 10 tan 22^° = -/2^-j = 10 (V2 - 1) = 4-14 miles. 85. (1) Separate each term into the difference of two cosines. (2) Second side ^'i^L^ll'i^^coi^ cos^ + 2cos2^ _ Bing(l + 2cos^) ~ cos ^(1 + 2^8 ^)'^^'^"^- 86 . First side ^ 2 cos"-J-^ cos ^^ + 2 cos^ ^ - 1 = 2cos|cos"-^^ + 2cos2j-l = 2cos|jcos^-^ + cos^l -1 = 2 cos I jcos -^ + cos —^\ - 1 = 4 cos- cos \^ cos -' - 1. 2 2 2 87. Put-^ =-A =-^ = ^, then sm A sm i> sin C first side = — ^ — ,— ^— ; ^ cos .i + . . . + , (sin (i5+C) sin (B-C) ) { sm^ T-...-r...^ = ^{sin(7?- (7)008^ + ... + ...} = -A:{sin(2?-C)cos(B + C) + ... + ...| = - 2^{(sin 22^- sin 2C7) + . .. + ...} = 0. 256 MISCELLANEOUS EXAMPLES. K. 88. First side = a cos 2^ + 6 sin 2d = a{l-2 sin^ 0) + 26 sin 6 cos = a + 2 sin {b cos - a sin 0) = a, since & cos ^ = a sin ^. 89 . T-iet logo b — x, so that a^ = &, logbC = y, by = c, log^a — z, c^ =a. Then we have a-c' — bv^ — a'^y^ ; :. xyz=l, or log^ 6 log^ clogca^l. We have log 8 = log 2^ = 3 log 2 ; whence log 2= -30103 ; log2-4 = log /3x8\_ V 10" ) ~ log3 + log8-l = 1-47712 + -90309 = -38021. log 5400 = 2 + log 2 + 3 log 3 = 2-30103 + 1-43136 = 3-73239. L tan 30°= 10 +log -^ = 10 - I log 3 = 9-70144. 90. cot {A +B)~-^ cot (90° - C) = tan (7 = - - ; cot .1 cot J5 - 1 _ 1 , cot A + cot B ~ cot C ' whence by multiplying up and rearranging we obtain the required result. For the second part, put A = 15°, i? = 30°, C = 45°. 91. First side r^ (1 + sin 2Ay^ + cos^ 2^ + 2 cos 2.4 (1 + sin 2.^) = (1 + sin 2 J )2 + (1 - sin2 2^ ) + 2 cos 2.4 (1 + sin 2 A) = (1 + sin 2^) {1 + sin 2^ + 1 - sin 2^1 + 2 cos 2A } = 2(l + cos2.4)(l + sin2.4) = 4cos2J(l + sin2J). 92. See Art. 150. 93. Wc have to prove that sin 9° sin 81° = sin 12° sin 48°. First side = - (cos 72° - cos 90°) 1 • 1O0 -s/o-l = _sinl8° = ^^. MISCELLANEOUS EXAMPLES. K. 257 Second side = - (cos 30° - cos GO'') 1/^5 + 1 1\ V5-1 2V 4 ~l) I 8 94. See Art. 136, Ex. 2. 95. L sin ^ > L sin 27° 45' by J^^^ x 60" ; whence ^ = 27° 45' 44". 96. By a well-known algebraical formula, .T^ + y^-\-z^ = 3xyz, when .T + t/ + z = 0; therefore we have cos^ A + cos^ B + cos-"' C= 3 cos A cos B cos C. Substituting - (cos 3,-! + 3 cos^) for cos^^, and similar results for cos^B, cos^C, we have 1 3 P -(cos 3^ +COS 3JS + cos3C')+^(cos^ + cos£ + cos C) = 3 cos y4 cos i? cos C, whence the required result follows at once, since P cos 4 + cos J5 + cos (7=0. / 49 24 97. cos yl = >/l - sin- J = . / 1 -,.,- = ± ^r^; but since A lies between 24 270° and 360°, we must reject the negative value; thus cos ^ = — . Hence sin 2 A =2 sin A cos /1=2( ~.7r)(s^) _ 336 ~ ~G25* ^ 1 - cos ^ Also tan — • = — . = cosec A - cot A 2 sin A _ _ 25 24 _ _ 1 - - y + y - - ^ . b 258 MISCELLANEOUS EXAMPLES. K. /3 Ji COS Tr / • yn r\\ o 98. We have » ^ (^__-^— j ; sin- 8 cos^ - sin - = 1 + sin 20, a 4cos2%in^ = l + sin2^, 2 sin ^ (1 + cos ^) = 1 + sin 2^ ; .-. 2 sin = 1, or = 30^ Again, by putting = 30°, we have 2cotl5° = (l + V--^)- = 4 + 2^3. .-. cot 15° = 2 + ^3, and tan 15° = 2 -V3. 99. We have log 3G0 = 2 log 2 + 2 log 3 + 1. .-. 2 log 3 = log 360 - 2 log 2 - 1 = 1-5563025- -6020600 = •9542425; .-. log 3 = -477121 3. Now log -04 = log 4 - 2 = 2 log 2 - 2 = 2 -60206. log 24 = 3 log 2 + log 3 = -90309 + -4771213 = 1-3802113. 2 log 6 = log - = log 2 - log 3 o = •30103 - -4771213 = 1-8239087. Again let log.^30 = .T, so that 2=^ = 30. .-. xlog2 = log30 = l + log3; . ,-l + l«g3.1-^^71^1^.4-90689. ' log 2 -30103 100. 1'iiia follows at once from Art. 134, Ex. 5. MISCELLANEOUS EXAMTLES. K. 259 101. cotia + (3)-^^^^^^^-^- x{x + x-^ + l)-l _ x'^ + x I cota + cot/3 (x^x-^ + \)\l+x) (1 + x) (x + x-i + 1)* _ X 1 _ 1 (x + x-1 + 1)* x-i(x + .r-» + l)^ (x-i + x-3 + x--)* = cot7. Therefore a + /S=7. 102. Let .i and a be given, and let i? be the right angle; thenc = acot^, h = acoseJ x^ + xy + /^ . ^ - cos C7=^ti^'±^^±^)^ _ 1 . 2x7/ 2 ' .'. C=120°. 105. cos3^+sin3^=4cos3.4-3cos^+3sin^ -4sinM = 4(cos3.-l -sin3yl)-3(cos^-siu^), which is evidently divisible by cos ^ - sin ^. See solution to Examples XII. c, 27. I I 5 20 106. We have cot ^ = tan "^ "^ ^' ^ ^^ \ 2 2 1 ^ ^~2* 637 2tan^ 2x? „^ .-. tanC= I =—L=^l. , , .C ^ 4 21* '-''^°-2 '-25 For the second part, it will be sufficient to prove that sin ^ + sin C = 2 sin JL 260 MISCELLANEOUS EXAMPLES. K. 2 tan ^ Now sin A = = -—- , on reduction. 1 + tan- — 20 1480 Similarly sin C =~, and sin £ = -^--, whence the result follows, 107. The first factor easily reduces to 2 cot 26, and the second to 2 — ; whence the product becomes 4cosec2^. COB 26* ' ^ 108 tan^l- A '-^)i'-<^) - ni^^ 108. tan .^ - ^ ^, f^^ _ ^^j - ^ (5^3 - log tan ^ = ^ {log 128 - log 003} = I -0634465 ; whence ^ = 24° 44' IG", and A = 49° 28' 32". 71 sin A cos A . . f^ n cos- A tan A - — . .. - tan A _, 1 - » sin-.-l v 1 - 7isin"-.4 109. tan{A-B)r , / It ain A coa A \ _ n tiin- A 1 + tan A { ;, r-^-- 1 + :; T-^- \1- n sm-^ ^ / 1-/1 sm-^ A = {1 -7i) tan .-I. 110. log 200 = 2 + log 2 = 3 - log 5 = 2-30103. log -025 = 2 log 5 - 3 = 2 -39794, log t/62^ = ^(log625-l)=.J(41og5-l) o o = •598626. Lsin30°=:10 + log('^^=10 + log5-l = 9-69897. Lcos45° = 10 + logf-4VlO + ^(^og5-l) V2y '2 = 9-849485. co t (A - 30°) _ c os {A - 30°) cos {A + 30°) ^^■^- ^ ^ tan (J + 30'^)" sin (.4 -30°) sin (^1 + 30°) cos 2.4 + cos 60° _ 2 cos 2 A -f- 1 cos 00° - cos 2A ~ 1-2 cos 2/1 2 + sec 2 A sec 2.4 - 2 ' MISCELLANEOrS EXAMl'LES. K. 2(U ,„, „ , ., „ 11 + cos 2a 1+ cos 2/3 1 - cos 2a 1-cor25) (2) Second Ride = 2 -j^ --— . -^- + - — ^^ . ,^ — ^j- = - { 2 + 2 cos 2a cos 2/3 } = 1 + cos 2a cos 2/3. 112. From a diagram it is easily seen that CD is equal to A(', and that froiJi the right-angled triangle ABC, AC = BC cos 'dO' = 132 "^^y 19 = 66x— = 114 yards. Also the perp. from A on BC = AC sin 30° = 57 yards. 113. Since a sin^ a + b + c sin A + sin B + sin C ^ . A A 2 sm — cos — A B C 4 COS - cos — cos — ^ a ^ [XII. d. Ex. 3.] a cot — + o cot -— - c cot — ^ A cos 2 a-\-b-\-c 1 B G cos 2 cos - + ... - r ..--1 ./.' „c\ ^ |cos-- +cos---cos---| ^i ~i c \ cos — cos — cos — 2 2 2 J Now cos2 1 + cos^ - cos- H = o \^ cos2 ^ + 1 -J- cos i> - 1 - cos r - = 2 |2cos2 2+2sm- 2 sin -^--^ = cos-2 |cos- + sm----| A \ . B + C . C~B] = cos2 |sin ^ +sm — [ A B . C — 2 cos — cos ~- sin ^ . Ji Ji a Whence tho required result easily follows. If. E. T. K. 19 262 MItSCELLANEOUS EXAMPLES, ^42 A Is (s - a) _ /40jv 2 " V he ~ V 75^ 114. Here cos ^ . , , —*/«,-„„ ■ •' ^ ' '- V 75x77 __2 2^/2 ~ ^o ~ ^10 ' log cos "- = - log 2 - - log 10 = •4515450- -5 = 1-9515450 log cos 26^ 34' = I - 95153 89 61 4 61 .-. - is less than 26° 34' by "., x 60" ; 2 -^ 632 ' that is, ^ = 26° 33' 54-2", or ^ = 53° 7' 48". 115. sin {A - 90°)= - sin (90° ~ A) = - cos .4 = - ^l^^^im^^ ^ _ ^6^ = -(±•8); but A is between 90° and 180°, therefore sin {A - 90°) is positive ; that is sin(^ -90°) = -8. cosec(270°-.4) = cosec(180° + 90°-J) = - cosec (90° - A) ^\ i'8y = ±1-25; but between the given limits cosec (270°-^) must be positive, that is, the required value is 1*25. 116. By Art. 168, log5C = -^",, log,(Z= " ; loga6' logaC hence the expression on the right = log,, b x , ; x i - loga/> logoC = logac/. MISCELLANEOUS EXAMPLES. K. 263 Iogio2-l-logio5 = -30103; Iogio8 = 31ogia2 = -90309. logg 10 = ,-i— - = 1- 1073098. P 32 loRjo (•032)'* = 5 log ~^^ = 25 log 2-15 [ = 7 -52575 - 15 I =8-52575. 117. First side = cos (360^ + 60= + .-!) + cos (00'^-^) = cos (60° + A) + cos (60° - A) — 2 cos 60° cos .4 = 008 A. For the second part, put .4 =45°. .T 118. Write t for tan '- , then the equation may be written 1 + ^2 - s^^ « cot /3 j-^^ = cos a ; .-. ^-(l + cosa) + 2« sin acot/3- (1 -cos a)=0; ,„ „ , ^ sin a 1 - cos a ^ f2+2cot/3- .f-, = 0, 1 + cosa 1 + cosa «2 + 2 cot/3 tan ^<- tan^^^O, f f+tan^cot|j U-tan^tan|j=0, [XI. d. Ex. 15.] , .T . a S a. ^ 8 tan - = - tan - cot ,- , or tan tan ^ 119. We have sinJ2£+^^^ sin £ w j?i _n_sin(24 + U)-sini? w + n sin (2^ + 7?) 4- sin B _ 2 cos {A + 7J) sin J "" 2 sin (7f"+7j) cos A = cot (J+L') tan J. 19—2 264 MISCELLANEOUS EXAMPLES. K. 120. Let X feet be the height of the tower ; then ' Z.4M; = 00^- /.ADB = 45°= lADB\ :. AD = AB = x', .-. AC = x -17. Kow from a ABC, .r2 + (x-17)''=532; .-. .r2-17x-l'260 = 0, (.r-45)(.r + 28)=0; 45 Again, tan A CB = ^3 ' , ^. 56 28 but tan 81° 48'= -62= ^ = jg; .-. z.4C;^ = 90°-31°48' = 58° 12'. 122. log 6=.^ log 36 = -778151, log 8 = log 48 - log 6 = 1-681241 - -77815.1. = -90309 ; .-. log 2 = J log 8 =-80103; .-. log 8 = log 6 - log 2 = -477121. Now log 40 = 1 + 2 log 2 = 1-60206. log ^^ = I (log 4 -log 30) = ^(2log2-log3-l) = 1-562469, on substitution. B-C h-c .A 9 , „^o 123. Wo have tiui --^^ = y> + c *'''* 2 " 10 ''''* ' ' log tan — ~ -^ = 2 log 3 - 1 + log cot 72^ = 1-4660186 log tan 16° 18' = 1-4660078 J^ ^ 60"= 1" 108 4687 ., :^;i^' = 16°18'l", ~+— =18°; .-. B^34 18' 1", C=V 4r 59". MISCELLANEOUS EXAMPLES. K. 265 124. (,1) COS ^ + cos 7? cos r =- cos 7? + (7 + cos 7? cos (7 = sin 7) sin ( ' ; .•. Second side = u- sin B sin C — b sin A . c sin A = be sin- A . (2) First side = c {b cos A+a cos 7i) + 2«6 cos G = c- + 2ab cos C = a- + b'-. - tan (^ - tan - 3 tan - 8 — a ^ 2 2 125. tan^ "- l+tan^tan^^ 1 + 4 tan- ^ a a it ., . tt a 3 sin - cos - ^ . 2 2 3 sin a cos2-- + 4:Sin^- 2cos2-- + 8sin-- 2 2 2 2 3 sin a 3 sin a l + C03a + 4 (1 -cos a) 5 -3 cos o' 126. sin (3G° + ^) - sin (3G^ - ,4) = 2 cos 36° sin A =^^^ sin A ; sin (72~ -A)~ sin (72^^ + ^) = 2 cos 72° sin {- A)= ~ '^''~^^ sin A. By addition we obtain the required result. 127. tan(?: 2 sin ^ _ ~ 3 _ 2 V9 The boundary line of Q is in the 3rd or 4th quadrant, hence the tangent is positive in one case and negative in the other. 128. (1) First side = sin 2/1+ sin ('^-27;') = 2 sin i^-^ + A- b\ cos C^- A - b\ . (2) tirst 8ide = 2 sin -^— cos -^.2cos -— cos = 2sin(^-0)cos*-'-^"^. 266 MISCELLANEOUS EXAMPLES. K. 129. Since u : b : c = amA : sinL' : sin C, we have {(l + b + c) {a+h-c) = 'dah, or (ii + ^>)'' - ^' = 3^^» that is !^^"'~^' = 1; .-. cosC = ^, and C = CO°. inai IS, .^^^^ .^ . 2 ' 130. Lpt «=^ = ^ ""^'^ c^^^. c« = Z;, then we have to prove px — qy. Now a^ = c'i, and «'' = r''; .-. aJ>^ = c^'^ = a^^ \ that is jja: = gy 131^ log 20-01 = 1-3012471 log 20-00= 1-3010300 di£f. for-01= 2171 .-. diff. for -0075 = '5 x 2171 = 1028 ; 4 .-. log 20-0075 = 1-3011928. 132. Let JD be the median from B ; then AB^-VBC- = 2{Aiy^ + BD"); that is, 4y + 81 = 2.r- + 32; whence x = l. 1 + sin A 133. We have 7- = ^' ^'^^' cos A :. (l + sin^)2=4(l-sin2yl), rejecting tlie factor l + sin ^, wliicli gives an inadmissible valne, we have 1 + sin ^1 = 4 (1 - sin A); . , 3 wlieuce sin A—^. 4(l-cos2^)-(l-cos4.J) 134. Fast side = -^—r tt-tt — jz :~7\ ^'^^' 4 (l + cos2.1)- (l-cos4i4) 8sin*-yl-2sin2 2.1 8 cos- A - 2 sin- 2 A sin" A (1 -cos- J) ^ , , , ,\ . „-., = tan^^. cos'2.4 (1 -sin- .4) MISCELLANEOUS EXAMPLES. K. 267 3 sin /( - 4 sin^ ^ + 4 cos'' A-S cos ./ 135. First side = ., . , — .— . .. , — , , < t o .; •*-^'^' 3siuvl -4 sin^^ -(4cos'*^ -3cos^) 3 (sin .4 -cos^) -4(sin^.^ -eos^A) ~ 3(sin.-l + COS.1) -4(sin-'.-l +cos^^-i) sin A - cos A 3-4 (sin- A + cos- A + sin A com A ) ~ sin A + cos ^ ' 3 - 4 (sm- A + cos'-^ A - sin A cos A ) tan yl - 1 3 - 4 (1 + sin^ cos^) ~ tan -4 + 1 ' 3-4(1- sin A cos A ) _tan^-l 1 + 2 sin 2A _ 1 + 2 sin 2A -^l + tan^ • l-2sin2^~l-2Biu2^ ^''^ ''^• 136. Put — ^ = -^. = k ; then cos^ cosl> First side = k cos A tan ^ + A; cos B tan S = A; (sin A + sin B) = 2a; sm — — — cos- ^+5 A-B^ A+B = 2^ cos —^ — cos — - — tan ^- — a a Ji — k (cos A + cos B) tan - — = (a: + 2/)tan— 2 - • 137. log 7 = log 24-5 -log 3-5 = -845098; log 5 = log 35 -log 7 = -69897; logl3 = log3-2o-log-25 = log 3-25 - (2 log 5-2) = 1-113943. ,««. TT . « 384 128 138. Heretan^=-=-^ = — . log 128 = 7 log 2 = 2-1072100 log 110 = 2-0413927 log tan ^ = -0058173 log tan 49° 19' = -0G5G886 dii!. 1287 1287 .'. prop' increase = _ _ x GO" = 30" ; .: ^=49° 19' 30"; ii = 40^^ 40' 30". 2G8 MISCELLANEOUS EXAMPLES. K. . d+a . ^-a '2 sin ■ sin — ^— 2 2 C08 a - cos 6 139. First side = ^.^-- ^^ ^- j^„- — -^ 2cos -^— cos — -— 2 2 cos o - COS a cos /3 _ 1 - cos /3 ~ cos a + cos a cos ^ l + cos/3 = tan2 ^ . 140. We Lave sin 6 sin ^ _ 2 sin $ sin cos - sin cos $ + sin "" cos^ 6 - sin- ^ sin { cos + sin ^ - (cos - sin g)} ~ 1-sin-^- (l-cos-» sin [cos + sin ^ - (cos0- sin d)] ~ (cos0 + sin ^y) (COS0 - sin ^) sin sin ~cos0- sin 6 cos0 + sin^* 141. ^^ e have -^ -^ , c (a^ 4- h-i + 2ah) {s - h) = h (a^ + c' + 2oc) {s - c) ; that U, n-^s (c - 6) - i^C! (c - Z^) + ?^c (r^ - /)2) + 2«/>c (f - ?*^ =- 0, or (f - h) {ah- he (s - <• - /v) + 2aZ>c } = 0, or {c - h) [ah -hc{a-s)-ir 2ahc] =0, or (c - ^) { ^' ''••^" + '"■•'* + ■^" '^i- } --= ; therefore Z>- c = 0, since tlie other factor evidently cannot be zero. 1 + COS^ .A r-v.-T J T-" m 142. (1) cot .4 + cosec ^ = . =--cot-; [XI. + sin C ^ . A A 2 sin — cos — \ . Ji+C B-G 2 sin — — cos 2 2 . A B + C cos— ^ cos-^2' B-G B+G cos — cos — , .,6^ 1-cos^ 2 2 tan^ - = 2 1 + cos B-C B + a cos - - + cos — g— ^ . B . C 2sm-sin- ^ ^, = Ti ^=^^=^^^2*"° 2- 2 cos- cos — 148. l^<^'t T l>e the radius of the circle, x, y the side of circumscribing equilateral triangle and hexagon respectively. Then from the figure of Art. 215, 2r .T = 2;-tan60° = 2r<^3; (/ = 2rtan 30-— -; whence xy = \r-={2ry-. 149. From the equation a'' = i'^ + c'-- 2ic cos ^, wo have on substitution and reduction 0-2-150^/2.^ + 10000 = 0; .-. (c - 100^2) ((•- 50 ^/2) = 0; .-. r = 100^2, or 50^2. MISCELLANEOUS EXAMPLES. K. 271 . ^ h sin A 150 3 Again sin B = —^ = -^^^^^ = ^^^^ , log sin li = log 3 - - log 10 = 1-9771218, log sin 71° 33' = 1-9770832 diff. 381 aiff. for GO" = 421; 381 .'. prop' increase = --— x GO" = 54" ; .-. B = 71° 33' 54", or 180" - (71° 33' 54"). 150. The series may be written (eosecx -cosec3a-) + (cosec 3x - cosec3-a:)+ ... + (cosec3"~-a:-eosec3"x), which reduces to cosec x - cosec S^x. 153. First side e „ S9 cos - 3 cos — "^ . 3^ sin - sm - .3^ d ^ . e d0 sin — cos ^ - 3 sin ^ cos — . 6 . -60 sm ,^ sin - sin|--2)-2sm,^cos- . d . -id sm - sin — - 2 2 sin ^ - (sin 2^ - sin ^) (cos d - cos 2^) 4 sin ^ - 2 sin 26 _ _4 siji^ (1 - cos 6) 1 + cos 6? - 2 cos- d ~ (1+2^^ 6) (1 - cos~0) 4 sin 6 1 + 2 cos^' 272 MISCELLANEOUS EXAMPLES. K. 154. We have, by Art. 135, Ex. 2, tan A + tan li + tan C = tan A tan B tan C ; .-. ^ + -^ + tan<:; = ^. ^.tauC; wlience tan (7= Also cos A 38 1 1 4 Jl + t&n-A /; " y 5' 1 1 1-J cos J5 — cos C = Jl + tan^B' I 25 ^'^ V^+144 1 1 33 s]\ + tan=* C "~ / 3l;iG'~ ^-J' V "^ 1089 the negative sign of the radical being taken in the third case since C is an obtuse angle. Again tan C = - ^ ; .-. tan ( 180= - C) = % , log 56 = 1-7481880 log 33 = 1^5185130 log tan (180;-C) =--2290741 . . , i„erease= '^J^j x GO" logtano9°29= -229^027 2888 diff. 1114 =23"; .-. 180°-C = 59°29'23"; that is, (7= 120° 30' 37". 155. We have sjnj/l -Ji) _ a2 - ^2 sin(^+ii)~<1^+62 _ sin2 A - sin^ />' sin- ^ + sin-' i/ ' • iiHilL:^- ''"^ ^-^ ~3 sin (^Hf^) ^ sin C ~ sin- .-1 + sin-^If ' .-. either sin(/l-7?)=0 (1), or 8in'(7 = sin"^ J +sin2B (2). If (1) is true,^ =B\ if (2) is true, we have c- = a- + t'. MISCELLANEOUS EXAMPLES. K. 273 156. The first pide 2A s (J be A s (s -a) + ...+ ) be j = 2{a(s-a) + b (.s- -b)+c{s- c) ] = {a + b + c)-- 2fr- - 262 _ 2c^ = 2bc + 2ca + 2ab - a^ - b- - c-. 157. We have cobC cos B cos {A + C) _ cos (A + B) sin C cos A sin B cos A sin (' cos A sin C cos A sin i? cos A = cot C - tan A - (cot i> - tan A) = cot C-cot jB. 158. log 3 = log 18 -log 6; log 2 = log 6 -log 3; log 11 = log 44 - log 4 = log 44-2 log 2. 159. (1) We have tan{60° + .4)tau{60°-.4) ^ 2 sin (60° + .4) sin (60° -J) _ cos 2.4 - cos 120^ _ 2 cos 2.-< + 1 ^ ~ 2 cos (60^^ + A) cos (60° - ^) ~ cos 2 A + cos 120= ~ 2^os22^ ' .•. second side = 2 sin A cos 2 A + sin J = (sin ?>A - sin A) + sin .-1 = sin 3^. (2) First side = 2 sin [A - B) cos (A +B) - ^ ^'^ "^"^^ ^ ' V / V -T- ''eos(J+i^) = 2sin(.4+£)sin(.-I-i>') = 2 sin2^ - sin2£. [Art. 114.] 160. tan^-^— ^^^:,^=:^-y- log2%=2•4712917 log 82:^ 1-0138139 4-3851056 xlQl x82 log 113 = 2-0530784 log 101 = 2-0043214 4-0573998 4-3851056 log tan - = 2 !l-6722942 1-8361471 log tan 34° 26' = 1-8360513 diff. 958 958 prop'. increase = ,^ x 60"= 21"; .-. (7=68° 52' 42". 274 MISCELLANEOUS EXAMPLES. K. 161, T^ct a be a side of the octagon, r the radius of the circle, then 8a — 'Iirr ; and we have area of circle ttt- Stt/- it area of octagon o 2 . f ''' """ JiQ. cot ~ o _ 8(^/2-1) _ •414 X 8 _ 4140 _ 1380 "" T ~ 3- 1416 ~ 3927 ~ 1309 " 162. We have 26 = a + c, or a=2h-c\ Ij2^(,2_„2 62 + c2-(26-c)2 4bc - Bb^ 4c-36 cos^ 26c 26c 26c 2c 163. ^Ve have a (sin cos a -f cos ^ sin a) = 6 (sin ^ cos /3 + cos 6 sin /3) ; .-. sin e {a cos a - 6 cos /3) = cos ^ (6 sin /3 - a sin a) ; a cos a - 6 cos /S that is, cot^= - . — ^ . — . ' 6 sin /3 - « siu a 164 Let a, 6, .4 be the given parts; then li = :r-. — 7 , which is the same * ^ bin ^'i. for each triangle. 165. Let NS be a horizontal line pointing North and South. Then if A' be the position of the kite, and KD is the vertical from A', we have S, B, c sin /3 (' sin /3 7), .1, .Y in a straight line, and BA^c. Also K'^=-^BYA=^in{a + ^)' csina sin /3 And AD = KA sni a = - . - — . sin(a + j3) 166. co8a + co8/8 + co37 + l = 2cos— ^cos^-2^+2cos2- ('^-2)^°^"2 = 2 cos ( TT - .' ] cos "" '^ + 2 cos- ^ = - 2 cos ^ cos , + 2 cos-^ 2 2 2 = 2cos||cos--cos--j- = 2 cos I I cos (,r - ""Y) - ^^^ "4- } 7 a + S = - 2 cos ' - cos — :-^ + cos 2 I 2 J « /^ 7 = — 4 cos - cos - cos ^ . 2 2 2 V} MISCELLANEOUS EXAMPLES. K. 275 167. First side = 2 cos 9^ coaA + 6 cos 3A cos A = 2 cos A { cos dA + 3 cos SA } = 2 cOo ^ { (4 cos3 -dA - 3 cos 3^) + 3 cos 3 A ] = 8cos.l cos33.-l. 168. r, + r.= -4_ + A = ^J^'L+lT ^J = r /~Z^)~ - s-a s-b {s-a){s-h) ' \/ {s ~ a) {.<, -h)' ^^^"^ r-2r3 + fV^ + r,r^ = s~ [XVIIL a. Ex. 24] ; .-. First side = ''^' /ZSEfl ~^(^-») M^:iiO_ 4;>2 Y (s - a) (s - b) ' (s-b) {s - * (.s - c) (.5 - a) obc abc 4 Js {s - a) {s -b){s- c) -^^ 169. cos2^^izi^^^ ^tan^^ l + tan^^ 2(1 + tan- 0) -.R. = - sm^ 0. 170. First side = -'^J^0° + ^) ^^^ (120° + J) eos^co8(60° + ^)cos(120° + 2f) _ sin^ cos 60° -co s (180° + 2^) cos A ' cos 60° + cos]i80° + 2A) sin A 1 + 2 cos 2A cos ^ • 1 - 2 cog \2A sin A + 2 sin A cos 2A cos A -2 cos A cos 2A sin A + (sin 3.-1 -sin J) cos A - (cos SA + cos ^) sin SA cos 3-4 = - tan SA. 171. We have y/ -^ = 7?; therefore sin (^ -i?) = sin i?; also sin(^+7i) = sin(l80°-C) = sin(7; .-. sin {A + B) sin (J - li) = sin O sin B ; tl^^'tiy, sin2^ -sin-iy = sini?sinC; 276 MISCELLANEOUS EXAMPLES. K. 172. If ". b be the sides of the triangle and square respectively, and R the radius of the circle, it is easy to shew that a = E^/3; b ^2R cos -io^K ^2. ,,, , c + h^ A sinC + sinjB, A 173. A\e have — , tan --=.-- . tan - ^'^' c-b 2 smC-smB 2 sin (.4+i>) + siuir A = -, — ^— ' — -. tan — sin (.•!+/>)- sin i) 2 2 sin r^ + i>M cos '- ^ 2cos(\^ + B] sin - / 4 tan tan(^2 + 7?j=-^— ^^tan-- = ^tan-, log 10 = 1 log 4= •6020() •3979400 log tan 3° 18' 42" ^ 2 7624000 logtan(^ + 7>-) = 1-1603469 ^^^ log tan 8° 13' 50" ^ 1-1603083 ^^""1^ ^"^^'^"^^ = 1486 ^ ^^ 380 =2-6". .-. ^ + ii=-8<^13'53", and 4 = 3° 18' 42''"; .-. 1^ = 4° 55' 11", C = 168° 27' 25". 174. l^y 3- well-known geometrical property, we have A C- + A ir- = 2AD- + 2DB-. .-. ACr--AB-^2 {A D- + Z)52 _ a B-) = 4AD.DB cos A DB = 4 AD . DB sin .47)7J cot ADB = {AD . 7)7? sin ADB) 4 cot .17)75 = 4Acot.4D7;, for A D . DB sin A DB - 2 (area ul" triangle . < 7)7^) = A. MISCELLANEOUS EXAMPLES. K. 277 175. tau 40° tan 80^ = ^ ^i"10:^80_^ 2 COS 40° cos 80° _ cos 40° -cos 120° ~ cos 40° + cos 120'' _ 2 cos 40° + 1 ^ ~ 2 cos40° - 1 ' .-. tail 20° tan 40° t«.n ftQQ - ^ " ^^" ^^^ ^Q° + ^^ 20° 2 cos 20"" cos 40'' - cos 20° sin 60° = v-no = tan60°. cos 00° 176. See Art. 144. 177. 2i2=^^ 2.=?:^ ^•a s 178. ^ sin 2? = - sin ^=1^^ a 7 2 log 3= -2385600 2 log 2= -0020000 ^8406206 log 7= ^450980 log sin ^ = 1-9955226 log sin 81° 47' = 1-995518 8 prop^. increase = ^% 60" 183 = 12-4' 38 V -^7^1'' 47' 12"; but since a^h there is another value of B Bupnle- mentary to this, viz. 98° 12' 48". auppie .-. C = 68°12'48", or 51° 47' 12". To find c, we have c"- -2b cos A . c + 1'^- a8=0, [Ai't. 150] that is c=- 24c + 143 = 0, (c-13)(c-ll) = 0; ••• c = 13, or 11. H. E. T. K. 20 278 MISCELLANEOUS EXAMPLES. K. 1 + siu 2a sin 2 a' + s in 2a + sin 2a' ~ iTsin 2a sin2a' - sin 2a - sin 2a' _ (1 + sin 2a) (1 + sin 2a') ^ (1 - sin 2o) U - sin 2a') = tan2(^ + a'\tan2(^^ + a'V 180. ^Vith the figure of Art. 199, let PC = x, /3--28°, a = 16°, a =16071 feet. 16071 sin 28° sin 16" Then ^ = - sin 12" log 16071 = 4-2060 log sin 28° = 1-67 16 logsinl6° = i;4lU3 3-3179 log sin 12° = 1-3179 log x = 4-0000 .-. a: = 10000 feet. 182. '^'e have 2 cos - J cos — — r= 2 sin (.4 + C') ^ . A+C A+G = 4 sin cos — - — ; .-. either cos 2^=^ (^)' A-C ., . A + C H or cos— ^ = 2sm .^ [-)■ From (1) -^=(2n + l)^, and from (2) by expanding each side and 2 2 dividing throughout by cos - cos - we obtain the other result. A + Ji A - n ,, . C ^ C 183. l^'i'st flide =■■ 2 cos ^^ - cos — ^ "^ ^'" 2 2 C . C ( A-B C = 2 sin- -^fos cos- MISCELLANEOUS EXAMPLES. K. 279 = 2sm— jcos — cos ^^ '\ = 4 sin ^ sin ^45° - sin ^45° - |) . 184 ^g ^ hc{s- a) ^ 4i; (g -g)^ ^^ (^±^^^''\ ■ '"i A a " \ a 7 ' .-. first side = 27? (^ + ^-^ + c-^-a-h _^ a + h^-c \ c.-r,/b c c a a h \ = 2i? - + - + - + - + - + -_3). \a a h b c c J 185. Since the points A, B, E, C are concyclic, z BED= L C; also IEBB= IEAC=~; .-. from aBD£, ^^ MsinO . ^ ' sm — 2 from aD£(7, DC = 2£±J1. sm^ ••- by addition a = 1^ (sin Z? + sin C) ; sin- ■*' "^~ , (" = 11 . 7 . 3 . 2 . = 2310 sq. ft. 7-1 = r^= A _ 2310 .•>• - a 33 A 2310 - h 42 = 70 ft. = 55 ft. A 2310 ro= = — .._ =00 ft. s - c 35 tLut is, 192. I)i-aw J^A', JjK' perpendicular to AB, AC respectively; tlien UK.AB + DK' . JC = 2A; A A c .AD sin + 6 . AD sin - = he sin ^ ; .-. /17)(?> + r) = 2/)cco8-. MISCELLANEOUS EXAMPLES. K. 281 193. (1) sin o(?- sill 3(?:=^/2 sin t?. .". 2 sin^ cos4.) = 2; , ^ cot ^ - 1 ^ .-. cot^+ =2; cot<9 + l cot2 ^ + 2 cot (? - 1 = 2 cot ^ + 2 ; cot 2^ = 3; that is, eot^=±y/3, and ^ = »7r±^. 194. The given relation easily reduces to cos2a=sin2/?, one solution of which is 2a = - - 2/S. 195. We have tan (a + d) tan (a - ^) := tan^ ^ ; tan- a - tan^ •'' ^ — ^ .. X .i„ = tan-/3: 1 - tan- a tan- d ^ ' whence tan^ (9(1- tan^ a tan^ /S) = tan^ a - tan^ § ; . , 2 /I _ (tan g + tan ^ ) (tan a - tan /3) (1 - tan a tan /3) (1 + tan a tan /S) = tan (a + /3) tan (a - /3). 2A 196. (1) We have i)^ = a .-. second side =-8A^=^ ( 4A ) =^^"'- (2) Second side ='^^+i^-lM^^-^' 4A- ^^_ /c_Y_l_ 4a-'-v2a; -p,^- 282 MISCELLANEOUS EXAMPLES. K. 197. I5y Art. 215 we have perimeter = 30r tau —= , where irr'^ = 1386. lo Now r2 = ^ X L386 = 7 X 03 ; .-. r = 21 ; .-. perimeter = 30 x 21 tan 12° = 630 X -213 = 134-19 ft. 198. Let A, B represent the foot of the pole in the two positions; C, H the top' of the pole on the coping and sill respectively; also let W be the foot of the wall. Then x + SW=ACsma, but SW=BS &m ^ = AG sin ^; :. a; = ^C (sina-sin /3). Similarly a = JC(cos j3-cosa); 2 cos '^ sm --- . ^ _ ^ = ot - — •*• a~ ^ . a + ^ . a-/3~^ 2 2 sm -2- sm -g- 199. See Examples XVIII. a. 18. Each of the three expressions will be found to be equal to r. o 41 200. (1) Let 8in-i - = d; then cos 2^-1-2 sin^ d = -^; 41 2 .-. cos-i , -25^2Bin-i_. 4y I 1\3 (2) 3 tan-i j = tan'i ^]^^/ ^ ,3x10-1 -47 = tan-i-g^,2-tani^ 201. (1) As in XL/. Ex. 11 we may prove that tan ^ + sec ^ = tan I 45"^ + -;- j . ^ 1 + COB A ^ A Also cot A + cosec A = — -. — .— = cot — ; sm A iS .-. (tan A + sec A ) cot -^ =- (cot A + cosec A ) tan (45° + ^ j • MISCELLANEOUS EXAMPLES. K. 283 (2) First side = 2 cos (A+B) cos (.1 - li) - 2 cos (A+B) {cos {A - B) - cos {00" ~ A+B)} = 2 cos (A + B) sin (.4 + 7?) = siu 2 {A -\- B). 202. By Art. 214, ^'^^^''^^^^-^lA^-- -7sin 25«- •" ' diameter of circle" ' " '''''■ ' log 7 siu 2r,a° = -8450980 + I-()8737:^J = •4824713, which ia greater than log 3. 203. We have c'^ = a^- + h'^ -2ah cob C = {a'^+6=)(cos2|+sin2^U2a6 G r = (a + 6)2 8in2 - + (a - 6)2 cos^ - z 2 = (a + 6)2sin2|(l + tanV) = (a + 6)2 sin2 - • sec^ ; (J :. c = (a + 6) sin - sec 0. 204. We have ^^^ ^ = 21^^1:58 ^^^^^^^CX 9 cos2 — - sin2 -^) = jjjCot33°20'; .-. log tan = log 2 - 1 + log cot 33° 20' = 1-30103 + -18197 = 1-48300 log tan 16° 54' = 1-48262 diff. 38 /. = 16<^5i'5O", log sec 16° 55' = -01921 log sec 16° 54' = -01917 diff. for 60" 4 ••. log sec 0= 01020. ii 1* 38 prop' increase = — ; x 60 46 = 50"; prop' increase for 50" = x 50" 60 = 3"; 284 Now MISCELLANEOUS EXAMPLES. K. c c c = {a + h)B'm- sec = 395 sin — sec 0, log395 = ]og79 + l-log2 = 2-59060 log sin ^ = 1-73998 log sec 0= ^01920 log c ='2-35578 /. c = 226-87. 205. 2 cos^ 2^=1 + cos 4^; .-. 2cos2^= V'-^4-2cob-4^. Similarly 2 cos ^ = ^2-|-2co72^ = V2 + \/2 + 2^os 4^'. 206. Lot and let Then Again or sm~^ = a, so that cosa= — -^zz, x/73 ^73 11 5 cos~^ —7^= =j8, f^o that Bin 3= , ~— JUfS s/146 sin (a + ^) = sin a cos /3 + cos a sin /3 3 11 8 5 73 "^ JT6 ' JUQ Jn ' ^146 73^2 ^2 = sm- = sm(^--gj; OTT TT Svr . , 1 « + /5=12-6 = 12-^^^ 2- ■r-l 2x-l ^ , .7TT + 2^r + l ^ _,23 tan-^ , ., ■, =tan ^j^; . ;i- - 1 2.r - 1 3b x + l 2.r + l 4.r2-2 23 tan-i jLl __ = tan-i —, , 6x ob .-. 36(2.r--l) = 69.r, 24.r2-23.r- 12 = 0; .. (:{.r-4)(8x + 3) = 0, that is, •'• = 3' ^'^-S- MISCELLANEOUS EXAMPLES K 285 2A (ibc 207. We have .r = — , J?=.-; a 4A 1 obc bx b"^ ,, , . 6a; CM az a^ + b' + c^ that IS, _ + _^ + = _ . c a 6 2iv 208. We have cos (a + ^)-cos ij - (a - ^)l , .-. a + ^ = 2/»7rdb ||-(a-<9)l ; Ti- the upper sign gives 2a = Imir + ^j , IT and the lower sign gives 29 = '2,mir - - . 209. With the notation of Art. 228, Sr- = B?-2Rr. If a be the base of the triangle, .-1 = 120^, J5 = C = 30°; .. r = 4i? sin 60° sin 15° sin 15°; =E2(4-2V3); .-. /Sfl=i2(^3-1) .-. ,S:i:a = v/3-l : ^3. 210. cos^ + cos£ + cosC=l + 4sin-^ sin — sin • i ^ a = l + -=l+-r, .". 4 (cos J + cos B + cos G) = 7. ^, , f,- cos(^-a) l + ?u 211. Since . - = - , sm(^ + a) \-vi dividendo and componendo, we have cos (6 -a)- sin (6 + a) . _' ^ ' — JJ^ cos (^ - a) + sin (^ + a) 286 MISCELLANEOUS EXAMPLES. K. By expanding the sines and cosines we obtain (cos d - sin 6) (cos a - sin a) (cos d + sin 6) (cos a + sm a) 1-tan^ /C0ta + 1\ ■ ,, VT l -P r. rr n or . — , -^ = "* — :: 7 • [^ee XI. h. Ex. 6, 7.] 1 + tan^ \cota-l/ "■ ' •" 212. (1) sin 0^- sin 3^ ^^2 cos 4^; 2 cos -1(9 sin ^ = ^/2 cos 4^ ; .-. cos 4^ = 0; whence 40 = (2;i+l)^, .1 ir or Bind = —r:.\ whence ^ = 7Z7r+ ( - l)"j. (2) l + sin2^=^ + *'^"^ 1-t 1 - tan ^ ' 2 tan ^ 1 + tan 1 + tau- 6 1 - tan d (1 + tan ^)- (1 - tan ^) = (1 + tan ^) ( 1 + tan^ d) ; .*. 1 + tau ^ = ; whence d = mr->r'—r- , 4 or 1 - taii2 ^ = 1 + tan2 (f; :. tan^ = 0; whence ^=:»7r. 213. We have 2 cos " — cos — - — = 4 sin^ -^ ; A-B ^ . C '^ A-B . A + B , . C G :. cos — ^r — = 2siu-, or 2 cos— sin — - — =4sin — cos— ; that is, sin J +sinl? = 2 sin C, or a-\-h — 1c. 214. With the figure on p. 18G, we have tanj3 = x, P.4^80ft., CA = 100 ft. Let BP = x ft., then tan^ = -^--,tan(^-^)^^^^ = -. tan 6 - tan /3 _ 4 " 1 + tan ^ tan /3 ~ 5 ' x + SO 1 ~100" ~9 4 J- -f bO ~ 5 ' "^"gooT .-. 5 (Ox + 720 - 100) = 4 (080 + x) ; 4r>x + 3100 = 3920 + 4a:; 41x = 820; or u; = 20. MISCELLANEOUS EXAMPLES. K. 287 215. (1) cot-i7 + cot-i8 = cot-i^-^ = cot-i^, 7 + 8 lo cot-i 3 - cot-i 18 = cot-i —l-^ = cot-i f f . 18 — o 15 2 (2) 4tan-il = 2tan-i— i-- = 2tan-i4 o ^1 12 ~52 2x5 , , 12 ^ .120 = tan-' ,„ = tau-i — - . 2 i-i'- 122 120_ 1 . A+o«-i^ + -1 1 . 1 ll'.>~289 ^ ,120.239-119 .'. 4 tan 1 - - tan ^ -— = tau-^ , , = tan-i — 5 239 120 1^ 119.239 + 120 ■^119 '239 ^ -119. 2.39 + (239 -119) , , , tt = *"^^ — 119:239Tl20— = *"" ^ = 1 • 216. Proceeding as in Art. 259, Ex. 2, we find that — lies between 2/i7r + ^ and 2n7r + — -; that is .-I lies between (8?i + 3)- and (Sri + 5)^. 217. First side .= ^ jl -cos f |^ + ^V 1+ cos ('^- ^V- • ^ . ^ 1 . ^ = sin -r%m.Q = -j~ sin ^. 218. From the two given relations we easily deduce _ sin d _ sin ^~sin(^ + 0V ^""sin{^ + 0)' .'. sin 6 : sin V MISCELLANEOUS EXAMPLES. K. 289 224. The relation given will be true if I.e. it a->rh-\-c a + c h-\-c a + 6 + c h (i-b — c b b + c-a a + c b + c * (i + c 6 + c' i.e. if b{b + c) = {h + c) {a + c) - a- - ac. From this we easily deduce ; =1, which is triae when C = GO°. ab 225. The solution of this example is merely an extension of that of Ex. 205. 226. ^^f* hfixp. m pin (a - 6) cos {a-6)=n sin 6 cos 6, m sin 2 {a -6) —n sin 26 ; sin 2 {a -6)- sin 29 _n-m sin 2 (a - ^) + sin 26 ~~ n + m ' sin (a - 26) cos a _n- m COS (a - 2^) sin a ~~ ?i + m ' tan (a - 26)= tan a , ^r, , 1 fn-in^ \ a - 2^ = tan~^ { tan a ) , \n-\-m } 6= - \a- tan~^ | tan a][ . 2 [ \n + m J\ 227. P"t X; for each of the equal ratios, then it easily follows that s = k (i + n-j. A /is - b) (s - c) /(l-m'^)m2(l + 7i2) vi Now tan -r = » / y- c — — K I iw~, — 2r^ •• /I — ■>\ — ~> m ^=2tan-^ — ; similarly J5 = 2 tan ^inn. n Again, A= Js{s - a) (x - b) {s - c) = k- sJ{l + n-)-{l-ni-fm.'n^ = A-2 (1 - ///'-) (1 + n-) mn = kcmn vinbc . h . = —z r, , since — :,— K. m^ + n^ VI- + /r 200 MISCELLANEOUS EXA:\IPLES. K. 228. ^cc figure and solution of Example IL page 190. TT 7 x-,T asinS Here h = CD = -~~ — , cos (2a +^) I _ £>;,' ^ asinaco3(a + /3 ) ~ ^~ cos ('2a + jJ) But 2a + jS + d = 90°; /. cos ('2a + /3) = sin 6* ; /. by substitution, /t = a sin/3 cosec ^, 2Z = 2a cosec 6 sin a cos (a + /S) = a cosec ^ {sin(2a + /3) -Rin/3} = a cosec 6 (cos 6 - sin j3). 229. We have -tan -r + cot 6 2 2 _1 l-cos^ cos ^_ 1 -cos^ + 2cos ^_l + cos^ ~ 2 * sin ^ sin ^ "" 2 sin 6 ~ 2 sin 6 2co82- cos , 2 2 4 sin - cos g 2 sin ^ cos'^-T-sin'^-r , ^ , 4 ^^ \ J \^ d = ,. = 7 cot T - T tan - . , . i) 6 4: -14 4 4 sin -r cos - 230. With the figure of Art. 214 we liave A D=—. '2})i Let dr^lA OD, then cos ^ OjB = 1 - 2 sin^ _ 2AD^_ , _ 2_ _ 2»i2- 1 ^ " J O"-^ ~ ~ 2»7- ~ ~~2 //7^ ' 27«2 .-. ^OIi = sec-i ,—;,—= . 2//(- - 1 231. With the figure of Art 208, Ex. 1, we have PN = tan 1" — radian measure of 1", ap])rox. __-7r^ 1 1 ^ ~ 180 ^ 00 ^ 00 ' MISCELLANEOUS EXAMPLES. K, 291 ^-, 180 X 00x00. ^ .*, 0N = inches 180 X GO X no ., = T^,rn — r. — ; .- miles 1700x3x I-Jtt = yf.- niiles = 3^ miles, nearly. 4a: 232. tan-iy = 2tan-i-^^— = tan-i — ^ ~ f ,.- = tan-i-^^L:i^^). l-Ga;2 + a;4' _ ix(l-x\ IT If T/=taii^, l-Ga:2 + a;'*=0, 1 ^ but .T = tan---{tan~i2/)=tan -, TT thus tan 5- is a root of x^ — B-r^ +1 = 0. o 233. We have tan^a = ^ "^°^ ;° =ig-; 1 + cos 2a 529 ' .-. tana=±— . The two values may be explained as in Art. 201, Ex. 2. 234. We have ging ^ s_in0 ^ sin (^ + 0) ah c ' sin ^ + sin _ sin (^ + 0) a + i ~~ c But a + 6 = 2c, .'. sin ^ + sin = 2 sin {d + 0), whence cos— —- = 2 cos — ^ M), i? + 0-d> d + d> or cos — — ^ = cos — 7-^ - cos — 2 2 2 = 2sin-3in^ (2). 292 MISCELLANEOUS EXAMPLES. K. Now COS d + cos = 2 cos —— cos — ;; = 2 (2 sin ^ sin ^\ 2 cos ^\- ,^7 (1) and (2) = 4 sin - sm 2 . -t sni - sni - , by (2) = 16 sin2 ^ sin2 1 = 4 (1 - cos ^) (1 - cos 0). 235. (1) sin7^ + sin^ = sin4^; .-. 2 sin 4^ cos 3^ = sin 40 ; .-. either sin 40 = 0, or co83(9 = -; that is, 4d = mr, or 30 = 2«7r±J. o (2) tana:- -^ +1-^3 = 0; ^ ' tan a; tan^a: - {>,/3 - 1) tan a: - ^/3 = ; (tan a: - ^3) (tan a; + 1) = ; .*. either tan j: = y/3, or tana;= -1; TT 37r that is, x=mr + -, or a: = 7t7r+— r-. 236. (1) sin3^ = sin3(180°-i? + C^) = sin (3G0° + 180° - 35+ (7) = sin 3B+C. AVe have only now to prove that 2 sin 3 {B + C) sin (B - (7) = 0, and tliis follows by separating each term into the dilYerence of two cosines. (2) It will be sufficient to prove that sin3^sin(Z^-C) = 0. S 3 1 . Now sin^/1 =- sin ^ — - sin3yl ; 4 4 .-. 2 sin».4 sin (2? - C) = - S sin ^ sin B-C - 7 2 sin HA sin B - G 4 4 = 0, by the first part of the question. MISCELLANEOUS EXAMPLES. K. 293 237. Theangle JCP=^-.^. AP sin (^-.4) . , , , /. -zr- = . -: = Sin 9 cot A - COS d, PC sin A PC sin B 1 PB sm{d + B) sin^cotii + cos^' .*. by multiplication AP _ni _ sin 6 cot A - cos 6 PB n sin 6 cot B + cos d ' whence sin 6 {n cotA-m cot B) = {m + n) cos ^, or (?« + ;?) cot ^ = 71 cot ^ -m cot 5. 238. The equation may be written a sin ^ - cos ^ + 6 = (I). Since a and /3 are roots of this equation a sin a -cos a + b=0, a sinj8-cos/3 + Z>=0, whence a and 6 may be found. Again from (1), (asin^ + 6)2=l -siu^A or {l + a2)sin2^ + 2a6sin^ + Z^2_i_0; since a, ^ are roots of this equation, sm a + sin /3= - 1 + a-' 26 Similarly we may shew that cos a-f-cos)3=- 1^, whence the required result follows. 239. Write s and c for sin ^ and cos d respectively; then = sV W3-M5 _ s3 + c3-(s5 + c5) _ >i^(l-s'^)+c^(l-c-) _ g-V + cV Uj s + c s + c s + c , . W5-W7 s^ + c^-is' + c^) s^n-s"-)+c'(l-c-) 1*3 6-3 +C-^ Ji-^ + C"^ = n —-^s-c^. s^ + C-* H. E. T. K. 21 294 MISCELLANEOUS EXAMPLES. K. 240. Let E, F be the first and second points of observation respectively; then KF=a, and BAD is a straight hne. Let :r = a side of the square base, then EA = AB = AD = x. Then if Z J i<'D = ^, we have tuting these values in (1) we obtain x = 241. (1) First side = AD^ = Ar^ + FD^--2AF.FDcosd (1). aJ2 But JD2 = a--, AF^=x:^ + a\ FD^=Ax'^ + a\ Also cos^ = i^8. Substi- o 2 1 - sin- A 1 - cos^ A cos'-^ A sin^ A s,\nA coaA sin ^ cos ^ = sin A cos A. . . , .^ /sin2yl + cos-^\ 1 . . Again, second side = I . — , -— \ = sm A cos A . \ sin A cos A J (2) Firstside= - .^ + sec-* 6 cosec'* = sin d cos^ ^ +C08 6 sin^ 6 = sin ^ cos d (sin2 ^ + cos^ $) = ^sin2^. 1 5^ ^6 242. We have 2 sin 4^ cos ^ = -4- 2 sin —cos — ; • c/, • o/. 1 ^ . 56 56 .'. Bin 5^ + sm 66 = ~ + 2 sin — cos — ; /. sin 3^= -, one solution of which is 3^ = 30°. 2- 243. tan-i 4^-tan-^^l^=2 tan-^ i^; m--n- n- m ,\ first side =2 tan"^ — + 2 tan"^ - m p -'( tan 1— + tan-^ - in p n a - + - = 2tan-^il-il; 1 _ '^l inp MISCELLANEOUS EXAMPLES. K. 295 .*. first side = 2 tan~i -^ = 2 tan~i r-f inp — nq M , , 2MN = taii~^ il/2-2^2- r ^ 244. Write for tan — , then the first side becomes s- a 2 ■ two similar terms. {s -a) {a- b) {a - c) vr r rs(b- c) { s - b) {s - c) {s-a){a-b){a-c) A^ (a-b) {b -c) {c -a) 1 {b-c){s^-s{b + c) + bc} ~ A {a-b)(b-c){c-a) Now ^{b-c) {s^--s{b + c) + bc}= -(a-b) (b-c) (c-a). [See Hall and Knight's Elem. Algebra, Art. 224.] Thus the first side reduces to — . A 245. Wehave (i-'—^\(i-iZ^) = 2; \ s-aj \ s-aj or (6-fl)(c-r/)=2(.9-a)2; be - ac - ab + a- = 2s^ - 4as + 2a^; .: be = 2s2 - 4as + a{a + b + c) — 2s^ — 4as + 2as ; s{s — a) 1 whence be 2 ' A 1 that is, cos — = — r^ , or A =90°. 2 ^2 246 Let .4 =58° 40' 3-9", 6 = 237, c = 158. Then as in Art. 197 we obtain 2 fJ)c A a=(b + c) sin 6, where cos^=: — ^— cos — , ^ ' ' b + e 2 ' cos ^ = ^-^1^11^ cos 29° 20' 1-95" 39o = -^ cos 29° 20' 1-95", o I 21—2 296 MISCELLANEOUS EXAMPLES. K. log 2= -3010300 log 3= • 4771213 2 I -7781513 •log 6: •3890757 2 log 2 - 1 = 1-0020000 log cos 29° 20'= 1-9401091 Subtract diff. for 1-95" 23 log cos ^= i-9315425 log cos 31° 20'= 1-9315374 ^x60" = 3-97^ Again 247. We have 51 .-. ^=31° 19' 56". log 395= 2-5965971 log sin 31° 19' = 1-7158092 di£f. for 50" 1937 log a = 2-3126000 log 205-4 = 2-3126004 /. a = 205-4. sin(4+£)_3sinj4_ coa(A+B) ~ cos A ' sin {A + B) cos ^ = 3 cos {A + B) sin A , sin (2i4 +i?) + sini? = 3 sin {2A + B) - 3 sin i5; 2 sin (2/1 +2?) = 4 sin 5. I whence. or that is Multiply by cos B ; then by separating the product on the left into the sum of two sines we obtain the required result. 248. First side = 2 sin {6 - a) { sin (^ - a) + sin {^2md -a-d)\ = 2 8in2 (^ - a) + cos (2^ - 2md) - cos {2nid - 2a) = 1 - cos (2^ - 2a) + cos {26 - 2md) - cos {2md - 2a). 249. I^et ADhQ perpendicular to BC and meet the circum-circle in E\ then Z 'bed = C, and a = DE. Now BB ^ ^ . DC ^ ^ — = tan C, and = tan B ; a a BD+DC a ^ „ , ^ -. = - = tan B + tan G. a a Similarly whence the result follows h c -=tan C + tan^, - = tan^ +tan J?; /3 7 MISCELLANEOUS EXAMPLES. K. 297 250. From the first equation, 3 sin-^ = 1 - 2 sin- /V — co.s*2/) ; nnd from the second equation, 6 sin A cos A -2 sin 2Ii = ; multiiDly each term by sin.-l, and we have 3 sin^ A cos A - sin 2B sin A =0. Substituting cos2i> for 3 sin-.i, we obtain cos 2B cos A - sin 2B sin .1=0; that is, cos {A + 2B)=Q; or A + 2B = 90°. 251. (1) Firstside=cot-.(-L-)+cot-(-_^3-^) ^ , cot 2x \ cot 3.r / = cot- . jZ_ + cot 3x cot 2x _ J /cot 3a: cot 2a; + 1\ \ cot 2x - cot Sx J = cot~^ {cota;) = a;. 1-x l-y (2) First side = tan-i "T ^ IJ^' \ ='--'' ^W^^^ ■^{l + a:)(l + y) l + xy y-x = sin-i =Rin-i VH-a:2 ^1 + 1/2* (l + xyf 252. See Art. 197. Let A be the position of the station, B and G the positions of the two points; then ^=49° 45', c = 1250 yds., & = 1575 yds. O /l 9"S(( V 1 'i?'^ Now a = 2825 cos d, where sin ^ = ooog ^^^ ^4° 52' 30", log 1250 = 3-0969100 log cos 24° 52' = 1-9577456 log 1575 =3-1972806 ,^ 1 __ _„„ 2 I 6-2941906 '"^^^"^* 2 ^ ^^^ - H52 3-1470953 1-9577163 log 2 =-3010300 log cos 24° 52' 30" = 1-9577163 3-4058416 log 2825 =3-4510185 log sin d = 1-9548231 = log sin 64° 19'. 298 MISCELLANEOUS EXAMPLES. K. Again log 2825 =3-4510185 log cos 04" Vy = rG368859 log a =;f08790l4 log 1224-3 z= 3 -0878878 166 4 142 240 7 249 .-. a =1224 -347 yards. 254. Multiply all through by 2 ; then First side = 1 + cos 28 +1 + cos 2{S -A) + two similar terms = 4 + 2 cos (2;S - .dt) cos ^ + 2 cos {2S - B - C) cos {B - C) = 4 + 2 cos (B + C) cos ^ + 2 cos A cos [B - C) = 4 + 4 cos A cos B cos C. 255. It is easy to see that this is the same as Example 1 in Art. 135. 256. We have R = - ^ , = 18 cosec 61° 15'. 2 sin .4 log 18 =1-2552725 log cosec 61° 15'= -057 1357 log 72 =1^3124082 log 20-530 =1-3123889 193 9 191 • . R = 20-5309. . A . B . a r= = 4H sin — sin — sin 2 2 2 Again, logE =1-3124082 log 4 =_ -6020600 log sin 30° 37' = 1 -7069667 diff. for30" =_ 1067 log sin 36° 37' = 1-7755801 diff. for 30" = _ 850 log sin 22° 45'=. ! -5873865 .-. logr = -9845932 log 9-6514 = -9845903 29 6 27 .-. r = 9-65146. 257. This follows from XVIIL c. Ex. 5 and XII. d. Ex. 12. MISCELLANEOUS EXAMPLES. K. 299 258. Let lAPB = a, lBrC=^, lPBC=y; then -^ = ^'" ^y - ") l^ = finJ^+T) . AB sin a ' BC sin/S ' but AB = BC, sin (7 - a) _ sin (^ + 7) sin a ~" sin ^ ' or sin7Cot a-cos7 = cos7 + sin7 cot^ ; .*. 2 cos 7 = sin 7 (cot a - cot j8), 2 cot 7 = cot a - cot /3 ; ,, , . 2 11 that is, ™ = -, — , since 7 is the supplement of the angle BP makes with the road. B+G B-C 1 -r, x-.\ /-f ^ ,\ ii cos ^ — cos- 259 T.,w.;..- fc»^-g + eo«<^)a + 2cos.-l) 2 2 (l + 2cos.^)(l-cos.l) ^ . _J 2 sm^ — 2 ^ i?-C .-f ^ B-C . B-^G 2 cos — - — cos— 2 cos - sm 2 2 2 2 c^ ' A A sin^l 2 sin — cos — 2t 2 sin B + sin G 6 + c sin^i a 260. From the fig. of Art. 219, we have 45 = ^^ ; Al^ s » , ., s-o, s-b s-c Ss-2s , /. nrst side = 1 1 — = = 1 s s s s 1 2/2 261. Let sin~i- = a, then cosa=— ^; o o „ • / »> 1 \/2 2^2 3 7^/2 = cos(sin-'3-jj-j), 300 MISCELLANEOUS EXAMPLES. K. 262. We have cot a (cot /3 cot 7- 1)= cot y3 + cot 7, .-. cot(^ + 7)=-i„ = cot a tan a 7r .-. /3 + 7 = 7i7r + --a, or a + j3 + 7 = (2/i + l)^. 263. Since tan A + tan B + tan G = tan A tan i? tan C, , ^ , .^ tanM + tan2£ + tan2(7 the first side = -r j — 7 ^-tt, — 7=r tan A + tan i? + tan G (tan ^ + tan B + tan C)^ - 2 tan ^ tan B -2 tan 5 tan C - 2 tan G tan ^ ~ tan A + tan i^ + tan G ^ , ^ 2(tan^ tan5+. .. + ...) = tan^ + tan2? + tanC ^— — — -— — - tan A tan B tan (; = tan A + tan 2J + tan C - 2 (cot ^ + cot B + cot G). 264 I^et Oj, Oo be the two points of observation, A and B the two objects,'so that AA0fi., = 4:O°, A 0..0^ = O^OJi = 2^°. Tlien Z Oi^O„= 1121°, Z O^BO.i = 22\°, and 0iB = 0i02 = l mile. O A sin 22J-° Now from the a O^A 0. , ,-V- = -— rio''i^ = *^an 22^°= ^2 - 1 ; ^ - 1 mile smll2^° " ^ .-. 03^=^/2-1 miles; /. ^^ = ^2 miles. Again, lip^, P2 be the perpendiculars from AB on O^Oo 2Ji+2J2=(Oi^ + <5i^) si» 45° = ^i? sin 45° = 1 mile. 265. This follows from the identity tan A + tan B + tan C= tan A tan i> tan C, where A + B + G=180°, by putting ^ = 20°, B = AO°, C=120^. 266. The equation may be written (2 cosec 2^)3 = 3 (2 coscc 2^) + -^^ , / 1 y .3 cos^^ \sin^cos^/ sin ^ cos ^ siu'*^' that is, l = 3sin2^cos2^ + cos6^, which reduces to (cos'^^- 1)'' = 0, whence d = inr. MISCELLANEOUS EXAMPLES. K. 801 267. When A + B+ C=1S0°, 1 - cos2 B + cos= A - cos2 C=sm-B + sin^ C - sin- ^ = siu2 B + sin (C + A) sin (C - /I) = sin£ {sin(C + ^) + sin(C- J)}, since C + A = 180°-B, = 2 sin i? sin C cos A. Wlien .1+7? + C = 0, l-cos^B + cos'^A-cos^G = s'm-B + sm{G + A)sm{C-A) = -sin£{sin(C + .I) + sin(C-^)}, since (7 + ^1= -/)', = - 2 sin £ sin G cos J . 268. We have cot^ -coti? = cot£-cot C, that is sin (^ - J) ^ sin (B-C) Sin ^ sm jB sin jB sin G sin(^-^ ) _ sin {B - G) sin(£+C)~sin(^ + ^)* Whence sin {A + B) sin {A-B) = sin (5 + (7) sin (i5 - C), sin2 ^ - sin2 B = sin2 B - sinS C ; that is, a--b'^=b^-c\ nan Q /i -^ o oa-2^-y I 4^ + 37-3a 6S-7y + a) 269. oecond side = 2 cos ^ — - -^cos -^- f-cos —^V 4(4 4 j 2a + 2/3 + 27 6|8 + 47-8a 6a + 48-8y = COS f ^ + COS '^— / + C0S ^^ '^ 4 4 4 + COS— ^ 4 cos| + cos ^^+-y_2aj+cos/^^+/3-27J + cos^-^ + a-2^j = first side. 270. Denote the radii of the three escribed circles by x, y, z respectively, then we have to shew that (?/ - z) (2 - x) (x - ?/) + («/ - 2) (2 + x) (X + y) + (2 - .r) (a; + y) (y + ^) + {x -y) {y ^-z) {z-V x) = (). Taking the terms in pairs, the expression on the left reduces to [y-z){'l{zx^-xy)\^{y-\-z){1{zx-xxi)\, or 2x(?/-2)(y + z) + 2.c(y + 2)(2-7/), which is identically equal to zero. 302 MISCELLANEOUS EXAMPLES. K. A 'iA 271. We have 32 sin - sin — = 16 (cos 2A - cos 3^). Now co8 2.-I = 2cos2yl-l=?^-l = ^, lo o Qi < -^A o . 4x27 3x3 9 cos A A = 4 CCfi' A -i) cos .4 = — 7. , ^ — = - -— ; 64 4 16 .-. 32sin2Sin- = lo(^^+j^j = ll. 272. Solving tho quadratic, we have tan Q=. - 1±;^2. Now ' ,./2 - 1 = tan ^ . [ Ai't. 25 1 . ] o - (V2 + l)= - cot -= - tan (^2 ~ 8 j " TT From the first result, we get ^=n7r+ - , o and from the second, 6 = mr ~(k'~q) = '^^~"^> TT TT both of which are included in (8n - 1) r =^ 7 • 2 1 7 14 7 24 273. (1) 2 tan-i - = tan-i _L. = tan-i ^^ = tan-i - = cos"! - , 2 3 1 3 3 2 24 4 tan-i - = 2 tan^^ — ^ = 2 tan~i j = tan~i = tan~i Q Tfi • -i24 = ^^^ 25 9 16 24 Thus each side of the identity = -^ . 3 sin 2a 6 tan a ( 3(1- tan^ a) / 3 tan a ^ ' 5 + 3 cos 2a ~ 1 + tan^ a ' \ 1 + tan- a ) "" 4 + tan^ a ' .-. first side of the identity = tan "^ -i — - -°^+ tan~^ ( -^-r- ) 4 + tan2a \ 4 / 3 tan a tan a , 4 + tan^ a 4 . . 16 tan a + tan*'' a = tan-^ ., ^ ., = tan-' — — — - — 3 tan- a 10 + tan'' a ~r(4 + ta^2-) = tan~^ (tan a) = a. MISCELLANEOUS EXAMPLES. K. 303 274. We have s{s — a) be ~ b^ + c\ ■ibc ' :. 2s (2s- 2a) = 62 + c2; (b + c + a){b + c -a) = &^ '+c^ h-^ + C' + 2bc -a' = b-^ + c2; a2 ■'• 2 " zbc, or that is, which proves the proposition, Orrr: W 1 • D ^ • * 119 Sin 50° 275. \>ehave smB=~sinA = ,^„ , a 97 log 119 = 2-0755470 log sin 50° = 1-8842540 179598010 log 97 = 1-9867717 log sin B = 1-9730293 diff- for 1' = 460, log sin 70° = 1-9729858 ^^^x60" = 57"- 435 460 .-. 5 = 70°0'57", or 109° 59' 3", both values being admissible since a*IC=180°-^^=90°+2; _ a _ g _ ■ ^1 ~ 2 sin I>'2C~„ A' 2 cos -5 ^//>c ii^ sin A sin 7? sin G ••• PiPiPi= ^ 7^ 1'= A B (T 8 cos - cos - cos - cos — cos - cos - = 812^ sin — sin -^ sin —=2rli'^, 2i Jt u since ^^ . A . B . C r=\B Bin — sin — sin — 281. This is a particular case of Ex. 13 in XVII. a. MISCELLANEOUS EXAMPLES. K. 805 282. The equation may be written h^sm^2e = {c-acoH2d)% or b'-{l- cos2 2^) -- c- - 2ac cos 26 + aP' cos^ 2(?, that is, {a- + h-) cos2 26- 2ac cos 26 + 0^ -h"==0\ therefore, by the theory of quadratic equations, 2ac cos 2a + cos 2/3 = a- + b- 2ac .'. 2cos2a-l + 2cos2y3-l = ^ whence cos^a + cos^^^ a2 + 62' a' + ac + W^ a2 + i2 • 283. We have c^ = a'^ + h- -2ab cos G Now sin A=-- sin C ~ c = 2 + 2 + ^2-2^2^2+72. ^^^^2^'^ = 4 + V2-V2(2 + v/2) = 2-V2; c=j2-^2. J2 V2^V2 1 V'JI72 ' 2 ^2 ' therefore .-1=45°, or 135°, and since a is not the greatest side the smaller value must be taken. Therefore 5 = 112^°. 284. We Pif ve sin 3^ = 3 sin ^ - 4 sin^ A ; 3 1 .*. sin^ A = - sin A - -^ sin^ A ; 4 4 3 1 .-. S sinS yi = - S sin ^ - -r S/sin 3^, 4 4 Now S sin -4=4 cos — cos — cos — , £i ^ a , ^ . - . _ . 3(.i+i?) 3(.^-7?) ^ . 3C 3(7 and S sm 3^ = 2 sm -^-— ^ cos — ^-— ^^ + 2 sm -- cos -— 2 2 2 2 - 2 sm I 270° - -— I cos -^^— r ' + 2 sm — cos —- V 2 ) 2 2 2 3C 3U-5) „ . 3C 3C = - 2 cos -^ cos ' — + 2 sm -r- ccs -r- o06 MISCELLANEOUS EXAMPLES. K. 3(7 f . 3(7 3 (.^-7)')] = 2cos-|sin--cos— 2— } = 2cos-^ <- cos — ^-cos-^— - — 'V 3^ 37? 3C = - 4 cos -^ cos — cos — ; ^ . „ , ^ ^ 7? (7 3.4 SB 3C .*. S sm^ A=S cos — cos -^ cos — - cos -r- cos —r- cos — - . 2 2 2 2 2 2 I 4 285. Take the third figure on p. 131, and first suppose that lAB^C=2lAB^C. Then it easily follows that A CB^B.^ is equilateral ; | .*. Z; sin yf = a sin 7? becomes &8in^ = ^a. Secondly, suppose that Z A CB^ = 2 i A CB^ . Then I A CB.^ = iB^- iA= i B^CB^ = 180° -2 iB^; .: 37?i = 180° + J, sin 37?i + sin ^ = ; or 3 sin 7?i - 4 sin^ B^ + sin A = 0. Substituting - sin A for sin 7> j , and reducing we obtain the required result. 286. If we write i^y in the place of x the resulting equation has tan- a, ta.n'^lS, tan- 7 for its roots. If in the last equation we further write z for 1 + y the resulting equation has sec^ a, sec'-^ ^, sec- 7 for its roots. After making the above substitutions the equation in z is z^ - {p^ + 3) z^ + i^p"^ - 2pr + S) z - {p - rf - 1 = 0; .'. sec2asec2/3sec2 7 — product of the roots = {p-rf+l. Otlierwiseo Let t^, t^, t^ be the roots of the given equation; then sec2 a sec2 ^ sec2 7 = (1 + 1^^) (1 + 1^) (1 + 1.'^) = 1 + 2 • "^ r. • 27r ^ . Sir , (?i-l)7r 2rsin-, 2rsin — , 2r sin — , 2r sin — : n n n n .'. the whole distance traversed is twice the sum of this series. . hi - 1) TT sin^^ — TT— ^ — sin Kow the sum of the sines = ^" '^ ' ■ [Art. 296.] IT sin— - 2n 1 J!i:4.(!LrlM 2 \n n f sm (TT TT \ . TT -T =^°^27.' ^^^27. whence the required result follows. TT 47r 1/ TT 7r\ 1/1 v/5 + l\ 3 + v/5 295. cos— COS^i^ = - ( COS- + COS- j =- ( - + ^ '- • 15 15 2 V 3 5/ 2 V2 4 / 8 27r Ttt 1 / TT BttN 1 /I v^5-l\ 3^v^5 1/ TT 37r\ l/V'5 + 1 n/5-1\ = 2 (^«« 5 + "^^ T j = 2 V-4 4- j Bit TT 1 cos-=cos-=:-; .*. multiplying these results together, we have 8 V 8 )\S j ~ 83 2^ V2/ ' 296. We have aa; + 6y + cz = 2 A. Now {a2 + &2 + c2) (0:2 + ij^ + z^~) - {ax + hy + czf = [hx - ayY + {cij - hz)~ + [ciz - cx)'^ ; or (a2 + Z>2 + c2) (x2 + 1/2 + 22) _ 4a2= (^x - aj/)2 + {cxj - hzf + {az - cxf. :. x^ + y^ + z- is a minimum when the expression on the right is zero; that is when hx = ay, cy = hz, az = cx; ^__y_z ax + by + cz 2A " a~b~ c ~ ~u- + b'^ + c'^ ~ a^ + b'- + c^' H. E. T. K. 22 310 MISCELLANEOUS EXAMPLES. K. 297. "^Vitli the notation and fi.t,'. of Art. 231, suppose AD and BC inter- sect in E, then r^^ is the radius of the escribed circle opposite to E in the A ABE; \ • r =■ A B a cos — cos — . A+B ' sin — - — a ^ A ^ B r. — = tan — + tan — ; r 'J, 2 'a a c A ^ B ^ C ^ D .-. f- — = tan - + tan — + tan - + tan — Ty Tc I 'l 2. I h d . ., , = — I — , similarlv. 298. Draw J 7/, AH' perpendicular to BG ; then Z PAH=:{90°-B)- [90° -C) = C-B, and AH = 2RsinBsinC, 2R sin B sin C AP = cos [C-B) ' 4 ±, . ^± (cos jB-G) ^p+. .. + ... 2R\sinBsinG h + I r-y T^ • • • t^ ' ' * C 1 (sin 2(7 + sin2. B) + ■.■ + ... G [Art. 135, Ex. 1.] 4ft * sin A sin B sin G 2^ Again, BA' = 2R cosBA'A = 2R cos C, since B, A',A,G are concyclic, .-. ^'if' = 2ii:cos5cos(7; 1 ^'P+- + _ J^ j cos((7-J) ) ' ~2ii tcosBcosC"^" f _ 1 (c o32C + cos2i?) + . .. + ... ~ 2R cos A cos B cos C 2 (cos 2.-1 + cos 2 .B + COS2C7) ~ 4cR cos A cos B cos C^ = Jl. icos^£cosBcosO+l^ [XXL cf. Ex. 9.] 2R ' COS ^ cos B cos C MISCELLANEOUS EXAMPLES. K. ^M 2t 1 — t" 299. Let tan ^=t, then sin = -- -., , cos ^ = — „; substitute these