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Un des symboles suivants apparaitra sur la dernidre image de cheque microfiche, selon le cas: le symbole — ^ signifie "A SUIVRE", le symbole V signifie "FIN". Maps, plates, charts, etc., may be filmed at different reduction ratios. Those too large to be entirely included in one exposure are filmed beginning in the upper left hand corner, left to right and top to bottom, as many frames as required. The following diagrams illustrate the method: Les cartes, planches, tableaux, etc., peuvent dtre filmds ck des taux de reduction diff6rents. Lorsque le document est trop grand pour dtre reproduit en un seul cliche, il est filmd d partir de Tangle sup6rieur gauche, de gauche d droite, et de haut en bas, en prenant le nombre d'images n^cessaire. Les diagrammes suivants illustrent la mdthode. 1 2 3 t : 2 I-"' 3 4 5 6 J A\' PLANE TRIGONOMETRY FOR THK LSr: OF COLI.ECKS AM) SCHOOLS. ■\Vrril EXAMPLKS, ]'1U)BLEMS AND TAHLES. BY I. J. BIRCHARD. M.A„ Ph.D.. Mathemnthal Ma.stn; Collc(,>ate Institute. lirantfnrd. Joint Author of " TiiR Hum School Alokhka.' ]..rl, T. and V.n If. ■' n TORONTO: WILLIAM BRIGGS. WESLEY BUILDINGS. 511T0N PUBii€ umm Kntkuki), accordiny: to Aot of the Parliament of Canada, in the year one thousand eight hundred and ninety-one, by William IJumos, Book Steward of the Methodist Boolv and l'ul)lisliing House, Toronto, at the I^partnicnt of Agriculture. timimn public uprak^ m 9 1959 e/i^\ PREFACE. IN tho following work on Trigonometry an attempt lias Vyoen made to supply the Canadian student with a text-bt>ok adapted to the requirements of Canadian schools. In examining the available works on this subject, the writer has found them to consist of two classes. The one class consists of large heavy treatises, lilled with ditHeult problems and references to the Difl'erential and Integral Calculus, which render them unsuitable for elementary instruction. The other consists of works which simi)lify the subject by the easy process of omitting all the difHculties, and are, therefore, insutHcient for hiying a sure foundation for advanced work. Between these extremes it should be possible to unite thoroughness and accuracy with simplicity and brevity, whilst keeping the whole well within the comprehension of the average intelligent student. To fullil these conditions is the object of the present volume. Trigonometry constitutes a very important part of a student's mathematical course. The study of mathematics embraces two chief divisions, viz., Algebra and Geometry. The former treats of numerical relations, and deals with symbols. The latter treats of relations of form, and deals with objective magnitudes. Trigcmometry unites the two. The meaning of the various symbols of tlie former, and the operations to which they are subjected, are exennilified by the diagrams of the latter, whilst the properties of the diagrams them- selves are inferred from operations with symbols. The different classes of symliols, those of quantity, of operation, and t)f function, are distinguished from one another; idea,s of limits are developed; the meaning of the infinitely great and the infinitely small is exemplified, the principle of continuity is illustrated, thus l>ringing into prominence the whole of the fundamental principles of mathe- matics. To secure the desirable results just enumerated, it is necess^iry to study the two departments, the syml)olical and the geometrical simultaneously, and to make constant reference from the one to the other. For this purpose the "line definitions" of the sine, cosine, IV PREFACE. I'tc, have buen givou in addition to tlio uhujiI "ratio" dotinitions. The values of tlio ratios for a coiisiduraMu nuuilter of ani^U'S huvo been deduced grcunetrioally. Various fornuil;e relatin^' to triaiiLjles, poly- gons, tlio ratios of compound angles, etc., are deduced both from symbols and diagrams. The theory will be found complete. The definitions and demonstra- ti(ins are perfectly general, though, in some instances, particular ca.ses are treated fir.st to ])rei)aro the way for general investigations, and also to fuinish material for exercises at each stage of the work. The demonstrations are given in the most convenient and concise form, and free from any mere explatiatory matter. Where such is retjuired it is given in separate paragrai)hs. This will bo found a great con- venience in preparing for examination. The exiunples are numerous, and illustrate every portion of the work. They have been selected and arranged with care, and consist of those which are necessary for future use, and such as have been jtroved to be proiitable exercise by the practical test of the class-room. Tedious and complicated problems have not been inserted. A peculiar feature of the work is the insertion in an Ai)pendix of a brief set of Mathematical Tables including the logarithms of numbers, the natural trigonometrical ratios and the logarithmic ratios calculated to live decimal places. These enable a student to choose his own method of solving a problem, to c()ni})are different modes ol solution, and generally to ol)tain a grasp of the subject which is imi)ossible .without their use. Considerable use has been made of the natural functions, whose use is more readily understcjod than that of the logarithmic functions. The accuracy obtainable by the use of five-place tables is suflicient to illustrate all parts of the theory, and more than sufhcient for all ordinary practical measurements. In the preparation of the work, originality has not been attempted. The constant aim has been to put the matter in a teachable form, and to adapt it to the wants of the student. The author thankfully acknowledges assistance from Mr. G. I. Riddell, B.A., Mathematical Master, Parkdale Collegiate Institute, Toronto, in reading the proof and verifying examples. From the great care taken to secure accuracy, it is believed that the errors have been reduced to a minimum, and that no serious difticulty will be experienced from that cause. I. J. BIRCHARD. BuANTFonr), 9/// May, 1891. CONTENTS. I CHAPTKU I. DEFINITIONS AND rtNDAMIATAI. CINCKI'TIOXH. Ihe Tngonon.etnoal Lino . . » Tl-e Trigonometrical Anglo Segniente,l'o«itivc and Negative 10 N)irectiou of Rotation . Atiilition of Segnienta Projections, Points and Jjucs Sum of Projections . . Co-ordinates of a ]\)int , Exami)lea— Exercise I. . H) I Addition of Angles i 1 ' Magnitude of Angles . 11 Kqualitv of Anides . 12 j Measurement of Angles 13 ! Exanii)les— Exercise II. I'AdR. 14 15 15 V\ 17 J 8 21 CHAPTER II. i' ANCI.KS MEAHUKKD V,\' CIKCULAR ARCS, Angles and An-s Interchangeable 24 Axioms Assumed 2.") J'erimeters of Polygons and C'ii-cles ofi Ratio of Circum. to Diameter . 27 Radians Lengths of A n's . . Kadians, Degree's, (hades Examples— Exercise III. CHAPTER III. TIIK TKIUONO.MKTUICAL RATIOS. (ieneral Definition ;^,-; Po.vitive and Negative Signs . .*](j Nature of Ratios 37 Constancy, of Ratios .... .'^9 Examples— Exercise IV. 40 1 ]<\-..r„^i^c x? • ' V-tt .* -^ ' • • • ■*" M'-\amples — Exercise ^ II Kelations between Ratios 41 P'xamples Exercise V. Formuhe connecting Ratios Examples— Exercise VI. . Linear Functions 2S 29 80 31 43 46 48 50 .11' VI CONTKNTS. CHAl'TKR IV. KATIOS OV I'AHTIcaLAK ANCSLKS. Ratios ior 45°, 30°, (10°, etc Changes in Signs . Trigonometrical K(|uatii»n8 Examples -Kxercisc V'lll. I'ractical Applicationa I'AdK. . 51 . 5(i . 57 . 59 . 01 PAdK. 03 Examples —ExerciHe IX. . . Ratios for Half and Double Angles 60 ExanipleH - ExerciHe X. ... 70 CHAPTER V. SOLUTION OK TKIANOLKH. Proportionjil Parts (ieonietrical Illustration Examples — Exercise XI. 73 i ForiHuliu for the Solution of 70 77 Triangles . . . . The Amhiguous Chhc . Examples — Exercise Xli. n 86 CHAPTER VI. I'HOPERTIKS OF CIKCLES, TRIANOKS A.NI) l'OLV(iONH. The Circumscribed Circle . . 90 Polygons and Circles .... 95 Inscribed Circle 91 Scct(»r.s ami Segments . . . 90 Escribed Circle 92 Area of Circle 90 Quadrilateral ill Circle . . . 93 | Examples — Exercise XIII. . . 97 CHAPTER VII. RATIOS FOR RKLATEO ANOMCS. Ratios for -A, 90^ + A , etc. . 1 00 Jlxamples — Exercise XIV. . . 105 Angles having the8ameSinc,etc. 100 Examples— Exercise XV, . .110 CHAPTER VIII. RATIOS OF COMPOUND ANOLES. -y Angle between Lines . . . .111 Sin(A±B),cos{A±B). . .113 Examples — Exercise XVI. . .117 Examples — Exercise XVII. . . 118 Changing Sums into Products and the Converse . . . .119 Geometrical Illustration . . . 121 Examples— Exercise XVIII. . 122 CONTKNTS. Vll CHAITKR IX. Ml'LTIPLK AND SCBMULTM'LK AXOLES. rA' w>' ..I- Liinit of cos r ooa ^ . Calculation of sill 10, los 10 .21/ '2 4 Tables of Sines, Cosines, etc. . '21.S Dip of Ilori/on . .ad inf. '2'20 . . . '2'2l K.xaniplea— Exercise XXIX. . 2'22 APPENDIX. MATIIEMATKIAL TARLKS. Logarithma of Numbers . Natural Sines, Cosines, etc. Answers and Results . . '2.38 Logarithmic Sines, Cosines, etc. 242 . '242 Numbers used in Calculations. 248 249 PLANE TRIGONOMETRY. CHAPTEK I. DEFINITIONS AND FUNDAMENTAL CONCBP'i. .^NS. 1. Plane Trigonometry in its i)riniary incaning, is the science whidi treats of the latious e.\;:iting between the lengths of the sides and the inagni' ule of the angles which torin a phme triangle. Tn its mure extended sigiuHeatioii it treats of angular magnitude in genei-;il. Its subject matter is Geometry, but the methods of investigation employed are cliiefly those of Algebra. THE TRIGONOMETRICAL. LINE. 2. In pure Geometry a straight line is considered only as the distance between two fixed points. In Trigonometry we fre(i[uently consider the mode in which the line has been formed as well as its magnitude. 3. When a point moves through any distance in a fixed direction it traces out (or generates) a straight line which is bounded by the initial and final positions of the moving point. When the point returns to its original position it generates a line of the same length as before, and in the same position, but in the opposite direction. These two lines are distinguished by representing the one by a jyositive number, and the other by the 2 t hi wssm 10 THE TRIGONOMETRICAL LINE. saino liumbcr taken neyatively. The nurnV^er represents the l(Mi<3'tli of llio line {i.*'., the amount of motion of the moving point), and the sign determines tiie direction in which it has been generated. 4.. A fixed point in a straight line, or a line produced is r>n.^pfl a point of section of that line ; and the lines intercepted be- tween the point of section and the extremities of tlic given line are called segments. Sometimes two or more points of section are taken, thus dividing the line, or the line produced, into a number of por- tions, each of which may be called a segment. 5. Segments are either positive or negative, according as they are generated by a point moving in the positive or in the nega- tive direction. In naming a S(?gment the direction is indicated by the order of the letters designating it; thus the segment AJi indicates that it is measured from A to B. 6. Two or more segments are added by placing the initial point of the second segment upon the final point of the first; the initial point of the third upon the final point of the second, and so on \ tlie sum of all the segments is the line reaching from the initial point of the first segment to the final point of the last segment added. I B Thus A n+nc^ AC. AIi + JiA=0. C JJ AC + CI) + DE = AF. AD + DC + CE = AE. E It will be observed that when a positive segment is added the final point moves forward, but when a negative segment is added it moves backward. A')ii/ straight line is tints equal to the alnebraio sum of all its seyineuts. le is PROJECTIONS. PROJECTIONS. 11 7. Tli(! projection of a point on a strai'^ht line is tlio foot of the peipendicul.il' from the point on the line. The projection of one strai<^ht line on another sttai^'ht line is that portion of the seco'.id line intercepted between perpen- diculars drawn to it from the extremities of the , the co-ordinates of P^ are a, h, those of 1\ are - «,/>, tliose of 7*3 are - a, - ^>, and those of J*^ are a, - b. For convenience of reference the points l\y P.^, /* , ^\, may be said to lie in the tirst, second, third, and fourth quadrants re- spectively. Y P2 M Pi a- , L L C'<- c,^ ^ . Pa M' P4> Y' I- rL.^.^ X V ; ". a I CO EXERCISE I. 1. li A^ B,C, D, E^ are points in order in a straight line, name the segments of AD^ BCy BD^ giving tlnee pairs for each, 2. In the previous example name the different lines of which AD may be considered a segment, name the remaining segment and state, in each case, whether it is positive or negative, 3. J 5C is a triangle ; show that the sum of the projections of AB and BG upon any straight line is equal to the projection oi AG upon the same line. 4. The sum of the projections of the sides of a triangle, taken in ordei', upon any fixed straight line, is zero. 5. Generalize the two preceding theorems. C. Draw a diagram showing the position of the points whose co-ordinates are, 2,3; 3, 2 ; 2, - 3 ; - 2, 3 ; - 3, -- 5 ; 4, ; 0, - 3 ; 0, 0. 7, Find the distance from the origin to each of the given points in the preceding example. .< .iKi 14 THE TRIGONOMETRICAL ANGLE. 8. Show that the first four points in Ex. Q all lie on the circumference of a circle, that the next two points are without tliis circle, the next within, and that the last is the centre. 9. The square of the distance between any two points on a plane is the sum of the squares of its projections upon any two straight lines at right angles to each other, 10. How is the position of a point affected by changing the signs of its co-ordinates? How is its distance from the origin aa'ected ? 11. A straight line, OP, revolves in a circle about 0, name the quadrants in which (1) the x co-ordinate of /* is positive, (2) X the y co-ordinate is positive, (3) the quotient, - , of the co-ordi- nates is positive. 12. Two equal straight lines, OP and OQ, at right angles to each other revolve in the same direction in a circle starting from the initial lines of the first and second quadrants respectively. Show that, if at any time, the co-ordinates of J* are a, b, the co-ordinates of Q ai'e — h, a. 13. A straight line revolves about its centre, what relation exists between the coordinates of the two ends? H the arms are of diflerent lengths, what relation exists? 14. Two equal straight lines, OP and OQ, revolve through equal angles in oi)posite directions, staiting respectively from the initial and final lines of the first quadrant; show that if, at any time, o, h are the co-ordinates of 1\ thD, then, since this is considered the positive direction of rotation, the angles AOli, AOC, FiOE, etc., are positive, whilst BOA, COA, EOH, etc, are negative. \ AhoAOB + BOC^AOC. AOn + nOA=^0. AOC + COD + BO E = A OE. AOC + COB + nOD - A OD, etc. It will be observed that when a positive ang' is added the revolving line moves forward, but when a negative angle is added it moves backward, 18. In pure Geometry the angular magnitudes considered are usually each less than two right angles, but it is evident that there is no limit to the angular motion of the revolving line, and consequently there is no limit to the angular magnitude which may be described. An angle may be infinite in n)agni- tude, the same as a line may be indnite in length, the former being generated by the continuous revolution of a line and the latter by the continuous motion of a point, 19. At the end of each complete revolution the revolving line will coincide with the initial line and the geometrical angle between them vanishes. At any other point in the course the geometrical angle will be the same in the successive revolutions, but the tri(/oiio)iietrical angle is assumed to include the complete revolutions as well as the fractional part of one revolution. i THE TRIGONOMETRICAL ANOLE. 17 1 20. Two geometrical angles are equal wlion the lines which bound one angle may be made to coincide with those which bound the other (as in Euc. I., 4), but this is not necessarily true of trigonometrical angles, for the following reasons : (1) Either line may be selected as the initial line. (2) When tlie initial line has been selected, tiie revolving '.ine may turn in either of two directions. (3) The revolving line may make any number of complete revolutions V)efore coming to rest. Thus, the three angles in tiie accompanying figure are geo- Uietrically equal, but the amount of rotation, as indicated by the spirals, by which they have been severally generated, is in each case ditlerent. A Moreover each angle may be generated by stai-ting from either OA or OB. Six different trigonometrical angles are thus represented by the same bounding lines, and by varying the number of complete revolutions the number of angles represented may evidently be increased to any extent. - .j' •I 21. The motion of the hands of a clock illustrates this part of the subject. Each hand, by its revolution, describes an angle by revolving about the centre of the dial. Tlu; initial line is ' that joining the centre of the dial to the mark for the hour of twelve. The minute hand marks the geometrical angle between its position at a given time and the initial line, and thus deter- mines the number of minutes since the hej/iniiiny of the hour or before its cotn])Iefiooi. The hour hand marks the number of complete revolutions which the minute hand has made since twelve o'clock, and the two hands combined mark the trigono- metrical angle described by tiie minute hand during this time. '."SiL^'t"'"' ' ^ 18 V ■. MEASUREMENT OF ANGLES. 22. The quarter part of a ooinpleto revolution is called a quadrant, and corresponds to the geometrical right angle. In describing a coniphite revolution in the positive direction the quadrants are called first, second, third and fourth, respectively, in the order in which the revolving line passes through them, and an angle is said to lie in a particular quadrant when the revolving line, after describing the angle, comes to rost in that quadrant. MEASUREMENT OP ANGLES. 23. To measure an angle we select a llxed angle as the unit of measure, determine how often this unit nmst be repeated to produce the given angle, and to this number prefix tlu; sign + or - , according as the angle has been generated in the positive or the negative direction. 24. The natural unit of angular magnitudes is a complete revolution, but this being too large for convenience, snialler units are derived from it in one of two distinct ways : (1) By dividing a complete revolution into an exact number of equal j>arts. (2) By measuring off arcs of a given length on the circumfer- ence of a circle of given radius. The former method is the more convenient for practical work, and the latter for theoretical investigations. 25. A complete revolution divided into four equal parts deter- mines the right angle, which is the basis of the ordinary system of measurement. The right angle is divided into 90 equal parts called degrees, the degree into 60 equal parts called minutes, and the minutes into 60 equal parts called seconds. We have, then, the following table : 1 revolution = 4 right angles i right angle = 90 degrees, written 90° 1 degree — 60 minutes, n 60' 1 minute 60 seconds, II 60" I f t . ) ti-Ws, i ] ♦^-l I . MEASUREMENT OF ANGLES. 19 From the division of the degree and the minute each into sixti/ equal parts, the; abcne is sonictinies known as tiie Sexagesimal Method ; it is also known as the English Method. 26. Another mode of subdividing the right angle, wliicli is theoretically more convenient than the former, is given in the following table : 1 right angle - 100 grades, written lOO*^ 1 grade = 100 minutes, n 100' 1 minute = 100 seconds, n 100" From the subdivisions being made into humlreds in each case, tiiis is known as the Centesimal Method. 27. The Centesimal Method was proposed by a number of French mathematicians at the beginning of the present century. Its theoretical superiority is at once evident, but there are many practical ditticulties which prevent its adoption. The results of many valuable observations, especially in astronomy, have been recorded, valuable mathematical instruments graduated, and trigonometrical tables calculated, all according to the Sexagesi- mal Method. The labor and expense involved in a change of system would be greater than the I'csulting gaiii , hence the only purpose this method serves is that of furnishing exercises in reduction for the student of Trigonometry. Other methods of .subdivision are also employed for special purposes. Thus, for reckoning time, we divide a complete revolution of the earth upon its axis into 24 equal parts, and call the time occupied in describing one of those parts an hour. Astronomers divide the ecliptic into 12 equal parts, called Signs of the Zodiac ; mariners divide the horizon into 32 efjual parts, called Points of the Com- pass, etc. It will be readily perceived that every systenj of angular measurement must be ultimately based upon the circle considered as one complete revolution. 28. When two angles taken together make up one right angle each is said to be the complement of the other, and when they together make two right angles each is said to be the supple- 20 MEASUREMENT OF ANGLES. ment of tho othor. In ordinary gnomotry conipleniontary angles must each Ih! 1(!SS than a right angh*, and sui)i)l('inentary angles must each be less than two right angh.'.s, hut these re- strictions are not observed in trigonometry. If J denote any angle whatsoever, then 90^ - yl is its complement, and 180° - A is its supplement. 29. The student will find but little ditticulty in performing the various reductions required. The following are a few easy examples : Ex. J. Express 13° 10' 15" as the decimal of a right angle. 60 15.000 seconds. 60 10. -J 5 minutes. 90 13.17083 degieos. Result ,14634259 ... of a right angle. Ex. 2. Express 43^ 15' 65" in degrees, etc. 436 15^ 65" = .431565 of a right angle. 90 38.840850 degrees. 60 50.451000 minutes. _ 60 ~27.060000 seconds. Result, 38° 50' 27".06 _.X V fa Ex. 3. Find the angles of an isosceles triangle, providing the 3 number of degrees in one of the base angles is — of the number of grades in its supplement. Let X denote the numl)er of degrees in a base angle. Then -rr (180 - x) is the number of grades in its supplement. 9 Therefore x ■■ :lof^(180-.) From which cc = 45. The triangle is, therefore, a right-angled isosceles triangle. H^ MEASUREMENT OF ANGLES. EXERCISE II. 21 1. Draw the boundary linos of tho following:]; angles and indi- cate the directions in which they have been described : 3 right angles. 4 right angles. 750 degrees. [) right angles. - 1 right angles. - 2.S50 degrees. - 3 right angles. 12r) right angles. 1000 grades. 2. If n denotes any int(^ger, positive or negative, draw the angles represented by l'2n+ j right angles, and (4n-f - j right angles. 3. The geometrical angle, AOH, contains 30°; write down a general expression for all the different trigonometrical angles contained by the same bounding lines. 4. Express the following angles as the decimal of a right angle: (1) 23" 17' 14". (2) 127° 15' 25". (3) 37« 14' 83". 5. Change the following decimals of a right angle into de- grees, minutes and seconds ; also into grades, minutes and seconds : (1) .07625. (2) 1.234506. (3) 3.0125. (4) .00075. 6. Find the complement and the supplement of each angle in the preceding example. Draw a diagram representing the various angles, and distinguish between the positive and the negative angles which occur. 7. How many degrees in the angle between the hands of a clock at 2 o'clock ? At half-past two ? 8. How many degrees in an angle of an equilaterial triangle 1 How many grades in an exterior angle ? 9. If the unit of angular measurement be ^V of a right angle, what number will represent tAxe angle of a regular hexagon ? • HA?y51lT0N PUBiJC LlBl?AKr 22 MEASUREMENT OP ANGLES. i i« '!■■ '1 " / 10. The Hiiglo of ;i regulfir pentagon is rrpresented by -t, how many dcgrci'S in tin' unit of Jingular nicusureniont? 11. If D l>c tiu^ nunilxT of degrees in an angle, and (i the number of t^raih-s in it, tluMil - J' '"'' /« , 12. Divide a right anghi into two parts sueh that the number of degree.s in one part is one-tentii the number of grades in the other part. 13. The ditterence between two angles is 2°, and the number of grades in tin^ second is greater by ;'> than the number of de- grees in the lirst ; find the angles^ 14. Find the times b(!tween 11 and 12 o'clock when the angle between the two hands is 110 degrc^es. 15. Through how n)any degrees does the minute hand of a clock move in one minute of tinu;'? 16. Through how many English minutes of angular measure does the earth turn on its axis in one minute of time? 17. The moon revolves around the earth in 29 davs and 2 hours; what is its average angular velocity per hour? 18. What trigonometrical angle does the minute hand of a clock describe between midnight and (1) 3.15 a.m.; (2) 4.50pm.? 19. The longitude of the Paris observatory is 2° 20' 9.45" east ; that of the observatory at Pulkowa is 30° 19' 39.9" east ; what is the time at Paris when it is 1'' 5'" 12* a.m. of Sept. 3id at Pulkowa ] 20. The circumference of a wheel is 10 feet; through what angle will it turn in moving over a space 3.^ feet"? If the wheel move forward at the rate of 12 miles per hour what is the angular velocity of the wheel per minute, taking a right angle as the unit of measurement 1 21. The front wheel of a bicycle is 15 feet in circumference volutions per minute and makes 100 revolutions per passu i ^« • V • • 'A r» ti « 7) MEASUIIEMENT OF ANCiLES. 28 cuKar rourso half a iniln in circumfcroncc ; throu^'h how iiniiy d«»<^rees will th' together contain ir)()', and the number of grades in 7> is greater by 15 than the number of degrees in A; 'i W li« f I I i! 1 l-(S- J ^it,^ Let ACB be a side of a regular polygon cicumscribed about a circle, ah the side of the corresponding inscribed polygon, let 1' and p denote the perimeters of the polygons, and let the number of sides of eacli be n. From the similar triangles OAC, Ouc, we have from Euc. VI., 4, OA AC Jln.AC r ac 2n.ac P Now, by sufficiently increasing n, the length of a side may be made indetinitely small, and, therefore, since the point A may be brouglio indefinitely near to C, OA may be made as nearly e(jual as w(! please to OG or 0", and consecpiently P may be made as nearly etjual as we please to ^>. But the circumference ANGLES MEASURED BY CIRCULAR ARCS. 27 of the circle lies botwoen P and ^j, therefore, it hcconies ulti- mately equal to each of them. 34. Tlin ratio of the circumference of a circle to its diameter is a Jixed number ; i.e., it is tlie same for all circles. Let ABC...., alir..,,^ 1)0 any two circles; place them so that they have the same centre ; let ^1, y>, C . . . . I'O th»^ aiiijjii- lar points of a regular polygon of v. sides inscribed in tlu; outer )ut a I let r unber ic, we ty be may leai ry )(; ftrence circle. Join OA, OB, OC . . . ., meeting the inner circle in ahc ...., then a,l>,c...., will be the angular points of a similar polygon inscribed in the smaller circle. Let 7* and p denote the perimeters of the polygons, D and (/ their diameters ; then from the similar triangles OAB,^ih, we have OA _AB n.AB _ /' C Oa ah n.iih p c the last equality becouiing true wlieti ?* is mad(» indefinitely great. '^ .! r I r ^ ^■'V •LA--': 28 ANGLKS MEASURE!) BY CIRCULAR ARCS. And since I) -^ 20 A, d^ 20i(, wo liavo C /) C c - = - ov , ~ . c d D d which proves tho proposition. 35. The vahie of tlie ratio of tho circumference of a circle to its diaiMoter is an incoinniensurablo number ; i.e., its vahie can- not In; expressed cither as a vuli,'ar fraction or a decimal. And since we ha\(; no convenient numerical expression for its exact value, it is customary to denote it by tt , the initial letter of the Cir(;ek word denoting circumference'. Its vahu! to 10 places of decimals is - ==0.1415020530. Approximate values frcujuently used are l]l, i'^;], and 3.1410. ~' f u 36. A radian is the angle subtended at the centre of a circle by an arc equal to the radius. Lot the arc, AJ>, of tlie circle whose centre is 0, bo equal in length to the i-adius AG, then the angle A OH is a radian. 37. AH radians are equal to (>.'ai aaittlicr. Since 27r times the radius of a circle equals the circumference, therefore tt times yl/i ^^^semicircumference or aic A B = ~~ of circumftrence 27r 1 therefore L AOB ^^ -- of a complete revolution Euc. VI. 33. — of 4 I'ight angles ^/ = of a rijiht anisic TT ^^ ANGLES MEASURED ]}Y CIRCULAR ARCS. 21) of i^ the ence, 1.33. li Hence a riulijiu is a fixed fraction of a ri''ht aiiulo, and con- .si!(iiiontiy does not depend upon the magnitude of the circle. Cor. TT radians make two riglit angk's, and one right angle is TT equivalent to - radians. 38. The circular measure of an angle is the number which denotes the angle when the unit is a radian ; or, The circular measure of an angle is the number of radians whicii it contains. 39. The circular measure of angles is usually denoted by a Greek letter, and the number of degrees by an English letter. In either case the letter alone denotes a nund)er, and only in connection with tlie angular unit does it denote an angle. Thus, denotes a number, just as letters in algebra denotts nundjers, but we frecjuently speak of " the angle 0," and, if so, we mean radians. The letter tt is restricted to the particular nuiid)er 3.1415. . . ., which is the number of radians in two right angles. Radians may be specially marked by '^, thus, 0% 2', etc., but this is seldom necessary. Of course, the '' used in this way must not be confused with an exponent. 40. Tojind the circular measure of the angle at the centre of a circle subtended by an arc of (jive 'i loiyth. Let AOC be the given angle, I the length of the arc AC^ r the radius of the circle, and the circular measure required. L AOC Then e= L A on arc AC arc AB I r Fig. of Art. 3G. Euc. VI. 33. This result should be memorized, both in the form just given and also in the equivalent form I = rO. V. ""^--rini I I Kl. i !t M ■ i "J J 30 ANGLES MEASURED BY ClRCULAU ARCS. Ex. 1. The angle at the centre of a circle 12 feet in diameter suhtended hy an arc 2 feet long is \ of a radian. Ex. 2. The length of the arc, radius 3 feet 2 in., subtending an angle whose circular measure is 1.235 is 1.235 x38=*G.930 inches. 41. To find the circular measure of an angle of a given number of degrees or grades ; and the converse. Let 6 he the circular measure of the angle, n the number of degrees in it, g the number of grades ; then since ^radians— 7i degrees— 7 gi'ades and TT „ =:180 ,. -200 1. Therefore e n 9 ■n 180 200 from which, if any one of ^, w, //, be given, the other two may be found. Ex. Find the circular measure of 2° 37' 25". T 37' 25" = 2.623G1 degrees = 2.G2361x:; — radians 180 = .04579 radians. 42. Problems in tiigonometry are solved by the same general principles as problems in algebra. Uidcnown quantities are represented by letters, the conditions of the problem are ex- pressed by equations, and the equations are solved in the usual way. The following is a simple example : Ex. Find the angle between two tangents to a circle 10 feet in diametei', providing the points of contact divide the circum- ' ference into arcs, one of which is 6 inches longer than the other. Let A /*, A Q, be the tangents, 0, the centre of the circle, 0, the circular measure of the angle A. Then, since the angles at P and Q are right, the angles at A and are supplementary. iieter uling limber ber of may be general ies are a,re ex- usual 10 feet circum- lir other, cle, Of (gles at bary. EXKIUiSE. 31 Tliercfoio, tlu; ciicuhir nicasui't' oF tlin ohfitsf. an^lo POQ is TT - 0, iind that of tlie ri'jlrx angle J'OiJ is IV - (tt - 6) or r + ^>. And since the radius of the circle is 5 fccr, tlui h'ligths of those arcs iii feet are ^^{ir - 0) and r)(7r + ^). Therefore, by conditions of the probhnn, 1 5(;r + ^)-r)(7r-^)= . from wliich -^ The angh^ is, th(»rofore, — of a radian, wliii li, expressed in 1 1 ^^O" degrees, is -- of = i2°.«SGt7, c, respectively; find the circular measure of the angles of the triaiii^les. t? 22. The circumference of one circle is just long enough to subtend an angle of one radian at the centn; of anoth(!r ; how many degrees in the arc of the larger circle whose length equals the arc of a radian in the smaller circle ? 23. 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 with- out the triangle, find the vertical angle of the triangle. 24. Two circles are described from the vertex of an isosceles triangle as centre, th(i one touching the base, and the other passing through the extremities of the base, and the difTerence of the intercepted arcs is -— of the difTerence of their radii ; find the vertical angle of the triangle (tt =^ 31 ). 25. Three circles, wliose radii are n, h, r, touch each other ex- 34 ANGLES MEASlTUEl) HY CIKCULAU AUCS. teriiiilly and their centres nvv, joined ; tlu; int<'rce|)ted arcs are all e(|ual ; find the circular measure of the angles of the triangle. 2C. The exterior angle of a tiiangle contains twice as many / grade's as one of the interior opposite angles contains degrees, and the sum of these two angles is 4 5 radians ; find the number of degrees in tin? remaining interior opj)osite angh; ( tt = 3| ). 27. The sides of an irregular pentagon inscrihed in a circle subtend angles at the centre in arithmetical progression ; the largest angle is five times the least and the largest side is a, find the length of the arc this side subtends. f CPIAPTER III. THE TRIGONOMETRICAL RATIOS. 43. Let tho stniight line OP revolvinfj; fi-oin 0A\ doscnl)0 any angle XOP ; take OXy and Y making a posftivo right angle ki\ i Fio. 1. y JC \~\ M O Fio. 2. X M V y ^Q Fjo. 3. X Fir,. 4. with OX, as axes of reference ; let .7-, y be the co-ordinates of any point P in OP, r the distance of P from the centre ; then the six ratios which exist between the numbers y, y, r, are called %•;. m THE TIU(iONOMKTKI( Al. JIATIOS. tli(! trigonometrical ratios of the .-ui^'le XOf. Tlu-y are dis- tiiiguislud rach l)y a se|>ar;it(! name, us t'ollous: 5^' ./^^ / X r !l X -sineof .A'O/'. ■■= cosine of XOP. = tangentof X()/\ r y r X X y ( -cosecant of XOI'. = secant of XOP. = cotangent of XOP. Tilt; naiiK's of those ratios are usually abhi-cviatod. Dtaioting tho anyle XOP by A, they are writteu thus : sin A, cos i4, <.aii A, cosec vl, sec A, cot yl. In this connection it is customary to define tho versed sine, and COversed sine, though these terms (Xo not denote? ratios. They are deiined and written thus : vers yl = 1 - cos .1, covers A = \ - sin A. 44. The direction of revolution is indicated as positive for each angle in the preceding Art., but if the line OP had revolved in the opposite dii-ection, and had come to rest in the same position, the ratios of the tici/dti ce angles thus described would in each case have been exactly the same. The ratios for any angle are tlie same as those for the remainder of a complete revolution described in the opposite direction. 45. The signs of tlu> numbers denoted by .r, y, r should be carefully noted. 1. r is always positive, since it denotes one of the bounding lines of the angle measured oi.twards. 2. X is positive when tl";e angle lies in th(^ lirst or the fourth quadrant; negative in the otluH' two. 3. y is positive when the angle lies in the lli-st or the second quadrant ; negative in the other two. For example, in Fig. 3, if the lem/fhs of OJ/and MP are three « I I 1> Ming Lu-th [ond liree I i TIIK TllKiONOMKTUICAL IJATIOS. 37 and four inclu's I'csjH'ctfully, then 01* is live inclu's, {iiul we luive SI «) «) — >J o 46. Tho stiidont should c ii'cfully oliscrvc : 1. Tlio trii,'ononu'tri<'ii| rii< ins, licini^' the j-iitios of tli(Mcnii;thH of liiHS, are ((/tsfrcct im nil>ri:-<, ;iiid as siicli niay Ix* treated us oi'diiiary nlgeltiaieal (juaMtities. Thus (sill i1)x(siii .1) (sin A)", which foi* convenifnee is usually Nvi'itten sin -A. 2. Tho name of a ratio nuist luncr ho separated from tho an-dc to which it lefers. Thus wo cannot assumes sin 'lA-'l sin ^1, for this would ho treating sin as a numerical fai-lor. Similar remarks iipply to all tlu; ratios. 47. Tin; (h^tinitions of the trigononieti-icai ratios hav(? heen gi\-eii in their most genera,! foiin, to serve as a sure hasis for all suhsefjuent investigations. JJnt the angles of a triangle, with which elementary trigonometry is chiefly concerned, are always consideied positive and always less than two right angles. With this simplification, the ratios of any angle may ])e.eadily written according to the directions in tlie following Art. 48. In either of the lines bounding a given angle take any point, and from it draw a perpendicular to tho remaining line (produced, if necessary), then 1. The sine of th*; angle is tho i-ati(t of tlu; opiioaifti side to the hypothenuse. 2. The cosine is tho ratio of tho nc!j to the adja- cent side. 4. The cosecant, secant, and cotangent are the reciprocals respectively of the sine, cosine, and tangent. r w i kl 38 THE TRIGONOMETRICAL RATIOS. In the above the perpendiculuc will ahv.ays \*o. positive, Imt the base will be positive for acute angles, (uid negative for obtuse angles. 49. The distinction between algebraical and geometrical magnitudes n>ust be kept clearly in mind. In algebra the symbols stand for munhers, and the latter, together with the unit of measurement, represent the magnitude. In geometry, the symbols represent the magnitudes directly, without any reference to number, they are mere names, and consequently do not admit of algebraical operations being performed upon them. But when the parts of a diagram have once been designated by letters used in a geometrical sense, it is frecjuently convenient to use the same symbols when algebraical operations are involved. Thus, if AB and CD are two straight lines, we shall frequently i li use such expressions as -— - to denote the ratio of the numbers which are the algebraical measures of the given lines. 50. Adopting this notation, we give a number of examples. From the given diagram, in which the angles at C and B axe right angles, we have . ,, CA ED „ nC BE Sin ADB = -p— = sin ADC, hence the sines of supplementary angles are equal. M ^./ THE TIlIGONOMETiaCAL KATIOS. 39 I DC DC Cos ADC = -7: . 'i-'itl *^'OS ADJi =-,-;, yot the cosines of supple- DA. JJA niontiiry angles arti iiof^ equal. For, with r(>ferenco to the angle ADC, DC is positives, being one of tht^ hounding lini^.s, hut with icgard to the .angle ADB it is negative, l)eing one of the hound- ing lines Dl>, produced backwards. Ilenco cos AD/i= -cos ADC, and the cosines of supplementary angles are e(|ual numeri- cally l)ut of opposite sign. In the same way, many other relations l)etween the ratios of complementary or su})plementary angles may easily be deduced. 51. The tri(jonomo.trl ral ra/ios for ', from whi(,'h perpendiculars are drawn to the opposite side. Then the three triangles POM, P^OM^, P.,0}L, have e;ich a right angle, and the angle at in common, therefore the third angles of each are e({ual and th(! triangles are similar. Therefore MP M, /^ .1/../',. OP OPy OP. P:uc. VI., \. Each of these fractions is by detinition sin AOP which is, therefore, constant for all positions of /* in each of the bounding lines. Similarly the constancy of each of the other ratios may be shown. Cor. If the sines of two angles be equal, both being acute or l)oth obtus<', tlu; angles themselves are equal. The same is evidently true for eaoii of the other ratios. r' i» I 1 ' i ■■ il i 1 /I 1 .4 T. m 1/ 40 THE TllIGONOMETUICAL RATIOS. EXERCISE IV. 1. A/tC is a triaiiglo light-angled {it C, and tlii> sides oj)posito J and yy aio 3 niid 4 r('sp(!ctively. Write down tlio trigouo- inotrieal ratios of the acute ansjfles. 2. In the preceding example prove the following : sin-' .1 + cos'-' J ^^ 1,1+ tan'-' A ^ sec'- J, 1 + cot'-' 7> =^ cosec'- />'. 3. In the liguie of Art, 50, if JW-^l, DC ^'.\, CA:^\. Find the numerical value of tan IL tan AD 11, tan Ji)(7, and — --. \. Draw an angle; wliose tangent is (1) jV, (2) - ,'l. Find the sine and the cosine of each of these angles. 5. A BCD is a scjuarc;, AC a diagonal, and tlu^ side DC is pro- duced to £. Write down the numerical values of the foUowinu' ratios : (1) sin CAD, (2) tan ACD, (3) sec AC/i, (1) cos ACB, (5) tan ACD, (G) cot ACE, (7) sin ACE, (8) cos ACE. 6. From the vertex of an ecjuilateral triangle a perpendicular is drawn to the ])as(; : find the ratio of its length to the length of a side. Of what angles is this ratio the sine and the cosine respectively? 7. Find the tangent of the angle wiiich a vertical rod inches long subtends at a horizontal distance of 3 feet, if the tangent of the angle subtended l>y a spire 300 feet distant be 1.125, tind the height of the spire. 8. In the triangle AC /I, C l)eing a right angle, AD is drawn to meet the base in D, and .l/>-=37, ^lC-12, CZ? = 35, CD=^o. Find the following trigonometrical ratios. (1) sin.l/>'C, (2) Bin ADC, {'^) cos ADC, (l)cos AD/i, ■ (5) sin /> vie, (G) Bin E AC, (7) sin E AD, ^) tun BAD, {d)tcxn A DE, (10) cot ADC, (il) sec ADE, (12) co&ec ADC. 9. In a right-angled triangle ACE, write all the ratios of each ( m % tmma RELATIONS BETWEEN THE RATIOS. 41 of the complementary acute angles A and Ji, and observe the relations which exist between them. ^ 10. Write the ratios of two supplementary angles and state the relations which exist between them. chos •rent tind [awn 1a jach i Relations Between the Ratios. 52. When one of the trigonometrical ratios of an angle is given, and also the quadrant in which the angle is situated, the other ratios may be found. We shall first gi\e some simple examples and afterwards proceed to a more general investigation. Ex. 1. — The sine of an acute angle is f. Find the other trigo- nometrical ratios of the angle. Let A BC be a right-angled triangle, and let the sides A C and Al> be respectively 3 and T^ units in length, then the angle li is tiie angle required. From the values given BC = V 5'- - 3'-= 4. Then cos B—\, tan B •J 4» sec B = 4, cot -fi = i), etc. Ex. 2. — Given tan A = , . Find the other ratios. a b' Draw the straight lines AC, CB, at right angles to each other, & and (t units in length respectively, join AB, then BAC is the 42 THE TRIGONOMETRICAL RATIOS. angle required. Then, since AB = '^ tr + h"^ we have a }> sin A Vd' + b' , cos A = V^r ^- i'^ Similarly the other ratios may be written. The two results given should be memorized VI If i ',1.: II EXERCISE V. 1. Find the value of the remaining trigonometrical ratios, having given (1) cos A = \ I (2) tan A = ^^. (3) sec A =3. 2rmi -^ (4) cot A = A, ^^5)-tan A = 2±Vz. (6) tan ^ = 3 cot A.- 2. The sine of an obtuse angle is }|. Find the cosine, tangent, and cosecant. 3. The tangent of an angle of a triangle is - %%. Find its sine and cosine. In this example, what is the use of the phrase "of . a triangle " % 4. In Ex. 2, Art. 52, how can we determine which sign to place before the radical in the values of the sine and cosine 1 Must the same sign be used in both cases ? 5. The sine of an angle of 70° is a Find the cosine and tan- c gent of 20°. 6. Given sin A=m cos A. Find sec A and cosec A. 7. If tan 6 — x, find the value of sin 6 and cos 6*, and prove sin'-' + cos" (l=\. i ■ 8. The perpendicular from the angle A, on the opposite side BC of a triangle is 3 feet, sin B — |, and sin C = f • Find the sides A B and A C. Show that two triangles can be drawn ful- filling these conditions. 53. Between tliree quantities only two independent ratios ^ < i, > 3 I M Sis I'- EXERCISE. 43 exist. Four other ratios may he written hut their vahios will clearly depend upon the values of the iirst two. This is the case with the six trigonometrical ratios of an angle which are formed from only three straight lines ; the last four may be exi)ressed in ttuMus of the first two, the sine and the cosine. Thus, denoting the angle XOP (Art. 43), by yl, the student can easily prove the followinii relations : 1 tan.l= -, cos A (3) sec A - 1 cos A (2) cosec A (4) cot A 1 sin vi' cos A sin A Again, the sides of the triangle OAfP are connected by the relation •'•■ + !/- = >' Euc. 1., 47, which furnishes tiiree more important formula Dividing in succession by >•-', <■-', //-, we get ■1 . .'-r. A ^ - y- {^) -. + ^ = 1, ,,1- ^i»- (6)1+^:=^: X' X' (7)-+. r- f or. or. or. cos'- ^1 + sin'- .1 1 '.; , 1 + tan'-. 4 =soc'-.I 1 + cot'-' A = cosec'- A . 'i .1 These three fonnuhe, being but different forms of the same equation, give but one independent relation among the ratios. We liave, therefore, in all five independent relations between six quantities, and consequently when tlu; value of any one is given, corresponding values of the others may be found. ]>ut one of these relations is of the second dcyree^ and couseciuciutly when this one is involved there will be two solutions. The full meaning of this double result cannot be given at once, but will become evident as the student proceeds with the subject. The seven formulie should be carefully memorized, I 'i n: 4*4 THE TUIGONOMETIUCAL RATIOS. i 54. The equations of Art. 53 enable us to express all tlie ratios iu terms of any one of them, the process being purely algebraical. As an exainpie, we Mill express them all in terms of .z 'ine. Fi.ai (5) from (1) and (5) tan .1 cos A — Vl - sin- A sin A sin A from (3) sec A = cosyl Vl-sm-.l _1 1^ co^^l~Vl_shi'M^ from (4) and (5) cot A = ^^ ^ ^1 - sin^ A am A and from (2) cosec A — . — . sm^l sin A Again, we will express them all in terms of the tangent by the tan A geometrical method. From the given diagram the ratios can be written by inspection ; thus : sin -4 = tan A Vl + tan'-yl =, cos ^1 1 p-, sec J = 'v/l+"tan''^l V I + tan- A , Vr+ tan- A , 1 cosec A = , cot xl= rzz~A. tan A ' ^-^^ ^ In employing the geometrical method, take as the unit of length the line which represents the denominator of the given ratio. The results are most easily obtained })y the geometrical method, but certainty of their universal truth can be obtained only by examining an angle in each quadrant in succession. 1 a- But I to ail fectll 55 ideni E.\ (se y i I TIIK TRIGONOMETRICAL RATIOS. 45 But the al<,'el)raical process, he'uv^ conducted without reference to any particular fi2;ure, gives results known at once to be per- fectly general, since the original 0(iuations are universally true. 55. The formula' of Art. 53 are extensively used to establish identities, solve eijuations, etc. We give a few exanipies. Ex. I. Sitnplify (sec A + cosec A)- - (tan A + cot A)''. (sec A + cosec Af - (tan A + cot ^1)- = sec- A + 2 sec A cosec A + cosec- A - tan- A -2 - cot- A — (sec- A - tan'-Vl) + (cosec- ^1 - cot- ^1) + :l sec A cosec yl - 2 = 1 + 1 + 2 sec A cosec A - 2 — 2 sec A cosec A. Ex. 2. Prove (sec - cosec ^^)(1 -I- cot + tan 0) — sin sec'-' - cos cosec'-' 0. (sec - cosec ^^)( 1 + cot + tan 0) \cos (I sin 0/ \ sm t) cos 0/ sin - cos sin cos (f + cos'-' + sin'- '.' cos (f sin ' sin ^^ cos ^/ sin'' - cos'' ^^ sin cos ^^ sin'- (f cos'-' ^y cos'-' ^/ sin'- ■ — sin sec'- ^ - cos cosec- 0. t. — Ex. 3. Given cot + cosec r-r 5. Find cos (f. 4. i.<^ H: We have cos + 1 ■ = 5 sin sin tlierefore cos ^^ + 1 =. l\ sin squaring, cos'- 0+2 cos ^ + 1 ^ 25 sin'' (f = 25 -25 cos- ^/ from which 26 cos- ^^ + 2 cos ^/ - 24 = factoring, (1 3 cos ^^ - 1 2)(cos // + 1 ) = therefore cos ^y=^ {n, or - 1. I It will be readily perceived that the above is an ordinary i'u) ^.^ , ^u o /. /- /J- W- :t^' /i r> -tC^-t-t- 1 46 THE TRIOONOMETIilCAI, UATIOS. ( (juadratic equation, the unknown (juantity bein<^ cos 0. In com- plicated examples the woik is facilitated by writing a single letter x for the tciven ratio. lO J '/i ^ EXERCISE ^1. Prove the following identities 1. sinyl.cot yl.sec yl = 1. 2. cos yl.tan -^.cosec ^ = 1. 3. vers -.1(1+ cos A) = sin- A. 4. covers -.1(1+ sin A) = cos'^ A. a. sin yl + vers"-' J = 2(1 -cos^l). G. cos'- A + covers'- A = 2(1 - sin A). 7. (tan A + cot A) sin A cos A = \. 8. (tan A + cot A )'-' = sec'-' A + cosec'- A. 9. sin'- A - cos- B = sin'-' B - cos'- A . 10. tan'" A - tan'- B ^ sec'- A - sec'- B. 1 1 . sin'' - cos* = sin'- ^' - cos'- (K ^ 12. tan'^y + sec*^/=l + 2tan'-Wysec'-^/. 13. tan* ^y + tan'- /y = sec* ^y - sec'- ^y. l-l. cot' + cot'- = cosec'* (I - cosec'^ 0. /;Ia^.'x^/(t. 9,: iU5. sin ^^ tan + cos <^ cot — sec ^^ cosec (1 - sin ^ cos 0). 1 6. sin'-' tan ^ + cos'- cot ^ + 2 sin ^v cos = tan ^ + cot = sec U cosec ^. 17. sin*"' X + cos*"' a; = 1 - 3 sin'- x cos" x 1 8. sin*^ X - cos" a; = ( 1 - 2 cos'- it- )(sin'^ x + cos'* x) = (2 sin'- a; - 1 )(1 - sin- aj cos' x). 19. (sin A cos 5 + cos A sin 5)'- + (cos A cos -6 - sin -4 sin Bf = 1 . 20. (1 +sin ^+cos^)'- = 2(l +sin^)(l +coSil). 21. (1+sin A - cos J)- + (l +cos A -sinil)^ = 4(l -sin A coSi4). tan' 1 - sin sec - tan sec ^ + tan I - sin ^^ 1 + sin d 32. Given tan + cot ^^ = 4 ; find tan and sec 0. 33. Given 3 tan- ~ \ sin'-' ^^= 1 ; find sin .and cot 0. 34. Given 3 sin'- - cos'- ^^ = G cot'-' ^^ ; lind cos and tan (7. V 35. Given 3 siir - cos'-' -f {\/{^ + 1 j sin ^/ - \ (3 - ^^5) ; find sin and cot 0. 36. If 12 sec'-' ^y = 6 + 17 tan ^V, find the value of tan + cot 0, 37. If tan ^ = 7-, find the value of a sin ^V + h cos ^A 38. If tan x + «7» cot x = a 4- ^>, find tan x. /39. If sin x tan v/ = tan a^ cot .»• cos v/ = cot/y, find tan a; and tanyT) Eliminate . 46. a;cos . 48 THE TRIGONOMETIUCAL JIATIOS. ( 56. Tt will he instructive to compare tli(! definitions of tlie trigonometrical ratios witii the nieanin^s attached to the terms sine, cosine, etc., hy the early writers on trigonometry. Let A BCD be a circle, AOCy HOD, two diameters at right angles to each othcM-, AP .any aic of this circle. From /*, either extremity of the arc, draw P}r perpendicular to the radius OA at the other extremity; draw also A7\ ^it, touching th(; circle at A, By and meeting the radius OP produced in 7', /; then J/P = sine of arc A P. AT = tangent of arc A P. OT = secant of arc AP. MA = versed sine of arc AI\ 0.}f=coHuui of arc AP, lit - cotangent of ai'c AP. Ot =: cosecant of arc A P. JiN =■ coversed sine of arc AP. ^^' re\^ .1 . I I A / • /• ^> Thus, the old system dealt with arcs and litifis, which we have replaced by angles and ratios, and it will be observed that the versed sine and coversed sine belong properly to this method. If the radius of the circle be unity, then the length of the arc AP becomes the circular measuie of the angle AOP, and the lengths of the various lines become its trigonometrical ratios. If r be the radius, the angle, then \ MP^rainO, AP=^rO, AT ==t inn 0. t : I # f \ |r"^ / RELATIONS HETWEEX THE RATIOS. 49 57, 'I'ho procodiii^ (lia^nuii contains the; piopcr construiBtion when tini .angle (or arc) lies in cither i\n\ fiist or tin; sci'ond ({uad- rant. The student should draw the corresponding^ H;:;ures for the third and fourth tjuadrants. Then, callinj^the radius of the circle r, and carefully distinguishinjj; between positive and ne<^ative diiections, as the point /* moves around the circle wo easily see tiie truth of tluj following : , / 1. The values of both sine and cosine lie between + r and - r for all values of the circular arc. 2. The secant and cosecant may each have any value which is not between + r and - r. 3. The tangent and cotangent may each liave any numerical value, eitlusr positive or negative. 4. Tiie versed sine and coversed sine are always positive, and may have any value Ix^twcen and 2r. 5. In the first ([uadrant all tlie ratios are positive, and the groups sine, tangent, secant ; cosine, cotangent, cosecant, are each in ascending order of magnitude. G. The sine and cosecant are positive in the first and second (quadrants (the semicircle above AOC), negative in the other two. 7. The cosine and secant are positive in the first and fourth quadrants (the semicircle to the right of BOD), negative in the other two. 8. The tangent and cotangent are positive in the first and third (piadrants (the (juadrants opposite each other), negative in the other two. 9. As the angle increases through the first quadrant the sine^ tangent, and secant, also increase; but the cosine, cotangent, and co.secant diminish. 58. The meaning of the prefix "co" found in three of the trigonometrical ratios may now be p' - -x. Since the arcs AP, PB, together make up a quadrant, they are said to be comple- mentary in the same way as the angles AOP, POB, which they subtend. If PN b(» drawn perpendicular to OB, then by defini- tion, Nl\ which equals OM, is the sine, Bt the tangent, and Ot i / / ./, r Mf,. I If Ife- IS 50 THE TIIIGONOMKTUICAL RATIOS. r L- tho socatit of tli(5 arc /V>, the complement of A P. Thus "cosine" is an iihhnniiition of "comphMiicnt sine," nieaninj^, •'sign of tlic cumj)K'n)ont." Siinihirly for the other ratios. 59. Since the cosine, cotangent, and cosecant of one angle are the sine, tangent, and secant of another (its complement), general properties of the former group, individually or collectively, will he similurly tiu(! of the hittt>r. Clood examples are found in Art. 57, or in almost any part of the subject. EXERCISE VII. 1. Using the Hneor values of the sine, cosine, etc., prove the following, in which (huiotea any circular arc, r the radius of the circle : (1) am" + cos- = r". (2) r- + tan'^ = sec- 0. (3) /•'- + cot- — cosec'- 0. (4 ) r sin = cos U . tan 0. 2. In the lig of Ai't. 50, prove the following: ( 1 ) <)M . T or-. . \.(2) A T . ju - or\ V (15) OM .AT-=Or. PM. (1 ) PM .1U = 0P. OM. State in each case the corresponding equations in te of the ratios of the angle A OP. 3. Prove PM . lit . 0T= 0P\ State the corresponding tht^orem with regard to the angle HOP. Express each in terms of trigo- nometrical ratios. ~ . 4. Find the length of a circular arc whose sine and cosine are each 5 feet in length. (1) sine and cosine each positive (2) sine positive, cosine negative. (3) sine and cosine each negative. (4) sine negative, cosine positive. 5, The tangent of a circular arc, radius 5 feet, is 12 feet in length ; find the sine, cosine, and cotangent. Find two arcs which satisfy the conditions and the difference in their lengths. I If Mr- y '^ CIIAPTEU IV. RATIOS OF PARTICULAR ANGLES. 60. With a few exceptions tho trigonometrical ratios ar*; in- commensurable quantities, and the calculation of their appi-o.vi- inate numerical values is a woik of some dilHculty and much labor, but their values for a few angles may l)e found from simple geometrical constructions. We shall frequently use their values in the surd form, in which they aie i; ost easily obtained, though in j)ractical work decimals alone are euiTiioyed. 61. In the study of trigonometry there are many facts and fornndie which must be committed to memory. This is most easily done by associating them with a geometrical figure. For example, the values of tin- ratios given in Arts. Gi'-Gf) can be im- mediately written whenever wanted by simply remembering the lengths of the sides of the triangles from which they were obtained. Also the angles being positive and acute, no distinc- tion of direction is netjessary. 62. To find the values of the triyonoiyietrical ratios for an angle ofW- r 52 RATIOS OF PARTICULAR ANGLES. & t^ Construct a riglit-caiif^led isosceles triangle ACB, of which C is the right angle, then each of the angles A and B is half a right angle. Euc. I. 5, 32. Take AC = CB as tiio unit of length, then AD=\/1. Euc. I. 47. Therefore sm 45" = jj, = ^^ ^'"^ ^^"^zl ^ ^2 tan 450 = ^1= I «« / V cot 45^ = -— - ^ 1 AC AB Ali .^ sec 45^ = — = \/2 cosec 45" = -— - = \/2 63. To find the ratios for aiujles of 30^ mid 6CP. .i\ h ►C-- Construct an (Mjuilatd'al triangle ABC, bisect the angle A 1)V the line AD, then the angles /I, B, C, heing all equal, each con- tains GO", and therefore HAD contains 30". Euc. I. 32. Take lU) as the unit of length, then BA=-'l, and yl/;=\/3. Therefore si n 60" = cos 30° = ^-r. - ^; cos GO ^ = sin 30" = ~-^-, = h AH 2 A li " tan 60" - cot 30" = —7. - v/3 ; cot 60" = tan 30" = --^ - -7-. 1)B AD y3 sec 60" = co&ec 30" = -— - = 2 ; cosec 60" = sec 30° = — - = -j^ BD AD v^3 r Cis gilt RATIOS OF PARTICULAU ANGLES. 63 64- To fnid the ratios for angles of 15° and 75^, Let ABC be the lialf of an equilateral triangle so that l. ACB = m', /- A/iC = m\ L. /iAC = 90\ Produce AB to D, making JiD = BO; join DC. Then, the angles BDC, BCD being equal, each of thein is half the angle ABC, i.e. \b\ and L DC A Euc. I. 32. Take AC as the unit of length, then BD - BC = 2, and AB = \/3. is 75° ^' l^ Also CD' = DA' + A C- - (2 + \/\\f + 1^ = .S + -1 v'3 - ( V^ + V'^Y Therefore Ci>=VO+ \/ 2. Ihen siulD"^ eo» , u^ = -^-^ = -^--^-^^ = V_^-V - • CO. 15»= sin 75»4^ = -l+ V^4 = V^^+yj CZ) v'6+v'2 4 tan 15°= cot 75' = ^^= T"- Vo =2- y'-'^ :ot 15°= tan 75°: AD AC --=2 + V3 sec 15° = cosec 75° = ^^^ = V"^' "^^'^ = t/6 - v^^ AD 2 + \/3 '^ '^ cosec 15°= sec 75^ CD AC = ^6 + ^2 When the sine of any angle is known, a construction similar to the preceding will give the ratios for half that angle. Thii^ principle is mo-e fully considered in Art. 73. , I » t '**W"m' "■' 'yp^ t 54 RATIOS OF PARTICULAR ANGLES. .^ t^ 65. To find the r alios for angles of IS^^ o(P, 5^-° and 72'^. Let ABD be an isosceles triangle liaving eacli of the angles at B and D double the angle A, Then _ BAD = ^ of 2 rigiit angles = 36^ Bisect BAD by AF, which will also bisect BD at right ;vngles. Then _ BAF= 18°, and angle AHF^ 72°. Bisect L BDA by DC, meeting AB in C ; denote AB by a, and BD - CD = CA by x. P'^ ' ^ L- \ From similar triangles A BD, DBC, AB : BD : : BD : BC, ov a:x: '.x'.a-x. Therefore or ^ from which (r - ax = x" -^ + 1=0 a^ a Euc. VI. 4. Then sin 18° = cos 72° = X _ -l ±\/5 />'/'_ 1 .r _ V5 - 1 I7y~ 2'rt~~l and cos 18°= sin "''4^=lN|«'-r = i-^'"+V6 t s at AF, 7 a, RATIOS OF PARTICULAR ANGLES. 55 .r The positive sign only is to be taken in the value of — , since x and a are both positive. Again, draw CB perpendicular to AD, Then AD is bisected, l GDE^ 36 \ and _ I)C£=rA' BE la 1 \/ry + l Therefore sin 54^^ = 008 36" = Cn 2'x v/5 - 1 and cos 54:'^= sin 36 VI) xSJ 'i = lVlO-2V5. The values of the remaining ratios may be deduced from those of the sine and cosine here given. 66. To find the radios of 0\ 90° and 180^. *. Let a straight line OP revolve in a semicircle ABC^ and let /*„ 7*2, /*3, /\ be different positions of P, from which perpendicu- lars are drawn to the diameter AOG as in the figure. From the triangles thus found the ratios of the corresponding angles may be written, excepting at the points A, B, C, where the triangle becomes a straight line. But the general definition, Art. 43, derived from the co-ordinates of the point /*, still holds. If r be the radius, the co-ordinates of A are r, 0; of ^, 0, r; of C, - r, 0. Then remembering that the revolving line is always positive, we get the following results : i 56 RATIOS OF PAIITICIJLAU ANGLES. ANGLE sino cosine tangent cosecant secant cotangent 0" -0. -1. = 0. 90^ 180^ 7t r r r r r T -1. r r « = 0. r ^, - '" 1 '— c = 0. r =- - 1. r r -/l. or ... 6 '" r = CC . -0. / - /• - r ■' L O -. 'H^O 67. The preceding article gives excellent illustrations of cer- tain algebraical operations. Thus, when we say taa 90" - we mean that as the point /*.. approaches /», the (juotient of the l(!ngth of the; perpendicular l\^ J/.> by the length of the base OJ/^ becomes greater and greater, and that before 7^ reaches B this quotient becomes greater than any finite quantity. Again, when 7*2 is indefinitely near to />' the tangent is indefinitely great and positive, but if 7*3 be brought indefinitely near to JJ the tangent is indefinitely great and ')ii'. The tangent thus passes in- stantly f rom + octo- CC, as the point /'passes thi'oufrh li, just as the cosine passes from + to - 0. When a variable quantity chaiKjes siyn it nuist first become either zero or injinite; the cosine is an example of the former, the tangent, of the latter. The reader should compare Art. 57, and observe how the linear tangent becomes infinite in hmjlh and changes direction, whilst the ratio tangent becomes infinite nnmerieaU y and changes si(jn as the point P passes from the first quadrant to the second. 68. It is important that tiie student should be able to state accurately in words, the changes which each of the ratios under- goes as the revolving line passes through each quadrant in \ RATIOS OF PARTICULAU ANGLES. 67 c-» t) i:-^(^> )f cer- of the ; OJ/;, this when t cancl Ingeut tes in- just (intity ,_, the litter. ^near whilst sips : 1. Express the different ratios which occur in terms of a single ratio. 2. Consider this ratio the unknown quantity, and solve the resulting equation by the ordinary rules of algebra. 3. Write down tlie stnallest positive angle whose ratio is known to have the value thus found. We give a few simple examples. Ex. 1. — Given tan ^/-fcot — 1^ to find (K Expressing all in terms of the tangent, we a.ive tan + = 2. tan Simplifying, tan''' 0-1 tan ^^ + 1 = extracting sq. root, tan - 1=0 or tan ^ = 1 therefore values of cos {:^n.±'^^ hs n is given different integral values, positive or negative. 7. Find all the values of tati (,.;r + J ) where n is any integer. 8. If A=45\ JI^30\ verify the forniuhe sin (A + li) = sin A cos Ji + cos A sin />', cos {A + Ji) = cos Aco&B- sin A sin JJ. 9. Prove the following identities : (1) sin 30-^ + sin G0'= v^o.cos 15^. (2) cos I - sin'' = y^^-sin (3) (sin GO '-sin 45^)(cos 45" + cos 30^) = sin- .W. (0 cos .36 '-sin 18^^ = 1 (5)4sin 18"cos3G'=l (G) sin 45° = 3sin 15^- 1 sin^ \'y\ (7) sin 36" cos IS^^sin 54^^ - 4 (8) 3 tan 1 5^^ + 3 tan- 1 5" - tan^' 1 5 ' ^ 1. *" 10. From the result of Art. 65 prove the following : TT 12 (1) tan 18" =. cot 72"= 1 - ^ y^5. (2) cot 18^ = tan 72" = \/5 + 2\7?r (3) sec 18" =cosec 72"= |2 - T y/.-). (4) cosec 18"-= sec 7 2° = v^5 + 1 . 60 RATIOS OF PAHTICIII.AII ANGLES. [ 1 (5) tail '3(j" -= cot 54'^ - y^f) 2 y/ ">. (6) cot 36" = tiin r^r = Jl + I ViJ- 1(7) sec 30'^ - coscc 54" = v^n - 1 . (8) cosec 3G ' - sec 51'^ :- U + ?- ^/r). 11. Solve the following e(iuatioiis : (2) 3 Hur.O = coii' 0. (4) cos ^/ - 3 cot ^/. (()) cosoc - 2 -= i sin fl. (cS) 3s(!cWy 10tan-^/ = 2. (10) 2 iimO + Hec-0^2. ( 1 2) cos- - sill- = (1) 2siiW/--taiWA (3) tan ^/ + cot // = -— . y o (5) cot^/-2 cos/^ (7^ Seconal! ^> - 2v/3. (9) sin'Wy + cos''/y-0. (11) sin^/ + cos^/- I',. (13) 2 sin tan ^y + 1 - tan + 2 sin ^^ '( 1 i) 3 cos- - sin- ^/ + ( v/3 + 1 )( 1 - 2 cos //) = 0. (15) sin (A - B) = \ and cos (J \- B) =. I (IG) tan (^ + 1>)^~V'^^ '11^^ t'^ii (^ - J') - 1- (17) tan (A + B)-^2- v/3, and cos (.1 - li) = ^^."J. (18) sin (^1 + n + C)=\, cos (yl + J] - C) = 1, tan (J - />') = 2-^3. 12. Find tan 7^", and tan 37i". 13. If the sine of an angle be greater than 1 ^iK.c^ii^^i i;i.,m — — -, and its cosine V'-' greater than \, between what limits does the angle lie ? 14. Trace the changes in the value of the following as (f TT TV changes from to — and troni to .iico 11 will v v\j "IT ri.iiv.1. j.Lv^iii _ fw TT I (1) 1 - COS 0. (2) cos - sin 0. (3) tan ^y + cot 0. 15. As the trigonometrical ratios change from positive to negative, or from negative to positive, state which of them can pass through the value (1) zero only, (2) infinity only, (3) either zero or infinity. I. 1 I i 0' ii st&m PRACTICAL APPLICATIONS 61 If), l)«!tonniiH! wliother the CMjuatioiis (1) sec^y- (^) ---Sr {(I + l>f aro possible! wIumi a and !> dcnoto une(i[ual numbers of the same si<,'ii. Also when they denote ecjual numl)ers of opposite sign. 17. Show <^'eom(>trii'ally that sin 2 J < 2 sin J, and that sin {A +Ji) < sin A +sin />. IS, ]<^limiiiate tl from the cciuations: ( 1 ) cosec - sin ^ n. (2) n cos" + h sin'- — m. sec - cos --= h. a sin'- — h cos- — n. 19. Eliminate I) and from the ecpiations : a sin'-' + /> eos"- " = )n. h sin'- + a cos- «/> = n. a tan ^/ = /> tan c/j. cosine as /y Ive to n can I either Practical Applications. 70. Thus fai" wo have Ihhmi engaged in determining the trigo- nometrical ratios of various angles from the known sides of right- angled triangles. AVe shall now icverse the process, and sliow that bv the aid of those results when one side and one angle are known, the remaining sides and iingle may easily be found. 71. The angles of a triangle are usually denoted by the capital letters A, I!, C, and tlu^ sides opposite them by the small letters (I, A, (', and in the case of light-angled triangles, C denotes the right angle. This notation will be followed unless otlierwise specified. 72. Given one side and one axf/Ie of a I'iyld- anyled ti'ianyle to find the other sides. From the right-angled triangle A BC, — = sin B .'. h = c sin Ji, so that when e and sin />' are each known the value of /* can be immediately B „ II i I* r .3 62 RATIOS OF I'AUTICirLAU ANGLES. — a cot A . ft cot />. . = n sec /j. found. Siiiiiliirly, each of tho follo\viii<^ r cosec Ji = c COS A —c cos // ' - h sec vl (E' L =i(i;ui/i- —hiMiA —iicoaecA Tl«o student should not attempt to meniorizo the ahove equa- tions lit(>rally, hut endeavor to accjuirc! facility in writin<.j them fi-om the figui'e. Ex. 1. — (livcMi c -^ 25, B— 15'^, find the other parts. Since A + A' - 90", .-. A - 90 ' - I fy = 75". To find a and A, we have a = ccoii/i, b^ci^xnJi Art. 72. = !i(V«± V^) . o ,U8 1 . ^ -•^' 'i> = 100, ]i = GO ", /> == 75", to find i>C and CA. l)(!note CA by x^ CD by y' Th en V/ + 100 „ 1 "^ - cot 60" - V3 2^^cot75"=2-v/3 if I I I *• KXERCISE. 63 from which id CA. $1 or and l£2 • _,,_v3)=i~-;^^ X y.i y.? lOuv/.-i _ ioov;K4 + 2v/.{) 2/ = .r(2-v/:i) = -^^^' = 8G.0O2. ' EXERCISE IX. 1. Given c = 20, 7^ = 30^, find a, h, and yl. 2. (Jivon « - 15, A ^- \t>^, (ind A and c. o. (liv(Mi (I '-- V^iO, 7/= 18', find h and r. 4. Given /> = —T-» ^ = 22. \", find <\ a, and the perpendicular y'2 from the right angle on the hypothenuse. f). Kind the length of the shadow of a vertical rod 4 feet long, when the sun is .'{0' degrees ahove the hoiizon. 6. Find tin; height of a perpendicular flagpole which subtends an an. Examine the effect of c changing B from an acute to an obtuse angle. Give also the value of tan B, using a diiferent construction. 19. From the top of a cliff 100 feet high th n ^les of depres- sion of two points on the horizontal plai ,ow in a st' ight line with the point of observation are 30° . id 15° respectively. Find how far the points are apart. I ;.JJflBji vn\ t^^hhb*" EXKllCISR. G;j )W 20. Kioiii the top of a tower tin; un()' the altitude of the sun, lintl the tangent of half the angle at the apex of the shad(jw. Jf this tangent be 20 , find tlu! altitude of tlui sun. 23. Standing straight in froiit of a house and opposite^ one coiner, 1 Iind that its hsngth subtends an angle whose sine is fi V-'j ^vhilf) its height, f) 1 feet, sul)tends an angle who.se tangent is 5. Find the length of the house. 24. The angular elevation of the top of a ti'et; on the bank of a river from a point on the other bank directly opposite is u, while ^ from a point at a distance d fi'om the former and pai-alhd to the stream, the elevation is p'. Find the biradth of the riser and the lieight of the tree. 25. At noon the altitude of the sun is 45 ', and the shadow c^f a tree standing vertical on a liillside slo])ing to the north at an an'de of 15 ' is 100 feet. Find the hei«dit of the tree. ight i 20. The sides of a triangle are m -\- 11, iii - )i, and \/2 [iir + ii'), y/r^ - 1 the sine of one an^le is » 4 Find the other an. Show tliat III) — —- and tlicnci^ that tan {A - />) = 1(1 L".). If n, h, c ar»> the Icn-'lhs of throe sti'aii'ht lines dr awn f )th one another. an( roni a j»oint niakini; e([ual anjjfu's with stiaiijht linens lie drawn joining tlu'ir exlicmities, the wh»)le triangh^ Unis fornu'd is— 7— { + f'c + role 'JO feet long which is in I'lined to the south at an angle: of ir>" from thc^ veilieal. Find also its length when tho sun is in the south and GO ' ahovo t)"} hoi"izon. Ratios for Half and Double Angles. 73. Givcti fh'! rntlo.i for tnhi/ fiul the. ratios for ItalJ that auyh' ; timl conrfrscl//. / nljL^ Z\a A L -\l R O ,. M N At the eentie O of a senncircle NPR, whose radius is a unit, make the angle NOP e([ual to tho given angle, draw I'M per- pendicular to HNnmX join ]U\ FN. Then KDJVis a right angle, and 7*A*X is half the given angle. Euc. III., 'M. Denote NO P by 2/1 , tlien angle PUN = MPN = A . Then since RO^-OP=\ tiierefore 3//* = sin 2/1 and (9i)A=cos 2 A MP MP sin 2yl and tan A RM RO + OM l+co8 2il (1) \ K ^Z »1 I RATIOS FOR HAF.F AND DOITIJLK ANGLES. siri^ 2A l-c(.s-LM 1 s L'/l ^ ' ( I + cos 2 /I )■' ( 1 + cos -2 Ay 1 + cos 2 A \~i'OH2A 2 C7 (2) Tlieu sec" ^-1 - 1 + tiili" A 1 + 1 +COS '2 A 1 +COS "J /I 1+cos 2 A or fos- yl ^-. . ( (3^ . „ , 1 -cos 2/1 Similarly sur A - (1) Suhtracting (4)froiii (.'{) ^ . / . <:!-^ ,; A- cos 2/1 ~ cos- jI sill'-' /I. llearraii;,'iiig (.'{) and (1) • ,'/''> - "- ' ■ .••^^ ' 1 (•'•') cos 2yl - 2 cos- A 1 («) = 1 -2sijryl. (7) From ( 1 ) sill 2A -- ( 1 + cos 2A ) tan A and from (.5) =2cos-J.- cos A Then tan 2A = 2 sin A cos A. sin 2A 2 sin ^ cos A 2 tan yl cos 2yl cos'- A sin- ^1 1 — tan'" A (8) The last stop hcin;^ ol»tainod l>y dividinj^ numerator and de- nominator by cos'" yl. Each of tho pr(!ceding results siioulil be menioriz:!d. 74. It will l»o instructiv(i to obtain the precedinif formulm direct from the figure. The radius Ixunj^ unity we have Therefore ^^-tanyl--^. PM MN MK _ON-OM 1 -cos 2/1 " IiO + OM~ l+coa2T (1) (2) ■ m m m i.-:Jt^& 68 RATIOS OF PARTICULAR ANGLES. Expressing the lengths of the various lines in terms of the radius and tlie ratios of the angle A, we have JiM= RP cos A = {RiY cos yl ) cos yl = 2 cos^ A, MN= /\r sin A - (/i'i\^sin A) sin A^-1 sin- A, MP = PN cos A = {RX sin yl ) cos yl = 2 sin ^ cos yl . Then or, from ('i) RAf^ RO + OM 2 cos-yl = 1 +COS 2yl. (3) (6) 2 And or, from (4) MN=OX -OM 2 sin- A = \ - cos 2.4. (7) ! •^ i Again expressing J/ successively in three different forms, we ha\ o 0M= \ {HM-MN) = RM- RO - OX- MX, from which And or cos 2/1 = cos- A - sin- A = 2 cos- /< - 1 = l-2sin2y(. /W=7?/'sinyl = RN sin yl cos i4 sin 2A = 2 sin ^ cos.^. (8) (9) CAy<^ 4- ri/-i^.'t _ ^ /' / ^^ ■ ■ ■ w^ * -^- ^l p b ^1L ^ b ^>it RATIOS FOR HALF AND DOUBLE ANGLES. M. 09 ^ yK 'vi Also from (1) tan 1A 2 Mr 2 MP 2^M RM-MN 2^ MP ITM 2 tan // ^/L 1~ MN \-t-Mx'A RM (10) r- ^ c c. ^^J / _ ^t/. Similarly various other forniuki) may be obtained. As an additional exorcise we give a second diagram in which the angle 2A is obtus(;. Tt will be found that th(; preceding proof app'ies without change, simply remembering that in this case OM is negative. 75. The truth of the preceding demonstrations does not de- pend upon the symbols employed to denote the angles. Thus, we might have denoted the angle NOP by y/, then NRP would have been , and our results would have appeared in a slightly different form. A sin A Thus tan 1 + cos A • ^ o • ^ ^' * sin A = 2 SHI cos , etc. The student should l»e familiar with each form. The formuUe are true for all values of the angh,', though the proof given evi- dently restricts the angle NOP to l>e less than two right angles. Complete proofs will be given hereafter. I * i 70 RATIOS OF PARTICULAR ANGLES. Ex. 1. — Prove X tn.n A {\ - i^n" A) ;— • = sin 4 A. (1 +tair .-/)-' i tan A{\- tan- y/) _ 4 sin A 1 - tan" A (1 +tanV7)''' ~ cos//' secM = 4 sin A cos' yl ( 1 :;— , ; ^ cos- A = 4 sin A co^ A (cos- A - sin'^ y/) = •2 sin 'lA.coiilA = sin 4^. Ex. 2. — Given coscc /i* — sin ^^ = tan - ; find cos 0. We have or tlierefore or from which 1 . sin snwy= sm 1 + cos Art. 7X (1) 1 - sin- = cos'- - - sin- (f 1 + cos I - cos'- 1 + cos (f = 1 - cos cos'" + cos ^/ = 1 yry - 1 2"" cos ■■ EXERCISE X Prove the following identities : 1. sin 2A cot yl = 1 +cos 2A. 3. cot A I tan A = 2 cosec 2 A 5. cosec 2 A + cot 2A — cot A . 2 tan A 1 + tan"- A 1 1 7. sin 2.4 = 2. cos' A - sin' A =■- cos 2^. 4. cot A - tan A ^ 2 cot 2^4. G. cosec2i4- cot 2A = tan A HCC^ A 8. sec 2^ = 2 - secM* 9. :; ^— i- , ■ ,=tan2/l. 10. 4 cosec^ 2^y= -""^^^1^. 1 - tan A 1 +tan A cosec' . sin"* + cos" r= (sin + cos 0){ 1 - ! sin 2^^). 1 7. sin (soc + cosoc ^/)( 1 - tan 0) = 2 tan (f cot 2^>. IS. sin'-' A - cos- A cos 2 A* = sin'-' /> - cos 2.1 cos'-' /». 19. 2 sin'- A sin'-' /i i 2 cos'-' ^1 cos'-' />'-= 1 4- cos 2.1 cos 2/;. '20. coscc 2.1 + cot I A — cot A - cosec l.I. 21. cosec 4.1(2+2cos 2^1 + .'? cos 1.1) - cot A + cot 2.1 + cot iA. 22. In a triangle, riglit-anglcd at C, show that A (I c - h . A \c - h tan =- = — , sm — . — -. '1 ij+c « 2 \ 2o 23. If tfiii ^'-- , show that uoo^ 2i'' + />sin 20 — a. a 21. ]f tan 2/1=" lind tan.l. 25. Prove geonjctiically cot A = sin 2yl cot" ^1 - 1 r^cos 2 A ~ 2 cot 2/1' rt cos , ., (I -\- h ., f/> 2G. Ji cos — — , show that tair — = r t^^'i" , • a - b cos , we require the numerical value of one or more of its trigonometrical ratios. 2. Ifaving given the numerical value of one of the trigono- metrical ratios of an angle, we nnpiire the magnitude of the angle. To facilitate these processes we have given in the Appendix the values of the ratios for angles between 0' and 4a ', at inter- vals of 10'. From these we can obtain the ratios for any given angle, or the angle which corresponds to any given ratio, by metliods which we proceed to explain. 77. The trigonometrical ratios are functions of the angh^, i.e., they are quantities whose values depend upon the magnitude of the angle and change as the angle changes. In this connection we shall assume the truth of the following general principle : The change in value of the function of a variable in approxi- mately ])roportional to the chanye in the variable, providi7i(/ the change be sujficiently small. This is known as "The Principle of Proportional Parts." It is also sometimes called *' The Rule of Proportional Diflerences." The following examples will show clearly the meaning and application of this principle : 6 y' 74 SOLUTION OF TRIANGLES. Ex. /.- (Jivim the valuoa of sin 13'^ 10', unci sin \T 20", to tind the value of sin \T W lT)". From the tables we find sin 1 3-^20' sin 13" 10' from which difTxV, and to require the length oi jmi. Now, assitniinf/ PQ to be a strai(/ht line, from tiie similar triangles J'jJr, PQs, we have The length oi pr thus found added to /M/ gives the length of pm required. The length of Pr may Ije similarly found and subtracted from OM for the cosine of A Op. i iily from »<• fouiul I I I SIDES AND ANOI.ES OF A TRIANGLE. 77 1 -) Tn tlio preceding examples V()(^ represents an an«j;le of 10', and since a complete revolution, or .'500, (-ontains 'JUJO such angles, w«? liav', giving two solutions. 12. Uiven tan }, (A li) ="~ ~, cot , arid A + 11= 105° 40', <(~-i)y 6- 3 ; find A, /i and C, which ar(! the angles of a triangle. Sides and Angles of a Triangle- 80. The sides ofiuuj trirmi//e are pruporlloiutl to the sines of the owusite ani/les ; or in sijmbols p^ u b c sin A sin li sin C f ^^ .•1L mmmm li! ' ]i (^ '^< ^ (3 C' if t <*- (L 78 fiOLUTlOM OK TKiAN(Ji-ES. Let AliC l»e any triangle, luifl from A draw AD at right angles to JiC, or to liC produced, and denote the sides opposite to the angles yl, li^ C by a, ^, c respectively. A A C B D B bsmC — AD~c sin /i sin Ji sin C a C Similarly, by drawing a perpendicular from Ji it may be shown that c a and therefore sin C sin A a b sin A sin Ji sin C* The above is known as " The Sine Rule." 81. To express the base of a triangle in terms of the sides and the base angles. With the construction of Art. 80 we have : In each figure c cos /?=-- length of JJD taken positively in Fig. 1 />cos C==length of (7Z> u n in Fig. 2 b cos C = length of CD taken negatively in Fig. 3 /> cos (7 = 0. Therefore in all cases BC=BD + DC (including direction) or a — coos J5 + 6cos C. (1) Similarly b — a cos C + c cos^ ft- (2) and c=b cos A + a cos^K^^ fi * ^^) V SIDES AND ANGLES OF A TRIANGLE. 79 f^t right [opposite \b - D .,1 •• 82. To eayresg the coshte of ait atujle. of a iriatujh in tei^iis of the Sides. Ill Fig. 1 III Fig. 2 III Fig. 3 Air-^^ JiC + CA" -21iC. CD, A li' = Hi •- + CM- + '1 liC . CI). A/{' = IiC' + CA'. Euc. IT., 13. V.uv. II., 12. Euc. I., 47. Now, in Fig. 2, h cos C = length of CD taken negatively, and in Fig. 3, /> cos C = 0. Therefore in all cases or cos C — c"' = a- + ir - 2 = — ' 2tv< (-') This may be quoted as "The Cosine Rule." f 83. To exjrress the siite, cosine and tanyent of half an awjfe of a triamjle in terms of the sides. D't s denote half the sum of the sides of the triangle, so that 1 v'«^ or i a + h + c = 2s, h + c - a = 2 {s - a) c + a -b = 2 (s - ft), a + h-c^2 {n - c). Then 2 sin'' - = 1 - cos A = l — = - 2 2hc 2hc - (<^-^> + c)(^' + ^> - '•) _ 2 {s - h) (s - e) ~ "be 2hc . A |(. /.)(.- r) G) I 80 SOLUTION OF TllIANOLES. Again, 2 COS' ^^ - 1 -H cos ^1 - I + — , r = ■ /, 2 '11 tc '2 he (h + <; + (t)(f> + C - (l) 2s{s-(l) '2/>r or cos From equations (1) and (2) wo fjet (2) tan A . A A ls-h)(s--c) ■ =sni ^ -rCOS , =^ -. 2 2 2 \ s{s-<,) (3) E(iuations (1), (2) and (li) may be quoted as "The Half-angle Forniulje." 84. To xjyress the sine of an angle of a triangle in terms of the sides. From e. 7W ^nC-DC and BE^BC + CE = (i- b =(i + h. tan .J (A - /?) _tan ^^^F'DF^AE^ tan i(J +B)~tsinAl)E~AD ' AD DF BD a - h Euc. VI., 4. AE BE a + b' This is often quoted as "The Sum and Difference Formula." 86>. To find the area of a trianyh. In the figure of Art. 80 we liave Area of triangle ABC^hliCAD = l.ab Bin C, (1) or in words, the area of a triamjle is equal to half the jyrodiict of any two sides and the sine of the amjle between them. ; f 82 SOLUTION OF TUIANULES. If for sin C we substitute its value from Art. 84, we get 2 Area of triangle A liC = \ (ih.—'S/s {s - (i){d - /')(.•< - c) = y/s {h - a){8 -77)(« - c) (2) which gives the area in tenns of the sides alone. This latter expression is frequently denoted l)y «S'. . . - (I b c Again from -; — -= ,- = - — 7- sni A sni Ji am (J we get c sin A a 1 > h^ c .sin B sin C ' sin C Substituting th(^se values in (1) we get c- sin A sin Ji area ot trianiile A ItC = . — ;^ , " 2 sui C which gives the area in terms of the angles and one side. 87. V/e purposely refrain for the present from giv::;,;" -Miy rules for ti\e application of the preceding formuhe. The t-u;»)ect will be treated in a future chapter. In the meantime we give a few examples. Ex. 7.— Given yl = 45°, 7i=60^ c= 10, to find //, and conse '-'^**«»B4>HB[1 i? 84 SOLUTION OF TllIANOLES. Having found the angles ^I au»l 7>, tlie side c umy l)e found from the foiiiiula c — a cos /> + l> cos A which gives c - 10 cos 34^ 47' 2" + G cos 108 ' 2' 43" = 8.2130 1.85862 = 6.35448. This result agrees with the former to live sigiiiticant ligures, a degree of accuracy much greater than is usually attained in practical woik. Ex. S. — Given yl, />, , C, c. Make CAX ecjual to the given .angle ^^1, ^C equal to A, and from C as centre with radius eijual to a descrihe a cii'cle ; join C to the point, or points, if any, in which this circle cuts AX, and the rcisult will be tlie construction required. a Fia. 1. X A ^ ^ Fig. 2. c. D li - W Fro. 3. X B Fig. 4. Different cases may arise, which we proceed to examine. The circle may not meet the line AX, may touch it, may cut it twic(^ on the same side of Ay or may cut it on opposite sides of A. Th sin .1. AVc have then the following results : No solution possible. One solution ; triangle right-angled. I, ah sin A {\)aA (Fig. 1). Two solutions ; C J //, CAB'. One solutioi; ; triangle isosceles. One solution ; CA li. Thus, when two sides of a triangle and tin; angle opposite oncj of them are given two tiiangles may sometinu'S be drawn fulfil- ling th(! riMjuired conditions. Thiii is consecpiently known as the ambiguous case in the .solution of triangles. 88. The jneceding results have been obtained from an inspec- tion of the geometiical diagram. It will now be very instructive to e.xamine tluj ecjuations which connect tlu^ sides and angles of a triangle, and observe how each of these results is indicated by synd)ols. We shall do this in two ways. From triangle C A Ji (or CA/i'), Fig. 3, we have, a' — b- + c- - 26c cos A. Arranging this as a (juadratic ecpiation to find c, we have c'-' lV> cos J . c -f 6- - ^r - 0. v. = It cos J ± V///-" cos- J -U' -f «'-' Solving, =^ h cos .1 + y/d- fr .sin-' A . Now, from Fig. .M, it will bo readily observed that CD = h sin J, I C/> = f/, and consequently /yZ>= /y'/>= v/rt'--/r sin- yl. The two values of c are therefore .1 A' and A/i', which is in Ijarmony with the fact that the originjil ecpiation belongs c(jua!ly to the two triangles. Again, from the 8ine Rule h sin A sin Ji a 86 SOLUTION OF TRIANGLES. ilU from which li may be found, then C is known since two angles are known, and then c may also be found from the Sine Rule. Now compare the three modes of investigation — the geomet- rical, the algebraical, and the trigonometrical : \. a \\ impossibility denoted. II. rt = 6sin A; the circle meets the line in one point; the quadratic has equal roots; sine ^= 1 ; a right-angled triangle. III. a>h sin A. (1) ah; the circle cuts the line on opposite sides of A ; one root of quadratic is positive and one negative; the obtuse value of B is inadmissible ; one solution. From the value of sin B alone we are unaV)le to determine whether there ai-e one or two solutions in the last two cases. The additional test required is furnished by Euc. I., 17. When a — h, A — li and both angles must be acute ; when a > b, A>B, .'. B must be acute, and in each case there is but one solution. The angle A has been considered acute throughout the investi- ffation. The student should examine the cases in which A is right or obtuse. EXERCISE XII. 1. Given A = 30", B 45'^, c- 20 ; find a, h and C. 2. Given A = 60°, 6 = 7 feet, c = 5 yards j find «, sin B and sin C. 3. The sides of a triangle are 5, 6, 7 ; find the cosine of the least angle, the sine of the greatest angle, and the area, TrruBBWBs" Eli KKCISE. 87 I fi c 4. Given n ■= 7, h = S, t* == 9 ; tintl sin . tos . and tan , . f). Two sides of a triangle are l.J and ir>, and the cosine of the included an<,'le is ;{]! ; Hnd the remaiiiinj^ side and the area. 6. The sides of a trian«^le are 21, 22 and 23; find all the angles. 7. The sides of a triangle arc 13, 37 and 40; find the least angle and the j)erpendicular on the longest side from the opposite angle. 8. The two sides of a triangle are 8 and 10 inches respect- ively, and the included angle is 38' 1.")'; find tlie remaining side and angles. 9. Two sides of a triangle are in the ratio of 2:10, and the included angle is IT)'^ ; find the remaining angles. 10. Given a - 2, /> - \/6, c = 1 + v^3 ; find A, B and C. 11. The sides of a triangle are .r- + .x'+l, 2x-\-\ and .x- - 1 ; find the cosine of the greatest angle and thence the angle itself. 12. If the angle J be acute and sin ^4 = J, 6 = 24 ; find the least value of a with which a triangle can be formed. Find Ji when rt— 16. 13. Given A = 18 \ a= 4, 6 = 4 + y'SO ; find the remaining parts of the triangle. 14. Given A = 15° a=4, 6 = 4 + ^48 ; find /A C and c. 15. Given (7r=:18°, rt-r = 2, r/c=4 ; find A and B. 16. The sides of a triangle are 8, 9 and 10 ; find the length of the line joining the largest angle to the centre of the opposite side. 17. The sides of a triangle are 25, 30 and 45 ; find the length of the bisector of the smallest angle, and the angle which the bisector makes with the base. 18. Two sides of a triangle are 20 and 32 rods respectively, and the area is one acre ; find the third side. 88 SOF-UTION OF THIAXni-ES. 'If 19. The base angles of a triangle are 22}/ and 111'/.'; show that its base is twice its height. 20. Which, and how many of the sides and angles of a triangle must be known l)efore the otlun-s can be found? State all the ditVerent cases which may occur, and refer each to a correspond- ing pi'oposition in J'^uclid. 21. From the fornmhe of Arts. 80 and SI, prove that in any triangle sin (/> -|- C) = sin Ji cos C + cos li sin C. 22. Krom tlu^ thr(;e ecjuations of Ait. SI, lind the values of cos A, cos /i iuu\ cos C, in terms of the sides. 2.'?. If sin A 2 cos // sin C, show that the triangle is isosceles. 21. M (I cos A—b cos y?, the triangle is either right-angled or isosceles. 25. ~li ('- - a- + of) + b'\ find cos (7, and thence show that the , . ^'m -^ area of the triangle is — -— . a 20. (liven sin C + cos C= , find Ji. 27. Given the value of A, h and a, show that the sum of the aieas of the two triangles which can be formed is ^ i^ sin 2yl, and the diflerenceof their areas is/> sin A\/(r li- sin- A. 2S. In a triangle, CD is perpenrlicular to the base, and CE bisecrts the vertical angh^ ; show that the angle ECD = ^ (A - B), and thence that tan AEC= , taii ^. a - 2 29. From the angle A of any triangle A liC a perpendicular is drawn to the base, and from D perpendiculars DE and DF axe drawn to AB and AE. Show that DE . cos C = 7> F . cos B and A E . EB . cos- C^A F . EC . cos^ B. 30. Given (a + h + c) {b + c -a) = ^br ; find A. 31. If ft + c:c' -fa :a + 6 = 4:5:6, find ^I, and if the area be GO-y/S, find a, b, c. i I 4 \i lit EXERCISE. 89 32. Three circles whoso radii are a, 6, c, touch each other ex- ternally ; find the area of tiie triangle formed by joining their centres. 33. Each of two ships lying half a mile apart observes the angle subt(MKlod by the other ship and a fort. Tiio angU^s are 5G' 19' and G3' 11' ; find the distance of each ship from the fort. 34. From the top of a hill I observe two successive milestones in the plain below, and in a straight line before mo, and find their angles of depression to be 5° 30', 14° 20'; what is the height of the hill ? 35. The angle of ehnation of a tower 100 feet high, due noith of an observer, was .50 ; what will be its angle of eleva- tion after the observer has walked due east 100 feet 1 3G. Prove that the perpendicular from C upon the opposite side of a triangle may be expressed by or sin I) + P sin A a + b 37. If perpendiculars be drawn from the angles of a triangle upon the opposite sides, show that the sides of the ti-iangle formed by joining the feet of these perpendiculars are a cos A^ b cos £ and c cos (7. CHAPTER Vl. PROPERTIES OF CIRCLES, TRIANGLES AND POLYGONS. 89. In this chapter will he ^iven a few f)f the simpler propo- sitions nilatin*^ to circhis, tiianghfi and polygons. 90. To Jhid the radius of the circle described ahout a given trianijla. Let ABC be the given triangle, the centre of the circle de- scrihed ahout it. Draw 01) perpendicular to liC and denote the radius OB by R. Then BD = \BC = ^ , and angh^ BOD - I BOC - A. Euc. III., 20. Then OB sin BOD ^ BD, i.e., li sin A = ^ , a r or Ii = 2sin A which gives B in terms of a side and the opposite angle. a a'lc ft he Again 7? 2sin A 2hc sin A iJS (I) (2) which gives H in terms of the sides alone. I ll 1 n il mthiMttmim (1) (2) CIRCLES, TRIANGLES AND POLYGONS. Also from ecjuatioii ( 1 ) we get at once h c a 2Ii 91 (3) sin A sin /i sin C which cives atiother proof of Art. 80. Also from - = sin A 2 A' wo get the important theorem : 77(H9 rdfio of any chord in n circin to the dinnielcr is rqiKif, to the sitie of the auf/ie at the circumference subtended by the chord, 91, To Jinl the radius of the circle inscribed in a (fiven triamjle. D 0^ Let ABC be any triangle, the centre of the inscribed circle touching the sides in the points D, E, F; then OD, OE, OF are perpendicular to BC, CA, AB, and OA, OB, OC bisect the angles 't» A C. Euc. IV., 4. Denote the radius by r, and the area of the triangle by *S^. Then *S'- sum of areas of HOC, CO A, AOJi = \{OD.BC + OE.CA + OF.AB) = ^ {ra + rb-\- re) r = ^{a+b + c) = rs. ^ A Therefore r = I ^ IMAGE EVALUATION TEST TARGET (MT-3) <. <^^^. t^- A^ h .<$• i 1.0 :^ii^ =1= •56 132 I.I M 12.5 1.8 1.25 1.4 1.6 M — 6" — ► ^ <^ /2 >>. °^ "> > or /A Photographic Sciences Corporation 33 WEST MAIN STREET WEBSTER, NY. 14580 (716) 873-4503 ,.^^ <^-^.^ V w- /. % 1^ \\ f^ 92 CIllCr.ES, TRIANGLKS ANO POLYGONS. Again, since AF=AE, 71 D - />'/; CD - CE th(;ref ore; J F + liD + CJ)^}^ (a + h + c) = «, from which A F = s - a. Then r=OF^ A F tan OA F= {s - a) tan ^ . (2) 92. An escribed circle '»f a triangle is a circle Avhicli touch( s one side of the triangle and the other two sides produced. 93. Tojiud the radius of an escribed circle of a triangle A Let A liC be any triangle, the centre of the escribed circle touching the side BC\ and the sides ^C and AB produced, in />, E, F, then OD, OE, OF are peipendicular to BC, CA, AB, and OA, OB, OC bisect the angle ^1 and the exterior angles at t ■ CIRCLES, TRIANGLES AND POLVlJOXS. 93 li and C respectively. Denote the radius by i\, and tiie area of the triangle by »S'. Tlion *S'=sum of areas of A OB, AOC less the area BOC = ^ (OF. A n + OE . AC - OD . BC) = ^ (^1 c + )\ h - >'i '') 0) r,= s - a Again, since 2.1/-'= J /'^+ .I/i'=(.l/>'+ /?/)) + (.1(7+ CZ)) = yly^f />'6' + C;J-L^s• the^efore AF=s. Then similarly. r, = 01''— A F tan - ---^ s tan V •) B , C ?'y = 6'tan ■ and r^=^ s tan 2 -^ 2 (2) (3) 94. To find the cosine and the sine of mi angle of a quadri lateral inscribed in a circle in teruis of its sides. Let ABCD be the quadrilateral ; denote its sides by a, h, c, rf, as in the figure, and let u + b + c + d= 2s, and consequently (t + b + c-d=2{s- d), a + h --.c + d= 2{s- c), etc. "'" ' *"■'' "'^"V Tn ni 94. CIRCLES, TRIANGLES AND POLYGONS. From the triangles A/iC and ADC, A C - «2 + 6^ - 2ah COS Ji Art 82. = c^ + ^ ~7"- — — . (1) Area of segment ABB = sector AOB - triangle AOB, (Fig. Art. 96.) r'- v' sin • 2 2 = ^{0- sin 0). (2) The student should examine this result when is greater than two right angles. Uii EXERCISE. 97 96). (1) (2) 0) EXERCISE XIII. 1. The sides of a triangle are G, 8 and 10 ♦'"et ; Had the radii of the inscribed, circuuiscril)ed and escrihed circles. 2. Find the ratio between the radii of the inscribed and circumscribed circles of an ecjuilateral triangle. 3. Find tlie angle subtended at the circumference of a ciicle 10 feet in diameter by a clnjrd 5 feet in length. 4. The diagonals of a quadrilateral are 18 and I'O feet in lengti.\, and contain an angle of 37^^ ; hud its area. 5. Find the length of the lines joining the centre of the in- scribed circle to the angles of the triangle. 6. Find the distances between the centre of the insci'ibed and the centres of the escribed circles. 7. Find tlu3 lengths of the lines joining the centres of the escribed circles to the angles of the triangle, and thence the dis- tances between the centres of the escribed circh^s. 8. Find the angles of the triangle formed by joining the centres of the escri!)ed circles. 9. Find the area of the triangle formed by joining the centres of the escribed circles. U\ Show that the /adius of the circle passing througli the centres of the escribed circles is double the radius of the circle circumscribing the original triangle. 11. If a series of triangles of the same perimeter be described about the same circle they will Isave ecjual ai-eas. 12. The two triangles in the ambiguous case may each be inscribed in the same circle. 13! If a be the radius of a circle circumscribing a triangle . _ b -a cos C theni2=^ 2 cos A sin C 14. Show geometiically, and also symbolically, that the area of a triangle is s (s - a) tan ' . m 1 Tf 98 CIRCLES, TRIANGLES AND POLYfiONS. 15. Show that the porpeiidicuhi.' from an angle of a triangle upon the opposite side is an liai-nionic mean between the radii of the escribed circles opposite the remaining angles. IG. Prove geometrically [s - a) tan _ = (,v - h) tan - = (s - c) tan , and thence deduce r = s tan ■ , tan -tan ■,. 2 2 2 17. Express a side of a triangle in terms of the radius of the insci ibed circle and the adjacent half angles of the triangle, and then deduce a b c r = Ji C C A A B' cot ~- + cot ^ cot ^ + cot ^ cot - + cot -- 2 2 2 2 2 2 9 '). 9 18. From the preceding example deduce n C s a sin ^ sin - 9 9 7* = A n c cot -I- cot +cot - Jl Z 4^ cos 19. In Art. 9.3, show that BF^s-c, FC = s-b, and thence prove rj = {s - c) cot -=(*?- 0) cot - . 20. Prove symbolically that r^ I cot - - tan - ) = c, verify the equality from the figure of Art. 93. B C and also 21. Prove r,= n a cos - cos - o 9 B C tan , + tan , 2 2 cos «^ -r. 1 1 1 1 , 22. Prove — I 1 — = — , and r, r^ + r., r^ + 7\ r, = s\ Ti r., 7-3 r 23. From two different expressions for the radius of the in- scribed circle, each obtained geometrically, deduce the formula for tan - in terms of the sides. t I ESBr? angle adii of t EXKRCISE. 99 24. From tho valuo of tan , obtained us in the precedinir ex- ample, deduce algebraically the values of sin \ and cos --. 25. From two expressions for the area of a triangle obtained geometrically, deduce the value of sin A in terms of the sides, and thence prove sin A = '2 sin - cos . 2G. From the value of sin A deduce that of cos yt, and then prove cos A = 2 cos- -1 = 1-2 sin- \ . 27. Assuming that the values of cos - and sin - arer««/, and o 2 2 that the value of cos A is numerically less than unity, prove Euc. I., 20, from each of these expressions separately. 28. From a -, and tlie fact that sin A + sin B sin A sin Ji sin C' > sin {A + B) obtained geometrically, prove Euc, I., 20 ; also prove the converse. 29. From a h -^ and A +7i+ C= 180^ deduce sin A sin B sin (f the values of sin A and cos A ; also prove c = a cos B-\-h cos A. 30. From c = a cos B + h cos yl, and two similar equations, prove Euc. II., 12 and 13. 31. Deduce the values of the ratios for tlie half angles from two different expressions for the radius of an escribed circle. 32. If r be the radius of the inscribed circle, and 1\ the radius of the circle inscribed between this circle and the sides containing the angle A, then 1 - sin 2 ra = r. 1+sin- i < CHAPTER VII. > t H ■ II RATIOS FOR RELATED ANGLES. 100. The connection lietwoen the ratios for complementary and supplementary angles has already been pointed out for such angles as occur in connection with triangles. Wo now proceed to generalize the results already obtained, and to investigate other formuhe of a similar character. In this and the following chapters we are especially careful to observe directions as well as maguiiude. 101. To compare tlie rnfios of tivo angles equal in magnitude, but described in opposite dln'Ctions. Y Y X X'- Take two straight lines, X'OX, VOY, at right angles to each other as axes of references. Let equal lines OP, OQ revolve through equal angles in opposite directions, starting from OX. Denote XOP by A, then XOQ h-A. Draw PI/, (>iV perpen- dicular to XX'. Then triangles POM, QON are geometrically equal. Euc. I., 26. RATIOS FOR RKF.ATEO ANfJLES. 101 And ixuco /' aiul (J are always on opposite sitlrs of XA'\ Init on tin; sarno siflo of I'Y', wv li,iv(; XQ _ ,1//', HUfl OX^O.U for all values of tlic angles involved. Then, Y() Ml* sin ( - A) -= sin XOQ - ^^ = ^^- - - sinX6>/' - - sin A, OX O.U ^, , cos ( - ^1) =^ cos A ()Q = Yi, ^ ^^^ '» ^^' ^^'^ ''^' Similarly pi'ove tan ( A) ^^ - tan J, sec ( - A) sec A, cot ( ^1 ) - - cot A, cosec ( - ^1 ) = - cosec A. 102. To compare the trlgononietrkal rov«i ^I'.V, and constMjuently OM antl Ji(J liave the same Ki^n. 1 hcrefoio Ji(J = O.V^ OM, and NQ ^0R= - MP. Thnri sin (90 ' + /I) - sin XOQ = ^^ = ^Ji= con XOP^coa A, cos (90 ' + /I) = cos XOQ = ^ = - .fJ-= - sin XOP - - sin A. Similarly prove tan (90"' + yl) = -cot J, sec(90' + J)= - cosec il cot (90^ + A)=- tan A, cosec (90 ' + ^1) -- sec ^ 103t To compare tlie t r if/ ono metrical ratios of any angle with tlioae of its complctnent. X X Let equal lines OP, OQ revolve through equal angles in oppo- site directions, starting from OX, OY respectively. Denote XOP by A, then XOQ is 90° -yl, and these angles are comple- mentary. Draw PM, QR perpendicular to OX, and QX per- pendicular to OY. Then triangles POM, (^OiV are geometrically equal (Euc. I., 26), Now when P is above XX', Q is to the right of YY\ thcM-efore MP atid XQ have the same sign ; and when P is to the right of YY', Q is above XX^, therefore OM and RQ have the same sign. ii\ 1 11 RATIOS FOR RKI.ATKI) ANfiFJX 103 Therefore J//' - XQ - OR, ami OM OX-^ RQ. Then sin (DO - J) - sin XOC^ = n7) ^ n /> ^ ^"^^ ^^'^^' ^ <^os A O R MP cos (00 ' - /I ) - cos .Y<^r^ -"no" /' "" ''"' ^^^^' " "*'" ^^ * Sitnihii'ly prove tan (i)0 ' - ^1 ) -cot yl, sec (90 - A ) =co!si-c yl, cot (90 ' - vl ) ^ tan A , cosec (90^ -A)^ sec yl . 104. To com/inre the triyonomelrical ratios of any anyle, with those of its sujtpfemeiU. Let equal straiglit lines OP, OQ revolve through equal angles in opposite directions, starting from OX, OX' respectively. Denote the angle XOP by A, then XOQ is 180' -yl, and these angles are supplementary. Draw 1*M, QX perpendicular to XOX', then triangles POM, QOX b-vg geometrically ecjual (Euc. I., 26). Now P and Q are always on the same side of XX', but on opposite sides of YY', therefore we have MP — XQ and 0X= - OM for all values of the angles involved. Then XO MP sin (180° - yl) = sin XOQ = --^ = — - =. sin XOP = sin yl, O N - OM cos(180'^--yl) = cos^r/(2 = -^-^^-^ -^^^.,- = cos XOP = - cos A. >Similarly prove tan (1 80'^ -y|)- - tan J, cot(180''- yl)= -cotyl sec (180'-yl)= - secyl, cosec(180°- J) = cosecyl. ^.cSsi-A r 104 RATIOS KOR REI-ATED ANGLES. i (! 1 ! i ' !'l a i i 1 1 li 105. To compare the trifjono metrical ratios of ISO'^ + A loith those of A. Let equal straiglit lines OP, OQ revolve through ecjual angles in the same direction, staiting from OX, OX' respectively. Denote XOP by A, then XOQ is 180'^ + A. Draw J//», QN per- pendicular to XX\ then triangles POM, QON, are geometrically equal (Euc. I., 26). Now since P and Q are always on opposite sides both of XX' and YY', we have XQ = - MP, and 0X= - OM for all values of the angles involved. Therefore sin (180^ + ^1) = sin XOQ XQ _ -MP ~OQ~'^OF - sin XOP ^ - sin J, cos(180° + ^) = cosZO(3=^=:-^^^= -co^XOP= -cos .4. Similarly prove tan(180" + yl) = tan J. cot(l80'= + J)--=cot^, sec(180°4-i4)= -seci4. cosec (180° 4- il) = - cosec A. i EXERCISE. 105 EXERCISE XIV. Provo the following (1) goometrically, (2) by the aid of pre- cedinsj' fonimht' : - cos 1. sin 100' = cos 10''. 3. cos 100"- - si 111 0\ T). tan 225^ = tan 45^. 7. sin240' = siu(-120") 8. cos 1 75" = sin 2G5^' = - cos 5^^. 9. sec 700^ - sec 20" = cosec 1 1 \ 10. cot 330' = tan ( - GO") - - cot 30". 11. tan 2. tan 100^= -cot 10'. 1. cos 300^ = sin 30^. 6. sin 225° = cos 135°. 30°. (-4) = cot V . 1 2- cosec o (-.+ :)= -sec^ \i A, /)', C be the angles of a triangle, prove the following : 13. sin /( = sin (7i + C), cos /> = - cos (C + A), tan C = - tan (A + B). 14. sin " =cos 7i+C' n ,COS ^^=i ni C + A C tan = cot A + n 2 ' 2 2 ' 2 2 1 5. sin (2^1 + 7? + C) = - sin A, tan (.1 + 2 /^ + C) = tan 7^. 16. sin {A+B-C) = sin 2C, cos {A-B + C)= - cos 2C. 17. tan {A - B - C) = tan 2J, cot (J ~ U + C) -- - cot 27^. B-C A + -in C-A B + 2C 18. sec — - — = cosec — - — , tan — ^, — = - cot — . 2 2 2 2 19. Given sin (90° — A)^ cos A, prove cos (90" - A) = sin A. 20. From the ratios of 90° + ^1 deduce those of 180° + ^. 21. From the ratios of 90° + .1 and - A derive those of 90" - A, 180° -.4, and 180° + ^1. 22. Given the ratios of 90° + yl and 90 -yl, derive those of 180°-^, 180° + yl, and -A. 23. Draw two angles, A and B, such that sin i4+8in^ = 0, and cos A + cos B = 0. 8 > i h i JOG RATIOS FOR RELATED ANGLES. 106. 2'o Jliul all the anyles which have a given sine, i.e., to solve tlie equation^ siii 0=^a. X X' Describe a circle of unit radius and draw the diameters XOX\ YOY' at right angles to each other. Lot OR, measured on OY (or Y'), represent a in magnitude and sign ; through R draw PQ parallel to XX\ nK'(>ting the circle in P and Q; join OP, OQ ; then any angle of which OX is the initial, and either OP or OQ the tinal line, but no other angle, will have its sine equal to a. Denote XOP by a, then XOQ is - - a. As the revolving line passes the points P and Q in successive revolutions in the positive direction we obtain the series of angles, a, TT — a, 2r + a, 3- - u, {ir + a', ^tt — a . Similarly from the negative direction we get, u, - 2- + i<, - ?>ir -a, — 1- + «, , etc. — TT r)7r a. etc. Now observe that 1. These series contain all integral multiples of -, V)oth posi- tive and negative. 2. To each ev(>ji multiple of -, a is added, but from eatih odd multiple, a is subtracted. .:^ . . I. \ I t llATIOS FOR llELATED ANGLES. QJ 107 Therefore ^ nir + { - l)"crf in which w denotes any integer, positive or negative, gives without excess or defect the series of angles reijuired. The same formula gives all the angles whose cosecant is the reciprocal of the given sine. 107. To Ji ml all the angles vliicli Imve a given cosine^ i.e., to solve the equation cos = a. r X' Describe a circle of unit radius and draw the diameters XOX', YOY', at right angles to each other. Let OM, measured on OX (or OX'), represent a in magnitude and sign; through J/ draw PMQ parallel to }'}"', meeting the circle in P and Q ; join OP, OQ ; then any angle described fiom OX to either OP or OQ, but no other angle, will have its cosine equal to a. Denote XOP by a, then XOQ is - a. As the revolving line passes the points P and Q in successive revolutions in the positive direction, we obtain tlie series of angles «, "2- — a, 2jr + a, 4r — fiven taiujoU, i.e., to solve the equation tan — a. X X' From in the horizontal lin(^ X'Xdraw 0^f ',\ unit in length in the direction indicated by the sign of n. Draw ^fP at right angles to X' X in the positive direction and of the length indi- cated by (I ; produce PO to Q^ making OQ equal O/^, and draw QN" perpendicular to X' X; then any angle descril)ed from OX to either OP or OQ will have its tangent e(jual to a. Denote XOP hy «, then XOQ is - + u. As the revolving line passes the points P and Q in the posi- tive direction we get the sei-ies of angles, a, TT -f a, 27r -f- (<, 37r -f «, 4?? + «, etc. Similarly from the negative direction we get, - 77 -f rt, - 27r -F a, - Stt + a, - 47r -f «, etc. Now observe that 1. These series contain all the multiples of tt, both positive and negative. 2. To each multiple a is added. \ i J >,' llATIOS FOR RELATED ANGLES. 109 Tlicrefof*! = inr-\-u, m wliicli n dcnoU's uiiy int(\iL,'(M-, ])().sitivo Of iiativt), gives without excess or defect tlu; series of {ingles recjuiicd. The same formula gives all the angles whose cotangent is ilie recipi'ocal of the given tangent. 109. In the foi-niuiu' of Arts. lOG-108, a represents tlie smallest positive angle wliose ratio lias the refjuired value. This restriction, howevei-, is not necessary. The angle « may be replaced liy any angle having the recpiired ratio, and the formuhe will still he true. In Art. lOG let ^ be any angle such that sill = v,\n p' = - n- + (" - I)"?- n- + ( - 1 )"+'•« =-',mr +{- !)'"«, in which 'tn = n^i\ according as n is even or odd. In either case 111 represents an integer and may con.seiiuently be replaced by ?^. The series of angles represented by 'inr -\- { - \)" a and ri7r + ( — 1)"^ are tlius identical. Similarly the otluir foriiiuhe may be shown to be universally true. 110. Since every numerical ([uantity has two s(juare roots, we see from Art. 51, that for any given value of the sine of an angle there are Y/ro cori'esj)onding \alues for the cosine, tangcmt, secant and cotangent, and the geometrical meaning of this douV)le sign may now be explained. In the diagram of Ait. lOG, 01i — . 8. sin-^^==i. 3. tanY^-l. G. cosec 0=2. 13. sin 2^y=:l 9. cos-^^=i. 12. tan- ^'' = tan' a. /{ 5. tan 2^0 = cot . 1 1 . sin- — cos- 0. 14. 2 sin A =tan A. . 1 G. tan// + cot ^/=r 1 . ^17. sec^ 0-2 tan" = 2. 1 8. sin + cosec = ^-. / 1 9. sin H.r + sin 3x = cos x. 20. cot - tan ^^ - cos + sin /V. ^21. sin 9^ + sin 50 + 2 sin' 0=^1. 22. sin {A+n)= ^, tan (^ - ^) = 1. 23. Prove that if a series of angles have a common tangent they are in aritinnetical progression. Is the converse true ? 24. If two angles have the same sign , show that either their sum is an odd multiple, or their diff(;rence is an even multiple of r. State a similar theorem with regard to the cosine. 25. Find all the angles which have both their sines and their cosines equal. i /. 26. The circumference of a carri.age wheel is 15 feet, and the carriage is moving forw.ard at the rate of 10 miles per hour; at what intervals of time will a point in the wheel which at lirst rested on the ground be at a height of I the diameter ? /,; > /( ■l'^^^ u )i il ": A ^/ • (•■ CHAPTER YIII. RATIOS OP COMPOUND ANGLES. 111. The angle wliich one trigonometrical line, PQ, makes with another, OX, is thus estimated. Move tlu! line OX parallel to itself until its initial point coincides with tlui initial point /* of PQ, and let PX' be the Hik; so placed ; thei\ the aiigle de- scribed by a line revolving from J'X' to PQ is th(; ungh- lequired. Similarly th(» angle X"QP h the angle M-hich the line Q/' makes with OA', and if PQ makes with OX the angle; //, it is evident that QP makes with the same lin(^ the angle r: + 0. 112. //PQ nxike v'ifh OX mi mif/Ifi 0, tho jvfojpction o/TQ, on. OX is PQ cos 0; aiidi/OY tiKikfirifhO^ a positive riyht auyhi^ the jyroJHctiuyi o/'PQ on OY is PQ si)i 0. Mli 112 RATIOS OF COMPOUND ANGLES. Through P and Q diaw RPT, OS, pfuallel to OX; PM, and (?/i?iV perpendicular to OX; then ^RPQ^O, J/iV== projection of J*(J on OX, ^W = projection on OV. Art. 7. Now by definition, Art. 43, we have PR . RQ . = cos (K ■ = sin 0. FQ ' PQ Therefore, for all values of 0, MN= PR = PQ cos 0, TS = RQ = PQ sin e. 113. One point in the preceding article requires careful atten- tion. When we speak of the angle at the point P we tacitly assume the direction from P to Q to be positive, but when we speak of the angle tz + nt Q, we assume the positive direction to be from Q to P. Clearly, then, we must not assume QP in the latter case to be the negative of PQ in the former ; each is positive in connection with its own angle. If we denote the length of PQ by /, we have proj. of PQ = I cos 0, proj. of QP = I cos (tt + ^) = -I cos 0, which shows the projection of QP to be the negative of that of PQ, as it should be. If, however, we choose to consider the line QP to be projected v" RATIOS OF COMPOUND ANGLES. 113 id [)t' t . / to be negative and denote it by - /, then PQ is positive and is the angle to be chosen. When, for any reason, a line has already been considered negative, this is the preferable mode of proceeding. 114. To find the sine and the cosine of the sum and the differ- ence of two angles in terms of the sines and cosines of the angles. Let the line OP starting from OX trace out successively the ungles XOQ ^ A, and QOP=B, then OP makes with OX the angle xi + B. From if, any point in OQ, draw MS, making a positive right angle with OQ and meeting OP in R, either line being produced if necessary ; then MS makes with OX the angle A + 90°. Now 0M= OR cos n, MR=OR sin B, Art. 112. cos (^1 + 90^') = - sin A , sin {A + 90°) = cos ^ . Art. 102. Since projection of OR on OX equals sum of projections of OM and MR on OX we have ^/^->^ e^-'- "" '"'^^' - Art. 10. OR cos {A+B)^ OM cos A + MR cos {A + 90°) = OR cos B cos A - OR sin B sin A. Therefore cos {A + B)^ cos A cos B - sin AsXnB. (1 ) Similarly projecting OR on F, we have OR sin {A +B) = OM sin A + MR sin (A + 90) = OR cos BsinA + OR sin B cos ^. Therefore sin {A+B) = sin ^ cos ^ + cos .4 sin B. (2) r 1 11 Ik ^ 114 KATIOS OF COMPOUND ANCILKS. E(iufitiuiis (1) and (2) being proved for all values of A and 7)', chan<^o /i into - />'. Then cos [A +(-/;)} ^ cos yl cos ( - />) - sin A sin ^ - II), or cos (/I - A') = cos A cos /i + sin A sin //. (^) And sin [ /I + ( - Z^} "= ^^'^ ^1 siu ( - />') + (;os ^I sin ( - Ji)r or. sin (A - li) = sin A cos /> - cos A sin //. (4) 115. The proof given in tlie preceding article is perfectly general, and Fig. 1, which re])resents hotli angles as positive, is alone necessary. It will, howcn-ei", be a valual)le exercise to adapt this proof to Fig. 2, which repi-esents the second angles as negative. Note then the following points: 1. O(^) is positive, being a bounding lino of angk^ ^1. 2. MS is })Ositive, because it makes a positive right angle with OQ. 3. MR is negative, b(H;ause it is drawn in a direction opposite to MS. 4. X'MS=A + 90' is the angle which must 1)6 chosen to pro- ject the negative line ^[R on OX. Art. 113. Then or OR cos {A - B) ■= OM cos A - MR cos (/I + 90) = Oil cos B cos A - OR sin B ( - Sin A), cos {A - />) = cos A cos B + sin A siii B. As an exorcise the student should draw the fifjure for various values of the juigles, positive and negative, and adapt tlie given proof to each. 116. The very iniport.ant formuhe of the preceding article may be proved in many different ways. We give another nu'thod, somewhat simpler than the former but less valuable, inasnmch as it is valid only for positive angles, such that A + B < 90'^ and A > B. RATIOS OF COMPOINH ANCLES. 11.') 117. To prove sin {A -f />) ^ sin A cos li + cos yl sin 11 cos (yl + B) = cos .4 cos S - sin ^1 sin />'. { Let the angle COD be denoted by A, and iJOi!' by /i; then the angle COE will be denoted by A + Ik In OK take any point P, draw /*J/and 7^<^ poipendicular to OC and Oi>; draw Qy and (/A* perpendicular to OC and /W. z- QPJi = 90^ - iY^T? = RQO = A. Then Now . ,, ,^^ MP MR+RP sni (yl + /y) = --- = 0/' ^>/' XQ RP + OP And cos (^ + B) NQ OQ RP ^luflJp^'pQ'Jrr = sin A cos /^ + cos A sin /i. ~ 0? ~ 6>7^ " 7>T' " 07 ox OQ RQ PQ (1) ay or P(yor = cos A cos iy - sin A sin /?. (2) lit) RATIOS OF COMPOUND ANGLES. r > * m 118. To prove sin {A ~B) = sin A cos B - cos A sin /? cos (/I ~ B)^ cos ii cos />' + sin A sin />'. Denote the angle COD by A and /)0^ by ^; then the angle COE will be denoted by A - l>. In OJS" take any point J\ (baw i^il/ and PQ perpendicul.vr to OC and OD; draw (^xV^ and (^A* perpendicular to OC and J//' produced. L QPR = 90 ' - PQR - i?(^i> = A. RP _NQ RP ~~Ol*~OP Then Now sin (/I MP MR '~1)P~ V. 0/' _^^(^ OQRP QP ~'OQ'~Or'"(JT''OP = sin -^1 cos B — cos ^ sin />'. (1) And ,^ „, OM ON + NM 'OP^'OP ON OQ QR QP ~'OQ'OP^'(JT*''OP — cos A cos B + sin -4 sin B. (2) EXKU(!ISE 117 119. From the fiindfiiiKMit.il formula' of Art. Ill, ii):iny im- purtiuit results art; easily ol>taiu('(l. We j^ive a few examples. sin (A + li) sin A cos li + cos ^ sin /? cos (J + li) cos ^1 cos li - sin A sin li tan A + tan li 1 - tan A tan li Ex. J.-~T.in{A + li) the last step bein<^ obtained by dividiiii^ both numerator and denominator of the preuedin<4 fraction by cos ^1 cos Ji. Ex. 2. — Sin '2A = sin (^I + ^1) = sin A cos ^1 + cos A sin i4 = 2 sin A cos A. Ex. ;?.— Sin (.1 + li + C) = sin {A + li) cos C + cos {A + li) sin C = (sin A cos /y 4- cos /I sin /)')cos C + (coSi'l cos/y-sin.l siny^)sinC = sin A cos Ji V C + sin li cos C cos ii + sin C cos ^ cos B - sin ^1 .s'li //sin C. EXERCISE XVI. 1. Given the sines and cosines of 45 ' and .'{0 ', prove sin 75' = '^"^-y^, cos 75^ = ^~~-r, tan 15^= 2 - x/3. 2. If sin A = cos Ji = j?, find sin (ii + Ji) and cos (-4 + Ji). 3. If sin yl = ;-|, and sin />'= ^% find sin (/I + li) and cos (vl - J5) (1) ^1 and yy both acute, (2) A acute and li obtuse. 4. If sin ^1 =— — , and sin /i— - —, and both anfflea lie in y;) ylO the first quadrant, then A +B = 4:b^. Find yl + H if each angle lies in the second quadrant. •5. Prove cos A + sin A = \/2 sin (A + 45) = \^'2 cos (A - 45°). . 6. Prove cos A - sin A = '\/2 sin (45° - yl) = V- cos (45° + il). 7. If tan yl = |, and tan Ji = ^, find tan (yl + B). 8. If tan yl = -^(j, tan ^ = iPn, find tan {A - ^). 9. If tan A = i, find tan 4yl, and thus show that yl is slightly greater than 11^°. 118 RATIOS OF COMPOUND ANGLES. 10. If tan 2.1 - 2 taii {A + B) tlicMi tan 7^ = taiv' A. 1 1 . Prove sin (J + II) cos ]» - cos (^1 + B) sin B - sin A . tan (a - p') + tan p' 12. Prove :, —r~,^i '- = tixnu. 1 - tan {a p) tan p 13. Write dowu other identities employing the j)iiMciple ex- (Mnj)liru'd in examples 11 and 12. 1 [. If tan ^= 2?^i + 1 and cot {" - (j)) = '2nt-, find tan •) + sin (.1 - /i) - 2 sin A cos B. 2. sin (.1 - B) - sin {A - B) = 2 cos A sin B. 3. cos (J - />') + cos (^1 + />•) = 2 cos A cos B. 1. cos (^1 - /)') - cos (A + B) ^- 2 sin A sin B. f). sin (J + /)') sin (.1 - />) ----- sin- A sin"-' B cos"-' Ji cos'- A. 6. cos (.1 + B) cos {A - li) = cos- .^1 - sin- B ^ cos- B sin- A. tan A + tan 7^ 1 - tan A tan 7/ tan il - tan B 7. tan (.1+7)')- 8. tan (A-B) = 1 + tan A tan 7>' , -r,x 1 + tan A 9. tan (.1 + 15^)=, ;. ^ '^ 1 - tan A 10. tan(J:-45) = tan A — 1 tan A+l 11. cot (.1+7?) = 12. cot(.l-7>') = cot A cot 7? - 1 cot A + cot B ' cot .1 cot ^ + 1 cot B - cot A T. sm (J + />) , ■ , „ sni (y1 - 7>) 1 3. tan A + tan ^ - \ /,. 1 1. tan .1 - tan £ = -^, /,. cos J. cos B cos ^1 cos 7i 15. cot A + tan A = 2 cosec 2.1. 16. cot .1 - tan A = 2 cot 2^1. T. cos(A B) ,, , „ cos(/l+i5) 17. cot A + tan Ji = .— \ - — -.. 18. cot A - tan 7? = ^^A-JL^J sin yl cos 7) sni yi cos ^ cot ^/ + tan ^ tan tan c/j + 1 1 + cot cot cos (0 - <;^) cot ^V - tan cot <"> cot - 1 . cos (^ + ) * RATIOS OF COMPOUND ANGLES. 119 20. tail 4- t;in f/) _ tan cot + 1 1 + cot tan sin {0 + ) tan - tan tan cot - i 1 - cot tan ~ sin {<^(]>)' 21. sin (^1 f JJ + C) = sin ^1 cos Ji cos C + sin />' cos C cos J + sin C cos .1 cos 7)' - sin .1 sin /> sin C. 22. cos (^1 + J> + C) = cos J cos Ji cos C- sin .1 sin />' cos C - sin />' sin C cos .1 - sin C sin ^1 cos B. OQ . / « , 7. />\ **'' -^ "+• tiiii />' + tan C - tan .1 tan />* tan C Jo. tan [A -j- 1) + ly) = 1 - tan^l tan/y - tan/* tan C - tan C tan^l' 120. For cojivenience of reference tlie important formuke of Art. Ill are here repeated. sin (A + J]) = sin A cos 7>' + cos A sin />'. (1 ) sin (.1 - />') ^ sin A cos 7^ - cos A sin 7)'. (2) cos ( J + /y) = cos J cos 7y - sin J sin Ji. {:\) cos (.1 - 7)) = cos ^1 cos JJ + sin .1 sin 7>. (4) From these, ])y addition and subtraction we easily obtain sin (J + /;) + sin (A - Ji) - 2 sin A cos li. sin (J -f />') _ sin (J - li) .= 2 cos ^1 sin Ji. cos (.!-//)+ cos (^1 + li) = 2 cos A cos /i. cos (.1 -B)- cos (^1 + Ji) - 2 sin ^1 sin B. Now let .1 +B^P and J - />' = Q, then ., = ^:±y .,,„,/;. ^^. Substituting tiiese values for .1 and Ji we get sin 7* + sni Q — '2 sni -!• cos -. 2 2 (? ■ (r>) (♦^) (") (8) sin 7^ - sin (? =::^ 2 cos — ^ sm — cos (? + cos r^'2 cos ^ cos - — -. •> 9. cos Q -cos 7^— 2 sin — ;^ — sin Q (9) (10) (11) (12) 1 J i f it ! I 120 RATIOS OF COMPOUND ANGLES. Again, re-arranginfi; formuhie (5) .... (8) we have sin A cos 7? - h {sin (^1 + B) + sin (A - V?)}. cos A sin />' = h {sin (A + IJ)- sin (^1 - B)]. cos yl cos 7i = \ {cos (yl - ^) + cos (A + B)}. sin yl sin /? = ^ {cos (yl ~ B) - cos (yl + B)]. (13) (U) (15) (16) \ 121. The importance of the preceding formulae is such that we repeat a portion of them in words, with a few observations, to assist the learner in committing them accurately to memory, 1. The sum of the sines of two angles is equal to twice the sine of half their sum nmltiplied by the cosine of half their difference. 2. The difference of the sines of two angles is equal to twice the cosine of half their sum multiplied by the sine of half their difference. 3. The sum of the cosines of two angles is equal to twice the cosine of half their sum multiplied by the cosine of half their difference. 4. The difference of the cosines of two angles is equal to twice the sine of half their sum multiplied by the sine of half their difference. 122, When the sine of a difference occurs care must be taken to observe the proper order in subtracting, and we observe : 1. The angles forming the difference have the same order on both sides of the equality when the difference of sines is involved as in (10), but have the reverse order on opposite sides as in (12), when the difference of cosin(?s is taken. 2. When the product of a sine and a cosine is to be trans- formed into a sum, subtract the cosine angle from the other and add the sine of the difference thus found to t\\°> sine of the sum of the two angles. Equations (13) and (14) are thus different modes of expressing che same relations. (13) (14) (15) (16) RATIOS OF COMPOUND ANGLES. 121 3. "When the cosine of a difference occurs as in (9) and (11), either angle may be subtracted from the other. Art. 101. 4. In transforming the product of two sines as in (16), the difference precedes the sitm. 123. It will be a valuable exercise for the student to prove equations (0). . . .(12) of Art. 120, geometrically. We give the proof of (9) as an example. 124. To prove gcometricalhj that sin /^ + sin Q = 2 sin . r+Q P-Q ) ^111 cos 9 Draw the angle XOQ = Q and XOP^P; take OP and OQ each a unit in length ; join PQ, bisect the angle QOP l»y OR which will consequently be at right angles to J'Q ; draw PM, RS, (;>ir perpendicular to OX, and R2\ (<>f/r parallel to OX. Then and angle (,0^ = '^.': = ^-« XOR = XOQ + QOR --= Q r " 2 ' P-Q P+Q "> t 4 i 1 i! ■ ' f' ' \4^ V I 122 Since and Then RATIOS OF COMPOUND ANGLES. rR=R(,> it easily follows that PT= RU, TR= UQ, and consequently PM+QN-^RS. sin r + sin Q = PM+ QN= -IRS on I? ■ ^'+^*^ = 20R sill — ; — ^ . r+Q P^Q = 1 sin — -— cos — ; — . 2 2 Similarly equations (!(•)» (H) and (12) may Le proved. The fundamental formuhe (1) .... (1) may also be proved from the same diagram. EXERCISE XVIII. Change the following sums and diilerences into products : 1. sin 7r)'' + sin 15^. 2. sin 75^ -sin 15°. 3. cos GO ' + cos 30°. 4. cos 30" - cos 60^. 5. sin 75V cos 30^ 0. cos 80^ -sin 30°. 7. sin3^ + sinfl. 8. cos 3^/ - cos . 12. 2 cos «.cos^. 13. sin 20 sin 2(/). 14. cos cos 36^. ,. . . , ,. . 'oo . 3/y 10 sin oc.sin ba. lb. sin — . sin —. Ji It 1 7. sin (yl + B) sin {A - B). 1 8. cos {A + B) cos {A - B). \^. cos(30>^i)cos(150" + ^). 20. 4 sin J sin (120° + yl) sin (240° + ^). . • Prove the following identities : „ , sin 3° + sin 33° , „^ ^_ cos 3° - cos 33° 21. :,^"-— 7T7^ = tanl8°. 22. -.-- .-- ^rrc^^^tan 15°. cos 3° + cos 33° sin 3° + sin 33° EXEUCISE. 123 23. 25. 27 cos 10' + sin 10' -tanr)ry\ 24, cos 25^ - sill 5"' = cot :\r^'\ coslO"-sin 10^ cos 5' - sin 25 ' sccGO"-sec40^ = 4sinlO\ 20. cot 50^ + tan 50^ = 2 seel 0\ 10 'Ml sin .;iu -— + S1U -- siu no sill 2^ysiii5^'. 2S. 29. 30. ^31. 32. 33. 34. 35. 3G. 37. 38. ^ 39. 40. 41. cos . 1 + cos ( 1 20 ' f J ) + cos ( 1 20^ - J ) = 0. 4 sin J sin (GO -f- J) sill (GO' -J) -sill 3^1. ' 1 cos J cos (GO " + J ) cos (GO ' - J) - cos 3^. 1 G cos 20 ' cos 40 ' cos GO' cos 80' - 1. 4 sin 20\sin 40\sin 80' -siu GO^ sin A - siu /> A 11 c(»s /*' cos .4 cos A + cos Ji 2 sin -il + siu // siu x\ + siu Ji A \- 11 cos />' — cos ^1 ~ =tiiu ■ — ^ ^ . . — -. cos A + cos h 2 sin A - siu h siu A +siu li t;in \{A + I>) siu A - siu Ji tiiu I {A - Ji)' cos A - cos Ji A + /> />' J tan- J, />' - tjiir i yl cos^yr+Tm7>' "^ '""'^ ~T" '^'^ "~2~ "" 1 - tail"'' yy7tiiir L i « 4 p , « - p' 2 sin a tail- ' +taii "' = — . 2 2 cos a + cos ^5 cot - „ +cot— - = V V 2 sin a cos jj — cos a sin J + sin (J + /^Hshi (J + 2/?) cos/+cos~(yl + y>'y+cosp'+2y.')" '^"'^ "^ '^" siu A 4 2 siu 3.1 + sin 5.1 siu '^A sill 3.4 + 2 sill 5.4 + siu 7-^1 siu 5^1* If (1 + tan A) (1 + tan ]i) --= 2, then tan (^1 + Ji) = 1, Solve the equations. 42. siu + siu 3^/ = 0. 43. .sin 10 - siu = sin 3^. 44. sin^/ + sin 26' + sin 3^y-0. 45. sin <^ + sin 9^ -sin 3^/ + sin 7(9. r * 124 RATIOS OF (COMPOUND ANGLES. 4 6. cos 10 - cos iO = sin 0. 47. sin- + sin'-^ 2<'/ = 1 . 48. sin « + sin {0 - «) + sin (20 + a) = sin (0 + «) + sin (2^ - a). ,,^ ,„sin,x' sin 3x sin 5ic , a.-lf/.^+Or. ^/., -3a. 4'J. It = = , then ^ = -^ . «i 0-, Or. a-x «, ^^ ^„ cos X COS (.»; + /y) cos (x + 2^^) cos (.*• + WO) 50. If = ^: = • = — , Hi a., O; prove ((i+n.y (l.^ + Oi a., tti Prove tlie following identities : 51. sin (36° + ^1) - t^hx {W - A) - sin (72^ + A) + sin (72° - A) = sin A. 52. sin(54° + vl) + sin(54°-^l)-sin(18° + yl)-sin(18°-yl) = cos J. 53. sin a sin ((i - y) + sin p! sin (}' - a) + sin y sin (a - (f) = 0. 54. cos a sin ((i - }') + cos pi sin (j- - a) + cos y sin (a - p') ~ 0. 55. sin (a - p') + sin (p> - 7) + sin (j' - a) . . a-(i . /5-7 . y-a = - 4 SHI — — ' sin —- sin — . J J J 56. sin (« + 13 - y) + sin {^ + y ~ «) + sin ( j' + '/ - (>') - sin (« + ^ + y) = 4 sin « sin ^ sin y. 57. cos (a 4- 15 - y) + cos ((5 + 7 - «) + cos (j' + « - (f) + cos (« + 1^ - 7) — 4 cos a cos |5 cos J'. 58. sin a cosec (a - ^) cosec(a - y) -\- sin|^ cosec(|3 - j')cosec(|^ - a) + sin y cosec (y -- a) cosec (y - p) = 0. ,' 59. COS' ic + cos- 2/ + COS" « + cos^ (« + 2/ + s) = 2 {I + cos (y + z) cos (z + x) cos {.v + ?/)}. 60. sin^ 03 + sin- 2/ + sin- ~ + siir (,'« + ?/ 4-;^) = 2 1 1 - cos () and cos (^1 + B) from Art. 114. These formuhe being universally true, any results algebraically deduced from them must also be universally true. 126. To exprpsii the ratios of 2 A in terms of the ratios of A. From Art. 1 1 4 we have sin (A + B) = sin A cos B + cos A sin B. cos (A + B) ^- cos A cos B - sin A sin B. In each of these formuke, for B write A. 0) (2) then or Also or Again sin (A + J) ==sin A cos A +cos A ain A, sin 2A=2 sin A cos A. cos {A + A):=eos, A cos A - sin A sin A, cos 2 A == cos- A - sin- A = 2 cos2 A - 1 = 1-2 sin- A. tan 2A '^"^ ^^ ^ ^"^ ^ cos tI 2 tan A cos 2 A cos- A - sin- ;4 1 - tan'- A' (3) (4) (G) The last step being obtained by dividing numerator and de- nominator of the previous fraction by cos- A. ■i^~ » 4 126 MULTIPLE AND SUBMULTIPLE ANOfLES. 127t To express the ratios o/3A in terms of tJie ratios of A. sin 3yl:=sin {^A + A) = siti 'lA cos A + cos 'iA sin A = (2 sin A cos yl) cos yl +(1 - 2 sin'-' A) sin yl = 2 sin yl cos- /I +siu A ~'l sin yl = 2 sin y1 (1 - sin'- y1) + sin A ~'l sin" yl = 3sin i1 -4 sin'yl. (1) cos ZA =cos (2il + yl) = cos 2^1 cos yl - sin 2yl sin A = (2 cos'- A - \) cos yl - (2 sin A cos J) sin yl = 2 cos'' A - cos A - 2 sin'- A cos yl = 2 cos' A - cos ^ - 2 (I - cos'-' A) cos yl = 4 cos-' A - 3 cos A. (2) tan 3i4 = sin 3yl 3 sin yl - 4 sin^ yl cos 3yl 4 COS'' A - 3 cos yl 3 tan A cos'^ yl - 4 tan^ yl 3 tan A sec- yl - 4 tan" yl 4- 3 4-3 sec'- yl cos'-^ yl 3 tan yl (1 + tair y|) - 4 tan'' .1 3 tan yl - tan'' yl 4-3 (1+ tan'- yl) 1-3 tan'-' yl (3) i ! 128. The process of Arts. 12G, 127 may evidently be continued so as to express tlie sine or the cosine of any multiple angle in terms of tlie powers of sines and cosines of the single angle. It will be observed in each case that the degree of the resulting expression is always equal to the coefficient of the multiple angle. Thus the ratios of 2yl are replaced by expressions of two dimensions, those of 3^-1 by expressions of three dimensions in sin yl, or cos A. Conversely, the squaie or cube of a sine or cosine may be replaced by sines or cosines of multiple angles. MULTIPLE AND SUBMULTiI'LE ANGLES. 127 (1) (2) (3) i; 129. The ratios of multiples of tt and - are sometimes useful. The truth of the following will be readily perceived, in which n denotes any integer. 1. sin ?irr = 0. 2. cos mr = {- 1)». 3. sin (2?i+l) "=(_!)". 4. cos (27i + 1) ~ = 0. -> 2 From these the values of the other ratios may be easily obtained. /^ 130. To find sin (;t7r + a) and cos (n7r + «) where n is any integer. We have sin {ntr + a) = sin n- cos a + cos nr. sin a Art. 1 1 4. = (-l)"sin«. Art. 129. Similarly cos {nir + a) = cos n- cos a - sin nir sin a Art. 1 1 4. = (-l)"cos«. Art. 129. 131. Given sin A = sin B, to compare sin niA and sin niB, where m is rt?i?/ integer. Let a denote the smallest positive angle whose sine equals that of A or B. Then A = mr + {- ly a, J] = r7r + {- If a, now sin ( - 1)" « = ( - 1)" sin a. Art. 101. Therefore sin mA = sin { />wi7r + ( - 1 )" ma } = ( - 1)'""+" sin vua. Art. 130. Similarly sin 7n5 = (- 1 )""'+'■ sin ma. The exponents of (-1) in the two cases are (m+l)n and {m + l)r. If rw,+ 1 be even, then both exponents are even, and sin niA = sin mB, but if m + T be odd, the exponents are odd or even, according as w and r are odd or even, and sin Wil = ( - 1)"+'' sin ?«/?. T wm i 128 MULTIPLE AND SUBMULTIPLE ANGLES. 132. To express tlie sq^iare or the atbe of a sine or a cosine in terms of the sine or the cosine of a muHiple amjle. From the equations cos 2A = 2 cos- A - I = 1-2 sill' A we got by rearranging cos'-' A 1 + cos 2 A 2 and . „ , 1 - cos 2 A sm- A = -— - — , . Similarly froin the equations, sin 'AA ^ 3 sin A - 4 sin'' A, cos 3^1 = 4 cos' vl ~ 3 cos vl, we get sin'' A = I (3 sin A - sin 3^1), and cos' ^1 =^ | (3 cos A + cos 3^1). 0) (3) (G) (") By means of tluise formillie any power of a sine or a cosine, whose exponent contains no factor except 2 or 3, may be raplaced by sines or cosines of multiple angles. 133. In transforming trigonometrical expressions it is usually easier to work with multiple angles than powers of the sine or cosine. We give a few simple examples. Ex. l.—^in' A + sin^ (A + 60'') + sin- (A + 120°) = ^, { 1 - cos 2 J + 1 - cos {2A + 1 20") + 1 - cos (2^1 + 240") I Art. 132, = I [3 - {cos 2 A + cos (2.1 + ?, iO")} - cos (2 A + 120'";] = I [3-2 cos (2^ + 120) cos 120'^ - cos (2/1 + 120")} = jl, since cos 120°= - i. Ex. ^.— Sin 3 A sin^' J +cos 3^1 cos'' A = \ {sin 3A (3 sin A - sin 3yl) + cos 3yi (3 cos A + cos 3A)} Art. 132. = I {3 (sin SA sin A + cos 3^1 cos ^1) + (cos= 3^ - sin'-^ 3/1)} = I { 3 cos 2 A + cos GA] = cos'2^. Art. 132. 1 itmmsm^ EXERCISE. 120 Prove 1. EXERCISE XIX the following; identities : sin 4 A = 2 sin 2 A cos '2 A. sill 2A yf^'l f). 7. 9. 11. 13. 15. in. 17. 18. 19. 20. 21. 90 2i. 25. 2G. 27. 28. tan A eot 2yl = 2. cos 4/1 = 1^.2, c.Qai2^. sin 2.1 1 + cos 2.^1* cot- ^1-1 2 cot A ' sec- A sec 2A= — 1 - tan'-' A 4. cot A = — (>. tan- A = 8. tan 2^ = 1 - cos 2 A' 1 -cos 2 A 1 +COS 2/1 2 tan .'^-^1 3 - tan2 A cot A - 3tanyl 10. cot 3/1 cot /I - tan /I cot' // - 3 cot A 3 cot" A I COS 3vl = (2cos2yl - l)cos.4. 12. sin 3A = (2 cos 2/1 + l)sin/l. sin .1+ sin 3/1 ^ „, ,, 1 2 sin' ^l 1 . . ,-^ = tan2/l. 14.-, - = . cos A + cos 3 A 1 + sin 2^1 sec 2/1 -t- tan2.^1 cos A + sill A = tan 2/1 +sec 2/1. cos /I - sin A tan- ^1 + cot- ^1-2(1 + 2 cof- 2 A). sin 3^1 cosoc A - cos 3.4 koc ^1 =- 2. 3 sin A - sin 3^1 - 2 sin ^1 (1 - cos 2/1). 3 cos A h cos 3.4 = 2 cos A (1 + cos 2/1). cos 3.4 + sin 3^1 = (cos A -- sin ^1) (1 + 2 sin 2/1). cos 3/1 - sin 3/1 ^ (cos A + sin /I) (I - 2 sin 2/1). cos ^1(1- tan 2/1 tan A) - cos 3^1 (1 + tan 2 A tan A). sin- 2 /f - 4 sin- yl _ cos- 2/1 - 4 cos" A +3 siir 2^1 + 4 sin- /I - 4 ~ cos- 2^1 +Tcos^7l "^ " 4 sin A sin (60° - A) sin (GO" + A)== sin 3 A. 4 cos /I cos (GO^ - .4) cos (G0° + /I) - cos 3^1. tan A tan (GO'^ - A) tan (GO" + .1) - tan 3/1. tan A + tan (GO ^ + A) + Urn (1 20" + /I) - 3 tan 3 A. cot A + cot (GO ' + ^1) + cot ( 1 20" + A) = 3 cot 3/1. I, ..Will 130 MUI/ni'LE AND SUI3MULTIPLE ANGLES. i cos(.^- .V) ^ COS (3x_?,) ^ _ Bin 2;t + sill 2i/ ^ •" 30. ?!^ ■^^>-+^'." (f*-±2^ = '. cos (X- + 2,). sm 2x' hsiu 2// tan + cot ^/ + 2 _ si ii^ {0 + 45°) taii7/Ti;oI 7y -i ~ si ii^(/7 -~ 4 5' ) ' 32. 1 - cos 3x' = ( 1 -- cos a;) (1 + 2 cos xf. ~" 33. cos 9.i; + 3 cos Ix + 3 cos 5.i' + cos 3.*; = 8 cos' x cos 6x 34. cos 6^y - 32 cos'' - 48 cos^ 0-\-\^ cos'- ^^ - 1 . 35. cos G^y - 1 - 18 sin- + IS sin' ^; - 32 siii« 3G. sin 50 = 7) sin ^^ - 20 sin^ ^y + 1 G siir' 0. 37. siu^ - cos 4^. 42. sin 3^1 sin" A + cos 3 J cos*'' A = cos' 2yl. 43. sin 3/1 cos"* Ji + cos 3/1 sin'' /I v^ \ sin 4/1. 44. Given {\-\-e, cos ^) (1 + « cos 2^^"2=^n| 1+e 1 e* 45. If sin ^ 1)0 the geometric mean of sin A and cos /I, then cos 2 />^ = 2 cos- (/I +45"). y-lG. If sin (« + |3) cos y = sin (« + ;') coi /9, then either « = (2n + 1 ) " , oi- |3' - J' = nr.. 47. If « sin 0-\-h cos = c = a cosec (9 + /; seo I 2 = 1 +sin-i'l. A Therefore cos -' + sin " = Vl + sin A, Similarly cos - - sin - = Vl - sin A. (1) (2) From which by addition and subtraction we obtain 2 cos ^^ = v/l + sin A-\-\/\ - sin A 2 sin ^^ ^ y'l + sin A - \/\ - sin A. (3) (4) Now since the numerical values of the surd expressions in (1) and (2) may bo tak(;n as either positive or negative, we shall have, in (3) and (4), four ditlerent values each for cos ^ and sin for each value of sin A. The signs which connect the surds in (3) and (4) denote the operation of addition or subtraction, and give no indication whether the following root is to be })Osi- tive or negative. This must be determined by the magnitude of the given angle as shown in Art. 139. MULTIPLE AND SUJLMULTIl'LE ANGLES. 133 137. To prove geometrically that for each value of cos A there A A* are two values each for cos and sin — . 2 2 Let XOP, XOQ be the smallest angles, positive and negative, which have the given cosine. Bisect these angles by 0/>, Oq, and produce these bisecting lines backwards as in tlie figure. Denote the angle XOl* by «, and the series of positive angles having the same cosine by A. Then A^a, 2--«. 27r-f«, 4r: - «, etc. A 2" a 9^' TT - a - + .. , a a •ir: - ^, , etc. — — -* ^ -ZO;;., A'O^y', XOi>', XOq,etc. Now cos AO;; = - cos AOr/ = - cos XOp' = cos X0<{, and sin AOy; = sin XOq' = - sin XOp = - sin A0«/. TT -^^ a . A . ,* lience cos ^ =±cos ,, sm ^, - ±sm -, which give the two = ±cos sin ^, - ±sin values required. The synnnetiy of the figure shows tl»3 same results to be true for negative angles. The proposition is, therefore, universally true. li "-"nbHP^' i:J4 MULTIPLE AND SUHMULTIPLE AN(;LES. W ■ i ■ \t 131 138. ^o pi'ove grj)iwtricul I y tlutt for eacli value of sin A there fire four tallies each for cos — and sin — . Let XOl\ XOQ be tlie angles in the first revolution which have the given sine. Bisect these angles by Op and O7, and produce the bisectors backwards as in the figure. T)(Miote XOP by a, and the series of positive angles having the same sine as XOr by A. Then A ■= «, ir - («, 2tt + a, ^tt - «, 4 n- + «, etc. A (I TT a a ., - a „ /< ^ +.. > -V o_ -- + 2' 2 etc. = .Y(9^>, A'Or/, XOp', XOq', XOp, etc. Now it is evident from the figure that the angles XOp, qOY, X'Op', (j'O V, aie geometrically ec^ual ; from w hich it easily follows that cos XOp = sin A'Oq = - cos XOp' = - sin XOq\ sin XOp = cos XOq ^ - sin XOp' = - cos XOq'. A a a a a Hence cos ---= ±cos , or ±sin ; sin = ±sia _, or ±cos — , J 2 J 2 J ^ whicli give the four values required. MULTIPLE AND SUBMULTIPLE ANGLES. 135 139. To determine the siyns ivhich must he taken ivith the radi- A A cals in obtaining the values of cos - and sin from the value o/ shi A. S70' Draw a circle, divide it into eight equal sectors as in the figure. As the angle is placed in each division in succession, consider the values of cos and sin ^ separately, whether posi- tive or negative, and which is numerically the greater, and thus determine the sign of the coml)ined expressions cos + sin ,^ and cos -sin - . If we mark the sign of the former on the outside of the circle, and that of the latter on the inside, the result will be as shown in the fii^ure. The sijjns thus obtained for any sections of the figure are those which must be taken with the corresponding radicals when lies in that section. The reader should carefully note the points for which the two expressions are zero, and where they conse(jucntly change sign. M Ill i 136 MULTIPLE AND SUJ3MULTIPLE ANGLKS. 140. The results of Art. 139 may also bo obtained symboli cally, and it is instructive to compare the two processes. Let a be the smallest positive angle whose cosine equals cos A. A = 2mr ia, A Then and cos - -^ cos Also sni - - — sin "■ . . a cos w- cos -^sin /in: sin - 2 2 (--l)"cos --. ^ ^ 2 (-±:^) '/ , a - COS ; "h cos nr: sin - 2 " 2 = ±(-l)"sin^-. (1) sin liu {'^) Equations (1) and (2) show that for each value of cos A there are two values each for cos and sin ; they also determine •J .J the sign to be prefixed. Again, let a denote the smallest positive angle whose sine equals sin A. Then A=n7: + {-\)" a. We must now distinguish the cases in which n is even or odd. 1. Let n be even and equal 2»i. A = 2m- + a, Then and cos V = cos (m. + ^) (-l)'"cos^ « 2 (3) Similarly . A sin - = sin = ( - 1 )»' cos -^. (4) i MULTlli.E AND SUJiMULTlPI.E ANGLES. 137 1 2. Let n l)e odd and equal 2/u + 1. Then and Similarly A ={2lH+ 1) TT- a, cos . = cos 1)1- H = ( - 1 )■•■ sin I. •J =(-irsm(|-;) = (-1)'" COS '^. (5) . A . SlU - :r::Sni 2 (6) Equations (3). . . .(G) show that for each value of sin A there are four values each for cos ^^ and sin ^ , and they also determine the sign to be prefixed in each case. 141. In the previous article a denotes an angle in the first quadrant, - lies between 0' and 45 ', cos ;^ and sin are both 2 Li -J positive, cos > sin ,y, hence equations (.">) and (4), Art. 136, give a 2 cos ;^ = >/! + sin ^1 f- Vl - sin yl, « 2 sin - = Vl +sin A ~V^ - sin J. Then equations (3) and (4), Art. 140, give cos =(-!)"' cos 2 (1) sin '-' =. _( Jl! 1^1+ sin i + Vl - sin ^ | (3) 2 2 ( J ;^ = ( - ir sin -" = ^";i^ [vu^uTa - vf^shTij (4) 2 2 2 \^ J IQ ju^^^ «p -. ■. iim if iM^ 138 MULTIPLE AND SUBMULTIPLE ANGLES. I ; 'I in wliich u is even, whilst (5) Jind (6) show that the connecting signs on tht; riglit must ho reversed when n is odd. In both cases VI is the integral part of , . ml Ex.— U± A - TjOO'^ - 37r - iO" ; tiien n is odd and m - 1. Then cos ^ - ( - 1)'" sin "-= - V^F+sin^^l - Vr^siiwl). This result should be verified by reference to the diagram of Art. 139. 142. To express tan in terms of tan A. In Art. ll'G, change A into - and we have, 2 tan h A 1 - tan- h A' tan A . ii'iv .1 ^1 + 2 tan I A - tan A = 0, tau Then from which ^ -l±/l+tan-^ tan - = — , . 2 tan A Thus, for each value of tan A there are two diflferent numerical A values for tan --,. The student should draw a diagram illustrat- ing the double value, and also examine the result symbolically, as has been previously done in the case of the sine and cosine. 143. The formuhe of this chapter may be employed to obtain the ratios of the angles formeily obtained by geometrical con- structions. Ex. L—To find the ratios of 45°. Since 45° is its own complement, we have sin- 45" = cos- 45" = 1 - sin- 45°. 'J'herefore 2 sin- 45°= 1, or sin 45° = cos 45° = -—. MULTIPLK AND SUBMULTll'LE ANGLES. 139 ting I il Ex. e.— To tind the ratios of SO' and 60'\ Since 30° and 60'^ are conipleniontary angles, we have cos 30° - sin GO - 2 sin 30^ cos 30^. Therefore cos 30^^ (1-2 sin 30') = 0. Now cos 30^ is not zero, ,*. sin 30'^ .\. Ex. 3.— To tind the ratios of 18°. Since 36° and 54° are complementary angles, we have sin 36° = cos 54°. Therefore 2 sin 18° cos 18°= 1 cos^ 18°- 3 cos \6^. Art. 127. Divide by cos 1 8°, 2 sin 1 8° - 4 cos- 1 8° - 3 = 1-4 sin- 18^. Therefore 4 sin= 18+2 sin 18° - 1 = 0. Solving for sin 18^ sin 18°= -V^^-. 144. In Art. 127, change A into and wo get Sin il = 3 sin - 4 sin' - . Let sin A=a, for sin write .*■, and rearrange the result ; we get A Hence it appears that sin may be found from sin A by the o solution of a cubic equation. Similarly, cos - may be found o from the value of cos A. If to a we give any particular numeri- cal value the roots of the resulting equation can always be found. The process, however, does not V)elong to Elementary Algebra, and is consequently seldom einployed. Other methods are adopted for obtaining the numerical values of the ratios, and when these have been obtained they may be used to solve cubic equations. I 140 MULTIPLE AND SUBMULTII'LE ANGLES, It is instructive to observe that the hisectiou of an angle and the solution of a quadratic equation are corresponding operations, and both are possible l)y elementary means. Also, the trisection of an angle and the solution of a cubic eciuation correspond, but neither is possible l>y elcmentaiy methods. .'i ; I »^#S EXERCISE XX. 1. Apply the formuhe of Art. 134 to lind sin and cos - for the following values of A : (1)30°. (2) -30^. (3)315". (4)378^. (5)132°. 2. Verify the results of the preceding example by the method of Art. 136. (^ 3. P'ind all the ratios of 9° from the value of sin 18° 4. Given tan 15° = 2- ■v/3, lind tan 7V' and tan 37i". 24 5. Given tan 2^1 = - -^, find sin A and cos A. Draw a figure showing how the different valutas are possible. 6. Given 2 sin vl = Vl +sin 2vl + Vl - sin 2yl, find the limits between which A lies. 7. Given 2 cos A= -y/l + sin '2A+Vl -sin 2A, find the limits between which A lies. Q T) .cos }f A sin i vl 8. Frove sec A = -— _,,^-— . + " , and show how to V 1 +sin A \/l - sin A determine the correct signs for the radicals. 9. Prove cos ^^^ = -^ I2 + V2+ ..V2 + -2 cos A ^ \2-V2-\- .. V2 + 2 cos A which con- tains n radicals. 10. Prove sin — = On which con- tains n radicals, and all connecting signs except the first being positive, ; !l ^ 11 i EXERCISE. 141 to I 11. Adapt tlie two preceding examples to find the perimeter and area of a regular polygon of '2u sides, and thenoe show liow the value of tt may be found. 1 2. Prove (t-os .1 + cos /if + (sin A + sin />)- = 4 cos- — ^^"^ . 1.3. Prove (cos A + coh 7)'/-' - (sin .1 +sin Jif ^ \ cos (.! + />) .A- li cos- 2 14. Obtain otln^r formula' similar to tin; two pr«'ceding by changing one or more of the connecting signs. 15. Express cot in terms of cot 0, and thence sliow that cot > 1 + cot U for all values of from to -. '1 X IG. (liven tan .'»! = (2 + \/3) tan -, find tan x. 17. Prove tan 142'° = 2 i v/2 - y^^^ - y^O in three diirerent ways. 18. Adapt tlie method of Art. 143 to find cos 18^. 19. Assume yl = 180" - 2yl, and thence determine the ratios for G0° 20. Assume dA = 180"^, and thence find the ratios for 36°. 21. In Art. 143, Ex. 3, if the negative sign be taken with the radical, of what angle will the resulting expression ho the sim;? 22. If cos a be one root of the equation 4x^-3x~a, tlien cos (120° + a) and cos (120 — a) are the other roots. 23. Fiom the preceding example deduce (1) cos a + cos (120^ + a) + cos (120 ' - a) :- 0. (2) cos a cos (1 20'^ + a) + cos (1 20*^ + a) COS ( 1 20^ - a) + cos(120"-«)cosa= -'l (3) COS a COS (120" + a) cos (1 20" - «)-^cos 3a. 24. Prove the three preceding identities trigonometrically, and thence write the equation whose roots are cos «, cos (120 -fu), cos (120" -a). r CHAPTEU X. I i» INVERSE NOTATION AND SUBSIDIARY ANGLES. 145. In the o(}uation tan (f — a, there are two quantities in- volved, representing an angle, and . ' Ex. S. — Given sin^^^rt, express the formula' for sin LV> and sin ^0 in the inverse notation. Since therefore and Then therefore, or Similarly fi-om we get sin "^ n c.ofi ^ VY~^'u^ <' = sin~' (I. sin 2^^ -^2 sin^; cos 0, 10 sin-' (2 sin cos 0) 2 sin-' a = sin '\2=3 sin 0-4- siu^^ 3 sin"' (/ =^ sin"' (3a - 4a^). (1) (2) ■^ ^nmm 144 INVERSE NOTATION AN' I) sriJSIDIAUY ANOI.KS. Ex. 4' — A'o prove tun ' n f tan ' /j=^i -'{ri} Let and then or. tan"' II = A, or = //, or ft ~ tan li; tan yl + tan B tan (id + B) = 1 - tan A tan 7/ tan(tan~'a + tan ' ft) == 7 therefore, tan"' a + tan" 'ft = tan"' n + b \-aV II + b 1 -aft This formula is of very frequent application. i 149. There is one point which requires careful attention. In the equation tan O — a, if we gave any one value to (f, we get but one value for it. For example, if — •1.'')" th; u a 1. If, Iiowever, we take the (Mjuivalent ecpiation tan"'(A — and give a a particular value, we get an intinite number of corresponding values for 0. Thus, if a=l, then ^y = 45", 225', or any angle denoted by / •) *d W *d r = COS •) therefore i-' 1 = 90,-. Sin"' i +COS-' i = 2n-±^^ (2) Now (1) and (2) do not give the same seiies of angles, hence they cannot both be correct. Upon trial it m ill be found that (1) gives only part of the admissible values; that (2) gives tlu? same v.alues together with others not adinissible, tliough both methods give the correct result when the angles have their least positive value. Combining the genei'al values of the angles taken singly, we liave sin-^^ + cos-' J = ?i- + (- 1)" " +2m7r±.^ .-=(2m + n)r±-^ + (-l)"^ = [r..-±2 + (-iyj^* (3) ill if:'' r- TP?^ ■I I4G IWKUSE NOTATION AND SUBSIDIARY ANCLES. I :! I which is tho corroct result HMjuircd. It will \)o, an instructive exercise for the student to search for tiie fallacy in (1) and (2). These methods give correct results when the aiiti functions are restricti^d to their least positive values, and aie frecjuently em- ployed lu such cases. The method (.*^) must be employed when the anti-functions are derived from dillerent ratios, as iu the given example. 151. In the solution of trigonometrical equations care nuist In- taken not to omit any real solution, and to exclude all roots not helonging to the original equation. I'lui following example will bo instructive in this particular : /3 E.v. Solve the ecpiation sirW^ + cos /^-- I , FFUST SOLUTION. Scpiaring given ecjualion \v— . ^'^^TTT' "*' ^^^ TTT* 1 ^ I w 5;: 0=2n.±^^..:ln.±^ (1) (2) INVERSE NOTATION AND SUBSIDIARY ANGLES. 147 TIirRD aOLTTTION. Transpose sin 0, squiirt^ and proceed as ])efore, and we obtain sin 0= ^ - = sni y-, or sin — , therefore 5:: 0^nr + {-ir~,ovnr:+{-\Y~^. (<5) Let us now examine the results of the three solutions and see whether the same series of an<,'les is given in each. Taking posi- tive angles oidy, we have from (1) ^=2 r'" + ( - ^)" '"i ) ^ ^^^' '''''^ ^^''^^ -^^''> ^''^'» "^35°. From (2) = 2n- ± .^ = 1 5^ 345 ', 375^ . . . or = 2nz± - =. 75 ', 2.^5 ', 435'^ .... 1 *j From (3) 5- = nz + {-\)'''~-=. 75'^, 1 05°, 435°, 465° .... = Mr: + ( - 1)" -^ = 15°, 165°, 375°, 525°. . . . or The only angles between 0° and 360° comn)on to the three solutions are 15° and 75°, and oi? testing these they ar(5 found to satisfy (he original equation. Testing the other angles we lind the following results : sin 195° + cos 105°= -sin 15° -cos 15° = V2 2 • sin 255° + cos 255°= - cos 15° - sin 15° = - ^. 2 Hence these angles belong to the equation sin + cos 0— - V — , w !) 148 INVERSE NOTATION AND SUBSIDIAKY ANGLES. Again sin 285^ + cos 285" = - cos 1 5^ + sin 1 5^ sin 345" + cos 315°= -sin 15° + cos 15'. Ifcnco iliese angles belong respectively to the equations sin - cos (I --- ± \2 (5) Similarly it may be shown that the angles 105', ir)5"l)elon{ lown respectively to the equations sin + cos fl — ±^ sjl (6) It will bo found by trial that equations (4), (5) and (G), when solved by tiie given methods, give precisely the same results as the original equation. The roots obtained from an equatioi after squaring do not, in general, all belong to the original equation. (See High School Algebra, Part II., chap. XII.) ^ 152. The preceding article shows that in solving trigonomet- rical functions it is desirable to avoid squaiing whenever possible. We, therefore, give another solution of tlie same equation by a method which avoids introducing roots not belonging to the given equation. '3 Given equation is Now sin + cos cos 45"' = sin 45 ' = 1 V/L> IMultiplying the lenns in succession by IIk'sc (-(juals we get sin cos 4;>"'+ <;os sin 4.)" = -.-, Ji or sin (/^ + 45 ) = sin (10 \ therefore + 45'^ = nr:+ {-\ }" " from wiiich 3 = {n-l)r: + {-\r-. Upon trial it will be found that the series of angles thus obtained, but no others, satisfy the given equation. INVERSE NOTATION AND SUBSIDIARY ANr.I,r:S 149 153. An angle introduced to assist in the solution of an etjualion, or to losoive a given expression into factors, is called a subsidiary angle. A suUsidiaty angle was employed in solving the equation in the preceding article. 154. We now give a few examples of the us(^ of subsidiary angles. Ex. 1. — Holve the e(|uation a cos d + h sin O — c. Aisumo then b sm - = tan sin CM3S [0 - (/)) 1^ V- ^;=cos'' -+tan-' - \/,r + b- « Since the cosine of any angle is always less than unity, the equation will be impossible when (-->(/■ + />'. Ex. 3. — Find the value of ^' for which the expression cos cos C as a complete square. = — ^^ — r cos -^ a + 6 2 = (a + />)- cos"^ will be found in any special case from the tables. In Ex. 3, it was necessary to assume a > b. otherwise cos cA = - would be impossible. In Ex. 4, a (J (a + f>)- is never less than 4a/>, and cos^ < 1, hence the value 2 given to sin is always possible. ENKHCISK, 151 Liiy to EXERCISE XXI. 1. Find the value of siii( - siu"^ ^ j,eos| siii"' A, tanf cot~^ 2. Simplify sin f sin"'- _^ + cos"'-- j, tan ( tan ' 4-tan~^-]. I \/'^ 13. Prove bin ' ^ —cos '— -— =oot"' \/'.\, when rac-li cxpres- sion has its least })ositive value. Examine the effect of removing this restriction. 4. Find the values of sin ( ^y - cos~' | and tan ( /^ + tan~' ^ j . 5. Express the equation sin~'.f = 2 cos"' .f in the ordinary notation. Also tind x and sin'' .c. 6. Prove sin"' x' + cos~' .»; = . . when the angles have their •J least positive value. Illusti-ate i)y a diagram. 7. Show that one value of sin"' — — + cot~' 3 is 45'. Find other values. 8. Prove the following when the angles have their least positive value : 15 5 1 :t (1) 2 tan"' ^ = tan-' ---. (2) tan"' , + tan"' ^. - -. ^ ' D 12 i 4: 1 1 — ^ 1 "^ (3) tan"' „ + 2 tan-' :r = T- (0 ^ot"' , + cot"' _ = '-^. ( o 4 4 / 4 1 0- 9 (5)cos-'^ + 2sin-'-^=-3. (6)eos-' — + cosec ^.,V/41_5. 4' 5 61 11 ,K 9. Prove tan"' . +sin"'-— = tan"^ --, or tan"' =-. Explain 13 45 75 D the two values by a diagram. ,1 ,1 ,1 ,1 " 10. Prove tan' ■+tan"' ^ +tan"' _ +tan-' ^ =m:+ --. 3 5 7 8 4 11. Prove tan ' tt - tan '6 — tan"' a - It 1 + afj '-•■■"*-■-"- ■■ ■ ■ Ill !.')2 INVERSE NOTATION AND SUHSIDIAUV ANGLES. , , ,, . , a- f> , /> — c , c — a 12. f rove tan"' -f tan"' ~ r- +tair ' =0, or n-. 1 + ah I +/ic 1 + ea 1 .'5. Fitul the values of tan (tail"' .'• + cot~' x) and sin (sin" ' .<• + cos~' .*•). Why has the latter two values whilst iho former has but one? 11. Solve the following equations : (1) tan-' 2./; + tan-';U' = 4 (2) 5 sin"' .V ^ COS"' x. (.'J) sin-^ 2,<;-sin-'.»"v/^ = «"^~'-'"- (0 wiii-^^os"' cot 2 tan"' a; = 0. ,v (.5) sin~^a; + sin '".=-. (G) sin' .r + tan 'x'— -. 4: 2 tan A (7) tan 'ffc + tan '/> — tan ' .f. (8) cot '.r + tan ^x—j. (9) tan -' {x-l) + tan ' x + tan"' (,c + 1) -: tan"' '3x. 15. Cliven tan .1 ^a, (\\prc'ss tlu^ equality sin 2^1 — . 1 + tan" A in the inverse notation. 2a 2h IG. Solve sin"^ ; + sin"' = 2 tan"' x. 1 + ir 1 + />- 17. By definition tan (tan~^a) — r/, and tan"' (ian a) = «. Does it follow that tan (tan"' a) ^ tan"' (tan (t) 1 , 18. Prove tan' {(v/2 + 1) tan o\ - tan"' {{y/'l ~ 1) tan O] -tan"' (sin 20). 19. If 2 tan' .f = sin"' 2/y, find the ecjuation I)etween y and x. 20. Solve tan- ' (x + 1) - tan"' (,r - 1) = cot"' (x- - 1). 'Ix 2x 21. Solve sin ^- :, + tan ' 1 + .t- I - X- — t:. 22. Solve the following e(|uations by means of a subsidiary angle : (1) smfl + \/^cosO=--\. (2) \/3 sin /? - cos fl='^2. (3) 3 cos ^/ + 4 sin = 2.5. '(4) 3 cos ic - 8 sin x = 3. (5) 5 sin X' - 1 2 cos u; = 13. (6) 5 cos x + Q sin aj = 8. / EXERCISE. 153 try 23. Express the following in factors by means of a subsidiary angle : I (1) 1 +sin A. (2) \+a cos A. (3) -1'--.^. ^ ' ' ^ M + a cos ^ 24. Use a subsidiary angle to deternnne the sign of the radical A 2 A . A /, ;—- in cos ^^ + sni - = y 1 -f sni A. or m 11 -1 -11 -Sin /^^-cos/' 25. Iracethf^ changes in the numerical value or-. sin cos as chang(!S from 0' to ::. 26. In any triangle prove a = (h - c) sec (f if tan (I — sin - . 27. Explain why ^/+'| =wr + (-l)"'l and -^^ =2?i7r±^, 4 o 4 6 give the same series of angles. 1 1 Iff 28. Given 3 tan~' , +tan~' ^7:+ tan"' =7, find x. 4 20 u." 4 29. Given tan~' r = tan"' 4-tan~' — t, find x. a~ I X tr - x-\-\ 30. Given see"' a + sec"' =sec"'6 + sec~' -, find x. a b 31. Given tan (cot //) = cot (tan 0), find 0. 1 rr 32. If sin (rr cos 0) = cos (- sin 0)y then (f = cos"^ ^ .. ■ + . . Verify this value in the given equation and examine the truth . 1 , 1 • , 3 of the result 6" = ±„ sin"' - . „„ -r, , « cos (/) ,rt-8in , 33. Prove tan"' .^ - tan"* 7- =• 1 - a sin f/» cos 34. Given y = tan - , prove X' = sin J//. U i ' ! I'f-j -fr if';;' CHxVPTEK XI. LOGARITHMS. 155. The logarithm of a number to a given base is the index of the power to which that base must be raised to equal the given number. Thus, if a" — N, then x is the logarithm of iV to the base a, and the preceding equation may be written a; = log,, N. Ex. — Find the logarithm of 81 to base 3, and of 1000 to base 10. Since 81 = 3S and 1000 = 10^, we have logaSl = 4 and logiol000 = 3. 156. The logarithm of 1 is for all bases, and the logarithm of the base itself is 1. For And a"= 1, therefore by definition loga 1 = 0. a' = a, therefore log,, « = 1. In both cases a may have any value whatever ; the proposi- tions are, therefore, universally true. 157. It should be carefully remembered that a logarithm is simply an index, or exponent, detached from the base quantity to which it belongs, and made to stand alone on one side of an equation. Thus the three equations, N=a*^ x = \ogaN, a= \/N, in which JV, x, and a respectively, stand alone, express the same relation of the quantities involved, and consequently any one of ithm )posi- Inn 13 [ntity )£ au I same iiie of LOGAlllTHM.S. 155 the c(jUiition3 may, at any tiiiio, he roj)lac('(l by oitlicr of the other two. Also, replacing x iii the first ('({uatio)! l)y its vjilue from the second equation, we get the important iilcnlity, iV=a^°>fa^' . 158. The suftlx denoting the base is frcciuently omitted. This may be done when it is perfectly clear what base is understood. It is also omitted in stating general properties which are true whatever base may be employed. Examples of the latter are found in the statement of the liOgarithniic Ijaws, Art. 159, and of the former in thdr proof where tlio base a is clearly under- stood. 159. The Exponential Ijaws may \)v. written, I. a' xa" = a'^«. III. (ff-)" =-- a'". From these wo derive the Logarithmic Laws ir. a^^av = n'-y. 1 * IV. {,ty =ay. m I. log vin = log m + log n. 1 1, log = log in - log n. 11 III. lo2 in" = n loji in. IV. log V^'H — log )ll. n These are simply a restatement of the former laws in a dif- ferent notation, as will appear from the following article. 160. To 2)rove the Loyarithmic Laws. Let m = a* n = a" then X = log„ vi, j = log« n. I. log mn = log rt*. a" = log f('+'' = x + y=^ log vi + log n. TT 1 ^"^ "' H. log - = log - = log «''"-- a? - y = log m - log n. a" IIL log vu" - log {a'Y = log (/"' = nx = n log m. IV. log v'm = log («')" = log «" = it' n 1 n JOg 7U. 156 LOGARITHMS. 161. The preceding laws should also bo remenibered in words as follows : 1. Tho loj^aritlun of a product is equal to tlio ouni of the logarithms of its factors. 2. Tlio loga*'ithin of a ((uotiont is Ofjual to tlui logarithm of the divid(!nd, minus the logaiithm of the divisor. 3. The logarithm of a power of a number is etjual to the logarithm of the numlxir multiplied by theind(!.x of the power. 4. Tim logarithm of the root of a number is etjual to the logarithm of the number divided by the indu.v of the root. We may also observe that l)y the use of h)garithms the oper- ations of nniltiplication and division ate replaced by those of addition and subtraction, whilst involution and evolution are « replaced by multiplication and division. 162. The following are applications of the logarithmic laws. No base is expressed, the results being true for all bases. Ex. 1. — Find the logarithm of 75 x 48 in terms of the log- arithms of 2, 3 and 5. log (75 X 48) = log 75 + log 48 = log (52 X 3) + log (2< X 3) = log h- + log 3 + log 2' + log 3 = 2 log 5 + 2 log 3 + 4 log 2. ^ , V 5 X v'20 Ex. 2. — Express log — ,-^=^— " m terms of log 2, log 3 and log 5. v'18v'2 log -~--r^:^ = log C 5 + log V 20 - log V 18 V 2 V18v2 = \ log 5 + \ (2 log 2 4- log 5) - 1 (2 log 3 + I log2) = ilog2-§log3 + Jlog5. LOOARITHMS. 157 arc « log- ancl og2) 163. Logarithms enable us to solve equations in wiiich the unknown quantity occurs as an exponent. Ex. L- We have that is -(jliven a' = hy to find x. log d' ^ log h X log il or a; = log 6 log a This result is true whatever may be the base of the logarithms used. If WQ employ the base d, we have at once by definition, x — \og^h, which agrees with the given result, since log„a= 1. Ex. 5.— Given w sim])le examples illustrat- ing the nature of such changes before investigating the theory in its most general form. Ex. i.— Since 64 = 2« = (2-)3 = 4' therefore log.^ G4 = 6, and log4 04 = 3. From which we observe that when the base is scjuared the logarithm nmst be divided by 2. 7 7 Ex. ^.— Since 2187 = 3^ = (37' = 27^. therefore loga 2187 = 7, and log,; 2187 = - This example shows that when the base is cubed the corre- sponding logarithm is obtained by dividing the former logarithm by 3. ti r^ ■IVI 158 LOGAUITHMS. Ex. 3.~ Let iV'= a% and let «" = A, then X = loga -^^ '^"^ 2/ = l^oa ^• Then tluMcfore lo^r,, X =' = 165. 7V> y/?i")(•=, log | \' rr '\///' :- \//'''. y//^|, in tenus of log , and log r. 8. Simplify log -^ - 2 log y + log — . *' " '"/ 1 o — ^ 9. Simplify log ~ ~ "-, and log ' 729 ^' 9-^ 27-*. V 24 v' 40 ' Ni 10. Given 8 (2^-')''= 4'+', find .r. 4"^ 1 :t 11. Given -— -, = ~ , 8 v '2 == 4", find x and y. 12 Solve equations (1 ) «"" = h. (2) a'. // - c. (3) «*+'. })"-' = c'. d\ ^\) 2' = 800. 4 (5) {a + hf («2 _ 6y-» ^ {a ~ hf\ (0) / = //. 13. Find two consecutive integers between which the values 01 the following logarithms lie, log3 95, log5 175, log,. 10, log, 2, log, ^ log^o .0004. < i Hv ■ !i I i r % M IGO .OGARITHMS. It. Given that N is an integer, and that logs -^^ > 2, and that log, W" < 3 ; find JV. 15. Find the logarithms of the numbers in Ex. 2 to the base 0, and compare tliem with tlie logs to the base 3. 10. If log,. N=x, find log A' to the base a'\ 17. If log N" to the base a" is x, of what number is x the logarithm to the base a 1 --^8. From the identity N=n^''^" '\ show that a'"s^ = 6'"^". 19. If loga iV= h, and log,, N—a, sliow that Is this o(juation true without the given condition ? 20. Simplify a'^a^ x a^oj?", Va^'^'^-, «io«' ^ ^ a''''« -. 21. Show that if a series of numbeis are in (/ 1\ their logs are in A P. 22. Sum the series log a + log or + log «r'- + . . . . to ?i terms. / 23. Prove log,, m"". log,, ii^ = log„»/i". log,, 7b = log,, n'. log,, w". 24. Prove log,, m'. log,, n". log,, r^ = log,, r". log,, v^^'log^ n* = log,, mMog, ?iMog„ r*. 25. Eliminate x from the equations «^ — ?/«, i' = n. W) 26. If X = log,, wi = log,, »A, then .r = log„,, mn = log,, - . 6 n 27. If .»' = log,, 7u = logft n — log,. ;>, then x = log m^ifp* to the base ^/'6V. 28. Prove 2 log sin yl = log (1 -t- cos A) + log (1 - cos A). 29. Prove log sin 2 A = log 2 + log sin A + log cos A. 30. Prove log cos 2 A ^ log (cos A +&mA) + log (cos A - sin A). 31. Prove log sin A - log cos ^ = log tan A. 32. Find the value of log tan 45°, and log (sec A + tan A) + log (sec A - tan A). 33. Simplify log tanyl - log ( 1 + tan yl) - log (1 - tan J) -i- log 2. \ M u DIFFERENT SYSTEMS OF LOGAUITHMS. IGI \ Different Systems of Logarithms. 166. Any i»ositive nunil)or difleieiit tVoiii unity may ho chosen as the base of a system of logarithms, hut only two ciliHerent systems are in conmion use. The lirst system, called Napierian logarithms, from their discoverer, Uaron iS'aj)it;r, has fur its base the sum of the series 1 _i_ _1_ 1 "^ 1 ■^1.2"*"l.2.3"*"l.2.3.-i ad. inf. This number is incommensurable, and consequently it can be only approximately expressed. Its value to eight decimal places is 2.718281828, and is usually denoted by e. They are also called natural logarithms, because logarithms to that base are the most easily calculated, '.'hey are always used in theo- retical investigations. The second system, called common logarithms, has for its base the number 10, the radix of our system ci notation, and is always used for practical work. 1 he great advantage of this choice of base will be easily perceived from Art. 171. r*. A\ A). U 2- 167. Writing the successive powers of 10, both i)ositivo and negative, 10 =- 1 .*. log 1=0. 10'= 10, .-. log 10-1. 10-1 = .1, ... log.l =-1. 10-i- 100, .-. log 100=2. iO -■•'-. 01, .-. h)g.Ol - - 2. io-'=ieoo. .-. log 1000-3. 10-^-. 001, .-. log .001- - ;i. From the above we observe : 1. The logarithms of numbers greater than unity aie j)Obiiive. 2. The lo^^arithms of numbers less than unity are iu:yalivi\ 3. Negative numbers have no icnl logarithms. 4. The logarithms of any num' r ix'tweeu two consecutive powers of 10 will be between the consecutive integers denoting those powers, i.e., it will be a whole number and a fraction (or decimal). ■^= r 162 LO(JAIllTUMS. It will 1)0 readily porceivod that whatever base may ho chosen, hub comparatively few nuinbei's will have exact logarithms. Thus iu the common system only the numbers 10, 100, 1000, etc , have exact logarithms. Tiie logarithms of all other numbers are incommensuiable. For example, tho common logarithm of G7 is the value of iv, for which lO-' — G7, but tiiis value can bo only ap- proximately obtained. Its value to 5 decimal places is 1.82G07, which means that 10"*-fi^^ = G7, or that lO'^-'^'^ C7"^*^ approxi- mately. To v(!rify this equality by actual multiplication would take more time than the student is likely to have at his disposal. 168. In dealing with negative logarithms it is most convenient in practice to express them in a form in which the integral part alone is negative. In such cases the negative sign is written over the integi-al part ; thus 2.75812 means - 2 -{-.75842. 169. When a h)garithm is expressed with its decimal part positive the integral part is called the characteristic, and the decimal part tho mantiSSa. Ex. 1. — The number 275 is between 10- and 10\ therefore log 275 is between 2 and 3. The value is found to bo 2.43933, in which 2 is the characteristic and .43933 is tlie mantissa. Ex. 2. — The number .0275 is between 10"- and 10~', therefore its log is between - 2 and - 1 ; its value is found to l)C -(1.5G0G7). In this case, however, the characteristic is not - 1, nor is the mantissa .5G0G7, for the decimal part is iwyatice. Transferring tho log to the proper form, we have - 1.5G0G7 - - 1 - .5G0G7 = - 2 -f- 1 - .5G0G7 = 2.43933, from which we see that - 2 is the characteristic and .43933 the mantissa. '" DIFFERENT SYSTEMS OF LOGAUITHMS. 163 \\ 170. We give a few simple examples to show the proper metliud of dealing with logarithms having negative characteristics: 0) 3.1468? ^.56347 AM335 (2) 1.24G38 5 4.23190 (3) 3 I 2.48250 1.49419 In Ex. (1) the subtraction is porfornud in the ordinary way until we reach the negative characteristic 2. There is I to carry from the preceding column whiih added to 2 makes 1, and this subtracted algebraically from 3 gives 4. In Ex. (2), multiplying in the ordinary way there is 1 to carry from the decimal to the characteristic ; this added to 5 times 1 gives 4. In Ex. (^J) we add 1 to the characteristic, and + 1 to the decimal; this does not alter the value of the given logarithm, and it renders the integral part exactly divisible by the given divisor. 171. To find the characteristic of the loijaritlnn of any number greater than nnitij. Let N denote the given number, and let it contain n digits in the integral part. Since 10""' is the smallest nunjber containing n digits, therefore 10'-^ N, 10", are in ascending order of magi>''tude, and consequently log X lies > between n - 1 and n. Therefore log N= (n - 1) + a decimal, or ?t - 1 is the characteristic of log N. Tlie characteristic of the logarithm of a number greater than unity is positive, and less by unity than the nauiber of digits in the integnd part. 1C4 LOr.ARlTIIMS. 172. To Jind the characteristic of the loyarithm of a decimal fraction. Let iV denote the given number, and let n denote the number of zeros between the decimal part and the tirt;t significant tigure. Since lO"'"*"'* is the least nund)er containijig only n zeros between tht; decimal part and the first significant figure, tiierefore 10-<"+".V, 10-'' are in ascending order of magnitude, and consequently log N lies betweeen — {n+\) and - tt. Therefore log A"= - (n + 1 ) + a decimal, or - (?4+ 1) is the characteristic of log N. The characteristic of the Int/arithm of a dcciinal fraction is ne(/ative^ atul numerically f/reatrr hy unity than, the iiuiuber of zeros beticeen the decimal part and the first siynificaut fiyure. 173. If two n^tmhers differ ouly in the posilioti of the decimal point, their loyarithms differ only in their characteristics. Let iV and iV" denote two such numbers of which K is the greatci', and let r denote the diU'erence in the nund>er of their decimal places, then and I^=N'xhy 10'- logiV^=logiV^' + log 10'' - log iV + r. Art. IGl, Now r is an integei*, and consequently its addition to log N' will change only the characteristic. Examples. — The characteristics of the logarithms of 37, 5, 87025,43 are 1, 0, 4 respectively, the;se numbers being each less by a unit than the number of digits in the integral part of the corresponding number. , DIFFERENT SYSTEMS OF I.OGAUITHMS. 165 01. N ri The cliaracteristies of the logarithms of .00, .:ir)7, .OOO.'M? are — 2, -1, — I icsjx'ftivoly, these nuiiiliers Itcinij each iK'gative and numeriually greater l»y a unit tlian the number of zeros he- tweeii the decimal point and the tirst signilicant ligure. Given then log 37S--')70 -^ O.'tTTTSTT log ;57.)^*2r)70 ^ l.r)777877 log ,00."»7fSJ.")7G -^- 3.r)777)S77, etc. The advantag(>s of the common system of logarithms arc now evident They are : 1. The charactcn'istic can li(> written at once by inspection, and conse(juentiy need not \)v legisteicd in the tables. 2. The mantissj'o are the same for all numbers consisting of the same digits in the same order, and consequently the niantisste of integers only need be given. Tlie ex})lanation of the method by which tables of logarithms are constructed does not lie within the scope of this woi'k, but the student may be informed that logarithms aie first calculated to the base e, and then transformed into common logarithms by the principle of Art. 105. 1 Thus log,, y 'Of so that when a table of Napierian logarithms has been formed it may be changed into a table of common logarithms by multiply- ing each by the constant factor, 1 1 log,, 10 2.^0258509 . .43429448. 'Oi which is called the modulus of the connnon system of logarithms- 174. In the Appendix is given a table of the logarithms of all integral numbers from 100 to 1000. As just explained, only the mantis«a* are registeicd, and since these ar<^ all decimals, the decimal point is omitted. Also when only the last three figures u. 106 LOGARITHMS. of tlio logarithm are given, tlie tirst two are to be supplied from the number next preceding which has the logarithm giviui in full. From the tal)le given, the logarithm of any number consisting of not more than five digits can be obtained by metliods which we shall now explain by means of examples. We again assume the principle of proportional parts, Art. 77. Ec /.—Find log SSf), log .0385, and log .'^8500. In the tables opposite 385 we find 5851(), which is the man tissa for each of the numbers, since they all consist of the same digits. Their characteristics are 2, - '2, and i respectively, llenco log 385 = 2.58546 log .0385 = 2.58546 log 38500 = 4.58546. Ex. ;.^— Find log 28763. From the tables we have log 28800 = 4 45939 log 28700 = 4^5788 from v/hich difl'erenoe for 100= 151 therefore dillbience for 63= 151 x .63 95.13. Then to add therefore W 28700 = 4.45788 difFerence for 63 = 95 loir 28763 = 4.45883. £x, o. — Find the number whose logarithm is 2.34698. From the tables we find, ' log 223 = 34830 log 222 = 34635 diff. forl= 195 given log = 34698 log 222 = 34635 ffiven diff. = 63 63 Then y^ = .32, which added to 222 gives 222.32, the numbet required. ^l.~..t_BJ^ii£;L'l. _ r'>. .. DIFFERENT SYSTEMS OF LOGAIllTFIMS. 107 E.r))lnnation. — Tii dealing with the mantissa wo take no notice of the chsiracteristic. Since the same number of decimal phaces is employed in each case, the decimal point is omitted. And since an increase of 105 in the logarithm •'ives an increase of 1 in the number, an increase of G3 in the logarithm gives an in- crease of —-T. •^'~ iii the number. Finally, we observe that the given characteristic is 2, a!id conse(iu(;ntly there must be three significant figures in the integral pait of the numlxM". Had the given characteris-tic l)e(m dirtercnt the position of tlu! decimal point in the final lesult would have been difl'erent ; otherwise the work woi'ld have been exactly the same. For example, 2.3 1G98 is the logarithm of ,022232. The mantissa 34698 depends upoii the se(|uence of digits 22232 ; the charac- teristic, 2, gives the position of the decimal point. 175. Logarithms are extensively used for simplifying expres- sions requiring tedious multiplications, divisions, or the extraction of roots. The general plan is to find the logarithm nf the given expression, and then find the number corresponding to this logarithm. .0084321 X (,'-,)& V/8.37 log 84400 -92()31 difi: for 21- iVo of 51 log 84300 = 92583 -- 1 1 nearly. difi: for 100= 51. • Then to log 84300 - 92583 • add dirt", for 21 = 11 therefor(! lo.' .0084321 =3.92594. iog (i^)4 =i (log 2 -log 15) = 1.70831. log V8.37 = \ log 8.37 = .4G13G. ~rwim III t ( I 1G8 LOGAKITHMS. Therefore log of given fraction -3.!)2r)0i + i.708:n = 3.17:289 .40136 Then log 1 19 - I 731 9 given log .- 172S9 log 148=170l'6 293 log lt8==1702r> 293 ) 20300 ( S9 2344 2860 2637 Then-fore 3.17289 = log .0014889. The given fraction = .0014889, which is tlu^ result re(juired. EXERCISE XXIII. 1. Add 2.07895, 3.67893, n.3178,5, .89207. 2. From 1.07638 take 4.25763, and from 3.48273 take 1.38405. 3. From .26875 take 2.39607, and from the result take 1 82753. 4. Multiply 2.37654 by - 5, and 3.20763 by 4. 5. Divide "2.34687 by 3, and by -J. 6. Divide 1.34067 by 2.56834, and by 2.68235. 7. Find by inspection the characteristics of the logarithms of 1 1827.54, 4.07, .0003, 125' .3765, ''82763. 8. The mantissa of log 8576 is .9332848, write down the logs of 8.576, 857600, .008576. 9. From the precedijig example write down the numbers whose logs are 2.9332848, 2.9332848, 7.9332848. EXERCISE. 169 10. How many positive integers are there whose logarithms to base 3 have 5 for ciiaracteristic 1 11, Find from the tables the logarithms of the following: (1) 28507. (2) 3.8762. (3) .0623. (4) .0075. (5) (1.05/. (6) V3T4Tb\ (7) (1.06)-J. (8) (.002)-3. 13. Find from the tables the numbers of which the following are the logarithms : (1)1.11613. (2)1.21650. (3)4.18037. (4)1.00042. 13. Writedown the logarithms of the numbers in Fx. 8 to the l)ase 100 and to the base .01. 14. How many digits are there in the integral part of (1.05)^^*^? < 15. Given that the integral part of (3.456)'«'»«^ contains 53856 digits, find log 345.6 correct to five places of decimals. 16. Given log 2^.3010300, find the logarithms of 8, 1;, 5, 800, v"2, .03125, -L, v'2 x v' 5. 17. Given log 3 = .4771213, find the logarithms of 30, \/dO, .3, 3, .01 v^24.3, (.081)-3. 18. From the known values of log 2 and log 3, find the loga- rithms of 2?, 1.44, 1.85i, (12)20^ J3*^^ Ni 750 19. From log 2, log 3, and log 7 =8450980, find the logs of ^'28 X v*'7.5, V.56^ v'.0049, (.0126)'' x (3.43)-i^(.044i)*. 20. Simplify lll^ X i^^-^ X ^ , given log 15193.64 = 4.1816618. 1.25 .81 ^5/1.2"' 21. Which is the greater, (1 .\)'\ or (H)'", given log 2 and log 3? 22. Find by logarithms the cube root of 7. 18 4 ■IBI r 1 1 1 t ^ 170 LOGARITHMS. 23. Kiiul l.y logarithms tho fifth root of 27.65. 24. Simplify v/^O x v 2.7 x {/ - 5 x 18 i. 25. Solvo tho following cciuations to three places of decimals : (1 ) 5^ .= 300. (2) 8' = 5 (3)'' 2. (3) i\Y = b\\. (i) 2^' X 3-'^ - 4.9. (5) 15^ - (2.5) '-\ (6) 3^ - 1000. 26 Tn how many years will any sum of money double itself at 3A per cent. 1 27. In how many years will any sum of money amount to 10 times as much at per cent, as at 5 per cent. ? 28. Find the edg«5 of a cube whose volume equals that of the earth, supposing the latter to be a sphere whose diameter is 7912 miles. 29. Solve equation {lY (125)'-Xr {\f"'^' iX)' to three decimal places. ' 30. Given 1' = ly, 3*= lOy, find .*; and y. 3' 31. Given .^ =4, 7^ ^3", lind ;<• and y. 32. Given log 1.44 = .1583025, log 10.2=1.2095150, find log 2 and log 3. 33. Find logy 270 and log^ 10 from the known values of log 3 and log 2. 34. (liven logg 2 = .693147, log^ 5= 1.609438, tind the common log of 6.4. 35. Find log^ 7 from tho known values of log,,, 7, log^ 2, loge 5, and thence tind log.j 7. 36. Given log, 3 = 1.098612, log, 6 = 1.791759, find log2 12. 37. Find approximately the value of 5 ^^ .,^, „. , ,, , „ .2 X .4 X .8 .... to 12 factors 38. Ynm the value or .3 X , 9 X 2.7 ... . to 9 factors * dHH CHAPTER XII. SOLUTION OP TRIANGLES BY LOGARITHMS. 176. In Chapter V. wo tlccluccd tli«! most important rolations between the sides and angles of a triiingle, and gave examples of their application. \Ve shall now show how logarithms are em- ployed to abbreviate tlie labor when numbers consisting of several digits aie involved. 177. The sines, cosines and tangents of all angles between 0° and 45^ are each less than unity and conse(|uently their loga- rithms fire negative. In practice it is found convenient to render all such logarithms positive by adding 10 to each. The loga rithms of the trigonometrical ratios thus increased by 10 are called Tabular Logarithujs, and are distinguished from true logarithms by the letter L. Thus, L sin 15^ means log sin 15°+ 10. The logarithmic cotangents being already positive, the 10 may be omitted and their true value given. 178. For convenience of reference we have given in the Appendix the tabular logarithms of the ratios most fre sill A. L SCO y|-20- L cos J L cot A = 20- L tail J. ^.f. i— Find tlu! value of L sin 2')' 13' 20". From tlio tables \\(! find : L sin 25'^ 20' = 9.031 .'i'i Then to L sin 2.r 10' --- 9.02805 A sin 25'^ 10' =9.02S(;.5 add diH. for 3' 20"= 89 Ditr. for 10' 208 Kcsult =9.62954. Ditf. for 3' 20"= 2G.8x3.^JJ 89. Ex. ^.— Find the value of L cos 22^ 37' 15". From the tables we find : L cos 22" 30' = 9.90502 Then from cos 22'' 30' = 9.90502 />cos22M0' =9.90509 take d iff. for 7' •45"== 41 Diff. for 10' = 53 Result Diff. for 7' 45"= 5.3 x 7|§ 41. = 9.96521 Ex. 3. — Find the angle whose logarithmic sine is 9.80537. From the tables we find : L sin 39'^ 50' = 9.80056 Given log = 9.80537 L sin 39^ 40' = 9.80504 L sin 39 ' 40' = 9.80504 Giv(ui diflf. = Diir. for 10'- 152. 33. Then f^^ of 10' = 2' 10" ; and 39^^ 40' + 2' 10" = 39° ^'^' 10" is the angle required. )fract- 280.'5 89 2954. 41 t)52l" 0"is RI(JHT-AN(JI,EI) Till ANGLES. 1 7'*^ Ex. 4. — Find tlio angle whoso logaiitlmiio cosine is 9.87900. From tho tables we tind : L cos tO^ 40' = 9.8799G L cos 40'' 40' = 9.87906 L cos 40' no' = 9.87887 Given log =9.87900 Diff. for 10'-= 109. Given diir. = ^. Then X.> "f 10' = 8' 48", and 40' 40' + 8' 48" = 40" 48' 48" is the angle reijuired. 180. When an angle is to ho found from its logarithmic sine, should the given value he greatir than L sin 45 ', we must find the angle whose cosine has the given value. The dillerence between this angh; and 90.' will evidently hi; tln^ angle nMjuired. An example is found in Ait. 194, where /i is to he found from L sin 7^ = 9.98550. We tind 9.98550 = /> cos 14' 41' :U", then 90^-14° 41' 3l" = 75^ 18' 29", the angle required. The same method may evidently he pursued with regard to any two com- plementary ratios. 181. When a given logarithmic tang(Mit is greater than L tan 45°, !.«., greater than 10, the method of the last article may be employed, but it is more convenient to reason thus : L tan (90^ -A) = L cot J = 20 - Z tan A. The value of L tan (90^- A) will be found in the tables from which 90- J is found, and then -^1 is known. This method is employed in Art. 18G, to find /> fiom its logarithmic tangent, which is greater than 10, and conse(|uently is not found in the table of tangents. The same im^thod evidently applies to any ratio and its reciprocal. Bight-Angled Triangles. 182. A triangle contains six elements, three sides and three angles. When three elements are given, one of them being a side, the others can ususilly be found. We shall begin with PS*,-**!-- hi I I t sin ^1 = 9.42376. (3) /> sin J =9.97028. (2) L cos A .-= 9.85623. ('1; /.tan ^ = 10.34598. (5) /. sec .4 = 10.07293. (G) L cot A = 10.18942. 3. Given Z tan 32° 32' = 9.8047447, diff. for l' = 2786, find /. tan 32^ 32' 32", and L cot 57^^ 27' 14". 4. Given /. cos 21^ 21' = 9.9691241, diff. for l' = 495, tind Z cos 21° 21' 21", and L sin 68^ 38' 25". 5. From Ex. 3, find tne angles whose L tan and L cot are 9.8047983 and 9.8048235 respectively. 6. From Ex. 4, find the angle wl^ose L sin antl L cos aio 9.9691037 and 9.9690884 respectively. 7. Find from the tables the value of L tan 38^ 25' 20", and verify the result by finding the logarithm of tlie natural tangent. 8. Prove that L tan A = L sin A - L cos A + 10, 9. Given L tan 15° 20'= 9.4380587, diff. fori' =4951, L cos 15° 20' = 9.9842589, diff. for 1' = 347 ; find L sin 15° 20' 35", L sec 15° 20' 41", L sin 74° 39' 18". 10. Given Z sin 10° 15' = 9.2502822, diff for l' = 6981, find the angle whose L cos is 9.2503940. 11. Show how to find the value of L sin 2A from the known values of L sin A and L cos A. Could the value of L cos 2il be found in the same way? 12. Siiow from an examination of the tabular logarithms that the sines and tangents of very small angles are proportional to the angles themselves. EXEIULSE. 177 aic lat to 13. Solve the triangle A/IC when tlie following parts are given, C denoting a right angle : (1) c = 432, A = IS' W. (3) c = 957.34, b .-= 240. (5) J -35° 15", a = 86.34, (7) « = 75.384, 6-14.82G. (2) c = 1234, iy.= 25' 19' 13". (4) e=598.2, 6 = 501.8. (r))J = 72^ 15' 20", f, = 365. (8)^3=391.4, /> = 89.62. 14. In a triangle right-angled at C, « = 250, 6 — 753 ; find the perpendicular from C on the opposite side. 15. In a right-angled triangle, the acute angles are propor- tional to 2 and 3 and the area is a quarter of an acre ; tind the sides in rods. 16. At 120 feet from the foot of a steeple the angle of eleva- fiou !,'^ the top was found to be 60^ 30'; tind its height. 17. From the top of a perpendicular rock 326 feet high the angle of depression of a i)oint on a horizontal plane below is 24°; lind the distance of the point from the bottom of the rock. 18. Two oljservers on the same side of a ])alloon, in the same vertical plane with it, and a mile apart, tind its angles of eleva- tion to be 15° and 65° 30' ; tind the height of the balloon. 19. From the bank of a river the top of a tower on the opposite bank is at an elevation of 54°; 35 feet from the bank its elevation is 49'; tind the breadth of the river, the points of observation and the tower being in a line perpendicular to the river. 20. The angle of elevation of a hill from a point due north of it is 53° 18' 27", and from another point due west of the former and distant from it 430.31 feet, the elevation is 49° 17' 18"; find the height of the hill. W' ;!l fi :i Z^SSf* 178 SOLUTION OF TUIANOLES IJV LOGAKITHMS. Oblique-Angled Triangles- IB?. We now give the solution of oblique-angled triangles. The methods einployed are applicable to all triangles whatsoever, but thos«; already given are simpler when one of the angles is known to be a right angle. There are four distinct cases which we shall discuss in order. M H . If • * II 188. Case I. — Uiven the three sides, n, ft^c. Euc. 1., 8. Fron. Art. 83 (3) tan ^ = |(-^-_"'^^.. Then L tan = - {log (.s- - h) + log (s - c) - log s - log(s - a)} + 10. A it The value of L tan being thus known, - may be found from B C the tables. Similarly and ^^ may be found. Having thus found the three angles, the accuracy of the work may be tested by adding them and comparing their sum with the proper value, 180°. £a:.— Given a = 372r), />.-4873, r = G258, (ind A, B, C. In this case s = 7428, .s - a = 3703, .v -h = 2555, s - c = 1 1 70. Then L tan ^ = ^ {log 25a5 + log 11 70 -log 7428 -log 3703} = ^ {3.40739 + 3.0G819 - 3.87087 - 3.56855} + 10 fr om w hich = 9.51808, ' ^IS'' 14' 45'. oryl=::3G°29' 30" OBLIQUE-ANGLED TRIANGLES. 179 B 1 Similarly Z tan — = -- {log (.v - c) + log [s- a) - log s - log (s -h)] + 10 = f [3.00819 + 3.50855 - 3.87087 - 3.40739[ = 9.07924, + 10 B from which ^ =25^ 32' 17", or ^ = 51" 4' 34". C 1 Again L tan ^-=-{3.50855 + 3.40739 - 3.87087 - 3.00819} + 10 = 10.01844 Therefore L tan ( 90" - -I- ) = 20 - 10.01844 (00"-f)=2( = 9.98150, Art. 181. from which C 90" - =43° 47' 2", or C=92^ 25' 50". 10 The sum of the three angles thus found is 180^, showing the work to be correct. 189. In the preceding article we might also employ the formula for sin — , or cos - , and if we were required to find but one angle either of these would serve our purpose equally weli. But the formula for tan requires only four logarithms to and all the angles, whilst either of the others requires six. The formula for cot —, or for sin A^ might also be employed, though the latter would require more logarithms. By using a table of natural cosines the formula for cos A^ Art. 82 (2), might be employed ; but since it does not consist of factors, it is not adapted for use with logarithu)S. m ^1 180 SOLUTION OF TRIANGLES HY LOGARITHMS. 190. Case //.—Given one side !ind two angles, <\ A, li, Euc. I., 26. Since A + Ji + C'= 180'', we have at onco C= 180° - (^ + B). Then from Art. 80, "* '' "" from which Then sin A sin JJ sin C* c sin A , c sin B a = , b = - sin C Gin C log a = log c+ fj sin A - L sin <7 log I) — log c + 7/ sin li - L sin C. From these equations c and i n)ay be found. Ex.-Qiven c = 338.G5, ^=53'^ 21', B = \SCf 27'; find C and a. We have C= 180" - (53" 24' + 6G° 27') = 60^ 9', Then log a -log 338.05 + L sin 53=^ 24' - / sin 60° 9', =- 2.52975 + 9.904G2 - 9.93819, 1^2.49618, from which a = 313.46. SC 191. Case ///.—Given two sides and the included angle, a, 6, C. Euc. 1., 4. We have yl + /y C , A + li. ^ — — - = DO — — so that IS known. •) '> '>. W W Ml From Art. 85, tan — -— = — cot A - // a - b a + f> C -> ' Then L tan — - — — log {(t. - l>) - log { = 982, ,^ ^ 1 7'^ 31)', A li Then Z tan ,^ log 40 4 - log 982 + L cot 1 7^ 39' ^ 2.()()0r)2 - 2.9921 1 + 10. 19733 =-10.17174 from which '-!— -^' := 50 ' 2' 40" and therefore Then or 2 •> J:- 128' 23' 40", yy-lO 18' 20". h sin C c = 'sin I'r from which log -log h + /. An C - L sin 11 - log 259 + L sin 3.^) ' 18' - L sin 10^ 18' 20" = 2. 1 1 330 -V 9. 70 182 - 9.44834 = 2.72078, 6- ---533.00. 192. The formula used in the preceding article was proved geometrically in Art. i>'^^ ; we now gi\e a symholical proof of it and two similar formuljo which are sometimes useful. From Then a a sin A - — -, we get , = -. sin A sin // ° 6 sin />'' (t. - h sin J - sin li 2 cos \ {A + />') sin \ {A - B) (t +1) sin J + sin />' 2 sin i {A + JJ) cos h {A - B) _ tan h{A ~ B) _ tan .1 (A - B) ~ tan 1(71 +//) " cot'^l 6' ' or A- n a-h C tan — — - = J cot - - 2 a + 6 2 (1) i!:; ;: f iwrii 182 SOLUTION OK TIIIAN'OI.KS BY I.O(JAHITHMS. I ■ i i ( Ai,'fiiii from -r a (' c « + A sill Ha sin li sin C* " '^ sin C sin A +sin Z^ Then e sin r L' sin .", (' cos .1 C sin \ C a + h sin J + sin li 2 sin J, (/I + /y)cos }, {A - li) cos \ {A - li)' {a +h) sin .1 C t'loiii which ( 8iniihirly we, can showc = cos ^ (yl - II) (ft - h) cos \ C ~sur['{A~^''ji) ' (2) (3) W(; nniy observo that from any two of tljosi? three formula* the remaining one may he ohtained. We shall use (li) to lind c in the example of the prticetling article. We have fiom which log c = log (a + h) + L sin I C L cos | {A - Ji) = log 982 + L sin 17" ;J9' - L cos 5G ' 2' 40" -2.99211 + 9.48173-9.71700 = 2.72678, c = 533.06 as before. This method has the advantage of requiring only two new logarithms, whilst the former recjuired three. 193. When two sides and the included angle are given, th(! third side is given at once by the formula c^ = rt" + 6" - '2(ib cos C, Art. 82 (1) which may be adapted to logarithmic calculation by using tin; result of Ex. 4, Art. 154. Thus, c" = a^ + Ir - 2af> cos C 2\/ah C — {a + h)- COS" />, if sin (f= y cos a + /> o' or c^{a + b) cos 0, i '•^nsmmmmi i »j» » OI}l,Uii;E-AN(JI,KI) tuian<;les. 183 (■5) Wo shall now iMiiploy this iiu'thod to solve the fxaiuplo in Art. 191. 2 \/nl, C Wohav(! ■iXii — , cos . u ^- h 2 C TluJli L sill a -^ log '1 + \ (log a -f log h) log (/< + /') + // cos ^^ =-. log 2 + \ (log723 + log-J51)) log 9SL' i- Acos 17 ' \Vy -!).!! 21 -JO or ^y = r)7 7'2G". Then log c -= log (n + h) + A cos //-!() - log <)S2 + L cos 57 ' 7' 20" - 10 = 2.72078 and c' = 533.00 as before. the. 194. C(tm /r.— Given two sides hikI the angle opposite one of them, ft, /^, A. luic. VI., 7. From th(! Sine Rule, sin /)'— , therefore L sin Ji - log h f L sin yl - log ff, from this ecjuation VV may be found. Then then Art. 80. or C= ISO'- (yl +y>), which gives C, h sin C sni /) log c = log />» + A sin ( ' - A sin />, from which <- may be found. J£;f.— Given a-- 379.5, A = 504.8, A = 10' 32' 10"; find A', C, r. 7. sin /j'- log 504.8 + A sin 40^ 32' 10" - log 379.5 = 2.75189 + 9.81288-2.57921 = 9.98550, from which H = 75 M 8' 29", or 104 41' 3 1 ". 1.S4 I SOLUTION OK 'li{lAN(il,KS I!Y l,0(JAKITIIMS. tl i , '■ ! 1 1 : ' Then C= 180^ - (A + //) = W 9' l.V, or :\ I ' 10' I.T, also log c - log 50 l.S + A sill G I ' 9' l-")" - L sin 7;) 18' 29" = 2.751 89 + 9.95122 - 9.9855G = 2.72055, thcreforo f = 525.48. Taking tin' otluM" v!ilu", that is to 2a cos Ji when />' has the acute value. This gives a test of the accuracy of the work. Thus a cos n = 379.5 x .253G2 = 9G.2498. Half the difference of the values of c = 96.25. This proves the work sufficiently accurate. 196. In the solution of problems in which the trigonometrical tables are employed we should be careful to select the method which is most appropriate. An. examination of the tables will show that for tlie natural functions when the angle is small, the sine and the tangent are the most appropriate. For the differ- ences in the cosines are too small and those of the cotangent are irregular, so that in both cases the principle of proportional parts fails. The reverse is, of course, true, when the angle is nearly equal to a right angle. For the logarithmic functions the principle of proportional parts fails for both extremities of ir IS' 'ju" I as tlio •itios are 3 student will be r is equal e. This ^metrical method Ihles will |uall, the he diffei- Igent are ortional laiigle is kmctions lities of EXKUCISE. 185 the quadrant, though it holds good throughout the central part. Wo, therefore, seek for methods of solution not involving very suiall angles nor those nearly ecjual to a right angle. For example, when tiie three sides of a triangle are given the angles may be found from one of the half angle formula*, or from the sine of the whole angle. If a particular angle is known to be small wo, should choose the latter method, but if it is very near a right angle we should choose the former. Again, suppose we desire to tind an angle from its cosine which is small so that the angle is nearly equal to a right angle. Then since cos A=n we have tan A |1 - cos il I I - w and this formula is free from objections. Such transformations, however, are rarely necessary. In practical work, measurements of length containing four significant figures, and angular measure- ment to the nearest minute, are considered very accurate. The results obtained from using the tables given in the ordinary way will more than satisfy these conditions. EXERCISE XXV. 1. The sides of a triangle are 350, 400, 450; find all the angles. 2. The sides of a triangle are 1263, 1359, 1468; find all the angles. 3. The sides of a triangle are 52.317, 24.659, 47.932; find the greatest angle. 4. Given ^ = 31° 13', £ 48° 24' 15", rt = 926.7; solve the triangle. 5. Given 6 -=7235, c = 1592, ^ =50°; solve the triangle. 6. Given if = 25^ 37', C= 15^ 23', c= 259 ; solve the triangle. 7. Given 6 = 354, c = 426, .1=49° 16'; solve the triangle. 13 \ f f • j»y*«r"^ IMAGE EVALUATION TEST TARGET (MT-3) // 1.0 I.I 1.25 2.5 2.2 ^ 1^ 12.0 ills 14 IIIIII.6 v] <^ % /a ^i >> y /^ i^ m Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 x^-Vn V \ \ % :^>. ' ^^ ^ <^ 1. 4is m?.. 18G SOLUTION OF TIUANGLES BY LOGARITHMS. Ill i 1 f! 1 ' J { r- 1 1 I ! If ■ ■I 8. Given a =156.5, c=:- 53.94, />'-15° 13' U"] solve the triangle. 9. Given (/ = 325, /;=.333, yl==52° 19'; solve the triangle. 10. Given c'=1249.6, a-397.3, A=S° 19' 35"; solve the triangle. 11. Given c = 432, a^l35.17, ^-18° 14'; solve the triangle. 12. Given 7^-37° 20', i'> = 4570, c = 75G3 ; solve the triangle. 13. Given C-152° 54' 20" c=1249.6, a = 397.3; solve the triangle. 14. Given 6-9268, c-6951, A^IG" 15' 38"; find 5 and C, from log 7 = .845098, L cot 8° 7' 49"= 10.8450980. 15. Given a =197, -6= 250, c = 448; find the angles. 16. Given J = 125° 30', b = 750, c = 250 ; find a without find- ing B or C. 17. In Ex. 16. find the segments of a formed by drawing a perpendicular from A. 18. Given a = 123.5, 6=167.38, c=250; find the area of the triangle. 19. T!ie sides of a triangle are in the ratio of 2:3, and the n/ contained angle is 60° ; find the remaining angles. If the area is 100 square feet, find the remaining side. 20. In a ti'iangle AD is drawn perpendicular to the base, and BD = 25, DC = A0,AB = 75; find the angles. 21. The base BC of a triangle is 37.54, the perpendicular from the vertex on the base is 100, L sin ^=9.68357 ; solve the triangle. 22. In the ambiguous case A, b and a have fixed values, the / latter being 88.34; the diflference of the two values of C is 26" 30'; find the diflference between the areas of the two triangles. 23. Two angles of a triangle are 7P 28' 6", and 50° 66' 10", the greatest side is 2264; find the least side. olve the mgle. ;olve the triangle, triangle, solve the B and C, ,hout find- ' drawing a irea of the and the the area EXERCISE 187 24. The sides of a triangle are 25.3, 40.7 and 50; find the radii of the inscribed, escribed and circumscribed circlec. 25. Two angles of a triangle are 37'' 20', and GS"" 40', and the radius of the inscribed circle is 100; find the sides of the triangle. 26. In the preceding example find the sides if the radius of the circumscribing circle is 100. 27. The radii of the inscribed, circumscribed, and one escribed circle of a triangle are 2.8284, 5.833G, and 7.071 respectively; find the sides of the triangle. 28. Given the sides of a triangle, show that the angles may be found from the equations cos A-B {a + h)&\nO . C c&mO 'l\/ab , sin ^ = 2y/aV where a-h = c cos 0. 29. Given the sides of a triangle, show that the angles may be found from the equations x + y = a, log (,«-2/) = log (c + 6) + log (c-i)-loga, log cos B = log X - log c, log cos C = log y - log Ij. base, and •pendicular solve the values, the Df C is 26" riangles. 0° 56' 10", W ■Ml.ll IHUjUMMMtipr''.'- ( I CHAP TEE XIII. PROPERTIES OP PLANE FIGURES. 197. In tlie preceding chapters simple examples have been given of the various parts of the subject usually treated in works on elementary trigonometry. We now propose to exteiid the principles already given to problems of a somewhat more difficult character. Sides and Angles of Triangles- 198. Many important identities have been established involv- ing the sides and angles of a triangle. Examples are given of the more useful transformations. Ex. 1. — Prove sin A -i-sin 7j>' + sin (7=4 cos - cos — cos — . 2i 2i 2i . , . ,. . ^ , . ^ A ^ , B^G B~G sm A + sin B + sni C = 2 sin , cos ^ + 2 sin — - — cos — - — -J 2 2 2 = Z cos -- COS !(■ yy+c + COS B-C r") A B C 4 COS — cos -^ cos -. 2 2 2 In the above, sin — - — is replaced by cos -, and sin — by B4.C . B + C'^ .A / 1 1 u cos — - — , since — — — and — are complementary angles, bucn interchanges are very frequently employed in transforming ex- pressions involving the angles of a triangle. "«*»»► RIDES AND ANGLES OF TRIANGLES. 1S9 ve been 3ated in ) extei>d at more Ex. 2. — Prove tan A +tan /i + tan (7 = tan A tan B tan C. ^, sin A sm iB + C\ tan A + tan /^ + tan 6 = + V. -, cos ii cos /> cos C 1 sin yl + cos A cos B cos CJ . , (cos /y cos c-cos(/; + ni = sm yl /v r (^ COS -.1 cos /> cos C j sin A sin />' sin C cos yl cos yy cos C ■- tan yl tan B tan C In the above sin (B + C) is replaced by sin A, and cos yl by cos {B + C), tliese being supplementary angles. i involv- given of Fx. 5.— Prove (b + c -a) tan - = {c + a — b) tan - ^ (<^ + b - c) tan ^^ , jj t^ ^ With the usual notation we have C B - C kos ) h ^ by I '-I Such lung ex- (b + c- a) tan .y = 2 (s - a) T- y^^^ = 2 P- ^^ '-^ The symmetry of this result shows that it will remain un- changed after any symmetrical interchange of the sides and angle. 199. In the solution of many problems it express the sides of a triangle in terms ems It is convenien pposite i t to les. B C Ex. 1. — Prove {b - c) cot - ->c{c-a) cot -- + ((/ - b) cot - = Let then a sin yl sin B sin C = k a = k sin A, b^k &\n-B. c = k sin C. m h 190 PROPERTIES OF PLANE FIGURES. And (b - c) cot ^ + (c - a) cot ^^ + () cot ^^ w ^ ^ = ^ I (sin Ji - sin C) cot ^^ + (sin C - sin yl) cot - I ^ 2 + (sin A - sin ^) cot — C\ 2/t/sin^ '-'/ yy-c . B+C . C-A . C + A sin -y- +sin — ^ sin — ^ 2>l- jsin- ^^ C .,C . A-B . yi + /n + sm-^sm-^| ... , . 2^ . ,7y) -snr - +sin- , - siir ^ +snr - sin- 2 2 2 2 = 0. 2J ^aj. ^. — If a?, b'\ c- are in A. P., then cot J, cot B, cot C are also in A. P Since «-, 6^, c^ are in A. P. we liave or yt^ gjn'yy _ ^•. ^gfj^j ^ _ g^^^ 7? + sin- C) or sin- B = sin (A + B) sin (yl - B) + sin (^ + B) sin C = sin C {sin (vl - 7^) + sin {A + 7^)} = 2 sin (7 sin A cos 7?. Dividing by sin A sin ^ sin C and interchanging sides, we get sin 7)' sin((74-i4) , ^ 2 cot 7i = -r-7,-. —7 = —-7^7--. = cot A -f- cot C, sin C sin ^ sin C sin A which proves the proposition. 200. If the terms of a ratio or the two sides of an equation are homogeneous functions of the same number of dimensions of the sides of a triangle, the sides may be replaced by the sines of opposite angles. The truth of this statement will be evident from the examples of the preceding article. A formal proof, however, may easily be given from Art. 34, of the High School v.^ Algebra, Part II. SIDES AND ANGLES OF TRIAN(JLES. 191 CI ?)cot-^j Bin 2 y cot C are sin C les, we get C, We give two further examples illustrating important principles. Ex. 1. — Find the greatest value of cos A cos li cos C in which A^ B^ C are the angles of a triangle. Suppose C to have any fixed value, then A + B is constant, and *ve have 2 cos A cos B = cos {A + B) + cos (A - />) for all values of A and B. But since yl + /i is constant and cos {A - B) is greatest when A - H is 0, therefore cos A cos B is greatest when A = B. Siniilarly for any value of yl, cos B cos C is groitt'st when /? = C, etc. llcace cos A cos /> cos C is greatest when A=B = C = 00°, and then cos A cos B cos C = ^. ^x. ^. — Factor cos- ^ + cos" B + cos^ C + 1 cos yl cos 7i cos T - 1, in which A, B, C are any angles whatever. cos- A + cos^ B + cos- C + 2 cos yl cos B cos C - 1 = (cos yl + cos B cos C)" + cos- J5 + cos- C - 1 - cos^ ^ cos- C = (cos yl + cos B cos C)'"^ - (1 - cos- ^) (1 - cos- C) = (cos yl + cos B cos C)' - sin- B sin^ C = (cos yl + cos i? cos C + sin B sin C) (cos yl + cos B cos C = {cosyl +cos(^- C)} {cosyl+cos(^ + C)} - sin B sin C) = 4 cos - A+B+C A-B+C A+B+C B+C-A cos COS — COS li equation lensions of lie sines of )e evident |nial proof, igh School If yl, B, C are the angles of a triangle this expression van- ishes. It also vanishes if any one of the four compound angles is an odd multiple of a right angle. 201. Since three of the six elements of a triangle, one of them being a side, are sufficient to determine the remaining elements, it follows that not more than three independent relations can exist between the sides and angles. Also from any three inde- pendent relations all the otheis may be found. The following .e^"' T 192 PROPERTIES OF J'LANE FIGURES. f are three different gjoups of such iclations from any one of which the otliers may be derived : h c f.JL_ = I. J. sin A siu Ji sin C' [a+B+C^-t:. {a = h cos C + c cos B, h = c cos A + cos A . {(I- = /r + c- - 2/>c cos A , /;-' =: ^2 + (/'^ - 2ca cos 7i, c'^^a2 + i2_2aicos C. 1. To derive II. from I. Art. 80. Art. 81. Art. 82. J<>om we have or, by substituting A + B + c = -, sin A = sin {/i + C) = sin B cos C + cos C sin ^, cos C + c cos B a, i, c, for sin A, .sin i/, sin C. Art. 200. 2. To derive III. from I. From A+B + C = 7:, sin A = sin {B + C), cos ^ = - cos (/? + C). Then sin- ^1 = sin'-^^ cos- C + 2 sin B cos C cos /i sin C + cos^ /i sin- C = sin^ B + sin- C + 2 sin />' sin C (cos /> cos C - sin 7j sin C) = sin-7i + sin^ C -2 sin ii sin C cos yl, or «2 _ /^2 ^ g2 _ 2/>c cos A. Art. 200. 3. To derive I. from II. From the three equations find eos A, cos B, cos C, in terms of the sides; thence find sin A, sin B, sin C, and the Sine Rule follows immediately. Ajzain from the same equavyions eliminate a, h. c, and we obtain 'I ■') ^> cos- A + cos'^ B + cos- C + 2 cos A cos B cos C - 1 = 0, f which Art. 80. Art. 81. Art. 82. P> ^rt. 200. -A'sin^C /; sin C) :'t. 200. terms of lie Rule liminate EXfillCISE. 193 a»id coiisoquently oiic of its factors, as givt'ii in the preceding article, must viinish. From the consideration that each unghi of a triangle is positive, and the sum of each pair is hiss than 180", we find that it must be cos i (A + Ji + C) which vanislies, for the TC value ~. Thus A + J) + C = n. Tiie student should make all the remaining transformations, no!\(^ of which present any ditUculty. EXERCISE XXVI. Prove the following identities in which A, />, C are the angles of a triangle : 1. sin yl + sin 7/ - sin C= 1 sin sin t;os . Jl ^ ^ 2. SHI A -sin /> + sin C = t sin cos sin -. ^ Z A 3. sin 2yl+.sin 27? + sin 2C = i sin A sin />' sin C. 4. sin 2^1 -sin 27i + sin 2C^ 1 cos A sin 71 cos C. 5. A B C A n a cot ^ + cot -- + cot ^ = cot - cot - cot -^ . 9 9 9 9 9 9 ABB C 6' A , tan - - tan — + tan ,- tan + tan -- tan - = 1 , \ 7. cot ^ cot B + cot B cot C + cot C cot yl == 1 8. 9. 10. I T) ^ ^ , . A . B . C cos A + cos B + cos 6-1=4: sin .^ sin - sin - . ^ ^ ^ cos 2 A + cos 2B + cos 2C + 4 cos ^ cos II cos (7+1=0. SI n(/>' + C-^) + sin {C + A-B) + Hxn{A+B-C) = 1 sin yl sin B sin C 11. sin^ A - sin'"^ 7> + sin^ C =2 sin yl cos B sin C, 12. cos-' - + cos'' B G B C COS' -^ 1 COS - COS o sin 13. 14. sin- A = cos- B + co.s^ C +2 cos yl cos 7i cos C. cos^ -<4 + cos'^ 7y + 2 cos A cos B cos C = sin- J + sin- B - 2 sin A sin 7^ cos C I* 194 PROPERTIKS OK PLANE FIGURES. In a triangle right -angled at C, prove the following IG. see 2// = a- + fr 2 sin A sin /^ sin- A - sin" // 1 9. a^ cos ii + //" cos B = abc. ^ a h 18. cosec 25 = — +-r-. 2 h 'la a- - o 9, sin A sin /^ 17. tan 2JJ = snr ^ - sin' /; 20. abr-H\n A sin li^AS'- In any triangle prove the following : A Ji a + h + c 21. cot -- cot - = — ~, . 2 2 a+b -G oo. tan .', />' h+ c-a tan .', yl c + (I - b' 23. or, tan /> a'- + 6^ - c^ tanC a:--b'^ + c^' a— h cos 7? - cos A c sm{A-B)_fr-lr sni C c" /* + c cos Ji + cos C 2G. 1 - cos A 1 + cos C 27. a sin {B-O+b sin (C - vl) + c sin (vl - B) = 0. 28. 2a (sin C - cos B sin ^) = i sin 2^1. 29. a (b cos C - c cos B) = i'- - c'"*. 30. {a" - b-) cot C + (/;-' - C-) cot A + ic"- o?) cot 7? = 0. 31. sin A (cos yl + cos 5 cos C) = sin 7i (cos B + cos (7 cos .4). 32. b (tan B + tan C) = « tan B sec C. 33. (« + b) cos C + (i + c) cos ^1 + (c + rf ) cos B = a + b + c. j 34. « cos yl + 6 cos B-^c cos C = 2a sin B sin C = 1h sin C sin ^4. 35. 2a6 cos C + Ibc cos yl + 2ca cos B = a' + b- + r. 36. a sec A -b sec 7i = sec C (b sec yl - a sec 7>). 37. «2 - 2a6 cos (G0° + C) = c- - 'Ihc cos (GO" + A). 1 38. The perimeter of any triangle is 2c cos -- cos - cosec - . 39. The greatest value of sin — sin -- sin — is - 2 2 2 8 40. If y sin'^ yl + a; sin'^ 7? = ;: sin^ -^ + 2/ sin- (7 = ic sin^ C + s sin^ -4 then a; y sin 2il sin 2j& sin 2C /' TRIANGLES AND CIIICI.ES. 195 y _ tail ^1 tail/) tail ^ taiwl tan />' tan C V_ V^ " tan li tan (J tan j{ tan 6' tan A tan />' " \ = sec il sec li soc C. ^ 42. If a, />, r are in A. P., then cos J, vers //, cos C are also in A. P. 43. If a- + hc, fr + cn, c- + nb are in A. P., then tan -, tan .^ , C - - tan -- are also in A. P. Ml 44. Factor 1 - cos'-' A - cos-' /? - cos'- C + '2 cos yl cos Ji cos C in which A, Ji, C are any angles wliatever. Triangles and Circles. 202. A triangle may be solved from various data. In geneial any three independent measurements, one of which is a length, are sufficient. We give two examples. Ex. 1. — Given two sides of a triangle and the line bisecting the included angle to solve the triangle. Let h, c be the given sides ; / the bisector of the angle A which meets the base in B. We have from which BD.DC = c.b Euc. VI., 3. h + c B* ij' sin^ A Then in triangle ABD wo have from the Sine Rule I BD ac sin B sin | A (b + c) sin h A^ or 1 = • P ] ' A 26c cos - ac sin Ji be sm A 2 Therefore {b + c) sin I A (b + c) sin ^ A b + c A (b + c) I Two sides and the included angle are now known, which is Case III. of the preceding chapter. i I 190 PROPEUTIKS OF TM.ANK FIGURES. Ex. 2. — Civcn tlio p{M|K!ii(li(;ul!ir iVoin tho vcd'tox of a triangle on tho baso, tin; bisector of tho vcMtical angle and tho lino Join- ing the vertex to tho centre of tlie base, to solve the triangh?. In triangle A B(\ let ^ Z) = d, AE^-e, AF=f and OA = OB = R, the radius of tlie circumscribing circle. lOAG= L OGA = L GAD = cos-» "^ e lOAF^ lOAE- lFAE = lEAI)-{i.FAD- lEAD) = 1^EAD- ..FAD Then and 2 cos' d — COS" From the triangle OA F wo have, OF OA on from which that is sin OAF' sin OFA sin FAD' OF sin OAF on sin FAD' sm I 2 COS"* COS A - COS' ■/) SUl I COS ' . I .( Till ANGLES AND TMKCf.ES. 197 no Joiu- n<4l(i From this equation A is dotcrniinod jiiid th(M\ li + C is known. Also tho aii^'lc! A' A/) may easily bo shown to l).5 ecjual to h {C - /y), and thus //ami C u\;\y \n', found. Tin; remainder of the solution presents no ditliculty. on=R, 203. Tho folloNvin<^ an^ a f(^v\' of the many interesting pro- perties connected uith the inscribed and oscriited circles of a trianj^lo. We assume tho results alieady proved in Arts. 91 and 93. 1. The centre of a circle lies on the line bisecting the angle between two intersecting tangents. The following sets of points are therefore coUinear : A,I,I,; B,1,L; CJA; I,,A,I,; I,,BJ,; I,,C,I,. i^SM.MW^^^SSr^r^ t ■V 'I If 198 PROPERTIES OF PLANE FIGURES. 2. Tangents drawn from the same point to a circle are equal. The truth of the following may therefore bo easily shown. \) AF =^AE r=8-a, A /-', =--AE,=^ s. Arts. 91, 93. 2) AF, = AE, = CE,- CA=^s - b. :\) A F = AE.i^ JiF., - nA=^s-c. I) FF, - A /; - A F = s - (s - a) - a. r)) F^ F., - A /'', + BF, -An= 2s -c--^a + h. 6) AF +A F, + A F, + AF, = a + b + r. 7 ) F,F =AF- A F, - (.s -a) - {s -h)-=h -a. 8) III = AI,- AI= {AF^ -AF) sec ^ = « sec ^. 9) /,C = A\(7 sec E,CI, - (*• - h) cosec ^^ . C C (10) /j/^j = /,C + C/, = (.s- - A) cosec + (« - «) cosec = c cosec 2 204. 7^0 exp7'ess the radii of the inscribed and escribed circles in terms of the radius of the circumscribed circle. From the triangle BIC we have, IC BG a a sin ^ B sin BIC sin I {B + C) cos | A' then -. -, . C a sin i ^ sin J C r = IC sin ^ = — ^~— — 2 cos ^ yl = 4 yr sin sin - sin — . J i3 a! Art. 90 (1). Similarly from the triangles BI^C, CI^A, AI^B^ we obtain, ,j, . A B C .r, . B C A Ti — 47C Sill Y, COS - cos - , 0^2 = "*« Sin - COS - cos —, 2t 2i 2i 2i u a JS A-O ' ^ ^ ^ r = 4:Ii sin cos cos -. J 2 2 TRIANGLES AND CIRCLES. 199 equal. 91, 93. ;cosec— . d circles . 90 (1). bam. A 2' 205. The rectan;/fe of the sognwnts of any chord of the circum- scrihed circle of a triam/le drawn throiu/h the centre of the in- scribed circle is equal, to ttoice the rfctane!i (k>s ; simi- le reniain- ,, J/, N, lie EXERCISE. 203 d external he - C" •pendicular he vertical angle, then c In , . Igle on the the area. Ingle ABC ; |s, and thus show that jot C. A 8. In any triangle prove a" = {h -{■ c)- - {be sin'- ^, and thence show how to solve a ti'iangle having given the base, the sum of the sides, and the altitude. 9. Solve a triangle, given the base, ditlerence of the sides and the altitude. 10. Ciivi'u the base, the vortical angle and the sum of the sides; solve the triangle, (From the Sine Rule cos A (/>'- (7) may be obtained.) 11. In the figure of Art. 203, show that the circle descrilxxl on //[ as diameter passes through /> and C, and that similarly described on IJ^ passes through .1 and B. 12. In the same figure pi'ove AEi — CE^^BD=^s-h. Find three other tangents of the same length. 13. Express the lengths of the various tangents from the vertices B and C in terms of the sides. 14. In the figure of Art. 210 show that A0^2R cos A, 0D = 2R cos Ji coa C. Examine the case in whicii one of the angles involved is obtuse. 15. The medians of the triangle ABC meet in G; show that the distance of G from the side BC is Ij Ji sin B sin C, and that 16. Prove that the length of the perpendiculars from the point of intersection of the medians of a triangle on the side are inversely proportional to the sides. 17. Show that the distances of the orthocentre from the sides of a triangle are inversely proportioned to those of the ctntre of the circumscribed circle. 18. Show that the area of a triangle is given by each of the following expressions : A ARC (1) s (s - a) tan - . (2) ^^ tan ~ tan - tan ^ . 4 ' 4 4 « \l :l ^• I il h 204 PUOI'Ell'I'IES OF PLANE FIGURES. o} sill B sin C 2s-. sill A sin li sin C ' sill A + sill /? + sin C ' a- + Ir + c" , 2^(/^o J 7i C 4) -— cos . cos - cos --. H + It -\- C J Ji J (6) ('^ + h + c)- (7) i (cot }, A + cot \ n + cot J, C) (8) 2A'- sin .1 sin B sin C. i (cot A + cot 7) + cot C)' (9) J> R{a cosyl + /^ cos B-^c cos C) = ^, 7i'-(sin 'lA + sin 27i + sin 2(7). 19. In any triangle prove the following relations : (1) n\ ^ tan- 7-.. r-x (2) 7»? tan A ahc Ir + c- - a'-^' (3) /'i + r._, + v^ ^ ^ (3 + cos A + cos 7>' + cos C). (4) 7t' \r-=^R (cos .1 + cos B + cos C). .1 B C s , , /' 1 1 1 (0) tan , + tan + tan ■ H — = 47i; + / + (I c 20 If the radius of the inscribed circle of a triangl* be equal to lialf that of the circumscribed circle the triangle is eiiuilateral. 21. The area of any triangle is lir (sin A + sin B + sin C). 22. Show that the angles of the triangle 7i7^,7, are i {B + C), ^ (C + A), I {A + B), and the sides a cosec - , b cosec - , <; cosec —-. 23. Show that the radius of the nine-point circle of a triangle is half the radius of the circumscribing circle. 21. The angles of the pedal triangle DBF are - - 2A, - - 2B, 71 — 2(7, and the sides are 'a cos A, h cos B^ c cos C. 25. The sum of the sides of the pedal triangle is iR sin A sin B sin (7, and the area is i be cos B cos C sin 2^1. 26. The radius of the circle incribed in the pedal triangle is 2R cos A cos B cos C, and the radius of an escribed circle is 2i? cos A sin B sin C. 27. Show that 7-, +r^ + r., - r = iR, and thence that the area of the triangle 7i7^7j is 2Rs. B C 2 cos 2- f_ + cot \ C)' aC. /i + sin2C). EXERCISE. 205 •)C -a 2* »1» be equal equilateral. sin C). }.{B + C), C , <; cosec —. jj if a triangle \2A, --25, Is 4^ sin ^1 triangle is id circle is the area of 28. Find the angles of a. ti'iangi(> whose sides are proportional to cos I A, cos I J>, cos h C. 29. In any triangle the sum of the reciprocals of the perpen- diculais from the vei'tices on the op})Osite sides is ecjual to the recipiocal of the radius of the inscribed circle. 30. If the sides of a triangle are in A. P., the radii of tlie esci'ibed circles are in H. P. 31. If the sides of a ti'iangle are in H. P., the areas of its escribed circles are also in H. P. 32. The rectangle of the segments of any chord of the circum- scribed circle of a triangle drawn throu'di the centi-e of an escribed circle is equal to twice the rectangle of their radii. 33. Show that the distance between the centres of the circles in the preceding example is given by the equation 01^= R'+ '2i\R. 34. If the points of contact of tlui inscribed circle of a triangle be joined the sides of the triangle then formed will be 2/* cos I A, 2r cos ;'; B, 2r cos }, C, and its area 2/-- cos h A cos .', /> cos }f C. 35. The sides of a ti'iangle are a+ />, /> + c', c + '^', show that the square of the radius of the inscribed circle is IvR. 3G. In the ambiguous case in the solution of a triangle when a, h^ A are given, show that the circles circumscribing the two triangles arc equal, and the distance between their centres is y'a'- cosec'-' A — li-. 37. If an equilateral triangle iiave its angular points in three parallel straight lines, of which the middle one is distant fi'om the outside ones by tained sctccl with 5 any two ; above it nstead of ",, we have lYand a, ^ ot a, BN inging all 216. A, B, C are the nn;/uhir poiiUx of a tnaiKjIe, tha lon<>lntintJie plane of the triaiKjh' at viliirh tlie sides AC, V>Vj suhfend givpii aiKjIrs «, p'. Jt is vv- qnired tojind the distancrs AD, IM). Let tlie angles CAD, CBD be denoted by .r, prospectively; then, since the sum of the angles of the quadrilateral ACDB is four right angles, and the sum of the angles ADB, ACli is fi+§+C, we have x + i/=2--a~^-C so that the sum of .r and y is known. Then from the triangles ADC, BDC we have _, ^, h sin X a sin ?/ ) J from which sm a sin p' sin X a sin a sin v/ /> sin ^ Now take a subsidiary angle - 1 sin X + sin 3/ ' tan + 1 tan (0 - 45^) as 210 MEASIIUEMENT OF IIK[(;HT.S AND DISTANCKS. from which tan '" ^^^ = tan ^ \ '' tau ( - 45"). Tliis o(juation fjivos tlin viiliu' of .'•-//, and siiicn ,7+1/ is known, x and y can be found. 'J'h<5 rcinaindtir of th , and height are — cot a - cot U cot a - cot U / 4. A spherical balloon whose diameter is d subtends an angle / a at the eye of a spectator, whilst the angular elevation of its centre is |3; show that the height of its centre is J(/ sin p' cosec . 5. At a distance a from a tower on a hoiizontal plane the angle of elevation of its summit is the complement of that of a flag stafF upon it ; show that the length of the flag-statf is %i cot I0.. G. A building is three stories high, and from the opposite side of the street the angle of elevation of the roof is double, and that of the second story is the complement of the elevation of the fiist story. Show that the width of the stieet is a mean j* proportional between half the height of the roof and the ditieFr,3 ence of the heights of the first and second stories. • 7. On the bank of a river there is a column 200 feet high > i EXKRCISE. 211 i :i + y is solution )W('r on ;i firy; find of 400 on 0, and clovation its slant an angle )n of its cosec . lane the liat of a ;-statF is opposite ble, and ation of a mean r» le (litiei^r,_) ?et high J" i' supporting a statin; .'50 f<'«!t high. To an ohser\ cr on the opposite bank the statue subtends the same aiigh^ as a man six feet high standing at the base of the colunni. Find the breadth of the river. 8. Two lailvvays intersect at an angh; of M5' 20'; from th<> point of intersection two trains start togcthei-, om; at the rate ot' 30 miles per hour, and at the end of 2^, hours they are 50 nnles apart ; find the rate of the second tiain. 9. A sj)here, radius 7*, on the top of a pole // feet high, sub- ^ tends an angle of 2a at a point in tluj hoii/ontal plan»; fiom which the elevation of the centre is p' ; show that the height of the pole is r (sin p' cosec a- 1). 10. The to{)s of three chimneys are in a lioiizontal line and at equal dis ' tees, <•, fiom each other. From a point of obs(!vation on the horizontal plane below their angles of eh'vation are «, ^, y\ show that their height is -\ 2c-* cot'^ « - 2 cot" [^ + cot- J' 11. ii and B are two inaccessible points on a horizontal ))lane and (7, D are two stations at each of which AH is observed to subtend an angle of SO*". AD subtends at C lO'^ 15,' and AC subtends at D 40° 45'. Show that CD = A li\^X 12. From a station, A, at the foot of an inclined plane, AB, leading up to a mountainside, 7?C, the angle of (!le\ation of C is 60°. The inclination of AB is 30°, its h'ligth is + ^?>). 13. The angular elevation of the top of a steeple at a place due south of it is 30°, at a place due west of it tiie elevation is 18°, and the distance between the stations is cot a -a cot ^ h — a 1"). The ani,'les of elevati- n of an object at three liorizontal stations A, />', (\ lying in a vertical plane passing through It, are \;^s 1:2:;^; if' A il — a, JiC = b, show that its height is ^ a , 9 j77;V(« + />)(:'>A-^0- 10. ABC is a horizontal line, C/)^ a vertical line, and DE subtends at A and B the same angle, a. If AC = f', BC = />, show liat DE = (a + h) tan a. 17. ABC is a triangle right-angled at C, and the side BC = ; if the angles of elevation of an object at A from B and C are 15° and 45*^, show that its height is (3 - 3 ). ^ 18. An object 2/> foot high, placed on the top of a tower, sub- tends an angle a at a place whose horizontal distance from the foot of the tower is b feet ; show that the height of the tower is b {VYc^i- 1}. 19. The angles of elevation of a tower from three points A, I>, C, in a straight line arc observed to be //, pi, j', respectively. If BC -a, AH -<\ .show that tlie square of the lieight of the tow<'r is (ihc a cot- n — h cot"-' p' -I- c cot'- J' 20. A tower standing on a horizontal plane is sui'rounded by a moat as wide as the tower is high. A person on the top of another tower whose height is a, and distance from the moat o, observes that the first tower subtends an angle of 45° 3 show that the height of the first tower is . a -c i**^. CES. r which leans to the south, ■ is ee horizontal 1 rough it, are ino, and DE N* JiC - />, show J side BC = a; ,nd Care 15° a tower, sul)- ice from the of the tower three points respectively, eight of the I'oiinded by the top of the moat f, 45'^; show EXERCISE. 213 21. A spherical balloon subtends angles '/, |9, y at three points J, B. C, in a straight Hue. If AB — a, BC = by show that its radius is ^ ah ((I + h) (/ cosec'- i,- - (a + I)) cosec"-' \ +b cosec" — 22. Three points, A, 7>, C, are situated that so J/> — G3, ilC = 4:4, BC = 7(). From a point, J\ in the same j)liine the sides AC and BC subtend angles of 20^ lU' and oO^ 20'; tind the dis- tance of 1* from A, B and C. 23. In the preceding example if tlu^ angles subtended by AB and BC s^re 89^ 15' and 130^ 45' respectively; tind the distance of r from A, B and C. 24. From a point on a hillside of constant inclination the altitude of the highest point of an obelisk at the top of a hill is observed to the '/ : c feet nearer the top of the hill it is p. Show that if be the inclination of the hill, the height of the obelisk is a sin (a - 0) sin (^ - 0) sin (pi - a) cos 25. From each of two stations on a horizontal plane, at a dis- tance, c, from each other, a pillar on a distant hill in the vertical plane passing through the stations subtends the same angle, and the angles of elevation of the top of the pillar at the stations are a and p'. Show that the length of the pillar is c cos (^ -f «) sin (p' - a) ' 2G. A vertical stick casts a shadow of length Jt from a lamp upon a horizontal plane. The horizontal and vertical distan*.es of the bottoui of the stick from its shadow are a and c respec- tively. If the stick subtend equal angles at the two ends of the (I he shadow, show that the heigiit of the lamp is ^ o- ' \ i / :t I CHAPTEli XV. CIRCULAR MEASURE AND RATIOS OP ANGLES. 217. We now give a few of the simpler propositions whicli show the relation between the circular measure of angles and their trigonometrical ratios. The reasoning is based upon the two assumptions of Art. 31, which the reader should carefully review before proceeding with the subject. 218. If he the circular measure of an arc in the first quac^ rant, sivi 0, 0, tan 0, are in asceuding order ofmayuitude. In the figure of Art. 33, let ach, AC B, be sides of regular polygons of n sides inscribed in, and described about, a circle of a unit radius. Denote the angle BOC by V; then ch = sin 0, Cb = 0, CB^ tan 0. Now the perimeters of the inscribed polygon, the circle, and the described polygon are in order of magnitude, and therefore / 1 \ "^ their I -^ I parts, viz., ch, Ch, CB are in order of magnitude, which proves the proposition. , 219. If d is the circular measure of an angle, the limit of -. — sm when is indefinitely diminished is uniig. We have sin 0, 0, tan in order of magnitude. Art. 218. Divide each by sin 0. e 1 Then 1, sin (>' cos When is indefinitely diminished, cos 0=^ 1. are in order of magnitude. n«9&il««— ■ CIRCULAR MEASURK AND RATIOS .OF ANGLES. 215 Hence -.-- which ahvuys lies between 1 and - L ^ust also hin f/ ^,Q^ If' become a unit when --^ 0. NGLBS. )ns which ngles and upon the carefully Cor. 1. — We have 7 :, = -• — ;, ^ cos = 1 wluui ^^ -= 0, s tan ^ sni ^ ' since sin and cos 0^ each become 1 when — 0. Cor. ;.>.-The reciprocals of -,",. and , ^-,, viz., ^^^nd *''" ^ sui tan each become a unit when = ^. h'sl quat^ if regular \ circle of ircle, and tlierefore agnitude, 'i of - .— silt Art. 218. tide. / /Cor. 3. — The limit of n sin when n is inderinitely great is 0. I Fo r ti sin . e sin n n ^^- = ^^, since is indefinitely small wlion u I is indefinitely great, and consequently the ratios of ' ^ ' '^ n unity. sm - to is n n ri I riM T ., n sill 7A . TT tor. ^.— Ihe hunt of is -— - For if is tlie circuit n 1 60 measure of m\ we have ^ = — . or n =— — ISO' - ' ir Then sin »fc" sin <9 sin (9 t: r . sin /? ~^ - 180^ "~^~- 160^180'^'''''^" r"^- Cor. o.— The limit of the ratio of the sine of an angle to the angle is the circular measure of tiie unit of angular measurement employed. « «: si P" ti' i /■■/ I 216 CIRCULAR MEASURE AND RATIOS OF ANGLES. "I! i J u E ll ti ; ; ' 1 II to r H 9 1 •:!- . ^ n B ■ ;,: .li ■ m ;..;' ■ ■ \\i 1 L \ ^ 220. The proposition of the preceding article is very impor- tant. We give another proof founded ujjon a different principle which will be found instructive. In the figure of Art. 33, the angle JJOC is — , where ft is the number of the sides ; denote this n angle by 0. By sufficiently increasing ?i the angle -, or 0, may ft/ be made indefinitely small, whilst the perimeter of the polygon becomes ultimately equal to the circumference of the circle. We have, therefore, . sin - 2nr sin ■ ^ c i sni I) u ')i pernnetcr or polygon n 'Itti' circum. of circle 1. Similarly from the exterioi* polygon we may prove tan - =\. 221. IfO in thr, circtddT nieasarH of a jMsitivi amjle le.ss than a rifjhi amjIe, she is (/r eater tJiait - . 4- Wo have sin = 2 sin ■ cos - =2 tan . cos'- := 2 tan 1 - sni- ') <) •) •> ') :) But , . tan > , , and sm - < ^ Art. 218. Substituting these values in the above we get sin vhic gives sin 0>0- Cor. — Writing - for we get sin — > -- - --. -ES. very impor- nt principle Alt. 33, the denote this , or Oy may bhe polygon circle. We on 1. tan U ~ less than a sin=:) Art. 218. CritCULAU MEASURE AND RATIOS OK ANGLES. 217 222. IfO is the circular measure of a positive a^ujle less than a rxyht angle, cos is greater than •^ ~ TTj ^^*^ f^ss than 1 - -j 2 2^16' We have cos ^=1-2 si.i^ ^'^ > 1 _ 2 (^T> 1 - ^-\ Also Hencf cos^/=.l-2sin^'^l-^", but90888 correct to eight places of decimals. ^^^ 2''>1 Similarly from Art. 222, cos 10' = .99999576 to thesanie degree of approximation. Cor i._The sines of all angles less than 10' are equal to their circular measun s as far as eight places of decimals. (.?7c^'^\7^^ '* ^T'"' ^"^ ""'"'^"'' ^^ ^^^«"^^« J««« than 600 (or 10 ) then sin n" = n sin 1" to eight places of decimals. ' 15 !l ■I 218 CIRCULAR MEASURE AND RATIOS OF ANCLES. Cor. tS. — If a 1)0 not greater than 10' and 2 (!OS a — 2-k, then k=2 (1 - cos a) = t .sin- ., — ( 2 sin j as far as eight dcciiual phices. :snr a. 1! ! 224. To find tlie sines of d seri>'« of omjles ichlch arc viidti- jdes of W. If in the identity sin (?i + 1) a + sin (vt - 1 ) a == 2 sin iia cos a we put 2 cos a--='2 - k, we get sin (n+l) a= 2 sin tia - sin [it - i) a - k sin ?ia. Let a:= 10' and for ?<, write 1, 2, 3, etc., in succession, Thus, sin 20' = 2 sin 10' - k sin 10' sin 30' - 2 sin 20' - sin 10' - k sin 20' sin 40' = 2 sin 30' - sin 20' - k sin 30', etc. These equations give the values of sin 20', sin 30', etc., in succession. We give a few stei)S of the work, which the student can easily continue, and conipaie his results with those given in the tables. sin 10' =.00290888. k sin 10' = .00000002. sin 20'- 581774. sin 30'= 872656. sin 40'= 1163531. sin 50'= 1454396. sin 60^= 1745249, etc. From Art. 223 (Cor. 3), A; = .00000846. Each result as it is rbtained is multiplied by k, and the product placed in the second f^olumn. It is then doubled, the previous result subtracted, and liom this is taken the hist product i:i th-3 second column ; this ^t remainder is the next result required. ^-sin 20' = 5. ^•sin 30' = 7. k sin 40' = 10. /•sin 50' = 12. k sin 60' = 15, etc. - A;, then irc viidti- na. tc. )', etc., in le student given in 002. 5. 7. 10. 12. 15, etc. It as it is the second acted, and umn ; this CIRCULAR MEASURK AND RATIOS OF AiNCiLES. 219 225. From the identity cos (m+ 1) « + cos (/A- 1) a~1 COS va cos a = 2 cos }i,a — k cos na, Nve get as before cos (w + 1 ) a — 1 cos tta — cos (w — 1 ) will usually be much less than the thousandth part of An, and the angle AUG will be very sn.all ; consequently PA = BP, and arc AC = PC approximately. Denote the height of AP in feet by A, the distance AC in miles by d, and assume i)'/'= 8000 miles. Then approximately d'' = 8000 x -A_ ^ ^h ^earl v Also, PC angle CO A = tan-^ — =tan- \SOOO\J2)' 222 ciucui.au measuuk and ratios of angles. 232. To sho?v that si7i 0>0 — where is the circular meas- (J ure of an acute angle. We have cos ,■ > I - - „ cos , > 1 - ^-t etc. 2 2-' 4 2^ mi <• sin 1 hei-efore — > or •-:)('-30-:;)- 1 ^'' /ill \ sin 0>(f- ~. Art. 2-J2. Art. 231. EXERCISE XXIX. 1. Show that tlio limit of the ratio of sin kO to k, when k is indellnitely diminished and is finite is 0. -^2. Show that tan is greater than + ^. o 3. Show that cot 0> ^ but< - . 2 4. Express tan in terms of tan . , and thence show that //' 0-' (P tan 0>0+ +_ + _+ 4 10 b4 i ?.(f 40 5. Show that tan 0<- but > :,. •J _ 0- 4 - 0' 6. From Arts. 218 and 221, prove cot fl< - -. ' ^ i 7. The chord of an arc subtending; an anjjle at the centre of the circle is 2r sin - and the chord of half the arc is 2r sin 7. 2 4 8. From Art. 232 prove the following rule for findin^^pprox- imately the length of the arc of a circle. From eight times the TSS^ssmm^m r ineas- rt. 2 -J 2. it. 231. hen k is )W that centre . sin -. 4 ipprox- bies the EXERCISE. 223 chord of half the arc subtract the cliord of the whole arc ; one- third of the remainder will b(! the length of the arc. 9. Find the dip of the horizon from the top of a mountain 1 J miles high, the radius of the earth being 1000 miles. 10. Having given that two points, each 10 foet above the earth's surface, cease to be visible from each other over still watiu' at a distanc(; of eight miles ; find tlu; earth's diametei'. 11. If fl be the dip of the horizon from the top of a mountain, and R the radius of the earth, show that tlie height is approxi- nwitcly \ li tan" 0. 12. Given that the moon subtends at the earth an angh; of half a degree, find the d. stance at which a circular plate of six inches in diameter must be placed so as just to conceal it. 13. Find the value of . , , when ^ is indefinitely diminished. sui bo ^ 14. If be the circular measure of a small angle of it", prove that approximately sin log n + log L sin n" - L sin 1". 15. When is small, prove that approximately (1) log sin ^/ = log 0+ \ log cos 0. (2) log tan = log - 5 log cos 0. 16. li p be the perimeter of a regular polygon inscribed in a circle whoso diameter is 1, and half the angle subtended at the centre by one of the sides, then Tt=p sec - sec , sec - 4 o ad. inf. i 11 ^ M ilil 1^ 11 \ EXAMINATION PAPERS. PAPER I. 1. Explain the diirorcnco betwcon tho tri<^onometricjil and the geometrical line. 2. The minute hand of a clock is ^\ inches lon^' ; in what length of time will its extremity move 2^ inches? 3. Show that sin* A + s^in- A cos'- A + cos* A = 1 ~ sin- A cos'- A . 4. Given tan 0= \/2, sin ; lind thi-ee difl'erent values for (f 5. In a triangle right angled at C, show that (r tan Ji ~ b' tan A = twice area of triangle. - ^ 1 + tan A 1 - tan A _ C. rrove — ; + -, —. = 2 sec 2 A. 1 - tan A I + tan A 7. The sides of a triangle are 11, 15, IG; find the cosine of the least angle. 8. Express log cos 45" in terms of log 2. 9. Show that the radius of the circle circumscribing a tri- angle right-angled at C is sec Ji, 10. Show that the least value of sin'* ;-' + sin * \ is , PAPER l[. 1. Explain the difference between t.'ie trigonometrical and the geometrical angle. 2. The angles of a triangle are in arithmetical progression, and the circular measure of their common difference is - ; express the angles in degrees. •'*rj.isiii»iSi-'->-' j»Ta.'jW!M||aw«*»«->;- s. ill and the ; ill what -' A eos" A . lues for - tail Ji - cosine of )iMg a tri- ll and the agression, ; express PAPER HI. nr> .'I. Prove (sin A + eos A)^ -\- (sin A cos A f = 2 sin J (;5 - 2 sin-yl). 4. Given sin A -fcosec A ~2 ■, find eos A. r». Show that the area of a triangle right angled at C is I (a- + b'-) cos A cos /A /. LI 1 1 . 1 + tan A ^^ b. Solve tiie e(iUiifiou , - - ^ 2 tan 2yl. 1 - tan ^1 7. The sides of a triangle an; 5, 12, l.'i; lind the radius of the inscribed circle. 8. l!i a ttiangle log sin t' = 0, and log sin li - I log Ji - log 2; find all the angles. 9. The sides of a triangl<< am 7, 21, 25; lind the radius of the largest esciihed circle. r 10. If sin 'm + sin~Si= : show that sin' >/t = cos"' n. PAPER III. 1. State accurately the meaning of r as used in trigonometry. 2. In a (juadrilateral A/iCD, the angle A =30", />' = G0«, C^Ijt; find the nundxir of degiees which must be taken from D and added to A, so th.it the figure may be in.scribed in a circle. A-V _ _, ., , /see yl - 1\ „ , /sin ,1. Prove cot- yl {, — .---r +sec-/l I, — \l+snwl/ \l+i sec A = 0. • 4. Given 8 cos* ^ - 8 cos' () + \^0 ; find cos 20 and cos 4//. 5. In any triangle shoAr that a cos B + h cos A is positive 6. Prove 2 cot 20 (tan + cot ^Z) = cosec' - sec'- //. 7. The sides of a triangle are 6, G + \/'2, G - \/2 ; find its area and the sine of the medium angle. 8. Two adjacent sides of a parallelogram are of lengths 2.") and 30, and the angle between them GO ; find the lengths of the diagonals. » ': \ ^ r 226 EXAMINATION PAPERS. :lh 11^ p s ! 9. Prove tan"' ] + '2 tan"' ?5=tan"' 1. Illustrate geomet- rically. 10. Prove log cos A + log cos (60^ + ^) + log cos (60' - yl) + log 4 = log cos 3yl. PAPER IV. 1. Name the quadrants in which the several trigonometrical ratios are respectively positi\e. 2 The perimeter of a sector of a cii-cle is equal to half the circumference of a circle ; how many degrees in the angle of the sector 1 3. Simplify {a + h) cos 180'^ + (« - h) sin 90^ + 21) tan 45". 4. In any triangle if A -^ GO ', then /> + c = y/a- + '3bc. 5. When the altitude of the sun is 22° 30', find the length of the longest shadow that can be cast upon a horizontal plane by a rod 1 2 feet in length. 6. Show that the area of ,a parallelogram is equal to the continued product of its diagonals and the sine of the angle be- tween them. 7. Solve the equation cot" <> + tan- = —. o 8. In a triangle right-angled at C, prove a hi: tan"' 7 - +tan ' " == < • b -{■ c a + c 4 9. Given 3'-^ 5^"' = 2-'^+' ; iind a*. 10. Prove that in any triangle tan 2^-f-tan 2i5-l-tan 2C = tan 2^1 tan 2 A' tan 2C. ■WWl PAPER V. 227 geomet- 60'' -A) metrical half the ingle of 45°. pngth of me l)V a to the ngle be- PAPER V. 1. State the independent relations which exist between the six trigonometrical ratios of the angles of a triangle, Which of them require proof 1 2. If tan 15°=^-"-"-, then .r= ] log, X 3. The distance between the extremities of the poipendicu- lars from any point in a circular arc on tiie radii through the extremities of the arc is constant. 4. Prove that the line which divides one of the angles of an equilateral triangle in the ratio 3 : 1 divides the opposite side in the ratio \/3 + 1 :2. 5. In any triangle if a, h, c are in harmonical progression, sin'"' A, sin'- Ji, sin'- C are in arithmetical progression. n r,- . , a- 1> . h - c . c -a . 0. (liven sm 1= — ,, sin w= , , sin ?i — - , prove that a + o u+ c c + a sec^ I +sec'' ni + &ec^ n — 2 sec /.sec lu.sec ?a+ 1. 7. If a, b, c, d are the sides of a quadrilateral taken in order and the angle between its diagonals, the area of the (piadrila- teral is | (, JiC, and is the angle BA C. Piove 00 = A0 taiv 0, 01'= OS tan'- 0, OR = OiJ tan'-' 0. 10. In the preceding example if OP=j>, OQ = q^ AC — c, ex- press p and a in terms of c and 0, vud thus show that S § i w.ii f f\ 228 EXAMINATION PAPERS. ti^ ■! ' ! PAPER VI. 1. Name the limits between which tlie tri-'onometrical ratios respectively lie. 2. In what length of time will the extremity of the minute hand of a clock nro\ e a tlistanc*; e(|ual to its own Icingth 1 A vei's A 3. Prove sec A + tan A ^ tan (1 5' + .', A), and tan ■ = . -- r • - ' 2 sni A V'G - \/2 4. Solve the equation cos"* - sin^ 0. 4 5. If in any triangle a + c h -, then 71 = 120". h+ c a - c 6. Solve the equation tan A + tan 2/1 = tan 3A. 7. Find the area of a pentagon each of whose diagonals is 10 inches. 8. ]f A cos (I + k sin ^^=^1, and / cos + vi sin 0—\, prove that (/ - //)•- + {vt - ky = (M - vikf. 9. If cot is an angle in the third quadrant, find sin c/) and cos . 10. A ring 12 inches in diameter is suspended from a point by six strings, each 10 inches long, attached to points in the circumference at equal intervals ; tind the angle between two consecutive strings and the angle each string makes with the vertical. PAPER VII. 1. In a triangle right angled at C, sin (45"'^^ ^)~ \^\.)'~' 2. Prove sin (A + B) : sin A +&'m B: : sin A - sin 7i : sin {A - B). 3. Prove sin A + cot A +sec A - cos A - tan A - cosec A = (sin A - l)(cot J - l)(sec A - 1). A 2' i T • 1 . /^ lA c + 6 4. In any triangle tan I + B) = tan \ w / c — rical ratios lie minute hi 1 vers A I sill A agonals is = 1, prove quadrant, in a point ts in the ween two with the 11 {A - B). ec A ec^-1). PAPER VIII. 229 5. The hands of a clock are 3 and 2 inches in length ; Hnd the first time after three o'clock when the cxtreuiitit-s of the hands are two inches apart. 6. In any triangle a~ b- c" a- + fr + c" 4 area— -^ -a —~ - - — -. tan ^1 U\\\ Ji tan C cot ^1 -f cot /i + cot C 7. Express the area of a triangle in terms of a side and the two adjacent angles. 8. In any triangle sin (^1 - J>) sin (/i- C) sin (C- A) sin A \. l()").r)2v^in 15" 12' sin 'lA - '* sin Ii\ [sin 'IB - ' sin c\ (sin 2C- ^ \/21.4 3x f.2 7037 9. J^ind log ^^^^^^ and log ^^^^^ .^^, ^^^^.^ a h c 10. If cos X = , — , co^y— , cos z 0+0 c+a then and tau'^ - + tan- "V + tan'- " 2 2 2 rt + 6' 1, X y z AUG tan . tan ^ tan . =tan . tan ^ tan ^. 2 2 2 2 2 2 PAPER VIII. 1. Can an angle be completely determined from the known value of one of its ratios ] Explain clearly. 2. A cube the length of whose edge is a is placed with its diagonal vertical ; find the height of each of the corners above the point of support. 3. If Xy y, z are the perpendiculars from any point within a triangle on the sides a, />, c; then cos 20 ^q cos ^/, obtain an equation independent of 0. 7. In the triangle A BC the straight line joining A to the middle point of BC is at right angles to AC; show that 2 (c- - rt-) 3«c cos A cos C = 8. \i AD, JiE\m drawn bisecting the angles of a triangle, and if r„ Vn are the radii of the circles inscribed in ABD, ABE, and S the area of ABC; then B A (I cos „ - COS ■=)■ 9. Solve the (>) - sin- {0 - c/,) = -^, eosec 2/7 + cos 2 c/) = 10. Prove vs log cos - + log COS - " + log cos \ + .... ad. inf. = log sin x — log x. J 4 fe PAPER IX. 1. Show that when the tangent of half an angle of a triangle is known all the other ratios of that angle may be determined without andjiguitv. « T . • . / ,, sin .V C ., , a-h C 2. In a triangle c = (a + b) ^-— , if tan 0= — — cot — . ° ^ ' cos ^ « + 6 2 3. The distances of the centres of the escribed circles of a triangle from the centre of the inscribed circle are as (n - n cos ^^ : (/•, /•) cos ;y : {r^-r) cos -. a 2 c DS A COS Ji. obtain an ^ A to the that a triani[»lp, JW, A BE, 1 X log X. a triangle eternunecl C cot-, rcles of a C PAPER IX. 231 4. Tn tho ambiguous case a, J), A, being given, if o,, c, are tlie third sides of the two triangles, show that tbe distance be- c< — c tween the centres of their circumscribing circles is ~r — . 2 sni A 5. The sides of a parallelogram are a and h, and tho angle between then) is 0; show that the tangent of the angle between ., ,. , . lah sin its diagonals is — ^ — — — . a- - Ir 6. Eliminate ) = a sin /^ 7. If D, B, Fure the feet of the perpendiculars from A,B,C, upon the opposite sides of the triangle ABC, the diameters of the circumscribing circles of the triangles AEF, BDI\ CDE, are a cot A, h cot 7>', c cot C respectively. 8. If tan (J4-/>' + C)-0, then A^]i + C = n-, and tan (2^1 + 2/)' + 2C) = 0. Hence prove trigonometrically that if ,x' + 2/ + - = .^"2/2, then 2;; ^xyz 2x - + 2.V T,+ Extend the same principle to obtain other similar identities. 9. If V'2/" + y^+~^' \^z- + zx + x-, V X- + xi/ + y' are the sides of a triangle ; then —- (i/z + zx + xij) is the area. 10. Three circles, two of which are equal touch each other, and a fourth circle lies between them touching each. The radius of the equal circles is ?•', that of the third circle is r, and is the angle between the lines joining the centres of the equal circles to to the centre of tho third circle. Show that the radius of the fourth circle is r (/•' + r)sin''^ \ r cos' I - /• sin"- 1 232 EXAMINATION PAPKUS. ttl I- ;;i 1 lli ! Examination Papers in Trigonometry usually furnish the Log- arithms required in the solution of the given problems. We give three such pape'-s, to render the student familiar with the form in which his work will be presented to him. PAPER X. TORONTO UNIVERSITY. 1. I low many digits in the integral part of (2 D)''^1 2. Simplify V 80 x \^2j, \/~^ x (18) s. 3. Sliow tliat the logarithms of the trigononu'tiical ratios need not be entered for angles greater tlrm 4r/\ Illustrate by using the last two values given beluw to find the values for other logarithmic ratios. 4. Adapt s=,iii J tan ^ A to logarithmic computation, and lind its logaritiini w aon A — 53 ' 0'. 5. Given /' - 1 23° 40', 6 =--100, c-GO; find A and C. 6. Given ^ - 1 1 ".i ' 10', h - 213.4, c = 2 13.4 ; solve the triangle. 7. Given «. = 200, b = 77A, C=4l° 50'; find the area. 8. If (sin + cos oy = 3 sin -f- sin 20 ; find in degrees. 9. If 1 + sin ^ = 2 cos \ O(cos, }r - sin I 0) ; find in degrees. 10. A person standing on one bank of a river observes that an object on tiie other bank has an angle of elevation of 45", and on going back 150 feet the corresponding angle is 30°; find the breadth of the river. 11. A vertical rod 10 feet long casts a shadow 7.74 feet long on a horizontal plane; find the sun's altitude. NUMBER. LOO. ANGLE. LOC. 1)1 KK. FOK OO" 20000 301030 tan 52° 15' 10.11110 26 30000 47712 sin 56" 20' 9.92027 41645 61950 siu 29° 57' 30" 9.69842 21340 32919 sill 41^50' 9.82410 17761 24946 sin W 28' 9.52278 51623 71284 tan 26 33' 9.69868 36 cos53'' 6' 9.77845 32 I the Log- We give tie form in icul ratios istrate by i for other ation, and e triangle. ;a. ^rees. n degrees. 38 that an 5°, and on i find the feet lonji F. FOH CO 26 36 32 PAPER XI. 233 PAPER XI. TORONTO UNIVEESITY. 1. Multiply 501.2G by .399. 2. Find what per cent. .;{!);M2 is of 78.492. 3. Simplify { V job X (30)" -} -^5.4. 4. (liven 2^-399; find x 5. Find L sin 24'^ 45' 15", L oosec 24" 45' 50", and tke angle whose L cos is 9.62204. 6. Given « = 589.17, /^= 195.75, C-52^; find A, />', c. 7. Given .4 = 12^ 43', C=90^ and the area=1995; find f 234 EXAMINATION PAPERS. i J! PAPER XII. QUEEN'S, TKINITY AND VICTORIA. 1. The circular measure of the angle C is —> s-ntl ^^ the sum ;^_ 1^ of A, 7>', C is -~ ', the number of grades in the difference of .1 and />' is 40 ; find the number of degrees in each angle. 2. Dethio the trigonometrical functions. Prove (1) cos A + sin ^1 = (1 +.tan A) cos A. ,„, A sin 2A cos A ^ ' 2 1 + cos 2^1 1 + cos ^1 { 3)./^ cot-' f^ ^),if^: siu ' ;<;. 3. Given sin a = .l; find cos f/, tan a, cot '/, sec ^/, sin 2a and cos 2a. 1. Calculate to three decimal places the value of tiie sine of an arc subtended by a chord whoso length is ^ of the diameter. 5, Prove the following identities : Sin. / 'K (1) -T — '■ — 2 cos 1x - 2 cos 4;« - 2 cos G.c= 1. sm X . tan a /sin aV- (2) sin ;' = ^-— — -, if y-, ;] +(C0S a COS ■/)'■= 1. tan j^ sni p -1 ^> <^ -I ^> c., (3) cot ^ + cot - + cot - — cot :; cot cot if r w J w .J ^ w -yl+i>' + a 6. Prove geometrically, (i) cos (.f + 7/) = cos ,»• cos ?/ — sin ;« sin y, when .r + _y> -. (2) cos X +1=2 cos- - . (3) From (1) deduce cos X + cos y ■■=r- 2 cos \ (.<; + y) cos -]- (.7j - y). 7. Make convenient for logaritlunic computation, ■ /i\ 1 i. i .„^ tan.r + tan ■»/ . .. 1 - cos 2a; (1)1- tan X tan ?/. (2) --. (3) - ^ ' '^ ^ ^ cot .f + cot 2/ ^ M + cos 2x PAPEU Xil. 235 of the sum ce of A and c a, sin za the sine of tliiinieler. A+B + a 1 - cos 2x 1 + cos 2x' 8. Prove, using the oidinaiy notation, that in any triangle n + h tan I (A + />') b^tairf{A Ji) (I) (.)sin::.J^-^)^-'^) a~b tan .V ^.i Ji) ' 'J. '\ be A " n c A n c (.5) /> cot , +''cot , 4-rtcot ■ ^c'cot +('cot + /> cot . ' ' ■' •) •) .) .) .) — — -J ^ w M (4) Area ot triangle = - — . L\cosec A + cosec />' + cos<'c (') {')) If in (1), A -A': -00' and JJ^C, describe the triaiighi and iind its angles. 9. Show liow to solve tlui ti'iangh; when (t, A, A an* given, and discuss the aml.'igiiity in full. If A and 8 ho the areas of tlu; two triani^les, prove A ' + 0- - -' A <-os 2 A --= 'j., ( A + 8)-. 10. Deduct; the expressions for the radii of the insci-ibed and circumscribed circles of a triangle whose sides are known. I'rove that the area of the triangUj of which the -sertices ai'(; tin; ]>oints of contact of the inscribed circle is to the arefi of the oiiginal triangle as the radius of the inscribed is to tho dianuiter of tiie circumscribed circle. 11. Find the angles in the following triangles : (1) a^-mP.2, /y=4522, C = 45° 7'. (2) a -343, i = 83;3, y.'-50\ 12. From a station on the top of a hill, three towers on a horizontal plane are found to suljtend equal angles at tlie eye of an observer, and the angles of depression of their bases are a, d', '/. Prove that 7i, p, (j, being the height of the towers, sin (pi -~ J') sin (;' - a) sin {a - []) u sin a + ]) sni p + = 0. q sm y NUMEER. 7000 2000 1190 LOG. ANGLE. LOC DIKK. FOR 60 8450980 3O103O0 0755470 cot 22^ 23' tan 7 ' 59' &in IS' 23' sin 25' cos 25' 10. .38 17047 9. 1468849 9.49S8245 9.6259483 9.9572757 3566 9176 3800 I m APPENDIX. MATHEMATICAL TABLES CONHISTINO OK I. LOCJAlilTHMS OF Nl'MlUOIiS. II. NATUIiAL SINES, COSINES, TANGENTS AND COTANGENTS. III. LOGAVJTIIMS, SINES, COSINES, TANGEN'JS AND COTAN- (JENTS. IV. NUMBERS OE'J'lvV USED IN CALCULATIONS. i| 238 LorjARITHMS. (1 !.* No. liOK. No. 1 JiOK. \o. Lot? \(>. I.o«. , No. LoK. ]()0 00000 115 10137 ' 190 27875 235 37107 ' 280 44716 I 4:{2 (> 4.35 1 2S1()3 6 291 1 871 2 S(iO i 732 '2 3.30 475 2 4r.025 3 01281 8 1702() 3 .55(5 8 (5r.8 3 179 4 7o;} 9 319 4 780 9 840 4 3.32 10.-. 02119 1.-.0 17009 195 29003 240 3802 1 285 45484 w.\\ 1 89S (5 22(5 1 202 (5 637 7 938 2 18184 7 447 2 382 788 8 03342 3 4(59 8 607 3 5(51 8 939 9 743 4 752 9 885 4 739 9 4(5090 110 04139 1.-.5 190.33 200 30103 245 .39917 290 46240 1 532 ('. 312 1 320 (5 094 1 3S!> 2 922 / 590 2 535 7 270 2 538 3 05308 8 SIK) 3 750 8 445 3 687 4 090 9 20140 4 9(i;{ 9 620 4 835 11.-. 0G(»70 11.0 20412 205 31175 250 397!)4 295 46982 44(5 1 08.3 (5 3S7 1 9(57 6 47129 7 819 2 951 t .597 2 40140 7 276 8 07 188 3 21219 8 . 80(5 3 312 8 422 9 555 4 484 9 32015 4 483 9 567 120 07918 105 2174S 210 32222 255 40654 300 47712 1 08279 () 22011 1 428 (5 824 1 857 2 r)3() 7 272 2 6.34 7 993 2 48001 3 991 8 531 3 8.38 8 411(52 3 144 4 09342 9 789 4 33041 9 3.30 4 287 12.5 09G91 170 23045 215 33244 260 41407 305 48430 6 10037 1 300 44(5 1 664 6 572 7 380 2 553 646 2 830 1 714 8 721 3 805 8 84(5 3 996 8 855 9 11059 4 24055 9 34044 4 42160 9 996 mo 11.394 175 24304 220 34242 2(i.i 42325 310 491,36 1 727 6 551 1 439 6 488 1 276 o 12057 797 2 635 7 (551 2 415 3 385 8 25042 3 K30 s 813 3 554 4 710 9 285 4 35025 9 975 4 693 135 13033 180 2552/ 225 3,-.218 270 43136 315 4983 1 a 354 1 768 6 411 1 297 (5 9(59 7 072 2 26007 (502 457 / 50106 8 9S8 3 245 8 793 3 616 8 243 9 11.301 4 48'i 9 984 4 775 9 379 140 14013 185 26717 230 36173 275 43933 320 50515 I 922 9:.l 1 301 (5 44091 1 651 2 15229 7 27184 2 549 7 248 2 786 3 534 8 416 3 736 8 404 3 920 4 836 9 640 4 922 9 5(50 1 4 510,-.5 LOGARITHMS. 239 i No. LoK. J 'JSO 447 K) [ 1 871 ) 2 4r.()2r) < .'{ 171) ) 4 :{;{2 L'S,') 4r.4S4 I () 037 1 ^ 788 S n3!> 1 !) 4(i(»!)0 'JOO 40240 [ 1 380 • > 2 .^)38 » :\ 087 \ 4 835 t 'J'l") 40982 G 47129 1 7 276 8 422 {) 507 ;m() 47712 1 8.-.7 2 8 48001 144 4 287 :?or) 484.']0 572 If 714 8 85;') i) 990 :iio 491.30 I 270 2 415 3 554 4 093 .".1.5 49831 () 9(59 / •WlOO 8 243 9 379 320 50515 1 051 2 3 786 920 4 51055 No. 1 Log. No. j Log. No. LoK. No. liOK. No. LoK. 70329 325 51188 .370 50S20 415 (il8()5 4(50 (5(5270 505 (i 322 1 !t37 iK)9 1 370 (5 415 7 455 2 5V054 7 02014 2 4(54 501 8 .587 3 171 8 118 3 558 8 5S0 9 720 5 287 \) 221 4 (552 (572 3.30 51851 375 57403 420 02325 405 (50745 510 70757 1 9.S3 (i 519 1 428 839 1 842 2 .52114 7 ((34 2 531 7 9.32 2 927 3 244 H 749 3 034 8 (57025 .3 71012 4 375 9 804 4 737 9 117 4 090 335 52504 .380 57978 425 (i2839 470 (57210 515 71181 034 1 58093 941 1 302 (5 205 7 703 2 200 / (i.3043 2 394 349 8 892 3 320 8 141 3 48(5 8 43.3 9 53020 4 433 9 240 4 578 9 517 310 .".3118 3S5 5S540 430 03.347 475 (57(5(59 520 71000 1 275 () 05! > 1 448 (5 701 1 084 2 103 7 771 •2 548 t H.52 2 707 3 529 8 8S.3 .3 049 8 !t43 .3 850 4 050 9 995 4 749 9 080.34 4 933 .345 5.3782 390 .59100 4.35 (5.3840 480 08124 525 72016 !t08 1 218 949 1 215 (5 099 7 54033 2 329 04048 2 305 / 181 8 1.-.8 3 4.39 8 147 3 395 8 2()3 9 283 4 .550 9 240 4 485 9 340 .350 54407 395 59000 440 64345 485 (58574 530 72428 1 .531 770 1 444 6 (504 1 509 2 054 7 879 'J 542 7 753 2 591 3 777 8 98S 3 (540 8 842 3 673 4 900 9 ()0097 4 738 9 931 4 754 355 55023 4(M) 00200 445 64830 490 (59020 535 728.35 145 1 314 933 1 108 (5 910 207 2 423 (55031 2 197 t 997 8 388 ■ 3 530 8 :28 3 285 8 73078 9 509 4 038 9 225 4 373 9 1.59 300 ,55030 405 00746 4.-)0 (55321 495 09401 .5-10 73239 1 751 853 1 418 (5 548 1 320 2 871 7 359 2 514 r* 1 (530 '2 400 3 991 8 61000 3 010 8 723 ,3 480 4 50110 9 172 4 700 9 810 4 5(50 305 56229 410 01278 4.')5 65801 5(K) (39897 545 73040 6 348 1 384 800 1 984 (5 719 7 407 2 490 7 992 2 70070 ^ / 799 8 .585 3 5!)5 8 00087 3 1.57 8 878 9 703 4 700 9 181 4 243 9 957 ■ J LOGARITHMS. I /"I No. log. No. liOg. No. l^OK. No. Log. No. Log. 550 74036 595 77452 640 80618 685 83569 730 863:52 1 115 6 525 1 686 6 632 1 392 2 194 7 597 2 754 ()})6 2 451 3 273 8 670 3 821 8 759 3 510 4 351 9 743 4 889 9 822 4 570 555 74429 600 77815 645 8(>956 690 83885 735 86629 507 1 887 6 81023 1 948 6 688 586 2 960 7 090 2 84011 7 747 8 663 3 78032 8 158 3 073 8 806 9 741 4 104 9 224 4 136 9 864 560 74819 605 78176 650 S1291 695 84198 740 86923 1 896 6 247 1 358 6 261 1 982 2 974 7 319 2 425 t 323 2 87040 3 75051 8 390 3 491 8 385 3 099 4 128 9 462 4 558 9 448 4 157 565 75205 olO 78533 655 81624 700 84510 745 87216 6 281 1 604 6 690 1 572 6 274 358 2 675 757 2 6;u t 332 8 435 3 746 8 823 3 691) 8 390 9 511 4 817 9 889 4 767 9 448 570 75587 615 78888 660 81954 705 84819 750 87506 1 664 6 958 1 82020 () 8^0 1 564 2 740 7 79029 2 086 / 942 2 622 3 815 8 099 3 151 8 85003 3 680 4 891 9 169 4 217 9 065 4 737 575 75967 620 79239 665 82282 710 85126 ^ -. ■. 87795 6 76042 1 309 6 347 1 187 6 852 7 118 2 379 413 248 910 8 193 3 449 8 47s 3 309 8 9()7 9 268 4 51S 9 543 4 370 9 88024 580 76343 625 79588 670 82()07 715 85431 7«i0 88081 1 418 6 657 1 672 6 491 1 138 2 492 7 727 2 737 552 2 196 3 567 8 796 3 802 8 612 3 252 4 641 9 865 4 866 9 673 4 309 585 76716 6:^0 79934 675 82930 720 85733 765 88366 6 790 1 80003 6 995 1 794 6 423 7 864 2 072 ^ 1 83059 2 854 / 480 8 938 3 140 8 123 3 914 8 536 9 77012 4 209 9 187 4 974 9 593 590 77085 635 80277 680 83251 725 86034 770 8S649 1 159 6 346 1 315 6 094 1 705 2 232 t 414 2 378 f- t 153 3 762 3 305 8 482 .3 442 8 213 .3 818 4 379 9 550 4 506 9 273 4 874 LOrJARITHMS. 241 LoK. -If) I r)io r)70 86029 088 747 8() 9S6 1 434 6 752 1 952 6 010 89042 2 487 4 802 2 999 t 091 8 098 3 540 8 852 3 9(5047 8 137 9 154 4 593 9 9'.; 2 4 095 9 182 780 802O9 825 91045 870 93952 915 96142 900 9S227 1 205 () 09S 1 94002 6 190 1 272 2 321 751 2 052 / 2.S7 2 318 3 370 8 803 3 101 8 284 3 363 4 432 9 855 4 151 9 332 4 408 785 89487 830 91908 875 94201 920 9()379 905 98453 6 542 1 900 6 250 1 420 6 498 597 2 92012 7 300 2 473 M 1 543 8 053 3 065 8 349 3 520 8 588 9 708 4 117 9 399 4 567 9 6;>2 790 89703 835 92109 880 94448 925 96014 970 98677 1 818 6 221 1 498 6 001 1 722 2 873 i 273 2 547 i 708 2 71)7 3 927 8 324 3 596 8 754 .3 811 4 982 9 376 4 045 9 802 4 856 795 90037 840 92428 885 94694 930 90848 975 98900 6 091 1 480 6 743 1 8!»5 945 7 140 2 531 4 792 2 941 989 8 200 3 583 8 841 3 988 8 99034 9 255 4 634 9 890 4 97035 9 078 800 903O9 845 92686 890 919.39 935 97081 980 99123 1 303 6 737 1 988 6 128 1 1(57 2 417 788 2 95036 i 174 2 211 3 472 8 840 3 0S5 8 220 3 255 4 52() 9 891 4 134 9 207 4 300 805 90580 850 92942 895 95 182 940 97313 985 99344 634 1 993 6 231 ' 1 359 (5 388 687 2 93044 1 279 ! 2 405 / 432 8 741 3 095 8 328 3 451 8 47(5 9 795 4 146 9 376 5 497 9 520 810 90849 855 93197 900 95424 945 97543 990 995(54 1 902 () 247 1 472 6 589 1 (»(t7 2 956 7 298 2 521 7 6.35 2 051 3 91009 8 319 3 569 8 681 3 (595 4 062 9 399 4 617 9 727 4 739 815 91116 800 93150 905 95665 950 97772 995 99782 6 169 1 5(K) () "l.i 1 818 (> 82(5 222 2 551 761 3 8(54 7 870 8 275 3 601 8 809 3 909 8 913 9 328 4 651 9 856 4 955 9 957 I I i 3 1 ! TRIGONOMETRIC FUNCTIONS. Angle. 8lll(>H. C'ohSim-k. TiiiiK<>iilN. I'otangciilN. Nat. Log. 1 Nivt. Log. Nat. Log. Xut. I'OK. 0^ 00000 - cc 1.0000 10.00000 O00(t0 - cc X + « 10' 291 7.40373 l.(o;)0 10.00000 291 7.46373 343.77 2.53627 20 582 7.70475 99998 9.99999 582 7.76476 171.89 2.23524 30 873 7.94084 996 998 873 7.940S6 114.69 2.0.5914 40 01104 8.0(5578 993 997 011(i4 8.06581 85.940 1.9.3419 50 464 8.16208 989 995 455 8.16273 08.750 1.83727 r 01745 8.24186 99985 9.99993 01746 8.24192 57.290 1.75808 10' 0'20;?0 8.30879 979 991 02036 8.30888 49.104 1.(50112 20 327 8.36678 973 988 328 8.36()89 42.904 1.63311 30 613 8.41792 9G(1 985 619 8.41807 38.188 ' '58193 40 908 8.46366 9:8 982 910 8.46385 34.;j(!8 .5.3615 50 03199 8.50504 919 978 03-201 8.50527 31.242 1.49473 03490 8.54282 99939 9.99974 03492 8. .54.308 28.030 1.4,')692 10' 781 8.57757 929 969 7S3 8.57788 20.432 1.42212 20 04071 8.60973 917 964 P4075 8.61009 24.. 54 2 1.. 38991 ;i0 362 8.63968 905 959 3(i6 8.64009 22.904 1.35991 40 653 8.66769 892 953 658 8.66816 21.470 1.. 33 184 50 943 8.69400 878 947 919 8.69l.-)3 20.206 1.30547 3^ 05234 8 71 880 P9S63 9.99940 05241 8.71940 10.081 1.28060 10' 624 8.74226 847 9:u 533 8.74292 18.075 1.25708 20 814 8.76451 831 926 824 8.76.")25 17. 109 1.2.3475 30 06105 8.78568 813 919 00116 8.78649 16.350 1 21351 40 395 8.80585 705 911 408 8.80674 15.005 1.19.326 50 035 8.82513 776 903 700 8.82610 14.924 1.17390 r 06970 8.84358 997r)6 9.99894 OG993 8.84464 14.301 l.l,-)5.36 10' 07266 8.86128 736 885 07285 8.86243 13.727 1.1.3757 20 650 8.87829 714 876 678 8.87953 13.197 1.12047 30 846 8.89464 692 866 870 8.89598 12.706 1.10402 40 08136 8.91040 608 856 08163 8.91185 12.251 1.08815 60 420 8.92561 644 845 456 8.92716 11.826 1.07284 6" 08710 8.94030 !HM)19 9.998.34 08749 8.94195 11.430 1.0.'-)805 10' (liK)05 8.95450 594 82.3 tiH)42 8.95627 11.059 1.04.373 20 296 8.96825 567 812 335 8.97013 10.712 1.02987 30 585 8.98157 540 800 029 8.98358 10.385 1.01(542 40 874 8.99450 511 787 923 8.99662 10.078 1.0033S 50 10104 9.00704 482 775 10216 9.009.30 9.7882 0.99070 «° 10463 9.01923 99452 9.99761 10510 9.02 h;2 9.5144 0.978.38 10' 742 9.03109 421 748 b(»5 9.03361 9.2553 0.966.39 20 11031 9.04262 390 734 11099 9.045:?8 9.0098 0.95472 30 320 9.053S6 357 720 394 9.05666 8.77C9 0.94334 40 609 9.064H1 324 705 688 9.0(5775 8.55.55 0.93225 60 898 9.07548 290 6S0 983 9.07858 8.3450 0.92142 TllinONOMKTlUC FUNCTIONS. 243 1.75808 l.()01I2 .63,311 .49473 AiiiCl<'. Hi IK'S. t'OKilK'S. Taiigciils. ('otaugonts. Xiit. Log. Nat. Log. Nat. Log. Nat. Log. R o 121S7 0.0S589 99255 9.99075 12278 9.08914 8.1443 0.91086 10' 47(1 9.00000 219 059 574 9.09947 7.9530 0.90053 20 7G4 9. 10599 182 643 809 9.10956 7.7704 0.89044 80 isonj 9.11570 144 627 13105 9.11943 7.5.9.58 0.88057 40 341 9.12519 100 610 401 9 12909 7.4287 0.87091 50 629 9.13447 007 593 758 9.13854 7.2087 0.86146 8° 13017 9.1435() 119027 9.99575 14054 9.14780 7.1154 0.85220 10' 14-2()5 9.15245 9S9S6 557 351 9.15688 0.9082 0.84312 20 493 9.1611(i 944 539 648 9.16577 0.8269 0.83423 30 781 970 902 520 945 9.17450 0.6912 0.82550 40 li'OiW 9.17807 858 501 15213 9.18306 0.5006 0.81694 50 SfiO 9.18028 814 482 540 9.19146 0.4348 80854 0° 15G43 9.19433 987C9 9.99462 158S8 9.19971 0.3138 0.80029 10' mi 9.20223 723 441- 10137 9.20782 0.1970 0.79218 20 1()'218 99i) 670 421 435 9.21578 6.0844 0.78422 30 505 9.21 7t)l 029 401 734 9.22361 5.9758 0.77639 40 792 9.22509 560 379 17033 9.23130 5.8708 0.76870 50 17078 9.23244 531 357 333 887 5.7094 113 ur 17;i(".r. 9.239G7 9S481 9.99335 17C33 9.24632 5.6713 0.75368 10' (ir.i 9.24G77 430 313 933 9.253(15 5.5764 0.74635 20 !)37 9.25370 378 290 18233 9.26086 5.4845 0.73914 30 18-2-24 9.20063 325 267 534 797 5.39.55 203 40 509 739 272 243 835 9.27496 5.2093 0.72504 50 795 9.27405 218 219 19136 9.28186 5.22.57 0.71814 ir 19081 9.28060 98103 9.99195 19J38 9.28865 .5.1446 0.71135 10' 300 705 107 170 740 9.29535 5.05.58 0.70465 20 652 9.29340 050 145 20O12 9.30195 4.9894 0.69805 30 937 906 97992 120 345 846 4.91,52 154 40 202-22 9.30582 934 093 (148 9.31489 4.8430 0.68511 50 507 9.31189 875 067 952 9.32122 4.77-29 0.67878 ii»° •20791 9.3178S 97815 9.99040 '21250 9 32747 4.7040 67253 10' 21070 9.32378 754 013 600 9.3.3365 4.0382 0.66635 20 oOO 9(J0 092 9.98986 804 974 4.5736 026 30 044 9.33534 030 958 22109 9.34576 4.5107 0.65424 40 9:8 9.34100 .500 930 475 9.35170 4 4494 0.64830 50 212 658 602 9(11 781 757 4.3897 243 i.r 22495 9.35209 97437 9.98S72 •23087 9.36336 4.3315 0.6S664 10' 7V8 752 371 843 393 909 4.2747 091 20 23002 9.36289 304 813 700 9.37476 4.2193 0.62524 30 345 819 237 783 24008 9.38035 4.16.53 0.61965 40 C27 9.37341 169 753 316 589 4 11-28 411 50 910 858 100 722 024 9.39136 4.0011 0.60864 If l1 I ; .!' r, M » 244 TRKJONOMETUIC FTTNCTIONS. Ingle. SilM-M, <'o.siii<>M. Tangents. <'otangenfH. i Nat. LoK. Nat. Log. Nat. Lor. Nat. Log. 14° 24192 9.38308 97030 9.98690 249.33 9.39677 4.0108 0.60.323 10' 474 871 96959 659 25242 9.40212 3.9617 0.59788 20 75(1 0.39369 887 627 552 742 3.9]:{0 258 :n) 2r.o:!s 8()0 815 594 862 9.41266 3.S667 0.58734 40 3'2(» 9.4o;uo 742 561 26172 7S4 3.8208 216 50 col 825 607 528 483 9.42297 3.7700 0.57703 15° 258S2 9.41.300 96.'>93 9.98494 26795 9.42805 3.7321 0.57195 10' 26103 768 517 460 27107 9.43.308 3.0S91 0.56692 20 443 9.42232 440 426 419 806 3.6470 194 80 724 ()90 363 391 732 9.44299 3.6059 0.55701 40 270(14 9.43143 285 356 28046 787 3.5056 213 50 284 591 206 320 3G0 9.45271 3.5261 0.54729 16° 27564 9.44034 96126 9.98284 ■2:675 9.45750 ;;.4874 0.54250 10' 843 472 046 248 .»s)o 9.46224 3.4495 0.53776 20 28123 905 05964 211 29305 694 3.4124 306 30 402 9.45334 882 174 (i2i 9.47160 3.3759 0.52840 40 CSO 758 799 1.36 938 622 ,3.3402 378 50 059 9.46178 715 098 S0255 9.48080 3.3052 0.51920 17° 20237 9.46594 95030 9.98060 30573 9.485.34 3.2709 0.51466 10' 515 9.47005 545 021 ,sitl 984 3.2371 016 20 793 411 459 9.979S2 31210 9.494.30 3.2011 0.50570 30 30071 814 372 942 530 872 .3.1716 128 40 348 9.48213 284 902 850 9.50311 3.1397 0.40689 50 625 607 195 861 32171 746 3.1084 254 18° 30902 9.48998 95106 9.97821 32492 0.51178 3.0777 0.48822 10' 31178 9.49385 015 779 814 606 3.0475 394 20 454 768 94924 738 33136 9..V2031 3.0178 0.47969 30 730 9.50148 832 696 460 452 2.9887 548 40 32006 523 740 653 783 870 2.9600 130 50 282 896 6 It) 610 34108 9.53285 2.9319 0.40715 10° 32557 9.51264 945.52 9.97567 34433 9.53607 2.9042 0.46.303 10' 832 ()29 457 523 758 9.54106 2.8770 0.45894 20 33106 991 361 479 35085 512 2.8.502 488 30 381 9.52350 264 4.35 412 915 2.8239 085 40 655 705 167 390 740 9.5.")315 2.79S0 0.44685 50 929 9.53056 068 344 3C068 712 2.7725 288 20° 24202 9.53405 9:i'.m9 9.97299 .30397 9.5.i;i07 2.7475 0. -13893 10' 475 751 s<;9 252 727 498 2.7228 502 20 748 9.54093 709 200 .37057 887 2.1.985 113 30 35021 433 «i67 159 .388 9.57274 2.6746 0.42726 40 293 769 .565 111 720 65S 2.6511 342 50 565 9.55102 462 0()3 .38053 9.58039 2.1)279 0.41961 i< i TKIC ONOMETlilO F U xN CTIONS. 245 Siii4>K. I'OsilK'S. Taiiu;«>iilK. 4'olaiiui>iilH. 1 AiikIc. Nat. ho^. Xiit. Nat. Lot,'. Nal.. I. OK. L.i^'. 21" ^5S;{7 9.55133 03358 9.97015 38r!8(i 9.58418 2.0051 0.41582 " 10' •Mh)S 7t)I 253 9.!)69»)() 721 794 2.5820 20(5 20 3/0 9.5608.') 148 917 30055 9.59168 2.. 5005 0.40832 SO Gf.O 407 042 808 301 540 2.5380 460 40 0-Jl 727 02035 818 727 909 2.5172 091 50 371'Jl 9.57044 827 767 40005 9.60276 2. 1'JOO 0.. 39724 374r.l 9.57358 :»27IS 9.90717 40103 9.60641 2.4751 ()..39.3.')9 10' 7;!0 0()9 000 065 741 9.61004 2.4545 0.38996 20 379! »!) 978 400 014 410SI 364 2.4312 63(5 30 3S2(;S 9.58284 388 562 421 722 2.4142 278 40 537 588 270 509 703 9.62079 2.3045 0.37921 50 805 889 1C4 456 42105 433 2.3750 5(57 2r 39073 9.59188 02050 9.9()403 42447 9 627S.- 2.3559 C. 372 1 5 10' 341 484 01930 349 701 9. 031 .35 2.3.309 0. .368(55 20 f.OS 778 S22 294 43130 484 2.3183 516 .SO 875 9.00070 700 240 481 8.30 2.2998 170 40 40141 359 500 185 8.8 9.64175 •2.2S17 0.35825 50 403 G4G 472 I'iO 44175 517 2.2037 483 2r 40r)74 9.00931 '.)i;>55 9.96073 44523 9.64S.")8 2.2460 0. .35 1 42 10' !)3'J 9.01214 230 017 872 9.65197 2.2286 0.34S03 20 41'204 494 110 9.95960 45222 535 2.2113 465 30 4 OS) 773 OOOJO 902 573 1 870 2.1913 1.30 40 734 9.62049 875 844 024 9.66204 2.1775 0.33796 50 908 323 753 786 40277 537 2.10O9 463 25' 42262 9.62595 IMKiSl 9.95728 40031 9.66867 2.1445 0.331.33 10' 525 8(55 .507 668 085 9.07196 2.12S3 0.32804 20 42788 9.631.33 383 609 47.341 524 2.1123 476 30 43(r,i 398 250 549 008 850 2.0905 i.-,o 40 313 602 133 488 4^0^.5 9.68174 2.0X10 0.31826 50 575 924 OO007 427 411 497 2.U055 503 2«" 43S37 9.64184 89870 9.95366 48773 9.6S818 2.0.503 0.31182 10' 41008 442 752 304 40134 9.69138 2.0353 0.. 308(52 20 359 098 023 242 495 457 2.0204 543 30 C20 9.-)3 403 179 858 774 2.0057 '}'}i) 40 8s0 9.6,")2n5 303 116 f.0222 9.70089 1.9912 0.2991 1 50 45140 450 232 052 587 404 1.9768 596 27° 45300 9.65705 80101 9.94988 50593 9.70717 1.9020 0.29283 " 10' 058 9.')2 S8008 923 51319 9.71028 1.94,86 0.28972 20 017 9.66197 835 858 088 339 1.9347 661 30 40175 441 701 79.3 52057 (548 1.9210 352 40 433 682 500 727 427 955 1.9074 045 50 C9J 922 431 660 798 9.72262 1.8940 0.27738 246 TRIGONOMETRIC FUNCTIONS. Augle. Nat. Log. 4'ox>iiu-.s. Nfit. Log. Ta Nat. Log. 4'otauKt;iilH. Nat. Log. 28" 40017 9.G7161 S8205 9.y4;)93 53171 9.72567 1.8807 0.27433 10' 47J0t 3!);-; 158 526 645 872 1.8676 128 20 460 (3.-13 020 458 9-20 9.73175 1.8540 0.26825 30 716 86() 87882 390 54296 476 1.8418 524 40 971 9.(J80!»S 743 321 673 777 1.8291 223 50 48226 328 (J(J3 252 5:i051 9.74077 1.8165 0.25923 20° 484S1 9.()So57 87462 9.94182 55431 9.74.375 1.8040 0.25625 10' 735 7S4 321 112 812 673 1.7917 327 20 989 9.69010 178 041 56194 9()9 1.7796 031 30 40212 234 036 9.93970 577 9.75264 1.7675 0.24736 40 495 450 86892 898 962 558 1.7556 442 50 748 677 748 826 57348 852 1.7447 148 30° 50000 9.69897 S6'J03 9.93753 57735 9.76144 1.7321 0.23856 10' 252 Q.70115 457 680 5S124 435 1.7205 565 20 603 332 310 606 513 725 1.7C90 274 80 754 547 163 532 905 9.77015 1.6977 0.22985 40 5100i 761 015 457 50297 303 1.6804 fc97 50 254 973 85866 382 691 691 1.6753 409 31' 51504 9.71184 85717 9.93307 60086 9.77877 16643 0.22123 10' 753 393 567 230 483 9.78163 1.0534 0.21837 20 52C02 602 416 151 881 448 1.6426 552 30 250 809 261 077 61280 732 1.6319 268 40 408 9.72014 112 9.92999 681 9.79015 1.6212 0.209S5 50 745 218 84959 921 62083 297 1.6107 703 32° 52992 9.72421 84805 9.92842 62487 9.79579 1.6003 0.20421 10' 53238 623 650 763 802 860 1.5900 140 20 484 823 495 683 63209 9.80140 1.5798 0.19860 30 730 9.73022 339 603 707 419 1.C697 581 40 975 219 1S2 522 64117 697 1.5597 303 50 52220 416 025 441 528 975 1.5497 025 33° 54164 9.73611 83867 9.92359 64041 9.81252 1.5399 0.18748 10' 708 805 708 277 65355 528 1.5301 472 20 951 997 549 194 771 803 1.5204 197 30 65104 9.74189 389 HI 66189 9.82078 1.5108 0.17922 40 436 379 228 027 608 3.-)2 1..5013 618 50 678 568 066 9.91942 67028 626 1.4919 374 34' 55019 9.74756 82004 9.91857 67151 9.82899 1.4826 0.17101 10' 66160 943 741 772 875 9.83171 1.4733 0.16829 20 401 9.75128 577 686 68301 442 1.4641 558 30 641 313 413 599 723 713 1.4550 287 40 880 496 248 512 69157 984 1.4460 016 50 57119 678 032 425 58S 9.84254 1.4370 0.15746 iaiiKt'iiiH. Loy. ■ 0.27433 i 128 ; 0.2082.") I 524 223 . 0.25923 I 0.25625 327 ; 031 0.247.30 - 442 148 0.23856 505 274 0.22985 t97 409 0.22123 0.21837 552 208 0.209S5 703 0.20421 140 0.19800 581 303 025 0.18748 472 197 0.17922 648 374 0.17101 0. 10829 558 2S7 016 n. 15746 TKIGONOMETKIG FUNCTIONS. 247 /illglO. Muv%, (OMliirM. Taiiitnils. 4'olniiK<'iits. 1 Nat. Lot,'. Nat. LoK. Nat. TiOK. Nat. 1-oK. 35° 573.'')8 9.75859 81915 9.91.1'?6 70021 9.84523 1.J281 0.154/7 10' 696 9.76039 748 248 455 791 1.4193 209 20 833 218 580 158 891 9.85059 1.4106 0.14941 30 5S070 395 412 009 71329 327 1.4019 673 49 307 572 242 9.90978 769 594 1.3934 406 50 543 747 072 887 72211 860 1.3848 140 8fi' 08779 9.76922 8(i|J02 9.90796 72654 9.86126 1.3764 0.1.3874 10' ;VJ014 9.77095 730 704 73100 392 1.36i0 608 20 248 268 55S 611 517 656 1.3597 344 30 482 439 383 618 996 921 1.3514 079 40 716 609 212 424 74447 9.87185 1.3432 0.12815 50 049 778 038 330 900 448 1.3351 0. 12552 37- t)01S2 9.77946 79SC4 9.90235 75355 9.87711 1.3270 0.12289 10' 414 9.78113 688 139 812 974 1.3190 026 20 645 280 612 043 76272 9.88230 1.3111 0.11764 30 876 445 335 9.89947 733 498 1.3032 502 40 01107 609 158 849 77190 759 1.2954 241 50 337 772 789S0 752 fCl 9.89020 1.2876 0. 10980 3S' C1566 9.78934 78801 9.89653 78129 9.S9281 1.2799 0.10719 10' 705 9.79095 622 554 598 51 1 1.2723 459 20 ('.•2024 250 442 455 70070 801 1.2647 199 30 251 415 261 354 544 9.90061 1.2572 0.09939 40 479 573 079 254 S0020 320 1.2497 680 50 706 731 77897 152 498 578 1.2423 422 31)° 02032 9.79887 77715 9.89050 80978 9.90837 1.2349 0.09163 10' 03158 9.80043 531 9.88948 81401 9.91C95 1.2270 0.08905 20 383 197 347 844 940 353 1.2203 647 30 608 351 162 741 82434 610 1.2131 390 40 832 504 76977 636 923 868 1.2059 132 50 61050 656 791 531 83415 9.92125 1.1988 0.07875 40° 04279 9.80807 76004 9.88425 83910 9.92,381 1.1918 0.07619 10' 501 957 417 319 84407 638 1.1847 302 20 723 9.81106 229 212 906 894 1.1778 106 30 945 254 041 105 85408 9.93150 1.1708 0.06850 40 65166 402 75851 9.87996 912 406 I.IOJO 594 50 386 549 661 887 86419 601 1.1571 339 iV 65006 9.81694 75471 9.87778 86029 9.9.3916 1.1504 0.06084 10' 825 839 280 668 87441 9.94171 1.14.30 0.05829 20 66044 983 088 557 955 426 1.1309 574 30 262 9.82126 74890 446 88473 681 1.1303 319 40 480 209 703 334 992 935 1.1237 065 £0 697 410 509 221 89515 9 95190 1.1171 0.(.»4810 ii '' 1 'Mil \ 248 TUKJONOMETUIC FUNCTIONS. 1 8IU(>it. I'OHlllfH. TaiiK<'i>l>'> 4'OtHUK<'lllH. Anglo. — — Nat. LoK. Nat. I. OK. Nat. Loj,'. Nat. LoK- 42' (i()Oi:{ 9,82.")r)l 71314 9.87107 !)(H»to 9.95444 1.1 IOC 0.04.K-)0 10' eriJi) ()!>1 120 9.80993 5(9 098 1.1011 302 20 314 S30 73924 879 91090 952 1.C977 0.04048 .SO 55!) 9(J8 728 703 (J33 9.90205 1.0913 0.03795 40 773 9.«.Hl(Hj 531 047 92170 459 1.08.'-jO 541 60 US7 242 333 530 709 712 1.0780 288 4;r (W.'oo 9.S.S.378 73135 9.8(»n3 93252 9.909(50 1.0721 0.0.3034 10' 112 5i:i 72937 29.-) 797 9.97219 1 ()(it!l 0.02781 '20 C24 ()48 737 170 91315 472 1.0599 528 30 835 781 537 050 896 725 l.05:;8 275 40 (ioow 914 337 9.S5930 95451 978 1.0177 022 50 •^56 9.84046 130 815 9i;ii08 9.98231 1.0410 0.01709 44 e94(i(5 9.84177 71934 9.8.)(i93 9(;5(i9 9.98484 1.0355 0.01510 10' 675 308 732 571 97i:}3 737 1.0295 203 20 SS3 437 529 448 70(» 989 1.02:t5 01 1 30 70(»J1 50() 325 324 98270 9.99242 1.0170 0.0()758 40 •29S 094 121 200 843 495 1.0117 505 50 605 822 7 541 288 O.O.SO.'U 0.02781 528 275 022 0.01709 0.01510 26:i Oil 0.0(1758 505 253 0.00000 |L0(1ARITII. 4071409 0220880 10524000 10036071 2485749 19042997 .342945 t)377843 •581226 >855716 ^982863 ANSWERS AND RESULTS. EXERCISE I. (I'AOK 13.) 1. A/i and y;/), AC and CD, AE and ED, etc. 2. AHy ACy AE; DJi and DC each negative, /> A' positive. 7. V13, V13, V13, -^13, V3I, 4, 3, 0. 11. (1) 1st and 4th. (2) 1st and 2nd. (3) 1st and 3rd. EXERCISE II. (PAfJK 21.) 3. {\n-\- \) right angles. 4. .2587409, 1.413906. ..371483. 5. 6" 51' 4.5", 111° 0' 19".944, 271^ 7' 30", 4' 13". 7« 62^ 50", 123K 45' 6", 301« 25\ 7' 50\ 0. 83"^ 8' 15", 173° 8' 15"; - (21^ C 19".944), 08° 53' 40".056: - (181° 7' 30"), - (91° 7' 30") ; 89° 55' 47", 179° 55' 47", 7. 00°, 105^ 12. 9°, 90«. 8. 00°, 1331K. 9. 10. 13. 25^ 27°; 05° 03°. 10. 4.^°. 14. 14,"i minutes after 11, or 20 minutes before 12. ■1 1 li 10. 15' 17. 30' 50".734. 19. Sept. 2nd, p.m. 11" 13"' 13.97' 21. 37.1°. 22. 21 inches. 18. 1170°, OOOO'^ 20. 120°, 422.4. 23. 45° 60° 75^ 24. A = 75°, Ji = 81°, C=r 105°, /; = 99°. 27. A = 00°, Ji = 80^ (7 = 40^ EXERCISE III. (Paok 31.) 1. (I) (2) r (3) (6) 17 203?: 04,800,000 144 (7) 758.325;: 27000,000' 180 (5) 1. (8) I Ml 250 PI.ANE TllKJONOMETHY. 2. (1) 180\ (2) m\ (:]) 180". (4) ISO" \^oo (r>) 20-. .'5. 1. 5. G. 9. (0) ..-. (7) 117" r ;u;". (s) loi o;;'ir.8. . TT" (1) 2.01« ft. (2) 13.09 ft. (3) .003515 ft. (I) 9.92103. (1) ll' = 4.7740. (2) 572.957 ft. (3) 2.715 ft. (4) 10 ft. (1)3 '.1831. (2)8".594. (3)343.774.(4)30'. 229". is:}, 254«.r)48, 4'-. S. () liours. 3;: 5;: 9;: (m 2) ;: _ 7;r 1 T'lT' To' " n '' ^^' ^' •^-^^^- ^^- 36- ^^r 14. 30 \ OO', 90". 15. 10, 24. 17. S2.S73 miles. IS. 400:1. 19. 473:489. 1800. " • 19-+ 1800' 21. iT(( -kI nc a + h + c «' + />+ c a + 6 + c* 00 22. ,,-9°.118. 23. "^ *>4 '^ '>5 2- -^^- 0- Tzhc 6c + ca + a6 20. 7')" 27 j:"'^. EXERCISE IV. (Pagk 40.) 1. sin ^ = COS H =^ f„ COS A = sin 7/ = i, tan ^ = cot /? = f . cot ^I — tan yy -- fi, sec yl = cosec Ji=% cosec yl == sec Ji --^ f;. •J i 4 * 2 •>. K. — ;i, -A, .1. 5) 4. (l)i%, ];}. (2),% -Hor.|,]s. •^'. (l)-^o- (^)l- (•^)V'^- (^) -72- ^•-^^- (^'^-^• (7) 1 (8) 1 0. ^-, GO", 30". 7. 2, 337.5. \/2' '"' V2' 8. (1) M, (2) H, (3) fk, (4) - A, (5) ,% (0) §?. (7) 5E?, (8) M^, ^)) 4^, (10) I'V, (11)- V^, (12) ii. 9, See ans. to Ex. 1 of thi^ exercise. 10. See Art, 50, )) 20-. ".8. ' '■ j.02io:J. 1) 10 ft. 4) 30". tniles. iOO' 118. . 3 12 b 13- |6) -1. 337.5. rt. 50, ANSWERS AND IIKSUI.TS. EXERCISE V. (Paok 42.) sine. cosino. tan. cot. 2v/2 251 SCO. cosec. (•'^) - jj— , J> -V-* oT/.r* 1 -t 8 • '-' r. 2S •• • 4 3, 3 2V2 ■ mr vr liiini »r - tr VI ■\- It' in + tr linn V/3, («) ^^. 1 1 9 _ 4 r.' •■5 r. Q H$> 20 y/'S 5. , '^ c a X 6. vw^+i; y'"'""^ , 7. . ■':__, -^^!_-. 8. 4.^, 3|. EXERCISE VI. (Pacjk 16 ) 32. 2±v/3, VGiV- \/r^ - 1 :n, ±A, ±,. 35 38. ^e, or h. , V5 + 2V/5. 3(). T^l or V- 3i. ±i, V'^- 37. V sec ^/ = 1, cos ^V t.-iu // cosoc ^^ -= 1. ,.,)^. (,)'Y' '^')^- (')''f-' r. »!»» 2 5 sr. . rv_ •'• li> l.i> ij » •'■•• an;,'!*'. 1. 1^5^ -45^ 2. 120' 150^ 210^ -30" anjirle. EXKRCISE VIII. (I'AOK m.) tan. sec. sin. 1^ V/2' 1 V2' -1, -^2. - -1, VA 2» sni. J. 1 V3' 2 1 1 1 V3' 1 2 2 2 ' V:5 tan. sec. OS. cot. cosec. 1 V2' 1 -1. V/2. 1 V2' -1, -v^^ 1 1 2 v/;r 2. 2 ' V-5, v-5 o » -V/3, -2. 3. 221" IV-^ V2. -^2-1, \/4-2v/2^, 67i° |V/2TV2, V- + 1. V4T2V2; cos. cot. cosec. |\/2 + V2» V2 + 1, Vt + ^\/2. angle. sni. -1, tan. sec. cos. cot. co. 4. 270° 5. 45°, 135°, 225°, 315°. cc , 0, cot. 0, VS 7. 1. \ i (K DSCC (I —i, )' 3r)rv/2 i cosec. v/2. 2 cosec. t + 2v^2. CC) \ -i ll. ANSWERS AND llEStTLTS. 25a 11. (I) 0^ G0°. (2) ±30^^, ±ir)0\ (.'5) 30', 60". (4) 90". (5) 90 , 30 . (0) IS , - r)4\ (7) 00. (S) ±30'\ ±15". (9) 135^, - 45^. (10) 22l^ (11) 15 , 75". (12) 18", 162". (13) 30 , 45', 150". (14) 30, 60". (15) A = 45", 7?= 15". (10) A = 52V\ Ji=^7V\ ( 1 7) J = 1 35' , y; :-. GO '. (1 8) A = 30", // = 1 5 ■ C = 45". 19 12. (v/3 - \/2){\/'2 - \),{\/^ - V-)(V- + !)• l'^- •t5"uM(l 00'; IC). (l)liMl)Ossil.l(', (•J)r().s,sil.l«'. 17. (1) Pos.sil)le, (2) impossible. 18. (1) {a- f>f + {ab'^f = 1. (2)^'*^-^;^^!!^^ 1. (f*--w)(r«--ji)^rr ^j^^^ ^ -^ 4 ^<. -f Oo^ EXERCISE IX. (I'AOK 63.) 1. «-10v3, ^'^lO, A =G0'. 2. /^-15(2 + V'^). c=15(v/fi+>/2). 3. b - VTO - 4^5," c - 2v/5"- V5. 4. c- V2 + V2, /-)• -/ 9. V3 8. 50 (v/5 -- 1), 2500 v/5 - 2^-'^. 10. 10(\/C-'\/2), 10(v/G + V'2) 11. 3(v/0 + x/2), 3v/2, !:(3-v/.U 12. 5 feet, 3 foot. 13. 0(v'3 + 2). 14. .30". 1 1 + V2 + V-^ IN 15. cot 22^'- cot 30" 2v/2 '^ ^ 17. Draw AD porpendicular to JiC, thvn h sin C = AD and c cos Ji = JiDy from which the results of this and the following example easily follow. 19. 200. 21. r\/2^V'27!i V^+V^^y 2rV2. I 22. ^, 45". 23. 170 feet. 00 ■wf mtmimmBatTn'iTn' -r»i»ii r> {| 254 24 PLANE TRIGONOMETRY. d d cot a 25. 50v/2. (cot" ^i' - cot" a^i cot" p' - cot-' a 2G. 90°, 72° 30. 10^8^+ -V^, V." (:V- - V^>)- EXERCISE X. {I>AGK 70.) -h±\/\r + lr r a 29. 2?tr sin , 2vt/- tan 7i 71 30. .1 7/r-sin — , /^y-Hai.''. 33. I, I" "^ f. 34. 20v/3, 30v/3. 71 /« \Ja- EXERCISE XI. (Paok 77.) .1: ^i= |'= 12' 1", or 18° 23' 24". 2. .ol672, 75° 38' 56". 4. 1.0634, 20° 28' 25". 6. 4.8053, 80° 53' 30". 8. 119° 14' 45". 10. 112° 29' 13", .92397. 12. yl = 71°6'. 7i = 34°35'. EXERCISE XII. (Pa«k 81!.) 1. a- 10.353, />-14.641, C=105\ 2. a=\3 feet, sin //= '-^^, sin C ^-YA 26 26 :^. h ?^', Gv/6. 1 n c V ^•^^"2=V6''°'2- = V21''^"2=^ a 5. 14, 8 4. 6. 55° 35' 2", 59° 47' 38", 64° 37' 20". 7. 18° 55' 28', 12. 8. 6.1926, 88° 38' 29", 53° 6' 31". 9. 161° 20' 55", 2° 9' 5". 10. 45°, 60°, 75°. 11. 120°. 12.8; 30°, or 1 20°. 1 3. Ji = 90°, C = 72°, c = 4 V5+TV5. ^- Vr.). n /:i, :J0v/3. ' 5G". ' 25". ' W. 32397. 7P6'. 34° 35'. C 1/5 ^2=-F- 37' 20". ' C 31". 11. 120°. 5TV5. ANSWERS AND KESULTS. o.-tr, 14. i/ = 45°or 135°, C=120°or 30; c. 2 (3x/2 + ^G), ir). ^ = 54°orl26°,/y=108°or36° «r 2 (v/2 + v/G). 10. IV190. 17. 20V/3 ; 54^ 44' 10". 18. 17.7G rods 26. 7y=.45°. 31. ., .GO- 28, 20, 12. 32. V./..(.+'^T^ 33.2525,2710. 34.815.85. 35 37-7'!^' EXERCISE XIII. (Pack 97.) 1- 2, 5, 4, G, 12. 2. 1:2. 4. 109.55G8. 1 •■». (s' - a) sec ^^ , otu 7. « cosec ■-, etc. 8. .] (A' + C), etc. EXERCISE XV. (PAOKllO) 1. (4?i+ 1) ''. 2 o,,_ Q - 3. 30'. t). « sec ^ , etc. 2'V sin ^1" 4. ?iT + 3r TT 5. nTT + - . G 12. 7iT±«. G. ?iT + ( _ 1 )" G 9- »^-±''. 10. 7i-+ " -* 4 /. 2«;r± ^.. II. n7:±~. 13. - !?iT4 { ~\y'' ' 2 I ^ ^ G f .-) r, (2>i+I)T 2(7/+ 7)' 18. n,T + (_ !)"■ 14. nrror 2w?r±^. 15. :^----'- i p. , 3 ^{p + f/)' impossible 17. (Gn±l)^ 19. (2..1);or;{..,(..i^.;;|. ^ 20.(4..3)^. 25. 2/i7r-fa. *- i^ ' 3 4/- 26. (6y*±l)ii5 seconds. ?! m 41 ii i ► 25G 2. 1, 0. ft .1 PLANE TRIGONOMETRY. EXERCISE XVI. (Pagk 117.) •'• V^y o.-.> «-i> tS5> - (1.V 4- •510. 9. ]^l 14. 2? 7. 1, :7/i 1 EXERCISE XVIII. (Faor 122.) 1. 2 sin 45° cos 30^. 2. 2 cos 45' sin MO'. 3. 2 cos 45^^ cos 15^ 4. 2 sin 4.5" sin 15°. 5. 2 sin 67r cos 7'". (i. - 2 sin 70' sin 10". 7. 2 sin 20 cos 0. 8.-2 sin 20 sin 0. 9. 2 cos ^^ cos'' . TT 11. sin {0 + ) + sin {0 - ). 10. 2 sin 7 cost—. 4 12 12. cos (« - /3) + cos (« + p'). 13. i {cos (2^ - 2' + cos2Yi). 20. - sin 3^. 43. !^ or 1/9..-^" 1 {2n.±;;}. 45. ?i7r, — , or W7r±- . 47. »i-±^, or(2n+l)^ 17. J (cos 2/>'-cos 2/1). 19. - 1(2 cos 2/1+1). , , »t7r ^ 2,T 44. — or 2n-±—. 4G. nr, or' |nT + (-l)"^|. 48. (10n±l)^,or(2/i+l);r±p. EXERCISE XX. (Pagk 140). 1. (3) ^\/2"-V2, -|\/2 + v^>. (4) _ 1^8 - 2VlO + 2v^ - I Vs + 2V10 + 2^57 (5) -1^10-2^5, -](^5 + l). 7. 1. 0'. 5°. 10'\ 3.S ^, cos ^ . - ; etc. lA-\/i (7) 1 - «// (2) 4 (4)±1, or±(l±y2). (9)0,or±i. IG. a + /; 1 - ah' 21. ±1. 19. y. \^-x^ ^ 1 + cr- 20. v/2. -». .El, (3) (2« ± I ). + 33° 7' 47". (4) „- or 2(„.- 69" 20' .W") (5) (2« + 1)^ ± 23° 25' 43". (6) r„,,«ssil.le. ; '< i ' 1 258 PLANE TUKiON'OMETRY. 23 . (1) \/2 SVC COS ('})- . j, whero tan cos ('A- \ )> ^\'lierc tan «/j-a sin vl. (3) (I \ '2 sin J (OS «/> sec (*/)- I, wlicn; tan ./'=-'< <''^« ^• 25. Given expression --cot (^/- iT)'). 28, IS;). 20. a, or «--f^-i-l. -"^O. ±(f/>. 31. tan-' 1 { {2u + \)-±\ {2n+\f--+ IC;. m v4 K n 1. 3. 5. EXERCISE XXII. (Pauk l.M).) 2, I, S, 10, -3, -n. 2. 4,ry, -1, -2. ^ 3, 1,0, -1, -3. 4. -ji, 32- 2n, n, 2. 6. 2^, - I {,, (^l- 2 log a + 3 log /> + log r, X log ^f - y log /> + ,:; log c, 8. 10. log ."). \ - ( i'-i log '^ + log />). 9. ,'i log 5 - ;i log 2 - ,\, log 3, log 3. 12. (1) jog h in log a ) 11. r=l, I • 7/ = 2 ^ v> J If 'i 2' log C 13. 14. 17. 22. 32. 2 1ogf log^rt^log^> log *^ -t- log // ^' log ) (4) 5 + 2 ( -'^ ). (5) - 1. (C.) '^-V « J «_ »_Z. ^ '' Vlog 2/ ^ ' ^ ' log « - log h 4 and 5, 3 ami 1, 1 and 2, and 1,-1 and - 2, - 3 and - 4. 2G. 15. 2,21, -I -1, ^ IG. ^^. h f iV.. 20. a'""'"-', a'"8^ «'•"•''•. ?». (u- 1) »i log « + ., - log r. 0, 0. 25. log h log //i ^ log ) and - 4. .r li ; AGK 1(58.) ^•^•^''•^- -• ^-^^^'^ 4.09808. 3. .04515 1:^11730, 12.83052. 5. 1.1,890, .413^8 1.74.328, .50038. 7.3,0, -4,-3,-1,0. .9332848, 5.9332818, 3.9332848 857.0, .08570, 85760000. iq. 48G (1) 4.45495. (2) .58841. (3) ,.79449.' (4)3:87500. {o).U^U. (0) .24857. (7)1.99102. (8)1.79931. ^'>''' (2).17G41. (3)15140. (4) .1001. 17. .2771 18. 19. 20. 23. 2.5. Half glv.a logs. ; half given logs taken negatively. '*'^- 15. 53855. 0030900. 1.0989700, .G980700, 3.9030900. .1003433 2.4948500, 1.5785580, .3835050. ^r^f'''^''''''''''^ '•'''''''' ^0457574, -lu, .tztbi 05. ' .3802112, .158302.5, .2070001, 4.4309740, 1.5058959 .7011513, ..3300548, 2.905^119. 151.9304. 1.9424. (1) 3.544. (5) .2528. 20.1 (2) 3.881. (G) 1.257. -1- (1^" 22. 1.9129. 24. -3.2229. (.3) 17.917. (4) .3710. 5.903. 2G 29, 31. 35. 2.80735 37. 36.554. ') years. 27. 242.7 y oars. 28. 0377.1 n.il 30. es. ^--10.736, y= -19.017. .8790, 2/ = .2028. 33. 2.5409, 1.430; 36. 3.58497. 38. .1023. 2G0 PLANE TllKJONOMETUY. EXERCISE XXIV. (Vauk 176) 1. (1) 9.r)l)646. (4) 10.4U273. 2. (I) ir)'^23' 9". (4) Gf)" 4.3' 50". (2) 9.i)GG42. (5) 10.07312. (2) 44" 5' 44' (5) 32" 17'. (3) 9.88390. (G) 10.49G54. (3) 71'" ir ir (G) 32" 53'. 3. 9.8048933, 9.804958.3. 4. 9.9G910G8, 9.9G90952. 5. 32^32' ll".5, 57'^ 27' 43".7. 6. 08^^ 38' 35".3, 21" 21' 43".2. 9. 9.42258G2, 10.0157G48, 9.984234G. 10. 79 ' 44' 50".4. 13. ( 1 ) « = 1 35. 17, /> - 4 1 0.3 1 , Ji^7 1" 4G'. (2) a- 1115.4, />- 527.77, yj = G4" 40' 47". (3) « = 92G.77, vl -75" 28' 53', 7,' =14" 31' 7". (4) a = 325.G4, A = 32" 58' 54", li = 57" 1 ' G"- (5) A:^122.17, c=149.G, yl = 54"45'. (G)«- 1140.7, r' = 1197.G, 7;= 17" 44' 40". (7) c - 7G.828, A = 78" 52' 25", Jl = 1 1" 7' 35". (8) c = 40 1.53, .1 ==77" G' 11", 7? =12" 53' 49". 14. 237.27. 15. 10.493,7.0237. IG. 212.1. 17. 732.22. 18. 537.19 yards. 19. 178.14. 20. 1000 feet nearly. 1. 48" 2. 52" 3. 28" 4. C- 5. 7;= G. j = 7. 7? = 8. A = 9. Jir= c= EXERCISE XXV. (Pa(ik l&'i.) 11' 23", 58" 24' 42", 73" 23' 55". 54' 5", 59" 7' I", 07" 58' 51". 2' 39", GO" 2' 35", 85" 54' 40". 100" 22' 45", ^>- 1337.2, f = 175S.8. 118" 53' 34", C --= 1 1" 0' 20", a = 0330. 0. 39", rt = 014.44, 6==: 793.G9. 53" 59' 3", C= 7G" 44' 57", a - 331.03. 157° 3' 31", C= 7" 43' 15", h = 105.41. 54" 10' 50", or 125" 49' 4". 73" 30' 4", or 1" 51' 50", c = 393.75, or 13.307. ANSWERS AND UESUITS. 261 9.88390. 0.49654. 1° ir 11". 12" 53'. 190952. ' 44' 50". 4. .1. .14. 10 11. 12. 14. 15. 18. 20. 21. ')•) 21, 25. 2G. 5. />• = 144^34' 4.5", or IH'^ W r," C- 1^7' ^''tO",o.. 152^ 54' 20"; A. 1590.1, or 882 72 />'= ri'Mf/, C^90", /. = 410.31. f."I.o.sil>lo. 13. ^-18^4G'5',./==^. i.r35",A^,8->7o ^-^12G\52' 11", C..3G'\52' 11". I.npossible. IG. U18.02. i;. 731.335, 18G GS5 »l''-'-2. 10. 79 G'24'", 40'5.3,3G'", 1G.414 71 33' 53", Gl'55'39". ^ =5" 55' 57", />':_- 28' 51' IG'", C- 145- 1-'" 47" r^207.2.. ^^31.754. " - -*' . ^'^^■^- 23. 1845.2. --8.8334, r,.15.GG8,n = 29.G15,v,.G4.043, 7^.25 122 279.11,428.7,442.41. 121.29, 18G.29, 192.25. 27.9,10,11. 85- 27' 3.3", 15.047 EXERCISE XXVII. (Paok 202.) 28. 90--^', m^-^\ is within the trian^.I. J/;6' Jf angles PA /;, PfJJ be denoted hy , and ., we shall find ./>==32'^7'50"',-^-'^" =160 08' 46"'. 7M = 23.656, /Vi=>)8.74, />C= 23.347. EXERCISE XXIX. (Paor wo ) '■'"'- '»•««« e. 12. ^inO u II 202 PLANE THlU(JNOMETUY. EXAMINATION PAPERS. (PAo^m 2. 7.', inin. PAPER I. 4. 0", 45'^, 180^^. 7 1^ 8. - 1 lo« 2. PAPER II. 2. 40', 00 \ SO. 4. 0. 7. 2. 8. 30', 0O\ 90 . PAPER III. 0. A=^{n+\) 9. 28. 2. ;}0. 1. ± -A-, 0. 7. a V 21, V'^r v'21. 8. 5 v 31, 5 \ 91- V -i 2. 65' 24' 30". 7. v-±'., or ^'~ ± ^ PAPER IV. 3. 0. 5. 12v/4 + 2v/2. 9 L^ _.. log 45 - log 8 PAPER V. 8. cot c = 1 1 a h' 10. 2) = c siti'' 0, q^c cos^ 0. PAPER VI. :>- TT 2. 9 niin. 33 sec. 4. 2/y = 2n-±--. 0. n-, or n::±-. 7. 50 (tan 30" + sin 30^') sq. in. 9. -iv/2-v"2, -lv2+v/2. 10. 2 sin-^Vo, s"^ s» PAPER VII. 5. 8.834 mill, after 3. ^ a"^ sin 7? sin C 9. 8 14709, 2.34797, ANSWERS AMj ui-;s(ri/rs. 263 i >«« 2. -])- 31,5\ Ul. 4 + 2v'2. 1 5 - iog~8" ^v. PAPER VIII. PAPER IX. 6. ■'" + /-2,-yc(,t,/,^c-. 1. 8. PAPER X. 2. I.KJi:,, ...77i. A i- tui ^ cos J, 1.17713. 5. C-. 29 ' r)7' ;}0", A -20'^ 22' 30" 6 7/-33-> .,(,', C = 33M0U1355;22. 7. 51G2 3 9. 2G-}3'54". 10. 75(v/3+I). 8. 19'^ 28' 17 11. 52^ 15' 37", 1. 200. 4. 8.G4. 6. ^ = 109'M7', /Llf^^ i:\ 7. '' = -'30, i= 133, 6- =130.3, 7/ =77 '17'. PAPER XI. ^- -'^'-^J- 3. .000078492. r>. 9.02193, 10.37791, 05' 1 1' 20". 8. 7" 12'. r nn ±3- sill' -1 ."» 6» c PAPER XII. 1. -<-118,//=82', C=70'. 3. .^.vn, ^.^vil,;Mn,- Ml, A VT17- 4. .994. 7. (1) cos (.'•+//) COS .7; cosy' (-)ta»-''t»»iy, (3)tau2x 8. (5)yl = 120', 7y=C=30^. 11. (1)^^7.5°2.5'54".7, 7;-..5'ro7' y' 3 (2)J = 18'^2;3'12".4, C'=ni'3G'47"*6,