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Les diagrammes suivants illustrent la mAthode. 1 2 3 1 2 3 4 5 6 •:-j - ^ >iJ .' V 7^ A TREATISE ON ELEMENTARY TEIGONOMETRY. W.U A TREATISE UN ELEMENTARY TRIGONOMETRY UY THE RRV. ,]. n. LOCK, M.A. FELLOW OF OONV.LLK AM, CA.U8 OOLLEOK, CA-IUCIDU. FOKMKRLT MASTER AT KTON S TKli EO T YI'HIJ K I) IT I OX SToronto THE COPP CLAKK CO., L,Mm:u Hontion MACMILLAN AND CO., Limited NEW YOKK: THE MACMILLAN COMI'ANV 1899 All riiihtu reserved 0^53/ FE Printed 1S82. New edition, 1884, new edition 1885, if^H' ^(///jow, 1886, new edition 1887. »<«; edition 1888. reprinted 1899. 1890, 1891, Afarf/i flwrf September 1892, 1893, 1895, 1897, 1899 The pi containi venienti series, for pri^ numeroi are not bridge few yea Tiie difficult, the stuc The be varie Eton, Mai In the indicated the Soiu FROM THE PREFACE TO THE FIRST EDITION. The preser.t Work on Elementary Trioonometry contains that part of the subject which can con- veniently be explained without the use of infinite series. It is intended either for class-teaching or for private study. Accordingly the Examples are numerous and for the most part easy. Those which are not original have been selected from the Cam- bridge and Army Examination Papers of the last few years. The Miscellaneous Examples are somewhat more difficult, and should in most cases be postponed until the student reads the subject for the second time. The order in which the chapters .r- read may be varied at the discretion of the Teacher. Eton, Match, 1883. J. B. L. In the Second Edition a short course has been indicated for the use of Students who wish to read the Solution of Triangles as early as possible. Such Vl PREFACE. Students are advised to omit every article that is marked with an asterisk. A double asterisk has been placed before those articles which should be omitted by all Students until they are reading the subject for the second time. An Appendix containing a description of the Vernier, the Theodolit(>, the Sextant etc. has been added. A Key by Mr Caur is now ready. CONTKNTS CUAP. I. n. lU. IV. V. VI. vu. VIII. IX. X. XI. XII. XIII. XIV. XV. XVI. xvn. xvni. XIX. XX. On MRAnURRMENT • • • • On ImCOMMKNHUIUBLK yUANllllKS . . . . On THK KkLATIOM BKTWKKN THK CjKCUMiiittt.NcK oy A ClBCLB AND ITB DlAMETEB On THK MKASUnEMKNT OF AnOI,E8 . Thk Tbiqonometbical Ratios On thb Ratios o» Ckutain Akolkh On thb Tbioonometrical Ratios or ma tuuK Angle On the Ubb or thk bioNs + ani- - On thk Use of + and - in TaKioNoiiKiuv On Angles unlimited in Maonitode. L On thb Ratios of Two Angles . On the Ratios of Multiple Angles On Angles unlimited in Magnitude. II. On Looab'thms On thb i d,4 OF Matukmaxical Tamlks . On the Relations betwbbn thb Sides and Angle, OF A TbIANOLB On the Solution of Tbianglks . . ', On thb Mbasurbment of Heights aud Dista^ncks On Tbunglbs and Circles On thk Abka of thb Cibolb, the Construction OF Tbioonometrical Tablks, etc. . Appendix Examplks for iixfixicisi; . Examination Papers * • • • • ANswiSKs TO Examples . l-AUK 1 14 38 44 n 66 86 90 101 117 133 144 ICl 177 191 206 221 230 241 253 275 2iJl 1. orete qi contain! We a have foil nieasure( how mnu 2. • mu8t ha I standard measurec Inumier < I standard 3. I land the l I it is meas |number is Examp ' the unit therefore W L. E. T. CHAPTER I. On Measurement. >Jizi ™;r:":,i! "": --r » '■■"•• ""^ - »-„™, „ field, whTClVX^tt "'*''''' '"'•'''" how mauy Bqu«re yatO^ it coutoi J. ' """"* ■"■*» "■• ".e«u^. Secondly, we „.„»t ha^ tte JL^'" „'"« nMwier of tinwa fK,» *i • , meamre or the landl It t ""'^'^^ ^' " ^"'^"*^*^ - *he number, and the mut 18 the concrete auantifv K "^"o^j [it m measured. quantity, by means of which iaampfel. A Une contains 261 feet H«r« ♦», jnumber is 261 and the unit a foot '^'^'•' ^' ^^ampfo 2. What is the measure nf 91 r«-i . the unit ? measure of 2J miles when a yard 2imUes«|x 1760 yards, ,, ^ '■4^«>0yards=4400xlyard, therefore the measure in dAni\ ^u , . . j^ J, ^, ~ '^""" » >'ara ih the unit 2 TRIGONOMETRY. ^ Example 3. What is the unit when the meaaui-e of a field of 10 acres is 242 '{ ^^^ 10 acres 10 acres -242 X —^^g— , •. the unit is 10 acres 242 =200 square yards. ExampU 4. If the unit be a yards, what is the measure of b miles ? h mile8=6 x 1760 yards, xa yards, /. h miles' a 6 X 1760 .♦. the measure required i» — - EXAMPLES. I. (1) What is the measm-e of 1 mile when a chain of 66 feet is the unit ? (2) What is the measure of an acre when a square whose side is 22 yards is the unit ? (3) What b the measure of a ton when a weight of 10 stc is the unit ? (4) The length of an Atlantic cable is 2300 miles and the length of the cable from England to France is 21 miles. Express the length of the first in terms of the second as unit. (5) The measure of a certain field is 22 and the unit 1100 square yards : express the area of the field in acres. (6) Find the measure of a miles when 6 yards is the unit (7) The measure of a certain distance is a when the unit is o feet. Express the distance in yards. (8) A certain sum of money has for its measures 24, 240, 960 when three difierent coins are units respectively. If the first .^;« id koif a unvAnAicm. wKftt are the others? ON MEASUREMENT. le unit 1100 3 wit .w' °*'""'' ""^ * ^^*^*y ^*« *^« ^""^ber of times which thut quantity contains the unit We may express the same thing in different ways quantiS to^h^ :r^^ ^' ^ ^'"^^^^ ^ *^« -«o Of that tHe qu^W't;rr::.f ^ ^^-*^^^ ^ ^« ^^^on that *6. Miis last statement is in the language of Arifch metic, and the word ' lijtion' i„ f« • i j , '^^' «IeV«t.L':"°'"* "* *'"*'<'" •««"»' ^»- of tie the Ji, '^'"'* " "■' "^^ "^ * '»"'» "ten 66 feet i, m ^^Z *'""!' ■*"" * "■"» ''°»*»i'' 66 feet, (m) Wlat w the ratio of 4 mile, to 66 foett (") What fiaction of 66 feet is 4 miles ? t.e^r''^ What ia the measure of <.,:ri. a, eet is ayards=3afeet=^x6feet, .'. the measure is 5^. (ii) How many tunes does a yards contain 6 feet? A8in(i),ayaxd8»^x6feet. Anmoer. -^ timea. (iii) What is the ratio of a yarrte to 6 feet? ^"»(i),ayaaxi8=^x6feet, . a yards 3a 6 feet ""T* . the reauinwl ««« ;- ^a 1— S TRIQONOMETRY. (iv) What fi-action is a yards of h feet ? A • /— \ « yards 3a .*. the reqmred fraction io -r- . « EXAMPLES. H (1) The ratio of the area of one field to that of another is 20 : 1, and the area of the first is half a square mile. Find the number of square yards in the second. (2) The ratio of the heights of two persons is 9 : 8, and the height of the second is 5 ft. 4 in. What is the height of the first? (3) The measure of a field with 3 acres for unit is 66. Find the ratio of the field to an acre. (4) One field contains a second of 2^ acres, 6| times. What is the measure of the first field in terms of the second ? What is the ratio of the first field to the second ? Express the first as the fraction of the second. (5) A certain weight is 3*125 of a ton. WTiat is its measiu« in terms of 4 cwt. ? How many times does it contain 4 cwt. ? What is its ratio to 4 cwt. ? What fraction is it of 4 cwt. ? (6) The ratio of a certain sum of money to 3 guineas is '^. Find its measure in terms of one pound. Find the unit when its measure is 22. (7) What is the measure of a miles when h chains is the unit? How many times do a miles contain c chains ? What is the ratio of a miles to d chains ? Vt uai. Iiw;uiuu 13 u tiiuca ui a, uHaxixc; r EXAMPLES. II. g ^ What is the unit when the measure of ^20 is W ? i>-25 contains a certain sum 4(W timpj, wkT- xl^^ The ratio of jfin +^ - -1 • ^^ * *^** ^ *hat sum ? «tuo oi ;fcdo to a certain sum la Anu titi. j. • ^, is found in terl of '"^ "***"" o^th' a-^ of » rectangle linear Jt "" " '^^^ "' "•« «""»ponding •Erompfe. Find in square feet ft,. „ sm&ce wlioae side is 12 feet "™ "' * '^'^■» The a«a is 12 x 12 square feet-144 x 1 square foot, • • 'he measure required is 144. 8. We shall apply this result to Euclid I. 47. the hjpoten^. '""''* '*• ■«P«ti™Iy i find the lengtt of Let X be the number of feef in +],« i. Then by Euclid I .!"f" " ""^ •'^P"*'""^ -feet=the sum of Ll^JT".^-^^ »» «"« «i?=the sum of the squares on DA .-. 4?8 sq.ft. «a2 sq.ft. + a2 sq.ft. ••• ^'^aa + a", Therefore the «q„ired length of the diameter U V2 . a feet. «f:Strr^-----7^-H.ht^.ea If the pole bH,^!^ »*^ !f "" ^1°^ to a peg iu the ground, from th^Lt of ^°;lre: *■"■ '"^ "«■'*' '*"'' *«"' f*^ the peg h. st«ft Jtt'"'S .' « !. T^f "*"'* "" *" '^ "8'' ""d tho mch from the t^rj'"' ^Vr^' "' ">« "<■« "'^-"' "iU of thert^t, "^ "^ "' *'" '""'^ *» tl" opposite side 8 TRIGONOMETRY. III. (6) A wall 72 feet high is built at one edge of a moat 64 feet wide ; how long must scaling ladders be to reach from the other edge of the moat to the top of the wall ? (6) A field is a quarter of a mile long and three-sixteenths of a mile wide : how many cubic yards of gravel would be required to make a path 2 feet wide to join two opposite comers, the depth of the gravel being 2 inches ? (7) The sides of a rectangular field are 4a feet and 3a feet respectively. Find the length of its diameter. (8) If the sides of an isosceles triangle be each 13a yards and the base 10a yards, what is the length of the perpendicular drawn from the vertex to the base ] (9) Show that the perpendicular drawn from the right angle to the hypotenuse in an isosceles right-angled triangle, each of whose equal sides contains a feet, is ~ . a ft. (10) If the hypotenuse of a right-angled isosceles triangle be a yards, what is the length of each side ? (11) Show that the perpendicular drawn from an angular point to the opposite side of an equilateral triangle, each of whose sides contains a feet, is ^ . a ft. (12) If in an equilateral triangle the length of the perpen- dicular drawn from an angular point to the opposite side be a feet, what is the length of the side of the triangle ? (13) Find the ratio of the side of a square inscribed in a circle to the diameter of the circle. (14) Find the distance from the centre of a circle of radius 10 feet, of a chord whose length is 8 feet. (15) Find the length of a chord of a circle of radius a yards, which is distant h feet from the centre. (16) The three sides of a right-angled triangle, whose hypo- tenuse contains 5a feet, are in arithmetical progression : prove that the other two sides contain 4a feet and 3a feet respectively. (9) * CHAPTER U. On IhoOMMENSCBABLE QnANTITIBlS ifc/ 1"^° "T'*" *"* '*''' "> >« eommensnrable when their ratio can be eicpressed as an omAm^i^ fraction: that 1. as a fra,rtion whose numerator and denominator are both whole numbers. Example. 4-95 and 81| are two commensurable number. Their ratio is 4#*^81?=iiiill " ' * 90x327* wi, ^^'.J^"" ""T^^™ ""^^ ^^^ ^ l>e incommensurable ration '^*'° ''*'***"' ^ '''P'''''^ ^ ^ arithmetical E^^nipU. V2 and 1 are two incommensurable numbers. For — ca ^-t be expressed exactly as an arithmetical quantity So ^3 and ^% are two incommensurable numbers. 11. One number alone is said to be an incommenmr- M numler when it is incommensurable with .n ~o "t^^:zt^:^' """^^^ --' ^ ™«^ - ^^Ztn^l::' ''' ^' *" -^ ^-^^^ are incom- Hi 10 TRiaONOMSTRY. 12. Two quantitiea are said to bo commensurable when their meaaurea referred to a common unit are com- mensurable. Example, A mile and a thousand yards are two commen- surable quantities. Their measures with a yard for unit are 1760 and 1000 ; and these are commensurable numbers. 13. Two quantities are said to be incommensurable when their measures referred to a common unit are incom- mensurable. Example. The side of a square and its diameter are two incommensurable quantities. For if the side of a square contain a feet, the diameter (see Example 3, p. 7) contains \'2 . a feet, and therefore the ratio of their measures is 1 : ^2. So that their measures are incommensiurable. 14. There is no practical diffictUty in dealing with incommensurable quantities. We can always find for their measures arithmetical expressions sufficiently accurate for all practical purposes. 15. A little consideration will convince the student that no measurement can in practice be made with absoltUe accuracy. For example : A skilful mechanic is probably satisfied if in measuring some material two or three feet in length the error in his measurement is less than the thirty-second part of an inch. (The thirty-second part of an inch is less than half the height of the smallest letter on this page.) That is to say, he is satisfied if he make no greater error than about a thouaatidth part of the whole lerigth to be measured. He would record such a measurement thus, 2 ft. 3|^ inches, which is 891 thirty-second parts of an inch. ON INCOMMENSURABLE QUANTITIES. H 16 We will suppose that the length thus measured is the side of a square, and that the workmai, wishes to know to tiie same degree of accuracy as his measurement, what 18 the length of the diameter of the square. He can find it thus. The diameter of a square = ^2 x its side, .-. the diameter of this square = ^2 x 891 thirty-second paits of an inch. Also ^2 = 1 '414 nearly, .-. V2x 891 = 1-414x891 = 1259-8. .-. the required diameter = 1260 thirty-second parts of an inch, nearly. The error being less than one thirty-second part of an inch. 17. ThB student will be able to see from the above example, that if the value of an incommensurable number 18 found to 4 figures, a very considerable degree of accuracy « attained. Also that a much greater degree of accuracy is attained for every additional figure. 18. It is no advantage in calculations such as the above to have the value of such quantities as J2 calculated to any greater degree of accuracy than the observed measurement For mstence, our oaJculation being correct as far as the whole numbers are concerned we should gain nothing bv using 1-4142 instead of 1-414 for J2. Ms would 'givl K!?4,f i^' ^ *^ .. — ^ P-^ ' an inch, a quantity -J "j^potuusiH wo smaii to be of any importance. 12 TRIQONOMBTRY. Example 1. The side of an equilateral triangle contains 2 feet ; find, correct to the ten-thousandth part of a foot, the length of the perpendicular drawn from an angular point to the opposite side. Here (as in Example 2, p. 6), let the perpendicular contain X feet, then ■m *« sq. ft. = 2* sq. ft. - 1' sq. ft., .-. «»=4-l-3, = 1-7320 &G.; .'. the length of the pei^endicular^ 1'7320 feet. Example 2. Find the length, correct to the ten-thousandth part of a foot, of the side of the square described upon a dia- meter whose length is a feet. Let X be the number of feet in each of the sides of the square. Then EXAMPLES. IV. .-. 2*3.02, .•V2.*-a, V2~ 2' IS .*. *• -|x(r4142)-ax-7071..., .% the length of the perpendicular -ax 707 1 . . .feet ♦EXAMPLES. IV. uSl 7;°^/°"';°* ^ *^« thousandth part of a foot, the length of the diameter of a square whose side is 7 feet. (2) Find, correct to a yard, the length of the diameter of a square whose side is one mile. K Jf? ^^^ "^"T* ^ *^^ hundredth part of an inch, the hypoten^ of a nght-angled triangle whose sides are 3a ijin and 3 ft. 4 m. respectively. ^ (4) Find, to the nearest inch, the side of a square whose area is 1000 square yards. ^ (6) Find, correct to the tenth part of a foot, the diameter of a square field whose area is ten acres. (6) Find, to the nearest inch, the side of a square field whose area is one acre. °4"«re neia 10 fll ^^^ ^^^ ^^'^^* ""^ *^ equilateral triangle whose side is 5-32 feet'^'* *^' ^''^^* ""^ ^ equilateral triangle whose side is oi ^^^J!^^^.^^ **^ * **^^® measures 24-6 inches and 41-3 inches ^ong two adjacent sides. What should the diameter measure 11 the table is rectangular ? °*«uii3 (10) Find, to the nearest inch, the diameter of a lawn-tennis ooiut whose length is 78 feet. anH 1 ^aaji^i. o« *•..* • that it is properly marked out' "" ^^^' '"^^"^^ ( H) •M CIIAPTER III. On thr Relation between the Oircumfbrenli. op a Circle and its Diameter. 19. The circumference of a circle is a line, and therefore it has length. We might imagine the circumference of a circle to consist of a flexible wire ; if the circular wire wore cut at one point and straightened, we should have a straight line of the same length as the circumference of the circle. 20. A polygon is a figure enclosed by any number of straight lines. 21. A regvlwr polygon has all its sides equal and all its angles equal. 22. The perimeter of a polygon is the sum of its sides. 23. If we have two circles in which the diameter of the first is greater than the dicuutst^u' of the sec:»nd, it is evident that the circumference vV <\xc i'j'st will ou greater than the circumference of the second. e 24. It seems, therefore, not unlikely that, if the dia- meter of the firat circle be twice that of the second, the circumference of the first will be twice that of the second. Ill TIIK CIRCUMPHRKSOU OP a CIRCLE. 15 26. Aud alm> uot unlikely that whatever be the ratio of the circumference of the fir«t circle to it. diameter the same will be the ratio of the circumference of the seisond circle to itA diameter. This suggents that it is not unlikely that the circum- ference of a circle = A times its diameter, where >fc is some number which is the same for ail circle. We shall presently prove that this is the case. 26. But although we can prove that the circum ference of a circle its diametoT " "" ^^*^^ nuroeri. al quantity, the method of calculating the value of this number is beyond the limits of aa elementary treatise. 27 We shall therefore simply state here, what is proved in the Higher Trigonometrj, (i) that this numerical value is incommens mble, (ii) that it is approximately 3- 141 59265 &c. 28. When we say that this number is incomme isurable we mean (cf. Chapter II.) that its exact value ca >not be stated as an arithmetical fraction. It also happens that we have no short algebraical ex- pression such as a surd, or combination of surds, which represeuta it exactly. So that we have no numen^l expression whatever, arithmetical nor algebraical, to represent exactly the ratio ot the cucumference of a circle to its diameter. Hence the universal custom ha.s ansen, of denoting its exact value by the letter tt. 16 TRIGONOMETRY. 29. Thus TT stands always for the exact value of a cer- tain incommensurable number, whose approximate value is 3*14150265, which number is the ratio of the circumference of any circle to its diameter. It cannot be too carefully impressed on the student's memory that tt stands for this number 3-14159265.. .&c., and for nothing else; just as 180 stands for the number one hundred and eighty, and for nothing else. * 30. We proceed to prove that the ratio of the circum- ference of a circle to its diameter is the same for all circles. The proof depends on the following important principle ; TAe length of the circumference of a circle is that to which the length of the perimeter of a regula/r inscribed polygon approaches, as the number of its sides is continually in^sreased. * 31. We take for granted that the straight line is the shortest line that can join two points. * 32. Let ABCDEF be any regular polygon inscribed in a circle. mi if t¥ THE CIRCUMFERENCE OP A CIRCLE. 17 Then the side AB is shorter than any other line join- ing AB, so that the side AB is less than the arc AB; therefore the peiimeter of the polygon, viz. AB-^BG + GD + DE + EF+FA is less than the sum of the arcs, that is, is less than the circumference of the circle. Now let each of the arcs AB, BG, etc. be bisectedin « P, y, 8, e, V, and let the lines Aa, aB, Bft, ^G, eta be joined ] then the figure AaB/iGy etc. is a regular polygon of twice as many sides as the first polygon. And since the sides Aa + aB are together greater than the side AB, it follows that the perimeter of the second polygon, viz ^a + a^ + ^;8 + /JC+Cy + y/) + etc. 18 greater than the perimeter of the first But the perimeter of the second polygon is less than the circumference of the circle, because each side is less than the corresponding arc. Hence the perimeter of the second polygon is nearer the circumference, but is less than the circumference. By bisecting the arcs of the second polygon we should get a third polygon, whose perimeter is nearer the circum- ference of the circle than the second. It is clear that by continuing this process, we can get a polygon whose perimeter is as near as we plea.se to the circumfei-ence of the circle. Hence, The I^th of the circumfere^e of a circle is that to which ^ngth of ths perimeter of a regv^ inscnbed polygon iipproaches, as the nvmber of Us »iA.o ,» -_..„.._77. . '^^ , -^ tie ct.^.^ IT, w/icc/tcKMcw increased. L. E. T, 3 18 TRIGONOMETRY. * 33. Prop. The ratio of the circumference of a circle to its diameter is the same for all circles. Let ABGDEF, abcdefhe any two cirolesi Let a regular polygon of any, the same number of sides be insciibed in each of them. Join Af B, Cy etc., a, 6, c, etc. the angular points of the polygons to the centres 0, o respectively. A Then AOB any one of the isosceles triangles in the first figure is equiangular with*, and therefore similar to, aoh any one of the isosceles triangles in the second figure. .*. AB : OA=ah : oa, [Euc. vi. 4.] and BC : OA=bc • oa, and so on. .-. AB + BC+CD + etc. : OA=ab + be + cd + eto. : oa; or, the perimeter of the first polygon is to the radius of the ♦ For at and at o, four right angles are each divided into the suTne number of equal angles, so that the vertical angles AOB, aob of the isosceles triangles AOB, aob are equal. TBB OIROUMPERBNCB AND THE DUMETBR. 19 But the «r«^«.«„„ of the circles aie what the^ ««e wheel Ho^s, ^:^ ^i:z:::^^,z "'^ "- ^ ^ ^-^ (8) A locomotive whose driving wheel i^ "i &^* i,- i. i. mstrument to record th^ r»,«, J i. ! '^^^ ^'^^ ^aa an 22 TRIGONOMETRY. V. (10) What is the diameter of the driving wheel of a locomo- tive engine which makes 4 revolutions per second when the engine is going at the rate of 60 miles per hour ? (11) The large hand of the Westminster clock is 11 feet long- how many yards per day does its extremity travel? How far does the extremity move in a minute ? (12) The diameter of the whispering gallery in St Paul's is 108 feet; what is its circumference? (13) Find the number of inches of wire necessaiyto con- struct a figure consisting of a circle with a regular hexagon inscribed in it, one of whose sides is 3 feet. (14) How 'many inches of wire would be necessary in a figure similar to that in Question (13), if the circumference of the circle were ten feet ? (15) Find how many inches of wire are necessary to make a figure consisting of a circle and a square inscribed in it, when each side of the square is 2 feet. (16) How many inches of wire are necessary for a figju-e similar to that in Question (15), when the circumference of the circle is 12 feet ? (17) Find the length of string necessary to string the handle of a cricket bat ; having given the diameter of the handle = U in the length of the handle = 12in., the diameter of the strings Ath of an inch. (23) CHAPTER IV. On the Measurement op Angles. 38. In elementary Geometry (Euclid I.— VI^ the angles considered are each always less than two right anglea » " For example, in speaking of the angle SOP in Euclid we should always mean the angle less than two right angles, ZlZt^r'"" '" ''" ""^-'^^0^ «-^r than the !!„ ^ ?j'"'r'*^. by the angle BOP is meant, not the pr^en^ rnclzn^um of the two lines OS, OP but the amount of turiUng which OP has gone through when" z^^op! '^""' ''• " ^ '"™^ "^^ ^^:''Z Exampk. Suppose a raco run romid a circular oouiae The SZ^ Th^if r;,^ " ""^^ ■"«'" "■»»* *be cent™ of Te. ^^L- ^ v^° ■^"^'^ *° *« "■» » tbree times round tte hne jommg each competitor to the centre would havrS desonbe an angle of 12 right angles. 6J ^T "^ "^^^ '"^^ ' competitor has described an angle of H nght angles, we record not only his present position, " t tht 24 TRIGONOMETRY. total distance he has gone. He would in such a case have gone a little more than one and a half times round the course. 40. Definition. The angle between two lines Oi?, (?P 18 the amount of turning about the point which one of P R the lines OP has gone through in turning from the position OR into the position OP. 41. The angle POP may be the ffeometrical representa- tive of an unlimited number of Trigonometrical angles. (i) The angle POP may represent the angle less than two right angles as in Euclid. In this case OP has turned from the position OP into the .position OP by turning about in the direction contnt^,/ to that of the hands of a watch. (ii) The angle POP may represent the angle described by OP in turning from the position OP into the position OP in the same direction as the hands of a watch. In the first case it is usual to say that the angle POP is described in the positive direction, in the second that the angle is described in the negative direction. (iii) The angle POP may be the geometrical repre- sentation of any of the Trigonometrical angles formed by any number of complete revolutions in the positive or in the negative direction, added to either of the first two angles. (We shall return to this subject in Chapter IX.) BXAMPLES. 25 EXAMPLES. VL Give a geometrical representation of each of the foUowinc ang es, the starting Une being drawn in each case from the tu n' mg ix>mt towards the right. 1. 2. 3. 4. 6. '6. + 3 right angles. + 6 right angles. + 4J right angles. + 7J right angles. - 1 right angle, log right angles. 7. - lOJ right angles. 8. +4 right angles. ^- 9. - 4 right angles. 10. An right angles. '11. (4n + 2) right angles. 12. -(4n + J)rightanglep. 42. There are two methods of measuring angles, (i) The rectangular measure, (ii) The circular measure. Rectangular Measure. wJf 'o^^^^^r ^"' ^^^^^" '"'"'"'"^ ^^ ^^''^'^ ^ith the nght angle (or part of the right angle) as unit unit^are-^' '''^''''' ^^^ *^' "^^' ^"«^' '' ^^'^^^^ ^^' ^ (i) All right angles are equal to one another, (ii) A right angle is practically easy to draw, (iii) It is an angle whose size is very familiar. «., J-^'-^ ?/ "^^* ^^^^ '' * ^'^'S^ '^'^Sle, and it is therefore subdivided for practical purposes. It is usual to explain two methods of subdivision, (i) The sexagesimal method, (u) The cent^mnl «»• 'io/»»»v,«i -_-ii- - j * • See Art. 65. I W III 26 TRIGONOMETRY. I. The Sexagesimal Method. 46. In this method the right angle is divided into 90 equal parts, each of which is called a degree ; each degree is subdivided into 60 equal parts, each of which is called a minute ; and each minute is again subdivided into 60 equal parts, each of which is called a second. Instruments used for measuring angles are subdivided accordingly ; and the size of an angle is known when, with such an instrument, it has been observed that the angle contains a cfertain number of degrees, and a certain number of minutes beyond the number of complete degrees, and a certain number of seconds Yhjond the number of complete minutes. Thus an angle might be recorded as containing 79 degrees +18 minutes + 36*4 seconds. Degrees, minutes, and seconds are indicated respectively by the symbols °, ', ", and the above angle would be written 79° . 18' . 36-4". * II. The Centesimal or Decimal Method. 47. The other method of subdivision is the Centesimal or Decimal. Here each right angle is divided into 100 equal parts each of which is called a grade; each grade is sub- divided into 100 equal parts, each of which is called a minute; and each minute is again subdivided into 100 equal parts, each of which is called a second. Instruments of observation would be subdivided accord- ingly, and any observed angle would be recorded as contain- ing so many grades -•■ so many minutes + so many seconds. :^ THE MEASUREMENT OP ANGLES. 27 Grades minutes and seconds are indicated reflectively by the symbols •. \ \ So that an angle of 26 UdesIlT minutes + 34-2 seconds would be written 36« 19* 34 -2^ of Ih!'^ ^* "T ^' ''^''''''^ *^'"* ^^' "^'^^^^ '•« «i»»pl7 that of the decimal system of notation. angi?' ''°" "'^' '" "^"P^^ = Vi/b - 1^ + T.8^^. of a right That is -3619342 of a right angle. This is equal to 3619342 of a grade Also to 3619-342 of a minute. Also to 361934-2 of a second. Example. Express 302« 2' 4-6'^ as the decimal of a right angle. Thisangle=?^+ 1__. 4-6 » . ^, , 100 ^ 10000 ^ 1000000 °^ * "«^* *°8^« =3-0202046 ofa right angle. *49. Hence, to express an angle given in grades, minutes oblTtwr.*'; '"'"'^^ ^'^ "^^^ '^^'^^ - ^-^ onlyTo observe that the>.. anc^ second decimal places are occupied pied by the minutes, and tl., fifth and sixth decimal places are occupied by the seconds. ""paces * 50. The same observation will enable us to express in Example Express 34650023 of a right angle in firadea minutes, and seconds. ^ grartea, 3 right angles = 300 grades. •46 of a right angle =46 grades. •00,50 of a right angle = 50 minutes. ^00,00,02,3 of a right angle-2-3 seconds. Therefore the angle is 346« 50' 2-3*\ 16 TRIOONOMETRY. ♦EXAMPLES. Vlt Express as the decimal of a right angle, (1) 63«2ri8r\ (7) 32«4'.V^ (2) 104« 26'99-r. (8) 1" 2' a--f ' (3) 2«l8'2r. (9) 69«tf7r. (4) 3« 29' 48-94''. (lO) 119«af0-4r. (5) 62f 4r. (11) ioo6« lar r (6) 1000« gr 12\ (12) 2« 26^ 4-ff\ Express in grades, minutes and seconds, (13) -367891 of a right angle. (19) 1-001 of a right angle. (14) 1-043021 of a right angle. (15) -012003 of a right angle. (16) -00102 of a right angle. (17) -0625 of a right angle. (18) 3-02125 of tt right angle. (20) -0101001 of a right angle. (21) 6-451 of a right angle. ^22) -023 of a right angle. (23) -0001 1 of a right angle. •(24) -00001 of a right angle. 51. An angle given in degrees, minutes, and seconds may be expressed as the decimal of a right angle by the u.sual method. Example. Express 39" 4' 27" as the decimal of a right angle. 60 ) 27 secon ds 60 Y 4-45 minutes 90 > ^39-07416666 etc. degrees •43415740740 etc. right angles Answer. -43415746 of a right angle. ON THE MEASURBMBNT OP ANGLES, S9 52. An angle given tw the decimal of a right angle may be expressed in degrees, minutes, and seconds by the con- verse of the above. ExampU Express -43415^40 of a right angle in degrees, mmutes, and seconds. * " e i •43416740740 etc. right angles _90 ^^ 3907416666600 degrees The last two figures would be 66 if we were to write down the recumng part to more figures. This gives 3907416666666 etc. degrees 60 that is or The result is 4-4499999960 minutes 4 '446 minutes 4*45 minutes 60 27 00 seconds. 39" 4' 27". *63. We have seen that an angle expressed as the decimal of a right angle can bo at once expressed in grades, minutes, and seconds. Hence an angle expressed in degrees, minutes, and seconds, can be expressed in grades, etc. by first reducing the angle to the decimal of a right angle. Example. Express 390 4' 27" in grades, minutes, and seconds. This angle is •43415t4d of a right angle, by Art. 51 and this «=43''4r 57'40'r\ *«54. An angle given in grades, minutes, and seconds can be expressed in degrees, minutes, and seconds by first ex- ht angle. ■ pressing the angle as the decimal of a riglit angle. ■M it* R Ls 30 ' V TRIGONOMETRY. secofr^'"' ^''"" ■'"" ^'^ ^^-^^^ ^ <^«g-es, minutes, and fro J Art^^sf '' ''''''^'' °' " "^^* ^S^^' ^^°^ i« 390 4' 27" * EXAMPLES. Vin. ril^^'T^^ "^ *^' ^'""^^°^ ^^^^ (i) «« tl»e decimal of a right angle, (n) m grades, minutes, and seconds ; (1) 80 15' 27". (4) 160 14' 19" (2) 60 4' 30-. (5) 1320 6.. (3) 9^0 5/ 15V ^g^ ^^^^ Express in degrees, minutes and seconds, (7) 1'3T50\ (10) 24«0'25-. (8) 8«76\ (11) i8«ri5". (9) 170« 45' 35^ (12) 35« 56. The decimal or centesimal system of subdividing a nght a^gle wa^ proposed by the French at the commence- ment of the present century; but, although it possesses many advantages over the established method, no one has been found wUling to undertake the great expense that would have to be mcurred in rearranging aU tables and all books of reference, and all the records of observations, which would have to be transferred from the old system to the new before the advantages of tie decimal system could be felt' Ihus the decimal system of angular measurement has never been used even in France, and in aU probability never will be used m practical work. III*? I m ON CIRCULAR MEASURE. On Circular Measure. 31 56. By the following construction we get an angle of great importance in Trigonometry. On the circumference of a circle whose centre is let an arc JiS be measured so that its length is equal to the '"^'"^ of the circle, and let /? and S be jo^ed to the centre! Pn7'' yVr ^^"""^ *° P'°^^ ^^^*- 60) that this angle »re equal to one another. * 68. We may state the same thing th.is-We «r. ,i„. * * prove that if we take anynumber of dilerenroil;^. t^l^ on the croumference of each an are equal in length Cie^tSr then the angles at the centres of these circles which st^™ these arcs respectively, will be all of the same oj circif t °f """'"'•• '^^ «^1« ^Wct at the centre of a o^cle stands on an arc equal in length to the «diua of the circle is called a Eadian. 60. yr^'tfu^tallSadiamareegualto^anol/^. Smce the Eadian at the centre of a circle stands on an »ro equal in length to the radius, .taJ^'orhlt! "' *"" 7" '"'^'" "* *'" ««"*«' »f ■> -cle Biiands on hall the circumference, m f 32 TRIGONOMETRY. i| I and since angles at the centre of a circle are to one another as the arcs on which they stand (Euc. VI. 33), therefore the radian is to an angle of two right angles aa the mdius is to half the circumference; that is, as the diameter is to the whole circumference; that is, in the constant ratio 1 : ir. Therefore the radian = ^Jl^S^l^g!?? TT That is, the radian is a fixed fraction of a right angle. But all right angles are equal to one another. Therefore all radians are equal to one another, o. e. d. 61. Thus the radian possesses the qualification most essential in a unit, viz. it is always the same. The student will find, in the theoretical pai-t of Trigo- nometry, that many expressions can be written more shortly when a radian is used for the unit of angle, than when any other unit is used. 62. Thus the reasons why a radian is used as a unit are : (i) All radians are equal to one another. (ii) Its use simplifies many formula in Theoretical Trigonometry. 63. The system of angular measurement in which a radian is the unit is called Circular Keasure. Therefore the circular measure of an amgU is the wwwt- her ofradiam which the angle contains. 64. A radian = - x 2 right angles, = ^il of 130° nearly, = 57-2957.... degrees. ON THE MEASUREMENT OF ANGLES. 33 The student should notice that a radian is a little less than an angle of an equilateral triangle. 65 Circular measure is, as we have said, used in theo- retool investigations, in which the angle under consideration IS almost always expressed bya letter. This is usuaUy one of theGVec^lettersa,/?,y...,0,^,^.... ^ Strictly speaking these letters represent numbers, i e measures; so that some unit of ar^le must bo ur^stood in such an expression as 'the angle $.' (Art. 2.) For this reason, when an angle is denoted by a Greek letter such as a, ^ y, etc., 0, ^, ^, etc., it is understood that circular meagre is the measure used, unless the contrary IS expressly stated. ^ So that 'the angle 0' means '^radians.' Similarly 'the angle ,r' means ',r radians' or '3-14159 radians,' thi't is two right angles. a. fTr ^\ ^\^'' ^" convenient, in using such letters f, /J, 0...>b, i, etc. to represent angles, to agree that the rlT n "'*' *'^^ '^^' ^^ letter'shalfbeaCt «nn,r; ,/^^^'^«^ «^*°^Pl«s it wiU be necessary to use some letter (c suppose) t« denote a radian. Thus 2« denotes 'two radians.' written /*"^t^'^''^'°^ '";'" ''^' '^"S^^ ^' «^°"I« written <9«. (Just as m speaJring of 'the angle ninety ' we tTft the^'f:?;''^"^*-^ Butifitiscl^^lyunde^'told thaj the angle ^' means 6 radian, there can be no am- biguity m the expression. L. K T, 3 ! 34 TRIGONOMETRY. 68. The student cannot too carefully notice, that unless an (mgle is obviously referred to, the letters 0, ,... a, ft, ., stand for mere wwmb&rs. Thus as we have said above (29) v stands for a nuniber and a number only, viz, 3* 141 59 , but in the expression 'the angle tt' that is 'the angle 314159 ' there must be Bonie unit understood. The unit tmderstood here is a radian, and therefore 'the angle tt' stands for 3-14159 ', that is two right a/nglea. Hence, wJien am. angle is re/erred to, iris a very convenient abbreviation for two right angles. 69. To express in degrees or grades an angle given in radians, we first express the angle in right angles, remem- bering that 2 right angles =7r radians. Ejcam/ple. How many degrees are there in the angle whose circular measure is 2 ? mu- 1 o J- o 2 1'ig'i* angles 4 . , . , This angle « 2 radians »=2 x — ^- ^ =- right angles. ir 4 X 900 3600 IT It 360 .'. the angle contains — degrees. If ■*70. If D, G and a be the number of degrees, grades and radians respectively in any angle, then 180 " 200 ~ ir ' For each fraction is the ratio of the angle to two right angles. EXAMPLES. IX. Example. Find the number of degrees in two radians Let D be the number, then D^ 2 180 "n-» 35 mvenient EXAMPLES. IX. 1. Express the following angles in rectangular measure. T- (3) 1«. 2c (1) rr. (2) (6) n (9) IOtt. •(4) 3« (5) 3'14159265«etc. (7) 6. .(8) •00314159'' etc. 2. Express the foUowing angles in circular measure. (1) 1800. (2) 3600. .(3) eoo. (4) 22^0. (5) 10, ^gj 67-2950 etc. (7) nO. (8) ^\ If .(9) A. * 3. Express the following angles in circular measure. (1) 33«33^33-5'\ (2) 50«. (3) u'^. (4) 1«. (7) n«. 4. Find the ratio of (5) r. (8) ^^. n (6) W\ (9) lOOOB. (1) 450to|r. /^ (2) 600to60«. ^/ (3) 25»tK>220 30'. /. (4) 24»to2«. \^ (5) l-75-to^-^.^2^ (e) ptoK '^' 3—2 ■««''■] 86 TRIGONOMETRY. *71. To prove that tJie measure of an angle at the cenifi'e of a circle in radians (i.e. in Circular Measure) is the ratio oftJie arc on which it stands to the radius of the circle. L \ R Since angles at the centre of a circle are to one another as the arcs on which they stand (Euc. VI. 33), Therefore any angle ROP at the centre of a circle is to the radian as its arc RP is to the arc of the radian. But the arc of the radian is equal to the radius, m\. e any angle ROP . , ^ its arc RP Therefore — ^i — ^-r: is equal to -^ -i. — . the radian ^ the radius itiS &l*C And therefore any angle ROP is equal to —^. — x (a radian). That is, the 7nea,swre a of an angle in radians is the ratio arc arc or, a = — t; — . radius _ _ *" _ **"^ ^ 180~200~^ " radius * Hence (cf. Art. 70) r^ = ^ X 200 IT radius ir Example. Find the number of grades in the angle subtended by an arc 46 ft. 9ia long, at the centre of a circle whose radius is 25 feet. The angle stands on an arc of 46| ft. and the radian, at the centre of the same circle, stands on an arc of 25 feet. .-. the angle=^ radians, ^^^^ ^ right angl es^ ~Tai^ '119« neai'ly. EXAMPLES. X. 37 ♦EXAMPLES. X. (In the answers ^ is used for w.) (1) Find the number of radians in an angle at the centre of a 3^ circle of radius 25 feet, which stands on an arc of 37i feet. /^ (2) Find the number of degrees in an angle at the centre of a ^ » cu-cle of radius 10 feet, which stands on an arc of 6. feet f^ (3) Find the number of right angles in the angle at the cen- tre of a circle of radius 3,^ inches, which stands on an arc of 2 feet. "^^ (4) Find the number of French minutes in the angle at the centre of a circle of radius 8 ft. 4 inches, which stands on an arc of V (5) Fiiidthelengthofthearc8ubtendinganangleof4Jradian3 at the centre of a circle whose radius is 25 feet. \t A' i> . 'I V' 4 feitUt' ''^ ^^^^' ^'^ "^ ^'^^^^^^ ^«^- - ^ circle of ^ ^^ (7) Find the length of an arc of sixty grades on a circle of ten feet radius. (8) The angle subtended by the diameter of the Sun at the eye of an observer is 32' ; find approximately the diameter of the Sun if its distance from the observer be 90,000,000 miles. (9) A railway train is travelling on a curve of half a mile • radius at the rate of 20 miles an horn-; through what angle has * it turned m 10 seconds ? (iO) A railway train istravellingonacurveof two-thirds of a , mile radius, at the rate of 60 miles an hour; through what angle 1 ^ has It turned in a quarter of a minute? M 38 TRIOONOMETRY. X. (11) Find approximately the number of English seconds con- tained in the angle which subtends an arc one mile in length at the centre of a circle whose radius is 4000 miles. (12) If the radius of a circle be 4000 miles, find the length of an arc which subtends an angle of 1" at the centre of the circle. (13) If in a circle whose radius is 12 ft. 6 in. an arc whose length is '6545 of a fot subtends an angle of 3 degrees, what is the ratio of the diameter of a circle to its circumference ? (14) If an arc 1'309 feet long subtend an angle of 7 J degrees at the centre of a circle whose radius is 10 feet, find the ratio of the circumference bf a circle to its diameter. (15) On a circle 80 feet in radius it was found that an angle of 22'> 30' at the centre was subtended by an arc 31 ft. 5 in. in length ; hence calculate to four decimal places the numerical value of the ratio of the circun.rerence of a circle to its diameter. (16) If the diameter of the moon subtend an angle of 30*, at the eye of an observer, and the diameter of the sun an angle of 32', and if the distance of the sun be 375 times the distance of the moon, find the ratio of the diameter of the sun to that of the moon. (17) Find the number of radians in (i.e. the circular measure of) 10" correct to 3 significant figures. (Use ff| for «-.) (18) Find the radius of a globe such tliat the distance measured upon its surface between two places in the same meri- dian, whose latitudes diflfer by l** 10', may be one inch. (19) Two circles touch the base of an isosceles triangle at its middle point, one having its centre at, and the other passing through the vertex. If the arc of the greater circle included within the triangle be equal to the arc of the lesser circle without the triangle, find the vertical angle of the triangle. (20) By the construction in Euc. I. 1, prove that the unit of circular measure is less than ^ EXAMPLES. X. 39 ^ (21) On the Slat December the Sun subtends an angle of 32' 36", and on Ist July an angle of 31' 32" ; find the ratio of the distances of the Sun from the observer on those two days. (22) Show that the measure of the angle at the centre of a circle of radius r, which stands on an arc a, is ^, where it r depends solely on the unit of angle employed. Find k when the unit is (i) a radian, (ii) a degree. /^" * 72. Questions couceming angles expressed in different systems of measurement are easily solved by ex^esHng each angle in right angles. Example 1. The Hum of the measure of an angle in degrees and twice its measure in radians is 23f , find its measure in degrees (7r=^). Let the angle contain x right angles. Then the measure of the angle in degrees =9().r, >» » n radians =— a?. w- m .'. 90;p + 2.|a:=23^, .*. 90x-^^3^x=^^, .'. 652d?=:l63, Th« angle is } of a right angle, that is 22f , nearly. ' '■M ■ i U, *0 TRIGONOMETRY. Example 2. The three angles of a triangle are in arithmetical progression, and the measure of the least in grades is to that of the greatest in circular measure as 120 : tt. Express each angle in degrees. Let the angles contain x-i/,x,a+y right angles respectively ; thoy are then in a. p. Their sum is 3a? right angles ; but since they are the angles of a triangle, their »um is 2 right angles j /. 3a;«2, •*• x=^. Again, the lea^t angle contains (a?-y)xlOO grades, aod the greatest angle coutaiua (x+i/) J radians, .-. 100(:p-y) :|(j;+y)=120 :n. .: 100(j?~y) = 60(a;+y), or, 40.1 = 1 60y, or, x=4i/. because x—^, or,y=J. .••^•-y=S-i=4, x^l ^+y=3 + J=t. Thus the angles contain ^, f , and | right angles respec- tively ; therefore the angles are 46°, 60*', 75". ♦* EXAMPLES. XI. (In the following examples the answers will be given in terms of »r.) (1) The sum of the degrees and of the grades in a certain angle is 38 ; find its cu-cular measure. (2) The difference of two angles is 208, and their sum is 480 ; find them. EXAMPLES. XI. 41 (3) One angle is double of a second, and the sum of their measures in degrees and in grades respectively is 140 ; express the angles in degrees. (4) Two angles are in the ratio of 4 : 6, and the difference of their measures in grades and in degrees respectively is 2^ ; find the angles in degrees. . (6) The difference between two angles is ^ , and their sum is 66 degrees ; find the angles. (6) If the three angles of a triangle are in arithmetical pro- gression, show that the moan angle is 60''. (7) The three angles of a triangle are in arithmetical pro- gression, and the number of grades in the least is to the number of degrees in the mean as 5 : 6. Find the angles in degrees. (8) The three angles of a triangle are in arithmetical pro- gression, and the number of grades in the greatest u to the number of degrees in the sum of the other two as 10 : 11. Find the angles in degrees. (9) The three angles of a triangle are in arithmetical pro- gression, and the number of grades in the least is to the number of radians in the greatest as 200 : Stt. Express the angles in grades. (10) If D be the number of degrees and G the number of grades in any angle, prove that G-D=\D. "* (11) If i^ be the number of English minutes and m the number of French minutes in any angle, prove that (12) If » cosine » tangent >» ..... MP perpendicular ^""^ m bs^i — NoiB. The order of the letters, in MP, OM and OP, indicates direction and decides their algebraical signs. [Art. 132.] I : THE TRIGONOMETRICAL RATIOS. 47 76. If A stand for the ap.gle HOB, these mtios are called sine A, cosine A and tangent -4, and are usually abbreviated thus : sinil, cos -4, tan -4. 77. There are three other Trigonometrical Eatios, formed by inverting the sine, cosine and tangent respectively, which are called the cosecant, secant, and cotangent respec- tively. '1 8. To define tJie three other Trigonometrical Ratios of an-'j angle. The same construction and figure as in Art. 75 being made, then the ratio /. V OP _ hypotenuse ^'""^ MP ~ ^i^endicular "^ ""^^^ ^^^ COSecant of the angle JROK i \ 9^ hypotenuse ^^^ OJ/" h^ » secant OM base )) (vi) MP perpendicular » cotangent „ 79. Thus if A stand as before for the angle ROE^ these ratios are called cosecant -4, secant A, and cotangent A. They are abbreviated thus, cosec -4, sec A^ cot A. 80. From the definition it is clear that 1 cosec A = secil = cot^ = smA ' 1 cos A 1 tanil 48 TRIGONOMETRY. 81. The above definitions apply to an angle of any magnitude. (We shall return to this subject iiiChapter X.) For the present the student may confine his attention to angles which are each less than a right angle. are 82. The powers of the Trigonometrical Ratios expressed aa follows : and so on. The student must notice that 'sin ^' is a dngle symbol. It is the name of a number, or fraction, belonging to the angle A ; and If It be at any time convenient, we may denote einA by a Singh Utter, such as « or ;r. Also sinM is an abbreviation for (sm A)\ that is for (sin A)x (sin A). Such abbreviations are used because they are convenient. 83. TJie Trigmiometrical Ratios are always the same/c the same angle. or T the li poinfa perpe: Tl contai mon; Th Th Bui to the a Thu the poll The: 84. ratios. numeria flame. W'e ] these rat Henc etc have Exami sin 30<>« L. E. THE TRIGONOMETRICAL RATIOS. 49 ' ^'^•fM iwrpendiculars to OE Ihen the three triangles OUP OhtP nwD" i contain a right angle, and they have the ».Ieft^ mon; therefore their third angfes must be e^ Thus the three triangles are equiangular. Therefore the ratios V-f ^ ^"P' OP ' OP' ' Op' "'''•' a" equal- (Eu. VI. 4.) But each of these ratios is E^Pfn^icular .^ , to the angU at ; that i,,, they are each «„ ROE theS^ron'^ir„:';;rr *''-.- "» ">« ^^^^^ -^ V r on « '»' ^^<' »<• ^^ be right foUowiog ratio, ffi i J^^ ^T "" '''"^ *" ^^ "^ «=« 4—2 Ml I 52 TRIOONOMBTRY. XII. (6) Let ABC be a right-angled triangle such that J 5= 5 ft, i9C?-3 ft., then AC will be 4 ft. Find the sine, cosine and the tangent reaiHJctively. ' III the above triangle if A stand for the angle at A and B for the angle at B, show that sin^^ + 0082^1 = 1, and that siu2 5 + C08»^=l. (6) If ABC be any right-angled triangle with a right angle at C, and let A, B, and C stand for the angles at Aj B and C respectively, and let a, h and c be the measures of the sides oppo- site the angles A, B and C respectively. Show that sin ^ =- , cos il = - , tan A —r Shc-v also that sin' A + cos^ A==\. Show also that (i) a=c . sin il, (ii) 6=c . sin B, (iii) a-c . cos jB, (iv) 6=c . cos A, (v) sin il =cos jB, (vi) cos il =sin 2?, (vii) tan A »cutJ7. (7) ITio sides of a right-angled triangle are in the ratio 5 : 12 : 13. Find the sine, cosine and tangent of each acute angle of the triangle. (8) The sides of a right-angled triangle are in the ratio 1:2: ,^3. Find the sine, cosine and tangent of each acute angle of the triangle. (9) Prove that if i4 be either of the angles of the above two triangles sin* A + cos^ -4 = 1. On (53) CHAPTER VI. Ox THE Trigonometeioal Ratios of Certain Angles. col laJ^ Trigonometrical Ratios of an angle are rm W oai quaniUus simply, as their name ratio implies. They are m nearly all cases incommensurable numbers. ^ Their practical value has been found for aJl andes between and 90», which differ by 1' • and a \i.7 * !? valufiH xinll \.^f J • ^ ' " * "^* o^ these values will be found m any volume of Mathematical Tables MathL?c«rT !:,"r!^«^^«^ *« ««^ - ^opy of Chamber.' Mathematical Tables for instruction and reference. 88. The finding the vaJues of these Ratios has involved a large amount of labour; but, as the results have be^ d'ot no! fL^^'^"' *'J "^"^^ ''' TrigonometriiTL^: does not form my part of a student's work, except to ex emplify the method employed. ^ Rj^\f^ ^"""^'"^^ """^^^^ ^^ fi^^g Trigonometrical be fouTd in r "^^ '''*"^^ ""«'^« "^««« R'^tio^ can oe lound in a simple manner. 90. 45". Tojlnd the sine, cosine and tangmt of an angle of theLfTf r^\f ^ right-angled triangle be i5", that is, on. c^gxc ux uii isosceles right-angled triangle. i i ■ 54 TRIGONOMETRY. I: A >>. m m M Let Let POM be an isosceles triangle such that PMO is a right angle, and OM = MP. Then POM = 0PM = 45". Let the measures of 0}f and of MP each be m. the measure of OP be rB. Then x' = m' + m' = 2m* ; .'. x= J2. m. Hence, sin 45" = sin POJI/ = 5? = "^ - _1 OP J2.m~J2' OM _ m _ 1 OP ~J27^-J2' cos id" = cos POM = tan 45" = tan POM^. ^' = !^* = i OJf m 1 1. 9L Tojimd the sme^ cosine and tangent of 60°. In an equilateral triangle, each of the equal angles ia 60", because they are each one third of 180". And if we draw a perpendicular from one of the angular points of the triangle to the opposite side, we get a right-angled triangle in which one angle is 60". Let OPQ be an equilateral triangle. Draw PM per- pendicular to OQ. Then OQ is bisected in M. Let the measure of OM he m; then that of OQ is 2m and therefore that of OP is 2m. Let THIQONOMETRWAL RATIOS OP CERTAIN AmiES. 65 i Then Hence, cos 60" = cos POM= ~- ^ OP ~ 2^1 1 2' tan 60« = tan POJ/ = ^ = n/1:J» ^ x/3 _ OM m ] ~ V ^' 92. ro/Tk/ ^/ifi «tMe, cosine and tangent of 30° With the same figure and construction as above, we have the angle OPJf= 30», since it i« a half of OPQ, i. e. of 60° Hence, sin 30° = sin 0P3f = — ^ _^ _ 1 PO 2m ~ 2' cos 30° - cos 0PM = ^'^ = \/3->^ _ >/3 /*0 2m ~"2"' . tun30°=.tanOi^Jf ^^ "^ _ 1 its I 50 TRIOONOMETR?, •*93. To Jindlli«8in« of IS*. In the 10th Prop, of EucUd IV. a triangle is descrihed auch that each of the angles at the base is double of the third angle. [See also Example (6) p. 140.1 Let POQ be such a triangle, and let the vertical angle I*OQ contain n degrees ; then n + 2n4-2w=180, w = 36. Draw OM perpendicular to FQ, bisecting the angle POQ. Then since ^0P = 36", MOP= 18". Also PM= MQ. Let the measure of MP be m, and the measure of OP be 00. From OP cut off OH = PQ. Then by Euclid IV. 10 PO.PR=P^. .'. x{x-2m)=={2my, .'. x'- 2mx + m' = 4m' + m' = om-y .'. x-m= J5.m, .'. x= J5 . m + m, m 1 J5-1 .'. mi\8' = 8m MOP = —=. OP J5.m + m J5 + 1 TRIQONOMETRJCAL JUTJOS OF CERTAIN ANOlEa, 57 94. Tojmd th« nne, cosine and tangent o/O". By this is me^nt,~ToJind the values, if any, ta which ths Tr^ancmetncai Ratios of a very small angle a^^oach, as tlie angle is continually diminished. — t'* 6 Let nor he a small angle. Draw PAf perpendicular to O/i, and lot or be always of the sumo length, so that P lies on a circle whose centre is 0. Then if the angle HOP be diminished, we can see that MP is diminished also, and that consequently ^ which 'ro^^ T' ^f ^"^^«^*^^- ^d' by diminishing^h'e angle ^OP sufficiently, we can make MP as smaU as we please and therefore we can make sin HOP smaller t/ian way assign- able number however small tJuit number may be. Thus we see that the value to which sin /?C>/^ ap- proaches as the aiigle is diminished, is 0. This is expressed by saying, sin 0" = 0. . . . i Ag«n, as the angle HOP diminishes, OM apj,, .aches OP in lengll.; and cos ROP, which is ^^, .pp,„„h,, ^ OP valueto^,i.e. to 1. This is expressed by saying, cosO"= 1 Also, tan ROP is -^^ - and we have seen that MP ap- preaches 0, while OM does not ; .-. tan i?OP approaches 0. This is expressed by saying, tanO" = i{{ UJ. 58 TRIGONOMETRY. 96. To find the sine, cosine and tangent of 90". B7 this is mea,nt,^Tofind the values, if any, to which tJie Trigonometrical Ratios of an angle apjytoach, as tlie angle approaches a rigJu angle. Let HOUhesk right angle = 90". Draw HOP nearly a right angle; draw PM perpen- dicular to OP, and let OP be always of the same length, so that P lies on a circle whose centre is 0. Then, as the angle POP approaches to POU, we can see that MP approaches OP, while Oi/ continually diminishes. Hence when POP approaches 90", sin POP which is — approaches in value to ^, that is to 1 , i.e. to 1. Hence we say, that sin 90" = 1 Again, when POP approaches 90", cos POP, which is OM Q ^ , ai)proaches in value to — , that is to 0. Hence we say, that cos 90" = .. . ;: TRIGONOMETRICAL RATIOS OF CERTAIN ANGLES. 59 Again, when ROP approaches 90», tan ROP which is ^ OM approaches in value to ^^ a quantity which approaches * But in any fraction whose numer...or does not diminish, the smaller the denominator the greater is the value of that v2 f ;w *^' ^«^«"^i"-tor continually diminishes the value of the fraction continually increases. Hence, tan ROP can be made carger than any a^sign^U This is what we mean when we say, that tan 90° is infinity, or, tan 90°Loo m. 96. The following table exhibits the results of this Chapter. angle 0" 18» 300 450 600 900 sine 4 1 2 1 1 V2 V3. 2 1 2 1 00 cosine 1 V3 2 tangent 1 v/3 1 V3 an.i?lvwr* """^ ^'*^'' *^^^ *^^ ^^°^ '"^^^^^^^^ With the angle, while the cosme diminishes as the angle increases. -Also that the squares of the sines of OO, 300, 450 60" and 90" 7. "^I^tively 0, I I J and ^, and that the squares of the cofiines of the same ungles are |, J, ^ |, and 0. ^;h i#k 60 TRIQONOMETRY. i EXAMPLEa xm. If ^ = 900, B=m\ C=300, Z)=450, ^=180, prove the following : (1) 2.8m2).cosi)-8mA (2) 2. sin C. ' cos C- sin A (3) co825-sin«J5=l-28in»A (4) sini?.cosC+8inC.coa5=«sin^. (6) co8«2)-sin2 2)=cos^. (6) 4.sin«^+2.sin^=l (7) sin»5+coB«5=l. (8) cos2(7+8in*C=l (9) cos«2) + sin»Z)=l. (10) sin B cos C- sin C. cos i5=sin C. (11) 2(cos5.co8Z) + sin5.sini?)a=-l+cosa (12) 2 (sin Z) . (ios C- sin C. cos 2))2= i - cos G. (13) sin300=-6. (14) sin 450= -7071. ... (15) sin600 = -8660.... (16) tan 600=1.7320508.... (17) tan 300 = -5773.... ^^g^ sin 180= -3090... 97. The actual measurement of the line joining two points which are any considerable distance apart, is a very tedious and difficult operation, especially when great accu- racy is required; while the accurate measurement of an ayle can, with proper instruments, be made w tii compara- tive ease and quickness. 98. A Sextant is an instrument for measuring the angle between the two lines drawn from the observer's eye to each of two distant objects respectively. A TheodoUte is an instrument for measuring angles in a horizontal plane; also for measuring ^angles of elevation^ and ^angles 0/ depression.' 99. The angle made with the horizontal plane, by the Kne joining the obsei-ver's eye with a distant object is called ' TRIOONOMETRWAL RATIOS OP CERTAIN ANGLES. 61 (ii) its angle of elevation, when the object is ahove the observer : its aagle of depression, when the object is behm tne observer, f 100. Trigonometry enables us by measuring ceri^ain a^/.Mo deduce, from one known distance, the lengths of other diBtances: or, by the measurement oi ^ cQ75^18 _ 100759 25 -0723100 - 723r * . ^ 2518975 ,,„ • • ^''"7231" ==3^S "early. Therefore the cliff is 348 feet high. 63 EXAMPLES, xnr. The answers are given correct to three significant figures. . iL^\t "^^K ^^V^'^ ^ * ^""^°°*^1 ^i"« fro°» the foot of a column the angle of elevation of the top of the column is observed to be 450. What is the height of the column ? (2) At a point 200 feet from, and on a level with the base of to be 60» : what is the height of the tow n- ? (3) From the top of a vertical cliff, the angle of denression of a pomt on the shore 150 feet from the base ofthe cU^Tr served to be 30" : find the height of the cliff. ' (4) lYom the top of a tower 117 feet high the angle of de- hTH't e^^^^^^^^^^^ ';^ V-' ^^«^ is observed! b^'3^" now lar is the top of the house from the tower? ■ J 64 TRiaONOMETRY. XlY. (6) A man Hit high stands at a dist^^aco of 4 ft. 9iu V,m a Imp-post, and it is obnervod that his .i.adow h 19 ft* k.^ Fmd the height of the lamp. ^ ^ ^' 100 ft lol^' 'YZ ^^ ^ ^'' '° *^" ^"""Sht is observed to he 100ft. long, and at tao same tioie tha shadow of a hmv-mit Off h.gh ,s observed to be 3^3ft. Ic.,. i^ind ,,. ,,„« of ^I I'on of the sun, aud the height of the tow«r. ^ M 4? n 1 th« othor baak, a man walks at right angles to PQ to ^^'tV'l^'^/J^'^y^l he then^bserl the a^^ P^i(i: < l)» .i2o 17' : fmd the breadth of f l.a river. tan32«17' = -63176(>.. (8) A flagstaft^25 feet high stands ca the top of a house- ^m a pomt on the plain on which the hou.e stands the angled of elevatzon of the top and bottom of the ll.gstaff are observed rhetrt:fobrrs:^^^^^ " ^^^^ ^'r '"''' -' ''- '--- ^'-^ (9) From the top of a cliff 100 feet high, the p^gles of depres sionotwoshipsatseaareobservedtobeyo,,d300resw7: Id ir.' ''r"^'^; '^^' ^'^"^ ^""''^^ "^ *h« fo«t of the cliff; find the distance between the ships. ' (10) A tower 100 feet high stands on the top of a cliff; from a pomt on the sand at the foot of the cliff the angles of elevS of the top and bottom of the tower are observed to be 760 and 600 respectively; find the height of the cliff. (Tan 750=2+^) mi^^P ^ T" ""^^ *^°°S * «*'*^«^* ^^ observes at one ^ t"d t W T."" * f'f '" "^^°^ ^ ^Sle 300 ^th the road and that at the next milestone the an-U is 600 • how far is the house from the road ? (12) A man stands at a point J o-tht .nk ^JS of a straight mer and^ ,rves that the line joi> ^ , to a post C on Se opposite bu.^ makes with AB an au^. - of 300. He then goes 400 yanis along the bank to 5 and find : t i HO makes with^ Hu angle of 600; find the breadth of tha^riv Ttm^i^ irom BXAMPLB8. XIV. 65 an angle whose tangent is v^. Show that /)5=a yards. are observed to be 4<^o i r a«j ^.-^o i «/ * ®^ *^^ *"" of the hilL "^ ^ ^^ ^^ ^^ 47012' respectively; find the height tan 45«13'« 1-0075918, tan 47» 12'= 10799018. be respectively N.W and N iTI^f ti . 7- ' ^ ** "^'nogsto deterXe the'height omf Moil ""''"" "" ' "^^ *""'• the baUoon. « is 40 . Find the height of (17) An isosceles triangle of woo/1 ;« «!„ j triangle, 4 ito height m^Iw .t^.-. . ** *"" *«« "^ *» tangfntorha. J^g^^I l^.trs^t' '"^ "^ "^ lower exuemity in a ve^«a'- u" t ^ t^ TU"' ■"""' ''^ in the «me direction, find wlit^ h.^ ?^°" " »'™^» «». the h...n ^Len Zr^^^^r^T^: ^Z Z'Z III ^• ■ (' III if liif (66) CHAPTER VII. On the Relations between the Trigonometrical Ratios of the Same Angle. 102. The following relations are evident from the definitions : cosec^ = ^— 2, sece = — L, sm e* cos tf ' 103. To prove tan^ = ?i^. cot^ = 1 tan $' B*»t We have sin ^ = P?P£I^^lS!^ hypotenuse ' and cos = base hypotenuse ' . sin $ _ pei-pendicular 104. We may prove similarly oot = ^?^ . Or thus, cot 1 cos tan ^ ~ sin 6' G TmoomMBT^rct mtwb op tub sA^fs ,mu. 07 * 105. Euclid I. 47 irivea ni. f»,o* • triangle the square on ZX^^L^_Z^'''-?1^ squares on the perpendicular rdo::he"bl:, "" ^^ ''' or, (hypotenuse)' = (perpendicular)' + (base)'. (i) Divide each side of this identity by (hypotenuse)', and we get f^^:P2)5!^l^V=./Perpendicularv / base \^ Wpotonuse; [h^^^t^^^)-{i;^^^^. t^atis, l-sin'^ + cos'^. ('^"-^)'=(^^^i^)%(-)'. that is, sec' ^ = tan' +1. (iii) Di^dde each side of the same identity by (perpendicular)', and we get /Jiypotenuse_y /pe^^ Wpendxcula,; (p-^^?iSdi^) ^ (^,^^ *^'*'^"' cosec'^ = cot'^4-l. * 106. Thus the three results (i) cos'tf + sin'd=l (ii) l + tan'^ = sec'd (iii) l + cot'^==cosec'^ aiv each a statement in Triannnnwri^-i i-- —- o — ^Ji^^nvai language 01 iiuc. 1. 47. 6—2 i ^ 1 1 ; ( iWR W TRiaONOMETRY. 107. We give the above proof in a different form. To prove that eoa* + sin' = 1. Let JiOJS be anv o^r''- *> In OB take any point P, and draw PM perpciifHcular to on. Then with respect to 6, MP is the pei-pendicular, OP is the hypotenuse, and OM is the base ; . ,. MP' cosd=-^-^ We have to prove that sin' $ + cos* 0=1. Now Mf^ + OM' OP^ uP' ~ ^' since by Euclid I. 1. , MP"^ 0M*= OP*, Therefore cos'd + sin'd = 1. Similarly we may prove that l + tan"^ = 8ec«^, ^^ t^-'ifc 1 + cot"^ = cosec'A mao^oMsmc, j,,rm of the s.mb .^lb. m sin 1 "cos^* 1 sin $ coaec $ = --^_, . 860^ i&ixO^ ootO coaO* COS0 sin^' 8in»tf + cos«tfs:l, cot«^+l=.cosec*A oonleLtVr'''^ Trigonometrical identities it is often r.r/r^lirLt;'^^^' THgonometH^Hatlos .. Since tan^=?!Hj^ cot v< <^^ ^ '^°^-c3b^<^<'o«eo^-^-ji^, ?in^ cosil 1 we have to prove that sin J or that °^^"^S:3"c-^-^' "c^^.siHX'^dos^.sin^' ADU this 18 tniO Ka^n,. ..J_a J - - J — ^-w-iOB 5iii-^ -1-008=^5=2, I 1 ■ I •1'., H I ff' ■ • f ^H ■ ^^^■9 ij : 70 TRIGONOMETRY, 110. Sometimes it is more convenient to express all the other Trigonometrical Ratios in terms of the Hne only or in terms of the cosine only. Example {i). Pr^), flin). ' (18) cos3tf-8in3tf-(co8d-8in5)(H.8indcostf). (19) sine ^ + co8« tf = 1 - 3 ^.^g ^ ^^^^ ^ (20) (8in» e - cose 6) ^ (2 sin^ ^ - 1) (1 - sin^ e + sin* 6). ,g,v tan^ + tanJ5 ^ , *^^^^ c"50Tccti?=*^^-^"^- /ggx cot a -I- tan /8 (23) (24) (26) V («6) 1 - sin 4 , 1 4" cos ^ ir^3^=(cosecJ+cot^)«. 2 versin d - versin* ^ = sin* A verain ^ (1 + oos ^) = sin« 0. I 72 TRIGONOMETRY. .J^^'a ^\ *^' ^"«^^«°^«*"c*l Ratios of an angle can be expressed in terms of any one of them. 5 ««i oe angle fr^';.l" r^ ** **- ^^^'-""^ ""^ «/ an i ./TV Let i20^ be any angle A. We can take P anywhere in the line OE-, so that we can make on. of the lines, OP, OM, or ifi> any len^h we pleaTe Let us take OP so that its measure is 1, and let , be the measure of i^/>; so that sin J, which is ^, =J; or,,=sin^. Let X be the measure of OM. Then since Oi/"2= OP^ - ifi>2 Hence and so on. Note. cos^ = ^ = i OM sfl -_f__j tan^=IS = OM ^l-«2 VlT^i^ sin^il, siuil The solution of the equation a?2=i -^, giygg and therefore the ambiguity (*) must stand before each of the root symbols m the above. This ambiguity, as will be explained later on, is of great use when the magnitude of the angle is not limited When we limit A to be less than a right angle we have no use for the negative sign. and so TRWO^-OMETUWAL J>.TJOS OP THE SAME ^OLE. 73 In this case ia.nd=^^. Take P so that the measure of OJf is 1, and let t be the meaau. of MP; so that tan., which is f|. .[; or, .=tan.. Then we can show that the measure of OP is >fTTfi. Hence, sin e=~= ^ ^ tan tf and so on. 112. The same results may be obtained by the use of the formulfle on page 69. .*. COs2.= l.giii2^^ ,*. cos.=s>/i -sin^.. Again and tantf=!ii^= COS. sin. so on. v^l-sm*.' ■~T^: \l 1^ ' I I' m 74 TRIOONOMETRT. m EXAMPLES. XVI. (1) Express all the other Trigonometrical Ratios of terms of cos J. (2) Express all the other Trigonometrical Ratios of A in terms of cot A. (3) Express all the other Trigonometrical Ratios of A in terms of sec A, (4) Express all the other Trigonometrical Ratios of A in terms of cosec A. (5) Use the formulce of page 69 to express all the other Trigonometrical Ratios of A in terms of sin A. (6) Use the formulae of page 69 to express aH the other Trigonometrical Riatios of A in terms of the tan A. 113. Given one of the Trigonometrical Ratios of an angle leas than a right angle, we can find all the otii&rs. Since all the Trigonometrical Ratios of an angle can be expressed in terms of any one of them, it is clear that if the numerical value of any one of them be given, the numerical value of all the rest can be found. Example. Given sintf=f, find the other Trigonometrical Ratios of (9. Let ROE be the angle 6. Take P on OE so that the measure of OP is 4. Draw PM perpendicular to OR. .(6) •7 M hosuXk. m m Then since sin ^=jYso that ^=a\ ,„. . ,, Itt ,; ?r " ''' °^"^^"^« °^ ^^ --t be 3 I^et a; be the measure of OM- ••• ^'=42-32=16-9=7. Therefore the measure of OJ^ is ^7. ^'^"^e; cos(?=^-.n/7 tan(9=:^«J ^3^7 <5-^ %/7 ~7"' cot ^=^^ 3 • IS (1) -(2) (3) (4) (5) . (6) « (7) (8) EXAMPLES. XVII. Given that sin^ = |, fin, tan^ and coaecX Given that cosi?=j,fi,d sin i? and coti?. Given that tan^=|,fi,,,i,^^,^^ Given that sec^=4, find cot^andsin^. Given that tan ^=^3, find sin ^ and cos A Given that cot ^=-|,fi,d sin ^ and sec ^. (9) -<10) (") hwndk. Given that sin 61=^, find tan ^. amnthatta.«=|.fl„dsm«andoo,A Given that oos«=i,fi„asm«andcotA «, ana tat. '^^ 6, prove that (1 - a^) n + A2^ _ , "°-^=^'-^'»^=^.«-'othoo,uatJ!i::ii 76 TRIGONOMETRY. 114. The following propositions are important. PROP. I. To trace the change* in the magnitude of sin A as A increases from 0^ to 90". Take a line OiJ, of any length; and describe the quadrant RFU of the circle whose centre is and radius OR. Draw the right angle ROUy cutting the circle in U. Let OP make any angle ROP {=>A) with OR ; draw PM per- pendicular to OR. MP Then 8in^ = OP' When the angle A is O", MP is zero, and when A is 90^, MP is equal to OP ; and as A continuously increases from 0® to 90", MP increases continuously from zero to OP ; also OP is always equal to OR. Therefore, when ^=00, the fraction ~ is equal to ~, that is 0; when A^^Qf> the fraction ^ is equal to ^, that is 1; and as A continuously increases from 0° to 90", the numerator of MP the fraction ^ continuously increases from zero to OP, while the denominator is imchanged, and therefore the fraction ~ , OP which is sin A, increases continuously from to.l. TRIC PE a»A« Wii we hav Wh equal t( 900, i/'i Wh< Oi/' is OM con Heni 0; wher finity' (a 90", the while th( so that t from u (1) £ cos^ con (2) l from k (3) 1^ ishss from (4) T from to TBIGONOMETEICAL RATIOS OF THE SAME ANGLE. 77 PROP. II To trace th^ changes in th^ magnUude of tan A we wl*" *■'' "'""' ~"**™°«™ '»'' fis^ " in the tot artMei When the angle ^ is 0». J^/> ia zero: when A ia ano «» • Tub 'I:^ ■" ',■:• '"^"' -«-o-'7^-^ ?1 of;' yo", ifP increases continuously from zero to OP. When the angle A is QO, OJf is equal to OP; when ^ is OQO ^^oonr°^ r'.^ ^ continuousi; increases from Vtlgoo' OM continuously decreases from OP to zero. ' Hence, when A is Qo the fraction ^ is equal to ^, that is 0; when A is 900 the fraction ^ is equal to f , that is e circle whose centre is mei^ureTf J^'ti''*''''""™"" *° "^ '^''^ 0^- that the Man «m-^, ^ would be outadl 2 "T"u °^ °^ ^'^ ^'""^ p-K^ to oie would ::^^^:^itr''' "' '"' '"-' TZT^r' ""' ^^P«T-OA has been drawn whoEw sine I's 9. 80 TRIGONOMETRY. Example 2. To draw an angU whose cosine m b; where 6 is a» proper fraction and positive. Draw any line OR, and take OB so that its measure is unity. Describe the 4uadrant RP(7ot the circle whose centre is and radius OB. Take M in OR so that the measure of OM is b. OM will be less than OR because h is a proper fraction. /rru^T^^f perpendicular to OR cutting the quadrant in P. (The student will observe that if h were an improper fraction, M would be outside the circle altogether.) Then the angle BOP is the angle required. Forcosi2C?i> = ^=^»ft. Therefore an angle ROP has been described such that the cos/20P=6. Example 3. To draw an angle A such that tan^-o, where a IS any positive numerical quantity. TRl D] OM\a Dr measu angle] Fo] Th. of anoi right a Exa Exa\ Exar 118. tfio cosin [Thi« proved h liA Draw Then therefore Now. IhE. Tnrooi. .,.,TR,CAi nATios of tbe bamb Amis, ai oJ'i:!^' "■' ^^' -<"»>^e ^ in it so that the „e„u« of angle required. "°° *''* »"8'e ^^OP is the PortaniJOi._^_«„„ Thus an angle ROP has been d«wn so that tan iJOP^o. of aiij^^srztrrrxr t: "'"•''*»'»* ExampU 1. The complement of J is (900 - A). £:xample2. The complement of 1900 is (goo-iw- lOAO Forl900+(90'>-i900) = 900. Example Z, The complement of ^ is Z^? 5^\ Stt provKt] ^'^*^"^ '^ *'^ ^^^^^*^^« «^ ^' - will be If ^ be less than 90', let ROF be A (see lust figure). Draw PJ/ perpendicular to Oi?. llien since Pi^O = 90", therefore POM + OPJ/ - ono ,„. therefore OPJf = (90» - A). ~ " XT • MP Now,sm4 = _^ = oosOPJf=cos(90"-A • . e"" sm 0= - 2, or sm tf =^. The value - 2 is inadmissible for ain /J fi.^ +i, whose sine is numerioaUy gresti: tt„T(l^' jt^; " ™ ""«" .*. sintf«J. "** sin 300=^. .*. sin ^ersin 300. Therefore one angle which satisfies this Pnu«f.v« ^-^^ ^ -•- ««« 6—2 i'ij s^. ** Kt V^. ^-^ IMAGE EVALUATION TEST TARGET (MT-3) Co .V * ^ 1.0 £|iuiM u 1.1 S US 12.0 lies HI yg 114 11.6 i^^^^^^^E HHIlHIIH^B ^NI^^HHB o/M Hiobgraiiiic ^Sciences CarporaliQn ¥i^ <^ 23 WIST MAM STIMT WIISnR,N.Y. l4StO (7l«)t72-4S03 ^* TRIOONOMBTRY. Exam^ 3. To solve the equation cosec 6 - cot« tf + 1 —a We hare 1 + oot« tf-coseo» B. .'. cot'^o.oosec'^-1. and we get ooseotf-cosec3^+2-ia Let X stand for coaeo 0. Then a?-4;« + 2«=0. .-. x«-a? + J»+J + 2=»J, a?«=2, or*=! -1, .*. cosec ^«2, (I cosec30°a2, .*. cosec tf—coseo 300. therefore 30« is on« angle which satisfies the equation for 6. [Art 108] or Whence baft EXAMPLES. Find one angle which satisfies each of the following equations. / .(7) .(9) v^ sm6>--^2. 2costf>-sectf. 4oostf-3sec^s»0. 3sin^-2cos'^«0. 2cos^— V3oot^. tantf + 3oottfi-4. ^m (11) (13) 2Bin«^+^2oostf-2. (14) (16) 3tan«^-48in«tf-l. (16) (17) oos«^-V3co8tf + f-0. (18) C0B>tf+2sin>^~f8intf-0. y> (2) 4 sin tf- cosec ^. ;(8) 4sintf-3cosoctfa0. 3t&ntf>-icot<9. V28in^-.tantf. tantf>-3cottf. tantf+oottf»2. 48inStf+2sin^->l. 28in*tf+V88intf-S. MISOBLLANBOUa BXAMPLBS, 85 108] [ona. **MI80ELLANE073 l8TAin> T.T!<^ ttt . (1) Prove that 3 sin 60<» - 4 sin" 6(y>« 4 ooe" SQO - 3 oos 30». (2) Prove that tan30»(l+oos300+oo860»)-8m3(y>+8in60». (3) If 2 oo8« tf - 7 cos tf + 3 -0, show there is only one v&Iue of costf. (4) Find cos 9 from the equation 8 coa' tf - 8 oos tf + 1 -0. (5) Find sind from the equation 8sin'tf-108intf-i-3«0^ and prove that one value of tf is s . 6 (6) Findtantffroir the equation 12 tan" tf- 13 tan tf+3-a (7) If 3 coe« tf + 2 . ^3 . cos tf - 5^, show that there is only one value of cos ^y and that one value of tf is - . 6 (8) Prove that the value of sin* tf + ooe* tf + 2 . sin' tf . cos« tf is always the same. (9) Shnplifyoos*^ + 2.sinM.co8U. (10) Expresa sin* ^1 + cob« ii in terms of sin« A and powers of tan* A, ^ (11) Express 1 +tan* B in terms of cos B and its powera (12) Prove that ^4±??lf+ ?^5.^±?^-0 sm il - sm 5 ^ cosil - coe 5 (13) Express (sec il- tan il)«m terms of sin X (14) Trace the changes in the magnitude of cosec tf as tf in> creases from to ^. (16) Trace the changes in the magnitude of cot^ as ^ d»- oreaaes from ^ to 0. 2 Solve Bin(^+^)-^,cos((9-,^)=^. '51 I' I ii I* ( 86 ) //CHAPTER VIIL Oir THE Use op the Signs + and -. 120. The student is probably aware that, in the appli- cation of Algebra to Problems concerning distance, we some- times find that the solution of an equation gives the measure of a distance with the sign - before it. Example. Let M,N,Ohe places in a straight line; let the distance from if to iV be 3 miles, and the distance from iV to 0, 3 miles. t 4. 4L One man A starting from M, rides towards at the rat 10 miles an hour, while another man B starting simultaneo^^i / from iV, walks towards at the rate of 4 miles an hour ; If § be the point at which they meet, how far is Q beyond 1 Let P be any point beyond 0, and let x be the number of miles in OP. We wish to find ar, i. e. the measure of OP^ so that P may coincide with §, the point at which A overtakes B. When A arrives at i>, he has ridden 6+ a? miles. The time occupied at the rate of 10 miles an hour is —^ hours. When B arrives at P, he has walked Z + x miles. The time oocupied at the rate of 4 miles an hour is ^-^ hours. USE OP THE SIONS + AND - 87 When P is the point at which they meet, these times are equal, so that 6 + j? 3+J7 , -7q- ■ ~j— i whence *=■ - 1. Thus the required number of miles has the sign - before it; and we have failed to find a point beyond at which A over- takes B. 121. Such a result can generally be interpreted by altering the statement of the problem, thus : -- The line whose measure appears with the sign - before it, must be considered aa drawn in the direction opposite to that in which it was drawn in the first statement of the problem. ■4—^ •H Example. Taking the former example, let us alter the ques- tion as follows : If ^ be the point at which A overtakes B, how far is to the left of 0? Let i> be any point totheleft of 0, and let * be the number of miles in OP. We wish to find x (i.e. the measiu-e of OP), so that P may coincide with $, the point at which A overtakes B. When A arrives at P, he has ridden Q-x miles. When B arrives at P, he has walked 3-x miles. Proceeding as before, we get 6-j? 3-x -Jo" ™ ~4~ » whence a?=. + 1. Therefore if P is to coincide with Q (the point at which A overtakes /i\. OP mUBf ha nne. m^la *^ *l~ f.-e. -f r\ i ■jsaBan^Basii^. TRiaONOMBTRY. 122. The conaidferation of such examples as the above has suggested, that the sign - may be made use of, iu the application of Algebra to Oeometiy, to represent a direction exactly opposite to that represented by the sign +. Accordingly the following Rule, or Convention, has been made. RULR Any straight line AB being given, then Unes drawn paraUel UiABin one direction shall be posi- Mye; that is, shaU be represented cUgebraicaUy by their mea- sures with t/ie sign + be/ore them : lines drawn parallel to ^^4 in the opposite direction shall be negatiye; that is, shall be represented algebraically by their measures vjiUi the sign - be/ore them. 123. We may choose for the positive direction in each case that direction which is most convenient. Example. Let Zi2 be a straight line parallel to the printed Imes in the page, L i SL^ and let Unes drawn in the direction from Z to i2 in the figure that IS, from the left-hand towards the right, be considered pasttiv^ Then by the above rule, lines drawn in the direction from BtoL, that is, from right to left, must be rusgative. 124. In naming a line by the letters at its extremities, we can mdicate by the order of the letters, the direction in which the line is supposed to be drawn. Example, Let and P be two pointe in the line LR as in ttie figure, and let the measure of the distance between them be a. Then OP, le. the line drawn ftY)m to P, which is in the positive diieotion, is represented algebraically by +a. While />0,ie. the line drawn from P to 0, which is in the negative direction, :s represented algebraically by - a. ■? U8B OP TUB SI0N8 + AND -, 89 1 26. Hence the o letters at its extremities to represent a line, the student will find it advantageous always to paj careful attention to the order of the letters. £xampU. Let LRhe& straight line parallel to the printed linos m the page. Let J, if, C,D,Ehe points in X72, such that the measures of AB, BC, CD, DE, ore 1, 2, 3, 4 respectively. Find the algebraical representation of ^ (i) AC^ CB (u) AD + DC- BC. r^—S- 2 S g (i) The algebraical representation of ^4 C is + 3, the algebruical representation of CB is - 2. Hence that of AC+ CB ia +3-2; that is, + 1 ♦. (ii) The algebraical representation of AD ib + 6, that of DC w -3, and that of ^Cis +2. , « w -c/o Therefore that of AD + DC- BC= 6-3-2=+ 1. This is equivalent to that of AB. EXAMPLES. XXn. In the above figure, find the algebraical representation of (1) AB+BC+CD. (3) BC + CD+DE+EC. (6) AD + DB+BE. (7) CD + DB + BK (2) AB+BC+CA. (4) AD-CD. (6) BC-AC+AD-BD. (8) CD-BD + BA+AC+CE. * By ^^+(7B (attention being paid to direction), we mean .flo from A to C and from C to B.' The result ia equivale^ ^bL^ from A and stonmna •♦ » t . „„^i_i„. . . J^ ^ starting ( 90 ) CHAPTER IX. On the Use of the Signs + and - in Trigonometry. 126. In Trig^onometry in order conveniently to treat of angles of arty magnitude, we proceed as follows. We take a fixed point 0, called the origin; and a fixed straight line OR, called the initial line. The angle of which we wish to treat is described by a Une OP, called the revolving line. This line OP starts from the initial line OR, and turns about through an angle ROP of any proposed magnitude into the position OP. 127. We have already said in Art 41 (i) that, when an angle ROP is described by OP turning about in the direction contrary to that of the hands of a waicK the angle ROP is said to be positive; that is, is re- presented algd)ra%caUy by its meanire mth the sign + btfore U. (ii) that, when an angle ROP is described by OP turning about in the same direction as the hands of a watch the angle is said to be negative; that is, is represmted \lge- bratcaUy by its measure with the sign - before it. V8B OP THE SIGNS + AND - IN TRIQONOMETRT. 91 Example. (180»-.^) iudicatos (i) the angle de8cribem the f! Or, (n) the angle described by OP turning about 0, from the poe,t.„n Oa, in tl>e positive direction untillt ha, de^nb^ «. angle of 180" (when it hae turned into the ,K»ition oT^^d the angle - A into the position OP. Or, (iii) the angle described by OP turning about from the position OR, in the negative direction through the angle - A and then turning back in the positive direction through the angle ISO", into the position OF, ^ The student should observe that in each of these three w ^ of regarding the angle (180»-^), the resulting angle ROP is .a:. EXAMPLES. XXm. Draw a figure giving the position of the revolving line after it haa turned through each of the following angles. (1) 2700. (2) 3700. (3) 4250 .(4) 690». (5) -300. .(6) -3300. (7) -4800. (8) -7600 X^) '^. (10) 2run + l. (11) (2n + l),r + ^. (12) (2n + l),r-|. (13) 2nK-|. ^14) (2n+l) ^-|. Note, nir always stands for a whole number right anglea numhAr f\f *■ WO OS TRIQONOMBTRY. 128. It ib often convenient to keep the revolving line of the same length. i In this case tlie point P lies always on the circumference of a circle whose centre is 0, Let this circle cut the lines LOR, UOD in the points Z, JP, CT, D respectively. The circle RULD is thus divided at the points R, U, Z, D into four Quadrants, of which RU is called the first Quadrant. UL is called the second Quadrant. LD is called the third Quadrant. DR is called the fourth Quadrant. Hence, in the figure, ROP^ is an angle of the first Quadrant ^^^t u „ second Quadrant. ^^^B „ „ third Quadrant. ^^^4 n ff fourth Quadrant line nee tits USE OF THE SIGNS + AND - IN TRIOONOMKTRY. 93 129. When we are given the initial and the final line of an angle, we can at once decide the Quadrant of which it is. We cannot however decide the mafl^ltude of the trigonometrical angle. For we do not know how many complete revelations the revoking line may have made. In other words, when the geometrical representation of an angle (consisting of the positions of the initial line and of the final line) is given, we are not (without further information) given the Trigonometrical angle. The Geometrical angle is here taken to be the amonnt of turning in the positive direction, which wiU transfer the revolving line from the initial to the final line. 130. Let il' be the magnitude of a given geometrical angle ROP ; then, every trigonometrical angle having OH for its initial line and OP for its final line, id included in the expression nx360* + A.*; or, 271x180' + A*, where n is some integer positive or negative. For nx86(y» hidioatea a number of eompUte revolutions of the revolving line. When we know the magnitud f the trigonometrical angle, then we know what integer n ie. ExampU. In the figure on p. 92 the geometrical angle ROP^ con- tahis 186°. This angle is the geometrical representation of the angles * 496«, 866°, -225°, etc., for 495°=s360°+185°; 865°«720°+185°; -226°= - 860°+ 186°. N.B. Let $ be the circular measure of a geometrical angle HOP; then, every trigonometrical angle represented by HOP, is included in the expression nx27r+0; or, 2n?r+ ^. f! i M TliiaONOMETRY. EXAMPLES. XXIV. Bute b which Quadrant the revolving line wUl be after dewriblng the follgwjpg angles: ^ (1) 120O. (2) 840». (3) 490<». (4) >100«». -880». (6) -lOOO". («) (^) V'-' (8) 10r+'^ 4 (9) 9r-Jr. (10) 2n,r-J. (11) (2n + l)r + ^. (12) n,r + J. Shew that the following angles are represented by the saiuo diagram geometrically : (13) 365%726', -356M086', -716*. (14) -6S*, 295', 666', -426'. (16) 190°, 659', 3799°, - 621', 7001*. (16) e, 4,r + tf, -(2,r-d), -4ir + 0. (17) ir-a, 3n--a, -ir-a, (2»+l)ir-a. (18) ir + a, 3n-+a, -»r + a, (2n 4-1) TT + a. 131. The principal directions of lines with which we are concerned in Trigonometry are as follows ; I. that parallel to 0/?, the initial line. OR is usually drawn from towards the right hand, paraUel to the printed lines of the page. II. The direction at right angles to OH. III. The direction parallel to the revolving line OP, U USE OF THE SIONS + AND - IN TMQONOMETRY. 95 Accordingly we make the following nilei ; I. For lines parallel to the initial line OR, the poii- tlve direction is from to R; consequently the negative direction is from RtoO, II. OU li the poiltion of the revolving line after it hai deicrlbed from OR a right angle about in the positive direction of revolution; and for lines perpendicular to OR, the positive direction is from toU; consequently for lines perpendicular to OR the negative direction is from to U to 0. i I ! III. Any line drawn parallel to the revolving line in the direction from to P is to be positive, and eonaequently any line drawn in the direction from P to is to be negative. Note. The student must notice that the revolving line OP earriet iU poiitive direction round with it, so that the line 'OP' is alwayi poritive. The revolving line, as it turns about 0, nowhere under|oea any sudden change of direction such as would be indicated by a change of sign. 132. We said, in Art. 81, that the definitions of the Trigonometrical Ratios (on pp. 46, 47), apply to angles of any magnitude. We have only to remark that it is gene- rally convenient to take F on the revolving line; that FM is drawn perpendicular to the other line produced {f necessary; and that the order of the letters in i/P, OP, OM is an essential part of the definition. * 3fP The order of the letters P, M, in the expressions —^ etc.| is therefora of great importance. 96 TRIGONOMETRY, 133. We proceed to show that the Trigonometrical Ratios o£ an angle vary in Sign according to the Quadrant in which the revolving line of the angle happens to be. From the definition we have, with the usual letters, MP OM Bin IiOP = ^p, COB ROP^-^^iB,uROP = MP OP Fig. J. I. When OP is in the first Quadrant (Fig. I.). MP is positive because from Jf to P is upwards (Rule II. p. &5.) OM is poffUiAje because from to if is twvwrda the right (Rule I. ) OP is positive (Rule III.) USE OP THE SIONS + AND - IN TRIQONOMETRY. 97 Hence, if A be any angle oi'iliQ Jurat Quadrant, MP sin Aj which is jrp , is positive ; A !-• 1- • OM . COS Af which IS jyp , IS positive ; MP tan il, which is ^^, is positive. II. When Oi* is in the second Quadrant, (Fig. ii.>. MP \a positive^ because fro:, if to i* is upwards, OM is negative, because from to Mia towards tlie left. OP is positive. Hence, if A be any angle of the second Quadrant, MP sin Af which is ^^ > is positive ; COS J, which is jjp . is negative; MP tan ^, which is ^j^ , is negative. ni. When OP is in the third Quadrant (Fig. in.) MP is negative, OM is n«yaves )ves 7 to OM lugh VSE OF THE SIGNS + AND - IN TRIQONOMETMY. 99 which they pass, as P moves tlirough the first Quadrant in the opposite direction from U to R. Example 1. Firui the sine, cosine and tangent of 1 20^ 120» is an angle of the second Quadrant. Let the angle HOP be ISO^ (Fig. ii. p. 96.) Then the angle POL= ISO" - 120*'=60*. Hence, sin 1200= ^^ =8in e(fi numencalli/, and in t^ie second Quadrant the sine is positive. Therefore 8inl20o=^ fi^ 2 w- Again, cos 120o=^=co8 600, numerically, and in the second Quadrant the cosine is negative. Therefore cos 120°= - - fii\ 2 v"/" Similarly, tan 120»=-^3 (iii). Example 2. Find the sine, cosine and tangent of 225". 2250 is an angle in the third Quadrant Let the angle POP be 2250 (Fig. in.) Here the angle POL—22b^- 1800=-460. Therefore the Trigonometrical Eatios of 225"= those of 45* numerically; and in the third Quadrant the sine and cosine are each negative and the timgent ia positive. Hence, sin 2250= - ;^ ; cos 2250 - 4 ; tan 2250-1. N.B. The cosecant, secant and cotangent of an angle A have the same sign as the sine, cosine, and tangent of A respectively. 7—2 100 TRIGONOMETRY. EXAMPLES. ZXVI. Find the algebraical value of the sine, cosine and tangent of the following angles : (1) 1500. (2) 1360. (3) -2400. (4) 3300. (6) - 460. (6) ^3000. (7) 3250. (8) -I350. (9) 3900. (10) 7600. (n) _84oo. (ig) 10200. (13) 2nn -|. (14) (2n+l)n-^. / (16) (2^-1)7^+^. 136. In Art. 133 we have shewn how the signs of the Trigonometrical Ratios change according to the Quadrant in which the angle is ; we proceed to consider the changes in the numerical value of these ratios of the angle ROF as the revolving line OP makes a complete revolution. Example i. To trace^ the changes in the sign and magnitude of sin A flw A increases from 0^ to 3600. In the figure, let OP be of constant length, so that as ROP increases from 0° to 360° P traces out the circumference of a circle. Let PM be drawn perpendicular to BfOR, and let r stand for the length of OP j then r is unchanged in magnitude and is positive. R Now sin ^ = — , and In the first Quadrant as .4 changes from 0® to QQO, MP is positive and increases from to r ; .*. sin A is positive and increases from to 1. In the second Quadrant as A changes from 90" to I8O0, MP is positive and decreases from r to j .*. sm A is positive and decreases from 1 to 0. VARIATION OF TRIGONOMETRICAL RATIOS. 100^ In the third Quadrant as A changes from 18(>> to 27O0 MP is negative and increases from to -r- .*. siu A is negative and increases from to - 1.* In the fourth Quadrant as A changes from 2700 to 3600 i^P is negative and decreases from -r to 0; .'. sin A is negative and decreases from - 1 to 0. Example IX. To trace the changes in the sign and magnUvde oj tan A as A. increases from 0« to 360°. In the figure, let OP be of constant length ; so that, as ROP mcreases from QO to 3600, P traces out a circle. Let PM be drawn perpendicular to ROR. Now tan ^ =.:^ , and In the first Quadrant as A changes from QO to 90° MP is positive and increases from to OP A CM is positive and decreases from OP to OJ ^^^^^^^^re tan A is positive and increases from >,^, or to ~ or « OP^ »"*""• In the second Quadrant as A changes from 900 to ISO^, MP is positive and decreases from OP to 0,) OM\B negative and increases from to OP J ^^^^^^^^ tan A is negative and changesf from _ ^, or - ra to — , or 0. In the third Quadrant as A changes from ISQO to 270», MP is negative and increases from to OP A OMiB negative and decreases from OP to OJ ^^^^^^^re tan A is positive and increases from 7^ , or to ~ or « OP ' Q > "* ''^ . In the fourth Quadrant as A changes from 27O0 to 360« MP is negative and decreases from - OP to 0,1 CM is positive and increases from to OP, J ^^^^^^^^ tan A is negative and changes t from - ^, or - 00 to -, or 0. Note. The same theory wiU be true of angles from 0» to 360o traced onfc hv the revolving line after any number of complete revdutions ^ Example. Trace the changes in sign and magnitude of sin A as th« an».lA a mcreases trom n times 860» to (n+1) times 86OV ^ himli«"to "w «/r««^ spcaKingi«o/,ia«.<«; so that tan 4 as ^ changes Xs im^!""*^' *"''*'''"'^*' '^*^P* *^«° " J-'^P" from 00 to - 00. CoL i I 100, TRIGONOMETRY. EXAMPLES. XXVn. 0» to SlV ""^^"^^ '° ^'^ '^"'^ magnitude aa ^ increases from (1) cos^. (5) cosecJ. (9) sin^+cosil. (2) tan^. (3) cot^. (4) seci. (6) l-sinJ. (7) sinM. (8) sin J. cos ^. (10) tan^+cotJ. (11) sin ^- cos X ♦It will be observed that the sign and magnitude of the Trigonometrical Ratios of an angle are always the same in the same Quadrant, and when an angle A, whatever its initial magnitude, had passed completely through each of the four quadrants and has thus made a complete revolution each of its ratios has passed through every possible value- and as the angle continues to increase, this passing through the same set of every possible values, is repeated for each complete revolution ; hence Each of the Trigonometrical Ratios is an expression which depends upon A and is such that its values are repeated m exactly the same order as A passes through each successive 360<>, or sin A is a periodic Function of A whose period is 3600. ]^ the figures given on p. 298 we exhibit sin^, cos^ and tan^ graphically as periodic functions of whose period is 2». The value of sin d, etc. is the perpendicular distance of a point on the curve from ttie honzontal line. The distance of the foot of this perpendicular from the fixed point giving the value of e. The student should notice that the curve of the cosine is the same curve as that of the sme ; it is shifted a horizontal distance of i, to the left. The curve of the cotangent is the curve of the tangent reversed. If the page is looked at in a looking-glass and the tangent curve shifted a horizontal distance nf i» to i\^Q ri^^ht «- a^f f^- - . 01 the cotangent. ( 101 ) CHAPTER X** On Angles Unlimited in Magnitude. 137. Just as the definitions of the Trigonometrical Ratios apply to angles of any magnitude whatever, so every general Formula involving these Ratios is true for angles of any magnitude whatever. -^ " It is most important that the student should examine for himself into the truth of this statement. 138. The formulas 1 sin^ ' cosec ii = secil = 1 COSil COtil = 1 tauii' are really definitions ; and since the definitions apply, there- fore these formulsB are true, whatever be the magnitude of A The formulae tan^ = siuil cot -4 = COSii cos -4' "" SlDil' are deduced immediately from the definitions, and therefore they are true whatever be the magnitude of A. 139. The formulae sin'^l + cos'il = 1, 1 + tanM = sec' -4, 1 + cot"^ = cosec' il, are each a trigonometrical statement of Euc. i. 47, and depend only on the fact that MP^ OMy and OP are the sides of a right-angled triangle. That this is the case, whatever be the laagnitude of the angle -4, is evident from the figures on page 96. To be oisitted on a first fcadiug, except pp. 104, 105. 102 IBWONOUETHr. HO. In Art. 118 we proved that the sine of an anirle 18 equal to the cosine of ite complement, provided the angle ml rr ° •'"•' '"'■ "^^ "»" «"™ -»" «-■»?•- oil metl^od of proving the truth of this and other like formute. Whatever be the magnitude of the angle concerned. to^t;:"!!: "■" '™" ""' "" •"" '' '•' "'•'''="- -•'« °/ That is, to prove sin A =oos (QO" - A) ^^ cos ^ = sin (900-^). drr;;^^^^^^^^^ ^^^ ^^^^^ ^ ^ ^^ «^^-^ ^-m or is tf In describing (900-^) we shall consider that OF starting from Oi? turns first through 90" into the position OV^^Tl turns back from OU through the angle UOF={-A) So that /2(9P, the angle which OP describes from OR is always equal to UOP', the angle which OP' de.onbes ^-om OU n the ojjposite direction. '" " "^^ '"^ /I n 01 01 or pr« ON ANGLES UNLIMITED IN MAGNITUDE, 103 Henco, N'P, that is 0M\ is aXma/t equal to MP in maqni. Also it will be seen that when P is abme LOR, P* is to the H^H of UOD\ when P is fte^aw; LOR, /" is to the fo/f of UOD. Hence, Oi/" and JfP have always the aaww «^n. Therefore f^ = ^ always. "'» sin ^ = cos (900 ^ ^j^ fo^ all values of ^. Again, 0N\ that is .V'i", is always equal to Oi/ in magnitude. And P' is above or below LOR according as P is to the right or to the left of UOD. So that M'P and OM have cdwayz the mrm sign. Therefore or, OM M'P QP " Qp> always, cos il -sin (900 - ^j fo^ ^11 values of A. EXAMPLES. XXVm. Prove, drawing a separate figure for each example, that (1) sin 300- cos 600. ^g) sin650=cos250. :^(3) sin 195n=oos(- 1050;. (4) cos 275 =sin (-1850). (5) cos (- 270) =sin 1170. <6) cos3000=sin(-2l00). If A,ByG\)Q the angles of a triangle, so that A-\-B+C= I800 prove * /Kx ^ . B+C (7) cos-2=sin-p:. 2 (8) cos-g=sm— 2 . (10) BiUg-cua— ^. , 104 ^"* t TRIOONOMETRY. .u ^*^' J^f' ^"^^ *"«^^ *''® ^^^ ^ be the Supplements fche one of the other when their sum is two right angles. Thus (1800 - ^) ia the supplement of A th^unl^ ^!^ *^' ?*^'* °^ * *"^8^*' (^+^+0-1800, «o tnat {B + C)iB the supplement of A. - ^^ran,^, 2. To prcyve that the sine of an angU^tlu^ nm of its Z^Uti "^ """ "^ '''' "'^^^ - ^'"^ '"''^ ^ ^ That ia, to p-oiw sin ^ ^sin (180" - A) ioBA--coa(18-A). frolo^ *r T"^^ ""~ ^^ ""• ^^- OP starting In desoribing (180»-4) we oonsido. that OP starting ftom ^n>^*r l!^K*^ T '"'° *• '""'*">■' <'^' '"'1 «">» •»* trom OL through the angle LOPz=.{- A). and or, or. So that ROP, the angle which OP describes from OA is always equal to LOF, the angle which OF describes from OL in the opposite direction. ON ANGLES UNLIMITED IN MAQNITUDE. Hence, MP and M'F are alwayt equal in magnitude. AIho, i> and T** are .ilways both above, or both below ZO/2 So that MP and J^'/" are a/««»y« of the tame ngn. 105 Therefore or. MP M'F Qp ■■ Qp always, sin 4 -sin ^180» - ^), for all values of A. Again, OM and OM' are alwayt equal in magnitude. Also it will be seen that when P is on the right of UOD, P is on the Uft of UOD\ when P is on the left of CTOZ). p is' on right of rOZ). So that Oi/" and Oif^' are always of ojt>po»i/« «^n. Therefore cos ii = - cos (1800 - A\ for all values of A. OM OM' , op" --^always, or, EXAMPLES. TCTrnr Prove, drawing a separate figure in each case, that (1) sin 6(y»=sin 1200. (2) sin 3400=sin ( - 1600),. (3) sin ^-400)= sin 2200. (4) cos 3200= -cos (-1400) (6) cos (-3800)= -cos 5600. (6) co8l950= -co8(-150). If Af B, C be the angles of a triangle, prove (1) 8inil=8in(5+C). (2) 6inC=8m(A + B). (3) oo8^=-cos(^ + (4) co8ii= cos(C+^). 106 TRIQOMOMKTRY. *• Ex + J). Am before, we take two revolving lines OP anrl 0/> nu starting from OR describes the angle ^. 0^ sLj-nl fC" ni describes the angle (I800+ J). ^ ' '**^'°» ^"^ ^^ In describing (1800 + J) we consider that 0/- starting from from OL through the angle ii. So that /2(?P, the angle which OP describes fi^m 0/2 i« i^Te^rl*'.^^'^' ^'^ ^"«^^ "*^°^ OP' ^^rZlo^'oL m <.iie «am« direction. ^^^ is «5oa;« LOB^ ;. ,, ,,,^ /'O'^ ; and moe versd*. ^^ So that MP and Jf '/" are alway, of ojt>po«V« ^^» Hon.., ^ = -^'^.w OP ~oF *'^*ya, sin ^ ^ - sin (1800 + ^), for all values of A Sim>.i4)j:, 0:d dwaya^^rM' in magnitude. ♦^ t^"^- "uV^ ' . '^^ ^^^ °'' ^ *^® "Sht of UOD according as /> is to the nght or to the left of UOD. ^ ^ ' This will be more clear if the student observes that POP' is always a straight line. *" or, or. or. 'OV ON ANOLb'S UNLfMITBD tN MAONITUDS. 107 Henoe, OM OP OM' , ■^p always, or. cos id - - COS (1800 + A\ for all values of A. - ExampU 4. To prove siii il — - cos (90" + A)y and cos 4 -sin (900 +ii). As before, we take two revolving lines OP and OF. OP starting from OR describes the angle A ; OF starting hom OR describes the angle 90° + A. In describing (90°+ J) we consider that OF starting from OR turns first through 900 into the position OU^ and then on from OU through the angle A. So that ROP^ the angle which OP describes from OB^ is always equal to UOF, the angle which OF describes from OU in the »ame direction. Hence» N'F^ that is 0M\ alwaya=MP in magnitude. Also, P' is to the left or to the right of UOD according as P is above or below LOR. , So that MP and OM' are always of opposite sign. MP OM' 0P~" Hence, 18 or. -Qpf always, sin ^= - cos (900+ ^), f^r ^\ values of A. 108 TRIOONOMJSTRY. Similarly, OM always^ M'P in magnitude. Md P' is above or below LOU according as /> is to the right or to the left of C^OD. * or. So that OM and M'P' are always of the same sign. Hence, cos^=8in (900 + ^), for all values of ^. OM M'F , OP^-qP always, MISOELLANEOUS EXAMPLES. XXX. Prove, drawing a separate figure in each case, that (1) sin 600= _ gin 3400. (g) sin 1700- - sin 3500 (3) sin ( - 200)- - sin I6OO. (4) cos 3800= - cos 66OO. (5) cos ( - 226)= - cos ( - 450). , (6) cos 10050= - cos 11850- (7) am 600= - cos 1500. (s) cos 600=sin 1500. (9) sin225= -oos3150. - (lo) cos ( - 600)=sin300. If ^ + 5+ C be the angles of a triangle, prove that (11) sin /J = - sin (2^ +5+0. (12) sin^=-cos^-^i±#±?. ^(13) cosi5=sin^t3^C' 2. (14) cos C= - cos (/I + fl + 2^. (15) cos^^sin^-;?^. y(16) sin ^=- cos ^^. / 11 ON ANGLES UNLIMITED IN MAGNITUDE. 109 Prove the following statements for all values of A and of a. (17) mnA^-Bm{-A). (18) cos^=cos(-^). (19) sm^=cos(il-900). (20) oosil= -sm(^-90»). • (21) sina-cos^^ + aV (22) cosa=-sm(^ + a) (23) sina=-oos(^-ay (24) cosa= - sinT-g - a] . ^-'- (25) sin(|-a)«sin(| + a). (26) cos (»r + a) = cos (tt - a). (27) tan (900-^)= cot A (28) tan4= -tan(- J). (29) tan(900 + .l)=-cot(il). (30) tana= -tan(,r-a). (31) tan^=tan(1800+il). (32) cot^| - a)=tana. 142. We have seen (Art. 135) that there are many angles of different magnitude which have the same sine. If two such angles are in the same Qttadrcmt they are represented geometrically by the same position of OF^ so that they differ by some multiple of four right angles. 143. If we are given the value of the sine of an angle it is important to be able to find, Geometrically and Alge- braically, all angles which have that value for their sina no TRIGONOMETRY. Jimnt*^t "'"^ ooaatructioa, let the radius of the circle SULD be the umt of lensth; then the measure of OpTi flg.7. I" Prom draw on OIT a line OJT so that its n,easure is a. (Fig^II.) from according as a is positive or negative.! Through ir draw f,P, parallel to 10£ to cut the c rcle mPandP.. Join Oi',, OP,. Draw P,if„ Pj/- pj pendicular to ZOJe. rhon M^P,=. ON = mJ>„ ""* O^ for OP. = :^ = « = ^.^._pe^.. .„ hyp- ' OP, 1 - 0.^ - Ei: *"»• ^'^t- Hence ««^ angle whose initial line is OR, and whose final hne « erther OP, or 0P„ is an angle whose sine isT ^d no other angle has its sine equal to a, for ther« \b no other possible position for N. 146. To find an ea^emon to iruilude all angles h.,A„„ me given value a for sine. With the usual construction, let the measure of the radius of the circle BUZD be 1 ; on OU take JT so that the measure of^s_a^JhroughJ^r^^ ^^ • If a were greater than unity, ON would be greater than OU and the construction would fail ' ^ 1! tl g« lei #^ i \ ON ANGLES UNLIMITED IN MAGNITUDE. Ill BOL ; join OP,, OP,. Then every angle whose initial line IS OH and final line, either OF,, or 0P„ and no other, is an angle whose sine is a. Let EOF^ contain a radians ; then FOP, = TT- a. ^^>.l. %• 2.,. T-, -r^ Every angle whose initial line is OR and final line OP is one of those included in the expression 2m7r + a, where m is some integer, positive or negative. [Art. 130.] Every angle whose initial line is OF and final line OP is one of those included in the expression 2r7r + TT - a, or (2r + 1) IT - a, where r is some integer, positive or negative*. [Art. 130.1 Both of these expressions are included in niT + i-iya, where n is some integer, positive or negative*. This is therefore the required expression. Example. Find six angles between -4 right angles and +8 right angles which satisfy the equation sinA^^sin 180. Wehave[^^=n7r + (-l)'»^,or1^o=wxl800+(-l)«180. Put for n the values -2, -1, Q, i, 2, 3, 4 successively and we get the six angles -3600+180, -1800-180, 180, I8OO-I8O 3600+180, 5400-180 i.e. -3420,-1980,180,1620, 3780, 6220. * ,_. * ?°^ if n be even, let it be 2m, when ( - 1)^= +1 : if „ h« nd^.. iesu oe :ar+l, when (-l)»ri-i~ . x^ - ' I. E. T. M ■?f ? 112 TRIGONOMETRY. The student is recommended to draw a figure in the above example. Also to draw a figure in each example of this kind which he works for exercise. EXAMPLES. XXXI. (1) Find the four smallest angles which satisfy the equations (i)sinJ=i. (ii)8in^=-^. (iii)sin^-^. (iv)8in^--i. (2) Find four angles between zero and +8 right angles which satisfy the equations (i) sin ^ -sin 200. (ii) sm wo«^., fu^ left according as a is positive or negative. ON ANGLES UNLIMITED IN MAGNITUDE. 113 Through M draw F,MP, perpendicular to OB to cut the circle in Pi, P,. Join OP,, OP,. Then, ~ ior OP -^^ « ^^ base^ ^„ ^yj>-""^^'-OPri=OPrh^.^''^^'' Hence every angle whose initial line is OH, and final line either 0P„ or 0P„ is an angle whose cosine is a. And no other angle has its cosine =«; for there is no other possible position of M. 147. To find ths complete Algebraical Solution of the equation cos = a. With the usual construction, let the radius of the circle MULD be 1; on OR^ take M so that the measure of OM is a; through M draw P^MP, parallel to UOD; join OP,, OP,. Then every angle whose initial line is OR and final line either 0P„ or OP,, and .io oP,, OP^. Then and base « OM - 1 ~ ^* Hence every angle whose initial line is OR, and final line either OP^ or 0P„ is an angle whose tangent is a. And no other angle has its tangent = «. 149. To find tJie complete Algebraical Solution of tho equation tan $ = a. Let OR he the initial line; from draw on OR two lines OM,, OM^ whose measures are + 1 and - 1 respec- tively; from M^, M, draw perpendicular to LOR Unes M^P^, M^P^ whose measures are a and - a respectively; join 0P„ OP,. Then every angle whose initial line is OR and final line either OP,, or 0P„ and no oiAer, is an angle whose tangent is a. ' of the easure final / iho I two 5spec- lines join ? and wgle ON ANGLES UNLIMITED IN MAGNITUDE. 115 Let ROP^ = a ; then, ROF^ = fl- + a. Every angle whose initial Hne is OR and final line OP IS one of those included in the expression " 2m7r + a. [Art. 130.] Every angle whose initial line is OR and final line OP , is one of those included in the expression *' 2rir + {7r + a); or, (2/- + 1) tt + a. [Art. 130.] Both of these expressions are included in Thus the solution of the equation tan tf = tan a is EXAMPLES. yyYTT (1) Write down the complete Algebraical Solution of each of the following equations : (i) cos^=i. (ii) tand=l. (iii) tantfc=-i. (iv)tan<9=-V3. (v) costf^cos^. (vi) tantf^tan??. (2) Show that each of the foUowing angles has the same oosme : -120°, 240», 4800, -480». (3) The angles 60" and - 1200 have one of the Trigono- metrical Ratios the same for both ; which of the ratios is it ? (4) Can the foUowing angles have any one of their Trigono- metrical Ratios the same for all i -23«, -1570 and 157<». (5) Find four angles which satisfy each of the equations in (1). • For if n be even, this is the first formula ; ii n be odd it is the 116 TRIGONOMETRY. I I 160. We can now point out the use of the ambiguous sign ± in the formula cos ^ = * ^F-sin'tf. If we know the numerical value of the sine of an angle 6, without knowing the magnitude of the angle we cannot from the identity, cos" ^ = 1 - sin" ^, completely ^ie- termine cos 0, for we get cos^ = ^Jl-sin'O, This is a general formula, and we shall find that it represents an important Geometrical truth. 151. Given sin^ = a, we can say that is one of the angles represented by one or the other of the positions OP,, OP^ of the revolving line in Fig. I. on page 110. K we attempt to find the cosine of these angles we get two different values for the cosine; for ^ and —• OP, OP, although equal in magnitude, are opposite in sign. Hence, if a be the least angle whose sine is equal to a, we have cos^ = ± cos a = ± y 1 - sin' a. 152. The same result may be obtained from the formula ^ = nTT + ( - l)-a. For cos {n,r + ( - l)-a} is of different sign according as w is even or odd. EXAMPLES. XXXin. (1) If 6 be found from the equation cos^^^a, show geo- metrically that there are two values of sin 6 and of tan 6. (2) If e be found from the equation tan^=a, show geo- metrically that there are two values of sin 6 and of cos 6. f 3) If A be the least angle without regard to sign such that sin A— a, show that cos ^ = + Vl - sin^ A. (4) If ^ be the leas t positive angle such that cos 4 = a, prove that sin ^ = + ^/i - cosM. ( 117 ) i CHAPTER XI. On the Trigonometrical Ratios of Two Angle& 153. We proceed to establish the following fundamental formulse : sin {A + B) = sin A . coa B + coa A . ain B' cos {A + B) = cos A.coaB-ainA .aiiiB sin {A—B) = smA.GoaB- cos ^ . sin ^ cos {A-B)= cos il . cos ^ + sin ^ . sin ^ (i). Here, A and B are angles; so that {A +B) and (A-B) are also angles. Hence, sin {A + B) is the sine of an angle, and must not be confounded with sin A + sin B. Sin (A +B) ia Si single fraction. Sin -4 + sin 5 is the sum of two fractions. 164. The proofs given in the next two pages are per- fectly general, an will be explained below (cf. Art 169) ; but the Jigures are drawn for the simplest case in each. The student should notice that the words of the twQ proo& are very nearly the same, 118 TRIOONOMETRY. and that and OE respectively. D^7v/\vf f "f ' ""^^^ *" ^^ Now sin {A+£)^ sin i20/'= :^ ^^+ ^P OP ^ OP — ' KN HP ■^SI + ^;|S-^ Ol^UP NP OP^lXp ON' OP'^'WP'OP oi^.op'^mrop =sin HOU. cos £OF+cos MPN.lin EOF =sin ^ . cos 5 + cos ^ . sin ^. __0£ HN OP OP " OP OK ON HN NP NP' OP Also cos {A + 5) =cos ROF^ OM OK- MK OP" ^OK.ON HN.NP ON. OP 'W^OP-ON'OP .rz- '-<^^IiOF.cosWF-BmHP,V.sin£OF -oosul.cofl^-sin^.sin^. an/nvP^"'' ^u^'^^ ^°^ '"""^^ ^^^^' ^'^^auBe the angles OMP and 02^P are nght angles j therefore MPN and J»/nV ! , in the same segment; so that MPN=M0N~4 ^^ at TRiQONOMETlUCAL RATIOS OP TWO ANGLES. 119 To prove thai m^ '4 - ^) = ain ^4 . cos 5 - cos Ji . sin B, and that of»{A- B) = Go&A.coiiB + miA .axuB. N / H .' / / / \f /.S ^s^^^ K ''M A Let ROE be the angle A, and FOE the angle B. Then in the figure, EOF is the angle {A - B). In OF, the line which houiida the compound angle {A - B), take any point P, and from P draw PM, PN at right angles t^ OR and OE respectively. Draw NH, NK at right angles to MP and OR respectively. Then the angle NPH=^Qf>-HNP=HNE=ROE=A*. mw,am(A-B)^H\n ffnp=^-.^'^^ KN PH 'on . JON OP ■■^uROF^^^MzIS OP OP PH.V^ KN ON ON' OP "P^ .OP _ _ -sin ROE . cos FOE- cos HPN . sin FOE =-'in^ . cos 5- cos ^ . sin J5. PH PN PN' OP Also, co3iA-B)=ooaROF^^^-2K±^^OENff OP OP ~OP^~OP ^OKAmNH_^^OK ON NH PN O^.Op-^m.Op-ONOP'^ PN'OP =co3 ROE . cos FOE+ain HPN . sin FOE =cos ^ . cos J?+sin ^ . sin B. A^?JJ!^''^' ^ "^""^^ ^°®^ '°°''^ ^^^^' because the angles OMP and OiyrP are right angles; therefore MPN and MON together make up two right angles ; so that HPN= MON =A, 120 TMIOONOMETR Y. Example. Ffnd the value of sin 75«. 8in750-8in(45<» + 3()'') -sin 46", coH 30" f coh 45». ain 300 In/3 1 1 3* 2 ■^72*2 _ V3 + 1 n/2(\/3 + 1) " "272" " 4 • EXAMPLES. XXXIV. (1) Show that cos 76<>-.^-lli V2 * (2) Show that sin 160= -^"i 2v/2 * (3) Show that cos 150= ^^itl (4) Show that tan 750 = 2 + \/3. for foi (i -t).'' ^^ '"^ "" '''^' '"' ' '"'"^ '^^ '^^^ (^ ^ ^) ^ "~ cos (^ + 0) ' tand + cot0 ,. ,. ,^ .. cot,^-tan2 ^ =cos (5 - <^) . sec (6 4- <^). cot^d + cot <^ _ sin {6 + ) ' tan ^ . cot + 1 __ sin (6 + 0) tan 6. cot (f>-i ~ sin {d - ) ' l+coty.tan8 ^ cotyltana '-^^(y-^^)' 1 - cot y . tan 8 m ^j^ i=>i?^ cot V + tan 5 =tan(y-d). 122 (18) (19) (20) (21) (22) TRIGONOMETRY. tan y . cot d - 1 taiiy + cota""*^(y-*)- tany.cot a+1 cotd-tany~== *=*"(>+*)• COtd-coty COty.cotd+1 =*»° (y - ^). tan»n-t.«.n2p^ sm(« +^). sip („-^) cos2a.cos''/3 cot2 a - tan2/3 = cos(a+/3).cosfa-fl^ ' i-tan^71S52^=*^(«+/S)-tan(a-^). (24) (25) (26) (27) (28) (29) (30) (31) (32) (33) (34) «in(« + /S).sm(a-^)=sm«a-8in2/3=COS«^-C082a. co8(a+/3).cos(a-^)=cos2«-sm2^=cos'»^-.sin2a. 8m(J-450)=.?^^~<^s^ (38) (36) (37) (38) (39) ,- V2 V2.sin(^+450)=sin^+co8^. cos ^ - sin ^ = \/2 . cos {A + 450). cos {A + 450) +sm (^ - 450) =0. cos {A - 45«)=sin (^ +460). sin(tf + 0).cos(9-cos(^ + 0).sin(9=ain0. sm(d-0).cos0 + cos(tf-0).sin0=sinA cos(d + <^).cos(9+sm(tf + 0).sin^=cos0. *ap(^-0)+tan d) l-tan(^-0).taii0=*^'^- tan(^+0)-tan<9 I + tan (d + 0) . tan (9 "= **^ *^- 2sin(„ + ?).cos(^-?)=eos(a-;3)+sin(a+^). 2sm(?-a).cos(?+^)=oos(a-^)-sin(a + /3). C08(a+/3) + 8m(a - /3) = 2sin (?+a) . cos (^+^). co8(a+/3)-.in(a-^)«28m(?-a).cos(?-^). EXAMPLES. XXXV. 123 (40) sin nA . cos ^ + cos n^l . sin J = sic (w + 1) il. (41) cos (n - 1) ^ . cos^ - sin {n-\)A. sin^=cos nA. (42) sin»4.cos(n-l)^-cosnil.8in(n-l)4=8in^. (43) cos (w - .1) il . cos (n + 1) ^ - sin (n - 1 ) ^ . sin (n + 1 ) ii =cos2nil. 166. The following formulae are important : . t , „. tan A + tan B ^ tan ( J + i?) = - — _ 1 — tan A . tan B 1 /A n\ ^^ ^ — tan B tan (A-B) = ^ ' 1+tan^.tan;?. * /A Kov tan A+ I tan (A + < 5") = :; , , ^ ' 1-tan^' (ii). tan(^-45") = tan ^ — 1 1 + tan A' The proof of the first is given below. The student should prove the second in a similar manner. The third and fourth follow at once from the first two by putting 45" for B. Example. To prove tan (A + B)^ J^A+i^nB^ 1 - tan A . tan B (i) By using the results of Art. 164, we have tan («4 + 5) = -~-(^ -f ^^ = sin A . cos .g+cos A . sin B coH(A+B)^cosA.coaB-smA.BmB' Divide the numerator and the denominator of this fraction each by cos ^ . cos J?, and we get sin A . cos BcosA. sin B »»^ /A , m _ ^^^ -A • co^ ^ cos ^ . cos B cos .4 . cos .g sin ^ . s in!5 cos -4. cos J? COS^.COSjB tan^l+tan^ 1-tan^ .taui? v; . Q.E.D. 124 TRIQONOMETR Y. (ii) • * By Geometry. Make the construction of page 118 ; Then tan (4 + i5)-tan ROF^^^^SM±El OM OK-MK OK^ OK "Ok ^J\t' OKZOK M.VLV 0K~ OK ^ - HPToZ Jan^^ Now the triangles NOK and NPE are similar, HP NP = 77Af=*an5, OK OW .'. i^^n(i + g)- tan^+tan^ i-tan^.tan^- ^•^•°- EXAMPLES. XXXVI. (2) Iftan^=l and tan ^=-L, prove that tan(^+J?) = 2+V3. (3) Prove that tan 15»=2 - ^3. (4) If tan^=f and tan5=^, prove that tanM+J5)-l What is {A-{-B) in this case? v^-rx»;=-i. ■1. EXAMPLES. XXXVI. 126 (5) If tan A =m and tan B=- , prove that tan (A-k-B)=ai What is (A + B) in this case ? Prove the following statements : (6) cotM+J?)=??Mv°ot^ri ' cot^+cot5 ' (7) cotM-^=°°*^-«<^*^+l (10) tan(<9-^)+cot(^ + ?)=o. (11) cot(d-?) + tan(^+?)=0. (12) If tan a=^j- and tan ^=~^ , prove that tan(a + /3)=l. .,„» tan ( n +1)0- ta n »<^ ^ ^ l+tan(n+l)^teirn^'"**"'^- (14) *«^(^+l)+tan(l-n)<^ ^ ^ i-tan(n+l)0.tan(l-n)0°°**°^*^- (15) If tan a=m and tan j8 = w, prove that cos(a+/3)=_i^^=. N/(l+«»»)(l+n«) (16) Iftana=(a+l)andtaii/3=(a-l),then2cot(a-/3)=a« (17) Ifa+^+y-900, then ^^^^l-tenotan^ tana+tan^ ' 126 TUlOONOAfETRY, 157. Fix)m pages 118 and 119 we have ^{^-B)^amA.coa£-.coaA silt 7i i^ + B) = coaA.coBB-ainA.Bmfi cos (^ - ^) = cos ^ . cos i9 + sin ^ . giu /; From these bv addition and subtractio (i). •n We get 8iu(^+^)4.8in(4-i?) = 28in^.cos7? sin(^ + ^)-8in(^-^)=2eos^.sini9 cos(^ + J5) + cos(^-^)=2cos^.cos^ cos(^-^)_cos(^+^) = 2sin^.sin^ Now put S for (A + £), and put T for (A-.£} Then S-^T=2A, and S-T = 2B, so that ^ = ^Jt^ and B== S-T Hence the above results may be written sin^ + sinr=2sin^±^. cos^?" "^ sin^-sinr=2cos^^ sin ^^IlT cos ^ + cos 7^= 2 cos ^±1, cos ^^^ *cosr-cos^=2sin^^ ain'^--^ 2 • ^*° ~2~- ■..(iii). TRIQONOMETRICAL RATIOS OP TWO A HOLES. 127 lfi8. The formula (iii) are most important, and the student 18 recommended to get thoroughly familiar with them %n vx)rd8, thus : (1) TJie Hum of the sines of two amgles is equal to ttvice the sine of half their sum multiplied hj the cosine ofhalfthdr difference. (2) The sine of the greater of two angles mmuS tJie sine of the lesser is equal to twice the cosine of hilf their sum multiplied hij the sine of half their dif- ference. ^3) The sum of the cosines of two angUs is equal to twice the cosine of Imlf their sum multiplied by the cosine ofhalftlieir difference. (4) The cosine of the lesser of two angles minus the cosine of the greater is equal to twice the sine of half their sum multiplied by the sine of half their difference. 159. It will be convenient to refer to the formul» (i) as the '4,5' formulae, and to the formula (iii) as the 'ST formulae. , ' 160. The 'ST formulae can be proved directly from a figure. ^ We give the proof for the case in which ^ and T are each less than 90". On first reading the subject however the student should omit the next two pages, and all other alternative proofs^ L. E. T. 9 1 28 TRIOONOMETR Y. **a) To piweehat sin S-\- sin r=2 sin ^. cos ^::ir. In the figure, let MO£l be the angle S, EOF the angle T\ then /^^^ is the angle 8- T. Let OG bisect the angle EOF. Then the angle FOG=^ = ^^' ^ Also the angle IiOO=ROF+FOG=T+^^ =~— 2 2 * In OB take any point P, and from OF cut off 0$ equal to OP. Join P^, cutting OG in Z. Then OG' wHch bisects the vertical angle of the isosceles triangle POQ, bisects PQ at right angles. Draw PM, KL, QN at right angles to OB^ and draw QVW and KH parallel to NLM. Then by parallels, sinca PK=KQ, therefore PH=HW and WV=rQ. Hence, sin^'+sin^=^ + ^-^^+^« 0^ (?^ _ (LE+ffP) + (ZK- WH) OQ 2LK OQ 2LK.OK '• OK OQ '^^ ®*^ ^^^ • ^°^ ^^^ =2 sm — =r— . cos 2 2 . Q. £. D. TRWONOMETRWAL RATIOS OF TWO ANGLES. 129 (2) To prove co8^+cos2^=2cos^. cos^::!' 2 2 * With the same figure and construction, we have cobS+cobT=^^a.9^^9E±2K J.OL-ML) + {OL+LN) _ 20/, OQ ~ - -QQ ' 20Z.0K „ „^ = 2 008/20^. cos §(?j5r, OK . OQ (3) T'o jprow ■ 200.^. e<«^. Q.E.D. Sin .y -sin ^=2 Sin ^. COS ^±^ 2 2 In the above figure the angles VKQ^^ROK bxb each the com- Hence plenaentofZA'O .-. VKQ=ROK=?±^ OP OQ 0Q~ •> jZK+ffP) - (ZK- WH) 2 VK OQ = ~0Q' _2r^.KQ „ ^ ~ KQ . OQ '"^ ^°^ ^^^ • sin ^OA', (4) To prove ^2^iz^,^S+T_ Q.E.D. oos T- cos ^=2 sin -^ . sin S+T . S-T With the same figure and construction, we have go^T-cobS=^^9M-.ON-OM OQ OP 0Q~~' _ {OL+LN)-(OJ ^-^ML) 2LN OQ ''-OQ' ^27^. KQ „ . *" KQTW^^'^ ^^^^ ' sin^QA; =2sin- 2 sm 2 Q.B.D. 9-2 130 TRIOONOMETRY. EXAMPLES. XXXVU. Prove the following statements : (1) sin 600 + gin gQo^ g sin 450 . cos 15<>. (2) 8iii60«+sin200«28in40».cos200 (3) sin400-8inl0<»=2 cos 260. gin 150^ (4) cos^+cos;=2cos^.cosJI 12 12 (5) cos|-cos|=2sing.sii§. (6) sin 3^ + sin 5^ = 2 sin AA .cgsA. (7) sin 7 A - sin 5A=2 cos 6 A .sin A. (8) cos 5A + cos 9^ = 2 cos 7 A . cos 2A. (9) cos5J-cos4^=-2siu^.sin-. 2 2 (10) cos^- cos 2.4 =2 Sin- ai) (12) (13) sin 2tf + sin ^ 30 costf + cos2^~" 2"' sin2tf-sin^ Sd costf-cos2tf~^®*l- sia3^ + sin2^ B cos 2^ -cos 3^ ""2* . A sm-. (14; sin tf +sin _ cos tf + cos <^ oostf-cos0 sin0-sin^* (15) oo8(600 + J) + cos(60<'-^)=cosA (16) cos(450 + ^) + cos(450-^)=^2.cos^. (17) sin(450+J)_sin(450-^)=^2.sinvl. (18) cos(300-J)-cos(300 + ^)=sin^. ^ ^ cos0-coatf^°*~2~- sin ^ - sin ^"^ 8ind + sin0-'^*V-2~/'*^"( 2 j- TRIQONOMBTRICAL RATIOS OP TWO ANGLES U\ 161. It is important that the student should be thoroughly familiar v/ith the second set of formulffl on p. 126. Written as follows, they may be regarded as the inverse oi the % T* formuJaj. 2 sin ^ . cosif = sin (^ + 5) + ain(^ ~B), ' 2 cos il . sin .5 = sin (^ + i?) - sin {A-B), 2cos^ . cosi5 = cos {A + £)+cos{A-£), * ^' 2sinii.sin5 = cos (il - B) ~co^[A + B),. EXAMPLES. XXXVm. Express as the sum or as the difference of two trigonometrical ratios the ten following expressions : (1) 2 sin ^. cos 0. (3) 2 sin 2a . cos 3/3. >(5) 2 sin 3^. cos 5A (7) sin 4^. sin tf. (9) 2cosl0o.sin60". (2) 2 cos a. cos /3. (4) 2cos(a + /3).cos(a-/3). (6) 2 cos 1^ cos |. (8) cos^^sinf. (10) cos 450. sin 150. %(11) Simplify 2 cos 2^. cos d- 2 siu 4^. sin tf, te 6 (12) Simplifysiuf .cos|-sin^^cos?^ Z6 2 _ _ ^ (13) Simplifysin3d+sin2d + 2sin^.cos-. 2 2 (14) P«ve that Hio !i* .»m f+sin? . ™ ^.^„ a^.sinft 132 TRIQONOMETRY. * * MISCELLANEOUS EXAMPLES. XXXIX. (1) If tan a-i and tan /9« J, prove that tan (a +/3)« 1. (2) If tana =f and tan^=^, prove that one of the values ofa+/8 IS - m—\ (3) If tana= --- and tan/3- -L shew that one value 0f(«+/3)i8|. (4) Simplify (6) SimpUfy (6) Pi-ove that co s g - COS 5 a sin a + sin 5a * sin hx - sin Zx cos5a? + cos3«* cos ^+ cos 3^ cos2i4 cos 3^ + cos 6.(1 cos 4^ * (7) Simplify sin 3a?- sin a? sin3j?+sinjr cos 3d?+ cos X cos 3d; - cos 0? * (8) Simplify ^?^B^Mr-^i?^4)(^ At <^s 3^) •" (cos 4^+ cos 2^) (sin ^+Vin 3^)- (9) Prove that 2 sin 2a . cosa+2 cos4a . sin a=sin 5a+sin a. (10) Prove that cos 2a . cos a - sin 4a . sin a=cos 3a . cos 2a. (11) tan2il.tan3^.tan5^=tan5^-tan3^-tan:S4 (12) Solve 4 sin (tf +0) . cos {6 - <^)«=31 4 cos (^+ 0) . sin (tf - 0)= Ij • (13) Prove that sin .1. sin 2^+ sin 2^. sin 5^+ sin 3^. sin 10^ cos^ . sin 2^+sin24 . cos54 -cos3^ . sin 10^ "" ~ **" ^-^^ ^ 2siu5 2 (14) tan^:-^-tan^--^ = ( 133 ) i1 CHAPTER XIJ. On the Trigonometrical Ratios op Multiple Angles. 162. To express the Tiigonometrical Ration of the angle 2 A in terms of those of the angle A. Since sin (-4 + ^) = sin ii . cos ^ + cos ii . sin ^ j .'. sin (il + ii) = sin -4 . cos -4 + cos il . sin ^ ; .'. sin 2il = 2 sin i4 . cos il (1). Also, since cos (il + J5) = cos ^1 . cos ^ - sin J^ . sin 5 ; .% cos (il + ^) = cos il . cos il - sin il . sin -4 ; .'. cos2ii = cosM -sin'-4 (2). But 1 = cos* -4 + sin' A ; .'. 1 + COS 2il = 2 cos'^, and 1 - COS 2^ = 2 sin' A. The last two results are usually written cos 2ii = 2 cos" .4 - 1 (3)^ and coE 2 J = 1 - 2 sin' ^ U), K^' A /A n\ tan ^+ tan J? Again, tan (A + B)^ = — ; ; ® ^ ' 1 - tan ^ . tan i? ' ^ ' l-tan-4.tani4' . . n A 2 tan A .-. tan2.1 =, — ;-— (5). i — tan- A ^ ' 134 TRIOONOMETRY. m**To prove the '2A' formula, ffeonretricaUy, deac'rr sit- rt^ ,rvr ' r ^-^ ^^^- Join SP . PL. ^^ peipendiculttr to OS. ^JXploTpZ^L^'o^ff'''' Vr "•«■" »«'- -Ti' •^i^otJlOP=.A. mZ:vpk P'";; «^ = 0n .-. 0Z/.= ment of ir/'Z. :!; ^^^10^1!, "" '"'* "■" """?'»■ Hence 03/ LM-LO 2LM LO WP (1) sin 2^ =^^^-2^^ OP 2(9^ ^^ COH 2^ = OP OP 2LM.l'^ OP OP Then % fjp = 2 coa MLP . cos PLJi . LP.Z = 2C082^-1. LetOJ/'=OJ/: 20M=.M'M=LM- LM'^LM- MR lOP LR ~~LR-LR' 1 cos2A = ' Hence, H tan 2^ = "2~^=w>sM-8in2^. 20M LM^MR' 2MP ^^ ^^ mpTlm TRIQONOMETRWAL RATIOS OP MULTIPLE iNOLES. 135 164. These fivo formulio are very important, ahi2A^2HmA.iioHA (i\ cos 2.4 = cosV/1 - sin" y| ^ (o^^ 008 2vl = 2 cosM - 1 I (3)' 0O82il = l- 28ui'A ) (4)' 2tanii tan 2 A = l-tan»i (5), V. 165. The following result is important, sin 2 J 1 + cos 2 A 2 sin i4 . cos A 166. The student must notice that A is any angle, and therefore these formulaB will be true whatever we put for A. Example. Wri ^ m«ccad of ^ and we get and so on. ■(1), sinl «= 2 sin ^ . cos - cos4=cos2i-siu8:^ ^2), EXAMPLES. XL. Prove the following statements : (1) 2co8ec2J^=sec^.cosecJ, .Q. cosec2^ .«. 2-8ec2yl (4) cos" 4 (1 - tan2 J)=cos 2 J (5) c«t2^=^^!4r.l. At ^J\JV .£3 |i|i 136 .(6) (7) (8) (») (10) (11) (12) (13) (14) (15) (16) (17) (18) .(19) (20) (21) (22) TRIGONOMETRY, 2tan^ ,=sin25. l + tan^^' tan 5 + cot 5— 2 cosec 2A l -tan3^ l+tan2iS~"*^*^2^- cot 5 - tan 5=2 cot 25. cot25+l 35^^31 -sec 2A (^sm- + cos2J =1+Kin(9. (^sing-cosgj^i-sin^. cos2|^l+tan|J=:i+8inA sm«|^cot|-iJ=i_ain^, tan^ + lX , 2 I l+sin^ tan^-l l-sintf' l-cos^"'*^*2' ^~^"'^-tan2^ cosec ^ - cot )3= tan ^. cos2;p 1-tan^ l+sin2a? l+tan^r* cos^r 1-sin^ 1 + tanf 1-tan- EXAMPLES. XL, 137 (24) (25) (26) (27) (28) (29) (30) (31) (32) (33) (34) (36) (36) cosar 1 + sin a? COtg-l cotg+l 1+sin^+cosa? X 1 -' = cot - . 1 + sm a? - cos a? 2 cos^g + sin^g _ 2 - sin 2o cos a + sin a ~' 2 " cos^o-sia'a 2 + sin2a cos a - sin a 2 cos* a - sin* a = cos 2a. oos»a + sin'a= 1 + 3 cos" 2a cos" a-sm^as sin3/& cos 3^9 sin/3 " cos"^ _(3 + cosa2a)co8 2o = 2. cos3^ . sinas „ , IE? +^=2 cot 2A sin 4/3 sin2^ sin5j8 cos 5^ =2 cos 2/3. =4 cos 2/3. sin/3 cos/3 sm 5»r 12 cos Stt 12^ n- sin — 12 cos — 12 =2^3. /QhfV \af) tan (450+ J) - tan (450 - il)=2 tan 2A. tan (45<> - 4) + cot (46" - ^) =2 sec 24. ill' IQO ^° TRWONOMETRY. (39) sec^+tan^ = tan (450 + - ) sec ^ - tan A \ (A(\\ cos(-4+45'>) ^ ^ ^3i^(2r450^=sec2^-tan2A (41) tani?=-.?E:^?^2^ l+C08^ + COS2^' (42) f^n^- Sill2^ -jip^ l-cosi5 + cos2Z?* cot H-0- 167. bered ; The following two formula should be remem- sin 3^ = 3 sin ^ - 4 sinM ) cos3J = 4cos»^-3cos^/- .(vi). Note. The similarity of these two results ia «.r.f +^ n v^r .^^ """^ '^"'^' *°<* then oos34=oosO«=l n which c«e the formula gives oosa.=4oos0«-3o<«0». o 1=4-3, which IS true. *> v 1 ui The first formula may be proved thus : sin3^=sin(24 + ^)=8in2J.cos^+co824.8inJ =(2sin^.cos.l)cos^ + (l-2sin2^)sin^ =2sin4 .cos2^+sin^-2sin3^ «2sin^(l-sinM) + sin^-2sin3^ = 2 sm 4 - 2 sinS .i +sin ^ - 2 sinM =3sin^-4sin3j. The second formula may be proved in a similar manner. TRIGONOMETRICAL RATIOS OF MULTIPLE ANGLES. 139 Example. Prove that tan 3 A => ^>" ^ ~ ^ 4 l-3tan>^ • tan3^=tan(2^+^)= *^M±*??_i_ ^ ' l-tau2^tanJi 2tan^ . l-tanM+*^°^ 2tanii+tan^(l-tanM) l-tan2^-2tauM -- 3taiiii-tan3^ l-3tan2^ • emem- EXATVrPLES. ZLI. vi). cause second 00=1. OS or Prove the following statements : sin 3^ (1) (2) (3) (4) (5) (6) (7) (8) sin^l cos 3^ COSii =2 cos 24 + 1. 2cos2il-l. 3 sin 4 - sin 3 A ^ cos34+3cos4~ ^° ^• 3cot24-l ' sin 3 A - sin A cos 34 + cos i4 sin 34 -cos 34 sin4+cos4 sin 34+ cos 34 =tan 4. =2 sin 24-1. = 2 sin 24 + 1. cos 4 - sin 4 1 1 tan34-tan4'^cot4-cot34"°*^*^^' (9) / 3 sin 4 - sin 34 X8 ^ / sec 24 - 1 y \3 cos 4+ cos 34/ ~Vsec24 + l/ * /iA\ 1-008 34 „ ^^^^ T:^^3^=(1+2cos4)» 140 TRIOONOMETRY. *^M SOELLANEOTTS EXAMPLEa XLU. P«^)ve the following statements : ID "'fi^+cos^ ^ ^ cos i - sin ^"**" 2/!+ HOC 2^. tan - + 1 (2) .__^ _, t^jj ^j ^ g^^ ^^^ I -tan (3) ;"^;^^)-OB(n-l)„-.in2a,sinCn-l)„.eo.s(n.l)„. (4) «"^« + sm3 .. .."-^^ ■ coan + cos^~"^°~2" • (5) £252a+co522a c^^ j (6) ^^^ = lSM-vethatsin2^=cos3^; henceprovethat siul8»=^J 4 • (7) ?!°.-«±^j3±siii_(a+/3) « « sm a + sin /3 - sin (ar+^ = °o* 2 • ^°* 2 • -(8) «i'»2^.sin2i?=8in2(^+5)_gi^a(^.^. ^(d) cos44=8oos*^-8cos»^ + i. (10) tan 500 + cot SO" = 2 sec lO* .(11) «a3^=4sin^.8in(600 + ^)sia(600_^)^ (12) (<^t^-tan^)\cot^-2cot2^)-4cot^. (13) coa3a-sia;3.sin 5fl - cos 7a . sin3a+ sin/3. cos 6a"^"8in7a '® ^"^^Pendent of /i. (14) (cos*+cosy)»+(sin:p+sin5/)8-4cos«^~y. (15) 2cos2^.co8«i? + 2sin2>< sinS 7? i, ^«. -r^Bin ^.sin»^=l+cos2^.cos2A a6) cot^-tan^=a o 8 '(17) tan 4^= i*^ii(izi55ia l-6tan«tf+tan*^* MJUVELLANEOUS EXAMPLES. XLII. Ul (18) (19) (20) 2coHg = v/2T72. 2 cos IP 15' « V 2 + V2T\/2. sinil.sin2ii+Binil . aiii AA + win 2ii . Hiii lA Bin ^ . ooH 24 + Bin 2/1 . cos bAfi^i^JTooBM ^^^^ ^^' (21) "'" ^ + Bin (^ + ) + jjgl'gjlgt^) (22) (23) (24) (26) (26) (27) cos e + COS (^ + <^) + cos (#+20) "^ ^^" ^^ ^ ^'^• 2 cos' il - 2 sin' A = cos 24 (1 + cos'-' 2 A). (3 sin 4 -4sin3 4)a+(4oo8» 4 - 3coh 4)2= 1. sin 2a . con V -- tan = cot a - ttin a. (1 + cos 2a) (1+ cos a) 2 CQ ^(^-2)a.Co t na + 1 cot (» - 2)^ cotTiii If tan a = f and tan (i = f^ , prove tan (2a + 0) = I. Prove that tan - and cot - are the roots of the equation a''* - 2x . cosec 4 + 1=0. (28) lftan5==, prove that a * 168. The following examples are aymmetrical, and each involve more than two angles ; Example 1. Prove that sin (a T^+y)«sin a . cos/3, cos y +8iu^ . cos y . cos a + sin y . cos a . cos iS - sin a . sin /3 . sin y. 9in(o+^+y)=sin(a+^).cosy+cos(a+^)siny -sin a . COS ^. COS y + cosa.sin /3 . cosy + oosa . oos/3. siny-sin a . 8iu/3. siny «»sin a . cos ^ . cos y + sin /3 . cos y . cos a + sin y. COS a. COS /3~ sin a. sin /3. sin y. Q.E.D. r ''I U2 TRIOONOMETRY. Example 2. ' Prove that sin a + ain /3 + sin y - sin (a 4/3 + y) '4.8in^y.sin^ 2 2 . sm H^- . sin a + ^ 2 ' Nowsin«-am(a + /3 + y)=-2cos.?^±^±> ain ^^V And Hi.^ + smy-2«n^V.„os«-p:, [Art. 168] .-. sinafsiu/3 + siny-sin(a + /3+y) = 2sinto.eos'^-.2cos?2±J?±y 8i„/3 + y -2 sin '^jcos ^ - cos ?^^ =2sin^.2siny±f.sin?-t^ ^?>.sinr±^.,i,«_^^. [Art. 158J =4 sin -g-'.sm-^.sm Q.E.D. * EXAMPLES. XLia. Prove the following statements : (1) «os(«+/3+y)=cosa.co8/3.cosy-cosa. sin/3. siny -cos/3.siny.8ina-cosy.8ina.sin)8. (2} «"»(«+/3-y)=sina.cos^.cosy + sin/3.oosy.cosa - sm y . cos a . cos /3 + sin a sin /3 sin y. (3) cos(a-/3 + y)=cosa.cos;3.cosy + cosa. sin/3. siny . -^'^^/S-sioo-siny+cosy. sin/3, sina. (4} sm« + sin^-siny-sin(a+^-y) =4sin«-p.sin^V.,,,a^^ (6) sin(a-/3-y)-8ina + sin;8 + siny =4sin^.sin^>.sin^. (6) sin2a + sin2^+sin2y-sin2(«+/3 + y) «4 sin (^+y) , sin (y + a). sin (a +/3). EXAMPLES. XLIII. (7) Mn05-y)+sm(y-a)+sin(a-/3) (8) sinOS+y-a)+sin(y + a-/9) + sin(a+/3-y) -«i°(°+^+y)=48ina.sin/5.8iny. (9) «'»(«+^ + y)+8m03 + y>a) + sin(y + „-;3) -sin (a+^ -y)=4cosa . co8/3 . sin y. (10) cos* + cosj^ + cos2 + cos(a?+y + 2) = 4coe^^cosi±i'.cos^^. (11) cos2:p + cos2y + cos2« + cos2(^+y + e) =4 cos (y + «) . cos (2+ar) . cos (a:+y) (12) co8(y + «-,;)^cos(. + a;-y) + cos(a;+y-.) /^_ +<'«8(*+y^«)=4cos:p.cosy.cos«. 143 (13) cos2^ + oosV + cos22 + cos2(ar+y + «) / ■■^^^ + °o«(y + ^)-cos(« + a?).cos(a?+y)} A (14) Sin8^+8in2y + sin2e+8in2(^+y+^) -2{l-cos(y + «).cos(. + ar).cos(;p+y)}. ^^^5) cos2a?+cosV + cos2«+cos2(ar+y-«) =2{l+cos(a;-«).cos(y^«).cos(a;+y)}. (16) -o««.«m(^-y)+co8^.8in(y-a) + cosy.sin(a-^)=0. (17) «i^°.«n(^-y)+sin^.8in(y-a) + 8iny.8in(a-^)=0 (18) cos(a+^).cos(a-./3)+sinO + y)sin(/3-y) -COS(a + y).cos(a--y)=0. (19) co8(d-a).sin03-y) + cos(5-/3).8in(y.„) -cos(d-y).sin(j8_„)^0, m «-^-ifi^^.eosi±|r_^.eo8^J:|r_0.^^^ =cos2tf+cos20+co82;^ + 4co8(9.cos0.corv + l L. B.T. 10 ( U4 ) CHAPTER XIII.** On Angles Unlimited in Magnitude. II. 169. The words of the proofs (on pages 118, 119) of the *Aj B' formulsB apply to angles of any magnitude. The flgares will be different for angles of different mag- nitude. 170. The figure for t\ie*A-B* formulas on page 119 is the same for; all cases in whi h A and B are each less than 90". The figure given below is for the proof of the *A-\-B* formulae, when, A and B being each less than 90°, their sum is greater than 90". The words of the proof are precisely those of page 118. We may notice however that ^ ^ OP I OPJ~ OF 'OP'OP' and the rest follows as on page 118. ON ANGLES UNLIMITED IN MAGNITUDE. 144. NOTE. Fia. 1. Bm(A~Ji) = ^-^H-PH KN ON PH PN OP OP "ON' OP -pN' OP ^^ =amAooBB-coBABinB. Consider ^., KN is in the negative dim^tiou; ON ia in the negative direction, if we t.ke OE as the positive direotton. . • . ^ « . positive ft^tion. numerically equal to sin ROB; and sin iJO^ is positive , .-. ^=8in^. taction, numerically equal to 00. POB,>ri,l„l,i, ,1 • 2^ „ „ DCT ° "V, . . Q p=C08 Z». Ooneider ^, Pa i. „e^«,,, ,^ ^ ;„ ^^ ^,,.^^ __.^^^^ ^^^^^^ 0^ „ lniti.1 1™ .„. ^ .„ ^ „ .^ ,^,„, ,^„.,„„ „, ^^,__^_____ ^^^_^ ^^ Art. m], .-. _ h.„eg.ti,etaCi„„, „„„eri„,,^^„., ^ ^_,^^^^ ^_^ ^ C08/ir0iir,ortoooii.li OM-^tanegntivoi .-.^.o^^ Consider— 5 P^ is positive; OP is positive; and .-. ^^^inB. COS (A-B)^9E^9E±B^_OK ON NH PN OP OP -ON'OP+PN'OP' OK Oone.dcr ^, „ »,,„„ ,. „ . ^^,^ ^^__ ____^^^_^ ^^ ^ c««OB, .-. it to cul to cc^, »^ „ . „e^,„ ^«„„.^j,. ^ ,, ^ 10—2 i 144. Fig. 2. TRiaONOMETRY. ^ ' 0P~ OP ~ ON'OP~ PN'OP' KN Consider — ; KN is positive; ON is in the negative direction (OE being In the positive direction); .-. ^ is a negative fraction, numerically equal to ain HOE or sin A and sin^ is negative; ON. EN ON' ' sin A. So, ^ is a negative fraction numerically equal to cos JB, which is negative. JPB Consider ^; PE is in the negative direction ; PJ\r is in the positive direo- tion; for OE is the initial line and the positive direction of revolution is from F PH to E [Rule II. Art. 131]; .-. ^ Is a negative fraction, numerically equal to sin B, which is negative. PN . So, OP 18 a positive fhiction numerically equal to sin B which is positive. coaM B\ OM_OK+KM_OK ON NH PN ON' OP'^PN ' OP' OK Consider ^; OK is positive, and OiV is n^fative; /. the fraction is negative and numerically equal to cos ^, which is negative. So, j^- is a negative fraction numerically equal to cos B, which is negative. NJI Again, in ^, NH is in the negative direction, and PN is positive; .'. the ftraction is negative and numerically equal to s\n MPN\ or to sinJV^OJe; or •0 sin A which is negative. jjp is a positive flractionesin B. So ON ANGLES UNLIMITED IN MAGNITUDE. U6 171. Thus we have proved that the M, 5' formul» are true provided ^ and ^ each lie between 0° and 90". The student can prove them for any other values by drawing the proper figure. The '^, 5' formuliB are therefore true for any values whatever of the angles A and B. 172. By the aid of the '^, 5' formula we can prove the form u lee of A rt. 14 0. ^ - Example. Prove that sin (90" + A)=coaA. sin (900+^)=sin 900 cos A + cos 900 8in A, "Ixcos^ + Oxsin^, = C08 A. Q. B. D. EXAMPLES. XUV. Draw the figures for the first four of the foUowing examples j;i/^;i:t\ii"Sof "^^^' ^^^^ ^ ^^ ^-- ^^- ^^ betllo^'a^d^S^OO."^^ '^"^^^^' ^'^" ' ^' ' -^ ^^ 1800, and {A + B) hes between 180o and 270o. 270", and {A - B) hes between 1800 and A. Deduce the six following formuisD from the '^1, ^' formula (5) cos (900+^)=- sin X (6) sin (900-^)»cos^. (7) cos(900-.4)=sin.l. (8) sin(1800-^)=sin^. (9) cos(1800-^)=-cosX (10) sin(1800t^)=-sinJ ^JL^l ^^i^g that the foi-mula sin(^+5) = dn^. cos5 .tZ m'«, f ^ *?'u'' "" "^^^ ^^^ ^"d ^> dedu. the rest r>t th„ «^^ 5. formulea by the aid of t.L^ ra«.,Ua ^^ ^ ,^-, I -t of ? 146 TRIOONOMETRY. 173. We may Iso conversely prove that the * A^ B* lurmulee are true for angles of any magnitude by the aid of the result of page 107 (This is a vexy convenient method of proving the 'A, £' formula to be true for all values of the angles.) For, assuming that the *A, B' formulae are true for certain values of the angles A and £, we can show that they are true if either of the angles ^ or ^ be icreased by 90". ^ Example. Bin(9Qfi+A+B)=coB{A+B) [p. 107.] —cos ^ . COS 5 - sin il . sin By = sin (900+ il) . cos 5 - { - COS (900 + ^)} sin ^, =sin (900+^) . COS ^+cos (9Qf>+A) . sin A Or, writing A' for 90®+^, we have sin (^'+ J5)=Hin A' . cos B+ooa A' . sin B. We have proved (Art. 170) that the 'A, B' formulae are true for all values of A and B between 0° and 90*. And therefore, by what we have said above, they are true for all values of ^ or 5 between 0" and 180". And so on. Therefore the *Af B' formulse are true for any values whatever of the angles A and B. 174. It follows that all formuleB deduced from the * Af B' formula are true for angles of any magnitude what- ever. Thus the 'S, T' formulae (page 126) are true for angles of any magnitude. Also the formulae of the last Chapter for multiple angles, and all general formulae in the Ex- amples, are true for angles of any magnitude. EXAMPLES. 147 EXAMPLES. XLV. thu^iL ^Jf"^ *^^ ^.^"**' of sin 1800 and COB W from those of tne Bine and cosine of W>. f^n^? 7^« T'* ^ '' ^'^^' *^*° ^«0° *°d less than 2700 and tan^-i: findsin24 a .4 tAn iA. find sm 2tf and sin ZO. Y,.^ al.- , cos 3tf, and hence determine t which quadrant 3tf lies. (4) Prove that the different values of tf which satisfy the equation co^p6+ove that cos A ■COS A 2 ■ V 2 (2) When A lies between 180» and 5400, prove that A cos ^=v^ coaA A (3) Find sin - iu terms of cos A, when A lies between 180» and 360^. (4) Prove that when A lies between (4n +l)n and (4m + 3) n, n being a positive integer, cos -= - ./— COB A ~2 A . (5) Find sin - in terms of cos A, when A lies between 4n,r and (4?i4-2) tt, where n is a positive integer. 1 160 TRIGONOMETRY. 178. Since, J sin ^ . cos -^ = sm ^, and, • 9-4 ,A [Art. 166.] we obtain by addition and subtraction 8in'^ + cos'^+2sin^.cos^ = l+sin^, „...^ A A A sm-^ + cos»--2sin^.cosl=l.«in^. Therefore and . A sin ^ + cos A sin - - cos 1) =l+8in^, 2 J = l-sin^. Whence, . A A , sin ^ + cos -^ = ± VI + sin A . A A , sin--cos-^=± x/1 -sin^ .(ii). Adding we get, 2 sin | = ± Vl+sin^ * x/l ^sin~3 Subtracting we get, 2 cos ^ = ± Jl+^mA^ Jl-BmA (iv). Thus if we are given the value of sin ^, (nothing else being known about +he angle A\ we have four values for • -4 , - A sin -^ and four values for cos^ . 179. To prove GeometricaUy that, given the value of sin A (nothing else being known about the angle A), there are four values for sin jr^^ and four values fm- cos ~ . A 2 Let a be the least positive angle which has the given sine, and let /?0P„ FfiL in the figure each = a. ON 8UBMULTIPLE ANGLES. 151 U 1P4 (iii), (iv). Then A is one of the angles described by the revolving line OP, when, starting from OB, OP stops either in the posi- tion 0P^ or in the position OP,. The angle described is either (an even multiple of tivo right angles + a) or (an odd multiple of two right angles - a); a half of this is either (an even multiple of one right angle + J a) or (an odd multiple of one right angle - 1 o); so that a half of the angle whose sine is equal to that of a may be any one of the angles indicated by OR and the dotted lines OP^ and OP^, or OP^ and OP^ in the figure. In the figure, HOP^ = P^OU= LOP, = P,OD. And sin-^ may have any one of the values smUOP^, sin ROP^, sin POP^, sin POP^. These values are all dif- ferent, and are those given by the solution (iii). Hence, there are four values for sin ^ , of the nature indicated by the solution (iii); also four values for cos- (iv). li 102 TRIQONOMETRY. 180. If we know the magnitude of A, we can decide which sign to take in the formulse . A A /- ., sm -^ + cos 2 = ± n/ 1 + sm ii (i)^ . A A r sm^-cos^-^^Vl-sin^. .{ii> Example. When -^ lies between - 45« and +450, A 2 . A , A\ . sin -r-4-fina — I ia -m/um/' COS - is ffreaier than sin ^ and ia positive. and So that ^sin ^^+ cos ^ j is positive, and = + Vf+iml, / . A A\ ^^sm - - cos ~J IS w«grflrfm, and = - Vl - sinj. When - lies between +45<'and +135« sin - is greater than cos ^ , and is positive. So that (sin i + cos ^ j and (sin ^ - cos f\ are both positive. And so on. The fohowing diagram completes the above results. /X O'ijis .< ^^ '. QiJis -i / (O's (lijis \ ON SUBMUITIPLE ANOLEb. A 153 181. Since tan A = 2 tan 1 - tan» 4 A , ,A -*^^2 2 tan J^ -1=0, whence tan ^ = ~_1 * >/rTW^ 2 tan J 9 Thus, given tan^l we find two unequal rahies for tan one positive and one negative. 182. The student will be able by the aid of the fol- lowing figure to verify this result geometrically. P^ U 183. We may remark that in this figure P^OP, and P,OP„ are straight lines at right angles to each other. So that tan P^OIi=:- cot P^OJi ; or, tan P^OJi. tan P^OH = - 1. Hence, one value of tan ^ is the reciprocal of the other, and of opposite sign. So that there is always one positive value of tan ^, and one negative; one numerically greaf^^- than unity and the other numerically less than unity. TRIOONOMBTRY. Example. If A "^ igo", p-we t hat 2 tan J. ~ • Now tan ^ ^ tan 1900, which i n positivo ■, 1.^ ^„tan 9^, ^hioh 58 negative. Also ten - « Jiij^^^lTtanal 2 teT^i ' '^^« iiegaiive value of widch is :il-~^l±MI (1)^ state th. signs of (sin ^+cos^) and (si^-cosl) when - has the following values • A (2) Prove that the formulas which give the values of sin ^ and A 2 of cos - in terms of sin A, are unaltered when .1 has the viuues (i) 920, 2680, 9000, 4n,r + K or (4n+2)^-f^. (ii) 880, -880, 7700, -7700, or 4n,r-=|. (3) Find the values of (i) sin 90, (u\ coaflo /,-;;\ „:„ 010 (IV) cos 1890, (V) tan 202^0, (W) tan97ir ' ^ ^ ^"^ ®^ ' (4) If J = 2000, prove that 2 sin 2 = +^/l+iiE:i + vrrii^ (ii) tan:^ = lii±^^i±*5^) 2 tam2 • (6) If A lie between 270" and 3600, prove that (i) 2sin^=N^i-sin^-Vi+sin^. ten — s 2 (ii) ten^= t^+cosecil. (6) If A lie between 450<' and 6300 prove that 2 8in-=-Vl+8in^-s/l-sinX ON 8UBMULTIPLE ANGLES. 155 5", ^hioh I vaJue of i) -cos 346«, iOOOO. V!uues Jin 81* (7) Find the limits between which ^ must lie when 2 sin ^ = \/l+sinil - -JT^mA, (8) Given that A lies between 4500 and 6300, prove that 2 cos ^ = - Vl+sin^+-v/l-sinil. (9) If A Ue between n x 360» - 90<> and n x 3600 + 900 where n is a positive integer, prove that tan — 2 1+Vl+tanM tanul and that when A lies between nx3600+90» and nx3600+270'», then .o»^- -l-^^l+tanM 2 tan^l (10) Prove geometrically that if we are given the value of sin A, there are three different values for sin ^ , and wic different 3 values for cos A 3' (11) Prove geometrically that, if we are given the value of cos A, there are three different values for cos ^ , and six different values for sin - . (12) Prove that if we are given the value of tan A there are three different values for tan - . (13) Given the value of tan J, prove that there are /owr values each for sin -^ and cos - . 2 2 * (14) Given the value of sin A, prove that there are two values for tan - ^■■liij liB^ 166 TRIGONOMETRY. m.^\ ^' ^'>P«^*^°* i^ possible, in solving Trigone :S:^:^ "^^ to avoid squaring both Jes ol Z Emmple. Solve coa6=^k sin 6. If we square both sides of the equation we get cos2 tf=Psinad=,&2 (i _ cos^tf). •'• °'^s''<^=i-T7.2. or cos^=d.--_^._ Now if a be the least angle such that cosm fi^a the indices of the powers of 10 which are equal to (i) 453x650. (ii) (453)*. (iii) 6503x453^. (iv) ^''453. (V) .^45-3x^650. (yi) 4/453-x (650)3. (vii) ^45317650. (3) Express in powers of 2 the numbers, 8, 32, ^, ^, -125, 128. (4) Express in powers of 3 the numbers, 9, 81, |, -^, -1, ^. 1 90. Suppose that some convenient number (such as 10) having been chosen, we are given a list of the indices of the powers of that number, which are equivalent to eveiy whole number from 1 up to 100000 Such a list could be used to shorten Arithmetical cal- culations. Example 1. Multiply 3759 by 4781 and divide the result by 2690 Looking in our lut we should find 3759= 103 W60723 470. = 103-6796187^ 2690 = 103«9«23. ' ^^^^ Therefore 3759 x 4781 ~ 2690=103W6or23 ^ iQawosisr j. \c^mwQ = 103-6r60723+3-ur96187-3<297623=s 103"824838r The list will give us that 103'82<8387= 6680*9. Therefore the answer correct to five significant figures is 6680-9. Example 2. Simplify 3« x 2">-fv'r760l. The lUt gives 2 = 10-3«io3m 3 = io-4771213 ^nd 17601 = 10<24»378 '453. EXAMPLES. 163 Thus 3" X 210-f*yi760i = (10«71213)6 y^ (103010300)10^^io4124663r8\i _ X02M27278 x lOSflMSOOO^ lO*'*'"'^^^ _102'87i/\i/vrvn il |-}| i, I. ^A ' i\ 'in I 'k r4 170 TRIGONOMETRY. LIV. (6) Give the position of the first aignifi^arU figure in the numbers whose logarithms are ^^ y gure m tne -2 + -4612310, -1 + -2793400, -6 + -1763241. nuihL^'T *^' ^°''*'°'' '^ ^^' ^* signifu^ant figure in the numbers whose common logarithms are 4-2990713, -3040595 2-5860244, - 3 + -1760913, - 1 + -3180633, -4980347. (7) Given log^^ 2 = -30103. find the number of digits in the integral part of 8io, 2^\ l62o, a^oo. (8) Given that log 7= -8450980, find the number of digite in the integral part of 7^0, 49*, 343^, (V)2o, (4.9)12, (3.43)10. (9) Find the position of the first significant figure in ^2, (i)w (4^)20, (-02)4, (.49)6. (10) Find the position of the first significant figure in the numencal value of 050 one 20^ (-02)r, (.007)«, (3-43)^, (-0343)8, (-0343)1^. ^ 201 Prop. To prove thai if two numhera expressed inth^ decimal notation have the same digits (so that they differ only in the position 0/ the decimal point), their logo- nthms to the base 10 will differ only hy an integ&r. The decimal point in a number is moved bv multiplying or dmdmg the number by some integral power of 10. Let the numbers be 7^ and t*; then m = n x 10\ where A IS a whole number positive or negative; and log w = Iog (» X 10*) =log7i + logl0* = log w + k. Example 1. Let the numbers be 1-2346 and 1234-5. Then, 1234-5 = 1 -2345 x 103, therefore log (1234-5)=log (1-2345) + 3. ON LOGARITHMS. IJl Example 2. Given log 1-7692 = -24776, find (i) log 17692, (ii) log -0017692, (iii) 176-92. ' Here, logl7692 = log {(1-7692) x lO^} = log (1-7692) + 4 = -24776 + 4 = 4-24776. Iog-00l7692=log {(1-7692) X 10-3} = -3 + -24776. logl76-92 = log {(1-7692) X 102} =2-24776. 202. It is convenient to keep the decimal pmi, of com- mon logarithms always positive, because then the dedmal pa/rt of the logarithms of any numbers expressed by the same digits will be always the sama 203. The decimal part of a logarithm is called the mantissa. 204. The integral part is called the characteristic. 205. The characteristic of a logarithm can be always obtained by the following rule, which is evident from page 167. RULE. The characteristic of the logarithm of a number greater than unity is one less than the number of integral figures in that number. The characteristic of a number less than unity is negsr tive, and (when the number is expressed as a decimal ) is one more than the number of cyphers between the decimal point and the first significant figure to the right of the deci- mal point. 206. When the characteristic is negative, as for ex- ample in the logarithm -3+ -1760913, the logarithm is abbreviated thus, 3-1760913. '»a>^1^^\ll,. '^^^ cliaracteristics .., 36741, 36-741, -0036741, 3'6741 and -38741 are respectively 4, 1, - 3, 0, and - 1. l!P>fl 172 TRIGONOMETRY. ^f<^l^ 2. Given that the mantiasa of the logarithm of 86741 IS 6651610, we can at once write down the logarithm of any number whose digits are 36741. Thus and so on. log 3674100 «»6'5651510, log 36741 =4-5651510, log 367-41 = 2-6651510, log -.36741 =1-5651610, log -00036741 = 4-6651510, 1^ Am I 207. In any set of tables of common logarithms the student will find the mantissa only corresponding to any set of digits. \ It would Ob ;.>u8ly be superfluous to give the climac- teristic. 208. It is »;o3t important to remember to keep the mantissa always positive. . Example. Find the fifth root of -00066061. Here logj^ -00065061 =4-8133207, .-. logio(*00065061)^=:i(4-8133207)=i(-4+-8133207) =i(-5 + l-8133207)» - 1 + -3626641 = 1-3626641, and 1-3626641 =log -23050, .-. the fifth root of -00066061 = -23060 nearly. EXAMPLES. LV. (1) Write down the logarithms of 776-43, 7-7643, -0007764S and 776430. (The table gives opposite the numbers 77643 the figures 8901023.) ' (2) Given that log^o 69082=4-7714552, write down the logarithms of 5908200, 5-9082, -00059082, 590-82 and 5908-2. the ON LOGARITHMS. LV. I73 C3) Find the fourth root >f -0059082, having given that Iog5-9082 = -7714562; 4-4428638 = W o77"4 * (4; Find the product of '00059082 and -0277^4 hnuin„ • that -21431 =log 16380 (cf. Question 3). ' °^ ^''^'^ (6) Find the lOtJi root of 07704^ / ,^ *• ,x , . given that •8890102 = log 7-74l8. ^^' ^^^^"^ (6) Find the product of (-27724)2 and -n77ft4i /ca r^ Mens land 3 ;.7758288 = log 59680) ^ ^' ^'''- gtven has,, to amther syrtem with a difermt baee. If we are given a to of logarithm, calculated to a given Let a be the gi ven base ; let i be any other base Let m be ny number. Then the logarithm of m to the base a IS m the g,ven list. Let this logarithm be i Then We wish to find the logarithm of m to the base b. Let It be X. Then m = b\ Butm = a'; or. I . .'. a' = 6«; or, b==a'; X is the logarithm of 6 to the base a. Now the logarithm of b to the base a is given. For it is m the list of logarithms to the base a. Thus, or, i = log,6;or,a. = j-l^; log.^J^. loff, b ' i'l: L. E. T. 12 IMAGE EVALUATION TEST TARGET (MT-3) /> {./ ^ V ^ is .% ^ ^A 1.0 1.1 ^121 m ■» m 12.2 : itt 12.0 [[•25 |u Ui& ^ f uym Photographic _SciHices (Jorporation ^ a>^ v <^ ^. -«\"** 33 VnST MAIN STRHT WHSTiR,N.Y. MSN (716)S7a-4S03 v\^^ 174 TRIQONOMETRY, If in Hence, to ciilculate the logarithms of a series of numbers to a new base 6, we have only to divide each of the logarithms of the numbers to any given base a, by a certain divisor, viz. log,. 6. If then a list of logarithms to some baae e can be made, we can deduce from it a list of common logarithms, by multiplying each logarithm in the given list by ^ log, 10 • ExampU. Show how to transform logarithms, having 5 for base to logarithms having 126 for base. Suppose wt = ft', so that I = logj m. Now 126-.63, so that 3 = logj 126, and »»-6'*» = l26», sothat|=log,„m. Thus the logarithm of any number to base 6, divided by 3 («.e. by log, 125), is the logarithm of the same number to the base 126. ♦*210. The student will find that the logarithms of numbers cannot be calculated to the base 10 du^ctly. Thoy are first calculated to the base 2-7182818, etc., which is the sum of the series 1+-4.-1-J. 1^1 . . ^1^1.2 + rr2:3 ^ rxsTi"*"®**^ «^»*»/- This number is called e. And the constant divisor in this case is log, 10, and 1 1 log. 10 " 2^30268609 "" ''^3429448, etc. When this constant divisor is transformed into a mvUMier this constant nlultipUer is called a modulus. Example. Given that lofto 12=10791812, shew how to transform common logarithms to logarithms having 12 for base. Here i Qvvmsiz a, i £ • _i .". 10=12itrowi2=-i2i'2002,«tc. .'. 1()*=12**""2002 EXAMPLES. 175 EXAMPLES. LVI. Show how to transform (1) Logarithms with base 2 to logarithms with base 8 " 2 ^oganthxns with base 9 to logarithms with b^ 3 4 Sr^u^'^'^f'^^^^^g-it^n^^withbaseT 4 Logarithms with base 3 to common logs 6 Common logs to logarithms with base 3. (6 Given log,,2=3010300, find log, 10. 7 Given og,„7=-8450980, find log^lO. (8) Given log,. 2 =-3010300, find log^lO and log^, lo. * MISCELLANEOUS EXAMPLES. LVII (1) ^«»dlog,8,log,l,log,2.1og,l,log3,i28. W bJiow that the loearithms nf „ii numbeiB from 1 to 30 inTr^ u ^"""'P' "'^^^* ^^ *he log 2, loga and log ' ''' '"" ^' ^^^^^'^^^^ in tenns of (3) Show that the loffarithmq nf f k« « u i.«^*/«'""*"'"'""°^''^«"«^^«»^««- Find the mul^l^ed':;"^' *^ '"»" "" - («»^«2)". when it is (0 ^x3--7', (ii, 3--128V7.-, (8) Given Iog„ 7, find Iog,490. (») Given Iog„3, find log,270. (10) Given !<«,„ S. find ln» in 12—2 h rf 176 (11) Given logg9«o, logg6-=/i baae 10 of numbers 1 to 7 inclusive TRIOONOMBTRY. LVII. log,7-c; find the logs to (12) How many positive integers are there whose logarithms to base 2 have 5 for a oharacteristic ? (13) If a be an integer, how many positive integers ars there whose logs to base a havo 10 for their characteristic ? (14) Given log 2 and log 7, find the eleventh root of (39-2)«. log 1-9486= •289688. (16) Prove that 7 log ^ + 6 log | + 5 log | + log H-log 3. (16) Prove that 21oga+21oga« + 21oga»...+21oga»*=n(n+l)loga. (17) Prove that log,6 . log»a - 1 ; and that log.6 . log»c . lo&a- 1. (18) Prove that log/-- log^ . logjc . log^. . .log,r. H^(19) Given that the integral part of (3-456)i«»f» contains 63866 digits, find log 346-6 correct to five places of decimals. w(20) Given that the integral part of (3-981)i<»«w contains ■ixty thousand digits, find l(>g 39810 correct five places of decimals. (21) If the number of births in a year be ^ of the popu- lation at the beginning of the year, and the number of deaths 1^, find in what time the population will be doubled Given log 2, log 3, and that log 241 =2-3820170. (22) Prove that log « + log (« - a) - log 6 - log c- 2 log ,^A?^S . (23) Prove that log (a« + «») + log (a + «) + log (a - «) -. log (a* - ««). (24) Prove that log sin 4-4-log 4 + log sin J +logoo8 .4 +logcos2.4. ( 177 ) CHAPTER XV. On the Use of Mathematical Tables. f)^ ™n to .„ ..H TaMes ^hX^T^eTlir'' to !(»|»9, calculated to Beven significant figures ; These wiU be found in Cl»mbe«. i/„tt«„<^ y^ Brelllt :f ,f "'\«'«'P'«^ - gi™ the logarithm, to five fig;,re» of aU numbere from 100 to 999. This Tabk wm be found useful i„ questions involving certai^l^:"; nuraencai caicuiations. ~ '0 178 TRIOONOMETRY. III! 216. We have uaid that logarithms are in general incommensurable numbers. Their values can therefore only be given approximately. If the value of any number is given to seven significant figures, then the error (ie. the difference between the givm value and the exwt value of the number) is less than a millionth part of the number. Example, 3141692 is the value of «• correct to seven signifi- cant figures. The error is less than -OOOOOl ; for n is less than 3141693, and greater than 3-141692. The ratio of -OOOOOl to 3-141692 is equal to 1 : 3141692. The ratio of -OOOOOl to ,r is less than this; i.e. much less than the ratio of one to one million. i 217. An actual measurement of any kind must be made with the greatest care, with the most accurate instru- ments, by the most skilful observers, if it is to attain to anything like the accuracy represented by • seven significant figures.' Therefore the value of any quantity given correct to •seven significant figures' is exact for aU practical purposes. 218. We are given in the Tables the logarithms of all numbers from 1 to 99999; that is, of any number having five significant figures. A Table consisting of the logarithms of all numbers from 1 to 9999 J99 (le. of any number having amm significant figures) would be a htmdred times as large. 219. There is however a rule by which, if we are given a complete list of the logarithms of numbers having Jwe sig- nificant figures, we can find the logaiithms of numbers having six or seven significant figui'es. ON THE USB OP MATHEMATICAL TABLES, 179 EsampU. Suppose we require the logarithm of 4-804213. From the Tables we find log 4-8042--6816211, i.e. 4-8042 -10-«i<»«i- log 4-8043--6816301, 4-8043- iQ-^m^J 4 8043 whose logarithms are found in the Tables so that th. required loga^.thm must lie tet^^en the two given^gl^itt. Therefore we suppose that log 4-804213-6816211 +rf, i.e. 4804213- lO-^aiem..... 220. The EULE is aa follows Thn Aiiv tween thr«o «. v 'ouows. The differences be- tween three numbers are proportional to the coxresponC vided that the dtferences between the numbers are small compared with the numbers. Example. Given log 48042 = 4-6816211 log 48043=4-6816301, find log 48042-13. We may state the question thus Given log 48042 = 4-6816211 |og(48042+l)=4-68l6211 + -0000090. .An«? log (48042+ -13). Let log (48042 + •13)=4-6816211 +rf. Thus corresponding to the diflferencea 1 and -iq • .u numbers we have the differenws SXon ^ i '^ *^' logarithms. uiawences 0000090 and rf m the But these differences are in proportion -13 ' .-. rf=-__ times -000009 , , =T^ of -000009 = -000001 17 • /. Iog48042-13 = 4-6816211 + .()()oooin " = 4-68162227 = 4-6816223 (to seven figures). ii f M i: TRIOONOMBTRT. Pr J^i ^.\t''^^ ^^^'' ^ *^^ *^^« ^^^ «• «»e Bnle of PToportloiial Dlfferenceg. PaJl*'" ''^*'" "^""^ '^"^ *^' '^'^^^P^' ^^ Proportional 222. In Art. 197 we «.id that numbers art, not pi^por- fcional to their Logarithms. Hence the differences of numbem and the corresponding differences of their logarithms cannot be exactly m proportion. The rule is however true for all practical purposes. The proof of the rule belongs to a higher part of the subject than the present. 223. In the above example we said that 6-68162227 = 6'6816223; and for this reason. We are retaining only seven significant tigures m the decimal part of the logarithm. If we put 6-6816222 for 6-68162227 the 'error' is greater than -00000007. If we put 6-6816223 for 668162227 the 'error' is less than 00000003. Thus the second error is less than the first. In such a case, 1 must be added to the last digit which is retained, when the first digit which is neglected is 6 or greater than 6. 224. We give two more specimen examples ExampU 1. Find the logarithm of -004804213. We first find as before, by the rule of proportional differences, *^* log 4-804213- -6816223 .'. log 004804213=3-6816223. ON THE USE OF MATHEMATICAL TABLES, 181 Example 2. Fi^ the number whose logarithm ie 2-6354291 In the Table we find that •5364207-Iog 3-4310 /^ •nd •6354334-log 3-4311 1 Let -6364291-108 (3-4310+rf) !...... (Ui) Here we have three logarithms and three numbers. Subtracting •f.3.-,4334 -5.354291 j53r)4207 -.'J354207 •()(KX)127 •00(X)084 Tho c()rro«i)onding difibrcnccs uio (XK)! and d. Hence, c/=:^^84 •0000127 ' =iVr<>f -oooj = •0000661. Therefore from (iii) -6354291 =log (3-4310+ 000060...) = log 3-431066. Hence, 2-5364291 ==log 343-1066, or, the required nuciber is 343-1066. EXAiyrPTiTis Lvm. (1) Find log 7-66432, having given that log 7-6643 = -8839066, log 7-6544 =-88391 12. (2) Find log 664-123, having given that log6-6412=-7513716, log 6-6413= -7513792. (3) Knd log -0008736416, having given that log 8-7364= -9413325, log 8-7365 = -9413376. ^1 ^8* TIUOONOMETHY. LVUI, (4) Find log 6437128, haviug given that log 6-4371 --8086903, log 6-4372 --8086970. (6) Find log 3-72466, having given that log 3724fi- 4-67 10680, log 37246=4-5710796. (6) Find the number whose logaiithm ia -5686760. having given that -5686710 = log 37040, ^ -6686827 -log 3-7041. (7) Find the number whose 'ogarithm is 46602987. having given that -6602962=- log 4-6740, * •6603067 -log 4-5741. (8) Find the nuinber whose logarithm is 6-3P66938. havinir given that -3966874= log 2-4928, •3967049= log 2-4929. (0) Find the number whose logarithm is 46431 150, having given that '6431071-% 4-3966, * •6431 170 -log 4-3966. en- J«?K f ""^ ^^^ °'^^*' ''*'*^ logarithm is -7650480, having given that 3-7650436 -log 5689-1, 2-7650612 = log 56892. 225. The same Rule of Proportional Differences is used m the ca^ of angles and their Trigonometrical Ratios; ritiuns of thair Ratios. Thus the (small) differences between three angles are assumed to be proportional to the corresponding differences between the sines of those three angles; also, proportional to tJie cor^ponding differences between the logarithms of the sines of those angles. ON THE USE OF MATHEMATICAL TABLES. 183 .1^^^^' .K^'°^ *"** "^^""^ '''•" *^'^*>'« »«« than unity, a. »I80 are the tangenta of all angles between 0' and 45» n.n^^ '^«*"^»^«»« of these Ratios ,nust therefo,^ have negatwe characteristics. To avoid the inconvenience of havinff to nrin* f},«a. 2^tive oha™ot«n-,«o,. the who.e „„„bef ,^ Hd ^l' loganthms of the sine, co8i„e, etc, of «. angle. They are indicated by the letter ' L.' .in S^M,^ "'"/'' "■•''*"'''' *■■" *« "'•""''' 'ogari*". of WD Jl 15, and la equal to {log(ain3r IS")* 10) Thus in the Tables we find X sin 31" 15' = 97149776. Therefore log (sin SIMS') =97149776- 10-1-7149776. Example I. /Ynrf sin SI" 6' 26". The Tables give sr:;i0 6'„. 51^5333 /|x sin 310 7'- -5167824 tJ ^ «n3lo«'26"--6166333+rf ... /yix" Subtracting " ^ '* fi%^ ^""^ '''''' ^^^«^-- - ^. and the ~^)^m) '™^"^'"^ differences in the ^^'^■ii'i angles are 60" and 25" ; ••• «?=M of -0002491 = •0001038... Hence, from (iii) sir. 3 106'26" = -5165333+ -0001038= -5166371. I] I w ! V \l i ii ^®^ TRIGONOMETRY. 9'niZ7t *' ^"^^ """ *"*'^ ""^"^ logarithmic cosine i. The Table gives 9-76^7eil-Z cos 620 22' «. 9-78»9249-.Zoo8 52"21' (U), TTie cosine diminislies as the angle increases. Hence corre- spending to an increase in the angle there is a diminution of the cosine. Hence, let 9-7868083 = Z cos (62» 22' - /)) (\i{\ Subtracting the first tabular logarithm from the second the difference is 0001638. Subtracting the first tabular logarithm from the third, the difference is -0000472. Subtracting the first angle from the second, the difference is Subtracting the first angle from the third, the difference is By the Rule these four differences are in proiwrUon. Therefore /> = ^y^a^ „f gjy/ 17 3". Hence 9-7868083 - Z cos (52« 22' - 1 7") =»Z cos 62" 21' 43" EXAMPLES. LK. (1) Find sin 42" 21' 30" having given that sin 42* 21'= -6736677 8in42'»22'»-67i^8727 (2) Find cos 47" 38' 30" having gi> en that cos 47» 38'- -6738727 cos 47" 39'- -6736677. (3) Find cos 210 27' 46" having giveu that cos 210 27' = 9307370 cos 210 28'.= -9306306. cosine i» (i% (ii). Qce corre- ion of the ....(iii). 9cond the third, the erence is 9rence is BX AM PLUS. LIX. (4) Find the angle whose sine is (5666666 having given that -6665325 - sin 410 48' •6667493-8in41''49'. (fi) Find the angle whose cosine is J333333 having given that •3332684-oos 700 32' •333r)326-cos70°31'. (6) Find the angle whose cosine is 25 having given that -2498167-008 750 32' •2600984-.CO8 760 31'. 185 (7) Find L sin 45° 16' 30" having given that I sin 45° 16' = Z sin 450 17' = (8) Find L tan 27" 13' 46" having given that L tan 27» 13' = Z tan 27* 14'. (») FindZcot36«18'20" having given that L cot 36« 18' » Zoot36«19'- "9-8614969 = 98516220. = 9-7112148 -971 16264. 10-1339660 101337003. (10) Find the angle whose Logarithmic tangent is 98464028 having given that 9-8463018-Z tan 3504' ^^«*«^8' 9-8466705 =X tan 35« 6'. (11) Find the angle whose Logarithmic cosine is 99448230 having given that 9-9447862-Z cos 28o 17' 9-9448541 -Z cos 28M6'. (12) Find the angle whose Logarithmic cosecant is 10-4274623 havmg given that 10'4273638-Zcosec2l0 57' 10-4276774»Z cosec 21« 66'. m I :' 'Ii 186 TRIOONOMBTRY. I '■ 227. Problems m which each of the lines involved conteins an exa^t number of feet, and each angle an e:cact number of degrees, do not occur in practical work. As from time to time the skill of observers and of in- strument-makers has increased, so also has the number of significant figures by which observations have been recorded. Thus the want was felt of some method bj which the labour involved in the multipUcation and division of long numerical quantities could be avoided. At the end of the Seventeenth Century a celebrated Scotch mathematician. John Napier, Baron of Merchiston, proposed his method of Logarithms'; ie. the method of representing numbers by indices; 'which, ^,y reducing to a few days the labour of many months, doubles, as it were, the Kfe of an Astronomer, besides freeing him from the errors and disgust inseparable from long calculations.' Laplace. 228. We shall now give a few examples of the practical use of logarithms. ExampU I The sides containing the right angle C in a right ang ed tnangle ABC contein 3456-4 a and 4543 6 feet i^spl tively; find the angles of the triangle, and the length of the nypotenuse. ** Let a, be be the lengths of the sides of the triangle opposite theangles J, 5, C respectively. & vi^ ^ ON THE USE OF MATHEMATICAL TABLES. Then, a-3456-4 feet, 6=4543-6 feet. 187 tan J< 'b' Therefore log tan A^log^ =log « - log 6 =log 3466-4 -log 4543-5. In the Tables we end log 3456-4=3-5386240 log 4543-6 = 3-6573905. .*. log tan A = 3-5386240 - 3-6573905 a -8812336. .'. ^ tan ^=9-0812336. In the Table we find 9-8813144=Xtan37»I6' 9-8810522=Ztan370l6'. Whence by the rule of Proportional Differences 9-8812335 =Z tan 370 15' 42". and Therefore ^=370 15' 42' 5=90»-^=,620 44' )'42"1 [' 18"J (I). find It by the aid of logarithms thus ; + o , or we may -=cosec^=cosec 370l5'42", .% Iogc=loga+logcosec37«15'42" =logo+Z cosec370l5'42"-l0 =3-5386240+ 10-2179174 -10 =3-7665414 = log 6708-8, ,*. the hypotenuse contains 5708-8 feet. tri.!lSr "' '""' ^'"^ *^' «8lea and the third «de of the 188 TRIGONOMETRY. 229 There are some formula which are seldom used in pmctical work, because they are not adapted to logarithmic calculation. They are those in which powers of quantities are connected by the signs + or -. lenfhrf?' ^}'' ^^' *^J' '^P^' ^^ °^8^* ^^^« fo^d the length of the hypotenuse by means of the formula But we should have had to go through the process of calcu- latmg by multiplication the values of a^ and b^. For this reason, a formula which consists entirely of factors IS always preferred to one which consists of terms, when any of those terms contain any power of the quantities involved. \ .JV"" ^^^A ^^''^^ '^^^'P^"' *^' ^"°«*^ °^ **»« hypotenuse c and of one side a were given, then the formula ^-c2-a*=(c-a)(o+a) will give the length of b. For log6«=.log{(c-a)(c+a)}, or. 21og6«rlog (c-a) + log(c+a). And the values of (c+a) and (c-a) axe easily written down from the given values of c and a. EXAMPLES. LX. In the following questions A,B,C^ the angles of a right wt ^^^ f "'^'^f^ '' " "«^* ^«^^' ^'^ «' *' - ^ ?he lengths of the sides opposite those angles respectively. find^!l ^'''^'' *^* a= 1046-7 yards, c= 1856-2 yards, C^d(fi, log 1046-7=3-0198222, log 1866-2=3-2686248 Zsm 340 19' =.9-7610991, ' L sin 340 20' =9-7512842. BXiMPLBS. LX. 189 (2) Qi"- that a=8m feet, C=90«. and ^=34« 15'. &nic log843-2=2-9259306. Zooaeo 34" 1»'= 102496427 logl-4982--17S57. ' J3j Given that a=484« yani,, 6=4742 y^, and C-90. log484»=36852938, log4742=36759615 -Man45.36'=100090966, Ztan4o.37'=loo;93492. J4)^ G.ven that «=8762 feet. (7=90, and ^=37oi<., fi.d log8762=3-9426032, Zsin37» 10'=97811344 i;oo,37»10'=9-9013938, log 52934 =-72373 ' log 6-9823 =-843997. J'i «-» f-t 6=1694-2 chain, ^=90. and ^ = 18.47', logl6m=3-2289647, Zcot 180 47'= 10-4683893, log 5-7620= -76057. J6) Given that «=1072 chains, c=4849 chains, and C=900, log 6921 =3-7723951, log 3777 =3-6771470 log 4-729 =-67477. ' (7) Given that 4=841 feet, .=3762 feet, and (7=90«, find a Iog4603=3-6630410. log 2921 = 3-4655316 log 3-6668 =-66428. ' J'Jan^clr *''* ^^'''''' ^^^^°^> ^=«^71 stains, (7=900 log 7694-5=3-8861804, log 8471=3-9279347 Xtan420l5'=9-95824, Zcosec420 15'= 101723937 log 1-1444 =-05867. ' "-«™°«^ iuganthms from the Tablea J -Ml 1^ !•. B. T. 13 190 TRiaONOMETRY. * MISCELLANEOUS EXAMPLES. LXI. (1) A balloon is at a height of 2500 feet above a plain and its angle of elevation at a point in the plain is 400 36'. How far is the balloon from the point of observation 1 (2) A tower standing on a horizontal plain subtends an angle of 37019'30" at a point in the plain distant 369-5 feet from the foot of the tower. Find the height of the tower. (3) The shadow of a tower on a horizontal plain in the sun- light is observed to be 176-23 feet and the elevation of the sun at that moment is 33" 12'. Find the height of the tower. (4) From the top of a tower 163-5 feet high by the side of a river the angle of depression of a post on the opposite bank of the river is 290 47' 18". Find the distance of the post from the foot of the tower. (5) Given a=673-12, 6=415-89 chams, C=900, find A and B. (6) Given a=576-12, c=873-14 chains, C=900, find b and A. (7) From the top of a light-house 112-5 feet high, the angles of depression of two ships, when the line joining the ships points to the foot of the light-house, are 27<> 18' and 200 36' respectively. Find the distance between the ships. (8) From the top of a cliff the angles of depression of the top and bottom of a light-house 97-25 feet high are observed to be 230 17' and 240 19' respectively. How much higher is the cliff than the light-house ? (9) Find the distance in space travelled in an hour, in con- sequence of the earth's rotation, by St Paul's cathedral (Lati- tude of London=610 25', earth's diameter=7914 miles.) (10) The angle of elevation of a balloon from a station due south of it is 470 18' 30", and from another station due west of the former and distant 671-38 feet from it the elevation is 410 14'. Find the height of the balloon. 190 (i) On the four followim, nacM .^ „. m«ntfs«e of the logarithm! of^r„K^™f ' ^»"« °* «•• It i» only in observ.fi„ ' "g^-fioant figures then.03te.aU\t:: ~ S"""'^ "»-« ''"• a greater degree of aocuracv L .« • 1 «~umstance8, that by 4 significant fig„™, Kk'S T T '""'"^^ benoe the Table here UeTwill ^7 r '"■^' Toir "- ---- "^H.^b tiiti/: ^^1-t.hethenu.berofi^eheeinthediagon^, *^=3x (14-7)2, ••. «"n/3x14-7; «=iC477l2)+M6732 [from the Table] -•23856 + 1.16732 = 1.40688 TTAno *!, .. "^°^^^"'^°«^Jy [from the Tablel Hence the diagonal is 25-45... inches. ^ ^ampfo ii. nndthsvalu,of(4Zn9, log 4-37- -64048) log 4-38= •64147/ ^^^e^ence for -01 = -00099, .-. difference for -002 =-00020; .•. log 43-72-1.64068; .-. log(43-72)»=|xl-64068=:-9116... = log 8-156... rfrom *»«- rn 1 ' -J Therefore (4372)t_8-166... "" "°"^ 13—2 I 7*'' '.I li 190 (ii) TABLE OF THE LOGARITHMS OP ALL NUMBERS FROM 100 TO 1000 Na lOO lOI loa «03 104 lOS 106 107 108 109 110 III III 113 114 115 116 117 118 119 120 131 132 "3 134 "5 136 137 138 139 130 133 133 134 «35 136 »37 138 139 140 141 143 Log. 00000 00433 00860 01384 01703 03119 03531 03938 0334« 03743 04139 04533 04933 05308 05690 06070 06446 06819 07188 07555 07918 08379 08636 08991 09343 09691 10037 10380 10731 11059 "394 11737 H057 "385 13710 13033 13354 13673 13988 14301 14613 1493 1 15339 No. 143 144 145 146 147 148 149 150 151 ^53 '54 156 157 158 160 161 163 163 164 165 166 167 168 169 170 171 173 «73 »74 175 176 177 178 179 180 181 183 »83 184 i8s Log. »5534 15836 16137 16435 16733 17036 17319 17609 17898 18184 18469 18753 19033 193 1 3 19590 19866 30140 30413 30683 30951 3I3I9 31484 •JI748 330II 33373 33531 33789 22045 33300 «3553 33805 24055 24304 24551 34797 35043 25385 35527 35768 36007 36245 36483 36717 I No. 187 188 189 190 191 193 193 '94 195 196 197 198 199 300 301 303 303 304 305 306 307 308 309 3IO 311 313 313 214 215 3l6 317 218 219 220 321 333 333 334 225 236 337 338 Log. 186 36951 37184 37416 37646 37875 38103 38330 38556 38780 39003 39226 29447 39667 39885 30103 30330 30535 30750 30963 3"75 31387 31597 31806 33015 33333 33438 32634 33838 33041 33344 33446 33646 33846 34044 34343 34439 34635 34830 35035 35318 354 1 1 35603 35793 No. 229 330 231 333 233 234 235 336 237 338 239 340 241 343 243 244 245 246 247 348 349 350 251 353 253 254 255 256 ^57 258 259 260 261 262 263 264 265 266 267 268 269 370 371 LMg. 35983 36172 36361 36549 36736 36923 37107 37391 37475 37658 37840 38021 38202 38382 38561 38739 38917 39094 39270 39445 39620 39795 39967 40140 40313 40483 40654 40834 40993 41163 41330 41497 41664 41830 41996 43160 42325 42488 42651 43813 42975 43136 43297 No. 373 373 374 375 376 277 378 279 280 381 383 383 284 285 286 287 288 389 390 391 292 293 294 295 396 297 398 399 300 301 303 303 304 306 307 308 309 310 3" 3'2 313 314 Log. 43457 43616 43775 43933 44091 44248 44404 44560 44716 44870 45035 45179 45333 45484 45637 45788 45939 46090 46240 46389 46538 46687 4683s 46982 47129 47276 47422 47567 47713 47857 48001 48144 48287 48430 48573 48714 48855 48996 49136 49276 49415 49554 49693 LOGARITHMS. No. Log. 190 (iii) 49831 49969 50106 50243 50379 50515 50651 50786 50920 51055 51188 513^' 51455 51587 51720 51851 51983 52114 53344 S''!375 53504 53634 52763 53892 53020 53148 53375 53403 53539 53656 53783 53908 54033 54158 54383 5440. 54531 54654 54777 54900 55033 55145 55367 55388 I Lo^ I No. Log. 55509 1 405 55630 I ' :o6 55751 I 407 55871 1408 ^599' 1409 56110 J410 56329 1 41 1 56348 56467 56585 56703 56870 56937 57054 57171 57287 57403 575>9 57634 57749 57864 57978 58093 58206 58320 58433 I 58546 58659 58771 58883 58995 59106 59318 59329 59439 59550 59660 59770 59879 59988 60097 60206 60314 60423 60530 60638 60746 45 J ^853 J45; No. Log. I No. 60959 61066 61173 61278 61384 61470 61595 61700 61805 61909 63014 63118 63321 63325 62438 62.SS31 63634 62737 62839 63941 63043 63144 63246 63347 63448 63548 63649 63749 63849 63949 ,.- 64048 1 483 64147 484 64246 64345 64444 64543 64640 64738 64836 64933 65031 65128 65335 65331 65418 65514 I 65610 65705 65801 6.S896 65993 66087 66181 66376 66370 66464 66558 66653 66745 06839 66932 67035 67117 67310 67303 67394 67486 67578 67669 67761 67853 67943 68034 68134 68215 & ' 68485 68574 68664 68753 68843 68931 69020 69108 69197 69285 69373 69461 69548 69636 69733 Log. 69810 69897 69984 70070 70157 70243 70329 70415 70501 70586 70673 70757 70843 70937 71011 71096 71181 71265 71349 71433 71517 71600 71684 71767 71850 71933 72016 72099 72181 73363 72346 72428 72509 72591 72673 73754 72835 72916 72997 73078 73159 73339 73330 73400 73480 73560 i! j 'I: If .ii fiJl m I' i- i \ ii I ;! 'If 'i .11 I r 190 (iv) LOGARITHMS, i No. 545 546 547 548 549 550 551 55^ 553 554 555 556 557 558 559 560 561 563 563 564 565 566 567 568 569 570 57' 57a 573 574 575 576 577 578 579 580 581 582 583 584 585 586 587 588 589 590 Log. I No. 73640 737»9 73799 73878 73957 74036 74"5 74'94 74373 74-55' 744^9 74507 74586 74663 74741 74819 74896 74974 7505 ' 75'28 75 W5 75281 75358 75435 755 1 1 75587 75664 75740 75815 75891 75967 76043 761 18 7619a 76368 76343 76418 76492 76567 76641 76716 76790 76864 76938 , . . 77013 I 635 77085 1 636 591 59^ 593 594 595 596 597 598 599 600 601 603 603 604 v-35 606 607 608 609 610 611 613 615 616 617 618 619 620 631 633 633 634 Los. 637 638 639 630 631 633 634 77159 77333 77305 77379 774.^3 77535 77597 77670 77743 77815 77887 77960 78032 78104 78176 78347 78319 78390 7846a 78533 78604 78675 78746 78817 78888 78958 79039 79099 79169 79339 79309 79379 79449 795 '8 79588 79657 79737 79796 79865 79934 80003 80073 80140 80309 80377 80346 I No. 637 638 639 640 641 643 643 644 ^^45 646 647 648 649 650 651 653 653 654 655 656 657 658 659 660 661 663 663 664 665 666 667 668 669 670 671 673 673 674 675 Log. 676 677 678 679 680 681 683 80414 80483 80550 80618 80686 80754 80821 80889 80956 81023 81090 81158 81324 81291 81358 81435 8 149 1 81558 81621 81690 81757 81823 81889 81954 82020 83086 83151 82217 83282 83347 83413 83478 83543 82607 82672 83737 83803 83866 83930 83995 33059 83133 83189 83351 833 > 5 83378 I No. 683 684 685 686 6S7 688 689 690 691 692 693 694 695 696 697 698 699 700 701 702 703 704 705 706 707 708 709 710 711 713 713 7'4 715 716 717 718 719 730 731 732 733 734 735 726 737 738 Log. 83443 83506 83569 83633 83696 83759 83822 83885 83948 84011 84073 84136 84 1 98 84261 84333 84.^85 84448 84510 84573 84634 84696 84757 84819 84880 84943 85003 85065 85126 85187 85348 85309 85370 85431 8549' 85553 85613 85673 85733 85794 85854 85914 85974 86034 86094 86153 86313 No. 739 730 731 733 733 734 7.35 736 737 738 739 740 741 743 743 744 745 746 747 748 749 750 751 753 753 754 755 756 757 758 759 760 761 762 763 764 765 766 767 768 769 770 77> 773 773 774 Log. 86373 86333 86392 86451 86510 86570 86629 86688 86747 86806 86864 8692:$ 86982 87040 87099 87157 87216 87374 87333 87390 87448 87506 87564 87622 87680 87737 87795 87853 87910 87967 88024 88081 88138 88196 88353 88309 88366 88423 88480 88536 88593 88649 88705 88762 888 r 8 88874 LOOARITHMS. 190 (v) No. 775 776 777 778 779 80 781 782 783 784 785 786 Log. I No 88930 88.;86 88042 89098 89154 89209 89265 89321 89376 89435 89487 8954a 830 831 81a 833 834 835 836 Log. 787 89597 ^"" 89653 89708 89763 89818 89873 89937 89983 90037 90091 90146 90300 90255 90309 90363 90417 90473 90526 90580 90634 90687 90741 788 789 790 791 792 793 794 795 796 797 798 799 800 801 803 803 804 805 806 807 808 809 810 811 8ia 813 8r4 815 816 817 818 3 837 828 829 8:}0 «3' 833 833 834 835 836 837 838 839 840 841 843 843 844 845 846 847 848 849 91381 91434 91487 9»540 91593 , . 9'''H5 1870 No. 819 90795 1854 90849 " 9090a 90956 91009 91063 91116 91 169 9iaaa 1275 91328 91 9' 75 1 91803 91855 91908 91960 92012 92065 92117 93169 92221 92273 92324 92376 93428 93480 92531 92583 93634 93686 92737 93788 93840 I 92891 92942 92993 93044 93095 93146 93197 93247 93398 93349 93395 93450 93500 698 1 87 865 866 867 H68 869 Log. 864 9365' 1 909 I 873 873 874 875 876 877 878 879 880 881 883 883 884 885 886 887 888 889 890 891 893 893 894 895 896 897 898 899 900 901 90a 903 904 905 906 93703 93752 93803 93852 93903 93952 94003 94052 94101 No. Log. 910 911 91a 9>3 9'4 915 916 917 918 94'5i I919 94201 1 920 94250 1 921 93551 1907 93601 1 908 94300 94349 94399 94448 94498 94547 94596 94645 94694 94743 94792 94841 94890 94949 94988 95036 95085 95134 95183 95231 95279 95328 95376 95424 95472 95521 95569 95617 95665 95713 95761 933 923 924 925 926 927 928 939 930 931 932 933 934 935 936 937 938 939 940 941 942 943 944 945 946 947 948 949 950 951 952 95904 95952 95999 96047 96095 96142 96190 96237 96284 96332 96379 96426 96473 96520 96567 96614 96661 96708 96754 96801 96848 96895 96941 96988 97035 97081 97138 97174 97330 97267 97313 97359 No. 955 956 957 958 959 960 961 Log. 963 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979 980 981 983 983 984 985 986 98000 98046 98091 98137 98183 98337 98373 95809 1953 95856 1 954 |95- |95h 97405 1 987 9745» "" 97497 97543 97589 97635 97681 97727 97772 97818 97864 97909 97955 999 98318 98363 98408 98453 98498 98543 98588 98633 98677 98733 98767 988 1 1 98856 98900 98945 98990 99034 99078 99'23 99167 99311 99255 99300 99344 99388 99432 99476 99520 99564 99607 99651 99695 99739 99782 99826 99870 99913 99957 ilif, n If ./ •"'ii 'iLiil li^ 190 (vi) TRIQONOMETRY. E2A1IPLES. LXI.a. Prove that the foUowing statements are correct to four signi ficant figures : (1) -y 461-7-669. (3) (273)* x(234)i = 47-32. (7) (24-76)^ (•0046)^ ' = 8287. (11) 11=1096. , •3107. (2) ^^802-3-809. (4) (461)* X (231)* -65460. (0) <2iZ?)» ^3. (41-26)* 7 '89 (8) :Q^^^{^^'^^0)^='nQf>. C.) (?^)Ls.3.e. Solve the equations correct to 4 figures : (13) 10»-421 [a;-2-624]. (14) (fi)-- 3 [a;- 22-511 (16) (i«8)'^=2 [*=23-28]. (16) (««)«- 3 [^-28-011 (17) log37«+3=3-412[a?---8243]. (18) *-10^(31-2)[a?=31-48]. Example i. Fivd tU amount at Compound Interest ^'^„j a ^ t, , «« 5 per cent Compound Interest / "^^ '^ ^"^""^P^ ^' ^6^«f Let * be the number of jears, then »^-- S-r *"'"' '' *'' "' °'' ^"^"^ = 14-2. or *(log21-log20)-log2, • y^ -30103 30103 _ " -02119 -IITgr The principal which will amount to £l at th« «n^ r y-r^^at 4 p. cent Comp. Int. 1. [See Loc': ^Lt^:; l^a: Hence the required present value is £lOO{r+r>+r3+...+rio}, whore r^fgj «£1001Z^^ £100=^100 {1.(^)1,) x26-£100. log(m)"-ll(loglOO-logl04)=(2-.201703)xU -(l-98297)xll = l-81267 -log -6496... .-. the required present value =£100 x .3503... x 26 - £100. **»**" ^£810 about EXAIVrPT.ES. LXLb. 4 per L^'"^ *^' '^''"^'^^ '"*^* "'^ ^^^ ^o' 10 years at ^J2) Fmd the Compound Interest on £1 for 8 years at 5 per 3 ,Jl'"norj"^''"" r^" * '"^ '' ^'^'^ "^^ do-Wed at - ^ C4.4i, vompouiiu imprest? m > '^ ««^«a''» [Result, 2Z-^ illii I i ».. 190 (viii) TRIGONOMETRY. k (4) In how many jaors will a sum of money b« doubled at 4 per ceni Compound Iii '^t ? ' [Ruult, 11 L] (5) Find the present value of £100 to be paid 8 years hence reckoning Compound Interest at 4 per cent. [Hwilt, £7307.] (6) If the number of births in a town are 25 per 1000 and the deaths 20 per 1000 annually, in how many years will the population be doubled ? [After 140 t/car».] (7) On the birth of an infant £1000 is invested at Com- pound Interest in the Funds (3 per cent, payable half-yearly); calculate what it will be worth when the child is 21 years old. [Residt, £1869.] (8) In what time will a sum of money treble i'aelf at 3 i)er cent. Compound Interest payable half-yearly? ' [Result, 36-9 yean.] (9) A sum of 1 8hillii\(» lent on condition of 1 penny interest being paid monthly, accumulates at Compound Interest at the same rate for 12 years ; what will be then the amount? [Remit, about ;£5()66.] (10) A man puts by 2d. at the end of the second week of the year, Ad. at the end of the fourth week, 6d. at the end of the sixth week ; what sum would be put by for the last fortnight in *^® y^ ' [About 67,100,000 pe7ice.] (11) A train starting from rest has at the end of 1 second velocity -001 ft. per sec. and at the end -^i each "second iU velocity is greater by one-» Idrd than at the end oJ Vio precedir j second; find the velocity in miles per ho r/ tuc end of 26 8^«°<*8. [-678 miles per hour.] (12) The volume of a sphere is §frx(cube of the radius); "ud the diameter of the sphere which contains a cubic yard. [Result, 1-24 yards.] i i ') r:iid the present value at 4 per cent per annum Com- ^»%;i»v Interest of . Fellowship of £260 a year for six years, payabk half-yearly, the first payment being due in six months. [Result, £1322.] toubled at ^t, 17-7.J larahenoe , £7307.] 1000 and I will the 40 i/car».] at Com- Py early); re old. , £1869.] ' at 3 i>er •9 yeart.] r interest it at the ! ^5066.] 3k of the d of the might in petwe.] 1 secoud 3ond it** recediriir id of 26 sr hour.] radius); ard. t yards.] m Com- i years, onths. £1322.] ( 191 ) CHAPTER XVI. On thk Relations betwekn the Sides and Angles 0.V A Triangle. 231 The three sides dnd the three angles of any triangle, are called its six parts. ^ By the letters A, £, C we shall indicate ffeornetHcally, the three angular points of the tna, .,1. ABC • algebraically, the three angles at those anaul,.r n,r,f ' ■pectivelj. angular iOinta re- *h ^^, ^^'7.!.'"^'! "' *' ' ^^ '*^^" ^^^^''^^^ ^^'^^ °»e^"re. of dvely ^^ '^^'"*' '^" '''^^'' ^' ^' ^ '''^' '- 232. I. We know that, ^ + 5 + (7 = 1 80". [Euc. i. 32.] 233. Also if ^ be an angle of a triangle, then ^ may have any value between 0» and 180". Hence, (0 sin A must be positive (and less tb^ 1), (ii) cos A may be positive or negative (but must be numerically less than 1), (iii) tan A may Lave any value whatever, positive or negative. * • :*?: 192 TRIGONOMETRY. 234. Also, if we are given the value of (i) sin A, there are two angles, each less than 180*, which have the given positive value for their sine. (ii) cos A, or (iii) tan A, then there is only one value of A, which value can be found from the Tables. ABC A 235. - + - + _ = 90°. Therefore 1 is less than 90", and its Trigonometrical Ratios are all positive. Also, - is known, when the value of any one of its Ratios is given. Similar remarks of course apply to the angles B and C. Example 1. To prove sin {A-{-B)= sin C. A + B+C=\m .'. A+B=l8(fi^C, and .-. sin (^+J?)=sin (ISQO- (7)=sin C. [p. 104.] A+B C -s — ==C0S — 2 2 A+B Example 2. To prove sin zscos - Now ^+'^+?=m 2 . A+B 2'-^^'-% and .% sin^=sin (qOO-^^cos^. [Art. 118.] Example 3. To prove Now sin^+8in5+sin(7=4cos^ . cos|. cos-, sin^ +sin5=2 sin ^±? . cos "* ":?. and =2cos^.cos— g-. sinC=2 8in^.cos|. =2cos — g— .cos 5. 2i 2 [Art. 157.] [Art. 118.] [Art. 166.J [Art. 118.] SIDES AND ANGLES OF A TRIANQLE. 193 C .*. 8m^+sm5+sinC=acos-.cos^^^ «»-j^+2o<«|.o<«^+?, =2cos?(2cos^.oosD. f'^lWJ -4 COS -. cos |. cos?. ,).aD. EXAMPLES. LXn. (1) o<«^=J. (8) ooa^=-i. (3) ^^_j ^va the foUo«ing stat«nen^ a, i,, c bei.« the angles of a (7) smU+£+o=o. (8) cos(4+£+o=-i. (9) emil^J?, 2 = 1. (10) oos±t|±f=a (U) tiB.(4+5)=_tana (12) oot^^_t«n^. (13) oos(^+fi)=-c„sa (M) cos(^+iJ-0=-««20 (IB) tan^-cot5=coB0.8eo4.coseoA 06) S?4^=tan?.ta,.^-'» 2 (18) «in4+sm^-siuC=48in^.8mf.co8?. 2 j{ g f i 194 *(19) (20) (21) (22) (23) (24) (25) (26) (27) (28) (29) (30) TJliaONOMBTRT. LXII. sin^ -sm5+sinC=4sm I . cos 1 . 8in^. A B sin ^ . COB l+sin I . cos l+sin I', cos ^ a ^ B C = 2 COS -.COS-. COS |. COS ^+cos5+cosC-l=4sm^ . sinf. sin^. 2 2 2 co8»^+cos»f-cos^f=2cosf cosf.sinf. sin«^-sin«f+sin3f=i-2cos^.sinf.cosf. 2 2 -+sin -1 =4sin^.smf.8mf. (31) (32) (33) sin2^+sin25+sin2C=4sin4 . siaB.&mC. Bin A . cosil -sin^ . cos^+sinC . cos(7 =2cosil . sm5 . cosC?. ^{B+C-A)-Bm{C+A-B)+Bm{A+B-C) =4coSil . exaB . cosCl cos 2^ +COS 25+cos 2C= - 1 - 4 C0S4 . cos5 , cosa sin«^-8m25+sin2C=28in^ . cos5. sinC. cos(Z?+C-4)+co8(r+^-5)-cos(^+Z?-C)+l =4sin^.BmJJ. cosC. sin^ .cos|.cosf +sinf .cos^.cosi 2 +sm ^.cos^ .cos|=sm^ . sinf . sm?+ 1. tanil + taniB+tanC7=tan4 . tan5 . tauC. tan|. tail ^+tan §. tan ^+tan| . tan 1-1. aiDes AND AmiBs of a TRIANOLB. 196 then ;!; %iZ^ * '^'■'^'^ '^"8'« ^^ C. 90., Hence, sin A = sin (90° - J) = cos A ^K sin^ = fl = , cos j5, and so on. (See page 52.) prove EXAMPLES. T.TTTT In a right-angled triangle ABC, in which ve the Mowing statements. (1) UnA^c^B. (2) tani(=oot^+c<«a (3) 8m24=»i„2i,. (4) 0032^+003 2B.0. C* is a right angle, (6) sin2J = ?^. (7) cos2^=^'. (9) sin«| = ^-1? 2 2c • (6) cosec25=^ + i. 26 2a* (8) cos25=?!5j±l«E!^ (10) C082^=£±? 2 2B and j^ = cos ABD ; or, DB = c cos B, a=-CD + I)B=^bcoaC + ccoBB. Rg.ii. CD CA BD r = COS ACD ; or, CD = b cos C. j^ « cos ABD ; or, BD = c cos (180" - ^), .'.a = CD-BD = bcosC-ccoa (180" - ^) = 6 cos C + c cos B. Pig.iii. a=CB=bcoBC = 6 cos C + c cos ^. [For, cos ^ = cos 90" = 0.] Similarly it may be proved that, 6 = cco8ii + acosC; c = aco8^+6cosil. proportional to th, siSof^ "^^ "^"o'"' "'' «*» «-« acute ••^^=/«i^C't, and^i>.esin5. " . . 6 sm C = c sin ^ • sm^ sinC* Figa Here, 4^ = sin r^ ^ ^^ = sin (180"-^) osinct, and^Z) = csin^. .. 6sinC = csin^; or, -i = __i_ sm B sin C • rifria Here, ^ = sinr a AD I— ■ ., ^ * AC ^"^^t and —- — 1 = sm90' . o c Similarly, rt may be proved, that -JL. *_. . a i . ««tor, °S» ,Sl!!t"l*?.""-™«" "■»' ™« « a ««.n.H„., "Tjr " °'^ -ail ;:x:r: j{ - -'""-^ *^» **• x« H i'j 1 V 111 ir 198 TRIGONOMETRY. 239. IV. To prove that a' = 6» + c' - 26c cos A. Take one of the angles A. Then of the other two, one must be acute. Let B be an acute angle. From C draw GF , perpendicular to BA^ or to BA produced if necessary. There will be three figures according as A is less, greater than, or equal to a right angle. Then, .0 ^0 ja B e [Euc. II. 13] B e A B c A Pig.i. BC' = CA' + AB'-2.BA.FA', or, a''^b' + <^^2c.FA =^b' + c'-2cb COB A. [For i^^ = 6. cos -4.] Pig.il. B(T' = CA' + AB' + 2.BA.AF', [Euc. ii. 12] or, a' ='b' + d' + 2cb cos FAG ^b' + <^-2bccoaA. [For FAG ^^^ 180' - A.] Pig. iii. BG'= GA' + AB*; [Euc. i. 47] or, a" » 6' + c* - 26c cos A. [For cos A = cos 90" = 0.] Similarly it may be proved that 6" = c" + a' - 2ca cos jB, and that c' = a' + 6' - 2a6 cos G. 240. V. Hence, b' + c'-a* cos A 2bo „ J Z. . « -a COS -■ = / °\^ Nam ci A^x^ _9 ^ow, since cos ^ - *' + c» - a» >, ^' , and 2 sin'- 26o •*• 2«^^'^=-l-cos^=:l^*'+c«-aV, -^ „ ^5r~-f^^*«-240, 166J [Art. 166] -^ain,sincecos^ = 2co82i-.l. 2 ^' ••• 2cos»^ = i+cos^ = i + ^+c«-a8 iifil Pi ! ; : K iiOO TRIGONOMETRY. Similarly it may be proved that ca and that '^2 = \/~ i -a){8-b) C V s (s -c) ab NOTE. In taking the square root the positive sign only is admissible before the V, because the half of an angle of a triangle is less than 90°. 243. VII. Since i^ -h){s-c) he and therefore, 2 ~y ~br~'' cos . A sin -r- '""2 = ^1 COS -^ 2 2 \/{s-h){8-c) Vs (s - a) smce 244. VIII. Again, sin -4 = 2 sin -^ . cos ^ ; [Art. 166] .*. sin .4 = 2 /(^-&)(^-g) /£j 2 . 6c The letter *Sf usually stands for tjs {s — a){s- h) {s - c); so that the above may be written a ahc' Similarly, sin B 2S sin C b ahc c It should be noticed that this result gives an independent proof of III. ; for 2^ _ sin^ _ 8in5 _ sinO abc a ~ b o~' ssible before Since -J- = _£_ ,, ^ 8"i ^ sin C" ^®* ®^«^ o^ these fractions = , BinB sinC'— «J» and that then sin^l fcHcg-gg ^ d^ siu« B+d^ sin" C? - J sing ^ 26c 2c?2 8iii_gaiiiC' ' sin'^+sin^C-sin^^ 2cos^.sin5.sinC 2 sin ^ sin C 2sin^. sin (7 [Ex. 29, p. 194, since (A + B + q=^18(fi.] =coaA. 248. The formula a = d sin A sin B sin C is very frequently of use in solving examples. Example. Prove that a oosil+6 cos J5+c cosC=2a sin^ sinC Since a=c?sin^, b=dainB, c=c?8inC, the above may be written c^sin^i . co8^+c?sini?. cos5+c?8inC. cosC =2damA . sin5. sinC, or sin 2^ + sin 25+8in 2C=4sin^ . sinj5. sinC, which is Example (25) on page 194. 303 SIDJIS AND Amies OF A TJt/ANOlS. following formula : a = ft cos C + c cos ^ a h • A — -. — = S^ r ^1 «ftc 2bc ' 2S •(ii), (iii), COS Bin (vii), ^_ Ai;^ I 2 " V ~br~ , (in •* o + c 2 EXAMPLES. LXIV. In any triangle ABC prove the following statements : (1\ sin^ + gsin^ sin^ a + 26 ~' (2) s^-'^-m^s in^g sinac a^-m.b^ ^ ~^ ' (3) «cos^+6cos5-ccosC=2ocos^.cos5. (4) (a + fi)8in|'=ccoslr:?. (6) (6-c)cos^=a8in^Z^ (6) cos^ sin ^. sin /7' coaB coaC sin C. sin ^ sin ^. sin i?^"' iV" il-' 5 'I 'I ^ •I ir! 204 <7) (8) (9) TRIGONOMETRY. LXIV. aBin(Z?-C^+68in(C-J)+c8in(J-J5)-0. a — h COS B- COS A C " 1+C082 + 5i£'2+C'/'2)^3(^2 + J2 + ^^^ (16) If D is the middle point of BC, cot ADbJ' " ^ ~4^' (17) If fl?, e, / are the perpendiculars from A n r .u opposite sides of the triangle, then ' ' ^ °'' ^^^ «8in4 + 6sin5 + csinC=2(rfcos4+.cos5+/cosO «.„liT.!°i*^r^^^!°P'^« ^ "^« ^PPe-^x might be work.. .. .^« ■•ill:: If; 1 »fi ■\ '.: .;il, I 41 (I ( 206 ) CHAPTER XVII. On the Solution op Triangles. 250. Theproblem known as the Solution of Triangles may be stated thus: When a suffiderU number of the parts of a trumgle are given, to firid tl^ magnitude of each of the otJier parts. 251. When three parts of a Triangle (one of which must be a side) are given, the other parts can in general be determined. * There are four cases. I. Given three sides. [Compare Euc. I. 8.] II. Given one side and two angles. [Euc. I. 26.] [II. Given two sides and the angle between tJiem. [Euc. I. 4.] IV. Given two sides and the angle opposite one of **'^°'- [Compare Euc. VI. 7.] Case I. 252. Given three sides, a, b, e. [Euc. I. 8; VI. 5.] We find two of the angles from the formul» tan .^=_ /{s~b)(s-c) 2 V s{s-a) tan I = /{s-c){s-a) 2 \/—7{^) — The third angle C = 180 - il - 5. or, ON TEE SOLUTION OP TRIANGLES. 207 263. In practical work we proceed as follows : log tan - = log /(g-6)(a --c). 2 V 8{8-a) ' ^tan^.lO = niog(.-6)-.log(,.c)-Iog._log(._,)j. Similarly, ^ '^ f - 10 = i {log (. - c) + log (, - „) . log ._ log (, _ j,j 254. Either of the formulie sin - = /(»-6)(»-cj ^ .--_ _ 2 V ■ K ' °°° 2 ° V {7^ ""■*? »■«> he used as above. The sin^ and the cos| formula are either of then, as convenient as the tan ^ formula, when o„. of the angles only is to be found. If aU the angles are to be found the tangent fomuUa . convenient, because we can find the L tanS" twohalf angles from the same>.,„gs, viz. log ., log'(.-„t j:;r-Ti«:;tgXr' ^'^^ '»»- ('->• '^ <»-%. find'TZ^. °'™'' "-^''•«»' '=l««-^8. „=301-47 chains. He«, .=38305,.- a=107-70, .-a=i93-77, .-,=81-58. £tan|=10 + i{I„gl93-77+l„g8I-58-log383-05-l„g 107-70} [from the tables], fi whence | =310 45' 28-5"; .-. ^=630 30' 67", 208 Also B TRIGONOMETRY. Ztan-=l0+i(log8l.58 + logl07-70-log383.05-logl93.77} =9-6366287 =Z tan 180 59' 9-8"; .-. 5=37058' 20"; C ^\m^ - A - B^^m^ ^^l' , 255. This Case may also be solved by the formula cos^ = *!±flzii^ 26c • Let «=: 15,6=14,0=13. Then the greatest angle is 4. Now, cos A - 14^+132-1 52 _ 140 2x14x13 ~27T4la3'"^= "2^4^15. =cos 67° 23', nearly. [By the Table of natural cosines.] . *. the greatest angle =67" 23'. EXAMPLES. LXVI. ha4 IT""'""^ '''''-''' '-'''-'' y-'^' ^' *^« angle ., Iog674-10 = 2-8287243, log 321 -85 =2-6076535 log 160-83 = 2-2063671, log 191 -42 = 2-2819873' Ztan200 38'=9-5758104. Z tan 20^39' =9-5761934. g 193-77} aula IcuIatioD, ^hen the digits. to whose A. 5. )siries.] tngle Af •~v ON THE SOLUTION OP TRIANGLES. 209 are?84 ^6^2' w ^T'* "^"^'^ '' *'^ *"^°S^« ^^-e sides are 404, d76, 522 chains, having given that log 6-91 = .8394780, log 3-15 = -4983106, log 2-07 = -3159703, log 1 -69 = -2278867 Xtan 36046' 6"=9-8734581,Ztan31023'9" = 9-7853745 logl-0142 = -0061236, log 4-904 = -6905505, log 4-48 = -6512780, log 7-58 = -8796692 Xtan 14038' = 9-4168099, Z tan 15057'= 9-4560641 Z tm 140 39'=9-4173265, Z tan I50 58' =9-4565420.' - (4) 1^-4090,6=3850,0=3811 ,ards,find^,havinggiven log5-8755 = .7690448, log 3-85 = -5854607 log 1-7855 = -2517599, log 3-811 = -5810389, ZCOS320 15'=9-9272306, Z cos320 16'=9-9271509. log 3 = -4771213, Xcos360 42'=9-9040529 log 1-4= -146128, diff. for 60"= -0000942* log2=-30103, Zsin20»42-=9-5483585, ^=^^ ^-^^ -^ ^=^0, a^d «, given Here «^=352 + 2P-2x35x21 xcos500; .-. ^=5'* + 32-2x5x3xcos60''. =25 + 9 - 30 X -643, =14-71. •■" 7^^'^^ °®^ly J or, a = 26-74=about 26f feet, EXAUIPLES. LXVin. (1) Find B and C, having given that ^=.iOO, 6= I3i, (.=72. log 5-9 =-7708520, Zcot2(/> =10-4389341,' log 2-03= -3074960, Z tan 38^36'= 9-9021604,' Z tan 380 37'= 9-9024195. L. E. T. 15 ';h !; ' V. :. ' '-'! 214 TR^rOONOMETRY. LXVIII. (2) Findii and jB, having given that a =35 feet, 6=»21 feet, C=500. log 2= -301030, Z tan 280 11' =9-729020, L tan 650 = 10-331327, L tan 280 12' = 9-729323. (3) If 6=19 chains, c=20 chains, ^=60", find B and (7, having given that log 3-9 = -591065, Z tan 20 32* = 8-645853, L cot 300 = 10-238561, L tan 20 33' =8648704. (4) Given that o=376-375 chains, 6=251-765 chains, and (7=780 26', find A and B. L cot .390 1 3'= 10-0882755, log 1-2461 = -0955529, Z ban 130 39'= 9*3853370, log 6-2814= -7980565, Z tan 13040'= 9-3858876. (5) If a = 135, 6 ='105, C= 600, find A, having given that log 2 = -3010300, Z tan 120 12' = 9334871 1, log 3= -4771213, Ztan 120 13' =9-3354823. (6) If a = 21 chains, 6 = 20 chains, t/'= 600, find c. (7) Find c in the triangle of example (5). (8) In a triangle the ratio of two sides is 5 : 3 and the in- cluded angle is 700 30'. Find the other angles. log 2 = -3010300, Z cot 350 15' =10-1607464, Ztan 190 28' 50"= 9-5488864. Case IV. 261. Given two sides and the angle opposite one ot them, as 6, c, B. [Omitted in Euc. I; Euc. VL 7.] First, since sin C = csin B sin C sin i? ' G must be found from this equation. and, ON THE SOLUTION OP TRIANOLES. When C is known, A = IW-B~C hsinA 215 a = Which solves the triangle. sin^ 262 We remark however that the angle G, found from the tngonometrical equation dn r n ' whprA r \, o« ~ ^qua^ion fnn t=a given qvmitity, than 90^ I T "^ "^ ' '"'^«^^' ^^« *^« values, one les than 90 , and one greater than 90". j-^^t 2341 The question arises, Are both these values admissible? This may be decided as follows: f h« ^* ^1 "" """.* ^T *^"^ ^^°' ^ °^"«* ^« !««« than 90- : and the smaller value for C only is admissible. If 5 is less than 90" we proceed thus. • 1. If b is less than c sin £, then sin C, which = ^'^^'^-^ i. greater than 1 This is impossible. Therefore if \ ^ less than c sin B, thei^ is no solution whatever. C-900 'L?.r'' *" ?^^^' ''^^ '^ ^= ^' ^^"^ *^erefore O - 90 , and there is only one value of C, viz. 90". 3. If 6 is greater than c sin B, and less than c then B « less the. C, and C; may be obtuse or acut« 'l' "hf case C may have either of the values found from the eq^ tion BmC = -_ . Hence there are two solutions, and the triangle is said to be ambiguous. 4. If b is equal to or or greater than C, so that the smaller greater must value for C only is admissible than c, then B is equal to be an acute angle; and I';, HJ . !^' IM ■'■| 15—2 216 TRIGONOMETRY. fmi i 263. The same results may be obtained geonif^tri- cally- ConstructioEL Draw AB = c; make the angle ABD = the given angle B; with riontre A and radius = 6 describe a circle; draw AD perpendicular to BD. Then AD = c 6m B. 1. If 6 is less than csin^, i.e. less than AD, the circle will not cut BD at all, and the construction fails. (Fig. i.) 2. If b is equal to AD, the circle will touch the line BD in the point D, and the required triangle is the right-angled triangle ABD. (Fig. ii) 3. If 6 is greater than AD and less than AB, i.e. than c, the circle will cut the line BD in two points C,, C, each on the same side of B. And we get two triangles ABC^, ABG^ each satisfying the given condition. (Fig. iii.) 4. If 6 is equal to c, the circle cuts BD in B and in one other point C; if 6 is greater than c the circle cuts BD in two points, but on opposite sides of 2?. In either case there is only one triangle satisfying the given condition. (Fig. iv.) (omf^tri- ?Z) = the jcribe a jO he circle (Fig. I) the line I is the i.e. than C, each s ABG^, ) ' and in cuts ^2) ise there Ficr IV \ --0- -• •/ ON THE SOLUTION OF TRIANOLES. 217 264. We may also obtain the same re8.,If« «i v • caUy, from the formula ' algebrai- 6'=c' + a'-.2caco8A In this h, c, ^ are given, a is unknown. Write ^ f.r and we get the quadratic equation ' " ^ -^cco^B.x-b* ~ c". Whence, a:' - 2c cos 5 . a: -. c' cos' ^ = 6» - c' -f c' cos' 5 ' = ft'-c"8in'^; •'• *"=cco8J5±^i'»T7^i^r^^ Let a., a, be the two values of a: thus obtained, ihen a, = c cos 5 + Jb' - c» siV^) a, - c cos 5 - ^6^Tr^5^^|^^ J • deciL'^L\lw3T '^^ ^^'"^^'^^ ^ ^'-^^^^^^' -^ '^ tive^'so th?t «" '"' *'" "" ^' *'^" (*'-Csin«i?) is nega. live, so that a„ », are impossible quantities. 2. I[ * is equal to c sin 5, then (6' -c- sin- 5) = and a. = a. ; thus the two solutions become one. ' 3. If 6 is greater than c sin 5, then the two values a ccos^, 1.6. unless 6'-c"Hin»« « « r, f (Tsm if >c'cos'j5, i.e. unless A« , *' >c'. 4. «»« equal toe, then «, = 0; if J i,g,^ter than „ Ji -nese c^eo „, xa cue oi5iy available solution. i f ) . fr M / .!.; ^ i . ' 1 > i ' i j'l i 218 TRIOONOMETRY. 266. We give two examples. In the first there are two solutions, in the second there is only one. Example 1. Find A and C, having given that 6«379'41 chains, c» 48374 chains, and J?— 34<> 11'. X sin C= log + Z sin .5 - log 6 - 2-6846120 + 9-7496148 - 2-5791088 =9-8551180=Z8in45»45'; . -. (7= 45« 45', or, 1 BQO - 450 45' = 1 340 1 5'. Since 6 is less than c, each of these values is admissible. WhenC= 46«45', then ^ = 100<>4'. When C=134» 15', then ^1 = 11034'. Example 2. Find A and C, when &»483-74 chains, c=*379*41 chains, and ^ a 34011'. i sin C«» log c + Z si n 5 - log 6 =2-6791088 + 9-7496148 - 26846120 »=9-6441116=iX8in26«9'; .'. C= 260 9', or, ISQfi - 26" 9' = 1630 51'. Since h is greater than 0, C must je less than 90^, and the larger value for C is inadmissible. [It is also clear that (1530 51' + 34« 11') is > I8OO]. .•.(7=260 9', 4 = 119040'. EXAMPLES. LXIX. (1) If 5 = 400, 6 = 140-6 feet, a = 1706 feet, find A and C. log 1-405 =-1476763, Xsin400 =9-8080675, log 1-706 =-231 9790, L sin 510 18' =9-8923342, X sin 610 19' = 9-8924354. .379-41 EXAMPLES. LXIX. (2) Find B and C, having given that A=b(^ b a -97 ohaina, and that ' 219 ■119 chains. logM9«-075647, ZsinSOO =9-884254, log 9-7 =-986772, Zsiu 700 ^g-gyggsg' X sin 700 1'= 9-973032. (3) Find B, C; and c, having given that A«b(fi, b^rt o-=119 (see example (2)). ^ uvr, o-a/, log 1 -663 = -191 169, L sin 380 38' 24" = 9795479, L sin 880 3fy 24" = 9999876. that^^ ^'""^ ^' ^^^'""^ ^'^^"^ ^^^ '''^* '"°2*' ^=6^" »9'' ^^» log 2-6 = -3979400, L sin 650 69'= 99606739, log 2-4 = -3802112, Z sin 6I0 16' = 99429336,' Z sin 610 17' =9-9430028. JIZ^:^: ''"*' ■"" '^'^'"' ""> -^- ^ -" »»• logl-765 = -2442771, Z sin 720 4' =99783702, log 1-756= -2445246, Xsin 720 5'= 9-97841 il,* L sin 410 56' 10" = 9-8249V25, Zsin41056'26"=9-8249959; for other logs see example (4). the ^.tion Win he am^t^guT^^ltM^^^^^^^^ t^^ the obtuse-angled triangle in the ambiguous case : (a) ^=300, a= 126 feet, c=250 feet, O) ^=300, a=200 feet, c=260 feet, (y) ^=300, a=200 feet, c=125 feei log 2 = -3010300, L sin 380 41' = 9-7958800 log6-0389--7809578, Xsin8041' =9-17890o/ ,r| \ ' ♦•:; ^'V ■ ' : I 1 ■ ''\^l. 1 ; * !« I 'I i\ N 2^0 TRIGONOMETRY. 266. It saves a little trouble in practice when using . 1 i. 1 ^ sin A , .... , . . the formula a = —. — ^ , to write it thus ; a=6 sin -4 . cosec B. For then, log a = log 6 + Z sin ^ + Z cosec B - 20. Thus the subtraction of a logarithm is avoided. * 267. In the following Examples the student must find the necessary logarithms etc. from the Tables. ^MISCELLANEOUS EXAMPLES. LXX. (1) Find A when a=374-5, 6=«676-2, c=759-3 feet. (2) Find B wheti a=4001, 6 = 9760, c=7942 yards. (3) Find C when a =8761 -2, 6 = 7643, c= 4693-8 chains. (4) Find B when i4=86» 19', 6=4930, c=5471 chains. (5) Find C when 5=320 68', c= 1873-5, a=764-2 chains. (6) Find c when C=1080 27', a =36541, 6=89170 feet. (7) Find c when 5=74» 10', C=620 45', 6=3720 yards. (8) Find 6 when J? =100" 19', C=440 59', a =1000 chains. (9) Find a when B= 1230 T 20", C= 150 9', c=9964 yarda Find the other two angles in the six following triangles. (10) {7=1000 37', 6=1450, c= 6374 chains. (11) C=52<> 10', 6=643, c = 872 chains. (12) A = 76" 2' 30", 6 = 1000, a = 2000 chains. (13) C=540 23', 6 = 873-4, c=752-8 feet. (14) C=18"> 21', 6=674-5, c=269-7 chains. (16) il= 290 11' 43", 6-7934, a =4379 feet. (16) The difference between the angles at the base of a triangle is 17^48', and the sides subtending those angles are 105-25 feet and 76'75 feet ; find the third angle. (17) If 6 : c= 4 : 5, o =r 1000 yards and il = 370 19', find 6. The student will find some Examples of Solution of Triangles without the aid of logarithms, in an Appendix. SOLUTION OF TRIANGLES. 220 (i) •MISCELLANEOUS EXAMPLES. LXX b. ^ In the following Example, Logarithms are not required • they are not practicca Examples, but some of them require for their solution considerable neatness and ingenuity for which reason they are often set in examinations. - Part L (1) Simplify the formula In the case of an equilateral triangle. angS '^^' '''^'' ""^ ^ *"^"^^' are as 2 : V6 : 1+^3, find the the^^gle?' ""'' '' " *"'°^'' "** " '' '^^' 2(^^^-l). ^^ (4) Given C=l200,c=Vl9,a=2, find 6. (6) Given ^=60o, 6=4^7, c^.6 ^7, find a. (6) Given ^=450, B=e(y> and a=2, find 0. ^ j7) The sides of a triangle are a^ 7 : 8 : 13, find the greatest ^J8) The sides of a triangle are 1, 2, ^7, find the greatest fi i?i The sides of a triangle are as a : b : ^(a* + ab + b^) find the greatest angle. ^ tao + o), (10) When a : 6 : c as 3 : 4 : 6, find the greatest and lea^t angles; given cos 360 52'= -8. h" ana least .» *>! ani (11) If ffl=5jjiiles, 6^6 gle. 1008 49" 33' =-65.J 6 miles, c= 10 miles, find the greatest 1 ''f n T '1:; ;j I . .'. f - . 5- 11 :(,?! 220 (ii) TRIGONOMETRY. (12) If a=4, 6=6, c=8, find C; given that co8 54'»64'=-578. (13) The two sides a and 6 are ^3+1 and 2 respectively; C=30»; solve the triangle. (14) Given C= 18®, a= ^5 + 1, c--=V5 - 1, find the other angles. (15) If 6=3, C=120», C-V13, find a and the sines of the other angles. (le) Given A^lOb^, B:=^b\ c=V2, solve the triangle. (17) Given B = 75<>, (7= 30^, c = ^8, solve the triangle. (18) Given ^=45», 0:^^75, 6=^50, solve the triangle. (19) Given ^=80<», c=160, 6=50^3, show that of the two triangles which satisfy the data one will be isosceles and the other right-angled. Find the third side in the greatest of these triangles. (20) Is the solution ambiguous when B=Z(fij c= 150, 6=«75 1 (21) If the angles adjacent to the base of a triangle are 22^® and 11210, g^ow that the perpendicular altitude will be half the base. (22) If a=2, 6=4 - 2^3, 0=^6 (v'3 - 1), solve the triangle. (23) If ^=9», 5=450, 6=^/6, find c. (24) Given 5= 150, 6=^3 - 1, c= V3 + 1, solve the triangle. (25) Given sin 5 =-25, a =5, 6=2-5, find A. Draw a figure to explain the result. (26) Given C= 150, c = 4, o = 4 + ^48, solve the triangle. (27) Two sides of a triangle are 3^/6 yards and 3(^3+1) yards, and the included angle 45<>, solve the triangle. (28) If C= 30", 6 = 100, c = 45, is the triangle ambiguous ? (29) Prove that if A =45« and 5= 60» then 2c=a (1 + ^3). (30) The cosines of two of the angles of a triangle are \ and iJ, find the ratio of the sides, ( 221 ) CHAPTER XVIII. V. i ' ('. ■ 1 i (jy: liE Measurement of Heights and Distances. 268. We have said (Art. 97) that the measurement, with scientific accuracy, of a line of any considerable length involves a long and difficult process. On the other hand, sometimes it is required to find the direction of a line that it may point to an object which is not visible from the point from which the line is drawn. As for example when a tunnel has to be constructed. By the aid of the Solution of Triangles we can find the length of the distance between points which are inaccessible ; we can calculate the magnitude of angles which cannot be practically observed ; we can find the relative heights of distant and inaccessible points. The method on which the Trigonometrical Survey of a country is conducted afibrds the following illustration. lyt 222 ON THE SOLUTION OP TRIANGLES. 26i>. Tojmd t/ie distance between two distant objects. Two convenient positions A and B, on a level plain as far apart as possilple, having been selected, the distance between A and B is measured with the greatest possible care. This line AB ia called the base line. (In the survey of England, the base line is on Salisbury plain, and is about 36,578 feet long). Next, the two distant objects, P and Q (church spires, for instance) visible from A and B, are chosen. The angles PAB, PBA are observed. Then by Case II. Chapter xvii, the lengths of the lines PA, PB are cal- culated. Again, the angles QAB, QBA are observed ; and by Case II. the lengths of QA and QB are calculated. Thus the lengths of PA and QA are found. The angle PAQ is observed ; and then by Case III. the length of PQ is calculated. 270. Thus the distance between two points P and Q has been found. The points P and Q are not necessarily accessible ; the only condition being that P and Q must be vi8iJ)le from both A and B. >bjects. I plain as distance : possible tie survey I is about ;h spires, Case II. are cal- l by Case III. the P and Q Jcessarily must be MEASUREMENT OP HEIGHTS AND DISTANCES. 223 271. In practice, the points P and Q will generally be accessible, and then the line PQ, whose length has been cal- culated, may be used as a nevr base to find other distances. of ILL^"^ '^ '^"^'^ ^-^ ^ '''''''' ''^''' «^-^ '^ Voint Let^ be the point of observation ; P the distant object From B measure a base line BA of any convenient length m any convenient direction; observe the angles PAB PBA and by Case II. calculate the length of BP. Next observe at tl r It ^^,^l«^^*i-^' of Pi that is, the angle which the line BP makes with the horizontal line BM, Jf beinc the point m which the vertical line through P cuts the horizonUl plane through B. Then PM, which is the vertical height of P above B can be calculated, fur PM= BP . sin MBP. Example 1 The distance between a church spire A mui a , Z^ w known to be 1764-3 feet; Cis a distant spire. The Tarlff.^fZ'l-^ '^ "'^^^ '^^ ' '''''' ^-^^^ fj ^^A^^^ a triangle and we know one side c and two angles {A and B\ and therefore it can be solved by Case II. The angle ACB== ISO^ - 94» 54' - 66" 39' = 180 27'. Therefore the triangle is the same a« that solved on page 210. Therefore ^C=6118-9 feet. p iii i. i %mi M ;'! • l|5,i;;.( lilii f ^ll 1 Mil m 224 TRIGONOMETRY. Example 2. If the spire in the last Example stands on a hUl, and the angle of elevation of its highest point is observed at A to be 4® 19'/ find how much higher C is than A. The required bright x=AC.Bm 4P 19' [Art. 268], and AC is 5118-2 feet, .-. log a:=log {AC. 8in40 19') =log5118-2+Zsiii40 19'- 10 =3-7091173+8-8766150 - 10 =2-5857323= log 385-24. Therefore or =385 ft. 3 in. nearly. EXAMPLES LXXI. (Exercises xiv. and lxl consist of cds/ examples on this subject). (1) Two straight roads inclined to one another at an angle of GO**, lead from a town A to two villages B and C ; -B on one road distant 30 miles from ^, ac 1 C on the other road distant 15 miles from A. Find the distance from jB to C Ans. 25*98 m. (2) Two ships leave harbour together, one sailing N.E. at the rate of 1\ miles an hour and the other sailing North at the rate of 10 miles an hour. Prove that tho distance between the ships after an hour and a half is 10*6 miles. (3) A and B scce two consecutive milestones on a straight ruad and C is a distant spire. The angles ABC and BAG are observed to be 120" and 45" respectively. Show that the dis- tance of the spire from A is 3*346 miles. (4) If the spire C in the last question stands on a hill, and its angle of elevation at A is 15", show that it is '866 of a mile higher than A. (5) If in Question (3) there is another spire D such that the angles DBA and DAB are 45" and 90" respectively and the angle DAC is 45" ; prove that the distance from (7 to 2> is 2f miles very nearly. HeASUnBlfBWT OF BEIOHTS AND DISTANCES. 226 ToS^J^tf "^ *"° oonsecutiye milestones on a stmight ^Z? A'. » " "^"""^ "^ » •■<'»«' ™"''<' *»»■ both ^ ,^*^ The anglM C^a and CS^ are observed to be 36» 18' and ll^ar respectively. Show that Cis 2639-6 yards from i og 1760 =3-2488127 Z sin 36» 18'= 9-77233 W log 2639-8=3-42182 ^coseo230l8'=i:.:;36«6 J^ isi Xe^.tbrfi^rt;" i^T nt • -z^n that C is distant 1794 feet and 3140 f«t ^m ^ . r^*"^"^ tively and the angle JOS is ^ 17' sfw!^^ 1 '^ "^I^- fho !,•«« ^ • 1- 5 -°v^ iH oc 1/. bhow that the angle which the hne pomtmg fr, u. 4 to 5 makes with ACh 86» 88' 4!" og 1346 =3-1290461 z cot 29» 8' 30"= I0-2837m log4»34=3«931991 Ztan26.4' ,9"= 9-689866^ top'1 a dirfha *;wrc!,T:d^jr --r '- *"' be 610 50' „_j ^«o ^Q' !^^, ^^ *°^ ^^-^ ^e observed to ?«: C •! 5T6,8 feet '■ ''"™ *"'* '"^ *^™ fr°- log 61818=4-71193 z oosec 41» 18'= 10-1804882. a si^D O ^K*™ ?*1°"' ^ ""' ^ °° »'«"^' S'^a yards apart » ship 18 observed at sea. The angles BAG ART .~ S^ that the distance from ^ to the ship is 8522-7 yards, log 3742=3.5731038 Z sin 810 41'= 99954087 log8522.7=3-9306774 Xcosec 25045' =10-3620649. e i9?o vJ^' ^^^T ^*''''" *"^^ °^^^"^^ P^^ks is known to ^e 4970 ya^, and the angle of elevation of one of them IZ seen from the other is 90 14'. How much higheHs the Zf Th thesecond? Sin 90 14' =-1604555. ^^,.79^^ (11) Two straight railways intersect at an ande of fifto From their point of intersection two trains st^ one on o!^' line, one at th. rate of 40 miles an hour. Fi^TVhe 1 J .f^^ second tr^ that at the end of an hou^ th™ Z^fJ^' apart. Ar.. Either 25 or 15 miles an Lour. (1^26^) '' "^*' r H; t.l Mi: m 11 226 TRIGONOMETRY. LXXI. (12) A and B are two positions on opposite sides of a mountain ; C is a point visible from A and B ; AC and BG are 10 miles and 8 miles respectively, and the angle BCA is 60**. Prove that the distance between A and B is 9"165 miles. (13) In the last question, if ■ he angle of elevation of C at -4 IS 80, and at B is 20 48' 24" : show that the height of B above A is one mile very nearly. sin 80= -1391731 sin 20 48' 24" = -0489664. (14) Show that the angles which a tunnel going through the mountain from Aio By in Questions (12) and (13), would make (i) with the horizon, (ii) with the line joining A and (7, are respectively 6° 16' and 49° 6' 24". sin 60 16' = -1091 ; tan lOO 53' 36" = -192450. (15) A and B are consecutive milestones on a straight road ; C is the top of a distant mountain. At A the angle GAB is observed to be 380 19' ; at B the angle CBA is observed to be 1320 42', and the angle of elevation of C at J? is IQO 15'. Show that the top of the moimtain is 1243*5 yards higher than B. Z sin 380 19'= 9-7923968 log 1760=3-2455127 L cosec so 59' = 10*8064659 log 1 243*5 = 3*09465 Zsm 100 15' =9*2502822. (16) A base line AB^ 1000 feet long is measured along the straight bank of a river ; (7 is an object on the opposite bank ; the angles BAC^ and CBA are observed to be 650 37' and 530 4' respectively. Prove that the perpendicular breadth of the river at C is 829-87 feet ; having given Z sin 650 37'= 9.9594248, Z sin 530 4' =9*9027289 Z cosec 610 19' ^ 10*0568589, log 8*2987 = *91901. MeAsasBMrnr op hewets a«d dibtancbs. 237 *MIS0ELLA1I1!0US BXAMPMS. LXm. mil™ aatr^ee^ln °ff„:lT." "T^"* ^ "* *» «"« of three wu. tf ^r "^^^^^^^^^^^ be 360 20'. 20 minnti 1 * 11 , ^® ^^^^ ^« observed to ,lr J <^ '°° ^""^ (-« ^»" «>■) ("oaec 350 stf) .nUe, per hour. W i, the eWatioA .80 : s.:.TtltCX'>t^Tr .' a ^^2(l+-v/6)• duelwrriao t:iT V"' ^^ '' ^ «^^P^« -t a place station Jl^t; eet from it .h' ' '" .'" ""* °' *^^ ^^"-- Bzmi a teet from it the elevation is I50 ; show that the height of the steeple is f (3^ - 3-i) fegf f^. the f to the horizon, the observer finds that the altitude is 770. Prove that the vertical height of the rock above the first point of observation is 1034 ft. Sin 470=. '73135. (8) At the top of a chimney 150 feet high standing at one corner of a triangular j<^i, the angle subtended by the adjacent sides of the yard are 30^ and 450 respectively; while tnat sub- tended by the opposite side is 30". Show that the lengths of the sides are 160 ft. 86-6 ft. and 106 ft. respectively. (9) A flag statf h feet high stands on the top of a tower. From a point in the plain on which the tower stands the angles of elevation of the top and bottom of the flagstafl' are observed to be a and ^ respectively. Prove that the height of the tower Atan^ . , . Asin^.cosa ^ *® tan a - tauTj^ sin (a - 0) (10) From the top of a cliff A feet high the angles of de- pression of two ships at sea in a line with the foot of the cliflF are o and fl respectively. Show that the distance between the ships is h (cot ^ - cot a) f^et. (11) The angular elevation of a tower at a place due south of it is a, and at another place due west of the first and distant d from it, the elevation is 0. Prove that the height of the tower is d . (2sina.sin/3 ^^ _ _ 1 A — — - I — ■ 1^ — , •Joot^P-coVa* ' " Vsin(a-i3).Bin(a+^) (12) A. man stands on the top of the wall of height A, and observes the angular elevation (a) of the top of a telegraph post ; he then descends from the wall, and finds that the angular elevation is now /3; prove that the height of the post exceeds the height of , , sin/i . cosa the man by A^^— J. , 47^ and iclined at B altitude i the first Qg at one ) adjacent that 8ub- tha of the f a tower, ^he angles I observed the tower jles of de- le cliff are I the ships le south of I distant d tie tower is [ght A, and 3h post ; he ,r elevation le height of feASUJlJ,l,SI,T OP UBIOHTS AXD DJSTA^CSs 229 (J^rt! ^^*^' ^"^"^*' '^'^''*'*^" °^ ^^' «"«>°»its Of two spires ^008 .cosec03-„).co8ec(y-„)feet (16) A man walking along a straight road observes that fh« greatest angle which two objects subtend at his ^e isV Frl the phce where this happens he walks a yards the obiL /. the tower the fla^ra+jiff «„L a ^a^ng 2 ^ feet towards ue place where this happen ' he waits „ y^ aJZ V f" then appear i. a at^ight Une „.Ung an i^;!*";: tT Frove that the distance between the objects is ^^E^o^in^ cosa + cos/9 Tj-Jda tistffiip 16—2 ( 230 ) CHAPTER XIX. On Triangles and Oirct.es. 273. To jmd the Area of a Triangle. The area of the triangle A BG is denoted by A. /ij. L_^ / H. K ^ 'B a r Through A draw HK parallel to 5C, and through ABC draw lines AD, BK, GH perpendicular to BG. The area of the triangle ABG is half that of the rectan- gular parallelogram BGHK [Euc. i. 41]. BG.CH BC.DA Therefore A = 2 a . 6 sin C 2 (i). B^ or. But sin C = ^. %/«(«-») («-6)(»-c); .-. A = N/«(s-a)(s-6)(«-c) = 5^ (ii). f rough ABC the rectan- Euc. I. 41]. .(i). .(ii). Ott TRIANOLES AND CIRCLES. 331 274. Tojlnd the Radiut of the Circumscribing (Hrclo. Fig. I. r«t « e,rc e AA'CS be described about the triangle ABO Let it stand for its radius. Let be its centre. Join Jo and produce xt to cut the circumference in A: Join Ic Then Kg. i. the angles BAC, BAV in the same segmml aZT ' ,'■ l'.*^' ■"«'"' ^^''> ^^'<^ "" »upplemenZ also the angle BCA' in a semicircle is a right angle. Therefore -^ = sin C^'i? = sin C^i? = sin A, or, a 2^=»siu^; .'. 2Ii=.-, a sin A ' 275. Similarly, it may be proved that ^^'m^'' and that 2i?=^ HI sin 0' Hence, a sinui sin^-^iTC'^^^- ^^^'^^^;::^^:^ '^*-. <» »>." * *fi« i'.i illl j'.» 232 1 TRIGONOMETRY. 276. To find the radius of the Inscribed Circle. Let D, E, F be the points in which the circle inscribed in the triangle ABC touches the sides. Let / be the centre of the circle ; let r be its radius. Then ID = IE=IF= r. The area of the triangle ABC = area of IBG + area of ICA + area of lAB, And the area of the triangle IBG = |/Z> , BC = ^ . a, .-. area oiABC^ Jii> . BC + \IE . CA + ^IF . AB = ^ra+^rb + ^rc; or, A = ^(a + 6 + c) = |r.2« = r«. A -S- .•.r=— = -. 8 8 277. A circle which touches one of the sides of a triaLgle and the other two sides produced is called an Escribed Oirele of the triangle. ON TRIANGLES AND CIRCLES. 233 278. To find the radius of an Escribed Circle. Let an escribed circle touch the side BC and the sides AC, AB produced in the points i>., ^„ F^ respectively. Let /, be Its centre, r, its radius. Then The area of the tri ngle ABG = area of ABI^G - area of I^BC, = area of IfiA + area of I,AB - area of /,i?(7, A = K.^, . CA + ^I^F^ . AB- ^I^D^ . BG = ir,(6 + c-a) = ^r,(2a-2a) A ^ or 'I'i: III r,= »-a a-a 279. Similarly if r, and r, be the radii of the other two escribed circles of the triangle ABG, then S aS- — c * 280. The following results are often useful With the construction of the last two articles j the lines lA, IB, IG bisect the angles A, B, G respectively ; the lines l,A I,B, Ifi bisect the angles BAG, GBF^, BGE^ respect- ively ; so that All^ is a straight line. 234 TRIQ0N0METR7. (i) Two tangenis drawn from an external point to touch a circle are equal ; therefore AE = AF; BD = BF', CE = GD\ AE, = AF,', BD^ = BF^', GE^ = CD^. (ii) 2AE^ = AE^ + AF^ = {AC + CD,) + {AB + BD,) = AC+CB + BA=a + b + c = 28. (iii) 2AE ^AE + AF={AC-CD) + {AB - BD) = AC+AB-BC = b + c-a^2{8-a). Similarly, ii^ = J5i)= («- 6) j CD^CE={8-c). ^ (iv) BD^ = BF^ = AF,-AB = {8^c). Similarly, CE^ = CD, = (« - 6). (v) Hence, BD, = CD, and CD, = BD. (vi) FF,^AF,~AF = 8'-{8-a) = a. Similarly, EE^ = a. or. ON (vii) IL^Al TRIANGLES AND CIRCLES. 235 ^=FF,aeoi .-. //. = a A cos — . (viii) IF=AFt^^;or, r = (,-a)tani /,/; = ^i^,tan^; or, r, = 5tan^. (ix) AF =. IF cot ^ ^ and J3F= IF cot-' .-. c = AF + £F = r(cot^4.cotD; .*. r = oot^.cotf" »ini±f csm^.sm-^ C8in^.sm| cos^ (X) i^.^C'^180'-.^, .../,^C=90»>|. Hence, r.{cot(90»-.D^cot(90»-gj = a; or. r,= a cos — . cos -jr A ' cos^ (xi) r, = ^^, tan ^ , and r = AFi^n ^ ■ therefore 7 f V •* ii '■. 236 TRIGONOMETRY. * 281. To find an expression for the axea of a quadrir lateral maci'ihed in a circle, in terms of the sides. Let DEFQ be the quadrilateral ; and let d., e,f g repre- sent the sides. The angle F= 180"- 2). [Euc. iii. 22.] The area of GDE=^ \dg sin D. [Art. 273.] The area of EFG = ^e/sin F= ^e/sin (180" ~ D) = ^efamD. Thus the area of the quadrilateral is i {dg + ef) sin B. We must find an expression for sin D in terms of the sides. From the triangle EGD we have F6^ = d' + g'- 2dg cos D. [Art. 239.] From the triangle EGF we have EG' = e*+f-2efcoHF = «»+/» + 2«/'co8i>; [For F=lSO'-D] .'. d' + g'-2dgcoBD=ei'+f' + 2efcoaJ); I II qitadrir g repre- III. 22.] rt. 273.] IS of the t 239.] m-D] ON TRIANGLES AND CIRCLES, 237 2(dg + e/) cos D=^d' ■g'-e'-f ^{dg + e/y > r. Bin' D JJ^l±£fd!^_ttz^~fy The numerp tor of this expression is equal to {{2dg + 2e/) + (d'^g^ _ ,._/«^j ^^^^ ^ 2e/)-(d'^g^ _,._/.)j = {{d + i^)' - (a -/)'} X {(e +/)' - (d-gy) ^\i'^f) + {d-g)){{e^f)-{d-g)) -^^■'9^e-f)id^g^e^f){e^f^d-g){e^f^g-d). Let 2s stand for {d + e +/+ g). Then the above expression may be written (2s - 2/) (28 - 2e) (2s - 2g) (2s - 2d). Therefore sin'i) = iJgllJHgjI ^ (« -•/) (s -y) (dg + e/y or sm Hence the area of the quadrilateral DEFG which is |(c?gr + e/) sin Z) ^(s - '2^)(s-«)(s'^r7y^ 9)' I ff 238 . TriQONOMETRY. * 282. To prove that if I amd O a/te the cetiK-res of the inacribed and circumscribed circles of a triar^jlCf tht^a. Let 01 produced cut the circumscribing circle in E and K. Join lA, IC; produce lA to cut the circle ABC again in V. Draw the diameter VOW and join GW. Then the angle VIC = lAC + ICA = ^A +iC = i{A + C). The angle ICV = ICB + BCV==IGB-^BAV=\C-\'lA=^:^{A-\rC). Therefore the angle VGI= YIG, and .-. F/= VG. Now OE* - OP ^HI.IK [Euc. ii. 5] = VI.IA=^ VG.IA. [Euc. III. 35.] A But VG=-VW,BmGWV=2R.BmGAV=2R.sixi 2» and IA = . A' ^2 [See Figure page 234.] Hence A or i/A ' — OP = 2^ sin ^ X -' --- 2 . .d R'-0P=^2Rr. ON TRIANGLES AND CIRCLES. 239 f .|? EXAMPLES. LXXm. (1) Find the area of the triangle ^jJC when (i) a =4, 6= 10 feet, C=30«. (ii) &=5, c=20inche8, 4 = 60». (iii) c=66§, a=:16 yards, i?=l7o 14' [sin 170 14'=-29626]. (iv) a=13, 6=14, c= 15 chains, (v) a=10, the perpendicular from A on 5C=20feet. (vi) a-625,6=505,c=904 yards. Pi Jf ^ rif f .^ ^f " ""^ *^' ^"'^"^^^ ^°<^ ^^^'h of the Escribed Circles of the tnangle ^5C when a=13. 6 = 14, c=15 feet. (3) Show that the triangles in which (i) a=2, ^=600. (u) 6=§ . V3, ^=300 can be inscribed in the same circle. (4) Prove that A=^ ; find i2 in the triangle of (2). (5) Prove that if a series of triangles of equal perimeter are descnbed about the same circle, they are equal in aC (6) If^=600,a=V3, 6=^2, prove that the area =i (3+^/3). (7) Prove that each of the foUowing expressions represents the area of the triangle ABC : "preaenra abc (i) 4R- (iii) rs. (ii) 2i22sin^.sini?.sina (iv) Mr (sin A + sin £+ sin C). (v) Ksin^.sinC.cosec^. (vi) mcosec^cos^cos ^ 2 2 2 * (vii) (rr,r,r,)i. (viii) ^ («2 _ 42) gj^ ^ . «i„ ^ ^ ,^,^ ^^ _ ^^ ,1. ■ i; - ' Mi! If ii :P 240 TRiaONOMETRY. LXXIII. Prove the following utatemeuts : (8) If a, 6, c are in A. P., then ac=ijrE. (9) The area of the greatest triangle, two of whose sides are 50 and 60 feet, is 1500 sq. feet. (10) If the altitude of an isosceles triangle is equal to the base, R is five-eighths of the base. (11) B{BiaA+amB+BinC)==s. (12) 6c=4i22(co8^+co8-S.cosC). (13) If C is a right angle, 2r + 2R=a + b. (14) r2r^+r^ri + Y^=s^. 1 1 11 V ' 7>/» /«/» /»A Of R ' ho ca ah 2rR' A B B (16) fjcot -g =rjCot -^ =rjCot 2=r cot g- . cot ^^ . cot „ (17) ri + r2=ccot^. (18) 1 + i + i.i. (19) TiZl^'jqt^t, ^ ' a b To (20) Ti + r^ + r^-rr^AR. (21) a.6.c.r=4i2(«-a)(»-6)(«-c). (22) The distances of the centres of the escribed circles of the triangle ABC from that of the inscribed circle are 4ft sin - , 4ift sin ^ , 4R sin . (23) If il is a right angle, rg + rj = a. (24) In an equilateral triangle 3R=6r=2ry fOf\\ ^1 u. ^2 ''s ■^ ^ ( 241 ) * CHAPTER XX. On the Area of the Circle, the CoxsTRucriON of Trigonometkical Tables, Aa 283. Let r be the radius of the circle described about a regular polygon of n sides. Let be the centre : IfK one of the sides. ' /* ['»ii ^^v,. _----^ From draw OD perpendicular to BK, bisecting both ^^ and the angle JIOK. Then, since the polygon has n equal sides, n times the angle ffOi: make up four right angles .•.^Oir=^,and...i)Oir = ^. Hence, the perimeter of the polygon, which is n times HI^, = 2n . DH=. 2n . OH sin DOB the area = 2nr sin - ; the polygon, which is n times the area of BOK, = nOD . BD = nr COB - . r sin '^ s» nr" sin - . cos - . n n 242 rRiaONOMETRY. 284 i^jet r be the radius of the circle inscribed iu a regular polygon of n sides. Let bo the centre, UK one of the sides. » H ^ M f K From draw OM perpendicular to HK, bisecting both HK an-^ the angle HOK. Then since the polygon has n equal sides, n times the angle BOK make up four right angles, .-. ffOK= — and .-. MOH=^-. Hence n n the perimet : ' of thp polygor which is n times HK^ = 2w . MH^ In . OM t&n MOB - 27trtan-. The area of the polygon, wl on is w tiraea the a' oa ofEOK, IT = nr*tan n = the radius x half the perimeter. 286. Let a circle of radius r have a regular polygon of n sides circumscribed about it, and also a regular polygon of the Ban number (n) of sides inscribed in it. ON THE AREA OP THE C J ROLE, 243 The perimeter perimeter ( 'the tan n (1). 'Circumscribing polygon is; inscribed polygon is 2nrs The ratio of these perimeters ia 1 : cos ^T Thea^eaofthecirc ascribed poly .on I nr" tan ^ . The axea of the inscribed poly.n, -^ ^r-sin %o.J. The ratio of these areas is 1 : cos'^ '' "" r'^ 286. We must assume the two following axioms : (i) The circumference of the cirrl^ i. c • between the Derim«tpr nf • ^^ '" magnitude wie penmeter of a circumscribed anrl i\.S e inscribed polygon. '"'"oea and that of an Now, when n ig increasivl '^ ,*« j* • , , increased, ^ is dimm hed, and therefore (bjArt. 94) cos ^ approaches 1. Hence, as the number of the sidfis nf iV t. Art 285 is increased tb«,V ' • 7 ^^"^ Po^Jgons in ^ increased, their penmeters approach to equality approach the circumf^rtr^oTtllr^ '^"^^^^^ "^^^ Therefore, the circumference if a ..iV^u • xi. . L. E. T. (CY. Art. 32.) i : j-| f »iir m n 17 S44 TRIGONOMETRY. 287. In liko manner it follows, from (ii) Art. 286, that the area of the circle is the ' Ivmit ' of the area of a regular inscribed (or circumscribed) polygon when the number of the sides is indefinitely increased. Now, twice the area of a polygon circumscribed about a circle is equal to (the radius x the perimeter). [Art. 284] This is true however great be the number of the sides. It is therefore true when the number of the sides is in- definitely increased. Therefore it is true for the circle itself Hence, twice the area of a circle = the radius x the circumference. Or, the area of a circle = ^r (2irr) [Art. 34] = Trtr, EXAMPLES. LXXIV. (1) Prove the surd expressions of Ari 37 for the ratios of the perimeters of certain regular polygons to the diameters of their circumscribing circles. (2) Prove that the area of a regular polygon of twelve sides described about a circle whose radius is 1 foot is 3*215 sq. ft. (3) Prove that the area of a square described about a circle is f of the dodecagon inscribed in the same circle. (4) Find the perimeter of a regular polygon of 100 sides (i) when described about a circle of 1 foot diameter, (ii) when inscribed in the same circle. Ans. (i) 3*1 4263, (ii) 314108 ft. (5) An equilateral triangle and a regular hexagon have the same perimeter : show that the areas of their inscribed circles are as 4 : 9. "" """^ ^'^^ OF TSt' CROLE. 246 (10) The triangle formed fr^r*, pentagon, hexagon fnd deTafon Zcribl .r' '' ^ "^^- right-angled. ^ "iscnbed in the same circle, is (11) Prove that the area nf o« •-« i about a circle is equal to thr^roduct 7fh^ f '''''' '"^"^«^ perimeter of the polygon. ^^ ''^^'"^ ^^^ i^aJf the (12) The area of an irrefniloi. »^^l sides circumscribed aboutT d J ^^^'"^ ^^ ^ ^^«° °"^ber of «um of eveiy alternate^al ^ '^""^ *" *^« ''^^"^ >^ the It riL!t!";'''^ ^'"^^ -^osea.aisoneacreis39J,ds. is the hundred and Miil^'tJ ^''^ ^\ ^^'^'"^ °* ^^^^ a soUd cylinder. Find SZ^rl t"^ '"?' ^' ''^"^^ "P ^^^ cylinder, ^n,. 9-675in. ^^^^^^^^^^ *»»« diameter of the (16) The diameter of a mil «* ^ x . "ess of ' a carpet is the efehTw r^' '» ^ f^"' »°'l «>« thick- of the oarpet7!l^. X '"'"' ^' " «>» l«°«tl> 17—2 ^m 246 TRIGONOMETRY. On the Construction of Trigonometrical Tables. 288. To prove that, if be the circula/r measure of an angle less than a right angle, sin $, 9, tan are in ascending order of magnitude. 0-^ ::1L Let HOP be an angle (6) less than a right angle. Make the angle HOF' on the other side of OR equal to BOP. With centre and any radius OR describe the arc P'RP. Draw FT, FT to touch the circle at P and F. FT and PT will meet on OR. The line FP is bisected by OR at right angles at M. Then, since the circumference of a circle lies between the perimeter of the inscribed and circumscribed polygons, it follows that the arc FRF lies in magnitude between PMP and PT + TF. In other words, F'MP, the arc FRF, and FT ^r TF are in ascending order of magnitude. Therefore also their halves MP, RPy TF are in ascending order of magnitude. MP RP TP_ OP' OP' OP' And so also are That is, sin 6, 0, tan 6 are in ascending order of magni- [Q.VQ = — XT— is the circular measure of the radms_ angle referred to. VONSTmOTWN OF TRIOONOMETRIOAL TABLES. 247 289. Hence, 1, -1. J „ . ' sin ^ ' co^^ a^e in ascending order of magnitude. 290. To prove tluu the 'limif ofJ- ^k«,i, n . , « bmy the cvroul,^ ^^, ^f^,^ ^^ ^^^^^ ^ The value of ^^ „,, ^,^^^ j ^^ 1 ^^ ^ «ed, 00,. approaohes 1; and tHe ^Z! B becomes, the more nearly does « i , TWo«, by diminishing » sufficiently, we can make ^^ differ from 1 by less than any assignable quantity however small. ^ This is what is meant when it is said that HKe limit of 291. The student must notice carefully thaf fi h • the number of radians m rK. i "^eiuiiy that here is «i raaians m the angle referred to.* Rvample. Prove that t^ limit of -JL- ^hen^-n ' 180 •^ sin wO ' ^'^^ *^=0> w — . Let d radians =wO, then ^ = -iL . « 180^ »r 180' ••**=-;;:-» , W^- - - diminished, ^ is diminished also, and the limit of 8in^' when ^=0, is 1. Therefore the limit of -!L (^u:^. _180 e\ , ,«, sinnoi^^«»» 180 ^ \ , ion IT (ir.* 6^" ™ "" "~"'" «■»' » »""■" « -"" - . -U Of angle m ! ! •f* 248 TRIGONOMETRY. 292. To prove that if 6 be the circidar measiMre of a positive angle less than a right amghf sin 6 lies between 6 and It has been proved that sin tf is less than $ (i). And that 6 is less than tan 6. Therefore ^ is less than tan „ ; or, ^ cos^ is less than sin ^. ^ ^ J J 2 Now, sin ^ =2 sin ^ . cos ^ ; [Art. 166] (9 9\ 9 °°^ 9/ COS 9 2* 9 A i.e. greater than 9 cos' ^ > i-e. greater than 0-9 sin" - ^ 2 ,.i<^ sin 10". In the above, 9 is the circidar measure of the angle. The circular measure of 10", correct to three significant figures, is -0000484.... [Examples X. (17).] Let ^=-0000484.... Then, tf-;J^=^_;J (•0000484....)"= ^-•000000000000028... Hence, 9 and {9 - J^) are decimal fractions which agree in their first twelve figures at least. CONSTRUCTION OF TRIGONOMETRICAL TABLES. 249 And since sin 10" lies between these fractions, therefore tJie first twelve decimal places of sin 10" are the same aa tliose of the circular measure of 10". Hence, if the value of ,r be given to a sufficient number of decimal places, we can calculate the circular measure ot 10 , and therefore also sin 10", to 12 decimal placea 294. To show how to construct a Table giving the Trwo- myrmvncal Ratios of migles which form an oHthmetical m-o- gression having W for the common difference. In the identity sin(n + l)a + 8in(w-l)a = 2sinna.co8a, l!! ^^}^'\ ?^5 'T''' ^ *^^^' "^*^^ «^^« <^f ^11 angles at intervals of 10" to have been calculated up to n . 10". kno^!"' '^ ^"^ ~ '^ "' '"^ ""^ "^^ ^°' '^ [ = ^/^^iI5nF] are Therefore by the above formula sin (^ + 1 ) „ can be found Hence, since we know sin 10", the sine of 20" can be found ; and then the value of sin 30"; and so on. 295. When the sines of angles up to 45° have been calculated, the rest may be found from the formula siu(45» + 4)-8in(45°-^) = ^2.sin^. ' Also, when the sines of angles up to 60" have been cal- culated the remainder up to 90° can be found still more easily from the formula, sin (60° + A) - sin {600--A)1sZa The other ratios may be found from the following : cos^.sin(90°-^), tan^ = ^j, cot^ = tan(900-^), oosec A = -^-^ , sec ^ = cosec (90° - A). (*','; ~:,4 II 250 TRIGONOMETRY. 296. Smce3»=18»-15»,thesine8ofanglesdifferingby 3 or by any multiple of 3°, can be found independently. (See p. 120.) These values may be used to teatth^ accuracy of the Tables calculated as above. 297. The following fonnulae may be also used to test the accuracy of the Tables, 8in(36°+J.)-sin(36»-^)-sin(72''4.^)+sin(72»-^)=sin^, cos(36V^)+co8(36»-^)-cos(72»+^)-cos(72«-^)=cos^'. They are called fonnulaB of veriflcation. On the Limit op the Visible Hoeizon. 298. The surface of the sea is very nearly that of a sphere whose radius is 3957 miles. The height of the highest mountains on the globe is less than 6 mUes. Thus a point must be considered to be at a very considerable height above the surface of the sea if its height is a thousandth part of the earth's radius. 299. In the figure, let be the centre of the earth, PRF part of the surface of the sea, T a point of observation TR its vertical height. ' Draw TP, TF tangents to the earth's surface. Then PMF is parallel to the 'horizontal plane' at R. The angle TPM is called the dip of the horizon at T. It is the angle of depression of the most distant point on the horizon seen from 1\ It must obviously be a very small angle, since TR is so small compared with RO. ON me LIMIT OF THE VISIBLE BORIZON. 251 300. If SO be produced to out the circle again in L tten TF-^TR.TL [Euc. ,„. 36] ' =J'R(Rl+tm)^2tr.ro + tb: part ot UO, and therefore TR* will Ka mn.i, i thousandtJipartof 2^i?.i?a ''' *^'^ * Hence, the formula rP»=2r^.i?(9 i e T'P" ^ • t e^h-s radius x vertical heigh,^^, "give 1= Z of rP correct to at least three significant fig^es. Example. Three times the hmaht in {mi.-nf ,1.. „i ^ , Here, Tf^pz x 7914 miles. tet / be thr, n. n.ber of feet in BL, then the number of nales n. iez i„ sio ' '"' ^ •« th' ■»>»l>er of miles in T-P, then '«'"i/„^x7614 = f nearly. «.aD. * ■■ 5280 252 TRIGONOMETRY. EXAMPLES. LXXV. (1) Show that the limit of ^ R^ sin - (i.e. the area of a poly- gon of n sides inscribed in a circle of radius i2), when n=aois (2) Prove that the hmit of nr^ tan - , when w= qo , is ttt^. (3) Given that 77=3-141592653589793. ..prove that the cir cular measure of 10" is -00004848136811... (4) Prove that2 sin (720+^) - 2 sin (720- ^)=(^5 _ i) sin^i; and that 2 mn (36" + ^) - 2 sin (360 -A)= (^5 + 1 ) sin .4. (5) If a mast of a ship be 150 feet high, show that the greatest distance seen from its top is 15 miles nearly. (6) Prove that if the dip of the horizon at the top of a moimtain is 1^ 26' [=tan~i -025], the mountain is about 6530 feet high. NOTE. The definitions given in Arts. 75, 78 of the Trigonometrioal Ratios are now used exclusively. The NAMES tangent, secant, sine, were given originally to quantities defined as follows. Let ROP be any angle. With centre and any radius describe the arc RP. Draw PM perpendicular to OR and PT perpendicular to OP. (See Figure on previous page.) Then PR is called an arc, PT is the tangent of the arc PR, OT is the secant of the arc PR, MP is the sine of the arc PR. The name sine is derived from the word sinus. For, in the figure, PMP' is the string of the "bow" {arctis), and the string of a bow when in use is pulled to thr archer's breast. The co-tangent, co-secant and co-sine are respectively the tangent, secant and sine of the complement of the arc or of the angle. The sine, tangent, etc. of the angle are the same as the measures of the sine, tangent, etc. of the arc, when the radius of the circle is the unit of length. :i •f iQ n=cDiH APPENDIX. The Vernier, the Level, the Theodolite, the Sextant, THE Mariner's Compass. 301. The practical Surveyort has to measure distances and angles, and has also to make plans or pictures, re- cording the result of his measurements. For the measurement of distances the surveyor uses either rods, or chains, or tapes. Rods used in measurement are made of wood, or of metal or sometimes (when extreme accuracy is required, as m the case oi the measurement of the base line of the ordnance survey on Salisbury Plain) of glass. arel^ahtl' ^f *'"°^^f *«' ^^«° '^^^sei to changes of temperature. are liable to change of length ; hence for great accuracy, a Vurveyor must know the exact length of his measuring rod at lu ordirv emperatures; and when making a measurement, must note S TZT °^'r°' '' *'^ ^^^*^^* '' °^---ti-- The Inge of length caused by change of temperature is greater in a rod S metal than n a rod of wood. Hence wood is a vefy suitable materia or measurmg rods under ordinary circumstances A 7al made o cotton or hemp if used for measurement must be carefuUy p'otectei lTnX::7t r""'/^°*^™«^ - -^^^pes sensfbf;ttk :^rSi;^^rt?nrn:^^^^^ A tape of 66 feet can be easily stretched an inch or so. follUin^''^'in*afe waTtekl '^.t'^^ T^ ^' P*«^ori«J description «ucl. as fh. ■;i^i!i M il ''4 m ii TRIGONOMETRY. The Vernier. 302. A vernier is a simple instrument for increasing the accuracy of the measurement of a small distance by one significant figure. 303. Description of a Vernier. Suppose we have a rule (i.e. a measuring rod) of brass, graduated f to tenths of an inch. The vemier is a little slip of brass which slides along the rule. j This slip of brass is a little more than \^ inches long, and a portion of its length 1^^^ inches in length is divided into ten equal parts, by fine scratches on the burface of the metal Thus the distance between each scratch and the next is xVV of an inch or {^ + ^) of an inch ; i. e. this distance exceeds the distance between two scratches on the rule, by an hundredth part of an inch. 304. To read the Vemier. This will be best explained by an example. Suppose the length to be measured is ascertained to be 3 ft. 11-5 inches and a little over. This can be ascertained by the use of the rule (or measuring rod) Now let the rule be so placed that one end exactly coincides with one extremity of the length to be measured; then the other extremity K of the length to be measured will be between the ecratches on the rule mdicating 3 ft. 11-6 in. and 3 ft. 11-6 in. Now slide the CD vernier on the rule till its extremity D coincides with the extremity K of the length to be measured. t Le. having fine scratches upon it, each the tenth part of an inch from the next one. H Qcreasmg ce by one ) have a tenths of les along hes long, 8 divided ce of the B next is distance > rule, by »xplained I 3 ft. 11-5 iring rod). 3 with one ;tremity K n the rale vernier on K of the le next on& THE VERNIER. Rule, m It f , It will be observed that one of the scratchflp «« *u 3 ft. 11'56 inches nearly. For tl,= le-gth exo^ds 3 ft. U-s inches ky j„sl as mnoh as 6 spaces on the wmer exceed 6 tenths of an inch, that is by 6 hund,ed.r„, an 305. A vernier may be used to read the graduations rf a croular arc ; m wl>ioh case it is made curved so as to follow the line of the arc. 306. The student should notice that the advantage gained by the use of the vernier depends on the fact t ! the eye is able to judge with considerable accuracy when two scratches are cr are not, coincident. ^ to the student to discoveA./rh J^ZZ:^ '. I'll m ^ ■ I 1 1 1 ■ 1 ■ i' ' 1 wLiL IV TRIGONOMETRY. The Level. 307. A level is an instrument used for ascertaining whether a given straight line is or is not horizuutal. A- ^ G The essential part of it consists of a glass tube ABC^ in the form of the arc of a circle, which is closed at both ends and is nearly full of spirit. The t ibe, being not quite full of spirit, will have a bubble J) hi it. Thi'. b'al-ble, when the tube is at rest, will only rest at the highest point of the tube. The tube is fixed in a case of wood or metal which is so made that when the base of it FG is horizontal the highest point K of the tube is visible, and the bubble can be seen at rest in it at the highest point of the arc of the tube. This highest point is carefully marked on the tube. To ascertain whether any given line is horizontal it is only necessary to put the base FG of the case of the 'level' in the position of that line and watch the position which the bubble takes up when the tube is kept at rest in that position. If the bubble rests at the position marked on the tube the line is horizontal and not otherwise. 308. The student can easily purchase for himself an ordinary carpenter's level and can make experiments with it. THE TJIEOBOLITE. y The Theodolite. 309. A TheodoUte is an instrument fo, measuring the h^izonlal angle suDtenued at the pr ,n of observation by two distant visible objects. If v^^lo. "^ ^' *^° ''"^^' '''*'°* '^^''^*^ ^^«° '''^ « Pl^°« of obser. Let PM, g: perpendiculars be let fall from P and O resDectivelv to the horizontal plane passing through O. respectively Mn^T^ ^l """"^ ''^ ^'' '"^*'^^'^ "' ^ ^y ^ «^^ « i« the angle Jl/0^' [See Examples lxxvi. (16)]. [The angle subtended at by P and Q is the angle POQ.] 310. The essential part of a Theodolite may be de- scribed as follows. ^ Suppose two circular brass plates to be laid one on the other so that they are concentric and both are free to turn about an axis through their centre. Let the rim of the lower plate be graduated. That is on the rim will be marked 360 lines at equal distances mdicatmg degrees subtended at the centre. Each degree^u be and'^hef^ "^'°,°^'^*- '^' ^-^ling to the size of the ll and the degree of accuracy to which the instrument is to be read Let the rim of the upper plate have inscribed on it a vernier suitable for reading the graduations of th« n.i,.. run. ~" ^ ft " il > ■r'': , f ^ ■ a wmi .o.^X^, IMAGE EVALUATION TEST TARGET (MT-3) /. Cc?. 1.0 ^U2I2£ II ? ^ y£ ■ •I ^^^B 11:25 HIU 11.6 IIU4I nil I U«_ niUiUgl'cpiHJ .Sciences Corporation 33 WIST MAIN STillT WHSTIR,N.Y. 14510 (716)t72^903 4p> ? A% wishes irror (or 1 if the ror as in THE SEXTANT. ^^ of sfh/^ ■* " ''°"°'' °' "'" *«'^'"''P« '"'<' <^^ the line Make the angle C'DB equal to C2)5 Then an eye placed at C and looking directly at the no exactly what an eye placed at C, the eyepiece of a mirror were removed. and^Vr"'"' '*'\*^' '^^-"'^ ^^^ ^^*-««- the mirror and the axis of the telescope is half 01)0' fKo . between the direction of the Lis of th« . , ' T^^^ direction f„>n, 2> of the o£ctT ^'""^' '""^ *'« II. When the half of the object glass (or We .lass^ of a telescope is covered over fh^n »„ i , . ^ ^ ^ i uverea over, then an eye looking through I i it I i -i fii' il ^f . IJ il'^' (I Xll TRIOONOMETRY. m. When looking through the eyepiece of a telescope the eye looks at an image (which is a small picture formed by the rays of light coming from the object looked at, inside the tube of the telescope) of the distant object at which the telescope is pointed. The eyepiece or small glass of a telescope forms in fact a microscope with which the eye sees this image magnitied. .^ Thus in the figure, is the object-glass, E is the eyepiece or microscope ; / is the position of the image. / is called the focus of the object-glass. 316. To describe the arrangement of a Hadley's Sextant. Let the axis AB of a telescope be directed to the edge C of a plane mirror perpendicular to the plane of the paper whose section is CD, 'B'And let EOF bo the section of another plane mirror arranged so that light coming from a distant object Q THE SEXTANT. xiU »\ri.^.r.^"'*' '^"" ™'''°'*^ " '" the direction of the axis of the telescope AB. Then an eye looking through the telescope would see the images of two dUtant objecta P and Q These two images would be superposed. ^ICZ> the light from which is direct. dire^tionto'ti^Tj/ T"' "^ " ""*""' °^J^«' « '" "«> aireotion GQ the light from which is reflected. keeping always perpendicular to the plane of the paper and the angle through which it turns can be observed CnTb'r f""'}"^' ^^ '" t"™"! «o as to be pamllel to CO the reflected image and the direct image would be .^at of the same very distant objajt. The angle through which the mirror EP is tnmafl fVnn, the above position, nntU the image of Q V vS> ik Z te les^pe to half the angle betwl C And^ tL ^ half the angle subtended by P and Q at the obsemr's eye .. z zt orT.:i:.r '"" ■"""* '"^^ '"" " '"' ""'^o" »* -• For, produce OOioA'B', then the angle PCU- is Bxed. Draw Off perpendicular to EF: the aiurla h,«,.j .i. v i in an, niove^cu, is e,ual .o the ^.t rj^J-^r'J^^*^^ ^f. ^ li 1. «iv TItlGONOMRTRY. 317. The foUowiug is a picture of u Hadley's Sextant. AB is the telescope; Cis the fixed mirror; Fis the mirror which can be turned about an axis perpendicular to the plane of ABCF. FN is an arm by which the mirror F is turned. LL' is a graduated arc and N is a vernier for reading its graduations with the aid of the microscope M; K and T are the usual clamp and tangent screw ; // is a handle for holding the instrument. THK SEXTANT. „ the l.gh and the telccopo when looking at a bright object .uch a, the sun ; each glaa. ia on a hinge .o that it J be brought .nto the line of sight or turned back at pIcZ line of the reflected Ught or turned out of it at pleasure. "ttl- ' r ",""' ^'"^""'* ''°»- The observer hoZ l^n h". r • '^^ ''''"»'™"°" «'■"''>'» in hU noting the two distant objects, whose angle he wishes to take, are superposed in the field of the telescope. This he can do even if he cannnf ^> tk» • for «r.v l«„^k « •• • . "^nnot get the images to re/t tor any length of time m the field of view. andlTf "^'' ""^ "'?''' '"''"""^'^ ^^ ""' "^SO of the sun >M the bonbon can with a ,.xtant bo observed with con- twtn ,K r "' "'«•"• *'"' '"'S"''"- 'ii'fnce b<. tween the moon and a star can be observed under like circumstances. * ii! I ! !▼< TRrOOSOMKTRY. t 319. Below wo give a ^gure of a Mariner'B Oompasa. It consists of an ordinary magnetic compass with a card attached to the needle, this card is so arranged that when the needle is pointing along the magnetic meridian the pointer on the card is pointing due North. The Points of the Compass are figured on the card. They are N. =■ North. N. by E. = North by East. N. N. E. = North North East. N.E. by N. = North East by North. N.E. = North East. N.E. by E. = North East by East E.N.t. = East North East. E. by N. = East by North. E. = East. And so on. The angle subtended at the centre of the card by two consecutive points is = | of 90' = 1 1 ^•. EXAMPLES FOE EXERCISE. LXXVIa. ! t • i I. Doflne the terms Bine, cotangent; and prove that if A be any II lan^~|, and tin A aad cm A. for ™l„lT<,'be°wtnTa„'a ,'.'" "''^ "" """'""*« <" '-" — O rJ^'u 12"°^= taui'"" ■»• (1 W' - ^, = - 00, ^. ^ Prove (1) 8in {A + IJ). sia {A-Ii) = sin' J - 8in« 7; ; (2) ■^±**" ^ ^ tan J (^ + B) 8ini-8in5 tfivi{A-B)' 6. Prove that oosM - 008 A cos (60» + A) + sin* (80» - ^) = f . f«„**' Fiu^ *?.® 8'«*t«8t side of tho triangle of which one aidfl i. oim feet and the adjacent angles are 78« U'and 710 24' L .ii°7 J U' - S???^^ }°« f 274 ='46260733, LS8002^'C9%Zl86; ^°«^'-^^76=4.6260«36. oosL ^^^''" ^^ ''^'' trigonometrical ratios in term, of the 8. Prove «in(180 + ^) = -sin4; tan (90 + ^) = -cot 4. SOoLd^^^iVaTseSi: "'^'' °' ""^ *^« *^e'«« ^^^'^ "« -^Itiples of 10. Prove ta«a^- ^-ooa2i< 1+00824* ^11. If tan^. + 8eo4 = 2. prove that ein^^l, when^ is Ie38 than If BinA = 1t. prove that tan ^ j- s+J)=cot{W-^)V 28. Prove sin x (2 cos x-1) =2 sin -cos?? 2 2 * 29. Trace the changes in sign and magnitude of 2 sin g - si n 2g 2Bin^+sin2i9' as e changes from to 2ir. Bv- If the angle opposite the side a be Rno o„^ xt i. t .. remu.uing sides of the triangle, prove that ' ^ *' * ^® '^^ (a+6+c)(6+c-a) = 3&c. f ^ i expr'e^s tT^Z X^^^^Jtl^!^''^ "^r^^^^ «^ *- -«h* angles, number of degS ^t^t^wtleS" ra^r^ S ^ '""'^^ 82. Prove that tan-i«-tan-iy=tan-J £:iiL 1 + ary • 83. Prove sinar(2cos«+l) = 2oos|sin^. 84. Trace the changes in sign and magnitude of sing + 2Bin^ 8ind-28iuitf» as # changes from to 2«-. 'i, and tan i4 = A. Find sin il. 46. Prove (i) sin 24 = ,--!^. ^' 1-f-cot*^ (ii) Show that if -4 -t- B -i- 0= 90«, sin 2A + sin 2B -J- sin 2C=4 cos A cos B cos (7. 46. Find an expression for all the values of $ for which sin d-i- sin 29=0. 47. If in a triangle 6co8ii=acosB, show that the triangle ia isosceles. EXAMPLES FOR EXERCISE. LXXVIa. 257 me-Siref/j"" (SilST;.!"'"!,""''' ?'"• ,"' ""' '"ele whose circular T.^ " ' ' ' ^''' "'"*' oireulor measure is 5 60. Prove that (sin 300+cos300)(8iii 120o+cos 1200) = sin 30o. 61. Prove the formula : (1) C08> (o + /3) - sin» a = cos /3 cos (2o + /S) ; (2) 1+ootocot Ja=co8ecacotia. 62. Solve the equations: (1) 6taii>x-Beo»»=llj (2) Bin 60 -sin 39 =^2. cos 4». .h.i''^,e'^'i.i?ri'i^isi«"xris"£'ii'/'-' '^ '»""'• -^ 64. Given that Bin400 29'=0'6492268, sin 40» 30'= 0-6494480, ^^^ sin-i(0 '6493000). 66. Express in circular measure (1) 10', (2) i of a right angle. U the angle subtended at the centre of a circle bv the side of u. regular pentagon be the unit of angular measurement by what number 18 a right angle represented? ««*«uremeni, oy wnat 66. 67. 68. (1) 69. 6 feet. 60. find If sec o= 7, find tan a and coseo a. Prove the fonuulse: (1) cos''(a-/3)-sin2(o+/S)=cos2ocos2/3; (2) l + tanatania=seca. Solve the equations: 6tan3«+secaa:=7; (2) co8 6ff+cos330', 78» 10' respectively. Determine the SSl5vf^'^Jf^.*°^ ^^^ ^^®*- ^ si"^ 700 30' = 9-974 log io2 = 2-009. L sin 67. 2-3? Which is greater, 126» or the angle whose circular measure is 68. Establish the following relations: (1) cot^A - cos^ii =cot"^ coa^A ; (2) tan4+cot2J=coseo2i4; .3. cos(x-3y)-cos(3g-t/) ^^^ Bin2xH--iIE2P 2sm(x-y). 69. Show that for certain values of the angles 2oosi4=^H-sin^+y°^ean8onhese deduce the formula Bin (1800+ ^) = _3i^^, cos (1800 + ^) = . COS ^ 76. Prove the formula: (1) cot2j=cosec'^-l. Verify (2) + 6»-4a/.8in^siuZ/. If a = 123, JB = 29»17' r-iQfto « i 1 i«.. ' • *' - "^ . find c, hav n« civen log 123 =20899051, loaQ-L 'XmnLn log 3211 = 3.5060103 d?ff f^r 1 - 1 vo' L sin 16043'= 9-4327777 ~ ' (2) 93. 94. inviliab&!'' ""'* °' "^^"^'^ ----«. and prove that it is an whte;i;^^{^S.^-e.ntre^^a^ seoa'Ji ""'"*'"'*' "'^ ^^P^«-'°° f- all angles which have a given Write down the general value of sec-i ( - 9) 98. Prove that '' , 99. Prove that 2cod ./- - j. /rT~^ — z, i — mine the proper signa of the loo^Ht^^f^^loQ^. ''" ^* "^^ ^^^''■ 100. Prove that in any triangle 903 2^ C08 2jB 1 1 62 6«* 101. Provc^ (i) (ii) (iii) 2 cot 2^= cot if -tan J, sin-'?-sin-i^=sin-i|j^ cot {A + ij^} - tan [A - W) = -A££l-:1 ' 28in2J + i 19-2 i!' ■'II 262 TRIOONOMETRY. 102. Solve the equations 008(2« + 3y) = J, 0OB(8x + 2y)=4^3. 103. Trace the changes in the sign and magnitude of sin (r sin x) OB X increases from to 2ir. 104. Solve the equation tan~* X + tan-^ (1 - x) = 2 tan~* ojx - x\ 105. A ship sailing due north observes two lighthouses bearing respectively N.E. and N.N.E.; after sailing 20 miles the lighthouses are seen to be in a line due east ; find the distance between the light- houses in miles accurately to four places of decimals ; having given, log 2= -3010300, L tan 22030'= 9-6172243, log 11-715 = 1-0687423, and log 11-716 = 1-06877U4. 106. Prove (i) fen ^=^7(1::^). (i) (ii) •COStfy tan6d + tan3d . _. .^ i.- -TE—i — 5^=4 COS 26 COS 4d, tan 60 -tan 39 (iii) sin"* | + sin~^ ^=ain"^ ||. 107. Solve the equation 2sin' d- (l+s/3) sin 2(? + 2^3cos'd=0. 108. Show that in any triangle (i) asin(B-C) + 6sin(C-.4) + csin(^-B)=0, (ii) c(co8^ + cosjB) = 2(a + fi)sin«iC. 109. Prove that if D, E, F are the feet of the perpendiculars from A, B, C upon the oppoaite sides of the triangle ABC the diameters of the circumscribing circles of the triangles AEF, BDF, CUE are a cot A, booiB,ccotC respectively. 110. A man who is walking on a horizontal plane towards a tower observes that at a certain point the elevation of the top of the tower is 10" and after going 60 yds. nearer to the tower the elevation is le"; find the height of the tower in yards to 4 places of decimals, having given L sin IS" = 9-4129962, L cos e® = 9-9983442, log 25-783 = 1-4113334, log 25-784 = 1-4113503. 111. Prove thai when sin i4 is a geometric mean between sin B and cos B, then cos 2A=2 sin (450 - B) cos (45<> + B). 112. Prove (i) sin^ (co8 2il+co8 44 + cos6J)=sin34cos4il. (ii) 2sin-i4=coB-ii. 113. In any quadrilateral ABCD prove that ABoobA- BC cos {A+B) + CD cos D=AD, EXAMPLES FOR EXKnciSE. LXXVla. 263 114. Prove that in any triangle fit i^^^.z^SB^^-^'^^ from each of the ships, having given Istntt'^n^''^ ^^ "'« ^^^^^^n i Bin 620 25' 15" - g.ftqgnnr r: T ; „, ^^ ^ ^^ = 9-9852635, ^ -y 8990065, log 1-2197 = -0862530, log 1-2198= -0862886. "'"■-"-'"S^-^-eSS^^^^^^^^^^^^ 117. Prove (i) "°^+2Bin3^+Bm 5^ C084-2C0S34+C03"62 (u) cot-4=cot-i3 + cot-i|. 4sin^-3coseoJ 4coSii-3eeo^ ' 7Ji^.„^=.aV4*-M-f a Ma„g.,„, U2. m, and 1450 a farther distance c feet tho Rnmo^ a * ^^^ *^** ^'^en he has walked ^■. .how «.a. «.e inoatT«',&re t'o^T ""'^^^ cot-i(2cot-/3-cota). Ii«'Swe»°:i,o"tnd^&.""^'=' + 2"°i^ Vd-sm}^). when A lOA rt 124. and then a sin 4 +6 sin 5+c 9''n c=0, a cos J + A nna J5 -}, / '^=0 II ■ ir is* 2C4 TRIGONOMETRY. 125. A man aBOCi.t hi^'h and the an+co8l75»=0 (li) sin» 24» - sin' G<>= J (^5 _ i). ' 143. If i4+B + C+i) = lftO», then co«24-cos2£ + co82C-cos.Z) = 4sin(B + Oco8{C + 4)8inU + B) 144. Given log 85 = a, log 825 = 6, log 245 = c, ^^^ logs, log? and log 13. , ^*^-.„4 *fai? " going due East at the rate of 24 milefl An linn, when will It be 18 milea distant from a town which is on th« N p"; the train af a distance of 24 miles ? *'^° ^•^' ®' 146. then 147. then 148. If then If If 4 8in4tan(4-B) + 38eo4 = 4 8in»4 8ec^ tan id tan fi= 8. tan (it cot $) = cot (t tan 6), 4 tan tf = 2n + 1 ±^(4n' + 4n - 15). «=cos2a + co8o, y=Bin2o + sina, 2x=(x»+j/')»-8(x«+ya). 149. In any triangle 2R sin (7= 6 cos 4 + ^^(a' - 6» sin' J). 160. In a triangle right-angled at C tan-i-^ + tan-i -*-=![. 0-¥c a + c 4 162. It ein-iM+sin-i „_},, pro,e that »m-> m=cos-i b. 154. (oos»+sin»)(ooa2(?+sin2»)=ooa#+coB(S»-J,) li' ff^ ^1 I see TRIGONOMETRY, Sinuplify cos* (il + i?) + cob A-B)- cos 2A cos 2fl. 167. With two units differing hy 10" the measure of an angle if as 8 in to 4 ; find the units. 158. oos7»80'-J( -1 + ^2 + ^8)^/(2 + ^2). 159. In any triangle sin ^ /I + COB ifl - sin ^C _ 1 + tan i B tin \a + oo'ilir^BinliB ~ 1 + ton iC ' 160. In any triangle a', 6' - c», fc' + c' - 26c cos (D-C), are in o. p. Polve 0OB8^ + 8in3^=00B(? + sin^. If x>i8co80 + cos80. y := 8 sin - sin 8^, then x^ + I/l=4*. ooa« 19« ain' 86» - cos 86° sin 18» = ^. Qiven that oo8 0=-^ is one solution of the equation cos 9+ cos 89 a i, find the others. 166. If the diagonal BD of a ovadrilateral inscribed in a circle passes through the centre its area = (< - a) (• - d) s (« - b) (« - c). 161. 162. 168. 164. 166. then 167. 168. 169. 170. If tan (X+£) 3 Standi, sin (2^ + 2B) + sin 2^1 = 2 eia 2B, sin 18» + cos 18» = ^2 cos 270. Solve tan 39 + tan 29 + tan = 0. If i4 + B + C=90», then C08> 2 A + cos^ 2B + cob« 2(7 a:8 8in/i8inJ3BinC-Bin3J sin3£Bin8C + l. In any triangle 6c oob^ ^A+ca cos' ^B + ah cos' i C =» («)'. 171. Prove that 79=ir satisfies the equation 8 cos 9 cos 29 cos 30 = 1. 172. If -i+B + C=90«, then cos ^4 cos B cos C cannot be greater than * ^;8. 17b. 8in-» i ^5 + cot-i 8 = iv. W xt w;'tiiangle ais. ; ,^1 V Bin 9B + d' . 3(7+4 cos f i4 cos f 7? cos JC=0. 175. ^ if the cirde circumscribing an isosceles triangle is equal to the escribed circle touching one of the equal sides the triangle is right angled. EXAMPLES FOR EXERCns. LXXVIa. ftT 176. K-in, + .in«x«l,thenoo.'x + co.«,,i 177. ain 10'' sin 50» ain TO* = l. two together. ' •"* products of their coBiaei taken 181. Prove that lb . F.ndthevaIueaof sin2.ooa4.and tan. when lai„-:^^ of thf sanfe^gT^"^'""'^^ ^^' "^ 2. and 2 ain . coa . are always 184. If D ia the middle point of the side BC, then 186 HL ^^^'^^'+c'+2bccosA. circle from Ac; then' ^" '"'**^'^^ °^ *^« ««^t'« of the inscribed Imn _ 2abe r a + bhc' 187. Wsin-i« + sin-i/3 + 8in-^=,.then '* ^8in (g^-a)8in(._^) „ , 8in(7-o)Bin(y-iS) = ^* If^+^ + 6' = 90.then 190 ThT"^?-'^^'^ + «^'^oanuot be less than f circle i^ doubleKnL°L«i'b&r*'T"''."bi^« Polygon of a of Bides; what is that number? ^ ^^°° ^''''^^ *^« ^ame number 188. 189. h 268 TRIGONOMETRY. 191. Eliminate 9 from the equations m sin 2^ = n sin ^, 2) cos 2d = g cos <7. 192. Reduce 1 — cos' d - cos'

-2:.y in aform fitted for loganthmic calculation. y ^" » lorm ntted If Si fi! il m m ill \i PS 270 TRIGONOMETRY. (11) Solve the equations (i) sin 3^ + ^/3 cos 3^=1. (ii) sinmd=cosn^. ....V C08(/3 + x) msin/3 ,. . ^ ^^"^ ^%^-xr-^^a' (*^) tanm^=cotnA (v) tan ^+tan 2^4- tan 3^=0. (vi) cos 8d - cos 5^ + cos 3^= 1. (vii) cos d. cos 3^= cos 55. cos 7d. A-B =tan(0-^) cotg, where (12) In any triangle tan c^sstan"!^. (13) Prove that ft2 + c2 _ 26c cos (60<> + 4) = c2 + a=* - 2ca cos (60"+ 5). Hence prove that if equilateral triangles are described on the sides of the triangle ABC, the centres of circles inscribed in them will be the angular points of an equilateral triangle. (14) A round tower stands on an island in a lake. J, ^ are two points on the land such that AB is a feet and points directly to the middle of the tower. At A and B the baae of the tower subtends the angles 2a and 2^ respectively. Prove that the diameter of the tower is 2 ^T^'T ^ * . sm^-sma (15) At F the top of a tower of height A, the angles of de- .pression of two objects on th^ hv.izontal plane on which the tower stands are ^tt-o and J7r+a. Prove that the angle APB^2a, and that ^5=2A tan 2a. (16) On the side of a hill there are two places B, C inac- cessible to each other, but known to be at the same distance (a) from a certain station A also on the hilL At the lower place C the horkorUal angle between A and B is observed as well as the altitudes X, /n of ^ and B. Prove that the distance between i? and C is 2a |cos(X -/i)cos2 2- co8(X + ;i)sina|l . MISCELLANEOUS EXAMPLES 271 (17) In the ambiguous case rnvt^n n ^ t. ie two values of c, proveSe ioZ^TZdl^^] ^ "*» ^ "^ *^^ (i) ai + a2=2coosA (ii) a,a,:=. ?=acos0 +6costf ) prove that cos 2^ + cos20+cos2;, + 4cos^.cos<^.co8v + l=o hence [see E:c. (20), p. 143], prove that (^d=0*;^) = (2„,+ l)^ r^i ii 274 TItlQONOMETRY. (35) A circle ia drawn touching the inscribed circle and the sides AB, AC (not produced) of the triangle ABC. Another circle is drawn touching the circle and the stime side. Prove that the radius of the n* circle thus drawn is that the sum of the radii of all the circles, when n=oo , is |(cosec^-l). (36) If AO, BO, CO pass through the centre of the circum- scribing circle and meet the opposite sides on DEF, then AD BE^ CF~ R' (37) If A and ^ be the lengths of the diagonals of a quadri- lateral and 6 the angle between them, prove that the area of the quadrilateral \&\IJe^a\B. sm ^ and Z + r+ Z^ 1800, prove that (38) If-^= .fp = SLa J. sm X x^i^coaZ+zcoBY. (39) Two regular polygons of n and 2n sides are de- scribed such that the circle inscribed in the first circum- scribes the second; also the radius of the circle inscribed in the second is t^ th^t of the circle circumscribing the first ajs 3 -H V3 is to 4 \'9. Prove that w= 6. (40) Prove by the aid of a figure like that of Art. 282 that IiO^^B^+2Eri. (41) Uyz + zx+xi/=l, prove by Trigonometry that l-«« l-y>'*"l-«a (l-a^)(l-y2)(]-r^ay [See Ex. (33), (32), p. 194.] (" 1. : that Prov 2. ] Two and the diagonal 1. Dc the sine a If sin i tTheJ which are Mathemati The Ms gi) Elemei Elementar questions i were set in L. E. EXAMINATION PAPERa tl. PREVIOUS EXAMINATION. December, 1886. PAPER I. PAPER II. 2. Inyesligato the values of tan 4o» and sin CO' a PBEYIODS EXAMINATION, CAMBBIDGE. «^ttn«, 1887. PAPER I. «.e^-»f^«a°XnT;r„l\^S\°4«^- a*, ana sae^how .„ .na If Sin 4 = tan 5, prove ^ " g^^en. whi b ai^e^a^^^^^^^^^^ Previous Examination ThT^ ?'.' '' ^^«°°^' ^ GeS,an ^ ^"P°'' "« ^°^ (1887) ««S /;n 4i Mathematical subjects are S TT^n t^ • ||) Elementary Dynamics^ [see Loik's^' J"«?°r '*^ °' one Angle. Elementary Statics. The TrSo„^ J,^-^"""?'?' Z^'* Beginners] fiii questions in each of the twoCpSf Vh ''* ^^*•^« ^«* ^^^^ or wi were set m December. .ss6 oJ^?®t^' ^be. questions auoted ^hZZ £,_ jg ,p^ — "-' "^ . Prove that the sum of the cosines of two angles is equal to twice the cosine of h&l£ their sum multiplied by the cosine of half their difference. 4. Prove the following relations : (i) tan(4+B) = i^^^-. ' 1 - tan A tan B (ii) (l + sin4 + cosJ)»=2(l + sin^)(l+cosJ). ..... Bin34 + sin5ii (ui) cos 3A - cos oA :C0t^. 6. Prove that the logarithm of the product of two numbers is equal to the sum of the logarithms of the numbers. Having given log 2 =-3010300, log 7 =-8450980, find the logarithms of (1-75)3, (24 5)~i (3) (4) EXAMINATION PAPERS. «. Prove that in any triangle A£c W ^-bcoB G+c COB B. 277 ».. LeSf" "" '° -'™ . «-8le When o.e .Me „a .wo .„„« be complementarv Sho«, ♦vf , ^^^^^'g^t line with it are fonnTil a+b' IV. OXFOED LOCAL EXAMINATIONS. •^^ly, 1887. 1. What is.? J^^^«« Candidates. iTiiH::^:ri;;t;rf^----^-'^ and the Circular rneasure of C is I? ITr '^^ ^ ^ ^^-. each of the angles. ^ """'"^^^ <>^ degrees in each of the angles. 2. Prove that 2sinl^=i^f:;:^jj^^^j---__. and detemaine^hieh are the correct signs when ' 2700>^>1800. 8. Obtain the following formula: 1 «°«(f + 5)=co8^cos^-sin4sin5. (2) tan^ + tan£=sinr^ + R4 ° / ' (3) ^iii£±?h2«ni, + sin(«-^) co8(^ + j).2co8pi:^s^fr-5=tan/,; («; sin2,L.-.vthat EXAMINATION PAPEHS. 279 «olvf ^e ::;X:rf °" ^°' '^^ *^« -«•- -^ich h^e a J sine. -«.e, XTc*" '» -^ '"•-«'» « ». ». c b. .,e si.e, opposite .„ .., (1) ^}!^__amB sine a 6 --7- . (2) coaA+coaB + coBC = l+ t. Find the limiting value of (!^ sin ^\'' u,i,. ^- • creased. V^' n) ^^^^ « " indefinitely in- *6. Prove that, if e lie between ^ and - "• 4 4 • Shew that '=*"^''-**'»«'^+*tan»^-, • See Higher Trigonometry. VI. OXFOBD LOCAL EXAMINATIONS. July^ 1887. Senior Candidatks. coBi'ne o?TW",S»i'»' *" -' »' ^-^ " e,„al .0 .h. Find all tho voi,,„„ ^* « , ., I80», for which sin 6£ = iA 280 EXAMINATION PAPERS. a. (1) Prove that Qem^s A^^l + Bin2i4 a Jl-ain'iJ; and determine the signs of the roots when 2A is groiiter than 180" and loss than 270'>. (2) Verify the identities : (a) 8in4ils4ooa^ (Hinil-2Biu=*/l); (fi) {taniA + tan 2 A) (1 - tan* 3A tan' A) = 2 tan 3^ scc^ii. 3. Prove the following properties of a triangle ADG, of which the ■ides " re a, b, c, and the stimi-perimeter is « : (1) atea=J$ {$ -a){i- b) {« - c) ; ,_ bccoa A +caooB B + ab con C I*) bc + ca \-db asin^ + bsin B + e sin G {b + e)BiaA + {e + a)iiinB + (a-¥b)BmC' 4. AUG is an isosceles triangle, having the vertical angle A = 5ijfiS0'; on AB as base an isosceles triangle DAB is described, having its vertical angle D= 38" 40': the perpendicular from ^ upon ZiC=2a"76 inches: find the length of the perpendicular from D upon AD to four places of decimals. Loot 19» 20' =10-4548807; Zoos 28015' = 9-9449220; log 11376 = 4-0653514; log 30754=4-5663046, diff. for 1 = 119. 5. (1) If in a triangle the radius of the circumscribed circle is double of the radius of the inscribed circle, the triangle is equi- lateral. (2) Oompare the areas of two equilateral and equiangular polygons, each of n sides, one inscribed in ami the other circum- scribed about the same circle. VII. WOOLWICH. viber, 1886. [N.B. — Great importance will be attached to accuracy.'] 1. Explain what is meant by the circular measure of an angle, and shew that the circular measure of an acute angle is intermediate in value between the sine and the tangent of the angle. The perimeter of a certain sector of a circle is eiiual to the length of the arc of a semicircle having the same radius. Express the angle of the sector in degrees, minutes, and seconds. 2. Prove that the secant of any angle will be either greater than -J- 1 or less than - 1. Shew that secA= ^^^Jl + t&n.'^A, and explain the appearance of the double sign. 3. Express sin (A - B) and cos {A - B) in terms of the sines and cosines of A and B. Find the values of sin IS®, sin lOS", cos 1650, and tan 195o. BXAilWATtON PAPF.nS. 381 tant-r'"'" ' '°""""' """"""S .11 «ugl„. which ha™ . give- or o".'"" ^ ■" '"" '•' = '"" »'"> P''""' "«" ^ --usl b« a muUipte of 60. ™ i^ i^bl^tl 'S„f SS^ f"- • P»i"« 1 '00. ni i«f«.„„i„ t/ , r'i"'*^ biunga attached to its n rni,«,f„ . above its cc.itVe" V8Tx".quarriZ\'H''"f P".".'^'^ ^'^"^ '^ P^'"* 1 ^ot strings. e. Prove that 2tftn^ tlie Find from tlie (1) sin 2^ = ; l + tan2^ • (-) co3«4-co3^co8(G00 + ^)+sina(30-^) = } chan.etS:l?coTtt^^^ttnS^^J::M'oS^«' ^^ -«^-^-"' -^ root o? ofcm.'"^''' '"^^"'^ ^° approximate value of the seventh thro'o Hider''^' '^' ''''''' °^ '^^^ '^olo of a l^anglo in terms of the Prove that taut s?^;ii:^ru ?h?a^^ff ^^ "^ ''' ''' -^ 27- when\etsX:rta«trt:rnVtjSiVr^'*^T^°^^^^^^ a distant church makes an anSe of 30 w h If 'i'*'"^.^' "°« ^^'^^^'^ *« A mile lui-ther on the bearin?« nf iu ^'^^^^ direction of the road, and N.W. respectivelv S" .^ the windmill and tower are N E Windmill, and tttJS^, Xt^fZZll ^'^ *°"^^ ''^^'^^ circJJof Svl^ra'diuT °' * "«"^^' ^''^^'^ <>' « "^es inscribed in a perilje^^rtetrtr^C^ ^- ^^« -- VIII. SANDHUKST FUETHEB. December, 1885, Ld^th!r' *^'i *''° "^'^'^^^ ""^*« °^ «"e"'^r measure. J^ind the circular measure of 42o anrl f.n^ ti , , measure is *. « oi *- , ana hnd the angle whose circular 2. Assuming that Bin(J + £) = sinJcosB + co8Jsini?) fin^- . «08(^+-e)=cos4cosS-8in4sin7; • find m terms of the ratios of ^, the values Of ' ^n2A, co,2A, tan 2^, sin^A, cos i^ and tan i^. ill i I 282 EXAMINATION PAPERS. 3. Prove that l-tan2.4 tan»^= COR- B - sin? A cos^A cos'' B ' and from the equation 2t&n'^d=Bed^ d liud a general expression 4. Prove that in any triangle cos 4 + cos B = -i^^i-i sinHC. c ^ A lighthouse appears to a man in a boat 300 yards from its foot, to subtend an angle of 6o 20' 27-7"; find in feet the height of the hghthouse, having given L tan 6<»20'= 9-0452836, difference for 1 = -0011507; log 3 = -4771213. 6. Prove that in any triangle tan 2A + tan 2B + tan 2C=tan 2A tan 2B tan 2(7. Hence shew that iS x,y, z are three numbers such that x + y + zz=xyz, then x{l- 2/2) (1 -z^) + y{l- z^) (1 - a;'^) + ^ (l _ x^) (1 - 1/) = ixyz. IX. SANDHUKST FUETHER. June, 1886. 1. Express the value of the tangent, secant, and cosecant in terms of the sine of the angle, aud also in terms of the cosine of twice the angle. Find the values of the tangent, secant, and cosecant of 220 30'. 2. Prove the following: (sin24)»=2cosM (1- cos 2 J), tan 6^ cos 54 -cos 74 and if ' sin 74 - sin'54 ' a __ cos A b~ cobB' prove that a tan 4 + & tan i? = (a + 6) tan J (4 + B). 8. Shew that in any triangle (6 + c-a)tani4 = (c + a-6)tanJB = (a + 6-c)tania I a+b+c \ If R, r,, r», r, are the radii respectively of the circumscribed and three escribed circles of a triangle, shew that Rr, {« - a) = Rvt (s - 6) = iJr, (» - c) = ^abc, where 5 is the semi-perimeter of the triangle. 1 EXAMINATION PAPERS. 283 t tao 19.29'=9-5487„1, £ .anl9.t8.= 9.5488452. X. COLLEGE OF PEECEPTOBS, PUPaS' EXAMINATION. July, 1886. Shew that If ^ +B + c= 180« and .ha. iJtsttetumtriX «7ii°:r"'*''= 80, 60, 90 degrees ree^e™ ei;^ ' ' ° ""«'* °' » '"'»8fe •>» 4. SoIvecot'^ + tan*J=.U' ff,*v{n»*i. «. If . og. 243=5. ana. ''"'"'"' '""■"' ^• H^i-U. '°«.«^'>=1«"213, log„20=1.3010300, ii.d log„72 and 6. In any triangle, prove the formula tan^=± /HiZ^HlzSll and shew from it that it a-^ h-^\ 1 .1 angle. Can you give any expTanationVtIe 'dotwelign ? '' " "^^' beiJga.''"' *^^ ^''^ °^ ^ •"««-!- polygon of „ sides, each side being gwln. ff an° exUtfof for'.f ^' '''' '^^^ ^^^^^ed side C that if ^ and B each rSconstJnt^^^^^^^^^ 'fl'V'^ *^« result statement is true; 'that the areas oflS ^^f-r^ O y^^vies, Euclid's thfl HnnU/.o* ^il -/7T . . ^'^^^^ or Similar rectilineal f5mivl- .• f-„..,ro xaiiu ui laQjj, Homologous sides.' ' ' -o«~v<, toij,- ui 284 EXAMINATION PAPERS. 9. Standing at a certain point, I observe the elevation of a house to be 150 15', and that the sill of one of its windows, known to be 20 feet above the ground, subtends an angle of 20*' at the sauie point ; shew that the height of the house is nearly 22 yards ; logio 200= 2-3010300, L tan SO* 15'= 10-0800379, L tan 200 = 9 -5610659, log^o 6 -606 = 0-8200020. XI. OXFOKD AND CAMBEIDGE SCHOOL EXAMINATIONS 1884. 1. Define a degree, and the unit of circular measure; and find the ratio of the first to the second. In a triangle ABCy the circular measure of one angle (A) is a, and the number of degrees in another is B. If the ratio of -le number of degrees in the thu-d angle (C) to the circular measure oi B he B i a, prove that A, B, C axe in continued pro- portion. 2. Define the secarrt of an angle, and trace its changes in sign and magnitude as the angle increases from 90** to 270*. Simplify 1 1 ) sec^cosec^-li le. X cosec ^ - sec same tangent, their I cos ^ + tan® ^ sin ^ cos ^ cot'-^ ^ + sin 6 3. Prove that if two angles have the diflference is a multiple of two right angles. Solve the equation sec (sec® d + 2) (cosec d - sin 6) = 4. 4. Prove geometrically that when ^ +B is less than 90", cos (A+B):=: cos AcosB- sin A sin B. Prove that sm -1 3 + sin-^ 1 , . _, 3 T 3^ + smi^^- = -. 6. A hill is inclined at a angle 36" to the horizon. An observer walks 100 yards away from the foot of the hill, and then finds that the elevation of a point halfway up the hill is 18". Find the height of the hill ; and find the ratio of an error in measuring the distance walked to the consequent error in the height of the hill. ^ 6. In any triangle prove that sin.i _ sin.B _ sin C _ 1 a ~ b ~ 'V ~ 2^ ' where B is the radius of the circle described rouud ABC. Find 6, c, B. in the triangle for which JB=45", C=60», a =^2. EXAMINATION PAPEnS. 285 «ea'of rotli:" "'" " " "^"'- •'"'^s™ °f « «ite .• ami deduce the of the £ *" ""''"^ °f *e half a.g,e, of a Wang.o i„ .„„, usji ™at is the unit of „i.„,„ „,,,„„, ^^ .^ ^^^ ^_^.^ 'he ccula, „easu.. of tl:e let^^^SZVX'TuX^.r''^^ ( 22\ Fmd tan 30«, tan IW, tan 240«. ^ ''^• L!t 'e'Xr! "°'"«- °' '"= '«»ation sin ,=«;„,. (i) 2cos2^+V2sm^=2 (ii) (l + tan^)(l-sm2^) = l-tan^. 4. Prove geometrically • W «^»(;^-^)=sin^co8£-.eos^sin^ (ii) S?Ki^jt^)_l-tanJtan£ i*^ , '^ . ^^ , « a h ?• a a h h -a}. 8. Prove that if be the circular measure of an angle, the T -x. , . sin hriutmg value of — ^as is diminished is iinity. What is the limiting value of when n is diminished? n Questions 9, 10, 11 in Higlier Trigonometry. XIII. TRINITY COLLEGE. June 6, 1885. 9—12. 1. Define the sine and tangent of an angle. The base BC of a triangle ABC is trisected in Q and R. Prove that sin BAR sin CAQ=4: sin BAQ sin CAR. 2. Give a geometrical proof that (i) cos(^ + 0)=eos5co8 0-sin5sin0, (ii) coB^-coa0=2Bin^{0 + are each less than a right angle. Show that if <»of, (ft+a,)- ^"^^^^°^ 't> - <^os°^ sin < t> _ sin^0 cos - cos^0 sin cos tan a ~ cos tan /3 then will cos (a + 5) = sin°acosi8-cos2asin/3 _ sin" |8 cos a- cos' /3 sin a cos a tan 6 cos S tan fh EXAMINATION PAPERS, 287 '' ^'w.'"?''"'*^^^ ^"^"'^^^^'^ *he value of Show rtS::^l-,,:^i^*^™-^*-^:. tan.. provided *'°^'"*^°^iv:°l^=r^""°^*«^^-*2c, tau|^tanfB = cot|C. 4- From the results deduce th:?:mu1r''°''°^^^^°«^°^'^«d^+^ + C=.. (i) a8=62+c2-26ccoa^; (li) 0=6 cos C+c cos B. jl<0^lt^i^Sl&'^^Zi''r^^^ fr '»«-'- Wangle CD. Show that their dEen'et V^R iV ^"^ ''^ '» " «"» teangle by fomuls adapted t'ol^Sh*; ■^^'"^■^K Part« of the bet^e'7 *'* " ' "« *^ '»«'' »' the two given .ide. . „„,. „e R w ;> ^ J(^''5-l)''naJ(^/s■+l). one dd,Tnd*;„^';iL?' *" "''°'° '"-*»« - « '^"gle in term, of tc, .>,e a BC, *e. Show" tha't' «' r^VL^rSf ^rit. 1_ 1 1_ '•i >a ' ^~ ^ ~ **"i^ • 1- tan IB : 1 -tanjc. ^^^'^^^'l^^'^l^^^^ Of the circles 4 sin i^ Bin ^B sin i(7 : 1 Iwo circles of radii o an/i « «« j «.eeire„n.,eribtagoirele?oSuehV/LtS toe. ''sho"^£' "' - + -- -!L -^+2.g °^ ** ft Pi PiPft ija * 8. Show that tan-i ar - tan-J y = tan-i - - tan-i I . Eliminate o between ^ *' acot(.-a) = 6cot(^-«)=ccot(^-«) gue.^/o«. 9. 10, 11, 12 /,, H.V;i.r TrigommeU^. ' 288 EXAMINATION PAPERS, XIV. JESUS, CHRIST'S, EMMANUEL AND SIDNEY SUSSEX COLLEGES. June 9, 1885. 1—4. 1. Define the sine and tangent of an angle. A pyramid has for base a square of side a; its vertex lies on a line through the middle point of the base, perpendicular to it, and at a distance h from it; prove that the angle a between two lateral faces is given by 2/t J2a^ + 4/Ia sui a = a- + -ih- 2. Prove geometrically that (i) (ii) cos B - cos y4 =: 2 sin A + B sin A-B tan {A - B) ■■ 2 2 tan A - tan B i + tan A tan B ' If a, ^, 7 are in Arithmetical Progression, prove that 2 [ tan (g + /3) _ tan (a - / 3)) "^ , _ _ sin 4/3 (tan (/3 + 7) tan (/3 - 7)] "~ sin (a - 7) * 3. Investigate the value of cosine 72**. Solve the equation 32 sin^a; + 16 cos^a; + 12 sin 2x cos x + 20 sin2.c - 22 sin x = 15. 4. Find the value of tan 22 1». Trace the changes in xr-3— from to - , showing that 17 + 12V^ is a minimum, and 17 - 12^2 is a maximum value. 5. Prove that sin" a sin (/S - 7) + sin^/S sin (7 - a) + sin' 7 sin (a - /3) + sin (o - (S) sm (j3 - 7) sin (7 - a) sin (o +)3+7) =0. Also, if a, j8, 7 be the angles of any triangle, prove that cos 4a (cos 2/3 - cos 27) + cos 4/3 (cos 27 - cos 2o) + cos 47 (cos 2a -- cos 2/3) = 16 sin (fi-y) sin (7 - ^) sin (a - /S) sin o sin /3 sin 7. 6. Prove, with the usual notation, that the area of a triangle is equal to ^/s (s - a) (s - b) (» - c). Show that the sum of the areas of the two equilateral triangles, each of which has its vertices at three given distances from a fixed point, is equal to the sum of the equilateral triangles on these distances, and the difference of the aforesaid areas is three times the uiea of the triangle whose sides are the given digtanecs. EXAMINATION PAPERS. 289 and sides «"! ': ol^p^ov^' '"^ quadrilateral whose semiperimetev is ,, a + 7 ^' = (-^ - a) (« - o) (s - c) (. -d)- abed cos' Where «, 7 are opposite angles of the figure one;;erS^^^^^^^^^^^ floors of a hoase, a distant tower. From H,„ ,?.J ""f™ *e »"" sink mg behind seems to gr„.e the top t? the Ler''' r *%?■"'« !»«' Umb distance between the statioTran'd\htXVr„,et oltTun'"" ana -eure.tVtdt'T„ «::^ "•" "A'™^"' *» --"^ are radii of these cireles ^ " ^'^--2«'•, where r, Ji 'ieeorresponding- an area -^ intersection of these four circles has Question. 11. 12, 13 /„ ///,,,, Tnrjommetry. XV. MATHEMATICAL TRIPOS PART t .r 2-5 ^- ^^"^y 21st, 188F., Questions 1—6 ^/(/eftm. vii. Prove the following formuhe geometrically. W «^«^ + «i"i? = 2sini(.4+i^)cosH^-\ (ii) 5- = tan-x(l^) _,,,_, n - n ...'"" ™"'' + ™'''=™M.+.«-2sinasi„;5„os(, + ., vm. Ifc„t-..-cot-.(. + 2) = 15.,fi„d. If taiif^ (M) = tan^(|,|) 200 EXAMINATION PAPERS. prove that Bin ^=5 sin

• \n\ 10061C01 (8) •0102084 n (12) •0226048 (18) 86« 78' or. (14) 1041 3a2r. (Id) ii20'8r\ (16) l(T2(r. (17) 6»26\ (18) 3021 12' 5a'. (19) looiicr. (20) nr-r (21) 645«iO\ (22) 21 80. (23) r 10". (24) 10". VIII. Page 30. (1) ^09176 of a right angle = 9« 17' m\ (5?) •0676 „ r=6«75'. (8) 107875 „ =107«87'6a'. (4) •1804296l284667§ =18« 4' 29*. etc. (5) 1^467 ,y =146«7r77-t\ (6) -64 .„ =54«44'44-4". (7) 1«14'16". (8) 7»62'30". (9) 158'> 24' 29-34". (10) 21»S6'8-1". (11) 160 12' 37-26". (12) 31» 30*. IX. Page 36. 1. (1) 2 right angles or 180o. (2) } of a right angle. (8) - right angles. (5) 2 right angles. (7) — right angles. (4) - right angles. 4 (6) — , right angles. (8) -002 of a right angle. j#*r (9) 20 right angles. 2. (1) ». (2) 2t. (3) |. (4) f. (6) 180- <«^ ^"'- <7> 180'- <^> i«. (9) fgV 3. (1) I' . (2) I' (8) i^. <^) m' ^^^ 20000* ^^^ 200000" ^"'^ 200- (8) 1". (9) S». 4. (1) I (2) V. (3) 1. (4) 1^ . (5)H. (6),-^ ANSWERS TO THE £XA3fPl£s. 293 ... - ^' P*ges 87—89. m 838000 mile.." "m ■,„,(" »/ «• (10) 2>li degrees. nV ,Vf'!'"'=«''r degree.. (18) 1:81416. lU) lum' '"" «l>oul34jd,. "<" ^»«=«- (.rur:.. ^-r-. (23) (i) *=i, (ii, 4^100 (1) XI. Pages 40—43. B- (2' 83.. ,5.. ,3, 9^, ,5. (4,18.22,. SS4;:Ui^-^ -,.,<«) -.^81. • ' ' iOeOO- (17) (i) 120., 133-3.. ^'. („) i,5._ ,5^ 3, (iii) 1.60,1....^'. „„,,,,. ,21,„3Mii,'i.' — 2jr '^2) «"• (23) . right .„g,e. /oKi 9a + 106 , (28) _ ^8Q0y 19T+1800' (26) ^. (24) ^i^ (29) (27) ^ St T ^ ' 19 • 3~8 ' 2 • 10t+ 18000* (30) 9 or 16, XH- Pages 60—62. y^ r. •(").,. ,.)|. H- „^^.. '"'IC- Wg. (viii,^. „.,^^ ,.,^. OB ' oi AB' dC CD* CB (vi)^ BC 21—2 204 TRtOONOMETRY. (4) (1)^;,. ,.., BA AC (") EA °' DC Hi EC ^^BC- ^'^AE' , ^ AD Ali , .. BD (^) 7n O' AC ' <"') To ' AD (vii) DB AB BA __ CD ' °^ on ' "' CA DC , , ^^ , ., DB BC ^''^BD- ^""'^AB^'aC- ,^.., DA ,. , BA AC (5) Bin -4=1, coBA=i, tan^ = |; 8inB=»|, 0O8B = f, tanB = J. (7) Of the Bmaller augle, the 8ine = i^, oo8ine=H, tangent s-x^j. Of the larger angle, the Bme = H» cosine =^, tangent = V' \/8 1 (8) Of the smaller angle, the sine = i , cosine = -5- , tangent = — . /ft Of the larger angle, the sine— ~- , cosine = J, tangent = \/3. i XIV. Pages 63—65. (1) 179 ft. (2) 346 ft. (3) 86-6 ft. (4) 188-6 ft. (6) 7ift. (6) 60», 173 ft. (7) 63-17 yds. (8) 34-15 ft. (9) 78-2 ft. (10) 86-6 ft. (11) -866 mUes= 1524 yds. (12) 173-2 yds. (14) 878 ft. (15) 3733 ft. (16) V6 miles = 6465 ft. (17) 36 (18) 300. (19) about 523-6 miles. XVI. Page 74. (1) sm4 = Vl-coB«4, tanil=: ^ ^^^^ Beo ^ = . , cosec A = , .; , = . cos A' Ji- cos* A n f I cos 4 Jl - cos" A ' (2) sin ii = , (iosA = cot 4 sec ^/l + ootM' """^ 71 + °°*'*^ , ,^1 + oot'il . /r-; — 7i-7 A = — — .-. . coseOi4 = vl + oot*il. , tan^ = OOtil' COtil (8) sin wiA = , Jw^^A-l .1 . . 1 — j-j — r ^ = jv____^— J cos ii = , tan A = vaeo' 4-1, see id 1 seoil sec 4 — 7 — - , cosec il = ■■ . = , Vsec'^ -1 Vec' -^ - 1 A.\SW£/iii TO TUE EXAMPLES. 295 1 coseo A coseo A (4) sin /! = -—- , , 008 i4 SB V-"""*^' •« - A . . 1 cot A = ^00800=' ^ - r, Beo a = ~^^^^, Voosec' i4 - 1 ' >Jcoseo*A -1 (5) coa ^ = v'l^:iIHS:i: tan ^ = -^- Ji?.^ ,,, . _ \/r- Bin " A Bin A ^/l~sin-A Bin A' (6) 8in^ = --J?^ ... ,^3 ._ seo^ = ^lHrtiHr2, coseo il='^. + *»'i'^. tan A XVII. Page 75. 2v/2 1 ~3~' ^- (3) I. «. (7) -=L-. (8) ^ b Id] V^"'-^ 1 (1) I. s. 1 (2) (6) 4^ ?• XVIII. Page 77. (2) sec c mci eases oontinuously from 1 to oo 4 Zt.^^^?"' continuously from 1 to 0. (4) cot ^dmiinighes continuously from 00 too. XIX. Page 82. (4) Yes. (6) No. (6) Yes (7) (i) 600. (ii) .1000. (iii, 00. '(iv) -2G00. (v) II50. (vi) 4100. (vii) -J. (^iii) 2ir 3 • XX. Page 84. (1) 450. (2) 300. (3j 4go (5) 300. (6) 300. .^v 300 (4) 600. (8) 0*>, or 450. 2d6 TRIGONOMETRY. (9) 90», or600. (13) 90«, or46». (17) 80». (10) 60». (14) IS". (18) 80». (11) 45«». (15) 45». (12) 46«. (16) 45". XXI. Page 85. (4) 2±V2 (3) the value 3 is inadmissible. (6) I, ori. (6) i.orj. (7) the value - -^ is inadmissible. (10) l-3sin2tf+3sin*tf. (11) * ' 1 + sin^ (14) cosec 6 decreases continuously from co to 1. (15) cot 6 increases continuously from to oo . (16) 6^\, -\/3. (15) -J. V3 _1 p 1 * t. XXVII. Page 100. t.bi. give, .bo u.e ^ „?^„,: ^°^Xlr ""•• "" '^"' (2) tan^ (3) cot^ (4) eeo^ (6) cosec A (6) 1 - sin il (7) 8in«^ (8) sin A . QosA (9) sin A + qobA (10) tan ^ + cot ^ (11) &mA~coBA 298 TRIOONOMETRY. Fig. i. The Curve of the Sine. Fig. ii. The Curve of the Cosine. Fig. iii. The Curve of the Tangont. XXXI. Page 112. (1) (i) 30», 160», -2100, _33oo. (H) 450^ 1350^ _226», -316'. (iii) 600, 1200, _ 2400, - SOO". (iv) - 8O0, - 150o, 210o, 3300. (2) (i) 200, 1600, 3800, 6200. (ii) ^, ll ^ ^"^ , 1^. IS 4 4 4 ..... 8a- 13t 223r 27ir 7 ' 7 ' ~7~ * ~1~ ' (3) (i) d = «,r + (-l)»(-^). (ii) ^-b^2Tri^- <") ^=- 21og8+log7 ' (iii) x=^^bg7 4(log3+log7) 21og2+log3 ^ "' * 81og2 + 3(log3 + l. (8) Iogio7 log7)* (9) 1+ (10) 2log,^,3 IIW ^ * 3a+ g be ^ ' 6 + 1' 2ft + 2' 6+1' 6Ti' 26 + 2 6"+T l-log,o2- (12) 63-31=32. (19) 2-63855. (13) (o» - aW) integers. (14) 1-9486 nearly. (20) 4-59999. (21) 167 jears. LVIII. Pages 181, 182. (2) 2-7513738. (3) 4-9413333. (5) -6710750. (6) 3-70404. (7) 45740-26. (8) 2492837. (9) -000430658. (10) 6-689158. (1) -8839066. (4) 6-8086920. (1) -6737652. (4) 41" 48' 37". (7) 9-8515594. (10) 360 4' 23". LIX. Pages 184, 185. (2) -6737652 (5) 70»31'436". (8) 9-7114477. (11) 280 16' 27-5". (3) -9306572. (6) 750 31' 21". (9) 10-1338768. (12) 210 56' 41" ANSWERS TO THE EXAMPLES. 303 LX. Pages 188, 189. (1) 84M9'31.8". (2, 1498.2 ft. .3) 450 3G',«. (4) 6293.4 ft., 6982.3 ft. /5) .^fi./, . ^^ ^^ ^S"' iR\ ^900 1- . ^^ 576 -2 chains. (6) «»ch.™, ,7, 3666-8 fee.. (8, 42.13. >H44 cW„.. IJXI. Page 190. (1) 8842-9 ft. t9\ Qfti.7jft (J) G46-7mUcfl. (lO) looofj. liXVI. Pages 208, 209. (1) 41016'65-7". (2) 78082'12".620 46'18" (3) 29n7'16" SiosvQv/ ^ «o 10 . LXVII. Page 211. (1) 318-46 yds. /ox oq.qt • ,. /3) llfl9.fi«\i ^^ 28-87 inches, 81-43 inchefl. (3) 1192-66 yds. (4, 22415 ft. (6) 24-996 = 26 ft. nearly. 17.559ft., 650 69-42". IiXVlII. Pages 213, 214. (1) 108' 36' 80" .<110 9a'Q/v' <») ir.rZl'\m;' (^)9a.xr 49", 36.48.11-. V>) 20-6 ohams. /?) 100.7 (8) 740 13' 60", 360 16' 10". " LXIX. Pages 218, 219. 2 f °2''''"'''=''°"'''"' '>'^='28.4r39".<;=ii.i8'2r s -s::^:r;:;.Tif ■ :rrr " (5) ^ = 7204'48"B-4ioIVc!' ^^^'''- ^^^ ^^'l^'K^'. B=TO ; ; B '^ °''^ = 107066'12". ^6) ^18 ambiguous; 60-3893 ft. 304 TRIOONOMETRY. LXX. Page 220. The angles are given correct to the nearest second. (1) 28" 35' 39", (2) 1040 44' 39". (3) 82»20'48". (4) 43<»40'. (5) 1280 23' 13". (6) 106581ft. (7) 3487-6 yds. (8) 1728-2 chains. (9) 25376 yards. (10) 4 =660 27' 48", 5 = 12055' 12". (11) ^ = 92"12'68*,B = 35037'7". (12) 5=2901'40",(7=74065'50". (13) B-700 35'24"; or, 109024' 36". (14) B=51066'17";or,12803'43". (15) B = f.20 6'10";or,117053'50". (16) Very nearly 900. (17) 1319-6 yds. LXX b. p. 220 (i). (u). (1) GO%A=\, cosJ4=JV3. (3) 135<», 300, 160. (4) 3. (8) 1200. (9) 1200. i (12) 1250 6'. (14) il=54» or 1260, 5 = 1080 (16) (7=300, a=V3 + lt 6=2. (18) C=600 or 1200. (22) il = 1050, 0-600, B = i5o (24) il =900 or 600, (7=750or or 1500. (26) ^=450or or2V6(l + v/3). (27) 6O0, (30) 15:8^3:4^3+9. (2) 450, 600, 750. (5) 14. (6) 1+ J3. (7) 1200. (10) 900, 360 52'. (11) 1300 27'. (13) C=V2. 5=450, il = 105'>. or 860. (15) a=l. (17) ^=750, a=6=2V3 + l. (19) 100V3. (20) No. (23) iV3(V5 + l). 1050, a=2V2orV6. (25) SO" 1350, 5=300 or 1200, 5=2^2 (l+/v/3) 7i")0, 6 yds. (28) It is impossible. XjXXII. Page 229. (16) To find tb point JS in an unlimited straight line CE at which a finite straight line AB subtends the greatest angle, a circle must be described passing through A and 5, and touching the line CE in the point E, In (16) the centre of the circle lies vertically above E, and in the horizontal line through the middle point of AB. LXXm. Page 239. (1) (i) 10 sq.ft. (ii) 43-8 sq. in. (iv) 84 sq. chains =8-4 acres, (vi) 161872 sq. yds. (V) (iii) 14813 sq. yas. 100 sq. ft. (2) 4, 10^, 12, 14 ft. (4) b^t. ANSWERS TO THE EXAMPLES. ^^XXVl a. Pages 263—268. 305 (8) ^=nxl80»; or, 360«*60». (1) 8in^=f, cos^=|. (6) 4227-47 feet. (9) 800, 600. 900, 1200. to. have for Bine i J| 1 ^1 x o 1 y# A. ->/f, - i respectively. '' ^*' *' v*' 4' "» "ii (12) The other sides are 7654321 ft • lonfi-ft f* (16) 300 600. 900. etc. have for tan A /3 /, ,0 W3. 00 . - ^3. - J ^3 respectively * ^^' ^^^^ « • " ^3. - W3. 0. 168 168 32692 193 ' 195 • 193iri95 * (^^^ sec A = f ^2. iZhl^''^^'' ^^^^ °°«S«=16co8»a-20cos3« + 5co3a Both signs negative. (24) j^ ^"a + cos a. •W -4 radians ; 6*72956". sine. |;tan.|.cot,|;co8eo,f;8eo, |. (36) C= 180, a=c=--_l__ V{10 + 2^5)' (17) (21) (23) (25) (26) (31) W^deg.; 19098540. (37) -13200. (38) _i /« ,„q, ,. ,^ ,. -^-, (41) 1 foot. 1200, 300; or *2 feet eoo^flol (2n + l);r; or. 2n.=.i.. (43) -6300. (U) 1,5 'L. ^''^ 104028'39". (48) i ^6.^2} aLd 150. L^fi or. 10 f ^450"' "' '"'"«'• (49) 9«; 286o.28'.4M6". ^ '. ^^^6 . 45 . (69) 2^14 sq.ft. (60) 380. 25'. 32-725". 1-2 radians =5 76-394168. (61) (64) (66) 192 ft., 185 ft. and 9234 sq. ft. ^ (67) 2-3 radianss 131-7799260. (69) The proper formula is - Oa + sin ><\ tn • .x (72) 780 10'. 700 30^, 9334 sq ft ^^ ~ ^^^ " ^''^ ^^^ 79 A,^*^°V'"''*^" <7«) 1350. 150; or 450, 1050 (88) 1035-43 ft.; 765-4321 ft.*; 66O. (90) ij-i (95) 13-761 ft. (91) nT + (-l)nB (96) H. (89) 6-981 feet. (97) 2nir±f (94) 3210793. 306 TRIGONOMETRY. (99) (102) (104) (110) (116) (127) (184^ (189) (144) (146) (156) (161) (168) (182) (192) (196) (198) 2co8i/l-+s/(l + Bmil)-V(l-«n^)- x = l. (106) 11-7157 miles. (107) «ir+ Jtt; or.mr + Jir. 25-7884 yds. (116) 1 '219714 miles and 1 mile. Bin = |}J, coB= -M. tan= "x*- 4e±d = 2nir±57r; or, ^ (2n + 1) ir. 5 + ilog2 + iLBeo.4 = l-log2 + ^I.cosB + ^Lcosec. wt; or, nd^rir + {-lY{hir-^)- log5 = 2a-c, log7 = c-a, logl3=?>-4« + 2c. Ini(2V2±l)liours. (151) 60«, 45o, 135", 120». 1. (16Y) 30" and 40". 0=nwot2e=nv + iv. (164) coB^=i(l=*=\/5). 8in2