Skip to main content
#
Full text of "Elementary trigonometry"

i u- ^ ELEMENTARY TRIGONOMETRY BY W. E. PATERSOX, M.A., B.Sc. MATHEMATICAL MASTER, SIERCERs' SCHOOL AUTHOR OF ' SCHOOL ALGEBRA OXFORD AT THE CLARENDON PRESS LONDON, EDINBURGH, NEW YORK, TORONTO AND MELBOURNE HENRY FROWDE 1911 BY THE SAME AUTHOR SCHOOL ALGEBRA THIRD EDITION The complete book .... Without Answers, j 4s. With (Answers. 5s. In Parts — Part I (Chapters I-XXII : to In- dices and Logarithms, including Fractions and Quadratic Equa- tions) ..... 2s. 6d. 3s. Part II (Chapters XXIII-XXXIV: Progressions and Ratio to Loga- rithmic Series, Theory of Equa- tions and Theory of Numbers) . 2s. 6d. 3s. In Sections — Section A (Chapters I-XVII : to Quadratic Equations) . 2s. 2s. 4d, Section B (Chapters , XVIII- XXVIjLl: to Binomial Theorem for Positive Integral Index) , 2s. 2s. 4d. Section C (Chapters' XXIX- XXXIV: Series, Partial Frac- tions, Theory of Equations, De- terminants and Theory of Num- bers) ... ... 2s. 3d. 2s. 6d. Sections A and B: 'Elementary Algebra' ..... 3s. 6d. 4s. Examples only— Section A . Is. 3d. Is. 6d. PREFACE The common examination syllabus ' Trigonometry up to and including solution of triangles ' has caused most textbooks to treat the subject as if the sole use of Trigonometry were to solve triangles, and the practical examples deal almost exclusively with various forms of triangle-solving under the heading ' Heights and Dis- tances '. Further it is customary to define the trigono- metrical ratios by means of a right-angled triangle; this encourages the mistaken idea that the ratios are fundamentally attached to a triangle, and does not impress upon the pupil the fact that they are the pro- perty of an angle and of an angle only. In this book the trigonometrical ratios are introduced as functions of the angle. The trigonometrical proper- ties of the single angle are treated fully in the early chapters, and from the beginning the examples apply Trigonometry wherever it may be useful, to Geometry, Mensuration, Analytical Geometry, Physical formulae, &c. The right-angled triangle definitions are given in Chapter Y. This chapter contains, in addition to the usual matter, a short treatment of Plane Sailing in Navigation. It is hoped that the examples in Navigation will provide practice in the use of tables, and at the same time be of interest to the pupil. Other examples 260011 4 PREFACE in this chapter lead up to the formulae dealing with the ordinary triangle. The formulae for ^ A are proved by Geometry, indepen- dently of tlie addition formulae; thus the triangle is treated fully without breaking the sequence with a dis- cussion of the A + B and allied formulae. A chapter is devoted to Projection. This includes a discussion of Vector Quantities, their composition and resolution, and finishes with a geometrical treatment of Demoivre's Theorem. In the last chapter, the addi- tion formulae and the allied formulae are treated fully ; the projection proofs are used and recommended but the old-fashioned proofs are also given Throughout the book the student is given every oppor- tunity of developing the subject for himself. A large portion of the bookwork first occurs among the examples of earlier chapters. Also, when a formula has been proved, the proofs of others of the same kind are left for the student to supply. Thus, when siii^A + cos^^ = 1 has been proved, the student should have no difficulty in proving the connexion between sec^A and tsui^A; when sin(^ + ^) has been found, the student should himself find the expanded form of cos (J. + B), Sec. The sets of examples in the body of a chapter arc num- bered IV. a, IV. b, &c. ; these deal only with the matter immediately preceding them. The last set of examples in a chapter has no distinguishing letter and serves for revision of the whole chapter. There are also three sets of Revision Examples, l^ookwork is frequentl}^ set as an example, both in the Revision Sets and elsewhere ; only by constant repetition, oral or written, can the PEEFACE 5 bookwork be learnt. There are a few sets of oral examples; these are intended to fill up spare minutes at the end of a lesson and often bring out the weak points in a pupil's knowledge. The book contains nearly 1,000 examples ; it is not intended that any one should attempt all these, but it is hoped that they include a sufficient variety of types and a sufficient number of each type to meet all requirements. Many examples are taken from Examination Papers by kind permission of the following authorities : — The Controller of His Majesty's Stationery Office. The University of Cambridge. The Joint Matriculation Board of the Scottish Univer- sities. The Intermediate Education Board for Ireland. The Oxford and Cambridge Schools Examination Board. The Delegacy for Oxford Local Examinations. The Syndicate for Cambridge Local Examinations. The College of Preceptors. I am indebted to Mr. Norman Chignell, B.A., of Charterhouse, for many suggestions and for assistance in correcting the proof-sheets. It is too much to hope that the answers are wholly free from mistakes, and I shall be grateful to receive early intimation of any corrections that may be found necessary. VV. E. P. April, 1911. CONTENTS CHAP. PAGE Preliminary. Propositions in Geometry. Graphs. Logarithms. Greek Alphabet ... 9 I. Angles and their Measurement. The Degree. Positive and Negative Angles. The Grade. Connexion between Arc and Angle. Circular Measure. The Points OF THE Compass. Gradient .... 25 II, Definitions of Trigonometrical Ratios. Their Graphs. Inverse Notation. Construc- tion OF Angle with given Ratio . . 37 IIT. Elementary Formulae. Identities. Ratios OF Complementary and Supplementary Angles. Ratios of 0^ 90", 30°, 60°, 45^ The Small Angle 49 Revision Examples A .65 TV. Use of Tables. Harder Graphs. General Solutions of Equations. Use of Loga- rithms 70 V. The Right-angled Triangle Definitions of the Trigonometrical Ratios. Solution of Right-angled Triangles. Navigation: Plane Sailing. Ratios of the Double Angle and Half Angle 85 VI. Relations between the Sides and Angles of a Triangle. Area of Triangle. The Circles of the Triangle 107 Vn. Solution of Triangles 119 Revision Examples B 134 3,.^., CONTENTS CHAP. PAGE VIII. Projection. Vector Quantities. Composition AND Resolution of Vectors. Projection OF AN Area on a Plane. Geometrical Representation of Imaginary Quantities. Demoivre's Theorem 139 IX. Ratio of the Sum or Difference of two Angles. Sum or Difference of Two Sines or Co- sines. Product of Sines and Cosines. The Double Angle and Half-Angle . .155 Revision Examples C 175 Miscellaneous Problems (taken from recent Army Entrance and Civil Service Exami- nation Papers) 184 Examination Papers 190 INDEX 201 The following course of reading is suggested for l3eginners : — Chapter I, §§ 1-4, 9-11. Chapter II. Chapter III, §§ 21-31. Chapter IV, §§ 34-7, 40-2. Chapter V, §§ 43-5, 51-3. Chapter VI, §§ 54-8. Chapter VII. PliELIMINAEY CHAPTEE GEOMETRY A KNOWLEDGE of the following geometrical facts is required.* In this book these propositions are referred to by the numbers given below. Angles. Prop. 1. If a straight line meets another straight line, the adjacent angles are together equal to two right angles. Prop. 2. If two straight lines cut, the vertically opposite angles are equal. Prop. 3. The angle at the centre of a circle is double an angle at the circumference standing on the same arc. Prop. 4. Angles in the same segment of a circle are equal. Prop. 5. Angles at the centre of a circle standing on different arcs are in the same ratio as the lengths of the arcs. Triangles. Prop. 6. [a] The three angles of a triangle are together equal to two right angles. (&) If one side be produced the exterior angle equals the sum of the two interior opposite angles. Prop. 7. Any two sides of a triangle are together greater than the third. Proj). 8. Two triangles are congruent (i. e. are equal in every respect) if they have — (a) two sides of the one equal to two sides of the other, each to each, and the angle contained by the two sides of the one equal to the angle contained by the two corresponding sides of the other ; or (h) three sides of the one equal to three sides of the other, each to each ; or (c) two angles of the one equal to two angles of the other, * For proofs see Warren's Experimental and Theoretical Geometry (Clarendon Press), or any standard textbook. 10 GEOMETRY each to each, and a side of the one equal to the corresponding side of the other. Prop. 9. If two triangles have an angle of the one equal to an angle of the other, and the sides about another pair of angles equal, each to each, then the third angles are either equal or supplementary. Prop. 10. (a) If two sides of a triangle are equal, the opposite angles are equal. (&) If two sides are unequal, the greater side is opposite a greater angle. (c) If all the sides of a triangle are equal, all the angles are equal. Prop. 11. (a) If two angles of a triangle are equal, the opposite sides are equal. {b) If two angles are unequal, the greater angle is opposite a greater side. (c) If all the angles of a triangle are equal, all the sides are equal. Prop. 12. Two triangles are similar (i. e. their angles are equal, each to each, and the ratio of pairs of sides opposite equal angles is the same for all three angles) if they have — {a) their angles equal each to each ; (6) their sides in the same ratio ; (c) an angle of the one equal to an angle of the other, and the sides about the equal angles in the same ratio. Prop. 13. (Pythagoras' Theorem.) In a right-angled triangle the square on the hypotenuse is equal to the sum of the squares on the other two sides. Parallel Lines. Prop. 14. (a) If a line is drawn to cut two parallel lines, it makes (i) the alternate angles equal, (ii) the interior angles on the same side of it together equal to two right angles, (iii) the exterior angle equal to the interior opposite angle. (b) The opposite sides and angles of a parallelogram are equal. Area. The unit of area is the area of a square whose side is of unit length. Prop. 15. The number of units of area in a rectangle is equal to the product of the number of units of length in one side multi- plied by the number of units of length in the other. GEOMETRY 11 Or, more shortly : Area of rectangle = length x breadth. Proio. 16. The area of a triangle = \ base x altitude. The Concurrencies of the Triangle. Prop. 17. The lines bisecting the sides of a triangle at right angles are concurrent (i. e. meet at a point). The point in which they meet is the centre of the circle passing through the three vertices and is called the circumcentre. Prop. 18. The lines drawn from the vertices to bisect the oppo- site sides are concurrent. These lines are called the medians and the point of concurrency is called the centroid. Prop. 19. (a) The lines bisecting the angles are concurrent. The point of concurrency is the centre of the circle that touches all the sides, and is called the incentre. (&) If two of the sides be produced, the lines bisecting the ex- terior angles so formed and the line bisecting the interior angle contained by the produced sides are concurrent. The point of concurrency is the centre of the circle that touches the two sides when produced and the third side (not produced) ; it is called an e-centre. Prop. 20. The perpendiculars let fall from the vertices on the opposite sides are concurrent. The point of concurrency is called orthocentre. The Circle. Prop. 21. The straight line passing through the centre, at right angles to a chord, bisects the chord. Prop. 22. [a) The angle at the centre of a circle is twice the angle at the circumference on the same arc. (&) Angles in the same segment are equal, (c) The opposite angles of a quadri- lateral inscribed in a circle are together equal to two right angles. Prop. 23. The tangent at any point is at right angles to the radius drawn to that point. Prop. 24. (a) Two tangents can be drawn to a circle from any external point. (&) The parts of these tangents between the external point and the points of contact are equal, (c) The line joining the external point to the centre bisects the angle between the tangrents. 12 GEOMETRY Prop. 25. The ratio of the circumference of any circle to its diameter is the same for ail circles. This ratio is denoted by the symbol tv : its value is 3*1416 correct to five significant figures. Prop. 26. The area of a circle equals the area of the rectangle contained by the radius and a straight line equal to half the circumference. This is usually expressed in the formula : Area = tt r^. GRAPHS* Geometrical. If two straight lines are drawn in a plane, the position of any point in the plane can be determined by means of its distances from those lines. Y Fig. 1. It is usual to draw one of the lines horizontal and the other perpendicular to it. The customary notation is shown in Fig. 1. X'OX is called the axis of x ; Y'OY i% called the axis of y; is called the origin ; ON is called the abscissa of the point P; NP is called the ordinate of the point P. The abscissa and ordinate are called the co-ordinates t of the point P. * For a fuller treatment of Graphs see School Algebra published by the Clarendon Press. t Tliese co-ordinates are called Cartesian co-ordinates because they were first used by the French mathematician, Descartes. GRAPHS 13 The abscissa is said to be positive if drawn to the right, negative if drawn to the left. Similarly, the ordinate is positive if drawn upwards from N, negative if drawn downwards. The number of units of length in ON, preceded by the proper sign, is usually denoted by x, and the number of units of length in NP, preceded by the proper sign, is denoted by tj. In each case the sign + is often omitted. Thus, in Fig. 2, the co-ordinates of A are x = -4, y = 2, of C, x = 4:, y = ^, oi L, x = (), y = -7. 1 1 1 1 1 1 Y io 1 B 5 1 ID 1 c X A X O r~ 5 K 1 1 1 Gl E H 5 . i 1 1 1 V 1 LI 1 1 1 ^°, 1 1 1 i 1 1 Y| 1 1 1 1 i Fig. 2. Very often a point is described by writing the values of the co-ordinates in brackets ; e. g. the point H might be described as the point ( — 6, —4). Exercise. Write down the co-ordinates of all the points in Fig.- 2. Graphs of Statistics. The magnitude of any quantity may be represented by a straight line which contains as many units of length as the quantity contains units of its own kind. If two quantities are changing their values at the same time, 14 GRAPHS the simultaneous values may be represented in the same figure by taking horizontal lengths to represent one magnitude and vertical lengths to represent the other. 1 ~ 5 1 ' 1 1 i / 4 5 / / ' \ 4 1 \ \ / \ / / 3 5 1 \ / 1 i3 / « 3> / / c i 1 1 O / V / j 1 \ 1 1 S.\M. T. W. 1 Th. F. Sot. 1 i i 1 1 1 1 1 1 Fig. 3. In Fig, 3 the changing quantities are time and temperature ; and the dots show that at noon on Sunday the temperature was 29°, on Monday the temperature was 35, &c. In fact the figure conveys the same information as the following table : — Sun. Mon. Tues. Wed. Thurs. Fri. Sat. Temp. 29° 85° 42° 31° 27° 43° 50° If there is no information about intermediate temperatures, the points are joined by a series of straight lines. The figure now forms a graph. In describing such a graph we should say that the abscissae represent time and the ordinates temperature. Graphs of functions. If two quantities x and // are such that a change of value in the one causes a change of value in tiie other, then either of them is said to be a function of the other. GRAPHS 15 This is expressed thus : y =f{x), or x =f{y) where f[x) means a function of x. A graph can be drawn in which the abscissae are proportional to the values of x and the ordinates to the values of y. This graph is called the graph of the function f{x) or of the equation y=f{x). This maybe more easily understood by considering a few algebraical functions. Example I. Draw the graph when y = \x-% (Choose values of x which will make y a whole number.) y = \{^x-2) X -6-149 Zx-1 -20 -5 10 25 y _ 4 -1 2 5 - 1 - f^ 5 y ^ y ! / r^ 1 I>^1 1 ^ ! ' 5 u "51 "l O ^ f" - ! 1/ 1 ' ' ' 1 y 1 ! : I 1 /^ y 1 1 k- 5 i 1 1 Fig. 4. When the points corresponding to these values of x and y are plotted, it is found that they lie on the straight line shown in Fig. 4, It is also found (i) That any simultaneous values of x and y connected by the given equation are the co-ordinates of some point on this straight line ; (ii) That the co-oidinates of any point on the straight line satisfy the equation. It is found by experience (and can be proved from the geo- 16 GRAPHS metrical propositions on proportion) that, when x and y are connected by an equation of the first degree, the graph is always a straight line. Example II. In the same figure draw the graphs of y = x^-^x-V^ and r» = 2f/H3. Neither of these- equations is of the first degree, therefore neither of the graphs is a straight line. At least six points must be found on each. yr=x^-Zx\'l. (i) X -3 — 2 -1 1 2 3 4 5 x^ 9 4 1 1 4 9 16 25 -3a; 9 6 3 6 2 -3 -6 -9 -12 6 -15 12 y 20 12 2 JT 20 f I t -1 X II 4 3f 4 ^ 4t 1 :: ri i -L \ J i f: u 3 l" 7 4 A =+ r _ — — =-^ j^^-;^" -- -^ S^Z 5 ^ i ^ 5 lO 5 ^ -- — -___ Fig. 5. GRAPHS 17 (ii) y -3 -2 -1 1 2 3 4 2j/2 18 8 2 -3 2 8 18 32 X 15 5 -1 -1 5 15 29 The co-ordinates of every point on graph i satisfy the first equa- tion, and of every point on graph ii satisfy the second equation. Hence the co-ordinates of any points which are on both graphs, that is, the co-ordinates of the points of intersection, satisfy both equations. Fig. 5 shows, therefore, that the values x = '3, t/ = I'S, and X = 2'8, y = 1*6, are the solutions of the two equations. This graphical method of solving equations is very useful, but is, of course, only approximate. If more accurate answers are required, the graphs must be drawn on a larger scale in the neighbourhood of their points of intersection. LOGARITHMS * Fractional and Negative Indices. It is shown in Algebra that p _ 1 where p and g are any positive integers and x is any positive quantity, integral or fractional. A fractional index may be expressed as a decimal ; thus such expressions as 4'^^ 10*'°^ have a definite value. This value could in theory be found by reducing the decimal to a vulgar fraction and then replacing the power with a fractional index by a root, e.g. 10-301 This is obviously not practical. The value can be found by a graphical method which is easy but only approximate. Di-aw the graph of x = lO*'. * For a fuller treatment of Indices and Logarithms see School Algebra, Chapters XXI and XXII. _301_ , IQIOOO _ 1000 /j^QSOl \ 18 LOGARITHMS y •5 •25 •125 'lb •625 •875 1 X 1 3-16 1-76 1-33 5-62 426 7-49 10 The values of x are obtained as follows : — 10° = 1 by definition given above. 10-» = 10^ = VIo. 10-25 = 10^= V^\/ 10; similarly W^'-^^^ ^W\ W^ = 10-5+-25 = 10-5 X 10*25, &c. The graph is shown on a small scale in Fig. 6. It is seen that 10'^° is almost exactly 2 ; any other power of 10 can be found approximately from this graph when the index is between and 10. Definition of a logarithm. The logarithm of a number to a given base is the index of the power to which the base must be raised to equal the number. Thus 3^'= 9, therefore the logarithm of 9 to base 3 equals 2 ; this is written log3 9 = 2. In dealing with numbers the base is 10. In the remainder of this chapter it is assumed that the base is always 10, so that log 731 means logarithm of 731 to base 10. The equation x = 10^ may be written t/ = log x. Hence Fig. 6 provides an approximate means of finding the logarithm of any number between 1 and 10. Characteristic and mantissa. Consider a number, such as 4878. It means 4xlO.H3x 102 + 7x10 + 8. Also a decimal number, such as '0376, means A A _I_ _?_ 10 "^ 1Q2 "^ 10^ "^10^* If we use negative indices, this may be written •0376 = 3 X 10-2 + 7 X 10-^ + 6 x 10"*. Similarly 537-13 = 5 X 102 + 3 x 10^ + 7 X 10'' + 1 X 10-i + 3 x lO"''. It follows that 4378 > 103 but < 10\ •0376 > 10-2 but < 10-^ 587-13 > 102 but < 10», 4-37 > 10° but < 10\ •. log 4378 = 3 + a decimal ; •. log "0376 = - 2 + a decimal ; •. log 537-13 = 2 + a decimal ; •. log 4'37 = + a decimal. LOGARITHMS 19 We now see (i) that the logarithm of any number consists of an integer (which may be positive, zero, or negative) and a positive Fig. 6. decimal, (ii) that the integer is the index of the highest power of 10 contained in the number. The integral part of a logarithm is called the characteristic ; the decimal part the mantissa. Index 5 , 3 2 1 -1 -2 -3 -4 -5 -6 Number 3 7 8 9 )j 4 3 7 6- 5 2 ?> 3- 4 6 7 8 »> 0- 7 3 4 8 if 2 0- 20 LOGAEITHMS Consideration of the preceding table shows that the character- istic (i. e. the highest index) may always be found by the following rule : Count from the unit place to the first significant figure (i. e. the first figure which is not 0), the unit place being counted as nothing. The characteristic is positive or zero if the number is greater than one, negative if it is less than one. The mantissa is independent of the position of the de- cimal point. An example will make this clear. Given that log 4-376 = -6411, find log 4376 and log -004376. 4376 = 1000 X 4-376 but 4-376 = 10-"" since log 4*376 = '6411. .-. 4376 = 10^ X 10-<'^^i = 10s-6"S i. e. log 4376 = 3-6411 •004376 = 1 J-oo X 4*376 = 10-^ x lO-^^i = io-3+-64n .-. log -004376 = -3 + -6411. The negative sign of a characteristic is always placed on top and the + before the decimal is omitted. Thus log '004376 = 3-6411. To find the logarithm of any number. (a) Four-figure tables. The mantissa is found from tables, of which a specimen is given below. Logarithms. 1 2 3 4 5 6 7 8 9 I 2 3 4 5 6 7 8 51 53 54 7076 7160 7243 7324 7084 7168 7251 7332 7093 7177 7259 7340 7101 7185 7267 7348 7110 7193 7275 7356 7118 7202 7284 7364 7126 7210 7292* 7372. 7135 7218 7300 7380 7143 7226 7308 7388 7152 723s 7316 7396 I 2 I 2 I 2 I 2 3 3 2 3 2 3 2 3 ' i 7 5 ^ I 566 5 <i 6 55 7404 74r2j 7419 7427 7435 7443 7451 7459 7466 7474 I 2 2 3 5 5 6 Consider the logarithms of 5467 and '05467. By counting, the characteristic of log 5467 is found to be 3, and that of log '05467 to be 2. Both logarithms have the same mantissa. Look for 54 in t)ie extreme left-hand column. In the same line with 54 and under 6 we find 7372 ; this is the mantissa of log 546. Under the 7 in the small columns to the right, and in a line with 54, we find 6 ; this must be added to the last digit of the mantissa already found. Hence the mantissa is '7378. Therefore log 5467 = 3'7378 and log '05467 = 2*7378. LOGARITHMS 21 (b) Five-figure tables. Find the logarithm of 346'73. Proceeding as -with four-figure tables, we find that the mantissa of log 346 is '53908. ,Under 7 in the side columns, we find 88 ; this must be added to the last two digits already found. For a 3 in the fourth place we should add 38, for a 3 in the fifth place we add, therefore, ^^ x 38, i. e. 4 to nearest integer. Hence log 3467-3 = 2-53908 + 88 + 4 = 2-54000. To find the number corresponding to any logarithm. Method I. Reverse the process for finding a logarithm. Sup- pose the logarithm is 3*7271. Look in the logarithms for the mantissa nearest to 7271, but less than it. We find 7267, level with 53 and under 3 ; the first three figures of the number are 533. This leaves 7271 - 7267 = 4 ; in the right-hand columns 4 is found under 5. Hence the first four figures are 5335. The characteristic is 3, therefore the left-hand digit 5 represents 5 X 10^ ; hence the number is bd\p. The number is called the antilogarithm of the logarithm. Method II. If tables of antilogarithms are available, they are used in the same way as kcarithm tables. AxmLOGARITHMS. 1 2 3 4 5 6 7 8 12 3 4 5 6 7 8 9 -012 5023 5035 5047 5164 5284 5408 5534 5662 3058 3176 5297 5420 5546 3675 5070 508^ 5200 3321 3445 5572 5702 5093 5212 5333 3458 5585 5715 5105 3224 3346 5470 5598 5728 5117 5236 3358 5483 5610 5741 1245 6 7 8 9 n 5370 5493 5140 5260 5383 5508 5152 5272 5393 3521 3188 3309 3433 3559 1245 1245 1345 1345 6 6 6 6 7 8 10 II 7 9 10 II 8 9 lo II 8 9 10 13 5623 5636 3649 5689 1343 7 8 9 10 12 Look for *72 in the left-hand column ; level with *72 and under 7 we find 5333 ; in the small columns we find 1 under 1. Hence the first four figures are 5334. The characteristic is 3 ; as before, the number is 5334. Note. — The two methods give results differing by 1 in the last figure ; this shows that the number is between the two results. On using five-figure tables, it is found that the antilogarithm of 3'7271 is 5334-5. 22 LOGARITHMS Use of Logarithms. By the definition of logaritlim a = lO^oga, & = lO^os^, .-. db = 10log«x IQlos^ = iQlogrt+logft. Hence log (a6) = loga + log &. Similarly log ajh = log a — log &, log a'" = wi log a, log Vi/a = — log a. Thus, instead of multiplying, we may use logarithms and add ; dividing, ,, „ subtract; raising to a power, „ ,, multiply ; taking a root, „ „ divide. Note. — There is no process with logarithms to correspond with addition or subtraction with ordinary numbers. 1 T 17- ^ M 1 P 516-5 X -852 Example I. Find the value of —^- — • log of fraction = log SIG'S + log '852 -log 36500 = 2*7130 + 1-9304 -4-5623 = 2-6434 - 4-5623 = 2-0811 .-. fraction = -01205. Example II. Find the cube root of -1765. log of cube root = ^ log '1765 = ^ of 1-2467 = ^3 + 2-2467) = 1-7489. Notice carefully this method of division when the characteristic is negative. Hence V-1765 = -5610. LOGARITHMS Exercises. Find the value of (1) (2) (3) (4) (5) ^^319-2 X r756. •03056 X 0-4105. 3-142 x(71-43)^ ^x 3-142 X (9-67)3. 254-3 -r 0-09027. Ans. Ans. Ans. Ans. Ans, 31-37. •01254. 16030. 3787. 2817. (6) (-1136)^x^/81-86 ^x V2000 An?. •1874. 23 THE GREEK ALPHABET Greek letters are used so frequently in Trigonometry and other branches of Mathematics that it is useful to have the complete alphabet for reference. Name. Small. Capital. alpha a A beta ^ B gamma y r delta d A epsilon e E zeta c Z eta V H theta e e iota L I kappa K ^ : lambda X mu /* M nu V N xi i S . omicron o O pi IT n rho P p sigma a- 2 tau T T upsilon V Y phi * chi X X psi ^ ^ omega <o Q« CHAPTER I ANGLES AXD THEIR MEASUREMENT 1. Any angle such as BAC (Fig. I) may be thought of as having been formed by rotating the line AC about the point A from the position of coincidence with AB to its final position AC. Fig. I. This way of regarding an angle shows clearly the intimate connexion between angles and arcs of circles and this connexion leads to the usual method of measuring angles. 2. The Degree.* From very early times it has been the custom to divide the circumference of a circle into 360 equal parts or * 'The current sexagesimal division of angles is derived from the Babylonians through the Greeks. The Babylonian unit angle was the angle of an equilateral triangle; following their usual practice this was divided into 60 equal parts or degrees, a degree was subdivided into 60 equal parts or minutes, and so on ; it is said that 60 was assumed as the base of the system in order that the number of degrees corresponding to the circumference of a circle should be the same as the number of days in a year which it is alleged was taken, at any rate in practice, to be 360.' (From A Short Account of the Hisfory of Mathe- matics, by W. W. Rouse Ball.) 26 ANGLES AND THEIR MEASUREMENT degrees, each degree into 60 parts or minutes,* each minute into 60 seconds.^ The angle at the centre of a circle, subtended by an arc of 1 degree, is taken as the unit angle, and it, too, is called a degree ; it is divided into minutes and seconds in the same way as the arc degree. The notation used is shown in the following example : — 47° 15' 37" is read 47 degrees 15 minutes 37 seconds. If the line makes a complete rotation, thus returning to its original position, it has turned through an angle of 360°. A right angle is produced by one-quarter of a complete rotation, and is, therefore, equal to 90°. If two angles together equal a right angle, either of them is called the complement of the other. When the sum equals two right angles, either angle is the supplement of the other. 3. Positive and Negative Angles. In discussing the properties of a single angle it is usual to draw the initial line so that it is horizontal and to name it OA. If the rotating line moves in a direction opposite to that of the hands of a clock, the angle is said to be positive ; if in the same direction as the hands of a clock, the angle is negative. Fi- III. * Minutes derived from the Latin jmrfcs minutae ; seconds from the Latin partes fuinntae secundac. ANGLES AND THEIR MEASUREMENT 27 In Fig. II the line OP has made i of a complete turn, hence the angle ^OP = i x 360° = 45° ; in Fig. Ill the angle ^ OP is reflex * and is equal to § x 360° = 225°. If, in Fig. II, the line OP reached its position b}-- turning in the negative direction it would have made | of a complete turn so that the reflex angle ^OP in Fig. II = -315°. Similarly in Fig. Ill the obtuse angle AOP= -135°. 4. Angles unlimited in size. In Fig. II the line OP might have made one, two, or any number of complete turns, either positive or negative, and then have moved on to its final position : hence the angle AOP may represent 405°, or 765°, or —675°. All possible values are included in the general formula ^0P= 360 n + 45, where n is any whole number, positive, zero, or negative. Unless the problem under discussion allows the possibility of the angle being greater than 360°, it is always assumed that the angle is less than 360°. 5. The Grade. When the metric system was invented, the French Mathematicians introduced a new unit, the Grade, such that 100 grades = 1 right angle, 100 minutes = 1 grade, 100 seconds = 1 minute. This system never came into general use, even in France, and now exists only in old-fashioned examination papers. Examples I a. 1. Find the complement of each of the following angles: 32°, 47° 23', 75° 13' 14", 68°0'13", 27° 42' 18-6". 2. Write down the supplements of 75°, 68° 14', 115° 17' 48" 90°, 78° 24' 36". 3. The angles of a triangle are found to be 42° 13' 17", 73° 47' 5", 64° 0' 38". Is this correct ? 4. Two angles of a triangle are 17° 43', 92° 16'; calculate the third angle. 5. In a triangle ABC, 1{A + B) = 77° 29' and \{A-B) = 16° 25' ; find all the angles. * A reflex angle is an angle greater than two right angles, but less than four right angles. 28 ANGLES AND THEIR MEASUREMENT 6. Express in degrees, minutes, and seconds the angle of (a) a square, {h) a regular pentagon, (c) a regular heptagon. 7. Express each of the angles of question 6 in grades. 8. The magnitude of an angle may be expressed either as D degrees or G grades ; find the equation connecting D and G. 9. Draw the angles A and ^A in each of the following cases: («) A = 54°, (b) A = 414°, (c) A = 774°, (d) A = 1134°, (e) 234^ (/) -126°. 10. Through what angles do the hour, minute, and second hands of a watch respectively turn between 12^ 30' a.m. and 5h 3' a.m. ? 6. The ratio of the length of the circumference of a circle to the length of its diameter is the same for all circles. This constant value is denoted by the Greek letter -n (pro- nounced pi), so that if the circumference = c units of length, and diameter = cl units of length, then - = 77. d The value of tt can be found, correct to two or three significant figures, by actual measurement. By geometrical and trigono- metrical calculations its value can be calculated to any desired number of places. Correct to 5 significant figures, n = 31416. Correct to 6 significant figures, n = 314159. For mental calculations tt may be taken as 31. 7. By using Prop. 5, p. 9, problems dealing with the lengths of circular arcs may often be solved. Example. Find the length of an arc which subtends an angle of 49° at the centre of a circle whose 7-adius is 5 feet. arcylP _ angled OP semi-circumference 2 right angles ai'c AP __ 49° Fig. IV. i-^- "^5^ ~180°* The calculation is easily completed. ANGLES AND THEIR MEASUREMENT 29 ,,. ., , area of sector ^OP an^le ^OP ,t^ o^\ Similarly r^l = , .^, . 1— , (Pi'op. 26) "^ area of circle 4 rierht angles 1. e. area of sector AOP _ 49 2blT 360 8. Circular measure. By the method of the last section it is easily shown that the length of an arc of a circle, Att radius r, subtending an angle A° at the centre is r -— - . loO In many other formulae the fraction -— occurs in con- nexion with the angle A°. In theoretical work it has, therefore, been found convenient to use another unit angle, which simplifies formulae considerably. The radian is the angle subtended at the centre of any circle by an arc equal in length to the radius. Let x° equal 1 radian arc equal to radius _ angle of 1 radian _ x°^ semi-circumference 2 right angles 180° '•^* 7rr~180°' Since n is the same for all circles, it follows that the radian is the same for all circles and may, therefore, be taken as a unit of measurement. The number of radians in an angle is often called the circular measure of the angle. For this reason the symbol <^ is used to show that the angle is measured in radians, e. g. 2^^ means 2 radians. When the radian is the unit angle, it is customary to use Greek letters to denote the number of radians, and the symbol '^ is then often omitted. When capital English letters are used, it is usually understood that the angle is measured in degrees. 30 ANGLES AND THEIR MEASUREMENT Examples I b. 1. How many times is an arc equal to the radius contained in the semi-circumference ? Reduce 180°, 90°, 60°, 30° to radians. (Do not substitute for tt.) 2. Show by simple geometry that the radian is less than 60°. 3. How many radians are there in 10°, 75°, 138°, respectively ? Give the answers correct to 2 decimal places. 4. Express the angle of (i) an isosceles right-angled triangle, (ii) a regular nonagon, in circular measure. Give the answers in terms of 73-. 5. One angle of a triangle is ^ tt, another is -J- n ; what is the circular measure of the third angle ? 6. Find the length of an arc of a circle which subtends an angle 78° at the centre, the radius being 18 feet. 7. An arc of length 5 feet subtends an angle of 132° at the centre ; what is the radius of the circle ? 8. Find the area of the sector of a circle if the radius is 12 feet and the angle 40°. 9. What time does the minute hand of a watch take to turn through (i) 3000°, (ii) 3000 grades, (iii) 3000 radians ? 10. Fill in the missing values in the following table, which gives data about circular arcs. Length. 413 feet 220 yards 1 kilometre half a mile 11. Express in radians the angle of a sector of a circle, being given that the radius is 7 inches and the area of the sector 100 sq. inches. 12. Show that the length of an arc subtending an angle 0^ at the centre of a circle, radius r, is r6. What is the area of the corresponding sector ? 13. Find the circular measure of V and of \'\ correct to 5 signi- ficant figures. Kadius. Angle. (1) 5 inches 2 radians (2) 7-6 centimetres 74*6 grades (3) 314 degrees (4) 100 yards radians (5) 320 metres degrees (6) yards 5 radians ANGLES AND THEIR MEASUREMENT 31 9. The points of the compass. The card of the Mariner's Compass is divided into four quadrants by two diameters pointing North and South, East and AVest respectively. These are the Cardinal Points. Two other diameters bisecting the angles between the previous diameters give four other points, viz. NE., NW., SW., SE. The eight angles so formed are bisected and eight more points are thus obtained. These are named by com- -biningthe names of the points between which they lie, beginning with the cardinal point. Thus the point midway between E. and SE. is ESE. (East South-East). The sixteen angles now formed are bisected so that the circum- ference is finally divided into thirty two equal divisions. From their names the last sixteen points are called by-points. The point midway between N. and NNE. is called N. by E. ; that mid- way between SW. and SSW. is called SW. by S., &c. 32 ANGLES AND THEIR MEASUREMENT The angle between two consecutive points of the compass is also called a point, thus N. 2 points E. is the same as NNE. ; WSW. \ W. means | a point W. of WSW. The ordinary degree is sometimes used in defining a direction, for instance ENE. can be referred to as 221° N. of E. Similarly we may have 32° W. of N., 40° S. of W., &c. 10. Latitude and Longitude. The position of a point on a sphere can be defined by two angles, which may be compared with the abscissa and ordinate of plane geometry. These angles are easily understood by considering the special case of Longitude and Latitude. In Fig. V the meridian through Greenwich cuts the equator at A ; the meridian through P cuts at B. is the centre of the Earth. The Longitude of Pis the angle AOB and may be either East or West of the Greenwich meridian. The Latitude of Pis the angle FOB and may be either North or South of the Equator. Note. — A geographical or nautical mile is the lengtli of an arc of a meridian (or of the equator) subtending an angle of 1' at the centre of the earth. ANGLES AND THEIR MEASUREMENT 33 A ship travelling at the rate of I nautioal mile per hour is said to have a speed of one knot. 11. Gradient. It is usual to estimate the inclination to the horizontal of a road or hill by the distance risen veitically for a certain horizontal distance. Thus a hill might be said to rise 3 in 5 ; this would mean that if a horizontal line were drawn through a point B on the hill to meet the vertical line through a lower point A at C, then AC/BC would equal ?. The hill is said to have a gradient or slope of 3 in 5. Fig. VI. It is clear that in many cases it is easier to measure AB than BC; and some books take a gradient of 3 in o to mean a rise of 3 vertically for a distance 5 measured along the incline ; so that in the figiire AC/AB would be f. This latter interpretation of gradient is very common in books on Theoretical Mechanics. If the inclination is small, it makes no practical difference whicli interpretation of gradient is taken. It should be noticed that the angle is the same whatever units be used ; that is whether we consider a rise of 3 inches in 5 inches, 3 furlongs in 5 furlongs, 3 miles in '5 miles. This follows from Prop. 12 0. Examples I c. 1. Express in degrees the angle between (a) NNE. and E. by N. ; (/) S. 2 points W. and W. 2 points S. ; (&) W. by S. and SE. by N. ; (r/) 40° N. of W. and 30° E. of S. ; (c) ESE. and NE. by N. ; {h) NE. by E. and 1 point W. of N. ; (d) NNW. and SSE. ; (/) 30° S. of W. and ESE. ; (e) N. by W. and SW. ; (k) S. 2 points W. and W. 2 points N. In the following questions take the radius of the Earth to be 4000 miles. 34 ANGLES AND THEIR MEASUREMENT 2. Two places on the Equator are 300 miles apart, find the ditt'erence of their Longitudes. 3. Quito (Longitude 79' W.) and Macapa (Longitude SIJt W.) are both on the Equator, find the distance between them. What time is it at Macapa when it is noon at Quito ? 4. Find the distance between Poole (Lat. 50° 43' N., Long. 1°59' W.) and Berwick (Lat. 55° 46', Long. 1°59'W.). 5. Find the distance between Cape Breton Island (Lat. 45° 50' N., Long. 60° W.) and the Falkland Isles (Lat. 51° 32' S., Long. 60° W.). Oral Questions. 1. What is a degree ? How many degrees are there in the angle of a regular pentagon ? 2. How big is each acute angle of an isosceles right-angled triangle ? 3. One angle of a triangle is A°, another 30°, how big is the third angle ? 4. What is meant by a negative angle ? When screwing an ordinary screw in, is the turning in the positive or negative direction ? 5. Does the earth rotate in the positive or negative direction ? In which direction does the sun appear to move ? 6. Do you usually draw a circle in the positive or negative direction ? 7. The needle of a mariner's compass is deflected from its normal position through a positive angle 33i| degrees, to what point of the compass does it then point ? 8. Express the following angles in circular measure : 90°, 00°, 180°, 45°, 30°. (Give the answers in terms of n.) 9. What is the locus of all places having latitude 35° N. V 10. What is the locus of all places having longitude 15° W. V 11. It is noon at the same time at two different places, what do you know about their longitudes or latitudes ? 12. Give the latitude and longitude of the N. pole. Examples I. 1. In a triangle ABC, A = 43° 15', B = 07° 38', calculate the number of degrees in (i) the angle C, (ii) the angle subtended at ANGLES AND THEIR MEASUREMENT 35 the centre of the c ire urn circle by the side BC, (iiij the angle sub- tended at the centre of the inscribed circle by the side BC. 2. Express in circular measure, correct to 3 significant figures, (a) the supplement of I'S? radians, (/;) 74°, [c) the angle of a regular octagon. 8. Define a radian and a grade. If an angle, containing D degrees, may be expressed as either 6 radians or G grades, prove that 2)/180 = 6/tt = G/200. 4. The hands of a clock are coincident at noon, through what angle does the hour hand turn before they next coincide ? 5. Prove that whatever be the radius of a circle the size of the angle at the centre, which subtends an arc ec^ual to the radius, is constant. What is this angle called ? Show, by a geometrical construction, that it is a little less than 60°. 6. A wheel of a cart is 4 feet in diameter, through what angle does it turn when the cart moves forward 10 feet? 7. Explain how to find the length of a circular arc being given the number of degrees in the angle subtended at the centre and the length of the radius. ^'. Two places on the p]quator differ in longitude by 87° 16', find the distance between them, correct to three significant figures. (Radius = 4000 miles.) 9. Find the distance between a place, longitude 45° 17' E., lati- tude 0°, and another place, longitude 88°43'W., latitude 0°. 10. Through what angle does the Earth turn between 9.30 a.m. and 4 p.m. ? 11. When it is noon at Greenwich what time is it at (a) Calcutta (88° 15' E.j, (&) New York (74° W.), ic) Hawaii (156° W.) ? 12. The co-ordinates of two points P and Q are (7, 8), (9, 11) respectively, find the gradient of the line PQ. 13. Draw an angle AOP = 85^ in OP take 8 points P, Q, li such that OP = 1 inch, OQ = VI inch, OR = 2d inches. From P draw PH Sit right angles to OA, at Q draw QK at right angles to OQ, and from R let fall RL perpendicular to OA. -Measure OH, OK, OL and calculate, to 3 decimal places, the ratios OH/ OP, OQ/OK, OLjOR. Justify the result. 14. Explain what is meant by a radian, and find how many degrees and minutes it contains. Express in degrees, and also in radians, the angle of a regular polygon of 100 sides. c2 S(j ANGLES AND THEIK MEASUREMENT 15. An explorer reaches a latitude of 87° 28' 4«". Find how many miles he is distant from the pole, assuming the earth to be a sphere whose circumference is 25000 miles. 16. Find the gradient of a straight line joining two points whose co-ordinates are {x', y'} and (x", y"). Hence find the equation of the straisrbt line. CHAPTER II THE TRIGONOMETRICAL RATIOS 12. Definitions. Let OA the initial line be taken as axis of X, the axis of// being perpendicuhir to it at ; in the final P ./I N A \ Fi?. VI r. Figure VII position of the rotating line take any point P. shows four possible types of positions of OP. Let fall PX perpendicular to OA or OA produced, so that ON is the abscissa of P and PN the ordinate. Then as THE TRIGONOMETRICAL RATIOS NP ordinate y sine of AGP = cosine of AGP = GP radius r GN abscissa x GP radius r . . . ^^ NP ordinate y *\ a tangent of AGP = -—- = , — ^ = - • 7. a. GN abscissa x ^ These are the most important ratios ; the others are their reciprocals, viz. : G P radius r cosecant of AGP — secant of AGP — cotangent of AGP = NP ordinate y GP radius r GN abscissa x GN abscissa x NP ordinate y TWfollowing abbreviations are usually used: sin A instead of sine of AOP, cos A ,, ,. cosine of ^ OP, tan^ „ „ tangent of .4 OP, cosec A „ ,, cosecant of AOP. sec A ,, „ secant of J OP, cotan ^ ,, ,, cotangent of. f OP. Similarly, if ^OPismeasured in radians, sin 6, cos f/^ cosec \//, &c., are used. 13. Trigonometry was developed by Arabian and Greek astrono- mers who based their work on the circular arc and not on the angle. In the Middle Ages this early mathematical work was translated into Latin, and so the present names of the nitio were derived. The following section shows the reasons for these names. 14. Draw a circle with centre cutting the initial line at .1 and the perpendicular to it at B. Take a point P on the circumference of the circle. Draw the tangent at yl and product' OP to meet it at T. Draw PN perpendicular to ()A. AT wiiH called the tangent of the arc AP. OT, which cuts the circle, was called the secant of the arc AP. THE TRIGONOMETRICAL RATIOS 39 . NP was called the sine * of the arc AP. Clearly the lengths o^ AT, 02] AP change when the radius OA changes, even if the angle .lOP remain constant. But from similar triangles it is seen that, if the angle is constant, the ratios oi AT, OT, NP to the radius are also constant. Hence, as Trigonometry developed, it was seen to he advisable to divide hy * The word ' sine' is derived from the Latin sinus. If in Fig, VIII P^y be produced to meet the circumference at P', then PAP' resembles a bow (Latin arcus) of which PXP' is the string or chord (Latin chorda}. To use the bow, the string is pulled till JV touclies the bosom (Latin sinus^ ; hence PX is called the sine. NA is often called the sagitta of the arc. 40 THE TRIGONOMETRICAL RATIOS the radius and to treat the subject as depending on the angle AOF rather than on the arc A P. Thus we have angle AOP = '—^. — (when the angle is measured in radians), " radius ,^, sine of arc ^P iYP sin AOP = tan AOP = sec AOP =- radius OP tangent of arc AP _ AT _ yP radius OA ON secant of arc AP OT OP radius OA ON Now make a similar construction for the complementary arc BP. Then sine of the complement of the angle AOP __ sine of the complementary arc BP radius ^ XP OP _ ON ~ OP ' '' ' Sine of the complement of ' was shortened into co-sine. Possibly ' complementary sine ' was an intermediate stage. [Similarly, co- tangent and cosecant were derived. Since the values of the ratios depend on the values of the angle, the term Trigonometrical Functions is often used instead of Trigonometrical Ratios. Frequently the ratios are referred to as Circular Functions. 15. Ratios rarely used, in Fig. VIII N A is called the vorsine (i. e. versed sine) of the are AV. X'B is called t])e coversine of the arc AP. AP (not joined in the figure) is the chord of the arc AP. If we divide each of these by the radius we get the corresponding ratios of the angles AOP. These ratios are very rarely used. Another function that is now rarely used is tlie haversino, i. c. half the versed sine. 16. Projection Formulae. It is useful to remember that ON (i. e. the projection of the radius on the initial line) = r cos 6 and iYi^(i. e. the projection ot the radius on a line perpen- dicular to tlie initial line) = ;• sin 0. THE TRIGONOMETRICAL RATIOS 11 17. Polar co-ordinates. The position of a point P is determined if the distance OF from a fixed point is known and also the angle this distance makes with a fixed line OX through 0. The length is usually denoted by y and the angle by 6 ; these are the polar co-ordinates of P. In this connexion is called the pole. 18. Graphs of Trigonometrical Functions. The definitions of the last suction lead to an easy method of drawing the graphs. On page -12 the sine graph (i.e. the graph of the equation «/ = sin x) is given. It is obtained as follows : Step 1. On the extreme left of the paper (which should be ruled in squares) draw a circle with its centre at the intersection of two lines. Take the horizontal radius CA as initial line. Step 2. The perpendicular B'CB gives the angles 90° and 270°. The diagonals through C give the angles 45°, 135°, 225°, 315°. By stepping off chords equal to the radius, starting from A, the angles 60", 120°, 240°, &c., are obtained ; and, by starting at B, •the angles 30°, 150°, &c., are obtained. Only the points P, P' , . . ., on the circumference need be obtained as is shown in the third quadrant ; the radii are not needed for drawing the sine grajih. Step 3. Take a point as origin, some distance along the initial line, and, with a convenient scale, mark off abscissae to represent the angles 30°, 45°, 60°, &c., and, as far as space allows, mark off the negative abscissae. Stej) 4. Through the points on the circumference draw parallels to the initial line to cut the corresponding ordinates. These points of intersection are points on the graph. The ordinates of this graph* are proportional to the sines; if we divide by the radius, the actual values of the sines are found. The sine graph is shown on a larger scale in Fig. X. 19. The tangent graph. To obtain the ordinates for the tangent graph the radii must be produced to meet the tangent to * The giaph of the sine is a wavy or sinuous curve. Tlie name sine is therefore appropriate, although it is improbable that the originators of the name ever drew the graph. -.^ — __ _ — 7^ — 1 — _. / ? yt y 7^ y- ' t S I— * ^ i i \ :is. \, ^g S \ \ \ \ \ p J5 ~t / -»- -/- y i ± -7- it y I o r <5> \^ ' v . ^ - -- ^o ^ N ^ \ :;:: "::""\ -— o ._. ,^ 0> 4 7 ---- - ^ r- - -. - ^ ^ ^^ Ij;-- <-v:'^ ^^ ^^^ %-\ / A 7 , ;2;..^_^^_._/7 r< t..:-s\//.i:.\ ffl_-_ ^^^^ 5, ffl ^'lo ffi ^- V ^' / . :'7 " \: 7- - ,^r / -^ . i Z-_. A ^i --- - ±.-^^' _ ... - - -- ... ^.-. o^ fl> 1 1^ ; t i 1 i ; V ■- ! ^ \ i "" i\i i t4^ -^ i ^ 1 1 \ 1 . o 1 M . •*! ' ' \ i ! i ^1 i ( , , |v^ , 1 ■ ; : . j . ! j \ 1 \i ^ i '^ ill It- ^v^ mitx -1 — ^ 1 — ^ — r \ ' o 1 ■ \ ' 1 ' , fn, ! ; \ 1 i 1 1 \ i \ in 1 It it ^ -X-- "^ 1 1 : \ ' \ ^ a> do ri- <© i> ^- fo fu_\ ... . • . . . ; i 1 . . o 1 ^^— -" I < ^•^ ! .^ y^ / y / "^ X 7 it ' / ' ( ■ 1 / i '/i 1 / 1 1 / ' / i ■ ' 1 1 ' -l-i- \ ! I 1 iffl o 44 THE TRIGONOMETRICAL RATIOS the circle drawn at A. This is done in Fig. TX, and a few points of the graph are marked, but the graph is not drawn. The cosine graph. Since the cosine of x is the sine of the complement of x, the student should be able to modify the method for the sine graph so as to obtain the cosine graph. The secant graph. This is obtained b}^ marking off along the respective ordinates the corresponding values of CT (see Fig. IX). Examples II a. (Answers should be given correct to 2 significant figures.) By drawing to scale find the trigonometrical ratios of the following angles : 1. 30°. 2. 49°. 3. 79°. 4. 100°. 5. 78°. 6. 170°. 7. 250°. 8. 25°. 9. 300°. 10. 156°. 11. -80°. 12. 415°. Find the ratios of the angle ^40/' when the coordinates of P are 13.(4,3). 14.(4,-3). 15. (-4, -3). 16. (-4, 3). 17. (3,2). 18. (-7, -3). 19. (-5,4). 20. (63, -16). The following graphs should be drawn carefully and kept for use : 21-26. A graph on a large scale for each function, for angles from 0° to 90°. 27-32. A graph on a smaller scale for each function for angles from -360° to +360°. 33. The blanks in the following table are to be filled with the sign ( + or — ) of the respective ratios : Angle 0"-90° 90° -180° 180° -270° 270" -360° sine cosine tangent 34. If the gradient of a hill, inclined at A° to the horizon, is known, what trigonometrical ratio of the angle is known ? 35. Construct an angle whose (i) tangent is 1*45, (ii) sine is '75, (iiij cotangent is 1*45, (iv) secant is 2*7, (v) cosecant is 2"7, (vi) cosine is *75. Measure each angle in degrees. 20. Powers of the Trigonometrical Fimctions. The sc^uare of sin^ is written sin^^l ; and a similar notation is used for other powers and ratios ; thus, in general, sin" ^4 means (sin ^4)". THE TRIGONOMETRICAL RATIOS 45 Inverse notation. There is one exception to the above statement. Suppose sin A — a. then A is an angle whose sine is a. This is written A = sin ' ffv Similar!}^, tan"'«, means an angle whose tangent is a ; and so for the other ratios. If w^e wish to express -^ — - as a power of sin J., we must write (sin A)~'^. sin A Note. Continental mathematicians denote tlie angle whose sin is x by arc sin oc. Tlxis notation sometimes occurs in English books. Example. Determine, hy draivlng, the angle siir'^ 5. Step 1. Draw axes OA, OB. [ Step 2. Draw circle centre 0, radius 3 units. Step 3. Along OB mark off 07v equal 2 units. Step 4. Through K draw n parallel to OA cutting circle at FOP. Join OF, OP'. We now have two angles AOP, AOP' each of which has its sine equal to |. AOPifir.^AOP' 139°. Fig. XL 46 THE TRIGONOMETRICAL RATIOS It is always under.stood that siii^^rt means the smallest positive angle that has the sine equal to a ; and similarly for the other ratios. Examples II. 1. Find, by drawing to scale, the sine, cosine, and tangent of 30°, 45", 60°. Verify the results by calcidation. 2. The sine of an acute angle is ^-* ; find the cosine, tangent, and secant. 3. The sine of an angle, not acute, is f ; find the cosine and tangent. 4. The cosine of an angle is j%; find the sine and tangent. 5. Draw as many angles as possible having the tangent equal to -8. 6. Given that sin 63'= '89 find cos 63^ and cos 27°. 7. Find the value of sinM + cosM, it being known that sin A = -3907, cos A = -9205. Also find the values of tan A and sec A. 8. Given tan d = ^fi, find cot 6 and sec 6. 9. Draw and measure an angle A such that (i) sin ^ = —'5, (ii) cos J. = -"5. (iii) tan.i = -'5, (iv) sec ^ = - *5. 10. Find the value of sec- - tun^ 6 , when sec^ = r22l and tan 6 ■= '7002. Justify the answer by geometrical reasoning. 11. Are any of the following data inconsistent or impossible? Give reasons for your answers. (rt) sin A = I; [b] sec ^ = f ; (c) sin A = i, cos A = j ; {d} sin A = •4, cos^l = '6 ; (e) sin .4 = '6, cos A = 'S, tan A = '9 ; (/)sec^l = "35, tanyl = 1*35; ig) tan A — 1 ; [h) sin ^1 = 1; (/) cosec .4=1. 12. Prove, by means of the definitions in § 12, that cosyl = sin (90-^) and tan (^77 -0) = cot ^. 13. Find, by drawing to scale, (a) sin 36^ and sin 144' ; (h) cos 42° and cos 138° ; (c) cos 246° and cos 66 . 14. By means of graphs (or otherwise) test the following state- ments : («) sin (180-^) = sin yl ; (h) cos (180 + ^) = -cos^ ; (f) sin (90 + ^) = sin^. 15. By means of graphs find values for sin-^ *6, tan~^ 2"5 C0S~^ '34, C08~^ I'b. THE TRIGONOMETRICAL RATIOS 47 IG. Given sin36'' = -587;^, find cos 54", sin 144', sin 21G^ sin 324'^ 17. Given cos 53' = -6018, find sin 37% cos 127 , cos 233", cos 413^, cos 307". 18. Prove that sin 117" = cos 21\ 19. Is it possible to find angles to satisfy the following equations ? Give reasons. (i) tan^=l; (ii) cos ^ = :|^g ; (iii) sin ^ + cos ^ = 1 ; (iv) sin^ d + cos^ 6=1; (v) sec ^ = 31416; (vi) cosec ^ = i ; (vii) sin ^ = ; (viii) tan ^ = 100 ; (ix) cos <9 = 1 ; (x) sec <9 = 78. 20. Show that (i) sine and cosine cannot be numerically greater than 1 ; (ii) tangent and cotangent may be either greater or less than 1 ; (iii) secant and cosecant cannot be numerically less than 1. Why is the word numerically inserted ? 21. Find all the trigonometrical functions of 0"^ and 90\ 22. (i) Show that the straight line whose equation is )/ — mx makes an angle tan"^ m with the axis of x. (ii) What is the tangent of the angle made with tlie axis of x by the stmight line joining the two points {x^, yj and {x^, y^) ? (iii) Show that the equation of the line joining the two points , - . ?/ — Vi X- —x. (ivj If the equation of a straight line is y = mx + c, give the geometrical meanings of m and c. 23. Show that, if x be any numerical quantity, positive or negative, an angle can be found whose tangent is equal to x. Show what limitations in value, if any, exist in the case of each of the other trigonometrical ratios. 24. State concisely the changes in the sign and magnitude of sin ^ as ^ increases from 0" to 360^. 25. Define the cosine of an angle of any magnitude, explaining the conventions regarding the signs of the lines referred to in your definitions. Draw the graph of cos 6 from ^ = to ^ = .^ tt. 26. Define the sine of an angle and find by geometrical reason- ing the values of sin 45°, sin 90^ sin 135=. 48 THE TRIGONOMETEICAL RATIOS 27. Define the tangent and the versed sine of an angle ; and find the greatest and least values which each can have. 28. With ruler and compasses construct an angle whose cosine is ^; also an angle whose cosine is — ^. Calculate the sine of the latter angle to three places of decimals. 29. ABC is a triangle in which AN is the perpendicular from A to BC. U AB = 2'9 inches, AC = 2-5 inches, AN = 2 inches, find the values of sin 5, cos C, tan 5, cosec C Calculate the length of ^C correct to one decimal place. 30. If A, B, C are the angles of a triangle, express sin ^ (A + B), cos ?; (A + B), tan \ {A + B in terms of ratios of \ C. CHAPTER III ELEMENTARY FORMULAE 21. Reciprocal Relations. By definitions, sin A = -, cosec A =^-^ ' . '. sin A cosec A = 1 : 1 1 1. e. sm A = T- . cosec A = - — - • cosec A sm A In a similar way it can be proved that cos A sec A = 1. &c. tan A cot A = 1, &c. 22. Relations deduced from Pythagoras' Theorem (Prop. 13, p. 10). In Fig. YII, § 12, we have in all cases i. e. X- + 7/' = r-. Three sets of formulae are obtained by dividing in turn by r, x\ \f. f^ ' x^ iP" '^Divide by n, - + — = 1 ; but - = cos .1. - = sin A. r r Substitute, cos- A + sin- A = 1. The equivalent formulae must also be learnt, viz. : sin- A = 1 — cos^ J., sin^ = + Vl — cos'- A ; cos- ^4 = 1 — sin' J., cos^l = -j- Vl ~ sin"- -4. In a similar way the student should prove that tan^ A -f 1 = sec- A. and cot- A4 1 = cosec- A, 1216 D 50 ELEMENTARY FORMULAE 23. Relation between sine, cosine, and tangent. Substitute, tan A = 5; X ^. t r — X r tan A — sin A cos A is proved that cot A=: cos A sin A 24. Identities. By means of the relations proved in the preceding sections, any expression containing trigonometrical functions can be put into a number of forms. It is a useful exercise to prove that two expressions, apparently different, are identical ; such exercises serve to fix the relations in the memory and lead to facility in dealing with trigonometrical expressions. Example. Prove that scc^ A + cosec- A ^ sec- A coser A. [Express all ratios in terms of sine and cosine.] by § 21 Bthod I. L.H.S. _ 1 1 cos2 A sin2 A sin' A + cos'^ A %\\\^ A cos^J 1 sin'^ A cos' A 1 1 cosM ' sin'^l using Formula of J^ 22 = scc-yi co?L'C-yl. l>v § 21 Q.KD. ELEMENTARY FORMULAE 51 Method II. sec^ A + cosec^ •A = 1 cos^ A + 1 A sin^ A + COS' '■A sin'^ siii'^ A cos^ A 1 1 sec^A cose c^ A = sin^ A cos^ A ' 1 cos'^ A sin^ A sin^ A cos'^ A ' .•. sec^^ + cosec^^ = sec^^ cosec^^. Q.E.D. Method III. This method is clumsy, and should be used only if Methods I and II have been tried unsuccessfully. sec'^ J. + cosec'^^ = sec^^ cosec*^, j^ _1_ ^ 1 _ ^ 1 cos'^-4 sin^^ cos^^ sin'^^ ' i.e. if sin^^ + cos'^J. = 1. But sin'^^ + cos'^^ does equal 1 ; sec"^4-cosec-^ = 860^^-4 cosec^^. , Note. The introductory * if, or some similar conjunction, is vital to the logical statement of the work and must not be omitted. 25. Elimination. If two equations are satisfied by the same value of a single variable, there must be a relation connecting the constants of the equations ; this is also the case when n equations are satisfied by the same values of n—1 variables. In order to find this relation we eliminate the variable or variables. Example. Eliminate 6 from the equations si7i d = a, tan 6 = h. By formulae tan^(9= — sTi sin2^ Substitute h^ 1 1 l-8in2<9 a' l~a' ' I.e. - - — = 1. a^ Tt' The result is called the diminant of the original equations. D 2 52 ELEMENTARY FORMULAE Examples III a. 1 If sin A = ^§ use formulae to find the remaining ratios. Draw a figure to explain why some of the ratios may be either positive or negative. 2. Given that tan 6 = V find cot 6 and sin 6. 3. Find sec S in the following cases : (i) cos 6 = "7921 ; (iij tan d = 1'352 ; (iiij cosec 6 = 2'583. 4. Show how all the ratios may be found when (i) the cosine, (ii) the tangent is known. 5. Prove the following identities : (i) sin A cot A + cos A tan A = sin A + cos A : (ii) tan A + cot ^ = sec ^ cosec A ; (iii) sin 8 tan ^ + cos ^ cot ^ = sec ^ + cosec ^ - sin ^ — cos ; (iv) sec2 6 - cosec^ 6 = tan^ S - cot^ ; (v) l-2sinM = 2cosM-l; . . tan a + tan (:i _ siiux cos /^ + cos a si n 3 , '■^^^ 1 — tan a tan (:i ~~ cos a cos /3 — sin a sin j3 ' (vii) (1 -tanM) -r (1 ^tan^^) = cosM -sin^4 ; (viii) (sin A + cos A)"^ = 1 + 2 sin ^ cos A ; (ix) sin^^-cos^^ = (sin ^ — cos^) (1 + sin A cob A) ; (x) sin A cot A cosec A + cos A tan A sec A = sec A cosec u4. ; (xi) tan X -tan Y — (sin X cos Y—cobX sin Y) -^ cos JV cos Y; (xii) (tan J[ - tan B) -r (cot A-cot B) = - tan ^ tan 5 ; (xiii) cos^ 3 - sin* 6 = 2 cos^ ^ - 1 ; (xiv) (3-4 sinM) ^ cosM = 3-tanM. 6. Prove that versin ^ = 1 -cos ^, coversin ^ = 1 —sin A. 7. Show that the numerical value of sin^^ -^ (1 — cos^) di- minishes from 2 to as A increases from 0° to 180°, and illustrate your answer by a diagram. 8. Which is greater, the acute angle whose cotangent is 4, or the acute angle whose cosecant is | ? 9. Prove that, if 6 is an angle less than 180" for which 1 4 sin 6 = 7c cos 6, then cos 6 ~ 2k -i- (1 + l^) : and express tan in terms of k. 10. Eliminate 9 from the following : (i) sin d ■■= a, cos 6 ~ h; (ii) sin d-\ cos 6 = a, sin 8 - cos 6 — h ; (iii) sec 6 - tan B = a, sec 9 + tan = },; ELEMENTARY FORMULAE 53 (iv) a sin d + h cosS = p, a sin d-b C08 d = q; (V) a sin 6 + b cos 6 = p, «' sin 6 + 1/ cos 6 = j^'. 11. If rf (1— sin^) = h cos 6, prove that & (1 + sin ^j = « cos ^. 12. If a (sec ^+ 1 j = b tan 6, prove that 6 (sec d-l) = a tan ^. 13. If a; = a cos ^ cos (f), y = a cos ^ sin ^, z = a sin ^, eliminate ^ and 0. 26. Ratios of complementary angles. Let XOF^ A° I Fig, XII) and XOQ - 90-^° ; make OQ = OP, and let fall QK, PN perpendicular to OX. K NX Fig. XII. Then the triangles QOK, POX are congruent (Prop. 8c); so that KQ = OX and OK = XP. Hence sin XOQ = ^ ~ OP = cos XOP. i. e. sin (90 — A) = cos A. In a similar way it is proved that '^ cos (90 -A) = tan (90 -A) = Compare these results with § 11. What are the corresponding results when angles nro measured in radians ? 27. Ratios of supplementary angles. Mako XOP = .1^, and XOQ = 180 ->!'', so that QOK = A°. * The student is expected to complete thee formulae. 54 ELEMENTAKY FORMULAE Make OQ = OP and let fall the perpendiculars PiS', QK. Then the triangles QOK, PON are congruent (Prop. 8 c) so that 0K= Oy (in magnitude) and KQ = NP. But OK and ON are of opposite sign. Y Q ^\ p \ V ) N y X \ y 1 y^ Y ^ r~ < o ¥- W X Fig. XIII. Hence cos,XO^ ■=-- OQ - z9^ OP = - cos XOP, i.e. cos (180- A) = -cos A. In a similar way it is proved that sin (180 -A) = tan (180 -A) - [ OK ) * (where [OK) denotes the magnitude of OK with the proper sign prefixed) 28. Ratios of negative angles. Make XOP = + ^" and XOQ = - A". Then XOP = XOQ in magnitude. Make OQ = OP. Join PQ cutting OX at N. Then in the triangles PON, QON OP = OQ, ON is common, included angle NOI' = included angle NOQ. Nl' = NQ in magnitude, and ONP= ONQ, so that PQ is perpendicular to OX. * In writing it is usual to use tlu; symbnl OK to denote length preceded by correct sign ; it is more convenient to print {OK). ELEMENTARY FORMULAE {OX} DO Hence cos XOQ e. cos (-A) = cos A. Y OQ _ (ON) OP = cos A' OP, 0\ Q Fic^. XIV. In a similar way it is proved that sin ( — A) = tan ( — A) = The student should also work out the ratio of 90 + ^4, 180 + ^, 270-^, &c. 29. By means of the hist three sections the ratios of any angle can be expressed in terms of the ratios of an acute angle not greater than 45°. For example cos 139' = cos (180" - 41' j = - cos 4r, cos 246' = cos(-114'j = cos(114='j -cos(180'-6G") = -cos 66° = -cos (90' -24^) = -sin24^ 56 ELEMENTARY FORMULAE It is usually easy to work directly from the figure; thus in cos 246° = Fig. XV, where XOP = 246° and XOQ = 66°, (ON) OP OQ — cos 66° -sin 24°. Y N 1 Q K P 0/ X Fig. XY. 30. Ratios of 0° and 90°. If XOV =0,7' and A coincide ; so thnt XP = 0, OX = OP. Hence sin 0° = ^ - 0. If XOP ^ 90 , then PX fulls along the y axis and X coincides with the origin 0. In this case NP= OP and OX = 0. Hence ELEMENTARY FORMULAE NP 57 sin 90 "" = cos 90' OP ON 6F 1, ^ „ NP NP tan 90^=^^= -^ = 00 31. Ratios of 30°, 45°, 60°. Make XOP equal to 30", Fig. XVI. Let fall PX perpendicular to OX. Make XOQ equal to 30"" in magnitude, and produce PA" to meet OQ in Q. Fio. XVI. Then, by Prop. Sc, the triangles PON, QOX are congruent. It follows that the triangle OPQ is equilateral. * The syml>nl x means 'infinity', i.e. a nnml;er greater than any number we can imagine. Consider the value of 1/x as x gets smaller and smaller, 1=10. 4 = 1000, .^j -_. 1000000, As X diminishes, 1/x increases, and, by making x suificiently small, We can make 1/x exceed any assigned value however great. This is expressed thus: when x = 0, 1/x = co . Or more generally, if a is a constant, then a/x = oo when x = 0. 58 ELEMENTARY FORMULAE Hence also 0^ Hence sin 30*^ cos 30° = ' tan 30° = Similarly sin 60° = OP. PX = IPQ since PX = QN = \0P since PQ = OP ; 0.Y-' = OP'-PN^ = OP'- 1 OP'- = ^^OP"". VB 2 NP OP 2 Ni' 1 f = -866, OP ON •866, tan 60" = -v/3 = 1*732. As an exercise the stndent should find the values of the ratios of45^ sin 45^ = cos 45° = tan 45"" = Fig. XVIL 32. The very small angle. In Fig. XVll let the circular measure ol" the angle AOP be $. ELEMENTARY FORMULAE 59 Then arc AP - rO, NP - r sin 0, AT = r tan ^. Hence Area of triangle AOP =^ lOA . NP = |7-2sin^; Ai-ea of sector AOP = \r^6; Area of triangle^ or =\OA.AT But, if ^OP is an acute angle, Triangle AOP < sector ^ OP < Triangle AOT, i e. Ir^sin^ <\r'e i. e. sin S <6 This relation is true for throughout by ;• we have rBin^< rS < rtan^, < * ;^ tan 6, < tan d. any acute angle (Prop. 16) (§7) (Prop. 16) if we multiply NP < arc AP< AT. Fi-. XVIII. But, as the angle diminishes, these three lengths more and more nearly coincide ; and are practically indistinguishable when the angle is very small. This is shown in Fig. XVIII, which also shows that ON is indistinguishable from OA. Hence, when 6 is very small, there is very slight error in saying NP = arc AP = AT, and OiY = OA. 60 ELEMENTARY FORMULAE Substituting the trigonometrical values for the lengths of these lines, we have sin S = 6 = tan 6 and cos ^ = 1 , when 6, the circular measure of the angle, is small. This may also be expressed thus : The limit of or of ^^^ > when 6 is zero, is 1 ; or in symbols T sin ^ = . «i T tan <9 , ^ L —X- 1 an Jj -.- = 1.* 33. Error involved. Whatever be the value of ^, it has been shown that cos^^ + sin'^^ = 1. Using the above approximations, we have This last statement is true only when 6"^ is so small as to be negligible. Hence If ^ is so small that $- maybe neglected, we may say that sin S = e, cos ^ = 1, tan 6 = 0. It is shown in Higher Trigonometr}'- that sin i^' = ^ gives correct results if }r6^ is negligible. Example. // accuracy /.s required to four decimal places, find the sine of 1 degree. 1° = ToTj 77 radians = -01745 radian ■017452= -000295. (We are not, therefore, justified in saying cos r = 1.) •01745' = '000005. (This does not aft'ect the first four places so we may use the approximation sin 6 = 6.) Hence sin 1° = '0175, correct to four decimal places. Examples III b. 1. Write down the sine, cosine, and tangent of (i) 150°, 240°, 330°, 840^ (ii) 60°, 800°, 135°, 225"^; (iii) 180°, 270-^, 405°, 210°. 2. Find the secant and cosecant of G0°, 45°, 120°, 225^ 3. Use Ihe definitions of § 14 to find the ratios of 180-^1. * Fur explanation of the wonl ' limit " seo School Algtlra, I'art ii, p. 440. ELEMENTARY FORMULAE 61 4. Correct, if necessary, the following statements : Bin (180 -.4) = cos^ ; cos (270+ ^) = -cos^ ; tan(180 + ^)=tan^; sec (90-^) = sec ^ ; cot (90 + ^) = cot ^. 5. In a right-angled triangle the hypotenuse is 5 feet long and one of the angles is 60° ; find the lengths of the other two sides. 6. A ladder 25 feet long is leaning against a wall and is in- clined 45° to the horizontal ; how far up the wall does it reach ? 7. Find sin 1', correct to 3 significant figures. 8. Find sin 10', cos 10', tan 10' correct to 5 decimal places. 9. What angle does a halfpenny (diameter 1 inch) subtend at the eye when at a distance of 10 feet ? 10. A p3st 25 feet high subtends an angle of 30' at a certain point on the ground. How far from the post is the point ? 11. Find approximately the distance of a tower which is 51 feet high and subtends at the eye an angle Sfy'. 12. Prove that tan^ 60° - 2 tan^ 45^ - cot- 80° -2 sin- 80° - f cosec- 45°. 13. Find approximately the number of minutes denoting the inclination to the horizon of an incline which rises 5^ feet in 420 yards. 14. In any triangle show that cos(^ -\~B)=- cos C, sin (B + C) = sin J, tan {B +C)= - tan A . Write down the other similar relations. Oral Examples. Fill in the right-hand sides of the following equalities: 1. (i) sin2^= 2. (ij sec^J-tan^^ = (ii) sin 45' = (ii) cos 60' = (iii) cos 135° = (iii) tan d = (iv) tan^TT = (iv) sin (180 -^» = (v) sinyl cot^ (v) sec (90-5) = 3. (i; cos2 60° + sin^60" = 4. (i) cos (9 tan ^ = (ii) cosec'^ C = (ii) 1 - sin' x = (iii) cot^7r= (iii) tan 210' =- (iv) cos (180-^) = (iv) cos-^A = I v) cot^ d = i v) cos'^ i TT + sin'^ i T 62 ELEMENTARY FORMULAE 5. (i) 1 + cotM = 6. (i) cos-i ^ = (ii) sin ^ cot ^ = (ii) cos 225° = (iii) sin(180 + ^j = (iii) cos (90 + A) = (iv) cos^ 63° + sin^ 63 = (iv) cos(-^) = (v) tan 330° = (v) cos (180 -Z^j = 7. (i) tanirr - 8. (i) sin (360-^) - (ii) sec 60' - (ii) sin-i 2 = (iii) sin^ J^ + cosH^ = (iii) sec^ i tt — tan^ ^ rr = (iv) tan (|7r + ^) = (iv) cos 0° = (v) tan 135° = (v) cosec 120° = 9. (i) tan 150° = 10. (i) sec 150° = (ii) cos ^/ sin 6 = (ii) cos (360°-^) = (iii ) cos 90° = (iii) cos-^ ^Z = (iv) sec 240° = (iv) sin 77° cot 77" = (v) cot (180-^)= (v)tant7r = 11. (i) tan 1200°= 12. (i) 008^23° + cos' (ii) tan (180° + ^)= (ii) cos (270° + 5) (iii) tan|7r= (iii) sin-^ •4 + cos^(^= (iv) tan 15° cot 15° = . (iv) sin.(-(/)) = (v) tan-M-l)= (v) sin 225° = Examples III. 1. Prove, from first principles, that sin (90 + ^) = cos ^, cos(180 + ^) = -cos^, tan(360-^) = -tan^. 2. Show that sin (180-^) = sin ^, when A is (i) obtuse, (ii) between 180° and 270°, (iii) between 270° and 360°. 3. Show that cos (90 - ^) = sin J., when A is (i) obtuse, (ii) between 180° and 270°, (iii) between 270° and 360°. 4. Show that tan (180 + ^) = tan ^, when A is (i) obtuse, (ii) between 180° and 270°, (iii) between 270°, and 360°. 5. Give 6 different solutions of each of the following equations : (i) sin A=\\ (ii) sin A= \. \ (iii) cos Q = ^^ ; (iv) tan ^ = 1 ; (v) cos ^ = - ^ ; (vi) sin ^ = - ^ • 6. Show that all angles having the same sine as A are included in one or other of the forms: 180/? + ^, if n is an even integer, ELEMENTARY FORMULAE 63 180 71- A, it' n is an odd integer ; and that these are included in the single form 180 « + ( — !)" A where n is any integer, positive or negative. 7. Show that all angles having the same cosine as A are in- cluded in the form 360 n ± A, where n is any integer. 8. Show that all angles having the same tangent as A are included in the form 180 n± A, where n is any integer. 9. What do the forms of the three previous examples become when the angle is measured in radians ? 10. If a small angle equals A°, what is the value of sin A ? 11. Show that using the approximation sin ^ = ^ is equivalent to regarding a circle as a polygon with a large number of sides. 12. What do the following equalities become when the angle S is so small that 6^ is negligible ? (i) sin2^ = 2sin^cos^; "^' (ii) cos 2 ^ = 1 - 2 sin2 d ; (iii) sin (S + cf)) = sin 6 cos ^ + cos 6 sin 'p ; (iv) C = G tan ; (v) t'2 = 4^,.sin2,i^. 13. Two strings are tied to two pegs A and B in the same hori- zontal line, and knotted together at C; when the strings are pulled tight, it is found that ^C is 18 inches long and that the angles CAB^ CBA are 30^ and 60" respectively ; how far apart are the pegs and how far is C from AB ? 14. An inclined plane, length 4 feet, is inclined at 30" to the horizontal, what is the length of the base ? 15. A pendulum is held so as to make an angle of 30" with the vertical, what is then the distance of the end of the pendulum from the vertical line through point of support ? 16. Prove the following identities: (i) cot^ A cos^ A = cot^ A — cos'^ A ; (ii) sec"^ A — sm"^ A = ta.n'^ A + coi"^ A ; (iii) sin^ (cosec ^ — sin 6) = cos^^; (iv) (cos A + cosec A) (sin A + sec A) = 2 -f sin yl cos ^ -F sec A cosec A : (v) (cos J. -f sec A) (sin.l -^ cosec .^j = sin A cos .1 4 2 sec A cosec A : (vi) sec A - sin A tan A = cos A ; (vii) (sec A — cosec ^) (sin A + cos ^Ij 4 sec^ ^-1 cot .4 = 2 tan A. 64 ELEMENTARY FORMULAE 17. (i) If 6 and cf) dift'er by ^ n, prove that tan 6 tan = — 1 ; (ii) Show that the lines whose equations are, respectively, 1/ = mx and y = inx, are at right angles if mm' = — 1 ; (iii) Show that the graphs of the equations ax + hy + c = 0, ax + b'y + c = are at right angles if aa+hb' = 0, and are parallel if a/a' = b/V. 18. If tan 6 = b/a, find the value of a cos d + h sin 6. 19. If tan^ B = b/a, show that «/cos 6 + b/sin 6 = a^ + &3) 2. 20. Give a general formula for all values of A which satisfy the equation cos J. = — 1 . 21. If a sin^ 6 + b cos^ 6 ^ c and a cos'^ /9 + & sin'^ (9 = d, prove that « + & = c + rf. 22. From the vertex Cof an equilateral triangle ABC a perpen- dicular CD is let fall on AB ; DC is produced to E so that CE equals CA, and JLJ? is drawn. From the resulting figure find the sine, cosine, and tangent of 15° and 75°. 23. A is an angle between 180° and 270°, also cos ^4 ^ -'i\ find the value of cosec A + tan A. 24. Define the cosine of an angle of any magnitude and express the cosine of an angle between 180° and 270° in terms of each of the other trigonometrical ratios. If cos ^ = — 65? fin^^ sin^, sec B, coiB, and explain any double signs which occur in your answer. 25. Prove the following identities : (i) (sin Jl cos J5 + cos A sin B)"^ + (cos A cos B — ^m A sin BY= 1 ; (ii) sin^ B + cos*^ ^ = 1 - 3 sin^ B cos^ B ; (iii) cot .4— tan A = sec A cosec ^ (1 — 2 sin"^) ; (iv) (1— sin ^-cos A'f = 2 (1-sin A) (1 —cos A) ; (v) (2 cos J. — sec -4)-r (cos yl-sin ^)= 1 + tan A. (vi) (3 sin B cos^ B - sin^ B)^ + (cos" B-Sco&B sin^ Bf = 1 ; (vii) sec' B - tan' ^ = 1 + 3 tan^ B sec^ B ; (viii) versin (270° + ^) . versin (270°-^) = cos^ J. 26. Prove that cos (180° -yl) = -cos^, and cos (90° + .4) = -sin.l. For what values of A is tan A = v/8 and sec A = -2 ? 27. Solve for x the equations : (i) .rM 2 .;>• sec o 4 1 = 0; (ii) x"^ + 2 X cos a = sin^ a ; (iii) x"^ + (tan a + cot a) a; + 1 = 0. ELEMEXTAKY FORMULAE 65 28. Prove that the number of ^^ecolKl^^ in an angle whose circular measure is unity is 206,265. The moon subtends at the eye of an observer an angle of 30', its distance is 240,000 miles, find its radius. 29. If tan-^ = |, find versin 0, and explain the double result. 30. Eliminate from ( i) a tan + bcotd = c, a' tan ^ -f /^' cot ^ = c' ; (ii) a tan ^ + & sin ^ = r, a tan 6 + h' sin 6 = >:'. Revision E:samples A. i. Define the tangent of an angle. From your definition find tan 45^ and tan 135^, and prove that tan iItt-O) = cot 6. 2. A surveyor goes 10 chains in a direction 35^ S. of E., then 7'8 chains U^ E. of S. ; then 5'6 chains 10' N. of W. Find by drawing how far he is now from his starting-point. 3. Prove the relation 1 -sin-^ = cos- .4 for the case where A lies between 90 ~ and ISC. Show that (sin ^ + cos^j* = l -f 4 sin.4 cos .4 + 4sin^^-4sinM. 4. The gi-adient of a railway is 1 in 270 ; find the inclination to the horizontal to the nearest second. 5. When the sun's altitude is 60', find the length of the shadow cast by a vertical rod whose length is 10 feet. 6. Draw the graph of cos x between x = 15" and x — 135' with- out usinor tables. 7. Explain how to find the length of the arc of a circle of given radius, when the angle subtended at the centre is given in degrees. A wheel, radius 4| feet, rolls along the gi-ound ; what hori- zontal distance does the centre travel when the wheel turns through 157' ? 8. Why is the secant so called ? Prove that the secant is the reciprocal of the cosine. Given sec A = 2^, find tan A and sin A. 9. Show that the graph of the straight line y = 2x-Tj is in- clined to the axis of x at an angle tan"^ 2. Verify this by a care- ful drawing. 10. Trace the changes in sin ^ as 6 changes from 0" to 360' and exhibit these changes by means of a graph. 121C E 6Q ELEMENTARY FORMULAE 11. Find the smallest angle which satisfies the equation 3 cos^ + 2sin2^ = 0. Give also four other solutions. 12, If sin A = f, prove that sec A + 1/cotyl = 2. 13. What is a radian ? Prove the formula arc = rx 6. Show that if 6 is small, sin ^ = ^ api^roximately. 14. Show that tan (180 n + A) = tan^ where n is any integer. If tan 3^ =^3, state three possible values for A that do not differ by 360°. 15. Find the value of the expression cosec yl — j: cot yl, if sin A = ff (i) when A is acute, (ii) when A is obtuse. 16. Prove the identity 2 sin ^4 cos ^ = (2 tan J) H- (1 + tan^ A). 17. In a triangle ABC, C == 90°, AB = 15, sin ^4 = '37 ; find the length of^C and 5(7. 18. Criticize the following statements: («) sin2 (9=4; (h) sin 6 tan ^ = 1 ; (c) sin-^ ( - -3) = 170° ; {d) sin ^ + cos^ ^/sin 6 = tan 9. 19. Explain clearly what is meant by latitude. A place has latitude 30° N., what is its distance from (i) the earth's axis, (ii) the Equator, measured along the surface ? (Radius of earth = 4000 miles.) 20. Give a definition of cosine that applies to angles of any size. Prove that cos (180 — ^1) = — cos^. If sin A = l^ and A is obtuse, find cos A. 21.. Prove that (cos A cos 5 + sin A sin i?/ + (sin ^ cos 5 - cos A sin Bf = 1. 22. Draw the graph of y = seer from ;r = 0° to .r = 180°. 23. What is meant by the statement that tan 90° = oo ? Is sec 90° equal to tan 90° ? Give reasons. 24. Construct an angle A such that cos J = -5 and tan^l is positive. 25. Name the points of the compass between West and South. How many degrees are there in the angle between SW. by S. and S. by E. ? ELEMENTARY FORMULAE 07 26. Find the values of sin45% cts^^ J. tan \7t. 27. Prove by means of a figure that sin- A + tan- A = sec- A - coa- A. Is this true when the angle is measured in radians ? Give reasons. 28. Construct an angle such that its tangent = ^ and its versine is greater than unity. 29. Prove the identity cos^ J. - sin-^l = (coL-^-1) -h cosecM. 30. Find the value of tan from the equation 3 tan-^' = 2-v/3tan(9-l. Hence find three different values of ^ that satisfy the equation. 3L Write down six positive angles which have the same cosine as the angle a ; and find the positive values of 6 less than two right angles which satisfy the equation sin4^ = cos 5^, 32. Show how to find by calculation the value of sin 30' correct to four decimal places. Verify, by substitution, (i) sin 60' = 2 sin 30' cos 30' ; lii; sin 120' -sin 60' = 2 cos 90' sin 30' ; liiij cos 60' - cos 120' = 2 sin 30' sin 90^ 33. Prove the identities : (i) cosec- J. — cotan- .4 = 1 ; .. 5jfl3^nirf , 12 -13 cos 6i _ * "^ r2 + 13'cos e^ 5 - 13 sin ^' ~ 34. A steamer travels along the equator from longitude o7V W. to longitude 5' 30' E. in 4 days. ^Yhat is the distance travelled in nautical miles ? What was her average rate in knots ? 35. What is meant by the chord of an angle ? For which angle is the chord equal to unity ? Explain how to draw an angle when a table of chords is given. 36. Express the following ratios as ratios of angles not greater than 45' : sin 172', cos 412', tan 246', sec 76', cosec 147', sec 236', cot 138=, cosine 150°, sin 67', tan 102°. E 2 68 ELEMEI^TAEY FORMULAE 87. If the circumt'ercii'es ot* the quadrants of two circles be divided similarly to the right angles they subtend, what would be the radius of a circle divided according to the French scale, in which the length of the arc of one grade would be equal to the length of the arc of one degree on a circle whose radius was 18 feet? 38. Point out which of the trigonometrical functions are never numerically less than unity, and which may be either less or greater than unity. Express the numerical values of sin 135° and tan 150° with their proper signs. 39. If n be a positive v/hole number, show that the angles (2??. 180° + ^) and {(2;? 4 1)180°-^} have the same sine as y1. Express these in a single formula. 40. Distinguish carefully between (sin A)~^ and sin"^^. Show that cos-i |. + 2 sin"' i = 120°. 41. Trace the changes in sign and magnitude of the expression cos re -sin a; as .r increases from to 27r. Illustrate your answer by a graph. 42. A church spire, whose height is known to be 45 feet, subtends an angle of 9' at the eye ; find its distance approximately. 43. What is meant by tan"'' m ? If y= mx + c represents a straight line, state the geometrical interpretation of the coefficients m and r? What is the angle between the lines whose equations are ij = x-4:, y = a/Sx + 2'? 44. Show that the equation of the line joining the points 45. Find the equation of a line passing through the origin and (i) parallel to, (ii) perpendicular to, the line whose equation is y = mx + c. Deduce the conditions that the two lines whose equations are ax + hy + c = 0, a'x + h'y + / = 0, should be (i) parallel, (ii) per- pendicular. ELEMENTARY FORMULAE 69 46. Find the eijuatioii of the line joining the origin to the point F whose co-ordinates are {x', y'). Find the equation of the line perpendicular to OP and passing through P. Hence show that the equation of the tangent to a circle at the point x\ y' is xx ■\- yy' = v"-, the equation of the circle being X- + ?/- = ;•-. 47. If (;•, 6) are the polar co-ordinates of a point, what locus is represented by (i) r = 3, (ii) ^ = ^ tt, (iii) r cos = 5, (ivi r = 5 cos 6? 48. If (x, y) are the Cartesian co-ordinates, (r, 6) the polar co-ordinates, of the same point, what relations connect them ? Express the equations of the previous example in Cartesian co-ordinates. Express (i) x^ + y'^ — 4:X + hy — 1, (ii) 3x'-f4// = 5 in polar co- ordinates. 49. The sum of two angles is 3 radians, their difference is 10 degrees. Find each angle in degrees, assuming that 4377 = 135. 50. A ring, 10 inches in diameter, is suspended from a point one foot above its centre by six equal strings attached to its circumference at equal intervals. Find the angle between two consecutive strings. CHAPTER IV USE OF TABLES 34. It has been shown in the previous chapters that the trigonometrical ratios of any angle may be foancl roughly by drawing to scale or by means of graphs. By methods which are explained in more advanced books on Trigonometry, the ratios can be calculated to any required degree of accuracy. There are many collections of tables published, containing not only the actual trigonometrical ratios (the natural functions as they are called) but also the logarithms of these ratios. These collections differ slightly in their arrangement, but the following general remarks apply to most of them. 35. Since any ratio of any angle is equal in magnitude to the same ratio of some angle less than 90°, it is necessary to tabulate the ratios only for angles between 0° and 90°. Thus sin 156° = sin (180° -24°) = sin 24°, cos 215° = cos (1.^0° + 35°) = - cos 35°. But the tables may be made even shorter, for any function of an angle between 45° and 90° is equal to the complementary function of an angle less than 45°. Thus sin 76° = sin (90° -14°) = cos 14°, ^ tan69°= tan(90°-21°) = cot21°. This fact is used in two different ways. Some tables give all the ratios for angles from 0° to 45° ; so that if, for instance, sin 72° is required, it must be looked up as cos 18°. Other tables give the values of sine, tangent, and secant for angles from 0°to 90° ; in this case, cosine, cotangent, and cosecant must be looked for as the si]ic, tangent, and secant respectively of the complementary angle. The slight mental work involved is avoided by giving each column a " footing " as well as a heading. Thus '26892 is, in some tables, found on a page headed Natural Sines, on a level with 15° in the extreme left hanl column and under 36', i.e. '26892 = sin 15°36'. USE OF TABLES 71 Bat the same page has Natural Cosines at the bottom, "26892 is on same level as 74° in the extreme right-hand column and above 24', i.e. -26892 = cos 74°24'. A few minutes' inspection will make the arrangement of any set of tables quite clear. 36. Logarithmic Functions. Since the sine and cosine cannot be greater than unity, their logarithms cannot be greater than zero ; hence these logarithms have a negative characteristic. In order to avoid difficulties of printing it has been the custom to add 10 to all these logarithms, and to the other logarithmic func- tions. The values thus tabulated are called Tabular Logarithms and are denoted in writing by L, thus L tan 75^ = log tan 75'' + 10. Some of the modern tables give the ordinary logarithms with the negative characteristics. When tabular logarithms are used it is advisable to sub- tract 10 mentally and to work with the correct logarithm. 37. Interpolation. It is impossible to give the ratios for all angles. Four-figure tables usually give values for every 6', seven- figure tables for every 1'. Intermediate values may, in some tables, be found from side columns giving the differences^ as in the case of ordinary logarithms. If these side columns are not given, the method of proportional parts * must be used. This method is equivalent to assuming that the graph of the tabulated function may be treated as a straight line for portions lying between the points corresponding to two consecutive tabulated values. The practical use is easily followed from an example or two. Example i. Given that sm28°9'.= -4717815, ami sin 28^ 10' = -4720380, find sm 28=^9' 43". sin 28^ 10' = -4720380, "000004275 x 43 sin28'' 9' = -4717815. '00017100 Increase for 60" = '0002565 ; 1282 Increase for 43" = |^ x '0002565 '00018382 = '0001838; sin 28" 9' 43" =-4719653. ^ For a fuller treatment see School Algclm, Part IT, p. 376. 72 USE OF TABLES In practice the zeros are omitted as in the following example. Example ii. Given that Zo^ cos 73° 15' = 1-4058617, and log cos 73^16' = 1*4053816, ^/j72cl the angle ivhen the log cosine is 1-4056348. Denote the angle hy 73° 15' x'\ - log cos 73° 15' = T-4058617. log cos 73= 15' - 1-4058617. log cos 73° 15' x" = T-4056348. log cos 73° 16' - 1-4053816. Decrease for £c" = 2269. Decrease for 60"- 4801. X 2269 2269 ^^"^^ 60^4801- ^1 .r-28. 4801)136140 4012 171 .". required angle = 73° 15' 28" to the nearest second. Note. It is important to recollect that cosine, cotangent, cosecant, and their logarithms decrease as the angle in- creases ; consequently proportional differences must be suh- tr acted, not added. If the graphs of the functions are carefully drawn, it is seen that in some parts they approach much more nearly to straight lines than in others. It follows that the method of proportional parts is more accurate for some angles than for others. For a complete discussion of Proportional parts see Nixon's Elementary Plane Tngonometnj (Clarendon Press) or any advanced textbook. Examples IV a. Find, from tables, the natural function of the following angles, find the logarithm of the number found, and then look up the logarithmic function in the tables. There may be a slight dis- crepancy in the fourth decimal place. 1. sin 17= 15'. 2. cos 73° 47'. 3. tan 16° 39'. 4. cos 23° 19'. 5. sec 67° 15'. G. cotan44°5'. 7. tan 78° 53'. 8. sin 83° 43'. 9. cos 63° 28'. 10. sin 156° 17'. 11. tanl7G°16'. 12. cot 100° 10'. 13. cos 137° 42'. 14. sin 126° 37'. 15. tan 173° 14'. Explain carefully the difficulty that arises in connexion with some of the angles. USE OF TABLES 73 16. Find the Cartesian co-ordinates of a point whose polar co- ordinates are (i) 17, 16°; (ii) 25, 114°; (iii) 49, 227^ Find the angles less than 180' which are determined by the following data : 17. sin e = '8732. 18. cos A = _'8469. 19. sin 5= -9340. 20. logtan .4 = 1*7932. 21. Lcos^= 9-7432. 22. sec B = 2'5732. 23. logsin^ = 1-3465. 24. L tan .4 = 10-4385. 25. L cote = 10-7386. Find the sine, cosine, and tangent of the following angles, which are measured in radians : 26. ^n. 27. iV-- 28. 1-2. 29. ^;r. 30. Verify that sin 112^ = sin 70' cos 42' -f cos 70° sin 42'. ' . 31. Find from the tables the values of ■ I sinjV77 and sin 27' 18'/cos 32"' 45'. " ' — • 32. Employ the tables to verify the formula cot 24° 45' - cot 49= 30' = cosec 49' 30'. 33. Find the values of cos 110", cot 160°, sin 250". A quantity fi is such that /n = sin //sin ;• ; complete the following tables : I r H- 34. W 12" 35. 26° 18' 1-427. 36. 31° 52' 1-467. 37. 53° 49' 1-5. 38. Find the polar co-ordinates of points w^hose Cartesian co- ordinates are (i) (3, 7) ; (iii (-3, 7) ; (iii) (-3, -7) ; (iv) (3, -7). 39. The angle of friction e and the coefficient of friction fx are connected by the relation /^t = tane. Determine the missing quantity in the following cases : 6 40'' 15' i 17=39' I 47' 8' : | ,1 I -67 I -37 ' -50 40. In a circle of radius 17 find the lengths of chords subtending angles (i) 37°, (ii) 73", (iii) 143° at the centre. What are the areas of the corresponding segments V 74 GEAPHS 38. Graphs. Example. Draw the graph of 3 sin (^+30°)- 2 cos (a;-30°) from x =^ 0° to x =^ 120°. In other words, draw the gra2)h of ^j = S sin {x + 30°) - 2 cos (ic- 30°). X •500 15° 30^ 45° 60° 75° 90° •866 105° 120^ •500 sin (x + 30^) . •707 •866 -966 1-000 •966 .0. cos (x-30°) . •866 •966 1-000 -966 •866 •707 •500 •259 0-000 3 sin (x + 30°) 1^500 2-121 2^598 3-000 2^898 2^598 2-121 1-500 2 cos (.x-30°) 1^732 1-932 2-000 1-732 1^414 •1-000 •518 •000 y -•232 •189 -598 -966 1-268 1-484 1-598 1-603 1-500 The graph is shown in Fig. XIX. Use of Graph. Interpolation. The value of the function can be found for any intermediate value of the angle. From the graph it is seen that v/ = r07 when x = 50°, and y = 1'54 when x = 81°. Calculation shows that the correct values are 1'075 and 1*542 respectively. This is a useful method of testing the accuracy of a graph. Maximum and Minimum. When, as x increases, ij continually increases to a certain value and then decreases, that value is said to be a miaximum ; similarly, when j/ first decreases and then increases there is a minimum value. These maximum and minimum values are clearly shown on the graphs ; the corre- sponding points are called turning-points. From the graph the maximum value of 3 sin (.r + 30°) - 2 cos {x - 30°) is found to be 1*62, and the correi^ponding angle is OO"". GRAPHS 75 x S 1 - it I 1 t ^ \f\ o 1 ^ ^\r -^^ \^ J JV ^ Ul 4^5 I^^ 1 X-^ ^^di i v5 ° \^ ^ - ^x ^!s ^^ \^^ "^ ^s: ^ ' T ^^ ^ oK ^, o \ (0 ^ \ \ ^^ ^ \ - \. \ V h ib ^ •- -■ o n -__L 76 GRAPHS Kate of change of the function. The graph shows that, when X changes from 15° to 30°, the increase in y is more than when x changes from 60° to 75°; consequently the curve is steeper between 15° and 30° than between 60° and 75°. Thus the rate at which y changes compared with x is shown by the steepness of the curve. Join two points P and Q on the graph, and draw PK, QK parallel to the axes to meet in K. Then increase in v/ KP , ■ n , i i nr^r.- ^ — i' = = tangent of the angle PQK increase in x QK = tangent of the angle PQ makes with the axis of x. This is called the slope of the line PQ. When Q approaches indefinitely near to P, the chord PQ becomes a tangent. Hence The rate of increase of y at the point P is measured by the slope of the tangent at the point P. Notice that the slope diminishes in the neighbourhood of a turning-point and is zero at the turning-point itself. 39. Solution of equations. By finding where two graphs intersect or where one graph intersects the axis of x or a line pai'allel to the axis, equations can be solved just as in Algebra. Example. Solve the equation 3 sm (a;-}- 30°) -2 cm- (a; -30°) - r5. It is seen in Fig. XIX that the graph cuts the line whose equation is ?/ = I'S where x = 77° and x = 120°. These are, therefore, the solutions within the range of the graph. Examples IV b. (The graphs should be verified in the way that the example of § 38 is verified.) Draw the graphs and find the turning-points of: 1. sin^ X from it; == 0° to a; = 90°. 2. cos 1 X from x= -90° to x = 90°. 3. sin ^ ic -f cos ^J- a; from x = 15° to x — 135°. 4. I tan {x-m°) from x = 0" to x - 90°. 5. ^-sin e from ^ = to ^ - ^tt. 6. sec X — tan x from O'' to 90°. GRAPHS 77 7. cos- i ar-f sin- ] x from to 360\ 8. Draw the graph of sin .r + cos.r between o-- = and x = 360°. Solve sin 3-+ cos .r = '89, and find the slope of the graph at the points corresponding to these values of ,r. 9. Draw the graph of cos.r between the values of and 2 tt for x\ Show that an acute angle can be found to satisfy the equation X = cos X. 10. Draw the graphs from x = -1 to j^^ = -f 1 of (i) sin-^r, (ii) cos-^rr, (iii) tan-^a;. How are they related to the graphs of sin.r, cos.r, tana? respectively? 11. Draw the graphs whose polar equations are (i) r sin 6 = 11 ; (ii) r =10 sin ; (iii) ;• = 10 cos 6 : (iv) tan 6 = 2-45. 12. Find from your tables the values of cos 2 x for the values 0^ 10", 20% 30^ 40^ 50\ 60^ of .r. Draw the graph of cos2.r-cosa; as x increases from 0° to 60°. 13. Find, by drawing graphs of 2 sin .4 and sin 2.4, for what values of A, less than 90', 2 sin ^ — sin 2.4 = 1, 14. Find, by the aid of the tables, the values of sin ,r — tan2a: for the values 0=, 10^ 20", 45°, 60' of x. Make a graph to give the values of sin.r — tan2.r from ,r = to X = 60°. 15. Make a table giving the values of cos^ at intervals of one- fifth of a radian from ^ = to ^ = two radians, taking the radian as 57^30'. From your table plot the graph of t^cos^; and hence find for what value of 6, between the limits and 2, ^cos^ is greatest. 16. Plot the function ^ {sin ^-f. sin 2 (^ + 20°)} between <9 = 0^ and 6 = 180°, and find the maximum and minimum values of the function which occur within this range, and the corresponding values of d. 17. Draw, in the same diagram, the graphs of sin a- and 2 cos a; between x = 0° and x = ISO". Show how to find from your diagram an angle whose tangent is 2. 18. Taking tt as 8*1416 and using your tables, find the values of ^-sinf? when 3 = ^7r, ^tt, ^ tt, ^tt, f^rr, and -^ tt ; and hence make a graph to give ^ — sin ^ from 6 = to 6 = Itt. 19. Draw the graph of (i) sin~^a'-l-cos~^ic ; (ii) sin-'(l/a:). 78 SOLUTTON OF EQUATIONS Solution of Equations. 40. To solve a trigonometrical equation, (i) express all the ratios involved in terms of one ratio, (ii) find the value of tliis ratio by ordinary algebraical methods, (iii) find the angle from the tables, (iv) give the general solution. Example i. Solve 2 sin .r-f 3 cos a? = 2. Express in terms of sine, -^ + 3 V*! - sin^iP =2 — 2 sin x. Square 9 — 9 si n^ a; = 4 — 8 sin x + 4sin^ x. Transpose ISsin^aj-S sina7-5 = 0. Factorize (13 sin x + h) (sin x-\) = 0. .'. sin ^ = — j5 or 1. Substituting in the original equation, we find that : (i) If sin. r = — j%, cos^r = \'i = '9281. Hence the bounding line is in the fourth quadrant. From the tables it is found that cos 22° 37' = -9231. Hence the smallest positive angle satisfying the equation is 360"-22"37' = 337''2r. But we may add or subtract any multiple of 360° without altering the position of the bounding line ; hence any angle satisfies the equation whose value is 360° n + 337° 23', where n is any integer positive or negative. This is the general solution. (ii) If sin a; = 1, cos.^ = 0. The smallest solution is x = 90°. The general solution is 360°n + 90° or (4n + l)90°. Note. The same difficulty has arisen here tliat arises in Algebra when the original equation contains surds. After we have squared, the i-esulting equation is exactly the same as if we liad started with the equation 2 sin x — 3 cos x = 2. For this reason, after we found the value of sin 9, it was necessary to substitute in tlie original equation to find the corresponding value of cos x. SOLUTION OF EQUATIONS Example ii. Solve tan- ^ + 4 sin'^ A = 5. Express in terms of tan ^. 79 tanM + 4 tan^^ 1 + tan-^l tan^ ^ = 5. 4 log tan ^ = -6990. log tan ^ = -1747. = log tan 56° 13'. ^ = 56° 13'. Consideration of the fundamental figure shows tliat the general solutionis ^ = 180>i + 56° 13'. Multiply by l + tan^^ Take logarithms Use tables Hence a solution is 41. General Solutions. General solutions can always be obtained by mentally considering the possible positions of the radius vector that give angles having the same function as some angle already found. This is what has been done in the two preceding examples. It is, however, useful to know the formulae that give these general solutions. Find an expression for all angles that have a sine equal to sin oc. We have to solve sin 6 = sin a. Fig. XIX a. The bounding line may have either of the positions shown in Fig. XIX a. en 80 SOLUTION OF EQUATIONS Thus the line may revolve through n.r, where ii is even, and then go on CX, or may revolve through mr, where )i is odd, and th come back Oi. Hence ^ = n rr + 3c if n is even, or nrr — y if 71 is odd. These are included in the one formula e= 1177 + {-iyo(. If sin X = sin A, then x = ISO n 4 ( - 1 j" A. Exercises. In a similar way prove that i. 6 = 2mr±0(, when cos d = cos 0( ; X = S60 u ± A, when cos.r = cos^. ii. 0=1177 +CX, when tan^=tanCX; ;r ='180 n + ^, when tan .-c = tan ^ . Example. Solve sin 3^ = cosb^. This is the same as sin 36 = sin (i/r-5^); 3^ = n7r + (-l)"(lrr-5^). If n is odd Sd = mr-}>7T + 5e; 20 = 1 - n TV. Put 11 = - 2 i? + 1, then = ^; 77 - 1 rr, where ^; is any integer. If n is even 3 ^ = ??. tt + (In - 5 ^) ; Put n = 22) ^ = P^T^ + it^ ^' where p is any integer. The complete solution is d =2)rr- ^ 77 or jj Itt + ^tt. Examples IV c. Solve : 1. 2cos2(9 = 3(l-sin^). 2. sin^ + cos^=l. 3. sin + cos = 72. 4. 12 tan^ .4 - 13 tan A f 3 = 0. 5. 2 008^0-- 1 = l-sin^.r. 6. sin 3^ = sin 4^. 7. 3cot^<9-10cot2^+3 = 0. 8. 2sin .4 = tanvl. 9. tan^ + 3cot^ = 4. 10. sinj.r + ^) = cos(a;-^). 11. tanM + 4sinM = 6. 12. v'3tan2^ + 1 = (UV3)tan^. 13. cosec^ = cot ^+73. 14. cos(135° + ^)+sin(135°-^) = 0. 15. cos^ ^ - cos ^ sin A - sin^ yl = 1. 16. cos3^ + sin(9 = 0. 17. 3tan2 2^ = l. USE OF TABLES 81 18. tan2.r = tan2/;r. 20. 2cos2^-f3cos^-l =0. 22. l-7sin^--73 = 0. 24. 2tan2^ + 7tan^-f3 = 0. 19. 2sin2^-3sin^-2 = 0. 21. tan2^ + sec-^-2. 23. 3 sin (9 + 2 cos = 2. 25. tan^-2cotr' = 17. 42. Examples of the use of logarithms. Example i. Given that a ^ 250, h = 240. A = 72' o\ sin .4 a sin A n)i B find B wJicn We have 1. e. sin 5 ~V sin 5 a b sin A Take logs. log sin B = log h + log sin .4 — log r/ = 2*3802 -2-3979 + r*9784 = 2-3586 -2-3979 = f-9607 ; [.-. Lsin5= 9-9607]: .'. sinjB = sin66^ Hence jB = 180n' + (-1)» 66^ After a little practice the work may be arranged so that the logarithms are kept quite distinct from the remainder of the work. This same example is worked below to show the shorter method and the use of five- figure tables. Logarithms. sinB sin-B smA a"' 240 sin 72"^ 5' ^250 sin 65^ 59'. 2-38021 + 1-97841 2-35862 2-39794 1-96068 Example 11. The sides and angles of a irianglc are con- nected hij the relation tan \ [A —B) — — -, cot \ C: find A a)>d •^ ^ ^ a-\-h B irhen a 1216 242o, 6 = 164-3, C= 54^36' F 82 USE OF TABLES -^-^ '^-2 cot 27=18' 2 - 406-8 """" r 78-2 , , = tan 20° 26' .'. ^? = 20-26', by question ^^^ = 62-42-. Hence A = 83" 8', 5 = 42" 16'. Logarithms. 1-89321 + -28723 2-18044 2-60938 1-57106 The step in brackets is required if the tables do not give the cotangents. Since A and B are angles of a triangle, 1{A-B) cannot ec^ual any of the angles 180 n + 20° 26' (except when n = 0). so that there is no need to give the general solution. Example iii. // a, h, c are the sides of a triangle, (g y\ [g (A fan lA—^ ^ — -^ — —-^ iclierc s is half the sum of the V s(s—a) sides. Find A when a = 1762, h = 893, c = 1386. Logarithms. s = (1762 + 893 + 1386)-f-2 = 4041-f-2 = 2020-5, 305212 s-a= 258*5, 280243 s-h = 1127-5, + 5-85455 s-c = 634-5 ; 3-30546 ^ A 71127-5x634-5. ' * "^^^2 ~V 2020-5x258-5' 2-41246 -5-71792 .-. ^ = 49° 29', 2)0-13663 006831 A = 98° 58'. Examples I"V. 1. Use the tables to find the values of sin 52'', cos 140^, tan 220". cos 340", sm340^ 2. Divide sin52 ])y cos 52'; verify your answer liy finding the value of tan 52 ' from the tables. USE OF TABLES 83 3. Write down by using tables the values of sin 140', cos 160", cos 220°, tan 320°. 4. Find the smallest positive value of B which satisfies cos = sin {(4 m + 3) | tt + 0(}. 5. Find all the values of 6 which satisfy the equation 4cos6J-3sec^ = 2tan^. 6. Find the inclination to the horizon of an incline which rises 5i feet in 420 yards. 7. Solve the equation tan-t^ - (1 + ^/S) tan (9 + ^3 = 0. 8. Given that tsui^C = \/(s-a) {s-h)-=r s{s-c), find C when a = 32, b = 40, c = 66. 9. Solve the equations cos(2x + 3 //; = |, cos(3.r + 2 //) = ^^3/2. 10. Given log 2 = '30103 and log 3 = -47712, find (without the tables) L sin 60= and L tan 30° . 11. Find the acute angle whose cosine equals its tangent. 12. The current C in a circuit, as determined by a tangent galvanometer, equals G tan 6, where 6^ is a constant depending on the galvanometer only and 6 is the deflexion of the needle. Determine the ratio of two currents which give deflexions of 27' 14', 35' 23' respectively. 13. The length of a degree of latitude in latitude (p is (11 11-317 - 5-688 cos 0) 10* centimetres. Find the length at London (latitude 5r3rN.) and Melbourne (latitude 37"50'S.). 14. The length of the seconds pendulum in centimetres, at a place whose latitude is A, is 99-3563 --2536 cos 2A. Find the length of the seconds pendulum at Paris (lat. 48° 50' N.) and Calcutta flat. 22' 33' N.). 15. The acceleration of a falling body at a place whose latitude is X, when measured in centimetres per second per second, is 980-6056-2-5028 cos 2 X. Find the acceleration at Montreal (lat. 45° 30' N . i and Cape Town (lat. 33° 40' S.). 16. A quantity A is determined by the relation A = ^ ah sin C. Complete the following table : A a b C i. 17 43 IT 14 ii. 342-6 21-3 38-19 iii. 984-2 43-82 f2 43 21' 84 USE OF TABLES 17. Draw the graph of tan 6-0 from ^ i= to ^ = ^• Hence solve tan 6 = 6 + S. 18. Given that A and B, the angles of a triangle, are connected by the relation asm B = fesin^, find B when a = 181, b = 217, A = 34° 15'. 19. If 2i? = a'sin^, find the value of A when i? = 179*4 and a = 300. 20. Verify that cos 146' 43' - cos 56" 51' = - 2 sin 44° 56' sin lOP 47'. 21. Find the length of (i) the chord, (ii) the arc, subtending an angle 70° at the centre of a circle of radius 25 cm. Find also the area of the segment. 22. Find the length of the side of a regular decagon (i) inscribed in, (iij described about, a circle of radius 2 "7 inches. CHAPTER V THE RIGHT-ANGLED TRIANGLE 43. In the previous chapters we have had to deal with only one angle at a time, and have been able to draw one of the lines containing that angle horizontal. In applications of Trigonometry we often have to deal with several angles in the same example, and the lines containing them are drawn in various directions ; in such examples it would be difficult to apply the definitions of § 12. But it has been shown that the ratios of any angle can be expressed in terms of the ratios of an acute angle. In practice, therefore, it will often be found advisable to use the following definitions, which apply only to acute angles. In a right-angled triangle an acute angle is contained by the hypotenuse and one of the other sides which is called the side adjacent to that angle. The remaining side is called the side opposite. Then in Fig. XX 86 THE KIGHT-ANGLED TRIANGLE sin BAG = cos BAG = tan BAG = opposite hypotenuse ' adjacent hypotenuse ' opposite ^ adjacent ' cosec BAG = sec BAG cot BAG = hypotenuse opposite ' hypotenuse adjacent ' adjacent opposite These are clearly the same definitions as in § 12, the triangle BAC taking the place of the triangle PON; and the various formulae proved in Chap. Ill can be proved directly from the definitions of this section. 44. It is usual to denote the angles of any triangle ABC by the capital letters A, B, C; the lengths of the sides opposite the angles A, B, C are denoted by a, b, c respectively. Hence, in a triangle ABC, right-angled at C, a sin^ I.e. a = c sin A cos A = -, c tan^ &' i.e. b = ccos^ e. a = &tan^. Examples Va. 1. Prove, from the definitions of § 43, that (i) cos ^ = sin 2? = sin (90 — ^) ; (ii) sin A = cos B = cos (90 — ^) ; (iii) tan A = cotB = cot (90 - A.). 2. P N M Fig. XXI. THE EIGHT-ANGLED TRIANGLE 87 In the figure PNO, QMO, QKP are right angles. If 0N=o, NP=1, OM=G, MQ = o, find the values of sinPO.V, inn KPQ, tan KQP, sec QOM, cos KQO, cosec XRO. 3. If, in Fig. XXI, OP = 8, POQ = 30", QON = 45°, PQO = 90^, find the lengths of OQ, PQ, PK, QM, OM. 4. A circle is described on a horizontal diameter ^5 of length 10 inches; a point C is taken on the circumference, such that BC = 7, and CD is let fall at right angles to AB. Find the size of the angle BAC and the length of CD. 5. In a triangle, right-angled at C, a perpendicular is let fall from C to the hypotenuse ; prove, by Trigonometry, that this perpendicular is a mean proportional between the sides containing the ris^ht angle. D C B ^\ A Fig. XXII. In the above figure (which is not drawn to scale) AO is at right angles to DE, OC is at right angles to AG, OG is at right angles to AO and EF', also G is the middle point of AB. Use this figure in the following examples. 6. UAC= 10, CAD = 40°, find, if possible, the lengths of all the other lines. 7. If CD = S, AB = 24, find sin CAD. 8. If GF= 18, AE=b, OC = 5, find cos ACD and the length of^a 88 THE RIGHT-ANGLED TRIANGLE 9. If AB = I, CAD = 0, find CD and AO. 10. If CG = a, CGO = 6, find AD and AO. Fig. XXIII. With the ordinary notation for the sides and angles of a triangle, find in the above figure : 11. The length of AD when c = 70, i? = 49°. 12. The length of AD when 1= 42, C = 72°. 13. The length of BD when c = 76, B = 39°. 14. The length of CD when h = 114, C = 114°. (What geometrical fact does the negative sign in the result show ?) 15. Prove that the area of the triangle = i«& sin C; give the proof, also, when C is obtuse. Solution of Right-angled Triangles. 45. The angles and sides of a triangle are sometimes called the six parts of a triangle. The determination of all the parts, when only some of the parts are known, is called solving the triangle. If the triangle is known to be right- angled, the triangle can be solved if one side and one other part are known. Example i. A man, standinci 100 feet from the foot of a clmrelt stee2)le, finds that the inu/le* of elevation of the top * If a i)er.son is looking upwards, the angle Ins line of sight makes with the horizontal is the angle of elevation ; similar!}', if lie is look- ing downwards, the angle his line of sight makes with the horizontal is the angle of depression. THE RIGHT-ANGLED TRIANGLE 89 Ls 50°. If Ms cue is 5^ feet from ilie gyoiind, what is the height of the steeple ? [The figure should be drawn neatly but need not be drawn to scale.] In Fig. XXIY AE represents the steeple, BC the man; CD is drawn parallel to BA. [Mental. In the right-angled triangle CDE we know that CD = BA= 100 ft., angle DCE - 50^ and w"e wish to find DE. unknown side Here some ratio of known anglt = tan DCE, known side DE . BC i.e. /)i^ = 100 tan 50° feet = 100 xl'1918 feet = 119-18 feet. Therefore height of steeple = AD + DE= 124'68 feet. Example ii. The shadoir, cast by the sun on a horizontal plane, of a vertical pjole 10 feet high, is ohscrred to he 14 feet long ; find the altitude of the sun [i. e. the angle of elevation of the sun). 90 THE RIGHT-ANGLED TRIANGLE In Fig. XXV AB represents the pole, AC the shadow; so that CB is the direction of one of the sun's rays. loF^ [Mental. In the right-angled triangle BAC ^ve know 5^ and AC, and wish to find the angle ACB. ratio of known sides = some ratio of required angle.] — = tan ACB ; ^^^ ^^, ^^, ^ .^^^^9. .-. tan ACB = ]^ = -71429. 440 is diff. for 10'. .-. ^CJ5 = 35°32'. .-. 100 Sun's altitude = 35° 32' to nearest minute. 10_Q' _ 0' 44 Here sec BAC: Example ill. A ship C is observed at the same time from two coastguard stations A and B, 1459 yards apart. The angle ABC is found to he 90°, and the angle BAC to he 67° 14', ivhat is the distance of the ship from station A ? AC AB .'. log^O = logl459-+logsecG7°14' = 3-16406 + -41111 + 120 = 3-57637, i.e. ^0=3770-3. Distance of ship from A = 3770 yards to nearest yard. Fig. XXVI. ^ If tlie tables do not contain the secants, the working must be made to depend on the cosine. THE RIGHT-ANGLED TRIANGLE 91 AC 1 AB~ cos BAC log AC = log 1459- -log cos 67° 14 = 3-1641 _ 1-5877 = 3-5764, .e. AC = 3770. Example iv. Ttvo men, A and B, 1370 yards axmrt, observe an aeropJane C at the same instant and find the respective angles of elevation to he 40° and 67°. If the plane ABC is vertical, ralcidate the height of the aeroplane. Fig. XXVII. Let h feet be height of aeroplane. From triangle ADC, AD = h cot 40°. From triangle BDC, BD = h cot 67' ; but AD + BD = AB; .-. hcoti0° + hcot6r= 137. 1370 cot 40° + cot 67" 1370 1-61622' .-. h = 847-62 ; Height of aeroplane = 848 yards to nearest yard cot40°= 1-19175 cot 67°= -42447 Logarithms 8-13672 - -20852 2-92820 Examples V b. 1; The string of a kite is known to be 500 feet long, and it is observed to make an angle of 55° with the horizontal ; find the height of the kite. 92 THE RIGHT-ANGLED TRIANGLE 2. From the top of a cliff, 215 feet high, the angle of depression of a ship is observed to be 23° 20' ; what is the distance of the ship from the foot of the cliff? 3. From a point 56 feet from the foot of a tree the angle of elevation of the top is 73'' ; find the height of the tree. 4. The top of a conical tent is 9 feet above the ground ; the radius of the base is 5 feet ; what is the inclination of the side of the tent to the horizontal ? 5. The shadow thrown by a flagstaff is found to be 55^ feet long when the sun's altitude is 53° 15'; what is the height of the flagstaff? 6. I know that a certain tower is 144 feet high. I find that its elevation observed from a certain point on the same level as the base of the tower is 37° 16'. Find the distance of that point from the base of the tower. 7. A sphere of radius 4 inches is suspended from a point A in a vertical wall so that it rests against the wall. The string is 11 inches long and is in the same straight line as a radius of the sphere. Find the inclination of the string to the vertical. 8. From the top of a cliff, 254 feet high, the angle of depression of a ship was found to be 9° 28', and that of the edge of the sea 72° 40'; how far distant was the ship from the edge of the sea? 9. Two observers on the same side of a balloon and in the same vertical plane with it, a mile apart, find its angles of elevation to be 15° and 65° 30' at the same moment. Find the height of the balloon. 10. From the top of a tower, 108 feet high, the angles of de- pression of the top and bottom of a vertical column are found to be 30° and 60° respectively. What is the height of the column ? 11. A flagstafF, 80 feet high, is fixed in the centre of a circular tower 40 feet in diameter. From a point on the same horizontal plane as the foot of the tower the elevations of the top of the flagstaff and the top of the tower are observed to be 35° and 30"^ respectively. Find the height of the tower, 12. A river, the breadth of which is 2C0 feet, Hows at the foot of a tower, which subtends an angle 25° 10' at a point on the further bank exactly opposite. Find the height of the tower. THE RIGHT-ANGLED TRIANGLE 98 13. A person standing at the edge of a river finds that the eleva- tion of the top of a to'»ver on the edge of the opposite bank is 60'' ; on going back 80 feet he finds the elevation to be 45'^; find the breadth of the river, 14. From the top of a tower. 50 feet high, the angle of depression of a man, walking towards the tower, is noticed to be 30' ; a few moments after it was 45'. How far had the man walked between the two observations ? 15. Two p6.=!ts, 400 yards apart, at the sides of a straight road running E. and W., are observed to bear N. 20° E. and E. 20° N. respectively. Find the distance of the observer from the road. 16. Two points A and B and the foot D of a tower CD are in a horizontal straight line, and the angles of elevation of C, the top of the tower, as seen from A and B respectively, are 25" 46' and 35=25'. If the distance AB is 200 feet, find the height of the tower. 17. A vertical post casts a shadow 15 feet long when the altitude of the sun is 50= ; calculate the length of the shadow when the altitude of the sun is 32°. !>:. A vertical mast, having its base at A, is set up on a horizontal plane. B and C are points in the plane in a line with A, and such that the angular elevations of the top of the mast, when observed at these points, are respectively y and (3. If tan ^ = f , tan ,3 = f and the length of BC is 105 feet, find the height of the mast. 19. A man standing on a tower at a height of 80 feet from the ground observes that the angles of depression of two objects on a straight level road running close to the foot of the tower are 60° and 30°. If the objects are on the same side of the tower, how far are they apart ? 20. A, B, C are three points in succession on a straight level road, and P is another point so situated that the angles PAB, PBA, PCA are respectively 90°, 60°, and 45°. If a man walks at a uniform rate from ^ to JB in 25 seconds find, to the nearest second, how long it will take him, at the same rate, to walk from Bio C. 21. A ray of light passes through a hole A in a graduated horizontal scale AB in a direction perpendicular to the scale and is reflected by a vertical mirror which is distant 30 inches from the scale and makes an angle x'' with the incident (i.e. approaching) 94 THE RIGHT-ANGLED TRIANGLE ray. After reflection the ray makes the same angle with the mirror as before and shines on the scale at a distance 8 inches from A. Find the value of x. If the mirror now swings through an angle 1°, how far will the spot of light on the scale move ? Elementary Navigation. {The student sliould revise §§9 and 10 dealing with latitude and longitude and the points of the compass. [ 46. When a ship is sailing, the angle between its direction of sailing and the meridian the ship is crossing is called the course. If the course is constant, the ship is said to sail on a rhumb-line. The distance between two positions of the ship is then measured along the rhumb-line. The difference of latitude of two places is the arc of a meridian intercepted between the parallels of latitude passing through the two places. The departure ' between two meridians is the dis- Fig. XXVIII. tance between the two meridians measured along a parallel of lati- tude ; thus the departure between any two given meridians is not a constant but diminishes from the equator to the poles.* 47. A small portion of the earth's surface may be regarded as a plane ; for distances small, compared with the earth's radius, we may therefore use the formulae of Plane Trigonometry. Plane Sailing is the name given to that part of navigation which treats the surface of the eaith as a plane. On this assump- tion the meridians become parallel straight lines, the rhumb-line becomes the hypotenuse of a right-angled triangle of which the departure is the side opposite to the course, and the difference of latitude is the side adjacent. Thus problems on Plane Sailing are merely examples in the solution of right-angled triangles. * In navigation distances are usually measured in nautical miles ; a nautical mile is the length of an arc of a meridian (or the equator) which subtends an angle of 1' at the centre of the eartli ; thus a distance of 75 nautical miles is usually written 75'. THE RIGHT-AXCxLED TRIANGLE 95 Examples V c. (The distances are given in nautical miles.) 1. A ship sails SE. by S., a distance 81 miles; what is her de- parture and difference of latitude ? 2. A ship sails N. 49"^ 41' W., a distance 73 miles ; what is the departure and difference of latitude ? 3. A ship sails SSW. until its departure is 198 miles , what is the distance sailed and the difference of latitude? 4. If the course is 3^ points W, of N., and the difference of latitude 149 miles, what is the distance ? 5. A ship sails between North and West, maldng a difference of latitude 157^ miles and departure 79 miles ; what is the course ? 6. A ship sails westward 247 miles along the equator from meridian 16" E. ; what is now the longitude ? 7. A ship sails 247 miles eastward along the parallel 40" N. : what is the change in longitude ? 8. When a ship sails any distance (great or small) along a parallel of latitude, show that difference of longitude in minutes = departure x secant of latitude. 9. A ship, from latitude 54' 22' 10" N., sails 195^ miles I of a point S. of SE. ; what is now the latitude ? 10. Leaving latitude 49' 37' N., longitude 15' 22' W., a ship sails SW. by W. 150 miles ; find the new latitude and longitude. Fig. XXIX. 48. Parallel Sailing. If X' is the latitude, then the radius of the parallel of latitude (AT in Fig. XXIX) is cos X x radius of the 96 THE RIGHT-ANGLED TRIANGLE earth. If B is the radian measure of the difference of longitude of two places on the same parallel, the length of the arc between them is 6 cos X x radius of the earth. The radius of the earth is 21600 ,. , ., -^^ — nautical miles. a TV iience departure = - ?, — 6 cos X. When d is reduced to minutes, this relation iDecomes departure = difference of longitude x cosine of la,titude. 49. Middle Latitude Sailing. In Middle Latitude Sailing, the departure between two places, whose latitudes are X and X', is taken to be the departure between their meridians, measured at the latitude |(X + X'). On this assumption, departure = diff. of longitude x cos |(A + X'). 50. Traverse Sailing, If a ship sails on different courses, from A to B, from B to C, from C to Z>, &c., then, by the methods of Plane Sailing, the total changes in latitude and longitude can be worked out. This is called the method of Traverse Sailing. This method can only be used when the whole area traversed can be regarded as plane without introducing a great amount of error. Example. A ship left a j^osition in which Oporto Light (lat. 41° 9' N., long. 8° 38' T^.) lore W. hy N., 15 miles distant. Afterwards she sailed as under : Courses. Distances N.W. 70' S. by W. ^W. 55' E. 35' N.N.W. 42' S.E. 51'. Find her bearing and distance from the Light in her last position. We liave a sories of right-angled triangles to solvo, the hypotenuso and an acute angle being given in each case. In practical navigation special tables are userl, calU'd Traverse Tables. THE RIGHT-ANGLED TRIANGLE 97 The angle the hypotenuse makes with the meridian is taken iu each case. Oto A. Hypotenuse 15', angle 7 points = 18^ 45'. ri761 Diff. of latitude = 15' x cos 78= 45' 1-2902 -1 2-926' S. ^63 Departure = 15' x sin 78M5' 1-1761 = 14-71' E. r-9916 ri677 C Fig. XXX. Note that bears W. by N. from A, but A bears E. by S. from 0. A to B. Hypotenuse 70', angle 45=. Diff. of latitude = 70 x cos 45= = 49-497' N. Departure = 49-497' E. Bio a Hypotenuse 55', angle 1| points = 16= 52|'. Diff. of latitude = 55 cos 16= 52 V = 52-64' S. Departure = 55 sin 16° 52i' = 15-97' W. The other triangles are worked in the same way. •7071 1-7404 1-9809 1-7213 r7404 r4628 1-2032 98 THE RIGHT-ANGLED TRIANGLE Tabulate the results thus : Distance. Diff. of Latitude. Depai ture. N. S. E. W. E. byS 15' 2-93 14-71 NW. 70' 49-50 49-50 S.byW.iW. 55' 52-64 15-97 E. 35' 35 NNW. 42' S8-80 16-07 SE. 51' 36-06 36-06 88-30 91-63 88-30 99-21 68-10 68-10 3-33 31-11 We see now that the final difference of latitude from the light is 3-33' S., and departure 31*11 E. ; so that we have to solve a right- angled triangle given the two sides. 8*33 -5224 31*11' 1-4929 F0N=6°r. 1^95 ^^_ 31-11 1*4929 In Fig. XXX tan FON = cos7^( :>N 1*9975 = 31*29 1*4954 In her final position the ship bore G'^?' S. of E., 3r3 miles distant from the Light. To find thf lo7igitude of the ship. Latitude of 41° 9'N. Diff. of latitude for F 3*3' S. Latitude of F 4V 5-7' N. Middle latitude 41" 7'. Difference of longitude in minutes = departure cosine of middle latitude 31*11 1*4929 ~ cos 41° 7' -1-8770 = 41*30'. 1-6159 Longitude of i''= 8°38'-41*30' = 7" 57' W. THE RIGHT-ANGLED TRIANGLE 99 Examples V d. 1. Find the distance on the parallel between Cape Agulhas (lat. 34°50'S., long. 20M' E.) and Monte Video (lat. 34° 50' S., long. 56° 9' W.). 2. A ship steamed at the rate of 12 knots from Albany (lat. 35° 3' S., long. 118° 2' E.) to Cape Catastrophe (lat. 35° 3' S., long. 135° 58' E.}. How long did she take on the voyage ? 3. A ship sailed from Port Elizabeth (lat. 34° T S., long. 25°40' E.) SE. i^ S., until her departure was 397' ; find her final position. 4. Find the course and distance from Syracuse (lat. 37° 3' N., long. 15° 15' E.) to Fano (lat. 39° 52' N., long. 19° 19' E.). 5. A ship left a position from which Cape Clear (lat. 51° 26' N., long. 9° 29' W.) bore NE. by E. 12*5 miles distant and sailed South 150' and then West 290 miles. Find the bearing and distance of Cape Clear from the ship in her last position. 6. Find, by Middle Latitude Sailing, the departure between two places whose positions are 13° S., 50° E. and 20° S., 60° E. 7. A ship sails from 50° N., 50° W. to latitude 48° N., the distance being 157' ; find the new longitude. 8. Cape Ortegal (lat. 43° 45' N., long. 7° 6' W.) bore SW. JW. 12 miles distant. Afterwards sailed as under : Prue Courses. Distances. NNW.iW. 70' ESE. 85' NNE.fE. lor S. 50' wsw. 92' Find the final latitude and longitude. 9. A ship left the Texel (latitude 52° 58' N.) and then sailed W. by N. 34', S. by E. 45', W. by S. 35', SSE. 44', WSW. iW. 42'. Find the course and distance to Dungeness which lies 139' West of the Texel in latitude 50° 55' N. ' 10. A ship, latitude 17° 10' N., is making for a harbour, latitude 13°10'N., and 180' W. of the ship. She sails SW. by W. 27', WSW.|W. 30', W. by S. 25', W. by N. 18', SSE. 32', SSE.f E., 27', S. by E. 25', S. 31', SSE. 39'. Find the course and distance to the harbour. g2 100 THE RIGHT-ANGLED TRIANGLE 11. A ship left a position in which Heligoland bore ENE. 12', and then sailed NW. 24', S. by W. 20', NW. by W. 32', S. by E. 36', WNW.iW. 42', SSE. AE. 16', W.fN.45'. What is then the position of the ship ? Heligoland lies 54° 12' N., 7° 54' E. • 12. A ship sailed from Barcelona (41°25'N., 2"10'E.) SE. by E.|E. until she reached latitude 36°2rN. What was then her longitude ? 13. A ship left a position in which Sable Island (43°24'N., 65° 36' W.) bore NW. | W., distant 12 miles. Afterwards sailed as under : Courses. Distances. ESE. 72' SW.iW. 37' NNE. 42' E. 25' Required the latitude and longitude reached. 51. The Double Angle. In Fig. XXXI, the Sing\eBAC = A° ; onAB [i semicircle is described with centre 0, so that angle BOC =2A. Let fall CN perpendicular to AB. N B COS 2 ^ = ON OC _ AN- AO ~ OC ~ _AN__ "" OC m :hej — 1 hypotenuse of the triangle of ' 2 OC which AN is a side. J THE RIGHT-ANGLED TRTANC^I^ n A ^J Fill in the vacant places with the _ AN AG__ ~ ag'ab = 2cosM- 1. Exercises. In a similar way prove i. sin 2 ^ = 2 sin A cos A. ii. 003 2.4= l-2sin2^. Deduce iii. cos 2 ^ = cos^ A - sin^ A. . ^ ^ , 2tan^ IV. tan 2A = l-tan^^ V. sin ^ = 2 sin | ^ cos ^ ^ ; cos A = cos^ \A- sin^ \ A. vi. 2 cos*^ i ^ = 1 + cos A, vii. 2sin^^yl = 1-cos^. sin J. 1— cos^ viii. ■ ■ ... , , , /I— COS^ SI 111. tcin IA= . I, = ^ — 'NJl + cos^ 1 + cos>4 sin J. ix. Prove the formulae for sin 2^ and cos 2^ when 2yi is obtuse. X. Do these proofs apply to angles of any size? If not, between what limits do they apply ? Why is the ambiguous sign omitted in viii ? 52. Geometrical questions may often be solved by using Trigonometry. For example : If from a point outside a circle a secant and a tangent be drawn, the rectangle contained by the ivhole secant and the part outside the circle is equal to the square on the tangent. 102 I'HE EIGHT-ANGLED TRIANGLE In Fig. XXXII it is required to prove that rect. PA . PB = eq. on PT. r> Fig. XXXII. Let radius = r, OP = c, and angle OPB = ^, angle OAC = (/). PA = PC -AC = ccos6 — rcos(j). PB = PC+CB = PC+AC (Prop. 21) = ccos^ + rcos0 PA.PB = c' cos- e-7^ COS^ = c^ — i^ — c^ sin'^ 6 + 7-^ sin^ (p. But c sin ^ = OC from triangle OPC = rsincp from triangle OAC. Hence PA . PB = c" - r"" = OP^-OT^ = PT2 since OTP is a right angle. 53. Known results in Geometry are useful for proving Trigonometrical relations. Show that, in any triangle, tan^jA-B) _a-h tan^{A+B) ~ r7T7>* With centre C and radius CA (i.e. ?>), describe a circle cutting CB in E and CB produced in D. THE RIGHT-ANGLED TRIANGLE 103 Then BE = a-h, and BD= a + h. Join AD and AE. Through E draw i?i^ parallel to DA and meeting AB at F. Then the angle DCA at the centre = 180 - C = ^ + 5. So that the angle DEA at the circumference = \{A^-B). Fig. XXXIII. Also the angle BAE = BAC-EAC= A-\{A+B) ==\{A-B). Also the angle EAD, being in a semicircle, is a right angle. tani(^ + £) = ^, tan ^ (^ -P) = — - , since AEF= EAD = a right angle. AE Hence t an^(^-^) ^^i?^ tan|(J. + J5) ^Z) = ^— since -&i^is parallel to AD. (Prop. 12 «. a-b ~ a + b' Corollary. ^ + B+C=180^ .-. ^(A + B) = 90-^C. Hence the above result may be written tan4(A-B) = ?— ^cot-|C. ^^ ^ a + b ^ This formula will be used in a later chapter. 104 THE KIGHT-ANGLED TRIANGLE Examples V. In the following examples : A, B, C are the angles of a triangle ABC. a, b, c are the sides, s = half the sum of the sides ; R is the radius of the circumcircle. r is the radius of the inscribed circle. f\ is the radius of the escribed circle touching the side BC. A is the area of the triangle. D, E, F are the middle points of the sides BC, CA, AB, respectively. X, Y, Zare the feet of the perpendiculars let fall from A, B, C respectively on the opposite sides. is the centre of the circumcircle. / is the centre of the inscribed circle. K is the orthocentre. 1. Express in terms of the sides and angles the lengths of AX, BX, CX, AK, BK, CK. 2. Express the length of ^i) in terms of (i) a,b,C, (ii) a,b,B, (iii) a, b, c. 3. Show that a/sin A — b/sinB = c/sin C = 2R. Deduce that R = abc/iA. 4. Prove that r (cot IB + cot^C) = a. Write down the two similar formulae. 5. Prove that r = A/s. (No trigonometiy required.) Deduce that tan|^ = A-r {s(s-f/)}. 6. Show that BX = a — ?>cos c ; hence prove that c^ = a"^ + b^ — 2 ab cos c. 7. Prove that (i) A = ^ab sin C, (ii) A = rs, (iii) A = abc-i-iR, (iv) A = -v/s(s-«)(s-&)(s-c). 8. Prove that (i) sin^A= ^/{s~b) (s-c) -^bc, (ii) cos^^= \/s{s-a} -r-bc, (iii) tan ^A = y/{s -b){s — c)-TS{s — a). THE RIGHT-ANGLED TRIANGLE 105 9. Show that the triangles ABC and A ZZare equiangular ; hence prove that YZ = a cos A. 10. Two tangents are drawn from a point P to a circle of radius 10 cm. ; the tangents contain an angle of 43°. Find the lengths of the tangents and the distance of P from the centre. 11. A sheet of iron is shaped so that it can be rolled up to form a conical funnel 6 feet high with open circular ends 2 feet and 6 feet diameter respectively. Draw a plan of the sheet before rolling. What is the inclination of the edge of the funnel to the line joining the centres of the ends ? 12. A circle rolls without slipping along a straight line : prove that the co-ordinates of a point fixed to the circumference are such that x = a (^ — sin 6), i/ = a (1 -cos 6) ; the origin being taken at the point where the fixed point meets the straight line, and 6 being the angle turned through by the circle. 13. One of the angles of a right-angled triangle is the acute angle whose sine is §, and the length of the shortest side of the triangle is 10 feet. Find the lengths of the other two sides. 14. ^ is the highest point of a sphere with centre ; a particle slides from a position P, where the angle AOP = 6, to the position Q where the angle AOQ is 0. How much lower is Q than P and how much further from OA ? 15. The time t of sliding from rest down a length s inclined at 6 to the horizon is given by s = ^^^^sin^ where g is a constant. A circle is held with a diameter AB vertical ; prove that the time of sliding along a chord from the highest point A to the circum- ference is the same whatever be the inclination of the chord, and that the time of sliding from the circumference along a chord to B is also independent of the inclination of the path. 16. A plane, inclined at 20" to the horizon, is placed with the line of greatest slope pointing north. A line is drawn on the plane, pointing NNE. ; find the inclination of this line to the horizontal. 17. A man 6 feet high walks along a straight line which passes 3 feet from a lamp-post. If the light is 9 feet from the ground, find the length of the man's shadow when his distance from the point on his path nearest to the lamp is 10 feet. What is the locus traced out by the extremity of his shadow as he walks along the line ? 106 THE RIGHT-ANGLED TRIANGLE 18. If, in the previous question, there is a vertical wall parallel to the man's path and distant 2 feet from it on the side remote from the lamp, what is then the length of the shadow and the locus traced by its extremity ? 19. Draw the graph of 6 /sin 6 from ^ = to ^ = i tt. Use the graph to solve the following problem. A string 30 inches long is tied to the ends of a cane 35 inches long, thus forcing the cane into a circular arc. Find the radius of the arc correct to the nearest inch. 20. Find the length of a strap which passes tightly round two pulleys of radii 2 feet and 3 feet, their centres being 6 feet apart. CHAPTER VI THE TKIANGLE Several formulae connecting the sides and angles of a triangle have been proved in the examples of the preceding chapters. They are here gathered together for reference and proofs are given. Care should be taken that the proof applies when the triangle is obtuse-angled ; if it does not, a separate proof must be given. Relations between the sides and angles. 54. The angle formula. A + B + C = 2 right angles. a ^ ^ ^ sin A ~" sin B "~ sin C A The sine formula. (=211). Fig. XXXIV. Let be the centre of the circumcircle, and D the middle point of^a Join OB, OC, OD. Then, in the left-hand circle of Fig. XXXIV, angle 300 = 2'' angle BAC = 2A. Triangles BOD and COD are congruent ; (Prop. 8 a.) .'. BOD = COD = A. Also BD==^BC = ^ a. 108 THE TRIANGLE In the right-angled triangle BOD, BD = OB sin BOD, i.e. \a = R^mA\ Sin A In a similar way it may be proved that -X^ = 2E and ^ = 2R. sin B sm C sin J. sin^ sm C Hence Exercise. Supply the proof when the angle A is obtuse. Note. In using this formula the following algebraic result is often useful : If - =z - ^ - then each fi*action equals ; :.. 55. The cosine formula cos A = — ^r^. , and its 2 be equivalent a^ = b^ + c^— 2 be cos A. This can be proved very shortly by assuming Euclid II, 13 and 14 ; but it is better to base the proof on the theorem of Pythagoras. Let CZ be the perpendicular from C on AB, Fig. XXXV. ThenZC= &sin^, AZ=b cos A, and BZ=c-b cos A. BC^ = BZ^ + ZC\ a' = {c-h cos Af + (b sin Af = c'^-2bccosA + b^cos^A + b'^sm'^A, i.e. a'' = bUc'' -2 be cos A, h-i + c^-a' or cos^ = 2 he THE TKIANGLE 109 If ^ is obtuse, then in Fig. XXXV a, ZC = & sin (180-^) = &sin^, ^Z=Z>cos(180-^) = -&cos^, BZ= BA + AZ= C + {-bcosA} = c-hcosA. The proof is now the same as before. Exercise. Write down the corresponding formulae for cos B and cos C. 56. The Projection formulae c = b cos A + a cos B. In Fig. XXXV, BZ is the projection of BC on BA ; and AZ is the projection of ^C. AB = AZ+BZ, i.e. c = hcosA + a cosB. Exercises. Supply the proof when A is obtuse. Write down the other two corresponding formulae. 57. Area formulae The symbol A is used to denote area of triangle. (i) A = ^ any side x perpendicular from opposite angle. (Prop. 16.) (ii) A = ^ AB X ZC = ^ c . b sin A = ^ be sin A. (iii) A = ^/3 (s - a) (s - b ) (s - c). In Fis. XXXV. BZ ^ acosB =a X '-±!! "- . 2ca .'. (2c.Z(7)2 = {a-\-h + c){a — b + c){a + h — c) -h + c — o). Let 2s= a + b + c, then h-^c — a = 2(s — a) &c. ; so that 2c . ZC= ^/2s.2(s-a). 2 (s - &)T2 (s-c) ; A =iAB.ZC = >/s(s-a)(s-b)(s-c). Exercise. Show that 16 A2 = 2 (&2c2 + c-a^ + a' h") - (a* + h' + c'). no THE TRIANGLE 58. From these formulae others may be deduced. Example i. To show that in any triangle cos (A + B) = cos AcosB — sin A sin B. From sine formula «sin ^ — &siii A = 0. (i) From projection formula a cos B + h cos A = c. (ii) Square and add, a"^ + &" + 2 cch (cos A cos -B — sin A sin B) = c^ From cosine formula a'^ + h'^ — 2 ah cos C = c^ It follows that cos C = — (cos A cos ^ — sin ^ sin B). From the angle formula C = 180 -(^ + -5), i.e. cos C= — cos (^ + jB). Hence cos (A + B) = cos AcosB-~ sin A sin B. Example ii. In any triangle sin {A — B) = silt A cos B — cos A sin B. Multiply together equations (i) and (ii) above. a^ sin ^ cos 5 - h"^ sin ^ cos yl - ah (sin ^ cos 5 - cos A sin B) = 0. From Fig. XXXVI it is seen that «2 sin B cos B = BZ.ZC= 2'' triangle BZC, and h"^ sin ^ cos ^ = 2*^^ triangle AZC = 2'Uriangle .4'ZC, {ZA'= ZA, so that triangles CZA, CZA' are congruent). .-. ci^ sm B cos B -W sm A cos A = 2 ''triangle 5C/1' ^BC.CA' sin BCA' = ahsui{A — B). Comparing this with the result above, we see that sin {A-B) = sin AcosB- cos A sin B. This result can, however, be obtained more quickly. THE TRIANGLE 111 For I.e. sin BC A' ~ZrB~ sinjA-B) a cos B — b cos ^1 sin CBA' sin^ „ • / < T,N « sin 5 cos 5 < • D Hence sin (^ — 5) = cos^smJ? = sin A cos B — cos A sin B since a sin B = h sin A Example iii. To show that the area of a quadrilateial inscribed in a circle is \/{s — a) {s -b){s — c){s — d) where s — lUi + b + c + d). In Fig. XXXVII Area of ABCD = sum of triangles ABD and BCD = I ad sin A + ^lc sin {180 -A) = I (ad + bc) sin A. From triangle ABD, BD"' = a"^ -\- d"^ -2 ad cos A. Fig. XXXVII. From triangle BCD, BD'' = b'' + c'' -2 be cos {180 -A). Hence a^ + d'^-2 ad cos A = h"^ + c"- + 2 be cos A, i.e. 2{ad + bc)cosA = a'' + (P-{b''-hc''); 2{ad + bc){l + cosA) = {a + dy-{b-c)'', and 2(((d + bc){l-cosA) = {b + c)''-{a-d)\ 112 THE TRIANGLE Hence i {ad + bc)^ {1 - cos"^ A) = {-a + b + c + d){a-h + c + cl)(a + h-c + d){a + b + c-d), i.e. {i(af? + &c)siii^}2 = I {- a + b + c + d) I {a-b + c + d) l{a + b - c + d) ^{a + b + c- d) .'. Area of ABCD = ^/{s-a){s-b){s-c){s-d}. Examples Via. 1. From the three projection formulae deduce the three cosine formulae. 2. Prove that sin^ = sinPcos C+ cos^sin 0; and deduce that sin (B + C) = sin BcosC + cos B sin C. 3. Prove that cos(J.-5) = cosvl cos5 + sin^sin5. 4. Show that A = |- (&^ sin CcosC + c^ sin B cos B). 5. Show that A = i c^ {sin AsinB-^ sin (^ + ^)} . 6. Prove that sin A + sinB> sin C. 7. Prove that cot yl + cot5 = ccosec-B-^a. What third expression are these equal to ? 8. Show that R (i.e. the radius of the circumcircle) = s -^ (sin ^ + sin ^ + sin C). 9. Use the formula cos^ = l-2sin'^|^ to prove that sin 1^ = \/(s -b){s-c)-r- be. Write down the similar formulae for sin J 5 and sin | C. 10. In a similar way to that suggested in the previous example, prove that cos |^ = ^/s {s-a) -f be. Write down the formulae for cos 1 5 and cos I C. What is the formula for tan^yl ? 11. Given cf = 17, 5 = 12, 5^= 37° 15', find A. 12. Given a = 14, b = 13, c = 12, find the greatest angle. 13. Given a = 45, A = 45°, B = 60°, find b. 14. Given b=ll, c = 42, A = 72°, find a. 15. Given a = 176, b = 291, c = 352, find all the angles. (Choose a formula adapted for logarithms.) 16. Given « = 7, fc = 5, C = 49°, find e. 17. Given b = 9,c=10,C= 57°, find a. 18. By considering two forms for the area of an isosceles triangle, prove that sin ^ = 2 sin J A cos | A. t THE TRIANGLE 113 19. Two sides of a triangle are 3 and 12 and the contained angle is 30^ ; find the hypotenuse of an isosceles right-angled triangle of equal area. 20. Two adjacent sides of a parallelogram, 5 inches and 8 inches long respectively, include an angle of 60°. Find the length of the two diagonals and the area of the figure. 21. If in a triangle C = 60^, prove that l/(a + c) + l/{b + c) = 3/(rt + b + c). 22. On a straight line AB, 4 inches long, describe a semicircle, and on the arc of the semicircle find points P, Q, i?, S such that the areas of the triangles APB, AQB, ARB, ASB are 1 square inch, 2 square inches, 3 square inches, and 4 square inches respectively. If C is the centre of the circle, determine the sines of the angles ACP, ACQ, ACR, and ACS, and hence find, from the tables, the values of these angles. 23. If a quadi-ilateral can be inscribed in one circle and circum- scribed about another, show that its area is ^/ctbcd, where a, h, c, d are the lengths of the sides. The circles of the triangle. 59. It is shown in any Geometry textbook that (i) the centre of the circumcircle is the point of concurrence of the perpendicular drawn at the middle points of the sides ; (ii) the centre of the inscribed circle is the point of concurrence of the three lines bisecting the three angles ; (iii) the centre of an escribed circle is the point of concurrence of the bisector of the opposite interior angle with the bisectors of the two adjacent exterior angles. In Fig. XXXVIII, we have AQ = AR, (Prop. 24.) BP = BR, CP=CQ; .'. AQ-\-BP+CP=^. sum of sides = s. Hence AQ = s-a. Exercise. In a similar way, prove that BP = CF = CQ = QQ' = AQ' = PF = 1216 H 114 THE TRIANGLE Fig. XXXVIII. Examples VI b. Prove the following formulae : \. K = a^2&mA. 2. 7? = rt/;c-r4 A. 3. r = A/s. (Considerthesum of the triangles P/C, (7/.4,yi/B.) 4. r = rt^(cot^i? + cot|C). 5. ;-j = A/(.s-cf). 6. ri = rt^(tanJi?-t tan ^ C). THE TEIANGLE 115 Using the above formulae, prove the following relations ; 7. In a right-angled triangle R + r = ^{a + h). 8. l//'j + 1/^2 +1/^3 = 1/r. 9. l/r^+l/r^ = 2 -=-6 sine. 10. r7\ror2,= ^". 11- r r-^ = {s-h){s-c). 12. (aZ^c-fsin J.sin^sinC)^. 13. 27?/- = «&c-f(a + & + c). 14. 4i? sin yl sin B sin C = a cos ^ + 6 cos B + c cos C 15. tanf^ = V'(s-6j(s- c)-rs(s- a). 16. s2= AcotMcotii^cotiC. 17. If ABC is a triangle such that 2h = a-\-c, and ^ is the length of the perpendicular from B upon AC, show that tan | J. and tan \ C are equal to the roots of the equation 18. Show that the sum of the radii of the escribed circles of a triangle is equal to the radius of the inscribed circle together with four times the radius of the circumscribing circle. 19. Show that the area of the triangle formed by joining the centres of the escribed circles is Si^^cos-l^cosi^cos^O. 20. The sides of a triangle are 3, 5, 6 ; find the radii of the inscribed and circumscribed circles. 21. In an isosceles triangle the base is 100 cm. and the perpen- dicular from the vertex is 70 cm. ; find the radii of the inscribed and circumscribed circles. 22. A triangle is described with base BC = 5 inches and angle A = 70°. What is the radius of the circumcircle ? Find the dis- tance of the centre of the circumcircle from BC. 23. Find the radius of the circumcircle of the triangle ABC being given that BC = 7, CA = 6, and C = 60°. 24. If a = 32, & = 16, C = 42°, find R and r. 25. The area of a parallelogram having base 5*8 cm. and angle 123° is 37*7 sq. cm. Find the other sides and angles. Find the radii of the circles which pass through three of the corners of this parallelogram. 26. Two of the sides of a triangle are 7'5 cm. and 9'3 cm., the included angle is 37'. Find the radius of the circle which touches these sides produced and the third side. h2 116 THE TRIANGLE Oral Revision Examples. Complete the following identities and equations : 1. sin (270-^)- 2. cos2(9 = 3. 2 tan ^ cot ^ = 4. If sin 6 = 1,0 = 5. In any triangle &^ = 6. In any triangle R = 1. sm2A= 8. tan 225° = 9. If cos ^ = #, tan 6 = 10. A in terms of the sides = 11. tan-il = 12. secM-1 = 13. length of arc = radius x 14. sin'^5 + sin^(90 — J5) = 15. Definition of tangent. 16. In any triangle cosC' 17. In any triangle h cos C+c cos B = 18. In any triangle &csin^ = 19. In any triangle r = 20. tan I TT = 21. Definition of sine. 22. cos (360° -jB) = 23. In any triangle r^ = 24. In any triangle cos^ 25. What formula connects a, b, and B? 26. tan-'(--s/3) = 27. If cos/?= -^6 = 28. tan2 73l° + l= 29. abc = 30. 37° = ? radians. 31. cos2(^-45°) + sin2(^-45°) = 32. &sinC = 33. Express R in terms of the sides. 34. If sin (9 = sin 0^, then 6= 35. cos^= (in terms of sin |^), 36. Area of triangle = 37. aH c^ - 2 «c cos B = 38. cos 1200° = 39. Maximum value of 2 sin ot cos a = 40. cos^^-sin*^ = (in its simplest form). 41, A -r (.<;-«)= 42. tan (180-5) = 43. acos C + CC0S.4 = 4t4:. he sin A = 45. sui^ {A + B) + cos"^ {A + B) = 46. If cos^ = cosyl, then x = 47. In any triangle cos A = 48. tan 60° = 49. cos 2 ^ = 50. How many radians = >4° ? THE TKI ANGLE 117 Examples VI. 1. Prove that (a cos A - h cos B) -r (a^ - h') + cos C/c — 0. 2. Prove that c- = {a + hf sin- 1 C + {a- h)- cos- }, C. 3. In a triangle ^i?Cthe lines drawn from A and C, perpen- dicular to the opposite sides, intersect in 0. If the angle A is acute, show that OA == b cos A/sin B. Also draw a diagram in which A is an obtuse angle, and establish the corresponding expression for OA in that case. 4. Show that in any triangle the product of a side and the sines of the two adjacent angles is the same, whichever side be taken. 5. Find the area of a regular polygon of n sides circumscribed about a circle of radius r. 6. Regular polygons of 1 5 sides are inscribed in and circumscribed about a circle whose radius is one foot ; show that the difference of their areas is nearly 20 square inches* 7. ABCD are four points on a circle such that the angles BAC and BCA each equal 6. Show that AD +CD = 2BD cos 6. 8. If 2 cos B = sin A/sin C, prove that the triangle is isosceles. 9. If tan.-l/tan^ = sin-.-1/sin-i?, show that the triangle is isosceles or right-angled. 10. Express the sides of a triangle in terms of the angles and the semi-perimeter. 11. In a triangle ABC perpendiculars AD and BE are let fall on the opposite sides ; prove that the radius of the circle circum- scribing the triangle CDE equals R cos C. 12. If in a triangle the median bisecting the base AB is perpen- dicular to the side AC, prove that 2 tan A -f tan C = 0. 13. If ^; and q are the lengths of the perpendiculars from A, B on any arbitrary line drawn through the vertex of a triangle, prove that a'p"" -f &' 5' - 2«Z> pq cos C = w^ ^2 ^^^2 q 14. An isosceles triangle, vertical angle 35°, is inscribed in a circle whose radius is r65 inches. Find the lengths of the sides. 15. Show that in any triangle cos A cos B cos C ^ a^ ■\-}r A- (? a b c 2abc 118 THE TRIANGLE 16. If R is the radius of the circumcircle of any triangle and X, y, z are the lengths of the perpendiculars let fall from its centre on the sides, prove that R^-{x^ + y'^ + z')R-2xyz = 0. 17. The rectangular co-ordinates of the angular points of a triangle are (4, 5), (6, 7), (8, 6) ; determine the sum of the two smaller angles. 18. A rod AB, length 2 a, can turn about a hinge fixed to the wall at ^ ; it is supported by a string BC, length /, fastened to a point C on the wall at a height h above A. (i) If BC is horizontal, what is the inclination of the rod to the vertical ? (ii) If BC is horizontal, what is the inclination to the vertical of the line joining the hinge to the middle point of the string ? (iii) If the string and rod are inclined at 6 and to the vertical respectively, prove that (i) 2 a sin = Z sin 6, (ii) J cos 6 — 2 a cos (^ = /i. (iv) In the general case, what is the angle between the string and the rod ? Give the answer in terms of li, a, 6 or li, I, 0. (v) In the general case, what is the inclination to the vertical of the line joining the hinge to the middle point of the string? Give the answer in terms of h, a, (p. 19. Three equal spheres of radius 7 centimetres are fixed in a horizontal plane so as to touch each other ; a sphere of radius 6 cm. rests upon these three. Find the height of the centre of the fourth sphere above the horizontal plane, and the inclination to the vertical of the line joining the fourth centre to one of the lower centres. 20. Three equal rods of length 54 inches are fixed so as to fonii a tripod. If their feet are at the corners of an equilateral triangle, side 18 inches, find the inclination of each rod to the vertical. 21. In any triangle prove that the centroid trisects the line joining the circumcentre to the orthocentre. 22. Find the lengths of the sides of the pedal triangle of the triangle ABC. Find also the radii of the inscribed and circum- scribed circles of that triangle. (The pedal triangle is formed by joining the feet of the perpen- diculars let fall from the vertices on the opposite sides.) 23. If rt = 5 and ft = 4, draw a graph to show the value of c as C varies from 0° to 180°. Hence find the value of c when C = 40^ CHAPTER VII • SOLUTION OF TRIANGLES 60. It is known from Geometry that, if three parts of a triangle are given, the remaining parts can in some cases be found; and that, in other cases, relations between the missing parts may be found even though their exact values cannot be determined. When actual numbers are given, results can be obtained to a greater degree of accuracy by Trigonometrical methods than by drawing to scale. In all cases a formula is sought which shall contain the three given letters and one unknown letter. 61. Case I. Three angles given. The angle formula shows that A + B+C must be 180°. No formula contains the three angles and one side only ; but from the sine formula, viz. -. — - = -. — ., = -. — -, we can find the ratios sm ^-1 sin B sm C of the sides. 62. Case II. Two angles and one side given. The third angle can be found immediately since A + B+C= 180". Suppose a is the given side ; and it is required to find b. The formula must contain a, h, and two of the angles ; hence we use b _ a sin B sin A This is adapted for the use of logarithms as it involves no addition or subtraction. If the tables in use give the logarithms of the cosecant, it may be advisable to use the following logarithmic form log b = log a + log sin B + log cosec A. 63. Case III. One angle and the two sides containing the angle are given. Suppose a, b, C are the given parts. Then the cosine formula c^ = ci^ + b'^ — 2 ab cos C enables us to determine c. When c is 120 SOLUTION OF TRIANGLES determined, the remaining angles can be found by the sine formula. This method is of practical use only when the numbers involved are small ; the cosine formula is not adapted for the use of logarithms. It is usual, therefore, to use the formula proved in § 53,* viz. tan 1{A-B) = ^^ cot I C. This determines h{A—B)', also ^{A + B) equals the complement of I C; hence A and B are found by adding and subtracting. The value of c is then calculated by the sine formula. 64. Case IV. One angle and the two sides not containing the angle are given. Suppose a, h, A are given. Then we can determine c from the formula a^ = b'^ + c'^ — 2bc cos A. This is a quadratic equation to determine c, and it is seen that there is the possibility of two distinct values for c. This is also seen from the geometrical construction. On this account this case is usually known as the Ambiguous Case. If there are two values of c, there will be two values for B and for C. This is seen independently if the sine formula is used (as it usually is, on account of its adaptability for logarithms) : sin B _ sin A b a Suppose that this leads to the result sin B = sin x. Then B = x or 180 -ic. This shows that, if there are two solutions, those two solutions are supplementary. Hence one of the solutions will be obtuse. Preliminary geometrical considerations often show that there can be only one solution. (i) If the given angle A is not acute, then B must be acute and the obtuse-angled solution must be rejected. (ii) If a>b or = h, then A> B or = B; consequently B cannot be obtuse. Exercises. When a, b, A are given, show (i) from the geo- * Another proof is given on p. 103. SOLUTION OF TKIANGLES 121 metrical solution, (ii) from the cosine formula, (iii) from the sine formula, that (a) there is no solution, if a<h sin A ; (6) there is one solution only, if a = & sin ^ ; (c) there are two solutions, if a>68in^ but < &; [d) there is one solution only, if « > &. Point out the difference in nature of the one solution in {h) and [d). 65. Case V. Three sides given. Here again the cosine formula may be used, if the numbers involved are not inconveniently large. For logarithmic calcula- tion the formula for sin^^, cos|^, or tan|^l is used. These half-angle formulae are derived from the cosine formula. 2bin2M = l-co8^1* (§51) ~~ 2bc 2 he {a — b + c){a + b-c) ainiA- j (s-b)(B-c) srn.A-^ 2 Similarly cos ^ A = \ ~^ — ' Divide tan^A=J^^:^l^ ^ \ s(s-a) Of these three formulae it is best to use the tangent formula ; for the logarithms used in finding tan 4 A are the same as those required for finding tan i B or tan I C. If only one angle lias to be found, it is indifferent which formula is used. There is a simple geometrical proof for tan \ A. * In old books on Trigonometry the ' haversinc ' was used for solving triangles, and the values of log haversinc were tabulated in mathematical tables. The haversinc equals half the versed sine ; hence haversin -4 = J versin A = (1 — cos^)-^2 = sin^ \ A. The for- mula for solution of the triangle then becomes havei'sin A = (s—b) (s — c)-T-bc. 122 SOLUTION OF TRIANGLES In Fig. XXXVIII / is the centre of the inscribed circle, £" is the point of contact of the circle with AC. Then tan ^ = 42 2 AQ r s~a A (§59) (s-rt) ls -h){s-c) \ s{s — a) Examples Vila. (See p. 81 for arrangement of work.) In the following triangles when Case I. 1. A = 79° 20', B = 64" 10', find the ratios of the sides. Case II. 2. Ar= 58° 12', B = 64° 33', a = 385, find b. 3. ^ = 38° 24', C = 95° 5', c = 7-832, find a and b. 4. 5 = 63° 55', C = 48°27', c = b'16, find 6. Case III. 5. a = 409, b = 381, C = 58° 12', find A and B. 6. B = 23° 46', c = 9-72, a = 8*88, find A and 0. 7. a = -532, c = '259, B = 39° 33', find A and C. 8. ^ = 73° 15', b = 7315, c = 8013, find B and 0. Case IV. 9. A = 38° 14', a = '33, 6 = '44, find C. 10. « = 409, b = 385, A = 64° 32', find B and C. 11. 6 = 6-901, c = 5-749, 0=48° 27', findi?. 12. A = 73° 15', a = 7315, c = 8013, find B and C. Case V. 13. a = 17, fe = 13, c = 12, find the least angle. 14. a = 793, b = 937, c = 379, find all the angles. 15. s = 1410, a = 1437, b = 811, find all the angles. 16. s = 1437, a = 1410, b = 811, find all the angles. SOLUTION OF TRIANGLES 123 17. ft = 13, h = l, C= 60^ find A and B. 18. a = 32, h = 40, c = 66, find C. 19. a = 250, Z) = 240, A = 72° 4', find B and C. 20. a = 2 &, O = 120°, find ^, 5 and the ratio of c to a. 21. rt = 86, & = 63, c = 81, find the smallest angle. 22. & = 5, c = 3, ^ = 42°, find B and C. Oral Examples. State the formula to be used in the following cases : 1. Given a, h, C, find c. 2. Given a, h, C, find A and B. 3. Given h, c, C, find B. 4. Given b, c, C, find «. 5. Given c, a, C, find A. 6. Given c, a, C, find 5. 7. Given c, A, B, find C. 8. Given c. «, A, find &. 9. Given a, h, B, find C. 10. Given a, 5, ^, find c. 11. Given «, b, c, find C. 12. Given A, B, C, find a. 13. Given ^, C, b, find a. 14. Given a, b, B, find ^. 15. Given a, c, B, find C. 16. Given c. A, B, find &. 17. Given «, fe, r, find B. 18. Given &, c, A, find £. 19. What is the ambiguous case? 20. When a, c, A are given, what are the conditions that there should be no ambiguity? Examples VII b. Solve the following triangles : 1. a = h,b = l, C=30°. 3. a = 65, & = 68, c = 16. 5. a = 7, ^ = 120°, A = 45°. 7. 6 = 926*7, ^ = 48° 24', B S. a = 407-4, c = 115'9, A = 127° 45' 9. rt = 1263, b = 1359, c = 1468. 10. a = 53-94, b = 156-5, C = 15° 13'. 11. b = 457-2, c = 342-6, A = 73° 45'. 12. rt = 246-7, Z; = 342-5, B = 32° 17'. 13. c = 79-48, A = 54° 16', B = 85° 6'. 14. rt = 7-956, b = 10-35, c = 9-412. 15. b = 9463, c = 7590, C = 43° 47'. 16. a = 739, c = 937, £ = 146° 12'. 2. b = 4, c = = 3, C = = 60° 4. b = 8, c = = 9, c = :45°, 6. a = = 6, & = = 7, c = 5. 31' = 13' 124 SOLUTION OF TKIANGLES 17. c = 79-5, A = 35° 14', C = 117° 35'. 18. A = 89°, B = 18° 47', C = 72° 13'. 19. a = 87-6, b = 57'4, c = 46*8. 20. a = 79, c = 97, A = 2437. 21. A = 79°, C = 97°, R = 17-2. 22. b = 73-6, R = 57, a = 48*9. 23. ft2 + &2^34i^ sinC= 1, tanJ5 = fQ. 24. ^ = 42° 35', a = 83, b = 74. 25. a = 2-740, b = '7401, C = 59° 27'. Heights and Distances. 66. First a figure must be drawn, not necessarily to scale ; the known lengths and angles should be indicated in the figure. It may be necessary to solve, or partly solve, more than one triangle before the required measurement is found. The scheme for working should be carefully thought out before the work is actually begun. Example i. Wishing to find the height of a house standing on the summit of a hill of uniform slope, I descended the hill for 40 feet, and then found the height subtended an angle of 34° 18'. On descending a further distance of 60 feet, I found the subtended angle to be 19° 15'. Find the height of the house. Fig. XXXIX. Scheme.— In triangle ADC we know one side CD and all the angles ; so AC can bo found. Then in the triangle A CB two sides AC, CB aro known, and the included angle, hence AB can be found. HEIGHTS AND DISTANCES 125 From triangle ACD, AC CD sin ADC sin DAC i.e. ylC = 60sin 19^ 15' cosec IS*" = 76-182. From triangle ABC, h — a 3' tan ^(B-A) = , cot i C 36-182 Again, 116-182 ^{B-A) =45° 16', i{B+A) = 12°bV- A = 27° 35'. AB _ ^inACB 'CB ~ sin cab' 40 sin 34= cot 17^ 9' Logarithms. 1-77815 1-51811 -58559 1-88185 1-55849 + -51061 AB = 18' 2-06910 2-06514 -00396 1-60206 + 1-75091 35' 1-35397 1-66562 1-68835 sin 27' = 48-792. Height of house = 48-8 feet. Example ii. Wanting to Icnoiv the height of a castle on a rock, I measured a base line of 100 yards, and at one extremity found the angle of elevation of the castle's top to he 45° 15', and the angle subtended hy the castle's height to he 34° 30'; also the angle subtended by the top of the castle and the oilier extremity of the base line was 73° 14'. At the other extremity the angle between the first extremity and tJie top of the castle teas 73° 18'. Find the height of the castle. This requires a rough perspective figure of the whole, and sub- sidiary plane figures. A Fig. XL. 126 HEIGHTS AND DISTANCES AB represents the castle. C is the point in the same vertical as AB, and in the same horizontal plane as DE, the base line. The following magnitudes are known : DE = 100 yards. ACD and ACE are each right angles. ABC, ABB are known, therefore BBC is known. ABE, AEB are known. Scheme. In triangle ADE, DE and the adjacent angles are known ; hence AD can be found. AB can now be found from triangle ABD. A From triangle ABE, AB BE I.e. sin 73° 18' AB 2- + 1-9813 - 17415 log^Z) = 2-2398 Fig. XLII. From triangle ABB, AB .ID sin 34° 30' sin 79° 15' Height of castle = 100 yards. 2-2398 + 1-7531 1-9929 -1-9923 2-0006 HEIGHTS AND DISTANCES 127 Example iii. From the top of the Feak of Teneriffe the dip of the horizon is found to he 1° 58'. If the radius of the earth be 4000 miles, ivhat is the height of the mountain? In Fig. XLIII C is the centre of the earth, AB is Teneriffe ; BH is the tangent drawn from B to the earth's surface, so that H is the farthest point seen from B ; in other words, H is on the horizon. The angle between BH and BD (the perpendicular to the vertical) is called the dip of the horizon. Fi;?. XLIII, From triangle BCH, BC^ CH sec BCH, angle BCH = complement of CBH = HBD ; i?C = 4000 sec 1°58' = 4000x1-00059 = 4002*36 miles. Height of mountain is 2*36 miles. Note. Fig. XLIII is drawn much out of scale ; for small heights BH and BD are practically identical. Even for mountains the dip is very small, as in this example ; in fact, so small that we may use the approximation sine of dip = tan of dip = circular measure of dip. 128 HEIGHTS AND DISTANCES If E be the other extremity of the diameter through B, we have, from § 52, BA. BE = Bir\ i.e. h{2r + h)=- ct", whore r is radius of earth, h is height of place of observation, d is the distance of the horizon. Hence d = \/2 rh-\-h'^', = -s/2 rh ( \ + — — — ^... ) by the Binomial Theorem. \ 4 r 32 r' / So far the work is accurate ; usually h/r is so small that it may be neglected. Hence for ordinary heights Distance of horizon = v 2 rh. Exercise, (i) In the formula just obtained r, /;, and the distance are all expressed in the same units. By taking r = 3960 miles, prove that Distance of horizon in miles = •v/f X height of place of observation in feet. (ii) Show also that Dip in minutes = '9784 V height in feet. Examples VII. 1. Standing at a horizontal distance 100 yards from the foot of a monument, a man observes the elevation of its top to be 25° 35'. Assuming the man's eye to be 5 feet from the ground, find the elevation of the top when the man stands 50 yards from the foot. 2. OA and OB are two straight roads intersecting at and making with each other an angle of 85° 12'. vl is a house 1572 yards from 0, and 5 is a house 1129 yards from 0. Find the direct distance between A and B. 3. A man observes the angles subtended by the base of a round tower at three points A, B, and C, in the same horizontal straight line with the centre of the circular base, to be 2 a, 2 /3, 2 y respec- tively. Find the ratio of AB to BC. and find the diameter of the tower in terms of ^IC. SOLUTION OF TRIANGLES 129 4. A man observes that the elevation of the top of a tower is 37^ 40', and that the elevation of the top of a flagstaff on the tower is 43= 59'; show that the height of the flagstaff is one-fourth of the height of the tower very nearly. 5. Having given that the least side of a triangle is 17'3 inches, and that two of the angles are 63=20' and 72M0', find the greatest side. 6. If two sides of a triangle are 7235 feet and 4635 feet respec- tively, and if the included angle is 78=26', find the remaining angles of the triangle. 7. The base of a triangle being 7 feet, and the base angles 129=23' and 38=36', find the length of the shortest side. 8. Explain the ambiguous case of the solution of triangles. When a, h, A are given and the question is asked whether, from these data, two triangles, one triangle, or no triangle can be constructed, show that the question can be answered from a consideration of the roots of the equation x'^ — 2hx cos A-Irl? = c<}. 9. From each of two ships, a mile apart, the angle is observed which is subtended by the other ship and a beacon on shore ; these angles are found to be 52° 25' and 75° 10' respectively. Find the distances of the beacon from each of the ships. 10. A ship sailing due north observes two lighthouses bearing respectively NE. and NNE. After the ship has sailed 20 miles the lighthouses are seen to be in a line due east. Find the distance in miles between the lighthouses. 11. The angles A, J5, C of a triangle ABC are 40°, 60=, and 80° respectively, and CD is drawn from C to the base bisecting the angle ACB', if AB equals 100 inches, find the length of CD. 12. A man standing at a certain station on a straight sea-wall observes that the straight lines drawn from that station to two boats lying at anchor are each inclined at 45° to the direction of the wall, and when he walks 400 yards along the wall to another station he finds that the angles of inclination are 15° and 75° respectively. Find the distance between the boats and the perpendicular distance of each from the sea-wall. 13. From a house on one side of a street observations are made of the angle subtended by the height of the opposite house, first 121C I 130 SOLUTION OF TKIANGLES from the level of the street, in which case the angle is tan~^ (3), and afterwards from two windows, one above the other, from each of which the angle is found to be tan-^( — 3). The height of the opposite house being 60 feet, find the height of each of the two windows above the street. 14. A segment of a circle stands on a chord AB 10 cm. long and contains an angle of 40°. A point C travels along the arc ; for what value of the angle ABC is the chord CA three times the chord CB ? Verify by drawing a graph showing the chord CA as a function of the chord CB. 15. If the sides of a triangle are 1011 and 525 feet, and the difference of the angles opposite to them is 24°, find (correct to the nearest degree) the smallest angle of the triangle. 16. A ladder is placed against the wall of a room and is inclined at an angle Oc to the floor. If the foot of the ladder slips outwards from the wall a distance of a feet, and the inclination of the ladder to the floor is then /3, show that the distance which the top of the ladder will slide down the wall is a cot|(a + /3). 17. A man travelling due west along a straight road observes that when he is due south of a certain windmill the straight line drawn to a distant tower makes an angle of 30° with the direction of the road. A mile further on the bearings of the windmill and tower are NE. and NW. respectively. Find the distances of the tower from the windmill, and from the nearest point of the road. 18. A statue 10 feet high, standing on a column 100 feet high, subtends at the eye of an observer in the horizontal plane from which the column springs the same angle as a man 6 feet high standing at the foot of the column ; find the distance of the observer from the column. 19. It is found that two points, each 10 feet from the earth's surface, cease to be visible from each other over a level plain at a distance of 8 miles ; find the earth's diameter. 20. A plane, inclined at 33° to the horizontal, meets a horizontal plane in the line BC. From B a line BD is drawn on the inclined plane making an angle 27° with the horizontal plane. If BD is 18 inches long, find the height of D above the horizontal plane, and its distance from BC. Also find the angle BD makes with BC. SOLUTION OF TRIANGLES 131 2L A lighthouse was observed from a ship to be N. 23" E. ; after the ship had sailed due south for 3 miles, the same light- house bore N. 12" E. Find the distance of the lighthouse from the latter position of the ship. 22. Two streets meet at an acute angle; the one lies N. 51° W., and the other S. 48" \V. The distance from the corner to a chemist's door in the first street is 315 yards; and the distance from the corner to a doctor's door in the other street is 406 yards. Find the length of a telephone wire going direct from the doctor's house to the chemist's. 23. From a vessel at anchor two rocks are observed to the westward, the one (A) bearing WNW., and the other (B) W. by S. from the vessel. From the chart it is found that A bears NNE. from B and is distant 645 yards from it. What are the distances of the rocks from tlie vessel ? 24. Three objects A, B, and C forming a triangle are visible from a station D at which the sides subtend equal angles. Find AD, it being known that AB = 12 miles, AC = ^ miles, CAB = 46" 34'. 25. A tower on the bank of a river, whose breadth is 100 feet, subtends angles 22 1" and 67^" at two points A and B on the opposite bank of the river, whose distance apart is 6C0 feet, on a level with the base of the tower. Find the height of the tower. *26. A, B, C are three given stations, so that the triangle ABC is completely known. Show how to determine, by means of angles measured at a fourth station P, the distances PA, PB, PC, the four stations being all in one plane, the case for considera- tion being that in which P is within the angle A, and the points P and A on opposite sides of BC. If ABC is equilateral, and the angle BPC equals 60', show that 2 cos {(^0^ + BAP) + cos (ABP- BPA) =0. 27. A tower stands on the edge of a circular lake ABCD. The foot of the tower is at D, and the angles of elevation of the top of the tower from A, B, C, are 0(, ,3, y respectively. If the angles BCA, BAC be each equal to 6, show that cotan y. + cotan 7 = 2 cotan 3 cos 9. * This example is best solved by using the formulae of §§ 83 and 84. *i2 132 SOLUTION OF TRIANGLES 28. A mountain is observed from a place A to have elevation 15° 17' and to bear N. 24°29' W. From another place B which is 2347 yards north of A its bearing is N. 37° 2' W. Deduce the elevation from B. 29. The extremity of the shadow of a flagstaff 6 feet high, standing on the top of a regular j^yramid on a square base, just reaches a side of the base and is distant 56 feet and 8 feet from the extremities of that side. If the height of the pyramid be 34 feet, find the sun's altitude. 30. A man observes that when" he has walked c feet up an inclined plane the angular depression of an object in the horizontal plane through the foot of the slope is (X ; and that, when he has walked a further distance of c feet, the angular depression of the object is ^. Show that the inclination of the slope to the horizon is cot-^ (2 cot /3 — cot 0() ; and determine the distance of the object observed from the foot of the slope. 31. A straight flagstaff, leaning due east, is found to subtend an angle 0( at a point in the plain upon which it stands, a yards west of the base. At a point h yards east of the base, the flagstaff sub- tends an angle /3. Find at what angle it leans. 32. Four rods are loosely jointed at their extremities to form a parallelogram with sides 4 and 5 inches long. Two of the opposite corners are connected by an elastic string of length 7 inches. Find the angle between the string and the shorter side. If the length of the other diagonal be diminished by 1 inch, what does the angle become ? *33. Three posts on the border of a lake are at known distances from each other, namely 63 yards, 44 yards, and 76 yards. At a boat on the lake it is found that the two posts, whose distance is 63 yards, subtend an angle 89° 15', and the two posts, whose distance is 76 yards, subtend an angle 130° 45'. Find the distances of the boat from the three posts. 34. A base line AB is drawn 2 chains in length on a plane in the same horizontal plane as C the foot of a tree. The angles ABC, BAC are found to be 79° 56' and 78° 18' respectively; the angle of elevation of the top of the tree is found to be 19° 46' at A. Find the height of the tree to the nearest foot. * This example is best solved by using the formulae of §§ 83 and 84, SOLUTION OF TRIANGLES 133 35. A base line AB, 2527 links long, is measured on the sea- shore along the high water mark. is a point where a distant rock meets the sea ; the angles BAC, ABC are found to be 89" 15', 86" 21' respectively. The angle of elevation of the highest point of the rock, which is vertically above C, as observed at A, is 1°48'. Neglecting the curvature of the earth, find the height of the rock and its distance from A. 36. A hill slopes upwards towards the North at an inclination 14" to the horizontal. The sun is 15° W. of S., at an altitude of 47° ; find the length of the shadow cast on the hill by a vertical post 39 feet high. 37. If, in the previous question, the post is perpendicular to the surface of the hill, what is the lenorth of the shadow ? 134 REVISION EXAMPLES Revision Examples B. 1. Define the tangent of any angle, and prove from the defini- tion that (i) tan(90 + ^j= -cot^; (ii) tan(180-.4)= -tan^. Express the other trigonometrical ratios in terms of the tangent. 2. Show by substitution that sin 45" + sin 30° > sin 60°, and cos 30° - cos 45° < cos 60°. 3. Find the value of sin 45° without using tables. Solve the equation 4 sin ^ cos ^ -f 1 =2 (sin 6 -h cos S). Give the general solutions. 4. A man walks directly across the deck of a ship, which is sailing due North at 4 miles an hour, in 12 seconds, and finds that he has moved in a direction 30° East of North. How wide is the deck? 5. Show that in any triangle ABC, (i) sin A /a = sin B/b = sin C/c ; (ii) sin C {a cos B-b cos ^) = (a + b) (sin A — sin B). 6. Prove geometrically that cos 2^ = l-2sinM. Hence find the value of sin 15°. 7. The angle of elevation of the top of a spire seen from A is 30°, and it is found that at a point B, 115| feet nearer the foot of the spire, it is 60°. Find the height of the spire to the nearest foot. 8. Plot a curve giving the sum of 4sin^ and 3 sin 2^ from ^ = 0° to ^ = 180° ; and read off the angles at which the greatest and least values respectively of this sum occur. Estimate the slope of the curve when 6 = 90° and when ^ = 135°. 9. Define a radian. Express in degrees and minutes an angle of 1'36 radians. Find the number of radians in the angle of a regular decagon. 10. Prove (i) sin'^^ -f cos'^^ = 1 ; (ii 1 tan ^ -i- (1 - cot ^) + cot ^ -^ (1 - tan A) = sec A cosec ^ + 1. REVISION EXAMPLES 135 11. Draw the sine and cosine graphs, in the same figure, from ^ = 10° to <9 = 20°. From the graph find the angle which satisfies sin ^ + cos ^= V2. 12. Find an expression which will include all angles having a given tangent. Write down the values of tan 225^, tan 780", cot 1035% cot210\ Construct an angle, having given the cotangent. VS. Find a/cos A-^b/coi B + c/cos C in a form adapted to logarithmic calculation. 14. In any triangle prove that (i) a = 6cos 6'+ ccos ^ ; (ii) «(6cos C-ccosi?) = ^'^-c^; (iii) rcoalA = as\n^Bsin\C. 15. If the sides of a parallelogram are a, b, and the angle between them co, prove that the product of the diagonals is '^/a*-2a-b- cos CO + ¥. 16. A vessel is steaming towards the East at 10 miles an hour. The beanng of a lighthouse as seen from the vessel is 42^24' North of East at noon, and 25° 12' East of North 25 minutes later. Find how far the vessel was from the lighthouse at noon, and find also at what time the bearing of the lighthouse will be due North. 17. Assuming that a circle may be treated as a regular polygon with an infinite number of sides, show that the ratio of the circumference of a circle to its diameter is constant. What is the circular measure of the least angle whose sine is |, and what is the measure in degrees, &c., of the angle whose circular measure is '15708? 18. Prove by a geometrical construction that cos 2^ = cos^^ — sin^x4. Solve the equation cos 2 A = (cos^ + sin Af. 19. For what data will the solution of a triangle become ambiguous ? Explain this. Given J5 = 30°, c = 150, Z> = 50 ^/3, show that of the two triangles that satisfy the data one will be isosceles and the other right-angled. Find the third side in the greater of these triangles. Would the solution be ambiguous if ^ = 30°, c = 150, h = 75 ? 20. AB is a horizontal line whose length is 400 yards; from a point in the line between A and B a balloon ascends vertically, 136 KEVISION EXAMPLES and after a certain time its altitude is taken simultaneously from ^ and jB ; at ^ it is observed to be 64° 15' ; at B 48° 20' ; find the height of the balloon. 21. Find the radius of the circle circumscribing a triangle, in terms of its sides. If c^ = ci^ + W, show that this radius equals \ c. 22. Define the trigonometrical ratios of A involved in the equation cot ^ + tan A = sec A cosec A ; and establish its tmth by a geometrical constiiiction. 23. Prove that cos ' ^/{a -x)-^{a-h) = sin ^ ^y{x -b)-r- {a - h) = cot""^ \/{a — x)-7-{x — b). 24. Prove that sin 6 = tan 6-^ ^/l + tan2<9. Having given tan ^ = |, find sin 6, cos 9, and versin 0. 25. If 6 is an acute angle whose sine is j%, calculate the value of tan ^ + sec ^. What would the value be if 6 were obtuse ? 26. What is the angle between the diagonal of a cube and one of the edges at its extremity ? 27. Obtain an expression for all the angles which have a given tangent. Find all the angles lying between —360° and +360° which satisfy the equation 2 tan'^x = tan re — 1 = 0. 28. A circular wire of 3 inches radius is cut and then bent so as to lie along the circumference of a hoop whose radius is 4 feet. Find the angle which it subtends at the centre of the hoop. 29. A triangle ABC has angle A = 34°, a = ll'O cm., c = 7*8 cm. Calculate the perpendicular from B on h, and the remaining angles and side of the triangle. 30. In a triangle a = 14:, & = 37, c = 97 ; find the value of (i) acos^ + tcos^, (ii) cisinB — bBinA. 31. If ABC be a triangle, and 6 an angle such that sin = 2 \/ah cos J C -f (a + h), find c in terms of a, h, and 0. If a = 11, h=.2ry, and C= 106° 15^', find c. REVISION EXAMPLES 137 32. Find the area of a regular quindecagon inscribed in a circle of one foot radius. 33. Find an expression for all angles having the same sine as the angle Oc. Solve the equation sin {0(. + x) + sin {3 + x) = 0. 34:. An angle 0( is determined by the equations v^ = 2(/h, -b = tv sin a -Igf, tv cos 'X = a. Show that «2 tan^ Ot - 4 ha tan y + a'^-i hh = 0. 35. Criticize the proposition that three measurements are suffi- cient and necessary to determine a triangle uniquely in shape and size. 36. A square house, measuring 30 feet each way, has a roof sloping up from all four walls at 35° to the horizontal. Find the area of the roof. 37. Draw up a table showing in three columns the values of 10sin<9, 10 cos (9, and 8 sin ^ + 6 cos ^ for each 30° from 0° to 360°. From the table draw, in the same figure, the graphs of ?/ = 10sin^ and ?/ = 8 sin^ + 6cos^ ; and from the curves deter- mine approximately a value of 6 for which tan ^ = 3. 38. Taking the earth as a sphere of radius 4000 miles, find the distance London travels in an hour in consequence of the rotation of the earth. (Latitude of London 51° 30' N.) 39. ABCD is a quadrilateral in which AB and DC are parallel and 40 feet apart, and AB is 100 feet long. The angle DAB is 72° 30', and the angle CBA is 38° 15'. Find the lengths of AD, DC, and CB, and the area of the quadrilateral. 40. State the local time at the following places when it is noon at Greenwich. Cape Town 33° 56' S., 18° 25' E. Fiji 18° 0' S., 178° 0' E. Edinburgh 55°57'N., 3° 10' W. Singapore 1° 17' N., 103° 50' E. 41. Define the cosine and the tangent of an angle, and show how to express the tangent in terms of the cosine. Having given that cos^l = '8, and that A is Jess than 90°, find the value of tan .4; and by means of the tables find the value of Af both from its cosine and from its tangent. 138 REVISION EXAMPLES 42. Prove that, in any triangle ABC, sin B: sin C = b:c. In the triangle ABC the angle CAB is 50°, the angle ABC is 65°, and the side BC is 4 inches long. Find the length of the side AB. 43. Show how to find the height of a tree by means of a chain for measuring lengths and of an instrument for measuring angles. 44 Find an expression for all the angles which have (i) a given tangent, (iij a given sine. 45. Explain how it is that, tan ^ being given, tan 2^ is known ; but that, sin ^ being given, sin 2^ may have either of two values. 46. Prove that the area of a triangle is \/.'i {s-a) {s- b) {s — c). Show also that the area is ^ c^ -f (cot J. + cot B . 47. Find the radius of the circumscribing circle of the triangle for which A = 66° 30', B = IV 30', c = 200 feet. 48. A ship is sailing due East at a uniform rate : a man on a lighthouse observes that it is due South at 1 p.m. and 16^30' East of South at 1.20 p.m. In what direction will he see it at 2 p.m.? CHAPTEE VIII PROJECTION. VECTOES Fi-. XLIV. 67. If from the extremities of u line FQ, of definite length, perpendiculars PK, QL are let fall on a line AB, which may be produced if necessary, then KL is called the Projection of PQ on the line AB. Projections are subject to the same convention of sign as are abscissae and ordinates. Thus, in the above figure, KL is positive, but LK is negative. It follows that the projec- tion of PQ is not the same as the ^ projection of QP, so that the order of the letters in naming a line is of great importance when we are dealing with projection. When the direction of the line is to be taken into account as well as its length, it is called a directed length ; and we shall, in future, use the symbol {PQ) * to denote the directed length of the line from Pto Q. The number of unit> of length in that line we shall continue to denote by the symbol I'Q. Thus, in Fig. XLIY, the projection of (PQ) is (KL), and the projection of [QP) is {LK). Note. When we speak of the sum of directed lengths in the same straight line, tlie algebraical sum is always meant. Geometrically this means that we require the directed length between the starting- point and final point, and not the length of the actual path traversed. 68. If the length of PQ is I, and if 6 is the angle between PQ and the line AB, then projection of {PQ) on AB = I cos 0. This is usually written PQ. 140 PROJECTION. VECTORS Some care is necessary in applying this formula ; the safest plan is to keep ? and 6 both positive. Consider, for instance, the projection of {QP) in Fig. XLV. Imagine a line drawn from the initial point Q parallel to the line AB. Then it is seen that the angle between {QP) and AB is 6 + IT, while the length ^Pis I. Hence projection of {QP) on AB = lcos{d + 7r) = -I cos 6, Two other methods of treatment give the same result. A B Fig. XLV. In Fig. XLV the line QX is actually drawn parallel to AB ; but it is usually sufficient to imagine it. Then we may take the angle between (QP) and QX to be the negative angle XQP, i. e. — (n — 6); the length ^Pis positive so that projection of QP = lcos{ — Tr — 6) = — Zcos d. Or we may regard 6 as being the angle between {QP) and QX ; but this requires that the length of {QP) should be taken as —I, and so the projection of (QP) on AB = -/cos d. It will be found that, in all cases, I cos gives both the magnitude and sign of the projection of (PQ) on AB. Similarly, the projection of (PQ) on a line perpendicular to AB = I sin 6, 69. Proposition A. The sum of the projections on any PROJECTION. VECTORS 141 Jhie of two sides {AB), (BC) of (a triangle is equal to the project ioti of the third side {AC). Fig. XLVI. In either of tbe above figures (or in any other figure) projection of (.45) + projection of (BC) = {ah) + {hc) = {ac) = projection of {AC). Proposition B. The sum of the projections on any line of the three sides {AB), {BC), {CA) of a triangle is zero. Sum of projections of iAB), iBC), {CA) = {ah) + (be) + (ca). Hence on the line of i^rojection we start at the point a and finish at the same point, so that the distance between the initial and final points is zero. That is, the sum of the projections Proposition C. In any closed figure ABC ... HK, the stm of the projectioyis of the sides {AB), {BC) ... {HE) equals the projection of(AK). Proposition D. In any closed figure the sum of the pro- jections of all the sides taJcen in order in the same direction is zero. Propositions C and D are proved exactly in the same way as Pro- positions A and B. 142 PROJECTION. VECTORS Example. Prove that cos A + cos{120 + A)-\- cos (120 ~ ^) = 0. Fig. XLVII. Draw an equilateral triangle PQR, side a units. Draw a line OX inclined at an angle A to (QR). Then {RP) is inclined at .4 + 120 degrees to OX; and {PQ} is inclined at ^1 + 240 degrees. Project on OX; then, by Proposition B, a cos A + a cos {A + 120) + a cos (A + 240) = ; but cos (.4 + 240) = cos {360 -(120 -.4)} = cos (120-^); .-. cos A + cos (120 + A) + cos (120 -A) = 0. Examples VIII a. (These examples should be verified by drawing a figure to scale.) 1. Show that the projection of a line on a line parallel to itself is equal to the projected line, and that the projection of a line on a line perpendicular to itself is zero. 2. A line of length r, making an angle 6 with OX is projected on OX and at right angles to OX ; calculate the lengths of the projections in the following cases : (i) r = 5, ^ = 60°; (ii) r=-5, ^ = 120°; (iii) ;• = 5, ^ = 248° ; (iv) r = 5, d = 300° ; (v) r = -5, ^ = 330°. 3. Two rods AB, BC, of lengths 5 feet and 10 feet respectively, are joined together at an angle of 135°. The rods are fixed in PROJECTION. VECTORS 143 a vertical plane so that CB is inclined at 60'' to the horizontal, and the angle ABC is beneath the rods ; by projecting horizon- tally and vertically, find the inclination of the line ^C to the horizontal. 4. By projecting a diagonal and two sides of a square on a line making an angle .4' with one of the sides, prove that cos (.4 + 45'') = (cos A - sin A) -r >/2. Find a similar value for sinu-1 + 45"). 5. PQR is a triangle right-angled at Q, having the angle at P equal to A' ; FQ is inclined to OX at an angle B^. Prove by projection that cos {A + B) = cos A cos B — sin A sin B, and sin (A + B) = sin A cos B + cos A sin B. 70. If the projections of a line on two lines at right angles are given, the length and direction of the projected line can be found, but not its actual position. Let r be the length of the line and the angle it makes with one of the lines of projection. Then r cos S and rsin 6 are known ; suppose these values are x and y respectively, so that >'cos^ = x and rsin 6 = ij. Then r = a/x- + i/- and tan 6 = y/x. The projected line has therefore a definite length and a definite direction ; it is the simplest example of a group of quantities called vector quantities or vectors. 71. A quantity which possesses a direction as well as magnitude is called a vector. Yelocities and forces are examples of such quantities. The magnitude and direction can be represented by the length and direction of a directed straight line ; hence the properties of a directed straight line that depend only on its length and direction represent properties common to all vectors. 144 PROJECTION. VECTORS 72. Vector addition or Composition of Vectors. A displacement from A to B followed by a displacement from B to C produces the same result as a single displacement from ^to a B Fig. XLVIir. Or we may regard the displacements as being simultaneous. Suppose a point to start from A and move along AB, and while this point is moving, suppose the line AB to move parallel to itself, the point B moving to C while the point travels from A to B. The result of the two simultaneous displacements is that the point has travelled from A to C. Hence the vector (AC) is called the resultant of the vectors (AB) and (BC). Finding one quantity equivalent to two or more of the same kind is equivalent to the process of addition in Arithmetic. If we use the sign + to denote this process, we have {AC) = {AB) + {BC). If P, Q, and R are the respective magnitudes of the vectors represented by {AB\ {BC), and {AC), and if 6 is the angle between the directions of {AB) and {BC) (in Fig. XL VIII the angle ABC is the supplement of 6) ; then 112 ^ P2 + Q2 + 2 PQ cos 6. Similarly, if a number of vectors are represented by the directed lengths {AB), {BC), {CD)...{HK), then their resultant is repre- sented by the directed length {AK). 73. Besolution of vectors. In Fig. XLVIII the vector {AG) may be replaced by the two vectors (AB) and {BC). Viewed in this light they are called the components of the vector (AC). PROJECTION. VECTORS 145 When we talk of the component of a vector in a given direction, and no mention is made of the direction of the other component, it is understood that the other component is at right angles to the first. P Fig. XLIX. If (OP) in Fig. XLIX represents a vector of magnitude E inclined at an angle 6 to OX, then its projection {OX) represents the component along OX, and the projection (XP) represents the component perpendicular to OX. The vector is now said to be resolved along and perpendicular to OX. Resolving along OX, we find that the component is Rcosd. Resolving peqDendicular to OX, we find that the component is Psin^. 74. All the work of § 69 on projections can be applied to vectors and their components. For instance, Proposition C gives the following proposition : The sum of the components of any number of vectors in a given direction is equal to the component of their resultant in that direction. Examples VIII b. [In the following examples the letters P, Q, R imply that the vectors are forces ; the letters u, v, w imply that the vectors are velocities. When possible, figures should be drawn to scale to check the calculation.] 1. Find the resultant R in the following cases : (i) P= 17, <?= 13, e= 40"; fii) P= 17, (2= 13, 6^ 140'; (iii) P- 114. Q= 75, 6 = 65'; (iv) P= 123, (,^ = 496, e^nV. 145 PROJECTION. VECTORS 2. Find P when Q the other vector, 6 the angle between them, and R their resultant have the following values : (i) ^ = 176, i? = 249, e^ 72°; (ii) Q= 73, i^ = 193, 6= 110°; (iii) Q = 245, i? = 92, e= 130° ; (iv) Q= 36, 7? = 84, ^ = 20°. 8. Show that, if the resultant of three forces is zero, the sum of their components in any direction is zero. 4. Show that if three forces produce equilibrium (their resultant is, therefore, zero) they are parallel and proportional to the sides of a triangle. 5. A boat is being rowed due E. at a speed of 6 miles an hour ; at the same time a current carries it due S. with a speed of 3 miles an hour ; find the magnitude and direction of the actual velocity. 6. Find the resultant of velocities u and v inclined at an angle By when (i) 11 = 14, f = 16, 8= 180°; (ii) « = 14, i^ = 16, 0= 65°; (iii) u = U, v=lQ, e = 135°. 7. Vectors of magnitudes 7, 8, 9 respectively are parallel to three consecutive sides of a regular hexagon. Find the sum of their components (i) parallel to, (iij perpendicular to, the middle one of these sides. Hence find the magnitude and direction of their resultant. 8. Find the magnitude and direction of the resultant of four forces of magnitudes 5, 10, 15, 20 respectively, which act along the sides of a square. 9. A stream flows at the rate of 2 miles an hour. In what direction must a man swim in order that he may actually go straight across the river, his rate of swimming being 3 miles an hour? 10. A rod 5 feet long is hung by a string, attached to its two endS; over a smooth peg ; it rests, at an angle of 20° to the horizontal, so that the two portion^^ of the string are each inclined 35" to the vertical. Find the length of the string. PKOJECTION. VECTORS 147 Projection on a Plane. 75. If from every point in a line, atniight or curved, a perpen- dicular be let fall on a plane, the locuj; of the feet of the perpendiculars is called the projection of the line on the plane. If from every point in the boundary of a surface a perpendicular be let fall on a plane, the area bounded by the locus of the feet of the perpendiculars is called the projection of the surface on the plane. 76. The angle between a straight line and its projection on a plane is called the angle between the straight line and the plane. It follows that the projection on a plane of a straight line of length I, making an angle a with the plane, is I cos (X. Any two planes, not parallel, intersect in a straight line. If from any point P in this line two perpendiculars FA, PB are drawn to it, one in each plane, then the angle APB measures the angle between the planes. 77. If any plane surface, of area A, is projected on a plane making an angle a with its own plane ; then the area of the projection is A cos ol. *Step I. Consider a rectangle ABCD, having the side AB parallel to the plane of projection, and the side BC making an angle 0( with that plane ; then y is the angle between the plane of the rectangle and the plane of projection. Then, in Fig. L, abed is the projection of ABCD. D Fig. L. * A slight knowledge of solid geometry is assumed lu this proof. k2 148 PROJECTION. VECTORS Now Bb is perpendicular to the plane abed, and therefore to the line ab ; .-. Bb is perpendicular to AB ; but BC is perpendicular to AB ; .". AB is perpendicular to plane BCcb ; .•. ab is perpendicular to plane BCcb ; .-. ab is at right angles to he, i.e. abed is a rectangle. Hence art-a of abed = abxbc = ABxBC cos Oi = area of ABCD x cos CX. Step II. Consider a plane area with curved or rectilinear boundary. In the plane of the figure draw any line PQ parallel to the plane of projection. Then in the area we can inscribe a number of rectangles having the short sides parallel to PQ and the longer sides perpendicular to PQ. Fig. LI. The sum of these rectangles is less^than the original area, but may be made to differ from that area by as small a quantity as we please by making their width small enough ; and then the sum of their projec- tions will differ from tlie projection of the area by an even smaller quantity. Hence in the limit, when the width is indefinitely small, the sum of each set of rectangle;s will equal the area of the corre- sponding circumscribing figure. But the sum of projections of rectangles -- sum of rectangles x cos A ; .*. the area of projected figure =■ area of original figure x cos A. PROJECTION. VECTORS 149 Examples VIII c. 1. A pyramid VABCD has a square ba?e A BCD, side a, and the faces VAB, &c., are equilateral triangles. Find the length of the projection of VA on the base. Verify that the sura of the areas of the projections of the four faces is equal to a^. 2. A square house, whose side is 28 feet long, has a roof sloping up from all four walls at 40° to the horizontal, find the area of the roof. 3. Find, by projection, the area of the curved surface of a right circular cone, having height h, and semi-veitical angle 2 y. 4. From a cone 6 feet high a smaller cone 2 feet high is cut off. If the radius of the base of the small cone is TG feet, find the area of the curved surface of the remainder of the large cone. Verify your answer by projecting this surface on the base. 5. A circle with radius a is projected into an ellipse with semi- axes a and h ; show by projection that the area of the ellipse is TTCib. 6. The vertical angle of a conical tent is 67", and the radius of the base is 5| feet; find (i) the slant height, (ii) the area of canvas used, (iii) the content of the tent. 7. A pyramid on a square base is such that each of the other faces is an isosceles right-angled triangle, find by projection the angle between a triangular face and the base. Geometrical representation of imaginary quantities. 78. In Fig. LII OA is of length r. By the usual convention a line OA drawn to the right is B A' A Fig. LII. considered positive, so that {OA) represents -fr. If now [OA) is turned through two right angles, it takes up the position {OA') and, by the usual convention, {OA') represents - r. Hence the 150 PROJECTION. VECTORS geometrical operation of turning through two right angles repre- sents the algebraical operation of multiplying by -1. Let us consider what the operation of turning through one right angle represents. This is an operation which, if performed twice in succession, turns through two right angles, which represents multiplica- tion by — 1. But the algebraical operation of multiplying by aZ-I, if performed twice in succession, multiplies by —1. Hence it seems reasonable that the operation of turning a vector line through a right angle represents the algebraical operation of multiplying by \/-l ; that is, (OB) at right angles to (OA) represents r x V^— 1, i.e. -%/ — Ir. In future we shall denote \/—l by /. 79. With the interpretation of i suggested by the last section, cc + ii/ is represented by a vector line of length or followed by a vector line of length y at right angles to the first vector. X + iy ^ {0N)+ (NP) (Fig. LIII.) = (OP). (By vector addition, § 72.) Or, in words, or + iy is represented by the vector (OP), that is by a vector line of length >v/.r^ + f/^, with the positive direction. laking an angle tan"' 80. For our purjioses this statement is more useful if reversed, VIZ. (OP) =^ x + iy = }'cos6+ ir Bin 6 = (cos ^ + 1 sin B) r. PEOJECTIO^\ VECTORS 151 Or, in words, the vector line of length r, in direction ^, repre- sents the magnitude r multiplied by cos ^ + /sin ^. This gives the important result that turning through an angle 6 represents multiplication by cos^ + j sin^. Hence turning twice in succession through d represents multipli- cation by cos ^ + /sin B repeated twice ; i.e. turning through 26 represents multiplication ly (cos + / sin 6f ; but turning through 2 d represents multiplication by (cos2^ + /sin2^). Hence the suggested interpretation of -y/— 1 or /, leads to the identity fcos 26+ i sin 2 ^) = (cos ^ + j sin 6)-. If this is verified by algebraic multiplication and by the use of the ordinary formulae for cos 2^ and sin 2^, it will be found correct. Carrying on the argument in the same way, we deduce that (cos 7id + i sin n 6) = (cos 6 + i sin 6)'\ where n is any positive integer. Again, turning through a half 6 is an operation which, if re- peated, turns through 6, and, therefore, represents a multiplication which, if repeated, multiplies by cos 6 + / ?in 6 ; • Le. (cos 16 + { sin 16)=> (cos 6 + i sin 6} '^. / 6 6 \ - Similarly. ( cos - + / sin - j = ( cos ^ + / sin 6)" ; and ( cos - ^ + / sin - ^ ) = ( cos ^ -f / sin 6) i • a '1 / Lastly, turning through -6 cancels turning through 6, and, therefore, represents an operation which cancels multiplication by (cos (9+ i%\n6) ; i.e. {co8(-^) + /sin(-^)y = .cos^ + /sin^;-'. r!fimilarly, CCS { — n6) + i sin { — n6) = (cos 6 + / sin ^j~", where n is any positive quantity. 152 PROJECTION. VECTORS We have now clediiced from the geometrical interpretation of \/-l. that (cosn^ + isinn^) = (cos ^ + i sin ^)" for all real values of n. This is known as De Moivre's Theorem. Example. Use De Mo iv re s Theorem to find \^\. cos 2 nTT = 1, sin 2 « 77 = 0, where n is zero or any integer. Hence 1 = cos2«7r-f /sin 2;?7r ; .'. ^T = (cos2n7r + isin2«7r)i = cosf ^?7r + isin§n7r. If >?-0, >v/r= cosO + /sinO = 1. If n = 1, ^T ^ cos Itt + / sin §77 = -^ ( - 1 + / yS). If n = 2, ^r=cos^7r + /sin^n- = H-1-?V3). If ;^ = 3, v^l = cos27r + » sin2n- =1. For other values of n it is seen that the three roots are repeated, Hence De Moivre's Theorem shows that there are three different cube roots of unity. They are, of course, the three roots of the equation ic^ — 1 = 0. The student should verify that the same roots are obtained hy Algebra. Examples VIII d. Represent graphically and by imaginary quantities the follow- ing vectors : 1. (i) Magnitude, r = 25 ; direction 6 = y where a = tan-^g^^. (ii) „ r = 25; „ 6 = tt-OL; (iii) „ r = 25; „ 6 = tt + 0L] (iv) „ r=25; „ ^ = 27r-a; (v) ,. r = 25; „ 6=-0(. 2. Show graphically that [x + iy') + {x" - iy") = (.r' + x") + i (//' 4- y"). 3. Express the following in the form y (cos 6 -f / sin 6) : (i) 3+4/; (ii) 5 + 6/; (iii) 7-8/; (iv) -5-12/; (v) -5 + 12/; (vi) 8/. 4. Interpret geometrically (cos a + /sinOK) (cos 0(-t sina) r; and justify your interpretation. PKOJECTION. VECTORS 153 5. Show graphicall}^ that (cos 0( + -/ sin CX) (cos jS + / sin (3) = cos {0C + (3) + i sin {OC + /3). 6. Verify De Moivre's Theorem by calculation, when n = 2, 3, -1, -2. 7. Assuming De Moivre's Theorem, prove that v^ — l has three values, viz. -1, ^(1 + ^-3) and 1(1-7-3). 8. («) Prove De Moivre's Theorem by induction when n is a posi- tive integer. (b) Deduce the proof when n is not a positive integer, by methods similar to those used for the Binomial Theorem in Algebra."^ Examples VIII. 1. A man walks one kilometre in a direction 16 degrees North of East ; he then turns to the left, through an angle of 110 degrees, and walks one kilometre in the new direction. How far is he North and how far East of his starting-point ? 2. Show that «cos^ + 6sin^ can be expressed in the form r cos {d - Oc). Illustrate by a figure. 3. A number of rods are jointed together, and the two free ends are secured to two points A and B in the same horizontal line and distant c inches. If the length of the r^^ rod is «,. , and its inclination to the horizontal is 0,. (all the angles being measured in the same sense), prove that (i) 2 (ctr cos ^,.) = c ; (ii) 2 («,. sin 6,.) = 0. (See § 89, Example ii.) 4. Prove by projection that sin(90 + ^) = cos^ and sin (270-^) = -cos ^. 5. In what respects can a vector quantity be represented by a straight line? If three forces P, Q, R, acting at a point 0, are such that P/B\nQR= Q/smFP= R/sinPQ (where sin P^ denotes the sine of the angle between P and Q), show that the three forces produce equilibrium. 6. A man walks one kilometre in a direction A"" North of East, one kilometre in a direction making 120" with the first direction, and one kilometre at an angle 240° with thefirst direction. Draw a figure showing that he has now returned to the starting- * See School Algebra, pp. 407, 435, 4G5. 154 PROJECTION. VECTORS point ; and by considering the distances he has gone to the East and North write down two trigonometrical identities concerning the sines and cosines of A, 120 + -4, 240 + ^. 7. Suggest a geometrical construction which may help to sum the series : (i) cos a + cos (a + /3) + cos (CX + 2 /3) + . . .n terms ; (ii) sin (X + sin {0( + ^)+ sin (a + 2^) + . . .w terms. Deduce that both these sums become zero if «,3 = 2it. 8. A body which weighs 12 lb. is kept at rest by means of two cords, one being horizontal and the other inclined to the horizontal at an angle whose tangent is | ; find the forces exerted by the cords. 9. A mine shaft is 1650 feet in length. It slopes downwards at an angle of 45° to the horizon for a certain part of its total length, say x feet, and at an angle of 35° for the rest of its length. If the total depth reached is 1000 feet, obtain an equation for x, and hence calculate x. 10. A man playing five holes of a golf course first walks 260 yards due East, then 140 yards 20° South of East, then 300 yards due South, then 200 yards 40° West of North, then 220 yards 30° West of South, thus arriving at the fifth hole. Find how far the fifth hole is from the first tee. 11. The perpendicular from the origin on a straight line equals 2) and makes an angle a with the axis o( x; by projecting the co-ordinates of any point on the line show that the equation of the straight line may be pat in the form a; cos a + 1/ sin Oc = p. (This is known as the perpendicular form of the equation of a straight line.) Hence find the length of the perpendiculars from the origin on the lines whose equations are (i) Sx + 4ij = 7 ; (ii) 5x-l2ij = 2; (iii) x + 2t/ = 6. Verify by drawing to scale. 12. The co-ordinates of a point referred to rectangular axis OX, OY are x, y; referred to two rectangular axes OF, OW through the same point the co-ordinates are |, //. Prove by projection that ^ = xco^oc + ijs'm OC, where Oi is the angle between OX and or. Find three other similar relations connecting ^, »/, x, //, CHAPTER IX FORMULAE FOR (i) THE FUNCTIONS OF THE SUM OR DIFFERENCE OF TWO ANGLES, (ii) THE SUM OR DIFFERENCE OF THE FUNCTIONS OF TWO ANGLES, (iii) THE FUNC- TIONS OF THE DOUBLE ANGLE AND THE HALF-ANGLE 81. To express cos (A + B) in terms of the sines and cosines of A and B. Fig. LV. Let OX be the initial line ; and let the revolving line first turn, through the angle A to the position OA and then continue to turn through an additional angle B to the position OB. Then OB is the bounding line of the angle A + B. Along OB measure a length OP=r units. Project OP on the initial line, produced backwards if necessary also project OP on OA, produced backwards if necessary. Figures LIV and LV show two of the many possible cases. 156 FORMULAE FOR FUNCTIONS OF THE SUM In all cases the projection of (OP) on OX = sum of the projections of (OK) and (KP) on OX ; i.e. rcos{A + B) = (OK) cos A + {KP) cos (^ + 90) = r cos B cos A + r sin B { — sin A). . Hence cos (A + B) = cos ^ cos £ — sin A sin B. * Several proofs of this have already been given, but the earlier proofs have implied that A and B are together less than two right angles ; this proof applies whatever be the values of A and B. Exercises, (i) Deduce the formula for sin (A + B) by substitu- ting 90 -v4 in place of A, and -B m place of B. (ii) By similar substitutions deduce the formulae for cos {A — B) and sin(.4 — J5). (iii) By projecting perpendicular to OX, find the expanded form of sin {A -f B). (iv) Modify the construction so as to prove directly, by pro- jection, the formula for sin (A-B) and cos (A~B). (v) Complete the following formulae : cos {A -¥B) = cosU-5) = nn{A + B) = sin(^-^) = (vi) Learn these formulae in woids, as : cos sum = cos cos - sin sin. * When Fig. LV is being used it must be recollected that {OK) is negative, and tliat its inclination to OX is XOA not A'OK, see § 68, If (07v) is regarded as positive, its actual length is — r cos B ; but the angle is tlien JCOK, the cosine of wliich is —cos A. Whatever way it is taken, the projection of {OK) on OA' is found to be r cos £ cos A. OK DIFFEEENCE OF TWO ANGLES 15' 82. The followinrr proof does not involve any knowledge of projection ; its chief drawback is that it applies only to the case when A + B is less than a right angle. It is easily modified to suit any other given case. Let XOA = A and AOB ^ B ; then XOB = A + B, and in Fig. L VI is less than 90°. Take a point P in the bounding line oi A + B; let fall PN perpendicular to OX-, PK ,. „ OA; „ KL ,, ,, PX and. therefore, parallel to OX KM „ „ OX. ON OM-MN goh{A+B) = OP OP OM OP LK -\ OPj OM OK OK OP LKPK PK OP The spaces in the bracketed line (which does not appear in the completed work) are filled in with the hypotenuses of the triangles in which the respective numerators occur. Now angle LPK = 90° - LKP = LKO = A ; therefore LK PK = sin A. Hence cos (A + B) = cos A cos B - sin A sin B. From the same figure, prove that s'm(A + B) = To find the functions of A—B, the angle AOB must be made on the negative side of OA. The point P must be taken in the 158 FORMULAE FOR FUNCTIONS OF THE SUM Dounding line of A — B, and the construction and proof proceed as before. It is found that cos{A-B} = sin (.4 -5) = Exercises, (i) Prove the four formulae when A and B are each less than 90^ but ui + B is greater than 90^. (Make the same construction as when ^ + i>' is less than 90^, and pay careful attention to the signs of the lines.) (ii) Prove the four formulae when A and B are each obtuse and together greater than 270°. Examples IX a. 1. By using the formulae of § 81, verify that sin (90 — -4) = cos^ cos(90 + ^)= -sin^, sin (180-^) = sin .4, cos (180-^1 j= -cos^ sin(270 + ^)= - cos ^, cos (360-^) = cos^. 2. Express cosTO"" in terms of the functions of (i) 40° and SO"* (ii) 45° and 25° ; (iii) 95° and 15° ; (iv) 35°. 3. Express sin 40° in terms of (i) 30° and 10° ; (ii) 25° and 15° (iii) 70° and 30° ; (iv) 20°. 4. From the expansions of sin(^ + i^) and coii{A + B) deduce the expansion of tan(^ + J5) in terms of tan^ and tani?. 5. From the expansions of sin(^ — ^) and cos (^-^) deduce the expansion of tan {A - B) in terms of tan A and tan B. 6. Verify that sinO° = and cosO° = 1 by using the formulae for^-^. 7. Show that (i) sin (^ + -B) cos -B - cos (^ + 5) sin 5 = sin 5 ; (ii) cos {A + B) cos B + sin {A + B) sin B = cos B. 8. From the formulae for A + B deduce that sin 2^ = 2 sin -4 cos ^ and cos2^ = cos^vl — sin'-^. What is the value of tan 2^ ? 9. Find the values of (i) sin(^ + jB)-} sin(^-^) ; (ii) cos ( J + B) -f cos (.4 - £) ; (iii) sin {A 4- B) - sin {A - B) ; (iv) cos {A + B)- cos {A - B). Account for the signs of (iii) and (iv) from first principles. OR DIFFERENCE OF TWO ANGLES 159 10. Prove thiit sin A = sin }, (A + B) cos \ {A-B) + cos \ [A + B) sin 1{A- B). Prove similar results for cos A, sin B, and cos B. 11. From the results of 10 deduce that sin ^ + sin J5 = 2 sin \{A + B) cos \{A- B), and three similar results. 12. Prove that (i) cos- 6 -f cos- (p — 2 cos 6 cos (p cos ($ + (p) = sin- (d + cf)); (ii) sin- 6 + cos- — 2 sin ^ cos sin (^ + </>) = cos^ {B + </>). 83. Sums and difi'erences of sines or cosines ex- pressed as products. These formulae are most easily derived from the formulae of § 81, as suggested in Examples IX «. They can be proved inde- l)endently by projection. Make the angle XOA = .1 and the angle XOB = B.* On OA and OB take lengths OP, OQ respectively, each equal to r units. Join PQ. B Fig. LVII. Fig. LVIII. Bisect the angle QOP by a line cutting PQ at R. Then the angles BOP, ROQ each equal 1{A-B) ; and the angle XOP = l{A + B). From congruent triangles EP = RQ, and PRO is a right angle. The projection of (OP) = sum of projections of (OR) and (RP), and the projection of (0^; = sum of projections of (02?) and (RQ). * Notice the difference between tbi> construction and tbc construc- tion of § 81. 160 FORMULAE FOR SUM OR DIFFERENCE OF .•. projection of ( OP j + projection of (0^)^*2'^ projection of (OR), since projections of {HP) and (BQ) are equal but opposite. Projecting on a line perpendicular to OX, we have r sin A + r sin B = 2'^'' the projection of {OB). But {OR) is the projection of {OP) on the direction OR, i.e. {OR) = rco&\{A-B). rsin^ + rsin J5 = 2rco&\{A-B)^m\{A + B). Hence sin ^ + sin £ = 2 sin ^ (^ + B) cos \{A-B). From the same figure, by projection on OX, we have cos^ + cosjB = 2cos|(^ + jB)cos^(^-jB). Again, projection of (OP) -projection of (0$) = 2^^ projection of {RP). Hence sin^ — sin P = 2 cos | (A + P) sin \{A — B), cos^-cosP= -2sin|(^-P)sin J(^ + P). The proofs apply to all cases whatever be the magnitudes of A and P. The reason for the negative sign in the last of these formulae is obvious, for, if A> B, then cos A < cos P. It is useful to learn the formulae in words, it being understood in all cases that the greater angle is put first. sine + sine = 2 sine half sum cos half difference. 84. Products of sines and cosines expressed as sums or differences. In the formulae of the last section put i(^ + P) = X, \{A-B)=Y', so that A^X+Y, B = X-Y. Then sin(X+ r) + sin {X- Y) = 2 sin X cos Y, i. e. 2 sin X cos Y = sin {X+ F) + sin(X- Y). Similar results are obtained from the other formulae. If we replace X by ^ and Y by P, the formulae become 2 sin ^ cos P = sin (.1 + P) + sin (.1 - P), 2 cos .1 sin B = sin (.1 + P) - sin (.1 - P), 2 cos.l cos P = cos {A + P) -f cos {A - B), 2 sin A sin P ^ cos (.1 - P) - cos {A + P). These are more easily proved direct from the A-vB and A—B formulae. FUNCTIONS OF TWO ANGLES 161 In using these formulae it is usual (but not necessary) to put the greater angle first ; this shows why there are distinct formulae for 2 sin .4 cos i? and 2 cos .4 sin i?. Express the four formulae in words : Twice sine cos = sin sum + sin difference. Examples IX b. 1. Apply the formulae of §§ 83, 84 to the following cases and verify from the tables : (i) A = 70°, B = 30° ; (iii) A = 72°, 5 = 18°; (ii) A = 110°, 5 = 75° ; (iv) A = 78°, B = 46°. 2. Prove, from the formula for sin A + sin B, that sin 2 ^ = 2 sin 6 cos d, and, in a similar way, show that 1 + cos ^ = 2 cosH ^, 1 - cos JL = 2 sinH ^. 3. Prove that (i) sin ^ + cos ^ = ^/2 cos {A — 45) ; (n) sm A — cosB= - 2 sm ( 45 7,— 1 sm ( 4o ^ — j ; /•••x . • -^ ^ /.K A-B\ { ,^ A + B (ill) cos A + smB = 2 cos f 45 ^ — 1 cos ( 45 + 4. Prove that (i) sin 50° + sin 130° = 2 cos 40° ; (ii) cos 50° - cos 130° = 2 sin 40°. Verify these by squaring and adding. 5. Prove that (i) 2 cos 40° sin 50° = 1 - sin 10° ; (ii) 2 cos 40° sin 40° = sin 80°; (iii) 2 sin 64° sin 26° = cos 38°. Verify this last result from the tables. 6. Fill iu the right-hand side of the following : (i) sin /0° + sin50° = (ii) cos 30° -cos 110° = (iii) 2 sin 75° cos 10° = (iv) sin 37° + cos 24° = (v) 2 cos 84° cos 72° = (vi) cos 79° -cos 52° = (vii) sin 75° -sin 116°= (viii) 2 cos 80° cos 35° = (ix) cos 24° -sin 76° = (x) 2 sin 17° sin 48° = 1216 L 162 FORMULAE FOR FUNCTIONS OF (xi) 2 cos 73° sin 15° = (xii) 2cos U°cos 166° = (xiii) cos |7r-cos|rr = (xiv) sin 1 7r + COS J tt = (xv) 2 sin 43° cos 47° = (xvi) 2 cos 97° sin 46° = (xvii) sin 81° + sin 10° = (xviii) sin 49° -sin 53° = (xix) 2sin79°sinl5° = (xx) cos 43° -cos 216° = 7. Prove that 4 cos (75° + A) sin (75° - ^) = 1 - 2 sin 2 ^. Formulae for the double angle and half- angle. 85. It lias already been shown in § 51 that sin 2 ^ = 2 sin A cos A; cos 2A = co&'^A - sin^^ = l-2sin2^ = 2jiQsM-l. , The proof there given assumed that 2 J. is less than 180°. If we put A instead of B in the A + B formulae the same results are obtained ; thus they are true for all values of ^. The results can easily be proved independently by projection ; the proofs are the same as in § 81, A taking the place of B. 86. From the last section, by putting -^^ in place of A, we have cos'* ^A — sin'^ -| J. = cos A ; also cos'^^^ + sin^^ J. = 1. Add 2cos^i^ = l + cosX Subtract 2 sin^ ^A = 1- cos A. Hence cos^^ = + Kv^l + cos^) ; sin|^= +|(-v/l -cos^). If the value of A is given, there is no ambiguity of sign. If, for instance, A = 140°, then ^A = 70°, and the sine and cosine are both positive; if ^ = 264°, then -^.4 = 132°, and the sine is positive, the cosine negative. If the value of cos^ is given but not the value of A, the ambiguity cannot be removed. Suppose cos .4 = ^, then A may have any value of the form 360° n ± 60°. Hence I A may have any value given by 180° « + 30°. If we tabulate these values, we have r igle cosine sine 30° +|-v/3 +^ 150° -^^/S +1 210° -},^/3 -I 330° +^^3 -I DOUBLE ANGLE AND HALF-ANGLE 163 87. Tangent formulae. From the sine and cosine formulae the following tangent formulae are derived ; the proof of the first only is given : , , „, sinU+5) tan \A-{-B) = — — =r( ' cos (^ + 5) sin A cos B + cos A sin B cos A cos B — sin A sin B _ ^n A + tan B (By dividing throughout by ~l-tanAtanB cos^eo3J5.) Similarly, tan (^-5) = tan 2.^ = 1 A /I -COS A tan I A = + A z ^ - A' 1 + cos A 1-cosA sin A ,„ ,. ... , = + — -. — - — or + _ . (By rationaliznig. ) - sin A -1 + cosA ^ ^ °' 88. These may all be proved directly from the figures used for the sine and cosine formulae ; e.sr. in Fier. LVI. tan (^ + 5) = '^, MK^LP OM- MK OM LK LP "^ OM (By dividing so as to make the first ~ LK PK' term in the denominator to be L) PKOJl The triangles LPK, OKM are similar ; LP PK , ^ TT L / i T,\ tan A + tan B Hence tan(^ + ^)=:, — 7 tt ^• 1-tan^tan^ Exercises. Prove that tanyl + tan5 + tan(7 — tan^tan^tanC (i) tan(^4-5+C) = l-tanjBtanC-tanCtan^-tan^tan^ (ii) tan.i + ta„£ = '^^^^^'; cosJ-cos-B (iii) tan X- tan 5 sin(^--B) cos^ cos5 l2 164 FORMULAE FOR FUNCTIONS OF (iv) cot A + cot B = -. — -—. — =, ; sill A sm B {,)cotA-ootB =-^i'^Mz^'. sm A sm A Example. To prove that tan'^ g\ + tan ~ ^ ^ J y = tan~ ' ^^q • Let A = tan-^ ^g, B = tan-^ ^tf ; so that tan A = ^^, tan B — o|y . tan A + tan B tan (^ + 5) = 1 — tan A tan ^ ■9^ + - 1 ]L V _1- ■■■ 99 "^ 2'59 239 + 99 338 23900-240 23660 i. e. tan ^ t^^^ + tan"^ 239 = tan" -1 1 7(J' Examples IX c. 1. Prove that (sin A + cos Af = 1 + sin 2^, a nd (sin A - cos Af = 1 - sin 2 ^. 2. Assuming the values of sin 45°, cos 45°, tan 45°, deduce sin 90°, cos 90°, tan 90°. 3. Find a formula for cot 2^ in terms of cot^. 4. Show that sin ^A = 3sin^ -4 sin^ JL. Explain how it is that there are three values of sin J. when sin 3 J. is given. 5. Find the values of tan 22^°, tan OTf, tan 157i°. 6. Prove that 2sin-|^ = + -y/l + sin A ± ^/l-smA. Find sin ^.4 when sin^ = |. Illustrate by a figure. 7. Find cosl^ when sin^ = ^. Illustrate by a figure. 8. Prove that (i) sin 2^ = 2 tan^^(l +tan2^) ; (ii) cos2^ = (l-tanM)~(l+tanM); (iii) tan 2^ = 2 tan ^ ^ (1 - tanM). 9. Prove that sin ^A = 3 sin^l — 4 sin^^ ; cos 3^ = 4cos^u4-3cos^ ; tan 3^ = (3 tan^- tan^^) ^(1 - 3 tanM). DOUBLE ANGLE AND HALF-ANGLE 165 10. Show that (cos A + sin Af + (cos A - sin Af = 3 cos ^ — cos 3 A. 11 (rt). Show that sin^^ + cos^^ = + \/l + sin^, and sin ^A — cos ^ J. = + v^l — sin A. (b) Having given 4 sin 54°= ^/b + l, apply the formulae in (a) to find sin 27° and cos 27°, explaining how the ambiguities of sign are cleared up. (c) Show that 8(sinU2°-cos2 78°) = -/S + l. 12. Prove that fA R r — ^^'^^^tan^ + tan C— tan^tan5tan C '^ ~ 1-tan^tan C — tan C7tan^— tan^tan^ Deduce the formula for tan dA- What can be deduced ifA + B + C equals (i) 180°, (ii) 90° ? 13. If tan ^ = If and tan B = 2%, show that A + B = {4:n + l)l7r. 14. Show that cos ^ + cos 3 ^ + cos 5 ^ + cos 7 ^ = 4 cos ^ cos 2 ^ cos 4 S. 15. Find all the solutions of the equation sin 6 sin 3 ^ = sin 5 6 sin 7 6. 16. If tan^ = I, tan 5 = §, tan C = fj, and each angle is acute, prove that A + B + C = ^7r, 17. If tan^ = tanjcxtani^, show that tan 2 ^ = (sin OC sin ^) -r (cos y. + cos /S). 18. (i) If ^ = tan-4, find tan 2^. (ii) Show that 2 tan"^ | + tan-^ } = in. 19. Prove that cos2^-cos2^ = 2(cos-^-cos2 5) = 2(sin-5-sin^^). 20. Prove that .. _J sec4^ ^^^ a + bcosd~ {a + b) + {a-b)ta.n'^^B'' 1 1+tan^l^ ^ ^"^ acos^ + fesin^ ~ a + 26tan|^-atan2i^ * 21. Solve the equations (i) x2-^/2sin(i7r + Oi)a; + isin20( = 0; (ii) «2_2cot2/3.a;-l = 0.'' 22. Solve for OC and Fthe following equations : 2ag = V- sin 2CX, 2bg = V^ sin^ y. 166 FORMULAE FOR FUNCTIONS OF 23. A hemispherical shell of radius 16 inches rests with its rim on a horizontal table ; a rod is hinged to a vertical wall, 25 inches from the centre of the shell, at a point 5 feet above the table. The rod is in the same vertical plane as the hinge and centre of the shell, and touches the shell. Find its inclination to the vertical. Oral Examples. (a) (i) sin(P-^) = (ii) cosX + cos r = (iii) cos (90- 1 J+^) = (iv) sin 270° (v) 2 sin a cos /3 = (vi) cos^ 6 - sin^ 6 = (vii) tan (A-B) = (viii) sin ^ - sin C = (ix) cos245°-sin2 45°= (x) cos 2.1 = (6) (i) cos(C+.4) = (ii) sini? + sinC (iii).2sini(5-f Cji(P-C)= (iv) cos^ + cos(/) = (v) 2cos2 1(7-1 = (vi) tan25 (vii) sin (180-5 + C) = (viii) cos^TS^ + sin^TS^ = (ix) cos{^A + B + lA-B)= (x) sin (360-2 Cj = (c) (i) sin 2 5 = (ii)sin(P+^) (iii) cosin-oc + li) = (iv) 2siniCcosJC = (v) 1-2 sin^ B = (vi) cos^ | C- sin^ i C = (vii) cos2 1 C + sin^ A C = (viii) tan (B-C) = (ix) cosC-cos^ = (x) (sin j5 + cos 5)^ = (d) (i) sin^cosC-sin^sin C = (ii) cos(X-r) = (iii) sin 3^ = (iv) sin-^i? = (v) cos 5 + cos (7 = (vi) 2 cos 5 cos C = (vii) sin^ {B + C) + cos- {B + C) = (viii) 2 cos- ^ ^ - 1 = (ix) (cos 1^- sin 1 .1)2 = (x) tan (90 -C) = 89. The preceding formulae load to a number of useful identities in the cases where A + B + C=90° or 180". The method of dealing with these is shown in the following illustrative examples. DOUBLE ANGLE AND HALF-ANGLE 167 Example i. In any triangle tan H^"" ^) = r — ^^^ \ ^- [Here - — - gives the clue to the proof.] + c b c By the sine formula, -. — ^ = ~ — ;;,' *' ' sm j5 sm C sin B b I.e. . ^ = -, sm C c ^ , i T -i T sin 5 — sin b — c Componendo et dividendo, -; — ^^ -. — i^ = , ; 2cosl(B+C)sm^iB-C) _ b-c ^ 28in^{B+C)cos^{B-C)~ b + c' tan^(g-CO _ b-c ^ ^•®* tani(.B+(7j ~ b + c' hut A + B+C=180''; .'. IA + 1{B+ C) = 90. Hence tan |(5+ C) = tan (90 -M) = cotj^. b — c Substituting above, tan ^(-B - C) = cot ^A. This formula has been proved geometrically in § 53 ; it is usually proved by the method given above. Example ii. In any triangle ^^~~cUS^-\-€o^^ + cosC = 1 + 4 sin i A sin J B sin i C. L.H.S. = cos J. + cos^ + cos C = 2 cos * (^ + jB) cos 1 U - -B) + 1 - 2 sin^ -i C = 2sin|Ccosi(^-5) + l-2sin|Ccos^(^ + J5), since ^C = %-l{A + B) = l + 2sin|C(cosi^-^-cos^^ + ^) = l+4sin|^sin^5sin"?C. The symbol 2 cos A is sometimes used to denote cos^ + cos5 + cos C; and nsin^ to denote sin ^ sin 5 sin C. The above result can be written: 2 cos^ = 1 4-4nsin^^. 168 VARIOUS ILLUSTRATIVE EXAMPLES Example iii. In any triangle ^cos'^A = l — 2Yicos A. (Questions involving the sum of the squares of sines or cosines are usually solved by expressing these squares in terms of the cosine of the double angle.) 2 2 cosM = 2 cos2 A + 2 cos'^B + 2 cos^ C = 1 + cos 2^ + 1 + cos 25 + 2 cos^ C. (Note that one angle is left unchanged.) .-. 2 cos^^ = 1 + cos (^ + B) cos (A-B) + cos^ C = 1 - cos C cos (^-5) -cos Ccos (A + B), since C= 180 -(^ + 5) = 1 - cos C [cos {A ~ B)-cos(A + B)] = 1—2 cos Acqs B cos C. Example iv. Solve the eqimtion sin + sin 2e + sinS0 + sin 4:6 = 0. Rearrange sin ^ + sin 4 ^ + sin 2 ^ + sin 3 ^ = 0. Use formula for sum of two sines 2sin#^cos|^ + 2sin|^cos|^ = .*. either sin f ^ = or cosp+cos^^ = i.e. -^6 = nrr or 2 cos^cos ^^ = : i.e. 6 = ^nrr, or cos ^ = or cos^^ = i.e. ^=(2n + l)*7r or |^ = (2/i4- l)i7r. Hence the complete solution is ^=(2w + l)7r, (2n + l)i7r or fnTT. Example v. To prove that r = 4:B sin J A sin I B sin i C. From the figure of § 59, rcot^5 + rcot^C= a, /coshB cosiCX ^ „ . , r . l^ + -7-^-y, ) = 2i? sin ./, \sm|5 sin|C/ ' siniBcos^C+cos^^sin-JC ^^ . I.e. r —^ . ', p . , V, ^ = 27?sm^, sinf5sin|(7 ' sini(J5 + C) ^^ . , I.e. r . {1. . '^ = 27?sin^, sm|5sin^C i. e. r _._;o^i;^ = 4 7^ sin I A cos I A ; sinl^sin^C ''sincei(P+C) = 90-|^; .-. >• = 4 7? sin I ^ sin ^5 sin J C. Exercise. Prove that r^ = 4 7i? sin J ^ cos ^ 5 cos ^ C. I.e. VAKIOUS ILLUSTEATIVE EXAMPLES 169 Example vi. To show that the distance between the circum- centre and in-centre = \^{Ii- — 2Br). A In Fig. LIX, with the usual notation, BD = E sin A, DO = R cos A, BP= 1' cot IB, PI=r. Or- = {BP-BDf + {IP-ODf = {rcoi\B-R%\nAf + {i'-Rco^Af = i?2_2i;-(sin^coti5 + cos^) + »^(l + cot45) sin|^ sin^l^ ^0 nr. sinU+*5) ,^ sini^sinJ^sinlO (Substitut- smi^ sm'^i^ mgforr.) ^o or> cos^{A-C)-2smlAsm^C = U' — 6iir -. — , , sin ^5 since A + },B = ^0 + lA-lC siniB = R!^-2Rr. since i(^ + Cj = 90 -i5. This is more shortly proved by Pure Geometry ; but the method used here is a general method to find the lengths of lines connected with the triangle. Example vii. To p'ove that sin A + sin {A-\-B)-\- sin {A + 2B)-{- ... to n terms sin h nB sin {A + ^n—l B) sin I B 170 VARIOUS ILLUSTRATIVE EXAMPLES Let S* = sin ^ + sin (^ + i?) + . . . + sin (A + n - 1 B). Multiply by 2sinjP. Then 2smlB.S = 2sin^sini^ + 2sin(^ + ^)sini5+ ... +2sin(^ + n-l^)sini5. Use the formula for the product of two sines. 2sini^.5'= cos(^-ij5)-cos(^+|5) + cos (A + IB)- cos [A + IB) + .... - + cos (.4 + i2n- 35) -cos (^+12^1-1^) = cos(^-i5)-cos(^ + i2M-15); ^^m\)iB%m{A + \n-lB) sin^?i5 Note, Compare tliis with the formula for the sum of n terms of an Arithmetic Progression. Notice that A + \n — \ B = half the sum of the first angle {A) and the last angle (^4-n — 1 B), Examples IX. 1. Prove the following identities : (i) sin 3 ^ = 4 sin A sin (60^ + A) sin (60'' - A) ; (ii) sin 3 A sin^^ + cos ZA cos^ A = cos' 2 A ; (iii) (l-2sin2^)-f(l + sin2^) = (1 -tan^) f (1 + tan^) ; ,. . tan (45° + ^)+ tan (45°-^) ^ , (IV) r~ y-Ar^o-^Tx , 7TE5 7\ = COSeC 2 ^ ; ^ ' tan (45° + ^) -tan (45°-^) (v) sin {y->iZ-x) + sin {z + x-y) + dn{x + y-z)-^\n{x + y + z) — 4 sin a: sin y sine'; (vi) cot J^ — cot^ = cosec^ + cosec-|^; (vii) cos 4 ^ + 2 (cos ^ + sin J.)* = 3 + 4 sin 2 A ; (viii) sin ^ + sin 5 = sin (^ + -B) + 4 sin \ AQm\B sin \{A + B)', (ix) sin ^ - 3 sin 3 ^ + 3 sin 5 ^ - sin 7 ^ = 8 sinM cos 4 ^ ; (x) cos(^ + J?+C) = cos A cos B cos C — cos A sin 5 sin C — sin A cos B sin C — sin^ sin 5 cos C\ (xi) cos \ A (2 sin A - sin 2 A) = sin^ I A (2 sin ^ + sin 2 ^) ; (xii) cos A-vco%B + cos C + cos (^ + i? + C) = 4 cos ^ {A + B) cos I {B + C) cos },{C-\- A). SUM AND DIFFERENCE FORMULAE 171 2. If A, B, C be the angles of a triangle, show that (i) tan A + tan B + tan C = tan A . tan B . tan C ; (ii) sin 2 ^ + sin 2 5 + sin 2 C = 4 sin ^ sin 5 sin C ; (iii) sinH^ + sin2i5 + sin2iC + 2sin*^sin*5sin^ C= 1; (iv) sin ^ + sin jB + sin C = 4 cos \ A cos \ B cos \ C ; (v) cot A cot 5 + cot A cot C + cot 5 cot C = 1 : (vi) cot A + cot B + cot C = cot ^ cot B cot C + cosec .4 cosec B cosec C ; (vii) tan B tan C + tan Ctan J. + tan ^ tan B = 1 + sec A sec 5 sec C ; (viii) cos ^ sin (5 -C'j + cos 5 sin (6'-^j + cos Csin(^-5) = ; (ix) (tan A + tan B) (tan A - cot C) = sec^^ ; (x) tani5tan^C + taniCtanM + tani^tani5 = 1. 3. Show geometrically that sin(^ + 5) = sin^cos^ + cos^sini? when each of the angles A and B is between \tt and it, and A + B is less than frr. 4. Solve the equation cos 3^ + cos 2^ + cos ^1 = 0. 5. Find all the values of 6 which satisfy (i) cos(9 + cos2^ + cos3^ + cos4^ = 0; (ii) sin3<9 + sin4^ + sin5<9 = 0. 6. Solve (i) sin {A + 30°j = 1 -^ ^2 ; (ii) V'-3 sin A + cos A = ^/2 ; (iii) sin J. + cos ^ = 1 ; (iv) sin A + ^/2^ cos ^ = 2 ; (v) ^/2 (cos OX + sin 3a^) = 1 ; (vi) a cos6 + b sin 6 = c (put a = r cos 0(, h = r sin 0(). 7. Prove that (i) 2 sin-i ^ ^/2 = 90° ; (ii) 2tan-^| = tan-4. 8. In any triangle show that i? (sin 2^ + sin 25 + sin 2 Cj = 2r (sin^ + sin J5 + sin C). 9. In any triangle show that a^ COS. 2 B + b'^ cos2 A = a- + &'^-4rt& sin^sin 5. 10. Prove the formula (& + c) tan 1{B-C) = {b-c) cot hA. Write down two corresponding formulae. 11. Using the fact that 3 x 18° = 90°-2 x 18°, find the values of sin 18° and cos 18°. Give a geometrical method for determining sin 18°. 172 SUM AND DIFFERENCE FORMULAE 12. Simplify ., /sin 4^ cos4vl\ , , , .^ ,, V sm J. cos^ / sin 5^ -sin 3^ 2 sin 5 (9 + sin 3^ (") .^.^^ , ...o^ + cos5^ + cos3^ sin2^ cos5^ — cos3^ 13. D, E, F are the feet of the perpendiculars from A, B, C on the opposite sides ; P is the orthocentre. Prove that (i) ^P=2i?cos^; (ii) PZ) = 2PcosP cos C; (iii) perimeter of triangle DEF = 4P sin ^ sin B sin C. 14. State the general formula for all angles having a given cosine. Solve sin 3 ^ + sin 5 ^ + sin 7 ^ = 0. 15. Find sec {A + B) in temis of the secant and cosecant of .4 and 5, and prove secl05° = 72 (1 + ^3). 16. Prove that sinl8° = |(v^-l) ; and that sin^ 30° = sin 18° sin 54°. Show that in any circle the chord of an arc of 108° is equal to the sum of the chords of arcs of 36° and 60°. 17. Given cos ^='28, determine the value of tan ^.4, and explain fully the reason of the ambiguity which presents itself in your result. 18. Prove that cos~'ii; + cos~^y = sin"^ {x \/\ —if^-y^/X —x^), and solve the equation tan-i{(a:+l)-f(a^-l)}+tan-i{(a^-l)-^a;} = tan-'(-7). 19. Express sin3^ -r (sin2^ — sin^) in terms of cos ^. 20. Prove the identities : (i) (l + cos^)tan2i.4= 1-cos^; (ii) (sec ^ + 2 sin A) (cosec ^ - 2 cos ^) = 2 cos 2 A cot 2 .-1. 21. In any triangle prove that [h — c) cos ^A = a sin I {B — C). Ii A = 80°, a = 10, h-c = 5, find B and" C. 22. Prove the identity cos 2^; sin 3a7 = sin x cos ix + cos .r sin 2.r. 23. Solve the equations cos 2^ = cos(^-rt) ; cos 3^ = sin (6-^). 24. (i) If the equation of a straight line is put in the form ij = mx + c, what is the geometrical interpretation of m ? (ii) Show how to find the angle between two lines whose equations are y = mx + c, y = rnx + c\ SUM AND DIFFEKENCE FORMULAE 173 (iii) Deduce that the lines are at right angles if mm' = - 1 ; and parallel if m — m'= 0. (iv) Prove that the lines whose equations are ax + hi/ + c = 0, ax + Vy + c^O, are perpendicular if aa +W =0y and parallel if a/ a' = h/h'. 25. Find the angle between the lines whose equations are (i) Sx-4:y=b, 4x-2ij = 7; (ii) 4a; + 3?/ = 6, Sx-4i/ = 9', (iii) 2x— y =3, 4a7 + 5?/=l; (iv) 2x-y =3, 4x + 2?/ = 5; (v) 2a; + 4i/ = 5, a; + 2t/=3. In each case verify by drawing to scale. 26. Find the equations of the straight lines drawn through the point (3, 5), and respectively parallel and perpendicular to the line whose equation is 3ic — 4f/ = 5. 27. Find the equation of the straight line, parallel to the line whose equation is iccos a+i/sin a =_p, and passing through the point [x, y). Deduce that the length of the perpendicular from [x', y') to the line x cos 0( + y sin oc = p i& x' cos Oi + ?/'sin Oi —p. 28. Find in its simplest form the equation of the line joining the points {acos(CX + ^), &sin(CX + ^)}, {a cos (a -/3), &sin(a-/3)}. 29. Prove that sin 55° sin 15° - sin 50° sin 10° - sin 65° sin 5° = 0. 30. Show that in any triangle a^sin(J9-C ) &^sin(C-^) c^sin(J . -B) h + c c + a a + b 31. If 2cos^ = ic + l/a; and 2 cos = 2/ + 1/^/, prove that 2 cos {d + (f)) = xy + 1/xy and 2 cos {d — (p) = x/y + y/x. 82. If d + (f) = 240°, and versing = 4 versing, find the values of 6 and 0. 33. Draw a curve to represent the variations in sign and magni- tude of (sin^->v/3cos^)-7-(-/3sin^ + cos^), from <9 = to e = n. 34. If oc and /3 are the roots of « sin ^ + & cos ^ + c = 0, prove that cos-|((X4-3) _ COS |(CX-^) _ sin^(0( + /3) b —c a 35. Eliminate 6 and from a sin ^ + & sin = h, (i) rt cos ^ — & cos (p = k, (ii) , cos {d + (f)) = I. (iii) 174 SUM AND DIFFERENCE FOEMULAE 36. Eliminate 6 and when two equations are the same as (i) and (ii) in Ex. 35, and the third equation is (i) sin(^ + 0) = I, (ii) tan(^ + 0) = /. 37. Eliminate B and from the equations a cos{(b-hOC) sine/) . ,. x , • /, h cos(^-a) sm ^ ^ ■ v-r / 38. Expand sin 5^ in terms of sin^, and cos 6^ in terms of cos^. 39. If sinJ5 is the arithmetic mean between sin^ and cosyl, prove that cos2^ = cos2(^ + 45°). 40. If a cos ^ + & sin B = c, show that B = tan-^ h/a + cos-i cj{ Va" + ¥). 41. Find the maximum and minimum values of acos^ + Z>sin^ = c. Verify your answer when a = 3, Z> = 5, by drawing a graph. 42. Prove that (i) sin B + sin 2B + sin 3 ^ + ... to n terms sin|«^sin|«4- 1 B ^ ^ sinT^ " ' (ii) cos ^ + cos (^ + 5) + cos (^ + 2 i?) + . . . to n terms s,va.\nB C0& {A + \n~ 1 B) ~ sin \B (iii) cos a + cos (a /3) + cos(a + 23)+ ...to nterms = 0, if n^ = 2n. 43. Find the sum of n terms in the following series : (i) sin^ A + sin2 {A + B) + sin^ {A-v2B)+ ...; (ii) cosM + cos2(^ + jB) + cos2(^ + 2^)+...; (iii) sin^ sin 2^ + sin 2^ sin 3^4 + sin 3 J. sin 4-1 + .... REVISION EXAMPLES 175 Revision Examples C. (All the following examples are taken from recent Examination Papers.) 1. Find, without reference to the tables, the values of (i) sin 45° ; (ii) cos 150^; (iii) the tangent of the obtuse angle whose sine is 1/v/lO. 2. Trace the graph of the function cos^ + 2sin^ between the values and 180° of 6, and determine from your figure the value of 6 for which the function (ij is greatest, (ii) is decreasing most rapidly. 3. Express tan 6 in terms of sec 6. Show that (sin 6 - cos d) (sec 6 + cosec 6) = tan 6 - cot 0. 4. Prove that the sines of the angles of a triangle are in the ratios of the sides opposite them. 5. Solve the equation 2 cos a; + sin a? = 2. 6. In a right-angled triangle ACB, C being the right angle, the angle A is 35°, the side AB is 10 inches ; find the other sides. 7. If cos(^ + -Bj = cos^cos5-sin.4sin5, calculate cos(A + B) when A = 50° and B = 50°. 8. If OC is measured in radians, sin a = a-aV,3 + aV|5_-a7l2+ ... where j 5 means 1x2x^3x4x5. Find sin a correct to four significant figures when oc = 0'3. What is the angle OC in degrees ? 9. Define the tangent of an angle in such a way that your definition is true for all angles. If 6 be an acute angle, prove that cos (90 + ^) = -sin 0. 10. Arrange in. order of magnitude the angles 2 sin~^ -51, i cos-i '32, tan'^ 8-9. 11. Draw the graph of cos a; for values of x lying between 0° and 90°. Use your figure to solve roughly the equation .r = 100cos:r°, and verify your solution by the tables. 176 REVISION EXAMPLES 12. Given that sin 20° = O'Si and cos 20° = 0*94, write down the values of sin 160° + cos 160°, of sin 250° + cos 250°, and of sin 340° + cos 340°. 13. In any triangle ABC, show that (i) c = acos^ + &cos^ ; (ii) c2 = a2 + &2-2a&cosC. Find c when a = 5, & = 6, and 0=155° 31', having given cos24°29' = 0*91. Verify your result by a diagram drawn to scale. 14. Find to the nearest degree the angle subtended at a man's eye by a tower 50 feet high, when the man has stepped back 30 feet from the tower, assuming the height of his eye above the ground to be 5 feet 6 inches. 15. Write down a formula for sin^J. in terms of the sides of the triangle ABC and explain the notation. How is the formula modified when h — c^ Given that the sides are 100, 200, 160 units in length, calculate the smallest angle. 16. A and B are two acute angles but A + B \^ obtuse; prove that cos {A + B) = cos A cos i? — sin ^ sin B. Solve completely cos x + ^mx = cos Oi — sin OL. 17. Define the tangent of an angle, and show geometrically that tan^tan(90° + ^) + l = 0. 18. Draw a circle of diameter 1 inch. Draw a diameter AB and the tangent to the circle at B, divide either of the semi- circumferences between A and B into 8 equal parts, join A to the points of section, and produce the joining lines to meet the tangent at B. Measure the distances of the points so found from B, and use the results obtained for drawing the graph of tan ^4 from ^ = 0° to ^ = 90^. 19. Prove that the area of the triangle ABC is \a^ sin B sin C/sin A. Use this expression to find the area of the triangle when a = 106-5 yards, A = 56° 37', B = 75° 46.' REVISION EXAMPLES 177 20. A person walking along a straight level road running clue East and West observes that two objects P and Q are in a line bearing North-West, and after walking a further distance d he observes that P bears due North and that the direction of Q makes an angle A with the direction in which he is walking. Prove that the distance PQ is cZcos^/sin (^-45°). Find PQ when d = 1372 yards, and the angle A = 56° 31'. 21. (i) Show that (sin^ + cos^)2 + (sin^-cos^)2 = 2. (ii) Considering only values of A between 0° and 90°, find the value of A when sin^ cos J. has its greatest value, and show that the same value of A gives the greatest value of sin ^ + cos ^. 22. Let AD bisect the angle ^ of a triangle ABC, and let it meet BCin D -, show that BD sin B = CD sin C. Hence show that BD.AC= DC. AB. 23. (i) Show geometrically that sin {A + B) = sin ^ cos ^ + cos A sin B, when A, B, and A + B are each less than 90°. (ii) By means of this formula, and in view of the restrictions under which it has been obtained, show that sin 464° = sin 153° cos 311° + cos 153° sin 311°. 24. Find tan 6 and x in terms of a and b from the equations a sin ^ + & cos 6 = ^x, « cos ^ — 2 & sin ^ = 2 37. 25. An angle is made to increase gradually from 0° to 360° ; state briefly how the values of its sine and of its cosine change during the increase of the angle. 26. Calculate the values of A between 0° and 360° for which tan J. -2 cot ^ = 1. 27. A and B are two milestones on a straight road running due East across a horizontal plane, C an object on the plane. The bearings of C as viewed from A and B are 35° North of East, and 55° North of West respectively. Find, to the nearest foot, (1) the distance of C from A, (2) the distance of C from the nearest point of the road. 1916 M 178 REVISION EXAMPLES 28. Plot in relation to the same axis and origin the values of tana; and 2 sin a; for the values 0°, 12=^30', 37° 30', 50°, 62° 30', 75° of X, draw the graphs of tan x and 2 sin x, and find from them the values of x for which tana? = 2 sin a;. Give the general solu- tion of the equation tan a? = 2 sin x. 29. Prove that (cos A + sin A) -h- (cos A - sin A) = tan {A + 45°). 30. Prove for a triangle in which the angle B is obtuse the relation sin Bjh = sin C/c, and deduce the relation tan ^{B-C} = {b- c)/{b + c) cot I A. If & = 27*3 yards, c = IS'S yards, A = 48° 36'", find B and C. 31. Prove that in a triangle ?• = 4 J? sin ^ J. sin 1 5 sin | C ABC is a triangle ; B'C is drawn through A parallel to BC, A'C through C perpendicular to AC, and A'B' through B perpendicular to AB. Prove that the area of the triangle A'B'C is \ or cos- {B — C)-ir cos B cos C sin A. 32. (i) Find sin ^ + sin 5 in terms of functions of half the sum and of half the difference of the angles A and B. (ii) If ^ + 5 is between 90° and 180°, find under what circum- stances tan J. + tan 5 will be negative. 33. Find to the nearest minute the angle of a regular polygon of 17 sides. What angles less than 360° satisfy the equation 2cos2^-f llsin^-7 = 0? 34. Prove the identity tan^^ coi^A _ 1 — 2 sin-.l cos-.l 1 + tan^^ 1 + cot^^ sin J. cos ^ 35. Assuming the formula a^ = h'^ + c^ — 2 he co^ A, establish a formula for tan^^ in terms of the sides of the triangle, and find the greatest angle of the triangle whose sides are 13, 14, 15. 36. Prove that for any triangle ABC rt/sin A = &/sin B = r/sin C. U B = 39° 17', a = 4*2, and b = 3-5, solve the triangle fully ; draw a figure to illustrate your solution. 37. The angles of elevation of a vertical pole from two points on a horizontal line passing through its liase and 6 feet apart are a and /3 ; prove that the height of the pole is b/[cot 0( — cot^) feet. KEVISION EXAMPLES 179 38. From a point on a horizontal plane passing through the foot of a tower the angles of elevation of the top and bottom of a flagstaff 20 feet high, placed vertically at the summit of the tovrer, are 51*2^ and 47"3''. Find the height of the tower. 39. Prove that (i) sin (A + B) = sin AcosB + cos A sin B ; (ii) cos2^{l + tan2^)= l-tan-.4. Use (ii) to find the value of tan 15°. 40. Reduce the fraction a -r (cos^^ — sin-^; to a form suitable for logarithmic calculation, and perform the calculation when « = 10, A = 29' 55', and B = 15" 5'. 41. Prove that sin*^ .1 + cos'^ .^ = 1 for all values of A less than 180°. A and B are each less than 180^ sin ^ = '3900, smB= -9208, find four possible values of A + B. 42. Find from your tables the value to two decimal places of the expression sin^ + sin2^, when B is 10^ 20°, 30°, ... 90', and from these draw a graph of the expression on a suitable scale. 43. In a triangle ABC prove that (i) 2bc cos A = b- + c^- a-; (ii) cosM + cos2-B + cos^co35 = f, if C=60°. 44. In a triangle a = 12*76, h = 10-87, c = 8-37, find C. 45. Show how to find the distance between two visible but inaccessible objects. 46. In any triangle ABC show that four times the area equals {(("^ + b- + c^) -^ (cotan A + cotan B + cotan C). Show also that when C is a right angle this expression reduces to 2a&. 47. Prove the identities : (ij 1/sin 2.4 = 1/tan A - 1/tan 2.4 = tan A + 1/tan 2^ ; ,.., . ^ , sin ^^ + sin 4 J. (ii) tan ?>A = ^^-p- T-, • ^ ' co8 2J. + cos4^ 48. What is the meaning of tan-* x ? Prove that tan"' a? + tan"'?/ = id,n-'^[{x + y) -f {\-xy)]. Prove that 45° is one value of tan~^ -| + tan"' |- + tan"* j^jj. 49. Prove that (i) secM = 1 4- tan^ ^ ; (ii) cosec^-cot^ = tan^^. M 2 180 REVISION EXAMPLES 50. Construct an angle whose sine is 0'76. From your figure obtain the value of the cosine of the angle. 51. On squared paper draw graphs of tan^ and cot ^ between ^ = 10° and 6 = 80". From the graph, or otherwise, find angles which satisfy the equation tan 6 + cot ^ = 3. 52. Let D be the point in which one of the escribed circles touches the side BC of a triangle ABC. If the sides a, h, c of the triangle are given, find expressions for the radius of that circle and for BD and CD. 53. A tree which grows at a point A on the north bank of a river is observed from the points B and C on the south bank. The distance BC is 200 metres, the angle ABC is 46° 80', and the angle ACB is 58° 20'. Calculate the distance of A from the straight line BC. 54. Prove the formula sin|J. = \/{s — h) (s — c) -r he. If, in a triangle ABC, 2b = a + c, prove that sin|5 = 2 sin 1^ sin I C 55. Find the angles B and C and the radius of the circum- scribed circle of a triangle ABC in which A = 32° 42', a = 36, 6 = 44. 56. State De Moivre's Theorem, and, assuming it for integral indices, prove it for fractional indices. Write down all the values of (>/— l)o. 57. U A is an obtuse angle whose sine is i%, find the values of cos^ and tan^. 58. (i) Show, by drawing graphs of the two expressions sinu; and cos (.r + 90°), that sina;= —cos (a; + 90°). (ii) If sin a? = ^ \/2, find a formula which gives all the values of .r which satisfy the equation. 59. Prove that in a triangle (i) tan ^B = ^{s -c}{s-a}-^s {s — b) ; (ii) b cos B + c cos C = a cos {B — C). EEVISION EXAMPLES 181 60. If two sides of a triangle and the angle opposite one of them are given, show how to solve the triangle, and discuss by the aid of a figure all the cases that can arise. One side of a triangle is 20 inches long, the opposite angle is 34° 42' ; another side is 30"41 inches. Find the sides and angles of the two possible triangles. 61. Assuming the formulae for the sine and cosine of half an angle of a triangle in terms of the sides, prove that (i) r = ^/{s — a){s — b) ( s — c) -r s ; (ii) Z?= a/2 sin ^. 62. I observe the altitude of an airship to be 35°, and that of the sun, which is in the same vertical plane as my eye and the airship, to be 40^ The shadow of the airship falls on a tree on the same level as my eye and 500 feet in front of me. Find the height of the airship. 63. In any triangle prove that sin ^ - sin ^ + sin C = 4sin ^vi cosi^sin^C Assuming the formula for expanding ti\.n{A + B), find expressions for tan 2^ and tan 3 J. in terms of tan J.. 64. Make an angle AOC and bisect it by the line OB. From any point A in OA draw ABC perpendicular to OB, meeting OB, OC in the points B and C respectively, and draw AX perpen- dicular to OC. Use this figure to prove that (i) sin2J.<2sin^; (ii) tan 2 xt > 2 tan .4. 65. Prove that sin^^ + cosM = 1. Having given that the sine of an angle is '56, calculate its cosine. 66. Show how to construct an angle whose sine is "6. Find a value of x which satisfies the equation 4 sin ic + 3 cos a; = 1. 67. Given two sides of a triangle and the included angle, show how to find the remaining side and the other angles. Prove such formulae as you require. If a = 1097 feet, b = 781 feet, C = 31° 30', find c to the nearest foot. 68. A ship is sailing at the rate of 7 miles an hour. A man walks forward across the deck at the rate of 4 miles an hour 182 KEVISION EXAMPLES relative to the deck, in a direction inclined to the keel at an angle of 60°. Find the direction of his actual motion in space. 69. Prove the formula cos {A-B) = cos AcosB + sin A sin B. Show that if xy = a'^ + 1 then cot~^ {a + x) + cot"' (« + y) = cot"' a. 70. Find an expression for cos{Oi. + ^ + y) in terms of sines and cosines of Of, 0, and y. Prove the identity cos OC cos + 7) + COS /3 cos {y + OC) + COS y COS {01. + ^) = COS (ex + /3 + -y) -h 2 COS OC COS /3 cos y. 71. At what angle must forces of 4 dynes and 5 dynes act so that their resultant may be a force of 6 dynes ? 72. If B be the circular measure of an angle, prove that, as 6 is indefinitely diminished, the ratios 6 : sin d, 6 : tan 6 approach to the limit unity. A man standing beside one milestone on a straight road observes that the foot of the next milestone is on a level with his eyes, and that its height subtends an angle of 2' 55". Find the approximate height of that milestone. 73. Write down the values of sin 36° and cos 36° as given by your tables. Calculate the sum of the squares of these numbers to six decimal places, and explain why the result differs from unity. 74. Give definitions of the tangent and cotangent of an angle which is greater than 90° and less than 180°. Prove that (i) tan (180 - ^) = - tan ^ ; (ii) tan(90 + <9)=-cot^. 75. In any triangle prove that a/sin A = hj&m B = c/sin C. If BC be 25 inches, and CA be 30 inches, and if the angle ABC be twice the angle CAB, find the angles of the triangle ABC, and show that the length of the third side is 11 inches. 76. F, Q, R are three villages. P lies 7 miles to the North-East of Q, and Q lies 11^ miles to the North-West of B. Find the distance and bearing of 7' from B. 11. A point is moving with velocity 50 feet per second in a direction 60° North of East. Find the resolved parts of the velocity in directions East and North. EEVISION EXAMPLES 183 78. A man has before him on a level plane a conical hill of vertical angle 90^. Stationing himself at some distance from its foot he observes the angle of elevation Oc of an object which he knows to be half-way up to the summit. Show that the part of the hill above the object subtends at his eye an angle , tan0c(l-tan3() tan~ — • 1 + tan DC (1 + 2 tan y) 79. The latitude of London is 51' N., and the radius of the Earth 4000 miles. How far is London from the Equator measured along the Earth's surface, and how far from the Earth's axis ? 80. Prove that sin A + sinB = 2 sin l (A + B) cos I {A - B). Show that sin 10' + sin 20' + sin 40' + sin 50' = sin 70' + sin 80°. MISCELLANEOUS PKOBLEMS (The following examples are taken from recent Army Entrance and Civil Service Papers.) 1. I take measurements to determine the air space of a rect- angular hall : length 18*4 metres, breadth 11*8 metres, inclination to floor of diagonal of side wall 31 "8°, of diagonal of end wall 44°. Calculate the air space. More measurements were taken than were necessary. Check the measurements by deducing one of them from the other three. 2. The ancient Greeks measured the latitude of a place by setting up a vertical rod and comparing its length with the length of its shadow. Supposing observations taken at mid-day at the equinox (when the sun is vertical at the equator) to give ^ as the ratio of the rod to shadow at Alexandria, and ^ as the ratio at Carthage, find the latitude of each place. 3. The following method of determining the horizontal distance PR, and the difference of level QR between two points P and Q, is often used. A rod with fixed marks A, B on it is held vertical at Q, and the elevations of these points, viz. ACD = (X, BCD = jS, are read by a telescope and divided circle at C, the axis of the telescope being a distance CP=a above the ground at F. If QA = h, and AB = s, write down expressions for PB and QB. Find PB and QB when 0( = 6° 10', 8 = 1° 36', the values of a, h, and s being 5 feet, 2} feet, and 5 feet respectively. 4. Three balls, 5 cm. in diameter, lie on a floor in contact, and a fourth equal ball is placed on them. Find the height of the centre of the fourth ball above the i)lane of the other three centres. Find also the inclination to the vertical of any line that touches both the top ball and one of the lower balls. 5. The curved surface of a right circular cone whose semi- vertical angle is 45° is made by cutting out a sector from a circular sheet of copper, the diameter of the sheet being 5C cm. Deter- mine the angle of the required sector. MISCELLANEOUS PROBLEMS 185 6. If tangents be drawn to the inscribed circle of a tiiangle parallel to tbe sides of the triangle, show that the areas of the triangles cut off by these tangents are inversely proportional to the areas of the corresponding escribed circles. 7. A rod BC, of length 5*8 cm., rotates about B. Another rod CA, of length 8*6 cm,, has one end C hinged to the first rod, while the other end A slides along the line BO. By drawing the rods in various positions, find how the length o^ BA varies as the angle B increases ; and show BA as a function of angle 5 in a graph for one revolution of BC, showing the actual length of BA and repre- senting 30° by 1 cm. Write down an equation connecting the angle B and the lengths of the three sides of the triangle ABC. Solve the equation to find the length of BA when angle B is 35"". 8. The extreme range of the guns of a fort is 8000 metres. A ship, 14000 metres distant, sailing due East at 24 kilometres an hour, notices the bearings of the fort to be 20° 30' North of East. Find, to the nearest minute, when the ship will first come within range of the guns. 9. The face of a building is 136 feet long. A photographer wants to take the building from a point at which the face subtends an angle of 37°, and for this purpose he starts off from one corner of the building in a direction making an angle of 127° with the face in question. Find by calculation the distance from the corner at which he must take the photograph. Calculate the area of ground in the triangular space between his position and the face of the building. 10. From the top of a telephone pole three wires radiate in a horizontal plane. One wire, A, exerts a tension of 100 lb. weight ; the next, B, makes an angle of 90° with A and exerts a tension of 80 lb. weight ; the third, C, makes an angle of 35° with B and an angle of 125° with A, and exerts a tension of 90 lb. weight. It is required to equilibrate the three tensions by means of a fourth wire. Find its direction and tension. 11. A man passing along a straight road measures the angle between the direction of his advance and a line drawn to a house on his left. At a certain moment the angle is 36° 21'. He walks on 1500 yards and finds that the angle between the same direction 186 MISCELLANEOUS PROBLEMS and the line to the house is now 125° 36'. Find the distance of the house from the road. 12. Plot a curve giving the sum of 4sin^ and 3 sin 2^ from S = 0° to 6 = 180°, and read off the angles at which the greatest and the least values respectively of this sum occur. For the angle use 1 cm. to represent 10 degrees, and for 4 sin ^ + 3 sin 2^ use 1 cm. to represent unity. Also estimate the slope of the curve when 6 = 90° and when d = 135°. 13. A, B, and C are three buoys marking the corners of a triangular yacht racecourse round an island. The angles A, B, and C of the triangle ABC are found to be 75°, 63°, and 42° respectively. P is a flagstaff on the island, from which A and B can be seen, and the distances of P from A and B are found by a range-finder to be 650 yards and 585 yards resi3ectively, and the angle APB to be 187°. Calculate the length of one lap of the course. 14. Draw an angle XOP of 30°, making OP 2" long : through P draw PQ parallel to OX and in the same direction : produce XO to X\ making OX' = OP, and join X'P: cut off PQ = PX. Join OQ and measure the angle XOQ carefully. Now denote XOQ by 0, XOP by 6, and OP by e, and write down an expression for the length PQ. Deduce an equation for 6 and 0, and solve it for tan (p. Use your tables to evaluate (p when 6 = 30°, and compare your result with the measured value. It is said that the given con- struction trisects an angle. What is the percentage error for 30° ? 15. In running a survey the lengths of a series of lines are measured, and the angle each line makes with the direction of magnetic Noiih is measured by a theodolite. The data booked are given in the table below : — Line. Length in feet. Bearing. AB 433 29° 15' BC 521 89° 12' CD 352 182° 38' DE 417 233° 25' The angles are measured clockwise from the magnetic North direction. By an error the measured length of the closing line LA of the survey was not recorded, nor its bearing; from the data given in the table calculate these missing data. MISCELLANEOUS PROBLEMS 187 16. AOB and COD are two straight roads crossing one another at an angle of 57". A motor-car, travelling at the rate of 18 miles an hour along AOB, is 1500 yards from 0, when a man, walking at the rate of three miles an hour along COD, is a quarter of a mile from 0; car and man are both approaching 0. Find graphically the motion of the car relative to the man. Hence find the least distance between the car and the man, and when they are at this distance from one another. 17. In a triangle a = 10 cm., b = l cm., one angle is 95°. There being no restriction as to which angle of the triangle is 95", discuss how many distinct triangles can be made. Select any one case, and for this case calculate the remaining sides and angles. 18. X and Y are two fixed points in a straight line, P a point which so moves that cosPA^Z+cos PYX = Jc (a constant). Prove the accuracy of the following construction for obtaining the locus of P: With X and Y as centres describe circles of radius XY/k. From any point X in XY draw XAB perpendicular to XF cutting the former circle in A and the latter in B. Draw XA and YB, intersecting in P. Then P is a point on the locus. 19. A candle, C, is placed on the floor at a distance r from a point on a wall, and at the same level as the candle-flame, and the angle which OC makes with a perpendicular to the wall at 0, is 6. The illumination received on the wall at from the candle is known to be equal to Acosd/f- where ^ is a constant. If the candle be moved about on the floor in such a way that this illumination remains constant, plot on a diagram the curve described by the candle-flame. 20. Two small islands are 5 miles apart, and there is known to be a rock distant 3 miles from each. A ship is in such a position that the islands subtend an angle of 66^ at the ship. Calculate, to the nearest hundredth of a mile, her least possible distance from the rock. 21. Find by means of a graph two acute angles 6 for which 5sin2^ = 3sin(9 + 2-5. Find also the greatest value of 5 sin 2^ -3 sin ^ when 6 is an acute angle, and the angle to which this value corresponds. 22. The elevation of an aeroplane which is flying horizontally on a flxed course at a height of 150 feet is taken at two instants 188 MISCELLANEOUS PROBLEMS at an interval of 20 sees. At the first observation the elevation is 10° and the bearing is clue North, and at the second the elevation is 6|° and the bearing is N. 35° E. Find the course and speed of the aeroplane. 23. The strength of an electric current C is obtained from the formula C = k tan 6 where 6 is the angle read off in degrees on an instrument, and k is n constant. If an observer makes an error of 8^ in reading the angle ^, prove that the value of C thus obtained will be wrong by an amount equal to J^tt Ccosec2^S^. Hence find the error per cent, in C produced by making a mistake of -j^j degree when 6 is 60°. What value of B is likely to produce the smallest error in the value of C ? 24. If P denote the pressure of wind in lb. per square foot on a plane surface at right angles to the direction of the wind, and 2) denote the normal pressure of wind in lb. per square foot on a plane surface inclined at an angle 6 to the direction of the wind, the following formulae are used to determine the ratio p : P. (i)iV-P=(sin^)^'^^'-''''^-^; (ii) p/P = 2 sin 6/(1 + sin^ S). Compare the values of p/P given by these formulae for the values 10° and 50° of d. 25. A man walks due W. from a point ^ up a straight path inclined at 10° to the horizon. After walking 2 miles he reaches B, and turns up another straight path to the NE., sloping 15° upwards. He reaches C after walking one mile from B. What is the distance in a straight line from C to ^ ? What is the height of C above the level of A ? Taking the face of the hill ABC as a plane surface, what is the greatest slope ? 26. A flagstaff stands vertically on horizontal ground. Four ropes, each 56 feet long, are stretched from a point in the flagstaff', 50 feet above the ground, to four pegs in the ground, arranged at the corners of a square. Calculate the angle between two adjoining ropes. 27. Q is the centre of a circle of radius 10 cm., and QO is a radius. The seven points ABC ... lie on the circumference and the angles OQA, OQB, OQC... have the values 10", 20", 30°... 70°. Find by drawing or calculation the lengths of the chords OA, OB, OC..., and tabulate the results. MISCELLANEOUS PROBLEMS 189 Draw a graph to give the length of chord of the circle in terms of the angle which it subtends at the centre (for angles up to 70"). Show the chord's actual size, and represent 4 degrees by 1 cm. From jour graph find the length of the chord which subtends an angle of 48°. Make a triangle having one side of this length, and the other two sides 10 cm. long, and therefore having an angle of 48''. Check the accuracy of your drawing by measuring this angle. 28. A square made of jointed rods each 4 inches long is deformed into a rhombus having half the area of the square- Calculate the lengths of the diagonals of the resulting figure and check by drawing. If it is part of a lattice-work, the original height of which is 6 times the diagonal of one of these squares, find by calculation how much the height of the lattice-work could be increased if each square were reduced to half its area. 29. A straight rod AB, 3 feet 9 inches long, is held under water, A being 2 feet 6 inches and B 9 inches below the surface. Calcu- late (a) the distance below the surface of a point C on the stick which is 12 inches from A, (h) the angle which the stick makes with the surface of the water. If a parallelogram is held under water, show that in every position the sum of the depths of the 4 corners is 4 times the depth of the point of intersection of the diagonals. 30. If a closed loop of thread is placed on a soap-film that covers a ring of wire, and the film within the loop is joierced, the film outside takes up as small an area as possible and thus pulls the thread at A into a circle. Calculate the diameter and the area of the circle formed by the thread if length of thread forming the loop is 6 cm. If the ends BC of the thread are attached to the ring, and the film on one side of the thread is pierced, the thread again becomes a circular arc. If the thread BC is 6 cm. long, and the angle it subtends at the centre of the circle of which it forms an arc is 120°, calculate the length of the chord BC, EXAMINATION PAPERS OXFOKD AND CAMBRIDGE SCHOOLS' EXAMINA- TION BOARD. School Certificate, 1910. 1. Define the tangent of an angle. Construct an acute angle whose sine is "6, and find its cosine and cotangent. 2. Prove that cos (180 -a) = -cos Of. Arrange the angles Oi, /3, y in order of magnitude, if sin 3^ = -8211, cosi3 = -7738, tany= -O'GlOi, the angles being positive and each less than 180°. 3. What is the length of the shadow of a man, 5 feet 8 inches high, cast by the sun when its altitude is 55° 30' ? 4. Draw the graph of 10 + 10 cos 2a; for values of x between 0° and 60°. Find a value of x to satisfy the equation a; = 10 + 10 cos 2 0^°. [Take one-tenth of an inch as unit along both axes.] 5. Prove that in any triangle sin A/a = sin B/b. U A = 63°, B = 49°, a = 50 inches, find b to the nearest tenth of an inch. 6. Prove that ... cos ^ + sin ^ cos ^- sin ^_ 2 ^^ cos ^ -sin ^ cos(^ + sin^ ~ l-2sin-^' (ii) (sec 6 + tan 3) (cosec 6 ~ cot 6) = (cosec ^ + 1) (sec ^ - 1). 7. If 2 sin ^ + 5 cos ^ = 5, prove that tan^ = or 20/21. 8. Prove that sin {A — B) = sin A cos B — cos A sin B, where A and B are both acute angles and A is greater than B. ^ , , , sin 5 ^ + sin ^ , _ » „ ^ Prove that . ^ , ; — - = 1 + 2 cos 2 -4. . sind^ —sin J. 9. Show that in any triangle ABC b + c _ coslJB-C) a sin h A If fc + c = 24'8 cm., a = ir89 cm.,''yl = 39°, find B and C. EXAMINATION PAPERS 191 10. A lighthouse is observed from a ship which is steaming due N. to bear 62= W. of N. ; after the ship has sailed 10 miles the lighthouse is observed to bear 40° W. of S. Calculate the distance of the ship from the lighthouse when it was nearest to it. Higher Certificate, 1910. Part I. 1. Give a definition of cos^ that holds for all angles from 0' to 180°. Show that cos ( 180 -d)= - cos 0. 2. Show that secM = 1 + tan2^. Draw the graph of 1 +sin dx°, where x lies between 0° and 60°. 3. Construct an acute angle whose cotangent is 2, an obtuse angle whose sine is 3, and an obtuse angle whose secant is —3*5. Measure these angles as accurately as you can with the protractor, and verify your results by means of tables. 4. (i) Verify that 30'', 45°, and 60'' are solutions of the equation sin3a; + cos3rr = 2 cos 2a;. (ii) Show that (cosec^ + sec^j^ + (cosec A - sec A)^ = 2 cosec'^ (3 sec^A — 2). 5. Show that in an obtuse-angled triangle sin A/a = sin B/h = sin C/c. A man observes that the angular elevation of the foot of a tower on a distant hillside is OC, and that the angular elevation of the top of the tower is /3, and he knows that the height of the tower is h feet. Show that his horizontal distance from the tower is ^ cos (X cos /3 cosec (/3 — 3ii). Part II. 6. Draw the graph of cot.r between the values —180 and + 180 of X, taking the unit of x to be -^q inch and the unit of y to be one inch. Find an acute angle to satisfy the equation x = 60 cot a;''. 7. Show that sin (^-5) = sin .4 cos ^- sin 5 cos. 1, taking A and B to be acute angles of which A is the greater. If tana; = A; tan (^ -a;), show that {k-l)smA= {k-hl)3in{2x-A). Use this result and tables to solve the equation tana: = 2 tan (50° -a;). 192 EXAMINATION PAPERS 8. In the triangle in which a = 72 feet, B = 40°, and C = 55°, find c. 9. Find in terms of a, b, and c the radius of the circle escribed to the side BC of the triangle ABC. If Jj is the centre of this circle, show that aAI,''-hBI,^-cCI,' = abc. 10. AB is a diameter of a circle whose centre is ; on AB an equilateral triangle ABC is described, and a point D is taken in AB such that 1BD = 2AB ; CD is produced beyond D to meet the circle at E. Show that tan ADC = 7/^/3 and that sin OED = 3/^52. Hence, or otherwise, show that the error made in taking the arc BE to be one-seventh of the circumference of the circle is less than '2 per cent. Part III was beyond the scope of this book. OXFORD LOCAL EXAMINATIONS. JUNIOE. 1910. 1. (i) Find the sine of 60° ; (ii) If -4 is an acute angle, and cos^ = ^, find the value of 4 tan ^ + 5 sin A. 2. P and Q are points on a straight stretch of a river bank and R is a point on the other bank. If cot PQR = '32, cot ^Pi^ = '43, and the length of PQ is 15 yards, find the breadth of the river. 3. Draw the graph of sin (45° + 2a:) between x = and 180^ 4. U A, B, A — B are all positive acute angles, prove that cos {A — B) — cos AcosB + sin A sin B. 5. (i) A, B, C are the angles of a triangle; if tan^ = ^ and tan ^ = ^, find the angle C. ,.., „ ,- , cos5u4 + cos3^ , . (u) Prove that - — p— : — ir—r = cot A. ^ ^ sin 5 ^ - sm 3 ^ 6. Solve the equation cos2^ + sin^ =?= 1, EXAMINxVTlON PAPERS 193 7. Prove for any triangle that (i) rt/sin A = b/tiin B = c/sin C ; ( ii) {b + c) cos A + {c + a) cos ^ + (« + 6j cos C = a + 6 -I- c. 8. Find the angles A and i? of a triangle ABC in which « = 13, /> = 14, c = 15, having given: Iog2 = -o010, log? = -8151, Z tan 26= 34' = 9-6990, Z tan 29' 44' = 9-7569. Senior. 1910. 1. Find the tangent of 30"". Using the values of tan 30° and tan 45'', prove that tan 75° = 2 + -\/3. 2. A man on a straight level road observed two objects Pand Q {P being the nearer) in a horizontal straight line inclined to the direction of the road at an angle 0(. If tan Oi = -75, FQ = 400 yards, and the shortest distance of F from the road is 180 yards, what is the shortest distance of Q from the road ? 3. Prove that cos3^ = 4cos^.4-3cos.l. Find sin 18°. 4. If ABC is a triangle in which 6 = c = 5 inches and a = 8 inches, find the values of tan A and tan B. 5. Prove that cos2 j^ ^ cos2 B = sin2 {A + B) + 2 cos A cos B cos {A + B). 6. Prove that in any triangle c = (« + &) sin^, where cos ^ = 2 ^/ab cos | C/{a + &). In a triangle ABC, a = 36 feet, & = 4 feet, C = 55°. Using the above formula, find the third side, having given log 6= -7782, Z cos 57° 51' =9-7261, Zcos27°30' = 9-9479, sin 57° 51'= '8467. 7. Find the radius of the circle inscribed in the triangle ABC. C is the centre of a circle of diameter d, and A, B are two points on the circumference of the circle. If I is the length of the chord AB and S is the diameter of the circle which touches CA, CB and also the arc AB at its middle point, prove that 1/5 = 1/(^+1//. 1216 N 194 EXAMINATION PAPERS CAMBRIDGE LOCAL EXAMINATION. Junior. 1909. 1. Define the sine of an angle. What are the greatest and least values which the sine of an angle can have ? Prove that sin^ = cos A x tan J., and that sin A sin B cot B = cos A cos B tan A. 2. Construct an angle vrhose tangent is 1*45, and measure it with a protractor. Verify your results with the help of the tables. 3. Prove that (i) sin A = tan A/{ -/l + tanMj ; (ii) cos (90° + A)= - sin A. 4. Find by drawing graphs of sin^ and sin 2^ for what value of A, less than 90°, 2 sin ^ - sin 2 ^ = 1. 5. A vertical post casts a shadow 15 feet long when the altitude of the sun is 50° ; calculate the length of the shadow when the altitude of the sun is 32°, 6. Prove that sin ^ + sin i? = 2 sin 1{A + B) cos ^{A- B), and that tan 2 J = 2 tan A/{1 - tan-^). Show that sin ^ - 3 sin 3 ^ + 3 sin 5 ^ - sin 7 ^ = 8 sin^^ cos 4 A. 7. Prove that, in any triangle ABC, acos,B+ h cos A = c. Show also that ftan A + tan B) (tan A — cot C) = sec"^. 8. Show how to solve a triangle when three sides are given. Find the greatest angle of the triangle whose sides are 5'2 inches, 77 inches, and 9*1 inches. Senior. 1909. 1. Show that the ratio of the circumference to the diameter of a circle is an invariable quantity. Find to an inch the diameter of a wheel which makes 400 revolutions in rolling along a track one mile long. 2. Any positive proper fraction being given, show that there are two angles, one acute and the other obtuse, such that the sine of either is equal to this fraction. If the fraction is |, use the tables to find the angles, and the cosine and tangent of each. 3. Find by aid of the tables the values of sin it- -tan 2x for the values 0°, 10°, 20°, 30°, 45°, 60° of x. Make a graph to give the values of sin it' — tan 2 jc from ^=0 to X = 60°. EXAMINATION PAPERS 195 4. Show that sin A + sin 5 = 2 sin i (.1 + B) cos I (A - B). Prove also that (i) tan2^ = (l-cos2.4)^(l + cos2^); (ii) sin 55° sin 15° - sin 50° sin 10° - sin 65° sin 5° = 0. 5. Find the greatest angle of a triangle whose sides are 15, 21, 28 inches in length. Show that in any triangle a'^sin (^-C) &-sin(C-^ ) c=^sin(.l-^) h + c c-^a a + b 6. Find an expression for the radius of the inscribed circle of a given triangle. Determine to one place of decimals the length of the radii of the inscribed circle, and of the escribed circle opposite the greatest angle of the triangle referred to in Question 5. Questions 7 and 8 were outside the scope of this book. (The two following questions may be taken instead of 7 and 8, but considerably lower marks will be assigned to them.) A. Show that if .4, B, Care the angles of a triangle, tan A + tan B + tan C = tan A tan B tan C. Show also that tanA5tan^C + tan*CtanA.i + tani^tan|^= 1. B. Solve the equation a cos d + h sin 6 = c. Find all the solutions of sin ^ sin 3 ^ = sin 5 6 sin 7 0. COLLEGE OF PRECEPTORS. Cheistmas, 1910. H Hours. [Four-place tables of logarithms and of natural functions and square-ruled paper are provided. All diagrams should be drawn as accurately as possible.] Part I. 1. Define a radian, and find its magnitude in degrees to two places of decimals (tt = -->'). If the angle of an equilateral triangle were taken to be the unit angle, what would be the measure of a radian to two places of decimals ? 196 EXAMINATION PAPERS 2. Define the sine and tangent of an acute angle. Prove that sin'^^ + cos'^^ = 1. If tan^ = i\^, find the value of cos ^-8 sin ^. 3. Find, geometrically, tan 30°. If ^ = 30°, B = 45°, C = 60°, D = 90°, find the value of: (i) sin ^ cos 5 — sin BcosA; (ii) (tan^^ - cosec^^) / (cot C + cos D). 4. Use logarithms to find as nearly as possible the values of: (i) 3-142 X •9342/-00532 ; (ii) >/562'3/-00^984. 5. Solve, using the tables, the triangle in which C = 90°, « = 654, ^ = 38°45'. Part II. 6. Find all the positive values of 0, less than 360°, which satisfy the equations : (i) cos2^-sin"^ = 0; (ii) 4sin2^cos^^-sin^<9= |. Which of the following statements are possible ? (i) tan^= -2; (ii) sin^ = f. 7. Write down, without proof, the expansions of sin (.1 - B), cos{A-B). Find the value of ianA — B in terms of tan J, tan 2?. If tan^l = ^, tan J5 = -/, find tan (.4 + ^). 8. Prove that, in a triangle, a^ = &- + r-26fcos.4 when the angle A is (i) acute, (ii) obtuse. Deduce that tan lA = \/ (s -h) {s - c) -^ s [s - a). Find the greatest angle in the triangle whose sides arc 256, 389, 401. 9. AB is a horizontal straight line. A vertical straight line is drawn from B upwards, and in it two points P, Q are taken, such that BQ is five times BP. If the angle BAP is 30'\ calculate UmPAQ. EXAMINATION PAPKRS 107 LEAVING CERTIFICATE EXAMINATION (SCOTLAND). 1910. 1. Explain the circular measurement of angles. Express 30°, 50°, 166° 40' in radians. Express '0187 radian in degrees, minutes, and seconds, taking 77 = 3-1416. 2. Taking a horizontal inch to represent 10° and 5 vertical inches to represent the unit of length, plot, with the help of your tables, the values of tan 6 when ^ = 0, 10°, 20°, 30°, 40°, 50°. Plot also the values of sin 6 for the same angles, join both series of points by smooth curves, and thus find a graphic solution of the equation 5 (tan ^- sin ^) = 1. 3. State the relation which exists between the sine and cosine of any angle. Use this relation to find, and express in a diagram, all the values of 0^, less than 180°, which satisfy the equation 5 sin a + 6 cos'' 3^ = 7. Either, 4 a. A man walked 5 miles due North and then walked 6 miles in a direction 27° East of North. Find by a figure drawn to scale how far he now is from his starting-point, and in what direction he should have originally started in order to go straight to his final position. Verify your results by calculation. Or, 4 b. The sides of a parallelogram are 2 inches and 3 inches in length, and its area is 3? square inches. Find by a diagram the sizes of its angles and the length of its longer diagonal. Verif}- your results by calculation. Either, 6 a. Draw a circle of radius 2 inches, and inscribe in it a triangle ABC, such that ZB = .34°, ZC= 73°. Measure the lengths of the sides as nearly as possible. Calculate with the help of the tables the lengths of the sides to the nearest hundredth of an inch, and thus test the correctness of your di awing. Or, 5 b. State and prove the formula which gives tan(yl + i?) in terms of tan A and tan B. Apply this formula to find expressions for tan 2^, tan3yl, and tan 5 ^ in terms of tan A. 198 EXAMINATION PAPERS INTERMEDIATE EXAMINATION (IRELAND). Middle Grade (Pass). 1910. 1. Prove that sin^^ + cos^^ = 1, where A is an obtuse angle. 2. Find the value of the expression cosec^-gcot^, if sinyl = ^^, when A is acute, and when A is obtuse. 3. Prove the identity (1 - tanM) -r- (2 cosM - 1) = secM. 4. In a triangle C = 90°, c = 65°, tan A = '28. Find a and h each to two decimal places. 5. In a triangle a = b \/3, /; = 11, C = 150°. Find c and cos^. 6. In a triangle B = 45°, fc = 20, c = 4. Find sin C, and prove that the perpendicular from A on BC divides BC into two segments one of which is seven times the other. 7. Prove that the length of the perpendicular from the vertex A of a triangle on the opposite side BC is equal to «/(cot-B + cotC), considering the cases when both angles^ are acute, when one is right, and when one is obtuse. 8. Find the angles between 0° and 360° which satisfy the equation 6sin^-4 cosec^ + cot6^ = 0, being given cos 48° IT 23" = §. Middle Grade (Honours). 1910. 1. Show by a graph the values of cosec A for values of A between -90° and 360°. 2. If yl is an angle in the first quadrant, prove that sin A + cos A 4- tan A + cot ^ > sec ^ + cosec A. 3. Prove the identity 3 (sin ^ - cos ^)H (sin A + cos Af + 4 (sinM + cos^i) = 13. 4. The sides of a triangle are 37, 7, and 40. Find all the angles, being given that eo? 69° 25' 48" = ?; ; . 5. In a triangle a = \/5, h = -y/lS, C = 45°. Find c, and prove that cotyl = 2-v/f-l. 6. Find a solution between 180° and 270° of the equation 5(1+ sin .r) = — 3 cos .r, being given cos 28° 4' 21" = 1;. EXAMINATION PAPEES 199 7. Prove by drawing a line through B, making an angle x with the side BC, or otherwifse, that in a triangle ABC, c cos {B - x) + h COS {C + x) = acosx. P is a point on the hypotenuse AB of a right-angled triangle ABC. AP = x, PB = ij, PC = z. Find cos CPB interms of jr, y, and z. Find the sides of the triangle when x = S — ^/d, y = ^/3 + 1 , z= ye. Senior Grade (Pass). 1910. 1. Find the distance from the earth to the moon, assuming that the moon's diameter, 2165 miles, subtends an angle of 31' lU" at the earth. 2. Prove that tan \A = {\- cos ^)/sin A. Find tan 15° and tan 22i" without using the tables. 3. Find x if cos-^;r-f cot"' 2 = ^tt. 4. Assuming the formulae for the sines of the sum and difference of two angles, prove that sin A-sinB = 2cosl{A + B] sin \ {A - B). Find the corresponding expressions in factors for cos^ -cos B. 5. Find the solutions between 0' and 360^ of the equation cot 2. r- 3 tan re = 3. 6. In a triangle a = 183, h = 247, C = Sr 40'. Find A and B. 7. In a triangle A ^ 54° 80', B = 69° 20', a = 341. Find h and c. 8. Prove that in a triangle a cos B-bcosA= {a-- h'^)/c. 9. Prove that in a triangle rcot|^=s — a, where r is the radius of the inscribed circle, and s the perimeter. Senior Grade (Honours). 1910. 1. An arc 40 feet in length is taken on a circle whose radius is 35 feet. Find, to the nearest inch, the length of the perpen- dicular from the centre on the chord of this arc. 2. Prove the identity cos5^/sin^ + sin5^/cos^ = 2 cosec2^-4 sin 2 .4. 3. If cos x + cos y + cos z + cos x cos y cos 2- = 0, prove that tan I X tan J y tan Iz = ± 1 . 4. If X = cot~^ V'cos y — tan~' -y/cosy, prove that y = 2tan~^ -y/sin.r. 5. In a triangle A = 35° 20', a = 127, h = 104. Fin<l B, C, and c. INDEX INDEX A.bscis.sa, 12. Addition formulae, 155, Altitude of sun, 89. Ambiguous case, 120. Angle, 25. circular measure of, 29. measurement of, 25. negative, 26. .»^f elevation, 88. >of depression, 88. Antilogarithm, 21. Arc, length of, 28. functions of, 38. Area of triangle, 109. Axes, 12. Characteristic, 18. Chord of angle, 15. ircular functi measure, 29. Complementary angle, ratios of, 40. Co-ordinates, 12. cartesian, 12. polar, 41. Cosecant, 38. Cosine, 38. formula, 108, cos (A±B), 155. cos A ± cos B, 159. Cotangent, 38. Course, 94. Coversine, 40. Degree, 25. De Moivre's theorem, 152. Departure, 94. Difference of Latitude, 94. of two angles, ratios of, 15G. of two sines, 159. of two cosines, 159. Dip of the horizon, 127. Double angle, ratios of, 100, 162. Elementary formulae, 49. Eliminant, 51. Elimination, 51. Equations, solution of, 78. solution by graphs, 17, 76. general solution of, 79. Functions, algebraical, 14. circular, 40, trigonometrical, 40. Geometry enunciations, 9. Grade, 27. Gradient, 33. Graphs, general treatment, 12. of trigonometrical functions, 41, 163. harder trigonometrical, 74. Greek alphabet, 24. Half-angle, ratios of, 101, 104, 121, 162. Haversine, 40, 121. Heights and distances, 88, 124. -identities, 50, 64, 167, 168, 170. Imaginary quantities, 149. Indices, fractional, 17. negative, 17. Interpolation, 37. Inverse functions, 45, 164. Latitude, 32. Logarithm, 18. base of, 18. use of, 22, 81. tabular, 36. Longitude, 32. Mantissa, 19. Mariner's compass, 31. Minimum value, 74. Navigation : plane sailing, 94. parallel sailing, 95, middle latitude sailing, 96, traverse sailing, 96. 204 INDEX Negative angle, 26. ratios of, 54. direction, 13. Oral examples, 34, Gl, 116, 123, 166. Ordinate, 12. Origin, 12. Point, 32. Points of the compass, 31. Powers of trigonometrical ratios, 44. Product of sines and cosines, 160. Projection, definition, 139. of area, 147. propositions, 140. length of, 40. formulae, 40, 109. Proportional parts, 71. Quadrilateral, area of cyclic, 111. Radian, 29. Radius of inscribed circle, 114,168. of circumcircle, 114. of escribed circle, 114, 168. Rate of change of function, 76. Ratios, trigonometrical, 37. of 90-^, 53. of 180-^, 53. of -^,54. of 0° and 90^, 56. of 30°, 45°, 60°, 57. of 2 A, 100, 162. of i A, 101, 162. of 3 A, 164. ofA + B, 154. of A -B, 156. ofSA, 164. Rhumb line, 94. Secant, 38. Sector, area of, 29. Sine, 38. formula, 107. graph, 41. sin {A±B), 154. sin A ± sin B, 159. Slope of curve, 76. Small angle, 59. sine and tangent of, 60. >* Solution of right-angled triangle, 88. of triangles, 119. of equations, 78, 168. of equations by graphs, 17, 76. Sum of two angles, ratios of, 154. of two sines, 159. of two cosines, 159. Summation of a series, 169, Supplementary angle, ratios of, 53. Tabular logarithm, 36. Tangent, 38. graph, 41. tan iA±B), 163. Triangle formulae, 107. solution of, 119. Trigonometrical ratios : general definition of, 38. right-angled triangle, definition of, 86. Turning points, 74. Vectors, 143. addition of, 144. resolution Versine, 40. Five-figure Logarithmic and Trigonometrical Tables ARRANGED BV W. E. PATERSON, M.A., B.Sc. MATHEMATICAL MASTER, MERCERs' SCHOOL AUTHOR OF 'SCHOOL ALGEBRa/ 'ELEMENTARY TRIGONO.METRY ' OXFORD: AT THE CLARENDON PRESS LONDON: HENRY FROWDE, AMEN CORNER, E.C. AXD AT EDINBURGH, GLASGOW, NEW YORK, TORONTO AND MELBOURNE OXFORD : HOKACE HAET TRINTER TO THE UNIVERSITY These five-figure tables are intended to give results correct to four figures ; tlie fifth figure in tlie answer may be inaccurate. The decimal point is printed before all the logarithms of numbers ; it is hoped that this will obviate the common mistake of reading off logarithms instead of antilogarithms, and vice -versa. The trigonometrical tables are arranged so that, at one opening of the tables, all the functions of an angle may be found on the left-hand page and their logarithms on the right-hand page ; here again confusion is avoided. The characteristics of the logarithmic functions are the true characteristics ; no useful purpose is served by increasing them by 10. It should be noticed that, instead of dividing by a sine, one may multiply by the cosecant, &c., and, similarly, instead of subtracting the logarithm of a sine, one may add the logarithm of the cosecant, &c. In many cases this shortens calculation. For quick reference the last page may be used, which gives the trigonometrical functioas, to four figures only, for every whole degree up to 90° and the corresponding circular measure to five figures. R 1-0025 1-005 1-0075 i-oi . 1-0125 I -015 I -0175 I -02 1-0225 1-025 1-0275 I -03 . Logarithms of R for log R 00108438 00216606 00324505 00432137 00539503 00646604 00753442 00860017 00966332 01072387 01178183 01283722 Compound Interest R logR 1-0325 . , . . . -01389006 I -035 -01494035 I-0375 -01598811 I-04 . . ' . . . . -01703334 1-0425 -01807606 1-045 •01911629 I -0475 -02015403 1-05 . . . . . , -02118930 1-0525 -02222210 1-055 -02325246 1-0575 -02428038 I -06 -02530587 Constants used in Mensuration and their Logarithms 7r= 3-14159265 in = 1-57079633 iyr =052359878 •A 77 = 4-18879020 ^/n= 1-77245385 7f2 = 9-86960440 ^n= I -46459 1 89 7r/l8o = 0-01745329 logarithm 0-497150 0-196120 1-718999 0-622089 0-248575 0-994300 0-165717 2-241877 I -=- 77 1 -^ 477 -</ 6-r~ ^ \/ 3 -^ 4 TT \/l 4- 77 = 0-31830989 = 0-07957747 = 1-24070098 = 0-62035049 = 0-56418958 1 -^ 77^= 0-10132118 ^^= 2-14502940 180/77= 57-29577951 logarithm 1-502850 2-900790 0-093667 1-792637 1751425 1-005700 0-331433 1-758123 Naperian (or Natural) Logarithms e= 2-7182182 logjoe = -43429448 logg 10 = 2-30258509 logio -V = logg A" X logio e. log^ A' = logio ^V x logg 10 LOGARITHMS OF NUMBERS Mean Differences ^1 10 II 12 13 14 IS j'i) i6 7-^_i7 i8 J9 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 S3 54 •00000 •04139 •07918 •II394 •I46I3 P7609 •20412 •23045 •25527 •27875 •30103 32222 ■34242 36173 38021 •39794 •41497 •43136 •44716 •46240 •47712 •49136 •50515 •5I85I •53148 S4407 55630 56820 57978 59106 1 •00432 •04532 •08279 •II727 1-14922 •17898 •20683 •23300 •25768 •28103 •60206 •61278 •62325 ■^^3347 ■64345 65321 66276 67210 68124 69020 •30320 •32428 •34439 •36361 •38202 •39967 •41664 •43297 •44871 •46389 •47857 •49276 •50651 •51983 •53275 •54531 '55751 56937 58093 59218 l 69897 •70757 71600 72428 I -72509 3239 I -73320 •60314 •61384 •62428 •63448 •64444 •65418 •66370 •67302 •6S215 •69108 69984 ■70S42 ■71684 •00860 -04922 -08636 •12057 •15229 •18184 20952 '23S53 •26007 •28330 •30535 •32634 •34635 •36549 '3^3^2 •40140 •41830 •43457 •45025 •46538 •48001 •49415 •50786 •52114 •53403 •S4654 •55871 •57054* •58206 •59329 3 •01284 •05308 0899 r •12385 •15534 -18469 -21219 -23805 •26245 -28556 •30750 -32838 •34830 ■36736 •38561 •40312 ■41996 ■43616 45179 46687 •01703 •05690 -09342 •12710 •15836 •18752 -21484 •24055 -26482 •28780 •30963 •33041 •35025 •36922 •38739 40483 42160 43775 45332 46835 •60423 •61490 •62531 •63548 •64542 •65514 ■66464 ■67394 68305 69197 •48144 •49554 •50920 •52244 ■53529 54777 55991 57171. 58320 59439 60531 •61595 •62634 •63649 •64640 •65610 •66558 •67486 ■68395 ■69285 •70070 -70157 70243 -70927 -71012 •71096 •71767 ^71850 ^71933 •72591 •72673-72754 -73400 (-7^480 1 -731; 60 •48287 •49693 •51055 •52375 •53656 •54900 •56110 •57287 •58433 •59550 60638 61700 •62737 •63749 •64738 •65706 •66652 •67578 -68485 •69373 42 39 35 33 •o 28 27 25 24 23 21 20 19 18 17 17 16 15 15 14 14 13 13 13 12 12 12 II II II I I 10 10 TO 10 9 9 9 9 9 8 8 8 85 77 71 66 61 57 S3 50 48 45 43 41 39 37 36 34 33 32 31 30 127 116 106 98 91 85 80 76 71 68 64 6 59 56 54 52 50 48 46 45 170 155 142 131 122 114 107 lOI 95 90 86 82 7^ 75 72 69 66 64 61 59 43 57 33 44 32 32 31 30 29 29 28 28 27 26 26 25 25 43 42 41 40 39 38 38 37 36 35 35 34 33 212 193 177 164 152 142 134 126 119 113 107 102 98 94 90 86 S3 80 77 74 72 70 67 65 63 62 60 58 57 55 254 232 213 197 183 171 160 151 '•$^ i'35 297 270 248 229 213 199 187 176 167 i^r 1281150 1231143 ii7|i37 112 131 108 103 99 96 92 89 8 16 24 33 24 1 32 54 53 51 50 49 48 47 46 45 44 43 42 42 S3 81 78 76 74 72 70 68 66 65 63 62 60 59 58 56 55 54 53 52 51 50 125 120 116 112 108 104 339 309 284 262 244 228 21,4 201 190 180 86 10 1 411 49 40 97 94 91 89 86 84 82 80 78 76 74 72 70 69 67 66 64 63 62 60 59 58 57 48 I 56 172 164 158 149 143 138 132 128 123 119 IIS 111 108 105 101 99 96 93 91 89 86 84 82 80 79 77 75 73 72 71 31 3^ 3' 2C 2; 21 24 22 21 2C 19 18 16 16: 15' 14. 14. 13; 13^ I2( I. 121 11^ 114 II] 108 105 102 100; 57 95 93 90 88 86 84 83 81 79 69 78 68 67 65 64 76 75 73-^ 72 LOGARITHMS OF NUMBERS 5 6 7 8 9 I 2 ^ 4 ' 6 7 8 9 lO 02119 02531 •02938 03342 •03743 40 sT 121 162 202 242 283 323 364 II •06070 -06446 -06819 -07188 •07555 37|74 III 148 185 222 259 296 III 12 •09691 -10037 -10380 -10721 •11059 34|68 102 ^37 170 204 238 272 307 13 •13033 -13354 m -13988 -14301 32:63 i 126 1^8 147 190 221 253 284 14 •16137 -16435 -17026 •17319 29 59 118 177 206 236 265 15 •19033 19312 •19590 •19866 •20140 28 55 83 no 138 165 193 221 248 i6 •217481 -22011 ■22272 •22531 -227S9 26 52 is 104 130 156 182 208 233 17 •24304-24551 •24797 •25042 •25285 24 49 73 98 123 147 171 196 220 i8 •26717! -269^1 •27184 •27416 •27646 23 46 70 93 116 139 162 185 208 19 •29003 -29236- "•29447 •29667 -29885 22 44 66 88 no 132 154 176 198 20 •31175 •31387 •31597 -31806 •32015 21 42 63 84 los 126 147 168 188 21 •33^44 •33445 •33646 •33846" •34044 20 40 60 80 100 120 140 160 180 22 ■35218 -35411 •35603 •35793 •35984 19 38 57 77 96 H15 134 153 172 23 ■37107 -37291 •37475 •37658 •37840 18,37 55 73 91 no 128 146 165 24 •38917 -39094 -39270 •39445 •39620 i8!35 53 70 88 105 123 140 158 25 40654 1 -40824 •40993 •41 162 •41330 17 34 51 67 84 lOI 118 135 152 26 •423251-42488 -42651 -42813 •42975 16 I 32 49 65 81 97 114 130 146 27 •43933 •44091 -44248 •44404 -44560 16 31 47 63 78 94 no 125 141 28 •45484 •45637 -45788 •45939 -46090 i5j3o 45 61 76 91 io6 121 136 29 •46982 •47129 •47276 •47422 •47567 14; 29 1 44 58 73 87 102 117 131 30 48430 •48572 •48714 -48855 •48996 1 ^ 14 28 42 56 71 85 99 113 127 31 ■49831 P49969 -50106 -50243-50379 14 27 41 55 68 82 96 109 ^23 32 •51188 -51322 •51455 -51587 -51720 ^1,27 40 53 66 80 93 106 119 33 •52504 -52634 ■52763 -52892 -53020 13 26 39 51 64 77 90 103 116 34 •53782! -53908 .54033 -54158 1-54283 13 25 3^ 50 63 75 88 100 113 35 •S5023 1 55145 •55267 55388 -55509 12 24 36 49 61 73 85 97 109 36 ■56229 1-56348 •56467 1 -56585 1-56703 12 24 31 47 59 71 S3 95 106 37 ■57403. •57519 •57634 •57749 •57864 12,23 35 46 58 69 81 92 104 38 •58546 •58659 •58771 •58883 •58995 11,22 34 45 56 67 78 90 lOI 39 •596601-59770 •59879 •59988 -60097 II 22 33 44 55 66 76 87 98 40 1 1 •60746 -60853 ! -609591 -61066 -61172 II 21 32 43 53 64 74 8S 96 41 •61805 -^'909 -62014 •62118 •62221 10 21 31 42 52 62 73 83 94 42 •62839 -62941 •63043 ^63144 •63246 10 20 30 41 51 61 71 81 91 43 •63849 -63949 •64048 -64147 j -64246 10 20 30 40 50 60 70 79 89 44 •64836 •64933 •65031 -65128-65225 10 19 29 39 49 58 68 78 87 45 65801 -65986 •65992 66087 66181 10 19 29 38 48 57 67 76 86 46 •66745 -66839 -66932 -670251-67117 9 19 28 37 47 56 65 74 84 47 •67669 •67761 -67852 •67943 -68034 9! 18 27 36 46 55 64 73 82 48 •68574 •68664 •68753 -68842 •68931 9!i8 27 36 45 54 62 71 80 49 •69461 -69548 -69636 -69723 -69810 9;i7 1 26 35 44 52 61 70 78 SO 70329 •70415 •70501 •70586 ^70672 9 17 26 34 43 SI 60 68 77 51 ■71181 -71265 •71349 •7i433i^7i5i7 8,1/ 25 34 42 50 59 <^7 75 52 •72016 -72099 -72181 ■72263 •72346 8 16 ■^5 33 41 49 58 66 74 S3 •72835 •72916 -72997 •7307s •73159 8!i6 24 12 40 48 57 65 73 54 -73640 •73719 •73799 ■73878 •73957 8|i6 24 32 40 48 55 63 71 Mean Differences LOGARITHMS OF NUMBERS Mean Differences 1 1 1 2 3 4 I 2 3 4 5 6 7 8 9 55 •74036 •741 15 •74194 •74273 •74351 8 ^ 24 3! 39 47 55 6j 71^ 56 74819 •74896 •74974 75051 •75128 8 15 ^l 31 39 46 54 62 69 57 75587 75664 •75740 75815 75891 8 15 23 30 38 46 53 61 68 58 76343 •76418 •76492 •76567 •76641 7 15 22 30 1>7 45 52 60 67 59 77085 •77150 77232 •77305 '77119 7 15 22 29 17 44 51 59 66 6o •77?i5 •77887 •77960 •78032 •78104 7 14 22 29 36 43 51 58 65 6i 78533 •78604 •78675 •78746 78817 7 14 21 28 36 43 50 57 64 62 79239 •79309 •79379 •79449 •79518 7 14 21 28 35 42 49 56 63 63 79934 •80003 •80072 •80140 •80209 7 14 21 27 34 41 48 55 62 64 •80618 •80686 •80754 •80821 •80889 7 14 20 27 34 41 47 54 61 65 •81291 •81358 •81425 •8149 1 •81558 7 13 20 27 33 40 47 53 60 66 •81954 •82020 •82086 •82151 •82217 7 13 20 26 Zl 39 46 52 59 67 •82607 •82672 '^^7Z7 •82802 •82866 6 13 19 26 32 39 45 52 58 68 •83251 •83315 •83378 •83442 •83506 6 13 19 25 Z^ Z^ 45 51 57 69 •83885 •83948 •8401 1 •84073 •84136 6 13 19 25 31 38 44 50 56 70 •84510 •84572 •84634 •84696 •84757 6 12 19 25 31 37 43 49 S6 71 •85126 •85187 •85248 •85309 •85370 6 12 18 24 31 17 43 49 55 72 •85733 •85794 •85854 •85914 •85974 6 12 18 24 30 36 42 48 54 73 •86332 •86392 •86451 •86510 •86570 6 12 18 24 30 36 42 48 53 74 •86923 •86982 •87040 •87099 •87157 6 12 18 23 29 35 41 47 53 75 •87506 •87564 •87622 •87679 •87737 6 II 17 23 29 35 40 46 52 76 •88081 •88138 •88195 •88252 •88309 6 II 17 23 29 34 40 46 51 77 •88649 •88705 •88762 •88818 •88874 6 II 17 22 28 34 39 45 51 78 •89209 •89265 •89321 •89376 •89432 6 II 17 22 2\ ZZ 39 44 50 79 •89763 •89818 •89873 •89927 •89982 5 II 16 22 27 Zl 38 44 49 80 •90309 •90363 •90417 •90472 •90526 5 II 16 22 27 33 38 43 49 81 •90849 •90902 •90956 •91009 •91062 1 1 16 21 27 32 Z7 43 48 82 •91381 •91434 •91487 •91540 •91593 II 16 21 27 Z^ Z7 42 48 83 •91908 •91960 •92012 •92065 •92117 10 16 21 26 31 Z7 42 47 84 •92428 •92480 •92531 •92583 •92634 10 15 21 26 31 36 41 46 85 •92942 •92993 •93044 •9309s •93146 10 15 20 26 31 36 41 46 86 •93450 •93500 •93551 •93601 •93651 10 15 20 25 30 35 40 45 87 •93952 •94002 •94052 •94101 •941 5 1 10 15 20 25 30 35 40 45 88 •94448 •94498 •94547 •94596 •94645 10 15 20 25 30 34 39 44 89 •94939 •94988 •95036 •95085 •95134 10 15 19 24 29 34 39 44 90 •95424 •95472 •95521 •95569 95617 10 14 19 24 29 34 39 43 91 •95904 •95952 •99599 •96047 •96095 10 14 19 24 29 ZZ Z'^ 43 92 •96379 •96426 •96473 •96520 •96 5 Cy •970/5 9 14 1 9-. 24 2% zz 38 42 93 •96H48 •96895 •96942 •96988 9 14 19 23 28 ZZ Z7 42 94 •97313 •97359 •97405 •97451 •97497 9 14 18 23 2"^ z- Z7 41 95 •97772 •97818 •97864 •97909 •97955 9 14 18 23 27 32 36 41 96 •98227 •98272 •98318 •983^>3 •9S408 9 14 18 ^}< -7 32 36 41 97 •9H677 •98722 •98767 •9881 1 •98856 9 13 18 Zl ^7 31 36 40 98 .99123 •99167 •992 1 1 •99255 •99300 9 13 18 22 ^7 31 35 40 .99 •99564 •99607 •9965 1 •99695 •99739 4 9 13 17 22 26 31 35 39 LOGARITHMS OF NUMBERS 5 6 7 8 1 9 I 8 2 16 3 23 4 5 3TI39 1'. 47 7 55 8 62 9 55 •74429 -74507 •74586 •74663 -74741 70 56 •75205 •75282 1 -75358 •75435 -75511 8 15 23 31 38 46 53 61 69 57 •75967 •76042. -76118 •76193 •76268 8 15 23 30 38 45 53 60 68 58 76716 •767901-76864 •76938 -77012 7 15 22 30 37 44 52 59 66 59 •77452 •77525 -77597 '77^70 '777AZ 7 15 22 29 36 44 51 58 65 6o •78176 •78247 78319 •78390 •78462 7 14 21 29 36 43 50 57 64 6i •ySiSS •78958^79029 -79099 •79169 7 14 21 28 35 42 49 56 63 62 •79588 •79657 -72Z27 -79796 •79865 7 14 21 28 35 42 48 55 62 . 63 •^ozjj j -803461 -So^l^^ •80482 •80550 7 14 20 27 34 41 48 55 61 64 •80956! -81023 -81090 •81158 •81224 7 13 20 27 34 40 47 54 60 65 •81624 81690 81757 •81823 •81889 7 13 20 26 33 40 46 53 59 66 •82282 •82347 ,•82413 •82478 -82543 7 13 20 26 33 39 46 52 59 67 •82930 -82995 -83059 -83123 -831^ 6 13 19 26 32 39 45 51 58 68 •83569 -83632 •83696 -83759 •83822 6 13 19 25I32 38 44 51 57 69 •84198 •84261 ■^AZ27, -84386 •84448 6 12 19 25 31 37 44 50 56 70 •S4819 •84880 -84942 -85003 •85065 6 12 18 25 31 37 43 49 55 71 •^'5431 •85491 -85552 •85612 •85673 6 12 18 24 30 36 42 48 54 72 •^6034 •86094 -86153 •86213 ■m27z 6 12 18 24 30 36 42 48 54 73 •86629 •86688 -86747 -86806 •86864 6 12 18 24 29 35 41 47 53 74 •87216 -87274 ■^73i2 -87390-87448! 6 12 17 23 29 35 41 46 52 75 •87795 •87852 •87910 •87967 -88024 6 II 17 23 29 34 40 46 51 76 •88366 •^H2Z •88480 -88536 •88593 6 II 17 23 28 34 40 45 51 77 •88930 •88986 •89042 •89098 •89154 6 II 17 22 28 33 39 45 50 78 •89487 -89542 -89597 •■89653 •89708 6 II 17 22 28 33 39 44 50 79 ■90037 •90091 •90146 •90200 •90255 5 II 16 22 27 33 38 44 49 80 •90580 -90634 •90687 •90741 ^90795 s II 16 22 27 32 38 43 48 81 •91 1 16 •9II69 •91222 •91275 •91328 5 II 16 21 27 32 37 42 48 82 •91646 •91698 •91751 -91803 •91855 5 10 16 21 26 31 37 42 47 83 •92169 -92221 .92273 •92324 •92376 5 10 16 21 26 31 36 41 47 84 •92686 •92737 -92788 •92840 •92891 5 10 15 20 26 31 36 41 46 85 •93197 93247 -93298 -93349 ! 93399 5 10 15 20 25 30 35 40 46 86 •93702-93752 •93802 •93852 ^93902 5 10 15 20 25 30 35 40 45 87 •94201 1-94250 •94300 •94349 •94399 5 10 15 20 25 30 35 40 44 8a :24694 •94743 -94792 •94841 •94890 5 10 15 20 25 29 34 39 44 89 •95182 •95231 •95279 •95328 -95376 5 10 15 19 24 29 34 39 44 90 •9566s •95713 -95761 -95809 •95856 5 10 14 19 24 29 33 38 43 91 •96142 •96190 -96237 •96284 -96332 5 9 14 19 24 28 33 38 43- 92 •96614 -96661 -£6708 -96755 -96802 5 9 14 19 23 2S 33 37 42 93 •97081 •97128 •97174 •97220 -97267 5 9 14 19 23 28 32 37 42 94 •975431-97589 •97635 •97681 .97727 5 9 14 18 23 27 32 37 41 95 •98000 98046 98091 •98137 i 98182 5 9 14 18 23 27 32 36 41 96 •98453-98498 -98543 •98588 •98632 4 9 13 18 22 27 31 36 40 97 •98900 -98945 -98989 -99034 •99078 4 9 13 18 22 27 31 36 40 98 •99344, -99388 •99432 -99476 •99520 4 9 13 18 22 26 31 35 40 J 99_ •99782 -99826 -99870 '-99913 •99957 4 9 13 17 22 26 31 3<> 39 Mean Differences ANTILOGARITHMS Mean Differences 1 2 3 4 2 2 5 3 ~7 4 ~9 5 12 6 7 i6 8 19 9 •oo 1 0000 10023 10046 10069 10093 - 21 •01 10233 10257 10280 10304 10328 2 5 7 9 12 14 17 19 21 •02 1047 1 10495 10520 10544 10568 2 5 7 10 12 15 U 20 22 •03 10715 10740 10765 10789 10814 2 5 7 10 12 15 17 20 22 •04 10965 10990 1 1015 11041 1 1066 3 5 8 10 13 15 18 20 23 •OS 11220 1 1 246 1 1272 1 1298 I 1324 : J 5 8 10 13 16 18 21 23 •06 1 1482 1 1 508 "535 11561 11588 3| 5 8 II 13 16 18 21 24 •07 1 1 749 1 1776 1 1803 11830 11858 : 5 5 8 II 14 16 19 22 24 •08 12023 12050 12078 12106 12134 . 5 6 8 11 14 U 19 22 25 •09 12303 12331 12359 12388 12417 . 5 6 9 11 14 17 20 23 26 •10 12589 12618 12647 12677 12706 i ( 6 9 12 15 18 20 23 26 •II 12882 12912 12942 12972 13002 : 5 6 9 12 15 18 20 24 27 •12 13183 13213 13243 13274 13305 : 5 6 9 12 15 18 21 24 27 •13 13490 13521 13552 13583 1 3614 : 6 9 12 16 19 22 25 28 •14 13804 13836 13868 13900 13932 : 6 10 13 16 19 22 26 29 •15 14125 141S8 14191 14223 14256 3 i 7 10 13 16 20 23 26 30 •16 14454 14488 14521 14555 14588 : 7 10 13 17 20 24 27 30 •17 14791 14825 14859 14894 14928 : 7 10 14 17 21 24 27 31 •18 15136 15171 15205 15241 15276 ^ ^ 7 II 14 18 21 25 28 32 •19 15488 15524 15560 15596 15631 A ^ 7 11 14 18 22 25 29 32 •20 15849 15885 15922 15959 15996 A ^ 7 II 15 18 22 26 29 33 •21 16218 16255 16293 16331 16368 A 8 1 1 15 19 23 26 30 34 •22 16596 16634 16672 16711 16749 A ^ 8 II 15 19 23 27 31 34 •23 16982 17022 1 706 1 1 7 100 17140 A ^ 8 12 16 20 24 28 32 35 •24 17378 17418 17458 17498 17539 A 8 12 16 20 24 28 32 36 •25 17783 17824 17865 17906 I'jgA'j 4 [ 8 12 16 21 25 29 33 37 •26 18197 18239 18281 18323 18365 4 8 13 17 21 25 30 34 38 •27 1 862 1 18664 18707 18750 18793 A 9 13 17 22 26 30 34 39 •28 19055 19099 19143 19187 19231 A 9 13 18 22 26 31 35 40 •29 19498 19543 19588 19634 19679 5 9 14 18 23 27 32 36 41 •30 19953 19999 2004s 20091 20137 5 9 14 18 23 28 32 37 42 •31 20417 20464 20512 20559 20606 5 9 14 19 24 28 33 3» 43 •32 20893 20941 20989 21038 21086 5 10 15 19 24 29 34 39 44 •33 21380 21429 21478 21528 21577 5 10 15 20 25 30 35 40 44 •34 21878 21928 21979 22029 22080 5 10 15 20 2S 30 35 40 46 •35 22387 22439 22491 22542 22594 5 10 16 21 26 31 36 41 47 •36 22909 22961 23014 23067 23121 5 1 1 16 21 27 3^ 37 42 48 •37 23442 23496 23550 23605 23659 5 11 16 22 27 33 3» 44 49 •38 23988 24044 24099 24155 24210 6 1 1 17 7 T 28 33 39 44 50 •39 24547 24604 24660 24717 24774 6 1 1 17 23 28 34 40 45 51 •40 25119 25177 2523s 25293 25351 6 12 17 23 29 35 41 47 52 •41 25704 25763 25823 25882 25942 6 12 18 24 30 36 42 48 54 •42 26303 26363 26424 26485 26546 6 12 18 24 30 36 43 49 55 •43 26915 26977 27040 27102 27164 6 12 19 25 31 37 44 50 56 •44 27542 27606 27669 277Z1> 27797 6 13 19 26 32 38 45 51 57 •45 28184 28249 28314 28379 2844s 7 13 20 26 33 39 46 52 59 •46 28840 28907 2^97i 29040 29107 7 13 20 27 35 40 47 53 60 •47 29512 29580 29648 29717 29785 7 14 21 27 34 41 48 55 62 •48 30200 30269 30339 30409 30479 7 14:211 28 35 42 49 56 63 •49_ 30903 30974 31046 31117 31189 7 14I21I 29 36 43 50 57 64 ANTILOGARITHMS 5 6 7 8 9 I 2 2 5 3 "7 4 9 5 12 6 14 7 16 8 19 9 •00 10116 IOI39 10162 10186 10209 21 •01 10351 10375 10399 10423 10447 2 5 7 10 12 14 17 19 22 •02 10593 I06I7 10641 10666 1 069 1 2 5 7 10 12 15 17 20 22 •03 10839 10864 10889 10914 10940 3 5 8 10 13 15 18 20 23 •04 1 1092 I II 17 11143 11169 11194 3 5 8 10 13 15 18 20 n •OS 1 1350 1 1376 1 1402 1 1429 1 1455 3 5 8 II 13 16 18 21 24 •o6 11614 II64I 11668 11695 11722 3 5 8 1 1 14 16 19 22 24 •07 11885 II9I2 11940 11967 11995 3 6 8 11 14 17 19 22 25 •o8 12162 12190 12218 12246 12274 3 6 8 11 14 17 20 23 25 •09 12445 12474 12503 12531 12560 3 6 9 12 14 17 20 23 26 •10 12735 12764 12794 12823 12853 3 6 9 12 15 18 21 24 26 •11 13032 13062 13092 13122 13152 3 6 9 12 15 18 21 24 ^7 •12 13335. 13366 13397 13428 13459 3 6 9 12 16 19 22 25 28 •13 13646 ^1^71 13709 13740 13772 3 6 9 13 16 19 22 25 28 •14 13964 13996 14028 14060 14093 3 6 10 13 16 19 ^l 26 29 •15 14289 14322 14355 14388 14421 3 7 10 13 17 20 23 26 30 •i6 14622 14655 14689 14723 14757 3 7 10 14 17 20 24 27 30 •17 14962 14997 15031 15066 15101 3 7 10 14 17 21 24 28 31 •i8 15311 15346 15382 15417 15453 4 7 11 14 18 21 25 28 32 •19 15668 15704 15740 15776 15812 4 7 11 14 18 22 25 29 11 •20 16032 16069 16106 16144 16181 4 7 II 15 19 22 26 30 33 •21 16406 16444 16482 16520 16558 4 8 11 15 19 23 27 30 34 •22 16788 16827 16866 16904 16943 4 8 12 16 19 '-I 27 31 35 •23 17179 17219 17258 17298 17338 4 8 12 16 20 24 28 32 36 •24 17579 17620 17660 17701 17742 4 8 12 16 20 24 29 33 17 •25 17989 18030 18072 18113 18155 4 8 12 17 21 25 29 33 37 •26 18408 18450 18493 18535 18578 4 9 13 17 21 26 30 34 38 •27 18836 18880 18923 18967 19011 4 9 13 18 22 26 31 35 39 •28 19275 19320 19364 19409 19454 4 9 13 18 22 27 31 36 40 •29 19724 19770 19815 19861 19907 5 9 14 18 23 27 32 Z7 41 •30 20184 20230 20277 20324 20370 5 9 14 19 23 28 33 37 42 •31 20654 20701 20749 20797 20845 5 10 14 19 24 29 IZ 38 43 •32 21135 21184 21232 21281 21330 5 10 15 20 25 29 34 39 44 •33 21627 21677 21727 ^-^m 21827 5 10 15 20 25 30 35 40 45 •34 22131 22182 22233 22284 22336 5 10 15 20 26 31 36 41 46 •35 22646 22699 22751 22803 22856 5 II 16 21 26 32 37 42 47 •36 23174 23227 23281 ^ms 23388 5 11 16 21 27 32 38 43 48 •37 23714 23768 2ZZ21 23878 23933 5 II 16 22 27 ^^ 38 44 49 •38 24266 24322 24378 24434 24491 6 1 1 17 22 28 34 39 45 51 •39 24831 24889 24946 25003 25061 6 12 17 23 28 35 40 46 52 •40 2S4IO 25468 25527 25586 25645 6 12 18 24 29 35 41 47 53 •41 26002 26062 26122 26182 26242 6 12 18 24 30 36 42 48 54 •42 26607 26669 26730 26792 26853 6 12 18 25 31 17 43 49 55 •43 27227 27290 27353 27416 27479 6 13 19 25 32 38 44 50 57 •44 27861 27925 27990 28054 28119 6 13 19 26 32 39 45 52 S8 •45 28510 28576 28642 28708 28774 7 13 20 26 33 40 46 53 59 •46 29174 29242 29309 29376. 29444 7 14 20 ^7 34 41 47 54 61 •47 29854 29923 29992 30061 30130 7 14 21 28 35 42 48 55 62 •48 30549 30620 30690 30761 30832 7 14 21 28 35 42 50 57 64 •49 31261 31333 314OS 31477 31550 7 14 21 29 36 43 51 58 65 Mean Differences II Antilogarithms Mean Differences 1 2 3 4 I 2 3 4 5 6 7 8 9 •SO 31623 3T696 31769 31842 31915 ~7 IS ^ 29 37 44 SI 59 66 •SI 32359 32434 32509 32584 32659 8 15 ^3 30 3S 45 53 60 68 •S2 33113 33189 33266 33343 33420 8 15 23 31 38 46 53 61 69 'S3 33884 33963 34041 341 19 34198 8 16 24 31 39 47 55 63 71 •54 34674 34754 34834 34914 34995 8 16 24 32 40 48 56 64 72 •ss 35481 35563 35645 35727 35810 8 16 25 53 41 49 S8 66 74 •S6 36308 36392 36475 36559 36644 8 17 2q 34 42 50 59 67 76 •S7 37154 37239 37325 37411 37497 9 17 26 34 43 52 60 69 78 •S8 38019 38107 38194 38282 38371 9 18 26 35 44 53 62 70 79 •S9 38905 38994 39084 39174 39264 9 18 27 36 45 54 63 72 81 •60 3981 1 39902 39994 40087 40179 9 18 28 37 46 55 65 74 83 •61 40738 40832 40926 41020 41115 9 19 28 38 47 57 66 76 85 •62 41687 41783 41879 41976 42073 10 19 29 39 48 58 67 77 87 •63 42658 42756 42855 42954 43053 10 20 30 40 49 59 69 79 89 •&4 43652 43752 43853 43954 44055 10 20 30 40 51 61 71 81 91 •6s 44668 44771 44875 44978 45082 10 21 31 41 52 62 73 83 93 •66 45709 45814 45920 46026 46132 1 1 21 32 42 53 63 74 85 95 •67 46774 46881 46989 47098 47206 1 1 22 32 43 54 65 76 86 97 •68 47863 47973 48084 48195 48306 1 1 22 33 44 55 66 78 89 100 .69 48978 49091 49204 49317 49431 1 1 23 34 45 57 68 79 91 102 •70 SOI 19 50234 50350 50466 50582 12 23 35 46 58 70 81 93 104 •71 51286 51404 51523 51642 51761 12 24 36 48 59 71 83 95 107 •72 52481 52602 52723 52845 52966 12 24 36 49 61 73 85 97 109 •73 53703 53827 53951 54075 54200 12 25 37 50 62 75 87 100 1 12 •74 54954 55081 55208 55335 55463 13 25 38 51 64 76 89 102 114 •7S 56234 56364 56494 56624 56754 13 26 39 52 65 78 91 104 117 .76 57544 57677 57810 57943 58076 13 27 40 53 67 80 93 107 120 •77 58884 59020 59156 59293 59429 14 27 41 55 68 82 95 109 123 •78 60256 60395 60534 60674 60814 14 28 42 56 70 84 98 112 126 •79 61660 61802 61944 62087 62230 14 29 43 57 71 86 100 114 128 •80 63096 63241 63387 63533 63680 15 29 44 58 73 88 102 117 131 •81 64565 64714 64863 65013 65163 15 30 45 60 75 90 105 120 135 •82 66069 66222 <^^^374 66527 66681 15 31 46 61 77 92 107 122 138 •83 67608 67764 67920 68077 68234 16 31 47 63 78 94 no 125 141 •84 69183 69343 69502 69663 69823 16 32 48 64 80 96 112 128 144 •85 70795 70958 71121 71285 71450 16 33 49 66 82 98 IIS 131 147 •86 72444 7261 1 72778 72946 73114 17 34 50 (^7 84 101 117 134 151 •87 74131 74302 74473 74645 74817 17 34 Si 69 86 103 120 137 154 •88 75858 76033 76208 76384 76560 18 35 S3 70 88 105 123 140 158 •89 77625 77804 77983 78163 7S343 18 36 54 7~ 90 108 126 144 162 •90 79433 79616 79799 79983 80168 18 37 55 74 92 no 129 147 166 •91 81283 81470 81658 81846 82035 19 38 56 75 94 113 132 151 169 •92 83176 83368 83560 83753 83946 19 39 58 77 96 116 135 154 174 •93 85114 85310 85507 85704 85901 20 39 59 79 99 118 138 158 177 •94 87096 87297 87498 87700 87902 20 40 61 81 101 121 141 161 182 •95 8912s 89331 89536 89743 89950 21 41 62 83 103 124 144 165 186 •96 91201 91411 91622 91833 92045 21 42 63 84 106 127 148 169 190 •97 93325 93541 93756 93972 94189 22 43 65 80 108 130 151 173 195 •98 95499 95719 95940 96161 96383 22 44 6() 88 1 1 1 133 155 177 199 M 97724 97949 98175 98401 98628 23 45 68 90 113 136 158 181 204 12 ANTILOGARITHMS 5 6 7 8 9 ^\l 3 4 5 6 7 \' 9 •so 31989 32063 32137 32211 32285 7 15 22 30 37 44 52 59 67 •51 32734 32810 32885 32961 33037 8 15 23 30 38 45 53 61 68 •52 33497 33574 33651 33729 33806 8 15 23 31 39 46 54 62 70 '53 34277 34356 34435 34514 34594 8 16 24 32 40 48 55 C^3 71 •54 35075 35156 35237 35318 35400 8 16 24 32 41 49 57 65 73 •55 35892 35975 36058 36141 36224 8 17 25 33 42 50 S8 67 75 •S6 36728 36813 36898 36983 37068 9 17 26 34 43 51 60 68 77 •57 37584 17670 Z77S7 37844 37931 9 17 26 35 44 52 61 70 78 •58 38459 38548 38637 38726 38815 9 18 27 36 45 54 62 71 80 •59 39355 39446 39537 39628 39719 9 18 27 36 46 55 64 73 Sjj •6o 40272 40365 40458 40551 40644 9 19 28 37 47 56 6S 75 84 •6i 41210 41305 41400 41495 41591 10 19 29 38 48 57 67 76 86 •62 42170 42267 42364 42462 42560 10 20 29 39 49 59 68 7S 88 •63 43152 43251 43351 43451 43551 10 20 30 40 50 60 70 80 90 •64 44157 44259 44361 44463 44566 10 20 31 41 51 61 72 82 92 •65 45186 45290 45394 45499 45604 10 21 31 42 52 63 73 84 94 •66 46238 46345 46452 46559 46666 II 21 32 43 54 64 75 86 96 •67 47315 47424 47534 47643 47753 II 22 33 44 55 66 77 88 99 •68 48417 48529 48641 48753 48865 II 22 34 45 56 (^7 79 90 lOI .69 49545 49659 49774 49888 50003 II 23 34 46 57 69 80 92 103 •70 50699 50816 50933 51051 51 168 12 23 35 47 59 70 82 94 106 .71 51880 52000 52119 52240 52360 12 24 36 48 60 72 84 96 108 •72 53088 53211 ^ZZIZ 53456 53580 12 25 37 49 62 74 86 98 III •73 54325 54450 54576 54702 54828 13 25 38 50 63 75 88 100 113 •74 55590 55719 55847 55976 56105 13 26 39 52 64 77 90 103 116 •75 56885 57016 57148 57280 57412 13 26 40 S3 66 79 92 105 119 .76 58210 58345 58479 58614 58749 13 27 40 54 67 81 94 108 121 •77 59566 59704 59841 59979 601 17 14 28 41 55 69 Si 97 no 124 •78 60954 61094 61235 6\Z7^ 61518 14 28 42 56 71 85 99 113 127 •79 ^^Z7l 62517 62661 62806 62951 14 29 43 58 72 87 lOI 116 130 .80 63826 63973 641 2 1 64269 64417 15 30 44 59 74 88 104 118 133 •81 ^sm 65464 65615 65766 65917 15 30 45 60 76 91 106 121 136 •82 66834 66988 67143 67298 67453 15 31 46 62 77 93 108 124 139 •83 68391 68549 68707 68865 69024 16 32 48 63 79 95 III 127 143 •84 69984 70146 70307 70469 70632 16 32 49 65 81 97 114 130 146 •85 71614 71779 71945 721 1 1 72277 17 33 50 66 83 100 116 133 149 •86 73282 73451 73621 73790 73961 17 34 51 6S 84 102 119 136 153 •87 74989 75162 75336 75509 75683 17 35 52 70 87 104 122 139 156 •88 76736 76913 77090 77268 77446 18 36 53 71 89 107 124 142 160 .89 78524 78705 78886 79068 79250 18 36 55 73 91 109 127 145 164 •90 80353 80538 80724 80910 81096 19 37 56 74 93 112 130 149 167 •91 82224 82414 82604 82794 82985 19 3S 57 76 95 114 133 1^2 171 .92 84140 Hill 84528 84723 84918 19 39 58 78 97 117 136 i=;6 175 •93 86099 86298 86497 86696 86896 20 40 60 80 100 120 140 160 179 •94 88105 88308 88512 88716 88920 20 41 61 82 102 122 143 163 184 •95 90157 90365 90573 90782 90991 21 42 63 84 104 125 146 167 188 .96 922^ M70 -^ -24 92683 92897 93111 21 43 64 85 107 128 150 171 192 .n-it 94842 95060 95280 22 44 66 87 109 131 153 175 197 W > ^ ' 97051 97275 97499 22 45 67 90 112 134 157 179 201 ^- _.j993i2| 99541 99770 13_ 46 69 9^ 115 137 160 183_ 206 M ean ] Diffe renc 2S 0° NATURAL FUNCTIONS Differences are given for every lo'. Intermediate values can be found by n method of proportional parts ; e. g. : — To find tan 43° 56' and cos 37° 34' tan 43° 5o'= -96008 cos 37° 3°'= 79335 1 + diff. for 6'= 337 -diff. for4'=- 71 f .-. tan 43° 5^'= -96345 ••• cos 37° 34'= -79264 When there is no entry in the difference column, the value of the function char too rapidly for correct interpolation by proportional parts. C4reater accuracy is t obtained by expressing the function in terms of the sine and cosine. To find tan 67° 23' tan 67° 20'= 2-39449 Diff. for 10' = 1972 by proportional parts, diff. for 3'= 592 This gives tan 67° 23'= 2-40041. (The correct value is 2-40038.) Subtract differences when dealing with co-functions 0° 10' 20' 30' 40' ';o^ i'' 10' 20' 30' 40' 50' 2° 10' 20' 30' 40' so' 3° •00000 •00291 •00582 •00873 •o I 1 64 •01454 •01745 •02036 •02327 •02618 •02908 •03199 •0349Q •03781 •0407 1 •04362 •04653 ■04943 05234 cosine 291 291 291 291 290 291 291 291 291 290 291 291 291 290 291 291 290 291 D 00 34378 171-89 114-59 85-946 68-757 57-299 49-114 42-976 38-202 34-382 '31-258 28-654 26-451 24-562 22-926 2 1 -494 20-230 19-107 secant tangent -00000 •00291 •00582 •00873 •01 164 •01455 •01746 •02037 •02328 •02619 •02910 ■03201 03492 ■03783 ■04075 •04366 ■04658 04949 05241 D cotangent 291 291 291 291 291 291 291 291 291 291 291 291 291 292 291 292 291 292 ^^ cotangent 00 343-77 171-89 114-59 85-940 68-750 57290 49-104 42-964 38^188 34-368 31-242 28636 26-432 24-542 2 2 -904 21-470 20-206 I9081 tang(;nt secant I -00000 I -ooooo I -00002 I -00004 I -00007 I -000 1 I I 000 I 5 I ^0002 1 1^00027 I -00034 I -00042 I -0005 I I 0006 1 I -00072 1-00083 I -00095 I -00108 I -OO I 2 2 I -00137 cosecant 000 002 002 003 004 004 005 006 007 008 009 010 Oil 01 1 012 013 014 015 I -ooooo •99998 -99996 -99993 -99989 -99985 •99979 •99973 -99966 •99958 -99949 -99939 •99929 •99917 •99905 -99892 -99878 -99863 000 002 002 003 004 004 006 006 007 008 009 010 GIG 012 012 013 014 015 |9C 5< 4c 3C 2C IC 8S 5t 4C 3c 2C IC 50 4oi 30 20 10 87 0^ LOGARITHMIC FUNCTIONS The values given here are the true logarithms ; the characteristic is not increased by lo as in many tables. Differences are given for every lo'. Intermediate values can be found by the method of proportional parts. The differences for the logarithm of a function and of the reciprocal of the function are the same in magnitude but opposite in sign. When there is no entry in the difference column, the rate of change of the logarithm changes too rapidly for correct interpolation by proportional parts. The following rules may be used when the angle is small : — Log sine. Add &6S$S7 to the log of the angle expressed in seconds and subtract ^ of^tlie log secant. Log tan. Add 6-68557 to the log of the angle expressed in seconds and add ^ of the log secant. When the log sine is given, the angle is found in seconds by adding 5-31443 to the log sine and ^ of the corresponding log secant (found in the ordinary way). When the log tan is given, the angle is found in seconds by adding 5*31443 to the log tan and subtracting | of the corresponding log secant (found in the ordinary way). Subtract differences when dealing with co -functions log sin D log cosec log tan D |log cotan log sec D ] log COS d^ — 00 00 — 00 00 0-00000 0-00000 90° 10' 3-46373 2-53627 3-46373 2-53627 0-00000 001 o-ooooo 50' 20' 376475 2-23525 3-76476; 2-23524 0-0000 1 001 1-99999 40' 30' 3-94084 2-05916 3-94086 2-05914 -00002 ^; 11-99998 30' 40' 2-06578 1-93422 2-06581 jl-93419 0-00003 002 1-99997 20' 50' 2-16268 1-83732 2-16273 1-83727 0-00005 002 1-99995 10' r 2-24186 1-75814 2-24192 1-75808 0-00007 002 1-99993 89"^ 10' 2-30879 1 I -69 1 2 1 2-30888 i I -691 12 0-00009 00 i I -9999 1 50' 20' 2-366781 1-63322 2-36689 I-633II 0-00012 003 00 ■? 1-99988 40' 30' 2-41792, 1-58208 2-41807 1-58193 0-00015 1-99985 30' 40' 2-463661 1-53634 2-46385 1-53615 0-00018 ^l ■•9»«-'| 20' 50' 2-50504 I -49496 2-50527 1-49473 0-00022 004 004 1-99978 10' 2^ 2-S4282I 1-45718 2-54308 1-45692 000026 1-99974 88^ 10' 2-57757 1-42243 2-57788 1-42212 0-00031 1-99969 50' 20' 2-60973 1-39027 2-61009 I -38991 0-00036 005 ^-99964 40' 30' 2-63968. 1-36032 2-64009 1-35991 0-00041 o^!i-99959 30' 40' 2-66769! 1-33231 2-66816 1-33184 o-o(J047 0^^99953 20' 50' 2-69400' 1-30600 2-69453 j I -30547 o-(;<KJ53 006 ^-99947 10' 3" 2-71880' 1-28120 2-71940 11-28060 000060 i -99940 ST log COS j D log sec log cotan D 1 log tan log cosec D log sin 87= 3° NATURAL FUNCTIONS 05234 05524 05814 06105 06395 06685 06976 07266 07556 07846 08136 08426 08716 09005 09295 09585 09874 0164 0453 0742 1031 1320 1609 1898 2187 2476 2764 3052 3341 3629 3917 4205 4493 4781 5069 5356 5643 5931 6218 6505 6792 7078 736s 290 290 291 290 290 291 290 290 290 290 290 290 289 290 290 289 290 289 289 289 289 289 289 289 289 288 288 289 288 288 288 288 288 288 287 287 288 287 287 287 286 287 D 19-1073 18-1026 17-1984 16-3804 15-6368 14-9579 14-3356 13-7631 13-2347 127455 12-2913 11-8684 11-4737 11-1046 10-7585 IO-4334 10-1275 9-83912 956677 9-30917 9-06515 8-83367 8-61379 8-40466 8-20SSI 8-01565 7-83443 7-66130 7-49571 7-18530 7-03962 6-89979 6-76547 6-63633 6-51208 639245 6-27719 6-16607 6-05886 5-95536 5-85539 5 75877 tangent D -05241 •05533 -05824 •061 16 •06408 •06700 -06993 •07285 •07578 -07870 -08163 •08456 -08749 •09042 -09335 •09629 •09923 -IO216 -IO51O •10805 •I 1099 -II394 -I1688 -II983 -12278 -12574 •12869 •13165 •I 3461 •13758 •14054 -1435 I • 1 4648 -14945 •15243 •15540 •15838 •I 61 37 •16435 •16734 •17033 '^7Zr:s •17633 292 291 292 292 292 293 292 293 292 293 293 293 293 293 294 294 293 294 295 294 295 294 295 295 296 295 296 296 297 296 297 297 297 298 297 298 299 298 299 299 300 300 cotangent] D cotangent I9081I 18^0750 17-1693 16-3499 I 5 -6048 14-9244 14-3007 13-7267 13-1969 12-7062 12-2505 11-8262 Iit430i 11-0594 10-7119 10-3854 10-0780 978817 9-S1436 9-25530 9-00983 8-77689 8-55554 8 -34496 8-14435 7-95302 7-77035 7-59575 7-42871 7-26873 7^ii537 6-96823 6-82694 6-691 16 6-56055 6^43484 631375 6-19703 6-08444 5-97576 5-87080 5-76937 567128 tangent I 00137 I -00153 I -00 1 69 I -001 87 1-00205 1-00224 1-00244 1-00265 1-00287 1-00309 I -00333 1-00357 I 00382 I -00408 I -00435 I -00463 I -0049 1 ■I -005 2 I I -00551 1-00582 I -006 1 4 I -00647 I -0068 I I -007 I 5 I 0075 1 1-00788 1-00825 1-00863 I -00902 I -00942 I 00983 1-01024 I -01067 I -O I I I I I-OII55 I -01 200 I 01247 1-01294 I -01 342 I -01 39 1 I -01 440 I -01 49 1 I 01543 D 016 016 018 018 019 020 021 022 022 024 024 025 ^f026 027 028 028 030 030 031 032 033 034 034 036 037 037 038 039 040 041 041 043 044 044 045 047 047 048 049 049 051 052 99863 99847 99831 99813 99795 99776 99756 99736 ■99714 ■99692 ■99668 ■99644 99619 ■99594 ■99567 •99540 ■995 1 1 •99482 •99452 -99421 -99390 -99357 •99324 -99290 •99255 •99219 •99182 -99144 -99106 •99067 •99027 •98986 •98944 •98902 ■98858 ■98814 98769 ■98723 ■98676 •98629 ■98580 98531 98481 80^ LOGARITHMIC FUNCTIONS loo^ sin D log cosec log tan log cotan 1-28060 lo^ sec D log cos 3" 271880 1-28120 271940 0-00060 006 i -99940 87^ lO' 2-74^26 1-25774 2-74292 1-25708 ■ J -00066 008 1-99934 50' 20' 276541 1-23549 2-76525 1-2 347 S 0-00074 007 Of 18 1-99926 40' 30' 278568 1-21432 2-78649 1-21351 0-0008 1 1-99919 30' 4''' 2-80585 1-19415 2-80674 1-19326 0-00089 LHJO 008 T -999 11 20' ;o' 2-S2SI3 1-17487 2-82610 1-17390 0-00097 009 009 009 010 1-99903 10' 4^ 10' 284358 2-86128 1-15642 1-13872 2 84464 5-60243 1-15536 1-13757 000106 0-001 15 199894 1-99S85 86^ 50' 20' 2-87829 1-12171 2-87953 I -12047 0-00124 1-99876 40' 3<J' 2-89464 1-10536 2-89598 1-10402 0-00134 T r> 1-99866 30; 40' 2-91040 I -08960 2-91185 1-08815 0-00144 (J lU Oil 1-99856 20' Sf'j' 5-92561 1-07439 2-92716 1-07284 0-00155 Oil 1-99S45 10' 5" 2 94030 105970 2-94195 1-05805 000166 Oil 1-99834 85^ 10' ^•95450 1-04550 2-95627 1-04373 0-00177 r> T T 1-99823 50' 20' 2-96825 1-03175 2-97013 1-02987 0-00188 L) 1 1 012 013 012 1-99812 40' 30' 2-98157 1-01843 2-98358 1-01642 0-00200 1-99800 30' 40' 2-99450 1-00550 2-99662 1-00338 0-00213 1-99787 20' ;o' I -00704 0-99296 1-00930 0-99070 0-00225 014 013 1-99775 10' 6^~ 101923 098077 i-02162 097838 000239 i -99761 84° 10' T -03 109 0-96891 1-03361 0-96639 0-00252 1-99748 50' 20' 1-04262 0-95738 1-04528 0-95472 0-00266 014 014 015 r\T - 1-99734 40' 30' 1-05386 0-94614 1-05666 0-94334 0-00280 1-99720 30' 40' I -0648 1 0-93519 1-06775 0-93225 0-00295 1-99705 20' 50' 1-07548 0-92452 1-07858 0-92142 0-00310 oio T ; 1 -99690 10' 7=^ 1-08589 0-91411 108914 0-91086 0-00325 Ul3 016 199675 83^ 10' I -09606 0-90394 I -09947 0-90053 0-00341 016 016 1-99659 50' 20' 1-10599 993 0-89401 I -10956 987 966 945 926 908 889 873 856 840 825 811 0-89044 0-00357 1-99643 40' 30' 1-11570 971 0-88430 I-II943 o'88o57 0-00373 1-99627 30' 40' 8= 10' 1-12519 1-13447 I 14356 1-15245 949 928 909 889 871 854 837 821 805 790 0-87481 0-86553 085644 0-84755 I -12909 1-13854 i- 14780 1-15688 0-87091 0-86146 085220 0-84312 0-00390 0-00407 000425 0-00443 017 017 018 018 018 019 019 1-99610 1-99593 I -99575 1-99557 20' 10' 82° 50' 20' i-i6ii6 0-83884 T-16577 0:83423 0-00461 1-99539 40' 3^' 1-16970 0-83030 1-17450 0-82550 0-00480 1-99520 30' 40' 1-17807 0-82193 1-18306 0-81694 -00499 1-99501 20' 50' 9~ 1-18628 1-19433 0-81372 080567 1-19146 1-19971 0-80854 80029 0-00518 0-00538 019 020 020 1-99482 1-99462 10' 81° 10' 20' 1-20223 1-20999 776 762 748 0-79777 0-79001 1-20782 I -2 1 578 796 783 769 0-79218 0-78422 0-00558 0-00579 .021 021 021 022 022 D I -99442 1-99421 50' 40' 30' 40' 1-21761 1-22509 078239 077491 1-22361 I -23 1 30 0-77639 0-76870 0-00600 0-00621 1 -99400 1-99379 30' 20' ^0' 10"" 1-23244 1-23967 735 723 0-76756 0-76033 1-23887 1-24632 log cotan 757 745 0-76113 0-75368 0-00643 00665 log cosec 1-99357 1-99335 log sin 10' 80 log COS D log sec- D log tan 80= 17 10'' NATURAL FUNCTIONS M 17365 ■ I 765 1 ■17937 •18224 •18509 ■18795 19081 ■19366 •19652 •19937 •20222 •20507 •20791 •21076 •21360 •21644 •21928 •22212 •22495 •22778 •23062 •23345 •23627 •23910 •24192 •24474 •24756 •2503S •25320 •25601 •25882 •26163 •26443 •26724 •27004 •27284 •27564 •27843 •28123 •28402 •28680 •28959 •29237 286 286 287 285 286 286 285 286 285 285 285 284 285 284 284 284 284 283 283 284 283 282 283 282 282 282 282 282 281 281 281 280 281 380 280 280 279 280 279 278 279 278 D 5-75877 5-66533 5-57493 5-48740 5-40263 5-32049 524084 5-16359 5-08863 5-01585 4-94517 4-87649 480973 4-74482 4^68167 4-62023 4^56o4i 4^50216 4-44541 4-39012 4-33622 4^28366 4-23239 4-18238 4-13357 4-08591 4-03938 3-99393 3^94952 3^90613 386370 3-82223 3^78166 3-74198 3-70315 3-66515 3 62796 3-59154 3^55587 3-52094 3-48671 3-45317 342030 tangent 17633 17933 18233 18534 18835 19136 19438 19740 20042 20345 20648 20952 21256 21560 21864 22169 22475 22781 23087 23393 23700 24008 24316 24624 24933 25242 25552 25862 26172 26483 26795 27107 27419 28046 28360 28675 28990 29305 29621 29938 30255 30573 otangont 300 300 301 301 301 302 302 302 303 303 304 304 304 304 305 306 306 306 306 307 308 308 308 309 309 310 310 310 311 312 312 312 313 314 314 315 315 315 316 317 317 318 cotangent D 567128 5^57638 5-48451 5-39552 5-30928 5^22566 5-14455 5-06584 4-98940 4-91516 4-84300 4-77286 4-70463 4-63825 4-57363 4-51071 4-44942 4-38969 433148 4-27471 4-21933 4-16530 4-11256 4-06107 401078 3-96165 3^91364 3 •86671 3-82083 3-77595' 373205 3-68909 3-64705 3 60588 3-56557 3-52609 3-48741 3-44951 3-41236 3-37594 3-34023 3-30521 327085 tangent I-0I543 i^oi595 1-01649 1-01703 I •01758 1-01815 I 01872 1-01930 1-01989 I -02049 I -021 10 I -02 1 7 1 I 02234 1-02298 1-02362 I ^02428 I -02494 1-02562 I 02630 1-02700 1-02770 1-02841 1-02914 1-02987 I 03061 1-03137 I -032 1 3 1-03290 1-03368 I -03447 I 03528 I ^03609 1-03691 I -03774 1-03858 I -03944 I 04030 1-04117 I -04206 1-04295 1-04385 I -04477 I 04569 052 054 054 055 057 057 058 059 060 061 061 063 064 064 066 066 068 068 070 070 071 073 073 074 076 076 077 078 079 081 081 082 083 084 086 086 087 089 089 090 092 092 I) 98481 98430 98378 98325 98272 98218 98163 98107 98050 97992 ^7934 );875 9781S 97754 97692 97630 97566 97502 97437 97371 97304 97237 97169 97100 97030 96959 96887 96815 96742 96667 96593 965^7 96440 .96363 96285 96206 96126 96046 95964 95882 95799 95715 95630 73= 18 10= LOGARITHMIC FUNCTIONS log sin D 71] 23967 •24677 •25376 •26063 •26739 •27405 •28060 I 698 687 676 666 655 64s 635 625 616 = 3/S 582 -960 ^^^ ■33534 ,66 ■5ATOO. ^ •29340 •29966 ■305.8 •31 •3 ^-rlOO •3465 •34100!^ g . -J - -7 - T 537 •35752 •36289 'J' •37341 'i I 4/:> ■41300 ^g •^'^^' 464 ■-^''^'1458 •43591 :|^3 •44034 438 ■^^-^'"^ 433 •44905 li' ■45334/: 2^ -. . -_.o 424 log coscc 076033 075323 0-74624 073937 0-73261 0-72595 071940 0-71295 0-70660 0-70034 0-69418 0^688 1 1 0^682I2 0^67622 0-67040 0-66466 0-65900 0-65342 0-64791 0^64248 0-637II 0-63181 0-62659 ; 0-62 1 42 061632 0-61129 0^60631 0^60140 0-59654 0-59175 0-58700 0-58232 ;0-57768 io-57310 0-56857 0-56409 0-55966 0-55528 0-55095 0-54666 0-54242 0-53822 053406 46594 log COS D log sec log tan D log cotan •24632 •26086 •26797 •27496 •28186 •28865 ■-9S3S •30195 •30846 •31489 •32122 •32747 •33365 •33974 •34576 •35170 ■3S7S7 -36336 •36909 •37476 •38035 •38589 •39136 •39677 •40212 •40742 •41266 •41784 •42297 •42805 •43308 •43806 •44299 •44787 •45271 •4S7SO •46224 •46694 •47160 •47622 •48080 48534 733 721 711 699 690 679 670 660 651 643 633 62 075368 0-74635 0-73914 0-73203 0-72504 0-71814 0-71135 0-70465 0-69805 0-69154 0-68511 0-67878 579 573 567 559 6,8 067253 ^ ^ 0-6663 ^ 5^9 0.66026 601 , 0-65424 'f 0-64830 0-64243 o 63664 0-63091 0-62524 ^, 0-61965 ^^4 0-61411 :)47 541 535 530 524 518 513 508 503 498 493 488 484 479 474 0-60864 060323 0-59788 0^59258 0-58734 io-58216 '0-57703 0-57195 0^56692 0^56194 0-55701 0-55213 0-54729 054250 0-^3776 ^66 ^-53306 ^ 0^52840 0-52378 0-51920 051466 462 458 454 log cotan log sec D log cos o 00665 0^00687 0^00710 0-00733 0-00757 0-00781 000805 0-00830 0-00855 ©•00881 ©•00907 o -009 3 3 o 00960 ©•00987 0-01014 ©•01042 ©•01070 ©•01099 001128 ©•01157 0-01187 ©-01217 0-01247 ©•01278 001310 ©•©1341 ©•©1373 ©•01406 ©•01439 ©•©1472 001506 ©•©154© <J-oi574 ©•©i6©9 ©•©1644 ©•0168© 001716 ©•©1752 ©•01789 ©•01826 ©•01864 ©•©1902 00 1940 log cosic 022 023 023 024 ©24 ©24 025 025 ©26 ©26 026 027 027 027 ©28 ©28 ©29 ©29 029 03© ©3© 03© 031 032 031 032 033 033 033 034 034 034 035 035 036 ©36 036 037 037 038 038 038 u 1-99335 1-99313 1-99290 1-99267 1^99243 r992i9 199195 1-99170 1-99145 1-991 19 1-99093 I •99©67 i 99040 I -990 1 3 1^98986 T^98958 1-98930 i^989©i 198872 1-98843 1-98813 1-98783 1-98753 1^98722 I 98690 T^98659 i^98627 1-98594 i^9856i i^98528 I 98494 I •9846© 1-98426 1-98391 1-98356 1-98320 I 98284 i^98248 i^982i I 1-98174 I -98 1 36 I -98098 I 98060 log 50' 40' 30' 20' 10' 79^ 50^ 40' 30' 20' I©' 78^ 50' 40' 30' 20' 10' 77° 50' 40' 30' 20' I©' 76' 50' 40' 30' 20' 10' 75' 50' 40' 30' 20' 10' 740 50' 40' 30' 20' 10' 73° 73'' 19 17° NATURAL FUNCTIONS •29237 •29515 •29793 •30071 •30348 •30625 •30902 •31178 •31454 •31730 •32006 •32282 •32557 •32832 •33106 •33381 •33655 33929 •34202 34475 34748 •35021 •35293 •35565 •35837 •36108 ■36379 •36650 •36921 •37191 37461 37730 37999 38268 38537 38805 39073 39341 39608 39875 40142 40408 40674 278 278 278 277 277 277 276 276 276 276 276 275 275 274 275 274 274 273 273 ^71 273 272 272 272 271 271 271 271 270 270 269 269 269 269 268 268 268 267 267 267 266 266 3 42030 3-38808 3-35649 3-32551 3-29512 3-26531 3-23607 3-17920 3-I5I55 3-12440 3-09774 3-07155 3-04584 3-02057 3-99574 2-97135 2-94737 292380 2^90063 2^87785 2-85545 2-83342 2-81175 279043 2-76945 2^7488i 2^72850 2^70851 2^68884 2-66947 2^65040 63162 61313 59491 2-57698 55930 2-54190 2-52474 2-50784 491 19 2-47477 2-45859 "I tangent -30573 -30891 -31210 -31530 •31850 -32171 •32492 •32814 •33136 •33460 •34108 •34433 -34758 -35085 -35412 -35740 •36068 -36397 •36727 -37057 •37388 •37720 ■38053 •38386 -38721 -39055 •39391 •39727 •40065 -40403 -40741 •4108 1 •4 1 42 1 •41763 •42105 42447 •42791 •43136 ■43481 •43828 •44175 •44523 secant D cotangent D 318 319 320 320 321 321 322 122 324 323 325 325 325 1^7 2,27 328 328 329 330 330 331 332 33Z 335 334 336 338 338 338 340 340 342 342 342 344 345 345 347 347 348 cotangent D 3 27085 3-23714 3-20406 3-17159 3-13972 3-10842 3 07768 3-04749 3-01783 2-98869 2-96004 2-93189 2 9042 I 87700 2-85023 2-82391 79802 77254 274748 2^72281 853 2^67462 2^65109 2^62791 2 60509 2-58261 2-56046 2-53865 2-51715 2-49597 2-47509 2^45451 2-43422 2-41421 2^39449 2^37504 235585 2^33693 31826 2-29984 2'28i67 2-26374 2 24604 tangent I 04569 I ^04663 I -04757 1-04853 1-04950 1-05047 I 05146! 1-05246 1-05347 1-05449 1-05552 1-05657 1-05762 1-05869 1-05976 1-06085 1-06195 I -06306 I -06418 I -065 3 1 I -06645 I -06761 1-06878 I -06995 IO7115 1-07235 1-07356 I -07479 I -07602 1-07727 107853 I -0798 1 I -08 109 1-08239 1-08370 1-08503 I 08636 I -0877 1 I -08907 I -09044 1-09183 1-09323 I -09464 094 094 096 097 097 099 100 lOI 102 103 105 105 107 107 109 no 1 1 1 1 12 113 114 116 117 117 120 120 121 123 123 125 126 128 128 130 131 133 133 135 136 -^2,7 139 140 141 cosecant D •95630 •95545 •95459 •95372 •95284 •95195 •95106 •95015 •94924 •94832 •94740 •94646 •94552 -94457 -94361 -94264 -94167 -94068 -93969 -93869 ■93769 •93667 •93565 ■93462 93358 93253 93148 93042 •92935 -92827 •92718 •92609 •92499 •92388' •922761 •92164' -92050 1 -91930! •91822 •91706 •91590 •91472 91355 085 086 ,087 I' 1089 091 1091 092 092 J094 1 094 095 096 097 097 099 099 100 100 102 102 103 104 105 105 106 107 108 109 109 no 111 1 12 112 114 114 114 116 116 118 117 5^ 4C 3< 2C IC 6£ 5c 4c 3C 20 10 68 50 40 30 20 iO 67 50 40 30 20 10' 66' D 66^ 20 IT LOGARITHMIC FUNCTIONS loj; sin I D I log coscc •46594 •47005 •4741 1 •47814 •48213 •48607 401 406 •48998 ■49385 •49768 403 399 394 391 387 3S3 r^8i^^° 'f'368 f 4 36s ''^'^ 362 ;i99i ;235o 2705 jOI •505 ■5 SI )0^^3 50896' •53057 ^^8 53405 346 359 355 352 5375^ •54093 •54433 •54769 •55102 •55433338 ■55761 ... •56085 ^^^ •56408 ■•56727 342 340 336 333 33 323 319 317 317 57044 1/ 57978 58284 •58588 306 304 301 -■ ' -^ 299 ■S9I88 ^H .9/78 •60070 ~ •60359 •60646 •60931 ]o^ COS 289 287 285 053406 0-52995 0-52589 0-52186 0-51787 0-51393 0SI002 0-50615 0-50232 0-49852 0-49477 0-49104 048736 0-48371 0-48009 0-47650 0-47295 0-46944 0-46595 0-46249 0-45907 0-45567 0-45231 0-44898 044567 0-44239 0-43915 0-43592 0-43273 0-42956 042642 0-42331 0-42022 O-41716 0-41412 0-41 1 1 1 0-40812 0-405 16 0-40222 0-39930 0-39641 0-39354 0-39069 log SLC log tan D log cotan •48534 •489S4 •49430 -49872 •5031 1 •50746 •51178 •51606 -52031 •52452 •52870 •53285 •53697 •54106 •54512 •54915 •55315 •55712 •56107 •56498 -56887 •57274 •57658 •58039 •58418 •58794 •59168 •59540 •59909 •60276 •60641 •61004 -61364 -61722 -62079 •62433 •62785 •63135 •63484 •63830 •64175 •64517 •64858 439 435 432 428 425 421 418 415 412 409 406 403 400 397 395 391 389 3S7 384 381 379 376 372 369 367 365 360 358 357 354 352 350 349 346 345 342 341 ^^^0.51466 446 ^•5^"^^ 442 °'5°570 ^^ 0-50128 0-49689 0-49254 o 48822 0-48394 0-47969 0-47548 1 0-47 1 30 0-46715 o 46303 0^45894 o^45488 0^45085 0-44685 0-44288 043893 0-43502 0-43113 0-42726 0-42342 0-41961 041582 ^_^ 0-41206 ^^4 1 0.40832 0-40460 0-40091 0-39724 363:^-^9359 0-38996 0-38636 0-38278 0-37921 0-37567 0-37215 0-36865 0-36516 0-36170 0-35825 0-35483 i 0-35142 og cotan j D log tan log sec 001940 0-01979 0-02018 0-02058 0-02098 0-02139 002179 0-0222I 0-02262 0-02304 0^02347 0-02390 039 039 040 040 041 040 042 041 042 043 043 043 log cos 0-024 J 7 ' 0-02521 0-02565 0-02610 0-02656 002701 0-02748 0-02794! 0-02841 0-02889 0-02937! 002985 0^03034 0-03083 I 0-031321 0-03182 I 0-03233! 0-03283 0-03335 0-033861 0-03438! 0-03491 I 0-03544 1 003597 0-03651 I 0-03706 0-03760 0-03815 0-03871 003927 ' log coscc D 044 044 045 046 045 047 I 0461 047! 048: 048' 048! 049 049 049 050 051 050 052 0511 052 053 053 053 054 055' 054 055 056 o;6 -98060 •98021 -97982 •97942 -97902 -97861 -97821 -97779 •97738 •97696 •97653 -97610 ■97567 •97523 •97479 •97435 -97390 •97344 •97299 -97252 -97206 •97159 •971 1 1 •97063 •97015 •96966 -96917 -96868 •96818 -96767 •96717 -96665 •96614 •96562 •96509 •96456 96403 •96349 •96294 -96240 -96185 •96129 •96073 log sin 66^ 21 24° NATURAL FUNCTIONS 40674 ■40939 ■41204 ■41469 •41734 •41998 42262 •42525 ■42788 •43051 •43313 ■43575 j •43837 i •44098 i •44359! •44620 1 •44880 •45140 •45399 •45658 •45917 •46175 j •46433 •46690 •46947 •47204 •47460 •47716 •47971 •48226 •48481 •48735 •48989 •49242 •49495 •49748 •Soooo •50252 •50503 •50754 •51004 •51254 5 1 504 26s 265 265 265 264 264 263 263 263 262 262 262 261 261 261 260 260 259 259 259 258 258 257 257 257 256 256 255 255 255 254 254 253 253 253 252 252 cosecant 2^45859 2-44264 2-42692 2-41142 39614 38107 36620 35154 33708 32282 30875 29487 28117 26766 2-25432 2-24116 2-22817 2-21535 2-20269 2-19019 2-17786 2-16568 2-15366 2^i4i78 2^I3005 2-1 1847 2-10704 2-09574 2-08458 2-07356 2-06267 2-05191 2-04128 2-03077 2-02039 2-OIOI4 2-00000 1-98998 1-98008 1-97029 I -96062 1-95106 1-94160 ■^ secant 990 979 967 956 946 1) tangent 44523 44872 45222 45573 45924 46277 46631 46985 47341 47698 48055 48414 48773 49134 49495 49858 50222 50587 S0953 51320 51688 52057 52427 52798 53171 53545 53920 54296 54673 55051 5543 1 55812 56194 56577 56962 57348 57735 58124 58513 58905 59297 59691 60086 Dtangent 349 350 351 351 353 354 354 356 357 357 359 359 361 361 363 364 365 366 367 368 369 370 Z7Z 374 375 376 378 380 381 382 383 385 386 387 389 389 392 392 394 395 D cotangent 2-24604 2-22857 2-21132 2-19430 2-17749 2-16090 2-14451 2-12832 2-11233 2-09654 2-08094 2-06553 2-05030 2-03526 2-02039 2-00569 1-99116 1-97681 I -96261 1-94858 1-93470 1-92098 1-90741 I -89400 1-88073 1-86760 1-85462 1-84177 1-82906 1-81649 1-80405 1-79174 1-77955 1-76749 1-75556 1-74375 1-73205 1-72047 1-70901 I ^69766 1-68643 1-67530 I 66428 tangent 09464 09606 09750 09895 10041 10189 10338 10488 10640 10793 10947 11103 1 1 260 11419 11579 11740 11903 12067 12233 12400 12568 12738 12910 13083 13257 13433 13610 13789 13970 14152 14335 14521 14707 14896 15085 15277 15470 15665 1 5861 16059 16259 16460 16663 osecant D 142 144 145 146 148 149 150 152 153 154 156 157 159 160 161 163 164 166 167 1< 170 172 173 174 176 177 179 181 182 183 186 186 II 189 192 193 195 196 198 200 201 203 22 24° LOGARITHMIC FUNCTIONS 24° lO 20 30 40 50' 25° 10' 20' 30' 40 50' 26° 10' 40 50' 27° 10' 20' 30' 40' _Jo' 28° 10 20 30 40 50 29° 10' 20' 30 40' 50' 30° 10' 20 30' 40' 31° loR D log cosec 1-60931 2g3 039069 I-6I2I4 ,,g 0-38786 I -61494 ^-^0*38506 1-61773 276 ^'^"^^^7 1-62049 2-4 ^•'"'^'^ 1-62323 1-62595 1-62865 £•63133 1-633981 0-37951 0-37677 037405 272 270 268°-^7i35 265 264 0-36867 0-36602 I -63662 i^gj 0-36338 1-63924; 260'°'^^°^^ 1-64184^ g 0-35816 "64442|^^^iO-35558 •64698 256 1-64953!^^^ 1-65205 1 I -65456 1 1-65705 1-65952! 1-66197 r-66441 1-66682 1-66923 1-67161 1-67398! 1-67633 1-67866 I 68098 1-68328 i 68557 1-68784 I -690 10 1-69234 1-69456 1-69677 1-69897 1-70115 1-70332 1-70547 I -70761 1-70973 1-71184 2 252 251 249 247 245 244 241 241 238 237 1235 233 232 log; cos 0-35302 0-35047 0-34795 0-34544 0-34295 0-34048 0-33803 0-33559 0-33318 '0-33078 0-32839 10*32602 '0-32367 [0-32134 0-31902 230 "^ ^ ^on 0-31672 0-31443 0-3I2I6 0-30990 0-30766 0-30544 0-30323 0-30103 0-29885 0-29668 0-29453 0-29239 0-29027 0-28816 log sec log tan D log cotan 339 338 229 227 ;226 '224 '222 221 220 218 '217 j2I5 :2i4 212 211 1-64858 I -65197 1-65535 ,,, 1-65870:^^5 1-66204 ^^^ 1-66537 ^^-^ 1-66867 I -67196 1-67524 1-678501 330 329 328 326 f-68174'3^4 r-68497 \~,\ i-688i8 •^"' - r o 320 1-69138 I £-69457;^,^ £•69774 "f,; I -700891^ I -70404 '^j^ 1-70717 I -71028 1 1-71339'^" i-7i648'^°9 T4:955i307 1-72262 i-72567 "72872 73175 log sec log cos I T £-73476 \-7l777 1-74077 1-74375 £-74673 £-74969 1-75264 £•75558 £•75852 1-76144 £•76435 1-76726 £-77015 \j77Z^Z i-7759i| i -77877! 307 305 305 303 !30i! '300 1 298 298 296 295 294! 294 292 291 I 290 ;29ol 288^ 288, 286 03927 os6 -03983:057 0-04040:^^^ -04098.^ •^4156!^^ 0-04214^58 °'«4272 ^^ 0-04332 0-0439I 060 0^04451 i 061 0-04512 i^gj °-°4573'o6i! 004634062' °-°4696|^62 ^•°4758o63 0-04821 I ^ 0-04884 i°4 °'°4948,^6^ °°5oi2|o6s °-°5077|o65 0-05142^6^ -^•^^ 207 1066 •°5 273 1 067 •°5340^6; °°S407!o67 °-°5474lo68 ^•°5 542 1^68 0-05679^6^ 0-05748 I 005818 0-05888! ^ 071 0-05959 0-06030,^^^ 0-06102 ^^^ o-o6i74'^;3 o 06247 0^3 0-06320, r 074 0-06394! '^ o.o6^68|^;4 0-06543 l;^ 0-06618 r^5 '075 o 06693 log cotaii 0-35142 0-34803 0-34465 0-34130 0-33796 0-33463 0-33133 0-32804 0-32476 0-32150 0-31826 0-31503 0-31182 0-30862 0-30543 0-30226 0-2991 1 0-29596 029283 0-28972 0-28661 0-28352 0-28045 0-27738 0-27433 0-27128 I0-26825 0-26524 0-26223 0-25923 025625 0-25327 0-25031 0-24736 0-24442 0-24148 023856 0-23565 0-23275 0-22985 0-22697 0-22409 0-22123 f log tan I log cosec D log sin i -96073 T-96017 1 1-95960 ,£•95902 [£•95845 '1-95786 195728 : £-95668 1-95609 1-95549 1-95488 1-95427 I 95366 £-95304 £-95242 £•95179 1-95110 1-95052 i -94988 1-94923 1-94858 £-94793 £•94727 I -94660 1-94593 £-94526 £-94458 £-94390 £-94321 1-94252 1-94182 T^94ii2 1-94041 £-93970 £•93898 1-93826 i 93753 1-93680 1-93606 £•93532 £•93457 1-93382 i 93307 66° 50' 40' 30' 20' 10' 65° 50' 40' 30' 20' 10' 64° 50' 40' 30' 20' \o' 63° 50' 40' 30' 20' 10' 62° 7o^ 40' 30' 20' 10' 61° 50' 40' 30' 20' 10' 60° 50' 40' 30' 20' 10' 59° 59= 23 31° NATURAL FUNCTIONS 5 1 504 51753 52002 52250 52498 52745 52992 53484 53730 53975 54220 S4464 54708 54951 55194 55436 55678 55919 56160 56401 56641 56880 57119 57358 57596 S7^U 58070 58307 58543 S8779 59014 59248 59482 59716 59949 60182 60414 60645 60876 61 107 61337 61566 D 249 249 248 248 247 247 246 246 246 245 245 244 244 243 243 242 242 241 241 241 240 239 239 239 238 2Z7 2Z7 ^17 236 236 235 234 234 234 233 232 231 231 231 230 229 cosecant D 94160 93226 92302 91388 90485 89591 88708 87834 86970 86116 85271 84435 83608 82790 81981 81180 79604 78829 78062 77?>o?> 76552 75808 75073 7434S 73624 72911 72205 71506 70815 70130 69452 68782 68117 67460 66809 66164 65526 64894 64268 63648 63035 62427 934 924 914 903 894 883 874 864 854 845 836 827 818 809 801 792 784 775 767 759 751 744 735 728 721 713 706 699 691 685 678 670 665 657 651 645 638 632 626 620 613 608 n tangent D "60086 60483 60881 61280 61681 62083 62487 62892 63299 63707 641 17 64528 64941 65355 65771 66189 66608 67028 67451 67875 68301 68728 69157 69588 70021 70455 70891 71329 71769 72211 72654 73100 73547 73996 74447 74900 75355 75812 7(^7 i?> 77196 78129 i)tanj)«nt I) 397 398 399 401 402 404 405 407 408 410 411 413 414 416 418 419 420 423 424 426 427 429 431 433 434 436 438 440 442 443 446 447 449 451 453 455 457 460 461 463 465 468 cotangent 1-66428 1-65337 1-64256 1-63185 1-62125 I -61074 I 60033 1-59002 1-57981 1-56969 1-55966 1-54972 I 53987 I-53010 1-52043 I -5 1084 1-50133 1-49190 1-48256 1-47330 I -4641 1 I-45501 1-44598 1-43703 I-42815 I -41934 I-41061 I -40195 1-39336 1-38484 137638 1-36800 1-35968 1-35142 1-34323 1-33511 I 32704 I -3 1904 I-31110 1-30323 1-29541 1-28764 1-27994 tangent 994 985 977 967 959 951 943 934 926 919 910 903 895 888 881 873 866 859 852 846 838 832 826 819 812 807 800 794 7^7 782 777 770 D •16C63 -16868 •17075 -17283 -17493 -17704 •17918 •18133 -18350 -18569 -18790 -19012 •19236 •19463 •19691 -19920 -20152 -20386 -20622 •20859 -21099 -21341 -21584 -21830 •22077 -22327 -22579 •23089 ■2i2>A7 •23607 •23869 -24134 •24400 •24669 -24940 -25214 -25489 •25767 •26047 -26330 -26615 I 26902 20s 207 208 210 211 214 215 217 219 221 222 224 227 22i 229 232 234 236 237 240 242 243 246 247 250 252 254 256 251 260 262 265 266 269 271 274 275 278 280 285 287 24 LOGARITHMIC FUNCTIONS log sin 1-71184 £•71393 171602 I71809 I72OI4 I72218 172421 T72622 172823 173022 I73219 I73416 1-73611 173805 173997 1-74189 174379 1-74568 1-74756 174943 I-75128 I75313 1-75496 1-75678 1-75859 176039 1-76218 176395 176572 _ '6747 i -76922 1-77095 1-77268 177439 1-77609 177778 1-77946 1-78113 T-78280 1-78445 1-78609 1-78772 I 78934 log COS D log cosec ™ 0-28816 200 l^^ 0-28398 207 o 205 °''^'?3 ,4,0.27986 30], 0-27782 201 1 0-27579 ,01 0-27378 197 197 195 194 192 192 0-26978 jO-26781 1 10-26584 026389 0-26195 0-26003 0-2581 1 ;i9-362. ^8^10-25432 0-25244 0-25057 0-24872 0-24687 0-24504 0-24322 0-24141 0-23961 0-23782 0-23605 0-23428 0-23253 0-23078 0-22905 0-22732 0-22561 187 185 185 183 182 181 180 179 177 177 175 175 173, 173 1711 17O' 169 °-^-39i ^^^0-22222 0-22054 0-21887 0-21720 0-21555 0-21391 0-21228 167 167 165 164 163' 162 0-21066 i log sec log tan 177877 178163 1-78448 1-78732 I-79015 1-79297 i -79579 1-79860 r -80 1 40 1-80419 1-80697 1-80975 1-81252 1-81528 1-81803 1-82078 1-82352 1-82626 i'-^2899 1-83171 1-83442 1-83713 1-83984 1-84254 1-84523 1-84791 1-85059 £•85327 1-85594 1-85860 1-86126 1-86392 1-86656 1-86921 1-87185 1-87448 1-87711 1-87974 1-88236 r-88498 r-88759 1-89020 i 89281 log cotan 286 285 284 283 282 282 281 280 279 278 278 2771 276, 275 I 275 I 274 I 274' 273 272^ 271 271 271 270 269 2681 268 2681 267 I 266 2661 266 264 265 264 263 263 263 262 262 261 261 261 log cotan 0-22123 0-21837 0-21552 0-21268 0-20985 0-20703 0-20421 0-20140 0-19860 O-19581 0-19303 0-19025 0-18748 0-18472 0-18197 0-17922 0-17648 0-I7374 0-17101 0-16829 0-16558 0-16287 0-16016 0-15746 0-15477 0-15209 0-14941 0-14673 0-14406 0-14140 0-13874 0-13608 0-I3344 0-13079 0-12815 0-12552 0-12289 0-12026 0-1 1764 o-i 1502 0-11241 0-10980 0-10719 log tan log sec 0-06693 0-06770 0-06846 0-06923 0-07001 0-07079 007158 0-07237 0-07317 0-07397 0-07478 0-07559 007641 0-07723 0-07806 0-07889 0-07973 0-08058 008143 0-08228 0-08314 0-08401 0-08488 0-08575 log cos 077 076 077 078 078 079 079 080 080 081 081 082 082 083! 083! 0841 085 1 085: 085' 086 1 087! 087! 087! o,. 089 008664 ^gg 0-08752 0-08842 0-08931 0-09022 0-09113 o 09204 0-09296 0-09389 0-09482 0-09576 0-09670! 009765 U-0986I 0-09957 0-10053 0-IOI5I 0-10248 0-10347 log cosec 090 089 091 091 091 I 092 093 093; 094 I 094' 095 096 096 096 098 097 099 93307 93230 93154 93077 92999 92921 92842 92763 92683 92603 92522 92441 92359 92277 92194 92111 92027 91942 91857 91772 91686 91599 91512 91425 91336 91248 91158 91069 90978 90887 90796 90704 9061 1 90518 90424 90330 90235 90139 90043 89947 89849 89752 89653 log sin 52* 25 J" NATURAL FUNCTIONS •6is66 •61795 •62024 •62251 •62479 •62706 •62932 •63158 •63383 •63608 •63832 ■64056 •64279 •64501 •64723 -64945 ■65166 •65386 65606 •65825 ■66044 ■66262 ■66480 •66697 66913 67129 67344 67559 67771 679^7 68200 68412 68624 68835 69046 69256 69466 69675 69883 70091 70298 70505 7071 1 D 229 229 227 228 227 226 226 225 225 224 224 223 222 222 222 221 220 220 219 219 218 218 217 216 216 215 215 214 214 213 212 212 21 I. 211 210 210 206 208 208 207 207 206 I 62427 i^6i825 1-61229 I -60639 I ^600 5 4 1-59475 I •58902 1-58333 I-5777I 1-57213 i^5666i i^56ii4 I 55572 1-55036 1-54504 1-53977 1-53455 1-52938 I 52425 1-51918 1-51415 1-50916 1-50422 1-49933 1-49448 1-48967 I -48491 I -48019 I -475 5 1 1-47087 146628 1-46173 I -4572 1 1-45274 1-44831 I -44391 I 43956 1-43524 I -43096 1-42672 I -4225 1 1-41835 1-41421 1602 596 590 585 579 573 569 562 I558 552 547 542 536 '532 1527 522 517 :5i3 507 i503 499 494 489 485 481 476 472 468 464 459 455 452 447 443 440 435 432 428 424 421 416 414 tangent •78129 •78598 -79070 -79544 -80020 -80498 -80978 -81461 -81946 -82434 -82923 -83415 •83910 •84407 •84906 •85408 •85912 •86419 •86929 •87441 -87955 •88473 •88992 -89515 •90040 •90569 •91099 •91633 -92170 -92709 •93252 •93797 -94345 -94896 -95451 -96008 •96569 -97133 -97700 -98270 -98843 -99420 00000 J I -00000 cotangent D 469 472 474 476 478 480 483 485 488 489 492 495 497 499 502 504 507 510 512 514 518 519 523 525 529 530 534 537 539 543 545 548 551 555 557 561 564 567 570 573 577 580 D cotangent 1^27994 1-27230 1-26471 I-25717 I -24969 1-24227 1-23490 1-22758 I -2203 I I-21310 1-20593 I-19882 I-I9175 I -18474 I-I7777 1-17085 1-16398 1-15715 1-15037 1-14363 1-13694 I -1 3029 1-12369 1-11713 1-11061 1-10414 1-09770 1-09131 I -08496 1-07864 I 07237 I -066 1 3 1-05994 1-05378 I ^04766 I ^04 1 5 8 I 03553 1-02952 I-02355 I •o 1 76 1 I •o 1 170 1^005 8 3 I 00000 tangent j 26 D 764 759 754 748 742 72>7 732 727 72T 717 711 707 701 697 692 687 67S 674 669 665 660 656 652 647 644 639 635 632 627 624 619 616 612 608 605 601 597 594 591 587 583 289 1-2777S 1-2807 s 1-28374 I 28676 1-28980 1^29287 1-29597 1-29909 1-30223 1-30541 I -30861 1-31183 I -3 1 509 1-31837 1-32168 I 32501 1-32838 1-33177 1-33519 1-33864 1-34212 1-34563 1-34917 1-35274 1-35634 1-35997 1-36363 1-36733 1-37105 1-37481 i^3786o 1^38242 1^38628 1-39016 1-39409 1-39804 I ^40203 I ^40606 I^4IOI2 I 4 142 1 1-26902 1-27483 295 297 299 302 304 307 310 312 314 318 320 322 326 328 331 333 337 339 342 345 348 351 354 357 360 363 366 370 372 376 379 382 386 388 393 395 399 403 406 409 D 179- 180 181 I 18. 7 ) •78801 •78622 •78442 •78261 •78°79^8, -778971^82 •777151,84 •77531 ,8 '77347 •77162 -76977 -767911,8; 1 -766041,8^ 5 •7^417 ji88 ' •76229,^88 •76041 •75851 185 185 186 190 190 90 •7566 ■7S47I -75280. ^^ -75088 9; -74896 ;9- -74509 1 ^^ K •74314 1 4} •741201 -73924 •7372s -73531 •73333 196 196 197 198 - 198 •73135 8 •72937 200 •72737 200 -72537 200 -72137 / 00/ 201 -72136 2^_. •7^934 202 ■71732 ' ' ^ 20\ .71529 -^ •71325 -7II2I •70916 -^ -7071 1 5( 4^ 3< 2C 4] 5C 4^' 3C 2L IC 46 5<'- 204 40 J04 -^^ :o5 10 45 D 41 I LOGARITHMIC FUNCTIONS log sin 1 D log cosec 178934 ,61 1-79256 1-79415 1795731 I -7973 1 159 158 158 156 IS6 154 154 153 1-79887 1-80043 1-80197 £•80351 ^•80504, 1-806561^5- 1-80807' 1-80957: I -8 1 106 j 1-81254I T-81402 [ T-8i549| 1-816941 1-81839^ T-81983 1-82126 1-82269 1-82410 1-82551 1-82691 • 1-82830I 1-82968 I 1-83106^ 1-83242^ 183378 1-83513 1-83648 1-83781: 1-83914 1-84046 1-84177 T-84308 1-84437 1-84566 1-84694 1-84822 1-84949 151 ISO 149 148 148 147 145 14s 144 143 143 141 ! 141 140 139 138 138 136 136 135 135 133 133 132 131 131 129 129 128 128 127 0-21066 0-20905 0-20744 0-20585 0-20427 0-20269 0-20II3 0-19957 0-19803 0-19649 0-19496 0-I9344 0-19193 0-19043 |o-i8894 0-18746 10-18598 :o-i845 0-18306 0-18161 0-18017 0-17874 0-17731 0-17590 0-17449 0-17309 0-17170 |o-i703 10-1689 '0-16758 0-16622 :o-i6487 |o-i6352 ]0-i62i9 jO-i6o86 10-15954 0-15823 15692 15563 15434 15306 15178 15052 log COS I D I log sec log tan D 1-89281 I-89541 I-89801 T- 9006 1 T-90320 1-90578 1-90837 1-91095 1-91353 1-91610 T-91868 1-92125 1-92381 1-92638 1-92894 1-93150 1-93406 1-93661 1-93916 1-94171 1-94426 I -9468 1 1-94935 I -95 190 1-95444 1-95698 1-95952 1-96205 1-96459 I -967 1 2 1-96966 1-97219 1-97472 1-97725 1-97978 1-98231 1-98484 T-98737 260 260 260 259 258 259 2S8 258 257 258 'log cotan O-IO719 ■ 0-10459 {0-10199 ! 0-09939 j 0-09680 0-09422 0-09163 0-08905 1 0-08647 0-08390 0-08132 :% 0-07875 ^^^0-07619 I -99242 j 1-99495 ' 1-99747' 0-00000 0-07362 0-07106 ^^0-06850 i ^5 0.06594 12^. .0-06339 0-06084 0-05829 0-05574 0-05319 0-05065 0-04810 o 04556 0-04302 0-04048 0-03795 0-03541 0-032 0-03034 0-02781 0-02528 0-02275 0-02022 0-01769 O-OI516 0-01263 2 5 2' •^ 10-0101 I 253 L., 256 256 255 1255 i255 I254 I255 '254 254 254 !253 '254 253 '254 253 ,253 253 1253 i2S3 ■253 253 253 252 253 10-00758 0-00505 0-00253 0-00000 log cotan I D ' log tan log sec 0-10347 0-10446 0-10545' 0-10646 0-10746 10848 0-10950 O-IIO52 0-11156 O-II259 O-II364 o- 1 1469 0-II57S 0-11681 0-11788 0-11895 0-12004 0-12113 0-12222 0-12332 0-12443 0-12554 0-12666 0-12779 0-12893 0-13007 0-13121 0-13237 0-I3353 0-13470 0-13587 0-13705 0-13824 0-13944 0-14064 0-14185 0-14307 0-14429 0-14552 0-14676 0-14800 0-14926 0-15052 log cosec 1 D 099 099, loi ! 100 1 102 102 102 104 103 105 105 106 106 I 107 ;.o7i 109 I ! 1091 1 109 1 10 III 'hi; : 112 ! I 114! "4 iii6l ;ii6| 117; 117 118 119 120 120 121 122 122 123 124 1241 126 126 log cos 89653 89554 89455 89354 89254 89152 89050 8S948 88844 88741 88636 88531 88425 88319 88212 88105 87996 87887 87778 87668 87557 87446 87334 87221 87107 86993 86879 86763 86647 86530 86413 86295 86176 86056 85936 85815 85693 85571 85448 85324 85200 85074 84949 45^ 27 FOUR-FIGURE TRIGONOMETRICAL. TABLES Radians De- grees Sine Cosec, Tangent Cotan. Secant Cosine •OOOOO •0000 CO •0000 00 I -0000 I -OOOO 90 1-57080 •01745 I -0175 57-2986 -0175 57-2899 I -0002 •9998 1-55334 •03491 2 -0349 28-6537 -0349 28-6362 I -0006 -9994 88 1-53589 •05236 3 •0523 19-1073 •0524 19-08 1 1 1-0014 •9986 87 i-5i''44 •06981 4 •0698 14-3356 -0699 14-3OJ6 I -0024 •9976 86 1-50-98 •08727- 5 •0872 11-4737 -0875 11-4301 I -0038 -9962 85 I -48 - - •10472 6 •1045 9-5668 •IO51 9-5144 1-0055 •9945 84 i-46(.;' •12217 7 •1219 8-2055 •1228 8-1443 1-0075 •9925 83 1-44^^:. •13963 8 •1392 7-1853 •1405 7-1154 I -0098 •9903 82 1-431^7 •15708 9 •1564 6-3925 •1584 6-3138 1-0125 •9877 81 1-41372 •17453 10 •1736 5-7588 •1763 5-6713 I-OI54 •9848 80 1-39626 •19199 II •1908 5 '2408 •1944 5-1446 I-O187 .9816 79 1-77881 •20944 12 •2079 4-8097 •2126 4-7046 1-0223 •97 I 78 I--6i::5 •22689 13 14 •2250 •2419 4-4454 4-1336 •2309 -2493 4-3315 1-0263 •972 77^ 76 J -34:90 •24435 4-0108 I -0306 -9703 1-3264 •26180 IS •2588 3-8637 -2679 Vll^-i^ 1-0353 -965^ 75 i'3(^<-. •27925 16 -2756 3-6280 •2867 3-4874 I -0403 -961 4 1-291-- •29671 17 •2924 3-4203 •3057 3-2709 1-0457 -9563 73 "r274c •31416 18 •3090 3-2361 •3249 yo777 I-0515 -95 II 72 I..56C •33161 19 .3256 3-0716 •3443 2-9042 1-0576 •9455 71 |.239ifc •34907 20 -3420 2-9238 •3640 2-7475 I -0642 -9397 10 1-22173 •36652 21 •3584 2-7904 •3839 2-6051 I -07 1 1 -9336 */%- 1-20428 •38397 22 -3746 2-6695 •4040 2-4751 1-0785 •9272 ^9 r-18682 •40143 23 •3907 2-5593 -4245 2-3559 I -0864 •9205 ft? /-16937 •41888 24 •4067 2-4586 -4452 2-2460 I -0946 -9135 0^ /•15192 •43633 25 •4226 2-3662 •4663 2-1445 I-IO34 -9063 65 1-13446 •45379 26 •4384 • 2-2812 •4877 2-0503 I-II26 •8988 64 1-11701 •47124 27 -4540 2-2027 -5095 1-9626 I-1223 •8910 63 1-09956 •48869 28 •4695 2-1301 -5317 1-8807 1-1326 -8829 62 I -082 10 •50615 29 •4848 2-0627 •5543 I -8040 I -1434 •8746 61 I -06465 •52360 30 •5000 2 -0000 •5774 1-7321 1-1547 •8660 60 I -04720 •54105 31 •5150 I -9416 •6009 I -6643 I-I666 -8572 59 1-02974 •55851 32 •5299 1-8871 •6249 I--6003 1-1792 -8480 S8 1-01229 ■57596 33 •5446 1-8361 •6494 1-5399 i^i924 •8387 57 -99484 •59341 34 •5592 1-7883 •6745 1-4826 1-2 2 ■8290 56 •97738 •61087 35 -5736 1-7434 •7002 1-4281 1-2^08 -8192 55 -95993 •62832 36 -5878 1-7013 •7265 1-3764 1-236; •8090 54 -94248 •64577 37 •6018 I -6616 •7536 1-3270 I-2521 •7986 53 •92502 •66323 38 -6157 1-6243 •7813 1-2799 I -2690 •7880 52 -9075; •68068 39 •6293 1-5890 •8098 1-2 349 1-2868 -7771 51 -8901 •69813 40 •6428 1-5557 •8391 1-1918 1-3054 -7660 50 ■^72 > •71559 41 •6561 1-5243 •8693 1-1504 1-3250 •7547 49 •Si^i . i •73304 42 •6691 I -4945 •9004 1-1106 1-3456 -7431 48 •83776 •75049 43 •6820 1-4663 •9325 1-0724 1-3673 -7314 47 •82030 •76794 44 •6947 1-4396 •9657 I-0355 1-3902 -7193 46 •80285 •78540 45 •707^ 1-4142 I -oooo I -0000 1-4142 _707_i 45 De- grees •78540 Cosine Secant Cotan. Tangent Cosec. Sine Radians 28 ^ Ra4 •OI •OS, •o6l •o8j •io| •I2| •17;- •19 •20 •22) •24 .27; •29' •31 •33 •34' .36. •38 •40 •41; •43' •45 •47 •48; •5o< -^! •5S< •57.' •59j •61C '62i •641 •66^ •68c •69.^ •7 If •73: 75c •767 .785 N( 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. QiiK'^Ui ^eC'D LD D£t Is^inT'eaDK -rz.* •v/0. 7v>- ^1 OCT 5 'bJ-J^ LD 21A-50m-3,'62 General Library University of California Berkeley CTf^V^t> »^^- / r:%3 UNIVERSITY OF GAUFORNIA LIBRARY ^xp-jr.,^,.,.^.^.