!?v^3r^ LIBRARY UNIVERSITY OF CALIFORNIA ^ — > ^Accession ^o. 3 o ^ / J/- ^ Cla^s No. ^f^ wmm: ^•3^^: o^'-i ...»■'»' A "^^ ti — ^f"^ ( -r TEIGONOMETEY FOR BEGINNEES AS FAK AS THE SOLUTION OF TEIANGLES. Printed 1886. Second Edition 1887. Third Edition 1888. Fourth Edition 1889, Fifth Edition 1890. "9^ PREFACE. The present work is an abridgement of the more complete work on Elementary Trigonometry by the same Author. A few of the Articles have been rewritten and the order in one or two cases slightly altered. At the request of many Teachers a Table of the Logarithms of numbers from 100 to 1000 has been inserted. It will be seen [see Exercises xxxix. and XL.] that many interesting results may be obtained by the help of this Table. In the second Edition the Chapter on Logarithms was revised. In the third Edition 100 Easy Miscellaneous Ex- amples were added. In the fourth Edition a few corrections have been made, and a short Chapter on Triangles and Circles added. [UFIVBRSITT] CONTENTS. CHAP. PAGE I. On Measurement • 1 II. On the Belation between the Circumference of A Circle and its Diameter .• . • . 6 III. On the Measurement op Angles • . . . 8 IV. The Trigonometrical Katios ..... 18 V. On the Eatios of Certain Angles ... 26 VI. On the Trigonometrical Eatios of the same Angle 37 VII. On the Use of the Signs + and - . . . 48 Vin. On the Use of + and - in Trigonometry . . 61 IX. On the Eatios op Two Angles .... 62 X. On the Eatios op Multiple Angles ... 72 XI. On Logarithms 76 XII. On the Use of Mathematical Tables ... 86 Xni. On the Eelations between the Sides and Angles of a Triangle 102 XIV. On the Solution of Triangles .... Ill XV. On the Measurement op Heights and Distances 122 XVI. Triangles and Circles 128 Examples for Exercise 132 Answers to Examples 140 4, (UIH7EIISIT71 CHAPTER T. On Measurement. 1. It is usual to say that we have measured any con- crete quantity, when we have found out how many times it contains some familiar quantity of the same kind We say for example, that we have measured a line, when we have found out hoio many feet it contains. We say that we have measured a field, when we have found out how many acres or how many square yards it contains. 2. To know the measurement of any quantity then, we must have two things. First, we must have a unit, or standard of reference, of the same kind as the thing measured. Secondly, we must have the measure^ or the number of times the thing measured contains the unit, or standard quantity. 3. Hence, the measure of a quantity is the number, and the unit is the Concrete quantity, by means of which it is measured. Example 1. A line contains 261 feet; that is 261 times a foot. Here the measure or number is 261 and the unit a foot. EXAMPLES. L 1. What is the measure of 1 mile when a chain of 66 feet is the unit ? ^ c' 2. What is the measure of an acre when a square whose side is 22 yards is the unit? 3. What is the measure of a ton when a weight of 10 stone is the unit ? L. T. B. 1 2 TRIGONOMETRY. 4. The length of an Atlantic cable is 2300 miles and the length of the cable from England to France is 21 miles. Express the length of the first in terms of the second as unit. 5. The measure of a certain field is 22 and the unit 1100 square yards : express the area of the field in acres. 6. Find the measure of a miles when h yards is the unit. 7. The measure of a certain distance is a when the unit is c feet. Express the distance in yards. 8. A certain sum of money has for its measures 24, 240, 960 when three different coins are units respectively. If the first coin is half a sovereign, what are the others ? 4. It is explained in Arithmetic, in the application of square measure, that the measure of the area of a rectangle is found in terms of a square unit, by multiplying together the measures of the sides in terms of the corresponding linear unit. Example. Find in square feet, the measure of a square surface whose side is 12 feet. The area is 12 x 12 square feet = 144 x 1 square foot, .-. the measure required is 144. 5. We shall apply this result to Euclid I. 47. Example 1. The sides containing the right angle of a right-angled triangle are 3 ft. and 4, it. respectively ; find the length of the hypotenuse. Let X be the number of feet in the hypotenuse. Then by Euclid I. 47, the square described on the side of x feet ~ the sum of the squares described on the sides of 3 feet and 4 feet rcs;)ectively, ON MEASUREMENT. 3 .-. X- square feet = 9 square feet + 16 square feet = 25 square feet, .*. a; = 5. Therefore the length of the hypotenuse is 5 feet. Example 2. Find the length of the diameter of the square one of whose sides contains a feet. Let A BCD be the square, so that AB is a feet, and AD is a feet. Let the diameter BD be x feet. Then the square on DB = the sum of the squares on DA and AB. .'. ic- sq. ft. = a2 sq. it. + a^ sq. ft. .'. x- = a- + a^f .\x=jj2.a. Thus the required length ;j2.afeet= (1-4142+ .,.) x a ft. EXAMPLES. II. 1. Find the length of the hypotenuse of a right-angled triangle whose sides are 6 feet and 8 feet respectively. 2. The hypotenuse of a right-angled triangle is 100 yards and one Hide is 60 yards : find the length of the other side. 3. One end of a rope 52 feet long is tied to the top of a pole 48 feet high and the other end is fastened to a peg in the ground. If the pole be vertical and the rope tight, find how far the peg is from the foot of the pole. 4. The houses in a certain street are 40 feet high and the street 30 feet wide : find the length of the ladder which will reach from the top of one of the houses to the opposite side of the street. 5. A wall 72 feet high is built at one edge of a moat 54 feet wide ; how long must scaling ladders be to reach from the other edge of the moat to the top of the wall ? 1—2 4 TRIGONOMETRY. 6. A field is a quarter of a mile long and three-sixteenths of a mile wide : how many cubic yards of gravel would be required to make a path 2 feet wide to join two opposite corners, the depth of the gravel being 2 inches? 7. The sides of a rectangular field are 4a feet and Ba feet respec- tively. Find the length of its diameter. 8. If the sides of an isosceles triangle be each 13a yards and the base 10a yards, what is the length of the perpendicular drawn from the vertex to the base ? 9. Show that the perpendicular drawn from the right angle to the hypotenuse in an isosceles right-angled triangle, each of whose equal sides contains a feet, is ^ . a ft. 10. If the hypotenuse of a right-angled isosceles triangle be a yards, what is the length of each side ? 11. Show that the perpendicular drawn from an angular point to the opposite side of an equilateral triangle, each of whose sides con- tains a feet, is ^ . a ft. 12. If in an equilateral triangle the length of the perpendicular drawn from an angular point to the opposite side be a feet, what is the length of the side of the triangle ? 13. Find the ratio of the side of a square inscribed in a circle to the diameter of the circle. 14. Find the distance from the centre of a circle of radius 10 feet, of a chord whose length is 8 feet. 15. Find the length of a chord of a circle of radius a yards, which is distant b feet from the centre. 16. The three sides of a right-angled triangle, whose hypotenuse contains 5a feet, are in arithmetical progression ; prove that the other two sides contain 4a feet and 3a feet respectively. ( 5 ) CHAPTER II. On the Relation between the Circumference of A Circle and its Diameter. 6. The circumference of a circle is a line, and therefore it has length. We might imagine the circumference of a circle to consist of a flexible wire ; if the circular wire were cut at one point and straight- ened, we should have a straight line of the same length as the circum- ference of the circle. 7. A polygon is a figure enclosed by any number of straight lines. A regular polygon has all its sides equal and all its angles equal. The perimeter of a polygon is the sum of its sides. 8. If we have two circles in which the length of the diameter of the first is greater than the length of the dia- meter of the second, it is evident that the length of the circumference of the first will be greater than that of the second. It is in fact true that when the length of one diameter = (any number of) n times that of another diameter the length of the circumference of the one — (the same number of) n times that of the other. C TRIGONOMETRY. 9. Hence when diameter = ny. (another diameter), then circumference = 92 x (the other circumference), so that the ratio length of circumference length of diameter is the same for all circles. 10. The proof of the above statement is given in mo7e advanced works on Trigonometry. For the present the student must accept the following statements. T mi .. , circumference . . • /? i I. The ratio or number ^^ is a certain nxed diameter number. 11. It is an incommensurable number. III. It is 3-14159265 + ... II. When we say that this number is incommensurable we mean that its exact value cannot be stated as an arith- metical fraction. It also happens that we have no short algebraical ex- pression such as a surd, or combination of surds, which represents it exactly. So that we have no numerical expression whatever, arithmetical nor algebraical, to represent exactly/ the ratio of the circumference of a circle to its diameter. Hence the universal custom has arisen, of denoting its exact value by the letter tt. 12. Thus TT stands always for the exact value of a cer- tain incommensurable number, whose approximate value is 314159265, which number is the ratio of the circumference of any circle to its diameter. It cannot be too carefully impressed on the student's memory that tt stands for this number 3 '141 5 92 65... &c., and for nothing else ; just as 180 stands for the number one hundred and eighty, and for nothing else. EXAMPLES, 7 13. We may notice that ^ = 3-I4285I. So that ^ and ir diflfer by less than a thousandth part of their value. 14. Thus in a circle of radius r the circumference /oiiiKmc;/- \ ^ = (3-14159256 + ...) = 7r, or the circumference — 7rx2r = ~x2r, Example 1. The driving wheel of a locomotive engine is 5 ft. G in. high. What is its circumference ? Here we have a circle whose diameter is 5^ feet ; .*. its circumference = 7r X 5*5 feet, = (3-14159...) X 5-5 feet, = 17-278... feet. The eircumference is 17 ft. 3 in. approximately. Example 2. A piece of wire 1 foot long is bent into the form of a circle; what is the diameter of the circle ? Here the circumference = 1 foot, that i3 TT X diameter = 1 foot, .•. diameter = =t/^ x 1 foot TT = 14 inches = 3-8 inches, nearly. EXAMPLES. III. In the answers of the first 12 of the following examples ^^- is used for TT. 1, Find the circumference of a circle whose diameter is one yard. 2, Find the circumference of a circle whose radius is 4 feet. 3, Find the circumference of a 48 inch bicycle wheel. 4. The circumference of a circle is 10 feet ; find its diameter. 5. What must be the diameter of a locomotive driving wheel, that it may make 220 revolutions per mile ? 6. How many revolutions does a 36 inch bicycle wheel make per mile ? 7, How many more revolutions per mile does a 50 inch bicycle wheel make than one of 52 inches ? 8. A locomotive whose driving wheel is 5 feet high has an instrument to record the number of revolutions made. What number will the instrument record in running 100 miles ? 8 TRIGONOMETRY. 9. If the instrument in Question 8 indicates 3 revolutions per second, how many miles per hour is the engine running ? 10. What is the diameter of the driving wheel of a locomotive engine which makes 4 revolutions per second when the engine is going at the rate of 60 miles per hour ? 11. The large hand of the Westminster clock is 11 feet long; how many yards per day does its extremity travel? How far does the extremity move in a minute? 12. The diameter of the whispering gallery in St Paul's is 108 feet ; what is its circumference ? 13. Find the number of inches of wire necessary to construct a figure consisting of a circle with a regular hexagon inscribed in it, one of whose sides is 3 feet. 14. How many inches of wire would be necessary in a figure similar to that in Question 13, if the circumference of the circle were ten feet ? 15. Find how many inches of wire are necessary to make a figure consisting of a circle and a square inscribed in it, when each side of the square is 2 feet. 16. Find the length of string necessary to string the handle of a cricket bat; having given the diameter of the handle = l^ in., the length of the handle = 12 in. _, the diameter of the strings ^^th of an inch. CHAPTER III. On the Measurement of Angles. 15. In elementarj Geometry (Euclid I. — YI.) the angles considered are each always less than two right angles. For example, in speaking of the angle BOF in Euclid we should always mean the angle less than two right angles, ON THE MEASUREMENT OF ANGLES. ^ not an angle measured in the opposite direction greater than two right angles. 16. In Trigonometry, by the angle EOF is meant, not the present inclination of the two lines OB, OP but the amount of turning which OP has gone through when, start- ing from the position OB, it has turned about into the position OP, Exaviple. Suppose a race run round a circular course. The position of any one of the competitors would be known, if we remark that he has described a certain angle about the centre of the course. Thus, if the distance to be run is three times round, the line joining each competitor to the centre would have to describe an angle of 12 right angles. When we remark that a competitor has described an angle of 6| right angles, we record not only his present position, but the total distance he has gone. He would in such a case have gone a little more than one and a half times round the course. 17. Definition. The angle between two lines, OB, OP is the amount of turning about the point which one of the lines OP has gone through in turning from the position OB into the position OP, 18. The angle BOP may be the geometrical representa- tive of an unlimited number of Trigonometrical angles. (i) The angle BOP may represent the angle less than two right angles as in Euclid. In this case OP has turned from the position OB into the position OP by turning about in the direction con- trary to that of the hands of a watch. (ii) The angle BOP may represent the angle described by OP in turning from the position OB into the position OP in the same direction as the hands of a watch. In the first case it is usual to say that the angle BOP is described in the positive direction, in the second that the angle is described in the negative direction. (iii) The angle BOP may be the geometrical repre- sentation of any of the Trigonometrical angles formed by any number of complete revolutions in the positive or in the negative direction, added to either of the first two angles. (We shall return to this subject in Chapter viii.) 10 TRIGONOMETRY. EXAMPLES. IV. Give a geometrical representation of each of the following angles, the starting line being drawn in each case from the turning point to- wards the right. 1. + 3 right angles. 7. - 10\ right angles. 2. + 5 right angles. 8. + 4 right angles. 3. + 4^ right angle??. 9. - 4 right angles. 4. + 7i right angles. 10. 4n right angles. 5. - 1 right angle. 11. {in + 2) right angles. 6. 10| right angles. 12. - (471 + \) right angles, 19. There are two methods of meaburiug angles, (i) The rectangular measure. (ii) The circular measure. Rectangular Measure. 20. Angles are always measured in practice with the right angle (or part of the right angle) as unit. The reasons why the right angle is chosen for a unit are : (i) All right angles are equal to one another, (ii) A right angle is practically easy to draw, (iii) It is an angle whose size is very familiar. 21. The right angle is a large angle, and it is therefore subdivided for practical purposes. The right angle is divided into 90 equal parts, each of which is called a degree ; each degree is subdivided into 60 equal parts, each of which is called a minute ; and each minute is again subdivided into 60 equal parts, each of which is called a second. Instruments used for measuring angles are subdivided accordingly ; and the size of an angle is known when, with such an instrument, it has been observed that the angle contains a certain number of degrees, and a certain number of minutes beyond the number of complete degrees, and a certain number of seconds beyond the number of complete minutes. RECTANGULAR MEASURE. 11 Thus an angle might be recorded as containinj; 79 de- grees + 18 minutes + 30*4 seconds. Degrees, minutes, and seconds are indicated respectively by the symbols ^, ', ", and the above angle would bo written 79« 18' 36-r. 22. An angle given in degrees, minutes, and seconds may be expressed as the decimal of a right angle by the usual method. Example. Express Sd^ 4' 27" as the decimal of a right angle. 60 ) 27 seconds 60 \ 4*45 minutes 90 ) 89-0741G6G6 etc. de^^rees •43415740740 etc. right angles Ansioer. '43415746 of a right angle. Note. The French proposed to call the 100th part of a right angle a grade (written S^), the 100th jmrt of a grade a minute (written 3'), the 100th part of a minute a second (written 3"*). So that 1-437275 right angles would be read 143^ 72' 75'\ The decimal method of subdividing the right angles has never been used. ^EXAMPLES. V. Express each of the following angles (i) as the decimal of a right angle, (ii) in grades, minutes, and seconds : 1. 80 15' 27". T ? 4. 160 14' 19". 2. 60 4' 30". 5. 1320 6'. 3. 970 5' 15". 6. 490. Express in degrees, minutes and seconds, 7. -01375 right angles. X ' .. 10. '240025 right angles. \ 8. -0875 right angles. H. -180115 right angles. 9. 1-704535 right angles. 12. '35 right angles. 12 TBIGONOMETRY. On Circular Measure. 23. By the following construction we get an angle of great importance in Trigonometry. On the circumference of a circle whose centre is let an arc US be measured so that its length is equal to the radius of the circle, and let B and S be joined to the centre. 24. We are about to prove (Art. 26) that this angle liOS is sl Jlxed fraction of a right angle, so that all such angles are equal to one another. We may state the same thing thus — We are about to prove that if we take any number of different circles, and measure -on the circum- ference of each an arc equal in length to its radius, then the angles at the centres of these circles which stand on these arcs respectively, will be all of the same size. 25. Definition. The angle which at the centre of a circle stands on an arc equal in length to the radius of the circle is called a E;adian. 26. To prove that all Radians are equal to one another. Since the E-adian at the centre of a circle stands on an arc equal in length to the radius, and an angle of two right angles at the centre of a circle stands on half the circumference, and since angles at the centre of a circle are to one another as the arcs on which they stand (Euc. YI. 33), ON CIRCULAR MEASURE. 13 a radian radius 2 right angles semi-circumference diameter __ 1 circumference ir * Therefore a radian = - of 2 right angles, IT = a certain fixed fraction of ISO'*. 27. Thus the radian possesses the qualification most essential in a unit, viz. it is always the same, 28. The reasons why a radian is used as a unit are : (i) All radians are equal to one another. (ii) Its use simplifies many formulae in Theoretical Trigonometry. 29. The system of angular measurement in which a radian is the unit is called Circular Measure. Therefore the circular measure of an angle is the num- ber of radians which the angle contains. 30. A radian = | x 2 right angles, = 57-2957 degrees. 31. The expression ^ The angle 0' means that ^ is a number and some unit of an angle is implied. * The angle 180' implies the unit of angle a degree. When Greek letters are used the unit of angle implied is a radian, thus the angle = 6 radians, the angle tt = tt radians. When Roman letters are used the unit implied is a degree, the angle A = A degrees. 32. Just as W indicates 30 degrees, so we use a little c to indicate radians, thus 3*^ = 3 radians. 1^ TRIGONOMETRY. 33. The student cannot too carefully notice, that unless an angle is obviously referred to, the letters ^, ^, . . . a, ^, . . . stand for mere numbers. Thus as we have said above (Art. 12) tt stands for a number and a number only, viz. 3-14159 , but in the expression * the angle tt' that is * the angle 3*14159 ' there is some unit of angle understood. The unit under- stood here is a radian, and therefore * the angle tt * stands for 3*14159 % that is two right angles. Hence, when an angle is understood, iris a very convenient abbreviation for two right angles. 34. Let D and a be the number of degrees and radians respectively in any angle, then D _a For each fraction is the ratio of the angle to two right r.ngles. Example. Find the number of degrees in two radians. Let D be the number, then D_ _2 180 "" TT * TT Note. 2° indicates 2 radians. EXAMPLES. VL II Express the following angles in rectangular measure. 1. IT. 2. f . 3. 1°. 4. 3°. 5. 3-14159265° etc. 6. -• TT 7. e. 8. •00314159« etc. 9. IOtt. II. Express the following angles in circular measure. 1. 1800. 2. 3600. 3. 600. 4. 22^0. 6. 1^ 6. 57*2950 etc. 7. n\ 8. ^\ 9. A. EXAMPLES. VI. 15 *III. Express the following angles in circular measure. *IV. 1. 3. 5. 35. otlier as fore 338 33* 33-3*r 18. n8. Find the ratio of 37r 4 • 258 to 220 30'. lOQO 2. 6. 608. r. 2008 3. 6. 16-G8. 10\ 9, 10008. 450 to ■ 1-75" to • 600 to COS. 248 to 2«. V to 1«. Since angles at the centre of a circle are to one an- the arcs on which they stand [Euc. VI, 33], there- an angle IWP _ arc HP _ arc EF one radian arc 7i^aS' the radius * i Hence the an;]jle EOF = V -,-• — radians. the radius So that the circular measure of an angle (at the centre of a circle) is the ratio of its arc to the racSus. Example. Find the number of degrees in the angle subtended by an arc 46 ft. 9 in. long, at the centre of a circle whose radius is 25 feet. The angle stands on an arc of 4Gf it. and the radian, at the centre uf the same circle, stands on an arc of 25 feet. *i 1 40| T ,«- 2 right angles .*. the angle =-^ radians, =-}|J x — — , 1800 = 18^x1^ = 105-80 nearly. 16 TRIGONOMETRY. ^EXAMPLES. VII. (In the Answers -V- is used for ir.) 1. Find the number of radians in an angle at the centre of a circle of radius 25 feet, which stands on an arc of 37^ feet. 2. Find the number of degrees in an angle at the centre of a circle of radius 10 feet, which stands on an arc of bir feet. 3. Find the number of right angles in the angle at the centre of a circle of radius 3y\ inches, which stands on an arc of 2 feet. 4. Find the length of the arc subtending an angle of 4^ radians at the centre of a circle whose radius is 25 feet. 5. Find the length of an arc of eighty degrees on a circle of 4 feet radius. 6. The angle subtended by the diameter of the Sun at the eye of an observer is 32'; find approximately the diameter of the Sun if its distance from the observer be 90,000,000 miles. 7. A railway train is travelling on a curve of half a mile radius at the rate of 20 miles an hour ; through what angle has it turned in 10 seconds ? 8. A railway train is travelling on a curve of two-thirds of a mile radius, at the rate of 60 miles an hour ; through what angle has it turned in a quarter of a minute ? 9. Find approximately the number of English seconds contained in the angle which subtends an arc one mile in length at the centre of a circle whose radius is 4000 miles. 10. If the radius of a circle be 4000 miles,. find the length of an arc which subtends an angle of 1" at the centre of the circle. 11. If in a circle whose radius is 12 ft. 6 in. an arc whose length is -6545 of a foot subtends an angle of 3 degrees, what is the ratio of the diameter of a circle to its circumference? 12. If an arc 1*309 feet long subtend an angle of TJ degrees at the centre of a circle whose radius is 10 feet, find the ratio of the cir- cumference of a circle to its diameter. 13. On a circle 80 feet in radius it was found that an angle of 22^30' at the centre was subtended by an arc 31ft. 5 in. in length; hence calculate to four decimal places the nmnerical value of the ratio of the circumference of a circle to its diameter. 14. If the diameter of the moon subtend an angle of 30^, at the eye of an observer, and the diameter of the sun an angle of 32', and if the distance of the sun be 375 times the distance of the moon, find the ratio of the diameter of the sun to that of the moon. 15. Find the number of radians in (i.e. the circular measure of) 10" correct to 3 significant figures. (Use ff| for ir.) EXAMPLES, VII, 17 16. Find the radius of a globe such that the distance measured upon its surface between two places in the same meridian, whose latitudes diflfer by 1® 10', may be one inch. 17. Two circles touch the base of an isosceles triangle at its middle point, one having its centre at, and the other passing through the vertex. If the arc of the greater circle included within the tri- angle be equal to the arc of the lesser circle without the triangle, find the vertical angle of the triangle. 18. By the construction in Euo. I. 1, prove that the unit of circular measure is less than 60^. 19. On the 31st December the Sun subtends an angle of 32' 36", and on 1st July an angle of 31' 32" ; find the ratio of the distances of the Sun from the observer on those two days. 20. Show that the measure of the angle at the centre of a circle k a of radius r, which stands on an arc a, is — ^ , where k depends solely on the unit of angle employed. Pind h when the unit is (i) a radian, (ii) a degree. 21. The difference of two angles is ^ir and their sum SS^'; find them. 22. Find the number of radians in an angle of n\ 23. Express in right angles and in radians the angles (i) of a regular hexagon, (ii) of a regular octagon, (iii) of a regular quindecagon. 24. Taking for unit the angle between the side of a regular quin- decagon and the next side produced, find the measures (i) of a right angle, (ii) of a radian. ^ 25. Find the unit when the sum of the measures of a degree and of the hundredth part of a right angle is 1. 26. What is the unit when the sum of the measures of 9° and of •05 right angles is -^7 27. The measure of h right angles is a, find the measure of c degrees. 28. What is the unit when the sum of the measures of a right angles and of h degrees is c? 29.^ The three angles of a triangle have the same measure when the units are -^V of a right angle, ^ of a right angle and a radian respectively ; find the measure. 30. The interior angles of an irregular polygon are in a. p. ; the least angle is 120*' ; the common difference is 5<^ ; find the number of Bides. L. T. B. 2 ( IS ) CHAPTER lY. The Trigonometrical Eatios. 36. Let ROE be any angle (see the figure in Art. 37). In one of the lines containing the angle take any point P, and from P draw PM perpendicular to the other line OR, Then, in the right-angled triangle 0PM, formed from the angle ROE, (i) the side MP, which is opposite the angle under con- sideration, is called the perpendicular; (ii) the side OP, which is opposite the right angle, is called the hypotenuse; (iii) the third side OM, which is adjacent to the right angle and to the angle under consideration, is called the base. From these three, — perpendicular, hypotenuse, base, — we can form three different sets containing two each. The ratios or fractions formed from these sets, viz. ,.. perpendicular .... base ..... perpendicular (i) ^, ^ ^ , (n) i: 7 , (ill) u f ^ ' hypotenuse ^ ^ hypotenuse ^ ' base THE TRIGONOMETRICAL RATIOS. 19 and the ratios formed by inverting each of them, viz. hypotenuse , . hypotenuse , .. base perpendicular' ^^ ^ base ' ^'"^ perpendicular* will be found to be of great importance in treating of any angle BOB. Accordingly to each of these six ratios has been given a separate Qiame (Art. 37). Note. The student should obsenre carefully (i) that each ratioy such as , . , is a mere number; ^ hypotenuse (ii) that, as we shall prove in Art. 83, these ratios remain un- changed as long as the angle remains unchanged ; (iii) that if the angle be altered ever so slightly, there is a con- sequent alteration in the value of these ratios. [For, let ROEf ROE' be two angles which are nearly equal; WW Let OP=OP'; then OM is not = OM\ and therefore the ratios OM' OM Oht and -^TpT- are not equal; also MP is not=3I'P' and therefore the ratios MP .M'P' ^ .^ QP and --^ are not equal.] (iv) that by giving names to these ratios we are enabled to apply the methods of Algebra to the Geometry of Euclid VI., just as in Chapter I. we applied the methods of Algebra to Euclid I. 47. The student is recommended to pay careful attention to the follow- ing definitions. He should be able to write them out in the exact words in which they are printed. 9 9 20 TRIG0N03IETIIY. 37. Definition. To define the three principal Trigono^ metrical Eatios of an angle, E P V ' iW 7? Let ROE be an angle. In OE one of the lines containing the angle take any point P, and from F draw FM perpendicular to the other line OR^ or, if necessary, to RO produced. Then, in the right-angled triangle PM, the side MF, which is opposite the angle under consideration, is called the fer'pendicular. The side OF, which is opposite the right angle, is called the hypotenuse. The third side OM (which is adjacent to the right angle and to the angle under consideration) is called the hase. Then the ratio (i) (ii) (iii) perpendicular . hypotenuse MF OF 0M_ OF hypotenuse MF _ perpendicular 0M~~ "~ base is called the sine of the angle j^O-fi". cosme tangent The order of the letters in MF, Oi¥^and OP indicates the direction of the lines and (as will be explained later) is an essential part of the definition, 38. If A stand for the angle ROE, these ratios are called sine J, cosine A and tangent A, and are usually abbreviated thus : sin ^. cos^l. tan^. THE TRIGONOMETRICAL RATIOS, 21 39. There are three other Trigonometrical Ratios, formed by inverting the sine, cosine and tangent respectively, which are called the cosecant, secant, and cotangent respec- tively. 40. To define the three other Trigonometrical Ratios of any angle. The same construction and figure as in Art. 37 being made, then the ratio (iv) -T7T, = ypQ ^^^SQ .^ ^^^j^^ ^^^ cosecant of ^ ^ Mr perpendicular the angle ^0^. , . OP hypotenuse , (^>0i/= base " S"*'^^* » , .. OM base . . <"^) MF = perpendicular " cotangent „ 41. Thus if A stand as before for the angle ROE, these ratios are called cosecant -4, secant A, and cotangent A, They are abbreviated thus, cosec Ay sec A, cot A, 42. From the definition it is clear that cosec -4=-: — 7, sec-d= 7, coiA = - — -. sm A cos A tan A 43. The above definitions apply to an angle of any magnitude. (We shall return to this subject in Chapter VIII.) For the present the student may confine his attention to angles which are each less than a right angle. 44. The powers of the Trigonometrical Hatios are expressed as follows : (sin AYy i.e. ( , ^ , ) , is written sin*^, ^ \ hypotenuse / ' ' (cos Ay. i.e. f , 7^ ) , is written cos^ A. \hypotenuse/ and so on. (Of 22 TRIGONOMETRY. The student must notice that ♦ sin ^4 ' is a single symbol. It is the name of a number, or fraction, belonging to the angle A ; and if it be at any time convenient, we may denote sin A by a single letter, such as s or x. Also sin^yl is an abbreviation for (sin^)^, that is, for (sin A) X (sin A), Such abbreviations are used because they are con- venient, 45. The Trigonometrical Ratios are always the same for the same angle. M, A/; M P' f,p Take any angle EOE ; let P be an j point in OE one of the lines containing the angle, and let P', P" be any two points in OR the other line containing the angle. Draw PM perpendicular to OR, and P'M\ P"M" perpendiculars to OE, Then the three triangles OMP, OM'P\ OM"P' each contain a right angle, and they have the angle at common ; therefore their third angles must be equal. Thus the three triangles are equiangiilar. Therefore the ratios MP M'F M"F' OP \ OP' ' OF' are all equal. (Eu. YI. 4.) But each of these ratios is ^-^r^ with reference hypotenuse to the angle at ; that is, they are each sin ROE. Thus, sin ROE is the same whatever be the position of tlie point P on either of the lines containing the angle ROE, Therefore sin ROE is always the same. THE TRIGONOMETRICAL RATIOS. 23 46. A similar proof holds good for each of the other ratios. 47. Also if two angles are equal, it is clear that the numerical values of their Trigonometrical Ratios will be the same. We have already shown (Art. 36) that the values of these ratios are different for different angles. Hence for each particular value of A, sin A, cos A, tan A, etc. have definite numerical values. Example, We shall prove (Art. 54) that sin 300 = J= -5, cos300=^ = -8660..., tan300=-l = *577... 4^ In the following examples the student should notice (i) the angle referred to : (ii) that there is a right angle in the same triangle as the angle referred to : (iii) the perpendicular, which is opposite the angle referred to, and is perpendicular to one of the lines contain- ing the angle : (iv) the hypotenuse, which is opposite the right angle : (v) the base, the third side of the triangle. Example. In the second figure on the next page, in which BDA is a right angle, find sin DBA and cos DBA, In this case (i) DBA is the angle, (ii) BBA is a right angle in the same triangle as the angle DBA, (iii) DA is the perpendicular ^ for it is opposite DBA and is per- pendicular to BD, (iv) BA is the hypotenuse, (v) BD is the base. Therefore sin BBA, which is Pf '•P«54i<»ilF., = ^± , hypotenuse * BA ««« r\T> A -u' T, • hase BD cos DBA. which is , , = — . hypotenuse BA 24 TRIGONOMETRY. EXAMPLES. VIII. 1, Let ABC be any triangle and let AD be drawn perpendicular to BO, Write down the perpendicular, and the base when the follow- ing angles are referred to : (i) the angle ABB, (ii) the angle BAD^ (iii) the angle A CD, (iv) the angle DA C, 2. Write down the following ratios in the above figure; (i) BinJ5^D, (ii) cos^CD, (iii) tanD^C, (iv) sin^^D, (v) tani?^D, (vi) sin DAC, (vii) cos DC A, (viii) tan DC A, (ix) cos ABD, (x) sin /I CD. 3. Let ACB be any angle and let ABC and BDC be right angles; (see next figure). Write down two values for each of the following ratios; (i) ^in ACB^ (ii) cos ACB, (iii) to^n ACB, (iv) sin BAC, (v) coa BAG, (vi) ta.n BAG, 4. In the accompanying figure BDC, CBA and EAC are right angles. Write down (i) sinDJ5^, (ii) sini5i:J^, (iii) sin C^D, (iv) cos BAE, (y) cos BAD, (vi) cos CBD, (vii) tan 501), (viii) tan DJ5^, (ix) tan BE A, (x) tan CBD, (xi) sin DAB, (xii) sin BAE, EXAMPLES. VIIL TO 5, Let ABC be a right-angled triangle such that AB = 5 ft., /?C= 3 ft., then AC will be 4 ft. Find the sine, cosine and tangent of the angles at A and B respectively. In the above triangle if A stand for the angle at A and B for the angle at B, show that sm^A + cos^A = lj and that sin2i^ + cos2J5=:l. 6. If ABC be any right-angled triangle with a right angle at C, and let A, B, and C stand for the angles &t A,B and C respectively, and let a, h and c be the measures of the sides opposite the angles A, B and C respectively. Show that sin A =-. cos A=- , tan A=y , Show also that sin^^l + cos^^ = 1. Show also that (i) a = c.sin^, (ii) h = c , sinBf (iii) a = c.cosB, (iv) h = c, cos A, (v) sin^ = cosJ5, (vi) cos ^ = sin i?, (vii) tan ^ = cot ^. 7. The sides of a right-angled triangle are in the ratio 5 : 12 : 13. Find the sine, cosine and tangent of each acute angle of the triangle. 8. The sides of a right-angled triangle are in the ratio 1:2: ^3. Find the sine, cosine and tangent of each acute angle of the triangle. 9. Prove that if A be either of the angles of the above two triangles sin^A + cosM = 1. 10. ABC is SL right-angled triangle, C being the right angle. AB is 2 ft. and ^C is 1 foot ; find the length of BC^ and thence find the value of sin A^ cos A, and tan A. 11. ABC is a right-angled triangle, C being the right angle; AB = ij2 ft. and^C=l ft.; prove that sin^ = cos^ = sin^ = cosI^. 12. ABC is a right-angled triangle, C being the right angle; ^ C= 1 ft. and AB=f^/3 feet ; find AC and sin A and sin B. ( -^ ) CHAPTER Y. On the Trigonometrical Eatios of certain Angles. 49. The Trigonometrical Ratios of an angle are numeri- cal quantities simply j as their name ratio implies. They are in nearly all cases incommensurable numbers. Their practical value has been found for all angles between and 90^, which differ by 1' ; and a list of these values will be found in any volume of Mathematical Tables. It will be an advantage for the student to see a volume of Mathematical Tables that he may understand what is meant. It will not be necessary for each student to procure a copy, as in nearly all examples the necessary quotations from the Tables are given. A well arranged and useful set of Tables is that pub- lished by Messrs Chambers, of Edinburgh. 50. The finding the values of these Ratios has involved a large amount of labour; but, as the results have been published in Tables, the finding the Trigonometrical Ratios does not form any part of a student's work, except to ex- emplify the method employed. TRIGONOMETBICAL BATIOS OF CERTAIN ANGLES. 27 51. The general method of finding Trigonometrical Ratios belongs to a more advanced part of the subject than the present, but there are certain angles whose E/atios can be found in a simple manner. 52. To find the sine, cosine and tangent of an angle of 45°. When one angle of a right-angled triangle is 45°, that is, the half of a right angle, the third angle must also be 45°. Hence 45° is one angle bf an isosceles right-angled triangle. y-^ /^ m yV — Let POM be an isosceles triangle such that PMO is a right angle, and 0M= MP. Then POM= 0PM = 45°. Let the measures of OM and of IIP each be m. Let the measure of OP be x. Then Hence, sin 45^ = sin POM= ^^ = cos 450 = cos P03/ ,\ X— J2 . m. MP _ m _ J_ OP- J2,m~^^' OM _ m l_ 0P~ J2rm^'j2' tan45« = tanPOJ/: MP '' OM ' m 7n = L 28 TRIGONOMETRY. 53. To find the sine, cosine and tangent o/QO^. In an equilateral triangle each of the equal angles is 60", because they are each one-third of 180". And if we draw a perpendicular from one of the angular points of the triangle to the opposite side we get a right-angled triangle in which one angle is 60". Let OPQ be an equilateral triangle. Draw PM per- pendicular to OQ. Then OQ i^ bisected in M. Let the measure of Oi/be m; then that of OQ is 2m, and therefore that of OP is 2m. M Let the measure of MP be x. Then x' = (2my - in' = 4m' - m' = 3m^ .'. x= J2> . m. Hence, sin 6(F = sin POM = ^ = ^^ = ^ , cos60«=.cos7^6^i/ = OP OM OP '' y m 2m i-^ t2iiim'=:UnP0M= IfP _JS.m J3 = ^3. OM m 1 54. To find the sine, cosine and tangent o/SO". With the same fiojure and construction as above we have TllIGONOMETRICAL RATIOS OF CERTAIN ANGLES, 29 Hence, sm 30« = sin OFM =-777^= -^- = k , i^(y zm 2 tan 30« = tan 0PM J-^ = -^^ = — 55. To find the sine, cosine awes? tangent ^0". Let i?OP be a small angle. Draw PM perpendicular to 07?, and let OP be always of the same length, so that P lies on a circle whose centre is 0, Then if the angle POP be diminished, we can see that MP MP is diminished also, and that consequently yr-rj , which is sin POP, is diminished. And, by diminishing the angle POP sufficiently, we can make MP as small as we please, and therefore we can make sin POP smaller than any assign- able number however small that number may be. This is what is meant when it is said that the value to which sin POP approaches as the angle is diminished, is 0. This is expressed by saying, sin 0^ = i. Again, as the angle POP diminishes, OM approaches OP in length ; and cos POP, which is ^yp ? approaches in value to j-p, i.e. to 1. This is expressed by saying, cos (^ = 1 ii. MP Also, tanT^OP is vr^^; and we have seen that MP ap- proaches 0, while Oifdoes not ; . •„ tan POP approaches 0, This is expressed by saying, tan 0^=0 iii. 30 TRIGONOMETRY. 56. To find the sine, cosine and tangent ofW^. Let BOUhesi right angle - 90^. Draw BOP nearly a right angle ; draw FM perpen- dicular to OB, and let OP be always of the same length, so that P lies on a circle whose centre is 0. U M Then, as the angle POP approaches to EOU, we can see that if P approaches OP, while OM continually diminishes. MP Hence when POP approaches QO'', sin POP, which is ^- , OP . 1 . approaches in value to -x-p, that is to --, i.e. to-1. Hence we say that sin W^ 1 i. Again, when POP approaches 90", cosi^OP, which is ~rrp y approaches in value to jyn ? ^^at is to 0. Hence we say that cos90'' = ii. MP Again, when POP approaches 90", tan POP which is yr^ approaches in value to ■ ■ ■ r-r-, -. ;-. ^ ^ a quantity which approaches But in any fraction, whose numerator does not diminis^h, the smaller the denominator, the greater the value of that fraction ; and if the denominator continually diminishes, the value of the fraction continually increases. PRACTICAL APPLICATIONS. 33 Example 1. At a point 100 feet /row the foot of a tower, the angle of elevation of the top of the tower is observed to be 60^. Find the height of the top of the tower above the point of observation. Let be the point of observation ; let P be the top of the tower ; let a horizontal line through O meet the foot of the tower at the point M. Then 0M= 100 feet, and the angle MQP = m\ Let MP contain X feet. Then ^ = tanil/OP = tanG0<^=^3. •'• ioo=^^- .-. a; = 100V3 = 100x 1-7320 etc. = 173-2. Therefore the required height is 173-2. Example 2. At a point 100 yds. from the foot of a building, I measure the angle of elevation of the top^ and find that it is 23^ 15'; what is the height of the building ? As in Example 1 let the height be x yards. Then j^ = tan 230 15'. From the Table of tangents we find that tan 230 15' = -4296339. Hence a;=100x •4296339 = 42-96339. The height of the building = 43 yds. nearly, Ans, 34 TRIGONOMETRY, Example 3. A flagstaff, 25 feet liigh, stands on the top of a cliff; from a point on the seashore the angles of elevation of the highest and lowest points of the flagstaff are observed to he 47^ 12' and 4.5^ 13' respectively. Find the height of the cliff. Let be the point of observation, JPQ the flagstaff. Let a horizontal line through meet the vertical line PQ produced inM, Then gP = 25 feet, MOP = 4t7^ 12', MOQ = 4:6^ 13'. Let MQ=x feet; let OM=y feet. Then ^ = tan470 12', .•.^i^ = tan47« 12', - OAI y and ^ = tan 450 13', .-. - =tan450 13'. OM ' y ^ ^ ^.#. . x + 25 tan 47^2' Hence, by division, .•. - — = 7 ^--n-,Tr/» ' "^ ' X tan 40*^ 13' In the Tables we find that tan47<^ 12' = 1-0799018, and tan450 13' =1-0075918, 25 1-0799018 .1 + — = :l4 •0723100 X 1-0075918 "^1-0075918' X 1-0075918 100759 25 -0723100 _ 2518975 •*'^~ 7231 Therefore the cliff is 348 feet high. 7231 • = 348 nearly. PRACTICAL APPLICATIONS. / 35 EXAMPLES X. Note. The answers are given correct to three significant figures. I. At a point 179 feet in a horizontal line from the foot of a column, the angle of elevation of the top of the column is ohserved to be 45^. What is the height of the column ? ^ 2. At a point 200 feet from, and on a level with the base of a tower, the angle of elevation of the top of the tower is observed to be GO*^ : what is the height of the tower ? 3. From the top of a vertical cliff, the angle of depression of a point on the shore 150 feet from the base of the cliff, is observed to be 30" : find the height of the cliff. 4. From the top of a tower 117 feet high the angle of depression of the top of a house 37 feet high is observed to be 30^ : how far is the top of the house from the tower ? 5. A man 6 ft. high stands at a distance of 4 ft. 9 in. from a lamp-post, and it is observed that his shadow is 19 ft. long. Find the ^height of the lamp. 6. The shadow of a tower in the sunlight is observed to be 100 ft. ^ long, and at the same time the shadow of a lamp-post 9 ft. high is observed to be 3\/3 ft. long. Find the angle of elevation of the sun, and the height of the tower. ^ 7. From a point P on the bank of a river, just opposite a post Q on the other bank, a man walks at right angles to PQ to a point E so that PE is 100 yards; he then observes the angle PEQ to be 32^ 17' : find the breadth of the river, (tan 32« 17' = -6317667.) -I 8. I walk 1000 ft. away from a tower and observe the elevation of the top to be 15" 30' ; what is the height of the tower ? (tan 15" 30' = -2773245.) 9. A fine wire 300 ft. long is attached to the top of a spire and the inclination of the wire to the horizon when held tight is observed ^ be 40" ; find the height of the spire, sin 40"= '6428. 10. A vertical pole 30 ft. high stands on the bank of a river ; at the point on the other bank just opposite the pole the angle of elevation of the top of the pole is 21" ; find the breadth of the river. (cot21" = 2-6051.) II. A flagstaff 25 feet high stands on the top of a house ; from a point on the plain on which the house stands the angles of elevation of the top and bottom of the flagstaff are observed to be 60" and 45® respectively : find the height of the house above the point of obser- vation.^ 12. From the top of a cliff 100 feet high, the angles of depression * of two ships at sea are observed to be 45" and 30" respectively ; if the 3— li 36 TIIIG0N03IETRY. X. line joining the ships points directly to the foot of the cliff, find the distance between the ships. 13. A tower 100 feet high stands on the top of a cliff ; from a point on the sand at the foot of the cliff the angles of elevation of the top and bottom of the tower are observed to be 75^ and 60<^ respectively ; find the height of the cliff. (Tan 75^ = 2 + J 3) . 14. A man walking along a straight road observes at one mile- stone a house in a direction making an angle 30^ with the road, and that at the next milestone the angle is QO^i how far is the house from the road ? 15. A man stands at a point A on the bank^B of a straight river and observes that the line joining ^ to a post C on the opposite bank makes with AB an angle of 30^ He then goes 400 yards along the bank to B and finds that BC makes with BA an angle of 60^; find the breadth of the river. 16. A building on a square base ABCB has two of its sides, AB and CD, parallel to the bank of a river. An observer, standing at E on the other side of the river so that DAE is a straight line, finds that AB subtends at his eye an angle of 45^. Having walked a yards parallel to the bank, he finds that DE subtends an angle whose tan- gent is V2. Show that DB = a yards. 17. From the top of a hill the angles of depression of the top and bottom of a flagstaff 25 feet high at the foot of the hill are observed to be 45<^ 13' and A.V 12' respectively; find the height of the hill, (tan 45« 13' = 1-0075918. tan 47^ 12'= 1-0799018.) 18. From each of two stations. East and West of each other, the altitude of a balloon is observed to be 45", and its bearings to be respectively N.W. and N.E.: if the stations be 1 mile apart, deter- mine the height of the balloon. 19. The angle of elevation of a balloon from a station due south of it is W\ and from another station due west of the former and distant a mile from it it is 45". Find the height of the balloon. 20. An isosceles triangle of wood is placed on the ground in a vertical position facing the sun. If 2a be the base of the triangle, h its height, and 30" the altitude of the sun, find the tangent of half the angle at the apex of the shadow. 21. The length of the shadow of a vertical stick is to the length of the stick as V3 : 1. If the stick be turned about its lower ex- tremity in a vertical plane, so that the shadow is always in the same direction, find what will be the angle of its inclination to the horizon when the length of the shadow is the same as before. 22. What distance in space is travelled in an hour in consequence of the earth's rotation, by a person situated in latitude 00"? (Earth's radius =4000 miles.) \ B R A rT^ OF rnK NIVEH^TT CHAPTER YI. On the Relations between the Trigonometrical Ratios of One Angle. 63. The following relations are evident from the definitions : cosec - To prove tan^ = _1^ ^cos^' sin cotO== tan^' We have and . ^ perpendicular sin^= \^ , hypotenuse ',0 = base hypotenuse ' sin 6 perpendicular cos base = tan^ 64. We may prove similarly cot = cosO sin^' Or thus, cot^ = 1 cos tan sin 38 TRIGONOMETUY, 65. To prove that cos^ + sin^ ^ = 1. Let EOE be any angle 0. In OE take any point P, and draw FM perpendicular to OR. Then with respect to 0, MP is the i)erpendicular, OP is the hypotenuse, and OM is the base ; MP' '0 = OM^ OP' We have to prove that sin^ + cos^ 0=1, Now ,^ MP' OM' sm"^ + cos" = -g^2 4- -^2 1, [Euc. I. 47.] _ MP' + OM' _ OP' _ OP' ~0P'~ since MP'-^OM'^OP', Similarly we may prove that 1 + tan" = sec" ^, and that 1 + cot" = cosec" 0. 66. The following is a List of Formula with which the student must make himself familiar : cosec = ~. — 7, , sec cot (9 = tan^ = sin ' sin( tan ' cos sin"(9 + cos"6>-l, tan" ^ + 1 = sec" ^, cot" ^ + 1 = cosec" 0. COtO: 1 ' co7l9 ' cos^ 'slrT^' TRIGONOMETRICAL RATIOS OF ONE ANGLE, 39 67. In proving Trigonometrical identities it is often convenient to express the other Trigonometrical llatios in terms of the sine and cosine. Example, Prove tJiat tan A + cot A = sec A . cosec A. ^ ^ sin^ ^ , cos A ^ Since we have cos^ \j\JV J^ sin^' 1 sec A = J , eos A cosec ^^ilF^' tan ^ + cot ^ = sin^ cos^ cos^ sin^ s'm^A + cos^A 1 cos A . sin A " cos^ .sin^ seGA , cosec A, [Art. 65.] G8. It is sometimes convenient to express all the Hatios in terms of the sine only ; or in terms of the cosine only. Example i. Prove that sin^ 6 + 2 sin^ $ cos^ ^ = 1 - cos* d. By Art. 65, we have sin-^ = l-cos2^, hence Bin*^ + 2 sin^^ cos2^=(l - cos2^)2 + 2(1 - cos^^) x cos^^ = (1 - 2 cos2 e + cos^ ^) + (2 eos2 d-2 cos* 6) = 1-C0S*^. Q. E. D. Example ii. Express sin* ^ + cos* ^ in terms of cos^. sin* d + cos* ^ = (1 - cos2 6)^ + cos* d = (1 - 2 cos2 d + cos* 6) + cos* = l-2cos2^ + 2cos*^. Note. (1 - cos 6) is called the versed sine of ^, and is written versiu $, EXAMPLES. XI. Prove the following statements. 1. cos^.tan2( = sin^, - 2. cot ^. tan ^ = 1. 3. cos ^ = sin ^ . cot ^, ' 4. sec^ . cot ^ = cosec ^, »/ 6. cosecul. tan-4 = sec^. "'^ 40 TBIGONOMETRY. 6. (tan -4 + cot ^) sin ^. cos ^ = 1. 7. (tan A - cot ^) sin ^ . cos A = sin^^ - cos^^. 8. cos2^-sin2^ = 2cos2^-l = l-2sin2^. 9. (sin^ + cos^)2=l+2sin^ .cos^. 10. (sin^-cos^)2 = l-2sin^ .cos^. 11. cos4J5-sin4jB = 2cos25-l. 12. (sin2i? + cos2JB)2=l. 13. (sin2i5 - cos2 B)2= 1 - 4 cos^B + 4 cos^Z?. 14. 1 - tan* B = 2 sec^B - sec^B, 15. (secJ5-tanB)(secB + tanJ5) = l. 16. (cosec - cot e) (cosec + cot 0) = 1. 17. sin3 + cos3 = (sin ^ + cos ^) (1 - sin cos 0). 18. cos' - sin3 = (cos - sin 0) (1 + sin ^ cos 0), 19. sin«^ + cos«^ = l-3sin2^.cos2^. 20. (sin6 - cos6 0) = (2 sin2 ^ - 1) (1 - sin2 + sin'* 0). tan ^+ tan B . ^ . t> 21. —nr-, TB = tan ^ . tan B. cot il + cot J5 ^^ cot a + tan 8 . . ^ 22. X ; — To = cot a. tana tan a + cot /3 ^ *%« 1-sin^ , . , .,„ ^. 1 + cos^ , . . , ..„ 24. r = (cosec ^ + cot ^)2. 1-cos^ ^ ' 25. 2 versin ^ - versin2 = sin2 0, 26. versin ^ (1 + cos 0) = sin2 ^. Express in terms of (i) cos 0y (ii) of sin ^, 27. cos^^-sin*^. 28. (sin2^-co82^)2. 29. l-tan40. 30^ sins^ + cpss^. 31. tan2^ + cot2^. '' 32. 1 + cot*^. 33, l + cot2^-cosec2^. 34. 2 tan* ^ - 4 sin^ ^. TRIGONOMETRICAL RATIOS OF ONE ANGLE. 41 69. All the Trigonometrical Ratios of an angle can be expressed in terms of any one of them. Example 1. To express all the trigonometrical ratios of an angle in terms of the sine. Let ROE be any angle 0. We can take P anywhere in the line.OJEJ; so that we can make one of the lines, OP, OM, or IIP any length we please. Let us take OP so that its measure is 1, and let s be the measure MP s of MP ; so that sin Oj which is — — , = :. ; or, s = sin 0, Let X be the measure of OM. Then since OM^ = OP^ - MP^, .-. X^=l-8\ Hence MP OM Vl-s2 tan = = Vl - sin2 0y OM 8 sin \/r^2 ~ Vi-sm-^^' and so on. Note. The solution of the equation x^ = l- s^, gives and therefore the ambiguity ( ± ) must stand before each of the root sjmabols in the above. Tins ambiguity, as will be explained later on, is of great use when the magnitude of the angle is not limited. When we limit A to be less than a right angle we have no use for the negative sign. 42 TRIGONOMETRY. Examjile 2. To express all the other trigonometrical ratios of an angle in terms of the tangent. In this case tan 6 ■■ MP '' OM' Take P so that the measure of OM is 1, and let t be the measure of MP; so that tan 6, which is ^ , = ^ ; or, i = tan 0, Then we can show that the measure of OP is Vl + 1'-^. MP _ t _ tan "VlT^~Vl + tan2(?' Hence, OP . OM cos 6 = ^^ = - 'OP Vl + t2 Vl + tan-^i?' and so on. 70. The same results may be obtained by the use of the formulae on p. 69. Example. cos-^ + sin2^ = l, .». cos^^rrl -sin^^, .-. cos^ = Vl-sin2^. sin 6 A • ^ /J sin^ Again tan^=- cos(? Vl-sin^^' and so on. EXAMPLES. XII. Express all the other Ratios of A in terms of cos A, Express all the other Ratios of A in terms of cot A. Express all the other Ratios of A in terms of sec A, Express all the other Ratios of A in terms of cosec A. Use the formulas of Art. 66 to express all the other Trigono- metrical Ratios of A in terms of sin A. 6. Use the formulae of Art. 66 to express all the other Trigono- metrical Ratios of A in terms of the tan A, 1. 2. 3. 4. 5. TRIGONOMETEICAL BATIOS OF ONE ANGLE, 43 7 1 . Given one of the Trigonometrical Ratios of an angle less than a right angle, we can find all the others. Since all the Trigonometrical Katios of an angle can be expressed in terms of any one of them, it is clear that if the numerical value of any one of them be given, the numerical value of all the rest can be found. Example. Given sin ^ = f , find the other Trigonometrical Ratios of 9. Let ROE be the angle 6. Take P on OE so that the measure of OP is 4. Draw PM perpendicular to OR, P Then since sin $ = O v^7 M 9 = I ( so that -r-p - S ) ' ^^^ since the measure of OP is 4, therefore the measure of 3IP must be 3. Let X be the measure of OM ; then OM^ = OP^-MP^, .-. x = j7. Therefore the measure of OM is ^7. COS^: OM^Jl ~0P 4 ' tan^ = MP OM' Hence, 3__3y7 V7"~ 7 ' cot^ = sjl 1. If sin ^: 2. If cos B 3. If tan A 4. If sec ^:. 5. Iftan6>: 6. If cot ^=- EXAMPLES. XIII. =f , find tan^ and cosec^. = \^ find sin I? and cot^. — I , find sin A and sec A. - 4 , find cot 6 and sin 6. = ;^3, find sin 6 and cos^. 2 — ,_ , find sin 6 and sec ^. Vo 8, If tan d = a ^ find sin and cos 6. 7. If sin ^ = -, find tan ^. c a, find sin and cot 0. a, and tan = h, prove that (I - a^) (1 + Z^^) = 1. 9. If sec ^: 10. If sin ^ 11. If cos = /i, and tan = k^ find the equation connecting h and h. 44 TBIGONOMETBY. 72. To trace the changes in the magnitude of sin A as A increases Jrom 0^ to 90^. Take a line OB, of any length ; and describe tlie quadrant RPU of the circle whose centre is and radius OR. Draw the right angle ROU, cutting the circle in U. Let OP make any angle ROP {= A) with OR ; draw PM perpendicular to OR. Then . , MP sinA = -~^, When the angle A is 0^, MP is zero, and when A is 90'', MP is equal to OP ; and as A continuously increases from 0° to 90^, IIP increases continuously from zero to OP ; also OP is always equal to OR. MP Therefore, when A = 0^, the fraction y-^ is equal to MP jyp J that is ; when A = W the fraction -jyp i^ equal to OP Yyp , that is 1 ; and as A continuously increases from 0° to MP 90^, the numerator of the fraction -^-^ continuously increases from zero to OP, while the denominator is unchanged, and MP therefore the fraction -^ , which is sin A, increases con- tinuously from to 1. TRIGONOMETRICAL RATIOS OF ONE ANGLE. 45 73. To trace the changes in the magnitude of tan A 05 A increases from 0° to 90*^. With the same construction and figure as in the last article, we have "When the angle A is 0®, MP is zero ; when A is 90", MP is equal to OP ] and as the angle continuously in- creases from 0° to 90^*, MP increases continuously from zero to OP, When the angle A is 0°, OM is equal to OP ; when A is 90°, OM is zero ; and as A continuously increases from 0° to 90°, OM continuously decreases from OP to zero. MP . Hence, when A is 0", the fraction 77^ is equal to -— r-_ , OM ^ OM^ MP that is ; when A is 90°, the fraction -^7^ is equal to OP . . . -jr- , that is * infinity* (see Art. 5G); and as A continuously increases from 0° to 90**, the numerator continuously in- creases from zero to OPy while the denominator continuously MP diminishes from OP to zero : so that the fraction -^r^rj. , which OM is tan A, continuously increases from until it is greater than any assignable numerical quantity. EXAMPLES. XIV. 1. Show that as A continuously increases from 0° to 90°, cos A continuously diminishes from 1 to 0. 2. Trace the changes in the magnitude of sec^ as 6 increases from to - . 3. Trace the changes in the magnitude of sin ^ as ^ diminishes from 90° to 0°. 4. Trace the changes in the magnitude of cot ^ as ^ increases from to ^. , _^^^-g, 46 TRIGONOMETRY. On the Solution of Trigonometrical Equations. 74. A Trigonometrical equation is an equation in which there is a letter, such as 6, which stands for an angle of unknown magnitude. The solution of the equation is the process of finding an angle which, if it be substituted fur 0, satisfies the equation. Example 1. Solve cos^ = J. This is a Trigonometrical equation. To solve it we must find some angle such that its cosine is J. We know that cos 60^ = ^. Therefore if GO^ be put for 6 the equation is satisfied. ,*. ^ = 60^ is a solution of the equation. Example 2. Solve sin - cosec ^ + 1 = 0. The usual method of solution is to express all the Trigonometrical Ratios in terms of one of them. Thus we put - — - for cosec 6, and we get 1 sin 6 This is an equation in which 6, and therefore sin is unknown. It will be convenient if we put x for sin 0, and then solve the equation for X as an ordinary algebraical equation. Thus we get X or, x^+j-=l. Whence a! = - 2, or a5 = J. But X stands for sin 0» Thus we get sin ^ = - 2, or sin = ^. The value - 2 is inadmissible for sin 0, for there is no angle whose ' sine is numerically greater than 1. y^, ^xtcxA But Bm300=J. .•. sin^ = sin300. Therefore one angle which satisfies this equation for is 30<^. TRIGONOMEimCAL EQUATIONS. 47 EXAMPLES. XV. Find one angle which satisfies each of the following equations. 1, sin ^ = -7- . 2. 4 sin d = cosec 0, 3. 2cos^ = sec^. 4. 4 sin ^-3 cosec ^ = 0. 5. 4cos0-3secd = O. 6. 3tan^=cot^. 7. 3sin(9-2cos2^ = 0. 8. ^/2 sin ^ = tan ^. 9. 2 cos ^ = ^3 cot ^. 10. tan^ = 3cot + 2. ^3 . cos^ = 5^, show that there is only one value of cos 0. and that one value of ^ is 77 . o 8. Prove that the value of sin'* + cos'* + 2 , sin^ . cos^ is always the same. 9. Simplify cos^ A +2. sin^ ^ . cos^ ^ . 10. Express sin^^ + cos^.4 in terms of sin^^ and powers of sin^^. 11. Express 1 + tan'' in terms of cos and its powers. ,^ ^ ,. ^ cos ^+ cos 5 sin ^ + sin 5 12. Prove that -^ — ; r— ^ + -. 5 = 0. sm ^ - sm jB cos A - cos B 13. Express (sec A - tan A)^ in terms of sin A. 14. Trace the changes in cosec ^ as ^ increases from to Jtp. 15. Trace the changes in cot ^ as ^ decreases from ^tt to 0. ^16. Solve 2sm(0 + etc., is therefore of great importance. 56 TRIGONOMETBY, 88. We proceed to show that the Trigonometrical Katios of an an^le vary in Sign according to the Quadrant in which the revolving line of the angle happens to be. From the definition we have, with the usual letters, MP ■ 03f MP sin POP = ^,, cos BOP =^,tsin BOP =- I. When OP is in the first Quadrant (Fig. I.). MP is positive because from M to P is upwards (Rule II. p. 55.) OJM is positive because from to if is towards the riffht, (Rule I.). OP is positive, (Rule in.). Hence, if A be any angle of the^rs^ Quadrant, MP sin A, which is -^ , is positive; cos ^, which IS -TYp, IS positive; , ,. , , MP , tan J, which IS ^^>, is BATIOS OF ONE ANGLE. 67 II. When OP is in the second Quadrant (Fig. ii.). MP is positive, because from JIf to P is upwards, OM is negative, because from to M is towards the left OP is positive. Hence, if A be any angle of the second Quadrant, sm A, which is -j-p t ^^ positive ; cos A, which IS jYp , is negative; HI. When OP is in the third Quadrant (Fig. iii.) MP is negative, OM is negative, OP is positive. So that, if A be any angle of the third Quadrant, sin A is negative, cos A is negative, tan A is positive. IV. When OP lies in the fourth Quadrant (Fig. iv.) J/P is negative, OM is positive, OP is positive. So that, if -4 be any angle of the fourth Quadrant, sin A is negative, cos A is positive, tan J. is negative, 89. The table given below exhibits the results of the last Article. Quadrant ... I. II. III. IV. Sine + + - Cosine + •- - + Tangent ... + - + - The student should notice that for any particular Quadrant the three signs of sine, cosine, and tangent are unlike their signs for any other Quadrant. 58 TRIGONOMETRY, EXAMPLES. XX. State the sign of the sine, cosine, and tangent of each of the following angles : 1. eoo. 2. 1350. 3. 2G50. 4. 2750. 5. -W. 6. -9P. 7. -1930. 8. -3500. 9^ -10000. 10. 27i7r + j7r. 11. 2;i7r + |7r. 12. 27i7r - -^tt. 90. The NUMERICAL VALUES through which the Trigo- nometrical Katios of the angle ROP pass, as the line OP turns through Xh^ first Quadrant, are repeated as OP turns through each of the other Quadrants. Thus as OR turns through the second Quadrant from JJ to X, Fig. II. p. 56 {OR being always of the same length) MR and OM pass through the same succession of numerical values through which they pass, as OR turns through the first Quadrant in the o-pjiodte direction from U to R. Example 1. Find the sine, cosine and tangent of 120^. 1200 is an angle of the second Quadrant. Let the angle ROP be 120^ (Fig. 11. p. 5G). Then the angle POL = I800 - 1 200 = 600. MR Hence, sin 1200 = — -— = sin GQo numerically, and in the second OR ^ Quadrant the sine is positive. to Therefore sinl200='^ (i). Again, cos 120* = — — = cos 60^ numerically ^ and in the second Quadrant the cosine is negative. Therefore cosl200=-- (ii). Similarly, tanl200= -Js (iii). Example 2. Find the sine, cosine and tangent of 22o^, 2250 is an angle of the third Quadrant. Let the angle ROP be 2250 (Fig. iii. p. 56). Here the angle POL = 225^ - I8O0 = 450. Therefore the Trigonometrical Ratios of 225®= those of 45^ nu- rnerically; and in the third Quadrant the sine and cosine are each negative and the tangent is positive. Hence, sin 225^= - -1- ; cos 2250= - -775 ; tan 2250 = 1. ItATIOS OF ONE ANGLE. 59 91. The cosecant, secant and cotangent of an angle A have the same sign as the sine, cosine, and tangent of A respectively. EXAMPLES. XXI. Find the algebraical value of the sine, cosine and tangent of the following angles : 1. 1500. 2. 1350. 3. -2400. 4^ 3300. 5. -450. 6. -3000. 7. 2250. 8. -135. 9. 3900. 10. 7500. 11^ ^8400. 12. 1020^. 13. 2n7r+^. 14. (27i + l)7r-|. 15. (2n-l)7r+^. Find the four smallest angles which satisfy the equations 16. sm^ = i. 17. sin^ = -^. 18. sin ^ = ^-. 19. sin ^ = - J. Find four angles between zero and +8 right angles which satisfy the equations 20. Bin ^ = sin 200. 21. sin ^ = - -y^ . 22. sin^ = -sin^. 23. Prove that 300, 1500, -3300, 3900, -210o have the same sine. 24. Show that each of the following angles has the same cosine : -1200, 2400, 4800, -4800. 25. The angles 600 and - 1200 have one of the Trigonometrical Eatios the same for both ; which of the ratios is it ? 26. Can the following angles have any one of their Trigonometri- cal Eatios the same for all ? -230, 1570 and -1570. 92. Proposition. To trace the changes in the magni- tude and sign of sin A, as A increases from (f to 360°. Take the figure and construction of page 56. As A increases from 0° to 90^, MP increases from zero to OPy and is positive. Therefore sin A increases from to 1 and is positive. As -4 increases from 90** to 180°, MP decreases from OP to zero, and is positive. Therefore sin A decreases from 1 to and is positive. 60 TBIGONOMETRY. As A increases from 180° to 270°, 3fF increases from zero to OPy and is negative. Therefore sin A increases numerically from to 1, and is negative. As A increases from 270° to 360°, MF decreases from OF to zero, and is negative. Therefore sin A decreases numerically from 1 to and is negative. "^EXAMPLES. XXII. Trace the changes in sign and magnitude as A increases from 0^ to 3600 of 1, cos A 2. tan id. 3, cot .4. 4. sec^. 5. cosec^. 6. 1-sin^. 7. sin^^. 8. sin ^. cos ^. 9. sin ^ + cos ^. 10, tan ^ + cot ^. H, sin J.- cos ^. 93. Def. One angle is said to be the complement of another, when the two angles added together make up a Q'ight angle, Example 1. The complement of A is (90^ - A), Exam:ple 2. The complement of IW is (90o - 1900) = - lOOo. For 1900 + (900 _ igQO) = 900. Example 3. The complement of — is(^--j-j =- — . 94. To prove that the sine of an angle A is equal to the cosine of its complement (90°- A). Let A be less than 90°, and let EOF be A. Draw FM perpendicular to OR. [See figure, p. 20.] Then since FMO = 90°, therefore FOM+ OFM = 90°, and therefore OFM = (90° - A). MF ^ Kow, sin A = ^rp = cos OFM= cos (90° - A), q.e.d. EXAMPLES. XXIII. Find the complements of 1. 300. 2. 1900. 3. 900. 4^ 3500. 5. -250. 6. -3200. 7. lir, 8. -l^r. PiATIOS OF ONE ANGLE. 61 9. sin 700 = cos 20^. 10. 11. tan79« = cotlP. 12. If A be less than 90^, prove 13. cos^=sin(900-^). 14. 15. sec^=cosec(900-^). 16. Prove by drawing a figure in each case cos47n6' = sin420 44'. sec36o = cosec 54^. tan^ = cot(900-J). cot^ = tan(900-^). If A,B,C be the angles of a triangle, so that A-i-B + C = 1S0^, prove 17. cosi.4 = sini(^ + <^). 18. cosJB = sini(^-f C). 19. siniC=cos4(^+B). 20. sin ^^ = cos J (J5 + C). 95. Def. One angle is said to be the supplement of another when their sum is two right angles. Thus (180' - A) is the supplement of A. If A, B, Che the angles of a triangle, {A + B + C) ^ 180", so that {B + C) is the supplement of A. 96. To prove that the sine of an angle = the sine of its supplement, when the angle is less than 180". Let ROP be the angle A, take LOP' also = A, then EOF = (180^-^). Take OP^OP' and draw PM, FM perpendicular to POL, then the triangle POM, P'OM' are equal in all respects, since they are equiangular and 0P= 0P\ ^ MP M'P 4 ^^^^" Wp^Uf'^ that is, sin i?OP= sin ^OP'; or, sin^ = sin (180"-^). OM _ OW Also -Qp--Qpr> that is, cos ROP=^- cos POP' j or, cos ^ = - cos (180" - A). ^'-i TRIGONOMETRY. EXAMPLES. XXIV. Prove, drawing a separate figure in each case, that 1. sin 60^ = sin 1200. 2. sin340o==sin (- IW). 3. sin ( - 400) _ sin 220®. 4. cos 320® = - cos ( - 140'^). 5. cos (-3800)= -cos 5600. ^ cos 1950 = - cos(- 15'^). If ^, J5, be the angles of a triangle, prove 7. sin^ = sin(B + C7). 8. sin (7 = sin (^ + i?). 9. cos5=-cos (^ + C). 10. cos^= -cos((7 + I?). Prove by means of a figure that 11. sin (-^)= -sin^. 12. cos(-^) = cos^. 13. sin (900 ^ ^) ^ cos A. 14. cos (90o + J[) = - sin ^. 15. tan (1800 + ^) = tan J. CHAPTER IX. On the Trigonometrical Katios of Two Angles. 97. We proceed to establish the following fundamental formulse : sin {A-^ B)= sin ^ . cos ^ + cos ^ . sin B^ cos (A + B) = cos A . cos B — sin A . sin B sin {A — B) = sin A . cos B — cos A . sin B cos (A — B) = cos A . cos ^ + sin ^ . sin B^ Here, A and B are angles ; so that {A + B) and {A - B) are also angles. Hence, sin (^ + B) is the sine of an angle, and must not ^ be confounded with sin A + sin B. Sin (^ + ^) is a single fraction. Sin -4 + sin ^ is the sum of two fractions. The student should notice that the words of the two proofs of Arts. 98, 99 are very nearly the same. k..(i). RATIOS OF TWO ANGLES. 63 98. To prove that sin (A + B) = sin A . cos B + cosA. sin J3, a7id that cos {A + B) = cosA . cos ^ - sin ^ . sin B. Let BOm be the angle A, and BOF the angle i?. Then in the figure, BOF is the angle (A + B). In OF, the line which hounds the compound angle {A 4- B), take any point P, and from P draw Pi/, PN at right angles to OR and 0^ respectively. Draw NH, NK at right angles to MP and OP respectively. Then the angle NPH = 90' - II NP = i/iV^O = ROE =r. Af. Now • /. m • ^/^7^ ^^^^ Mil + IIP im IIP Bin (.1 4-P) = sinROF=-^ = — ^ ^Up ~0P KF.Ol^^ IIP.^Y KN ON IIP NP OP op"" NP ON . OP ' NP . OP ON = sin ROE . cos EOF+ cos /fPiV^. sin ^OP = sin A . cos B + cos -4 . sin P. Also 0^_^ ~ OP OP EN NP^ OP cos (il + P) = OiT.ON -,^^ OJf OK-MK cosPOP=-^- = — ^^-: ^iT.NP Oif OP' ' ON . OP NP. OP OiV^ • OP NP = cos ROE . cos POP - sin HPN. sin POP = cos ^ . COS P - sin ^ . sin P. t Or thus. On OP as diameter describe a circle; this will pass through 31 and N, because the angles OMP and ONP are right angles ; therefore MPN and iWOiY are angles m the same segment ; so that the angle MPN=MON=A. 64 TRIGONOMETRY. 99. To prove that sin [A- B)~ sin A . cos B — cos A . sin i>, a^icZ ^A(x^ cos (A—B)~ cos -4 . cos B + sin ^ . sin B, Let ^0^ be the angle A, and FOU the angle i>\ Then in the figure, EOF is the angle (A - B), In OF, the line which hounds the compound angle (^A — B), take any point P, and from P draw PM, PN at right angles to OR and (9^ respectively. Draw NH, NK bA, right angles to MP and OR respectively. Then the angle NPH= W - HNP = ENE = ROE = A^. Now • /. px • i.nn ^P MII-PII KF PH sm (^ - i?) = sin i^07^:= ^^ = —^^^— = ^^ - -^ NP im. ON OP PH.^V OP KN ON OP PH ON . OP NP ,0P OF ' OP FP ' OP = sin R OF . cos FOF - cos IIPF . sin i^O^ = sin A . cos B — cos ^ . sin B, Also cos (^ - ^) = cos ROF=-^ = ^— OZ. ON JT-^.NP NP . OP OF OF ~ OP'^ OP FH FP FP ' OP ON . OP ■ NP ,0P OF ' 01 _. - = cos ROE. cos PO^ + sin HPF . sin FOE ~ cos ^ . COS B + sin ^ . sin B, t Or thus. On OP as diameter describe a circle, this will pass through M and N^ because the angles Oil/P and ONR are right angles ; therefore the angles MRN and MON together make up two right angles ; so that the angle I[PN=MON=A, RATIOS OF TWO ANGLES. 05 Example, Find the value of sin 75<>. Bin 760= sin (450 + 300) = sin 450 . cos 300 + cos 450 . sin SO* "2^2 4 • EXAMPLES. XXV. -^ ^3-1 1. Show that cos 7oO 2. Show that sin 150 3. Show that cos I50 2y2 >_n/3-1 "2^2 • 4. Show that tan 750 = 2 + Js. 5. If sin ^ = 1 and sin5 = |, find a value for sin(il-f B) and fo- cos(^-£). 6. If sin -4 = '6 and sin B=z^^, find a value for Bin (A + B) .^nd for cos(^ + i?). 1 1 7. When sin i4 = -7^ and sin B = —== , then one vai: - \A-\-B) is 450. 8. Prove that sin 750= -9659... 9. Prove that sin 150 = -2588... 10. Prove that tan 150= -2679... 100. It is important that the student should become thoroughly familiar with the formulae proved on the last two pages, and that he should be able to work examples involving their use. EXAMPLES. XXVI. Prove the following statements. 1. sin(^ + B) + sin(^-i?) = 2sin.4.cosJ?. 2. sin(4 + B)-sin(4-i?)=2cos^.sinB. 3. cos(^+B) + cos(^-J5) = 2cos^ .cosi?. 4. cos (4-£)-cos(^ + J5) = 2sin^ .sini?. ^ sm(^ + B) + sin(^-jB) , . 0. T-A — rk 1-, — 77( = tani4. cos [A + B) + cos {A - B) L. T. B. 5 66 TRIGONOMETRY, n X .J. o sin (a + |S) _ , . ^ sin (a - /5) 6. tana+tani3= ^^ ^. 7, tan a -tan 5= ^^ ^. cos a . cos p cos a . cos /3 o X .X « cos(a-B) /> X X « cos(a + i3) 8. cot a + tan/3 = -; — ^ -^--. 9. cot a - tan i3 = -^ — ^ ^. '^ sin a . cos /3 ' Bin a . cos j3 10. tana + cotS=^2Kfr^). jj tang + tan ^^sin ((> + ») '^ COS a . sm j3 ' tan ^ - tan (p sin (^ - 0) 12. tan . tan + 1 _ cos (t* - ) 1 - tan 6 . tan "~ cos (0 + (p)' __ tan^ + cot0 ,^ , ,^ . 13. cot0-tan^ = ^"^(^-^)-^^^^^ + ^)' 14. cot^ + cot0_ sin(^ + 0) cot d-cot) . tan (p tan(g + 0)-tan^ =tan0. l + tan(^ + 0).tan^ ^ 2 sin (a + j) .cos (j8 - j j=cos (a - /3) + sin (a + jS). 37. 2 sin (g -aj .cos (^ + ^ j =cos(a-)3) -sin (a + jS). 38. cos (a + p) + sin (a-i3) = 2 sin (^ + aj . cos (^ + ^V 39. cos(a + i3)-sin(a-/3) = 2sin (J- - a\ . coa (j- - fij, 40. sin nA . cos ^ + cos nA . sin ^ = sin {n-\-l)A, 41. cos (n-1) A , cos A - sin (n - 1) 4 . sin A = cos nA, 42. sin 71-4 . cos {n - 1) ^ - cos nA . sin {n-1) A = sin -4. 43. cos {n-l)A, cos (?i + 1) ^ - sin (w - 1) ^ . sin (n + 1) ^ = cos 2nA, 101. The following formulae are important : , . ^. tan A + tan JB . tan (A + B)=-^ -, r. ^ ^ 1 - tan ^ . tan i? •(")• , . _,, tan A - tan B tan (A-B)= r, , ^ ^ 1 + tan ^ . tan JJ) The proof of the first is given below. The student should prove the second in a similar manner. Example. To prove tan (A + B) = ,^^^^'^^^— - . ^ ^ l-tan^.tan^ (i) By using the results of Arts. 90, 99, we have t&n(A + B) = ^^^( ^+^) _ sin^.cosjg + cos^ .sinB cos U +i^) "~ cos ^ . cos i? - sin ^ . sin 5 • Divide the numerator and the denominator of this fraction each by cos A . cos jB, and we get sin A , cos B cos ^ . sin B #Qr. /J -L p\_ ^Q^^ -cos-B cos ^. cos i? - ' cos A . cos B sin ^ . sin B ^ cos^I.cosjB cos^.cosi5 tan A + tan B 1 - tan A . tan B ' Q.E.D. 68 TRIGONOMETRY. EXAMPLES. XXVII. 1. If tan^=i and tanJB = J, prove that tan(^ + jB) = 4, and tan(^-J5)=|. 2. If tan ^ = 1 and tan B = -j^f prove that tan (A + B)=2 + y/3. 3. Prove that tan IS^ = 2 - y/B. 4. If tan ^ = f and tan B = ^V> prove that tan {A + B) = 1. What is {A + B) in this case ? 5. If tan A=m and tan JB = — , prove that tan {A-\-B) = tfj ■n\ COt^.COt^ + 1 - 7, cot (A-B) = — 7-^ TT- • " ^ '' cot^-cot^ 8. ,/. 7r\ cot^ + 1 cotO ^' C0t7 10. tan ^^ - ~ ^ + cot fd + '^\= 0. 11. cotfd--'^^+iaiife+^)=0. 12. If tan a = — ^ and tan 8 = ^r ^ , prove that tan (a + B) = l. tan(7i + l)0-tanr?<^ r^tand. ^' 1 + tan (n + 1) . tan iKp ^' tan(n + l)0 + tan(l-n)0 _. ^p. •^*- 1-tan (n + 1) . tan (l-w) ^-^^'^ '''^- 15, If tan a = m and tan j3 = w, prove that cos(a+jS) = 16. If tan a = (a + 1) and tan j8 = (a - 1), then 2 cot (a - /3) = a^. 1 - tan a tan j8 17. If a + /3 + 7 = 900, then tan 7 =- tan a + tan /3 RATIOS OF TWO ANGLES. 102. From pages 63 and 64 we have sin (A+B) = sin il . cos ^ + cos ^ . sin if sin {A '-B) = sinA. cos B-cosA .sinB cob(A +B)=co3A, cos^-sin^ . sinjB cos (^ --5) = cos ^ . cos^ + sin^ . sinJ5^ From these by addition and subtraction we get sin {A+B) + sin (^ - ^) = 2 sin ^ . cos B^ sin {A +B) - sin {A -B) = 2 cos ^ . sin 5 ^ cos \a +B) + cos \a - ^) = 2 cos ^ . cos B cos \A-B)- cos (^ + ^) = 2 sin ^ . sin B^ Now putts' for (^4-^), and put T for (A-B): Then S+T= '2A, and .S'- T= 2B, so that A = — ^ , and B = — ^ . Hence the above results may be written sin aS' + sin T= 2 sin sinAS'"SinT= 2 cos S + T S-T^ 2 S+T , cos , sm cos S + COS T=2 cos — ^— , ^cos T— cos aS' = 2 sin S+T . sm 2 2 2 S-T 69 (i). ... (iii). 103. The formulae (iii) are most important, and the student is recommended to get thoroughly familiar with them in w.ords, as on the next page ; * If A and B are each less than 90<>, then 5^, which is their surrij is greater than T, their difference. Therefore if S be less than 90^, cos 3 is less than cos T ; so that cos T - cos >S is positive. i 70 TRIGONOMETRY. (1) The sum of the sines of two angles equals twice the sine of half their sum into the cosine of half their difference, (2) The difference of the sines of two angles equals twice the cosine of half their sum into the sine of half their difference. (3) The sum, of the cosines of two angles equals twice the cosine of half their sum into the cosine of half their difference. (4) The difference of the f cosines of two angles equals twice the sine of half their sum into the sine of half their difference. \ Note. The difference of the cosines of two angles is the cosine of the smaller angle -the cosine of the greater angle. 104. It will be convenient to refer to the formulae (i) as the 'AjB' formulae, and to the formulae (iii) as the ^Sj I" formulae. X^EXAMPLES. XXVIII. Prove the following statements : 1. sin 600 + sin SQO = 2 sin 45^ . cos 150. 2. sin 600 ^ sin 20^ = 2 sin 400 . cos 200. 3. sin 400 _ sin iQo = 2 cos 250 . sin 15^ 4. cos - + cos- = 2 cos — .cos — . - X ^ r> . StT . IT 6. cos- -cos-=2sin^- .sm~, 6. sin ^A + sin 5 j; = 2 sin 4^ . cos A, 7, sin lA - sin 5 A =2 cos QA . sin A. 8. cos5^-f cos9^ = 2cos7^ .cos 2^. 9, cos oA - cos 4.4 = - 2 sin -^ . sin — , 3^ A 10. cos A - cos 2^ = 2 sin — . sin - . sin2g + sin(9 _^ 3^ sin2^-sin^_ 3^ - ^^' cos^ + cos2^~*^'^'2 • ^^' cosd-QOs2d~^'^^~2' IQ sin 3^ + sin 2^ d ^^' cos 2^ -cos 3^"^°^ 2* ^14. 15. 16. 17. RATIOS OF TWO ANGLES. sin 0+8in _cos^ + cos0 cos ^ - cos "~ sin - sin 6 ' cos (600 + J) + cos (600-^) = cos ^. coa(4:5^ + A)-hcoa(i5^'-A)=J2, cos A, sin (450 + A)- sin (450 -A) = ^/2. sin A. u & ^ 18. cos (300 - ^)- cos (300 + ^) = sin ^. 19. """■ 20. cos - COS d sin sin sin ^ ^ + m^ + 6in0 \2/ \2/ 105. It is important tliat the student should be thoroughly familiar with the second set of formulae on j). 69. Written as follows, they may be regarded as the inverse of the % 5^ 'formulae.' 2 sin ^ . cos B = sin (A + B) -i- sin (A-B)' 2 cos A . sin jB = sin {A + B) - sin (A - B) 2 cos ^ . cos jB = co§ (^1 + B) + cos (A - B) 2 sin ^ . sin J5 = cos (A-B) — cos {A + B) ^ .(iv). >C EXAIVLPLES. XXIX. Express as the sum or as the difference of two trigonometrical ratios the ten following expressions : 1. 2 sin 6 . cos 0. 3. 2 sin 2a . cos 3/3. 5. 2 sin 3^ . cos 5^. 7. sin 4.0 . sin 6, 9, 2 cos 100 . sin oQO. 11, Simplify 2 cos 2^ . cos ^ - 2 sin 40 . sin 6, 12. Simplify sm — . cos - - sin — . cos — . ia J J J )(, 13. Simplify sin 3^ + sin 2^ + 2 sin — . cos - . 14. Prove that sin — - . siip-j + sin — . sm — =sin 2^ . sm B, 4 4 4 4 2. 2 cos a . cos /3 4. 2 cos(a+j8) .cos(a-p). 6. 2 cos y . cos ^ . 8. cos y . sin Y • 0. cos 450 . sin lu^. CHAPTEE X. On the Trigonometrical Eatios of Multiple Angles. '106. To express the Trigonometrical Eatios of the angle 2A in terms of those of the angle A. Since sin (A -[■£) = sin ^ . cos ^ + cos A . sin B ; .-. sin (il + -4) = sin A . cos A + cos A .sin A ; .'. sin 2^ = 2 sin ^ . cos -4 (1). Also, since cos (A + B) = cos A . cos B — sin A . sin B ; . •. cos {A + A) = cos A . cos A — sin A . sin A ; .-. cos 2^ = cos""^ - sin^^ ,(2). Eut 1 = cos^J. + sinli ; ^ . '"^'^77+ cos 2^ = 2 cos'i^ it^J^ and 1 - cos 2^ = 2 sinM. f^M/uy^ . The last two results are usually written cos 2^ = 2 cos'^ - 1 (3), and cos 2^ = 1 - 2 sin^^ (4). tan A + tan B Again, tan (^ + ^) = i_,^^^,,^^^ i , . . . tan ^ + tan A 2tanul ,^. RATIOS OF MULTIPLE ANGLES. 73 107. These five formulse are very important, sin 2il = 2sinil .cos^ (1)," cos 2 A = cos^^ - sinM ' cos 2A = 2 co&'A - 1 cos 2^ = 1 - 2 sin^ J. ^ , 2 tan A tan 2 A -- 108. 1 -tanM* The following result is important, sin 2 A 2 sin ^ . cos J, ^ . - = tan^. (2) (3) (v). l+cos2^ 2 cos^J. 109. The student must notice that A is any angle, and therefore these formulse will be true whatever we put for A. Exaviple, Write -^ instead of A, and we get Bin^ = 2sm- . COS— — COS ^ = cos^ ^ - sin^^ ■(2), and so on. EXAMPLES. Prove the following statements : 1, 2 cosec 2 A = sec -4 . coscc A . 2 - sec^^ sec'^ A 5. cot2il = = cos 2 A. cot2J-l 2cot^ 7, tan ^ + cotJ5 = 2cosec25. ^ 9. cot B - tan B = 2 cot 2B. 11.' (sm-H-coSg-J =l + sm^. 13. cos2- M + tan^-j =l + sin0. 14. sin^|(cot|-iy = 2. 4. 6. 8. 10. XXX. cosec^ A = sec2^. cosec^ A -2 cosM (1 - tanM) = cos 2A, 2tanJ5 H-tan2J5 l-tan2J5 l + tan2jB cot2j? + l cot-^i?-l = sin2J5, = cos2Z?. = sec 22?. 12 . ( sm - - cos - J = 1 - sm 0. : 1 - sin d. 74 TRIGONOMETRY, sin ^ ^ 1 + cos j3 2 l+cos/3 2 cosec j3 - cot /3 = tan ^ . 1 + tan - cos a; 2 25. 38. 1 - sin a; _ ^ a: 1 - tan - cos a; cot X 2 + 1 1 - sin a; " cot X 2 -1 cos^a + sin^a 2 -sin 2a cos a + sin a 2 27. '- = 7. "^28. cos*a-sm4a = cos2a. cos a -sin a 2 ' >^ nn « . • fi l + 3C0s22a y^29. cos^a + sm^a = j • . . ^ (3 + cos2 2a)cos2a -^^30. cos^ a - sin^ a = ^^ j . 31 !^!if^_££il'3==2. 32. '=?^-f + ?^^=2cot2^. " sinyS cos^ sm;S eos/S '^ *^^' sin 2^ ^ sm^ cos^S '^ sin--^ cos "2 35. ^ ^=2V3. 36. tan (450 + ^) -tan (450-^) =2 tan 2.4. 37. tan (450 _ ,4) + cot (45^ - ^) = 2 sec 2^. tan2(450 + ^)-l . „. ^^ — =sin 2 A. tan2(450 + ^) + l 39. :S4^ = -(^^"-2)-K^^"-l)- ., ^ „ sinB + sm2B .„ . „ „ sin2B-8inB RATIOS OF MULTIPLE ANGLES, 75 110. The following two formulaB should be remem- bered : sin 3A = 3 sin ^ - 4 sinM | . .. cos3A =4cos^^ - 3cos^ J ^ ^' Note. The similarity of these two results is apt to cause con- fusion. This may be avoided by observing that the second formula must be true when ^=0^; and then cos 3^= cos 0^= 1. In which case the formula gives cos 0<^ = 4 cos 0<^ - 3 cos,^, or^=4-3, which is true. The first formula may be proved thus : sin BA = sin {2A + A)- sin 2A . cos ^ + cos 2A . sin A = (2 sin A . co^ A) cos ^ + (1 - 2 sin^^) sin A = 2 sin A . cos-^A + sin A -2 sin^A = 2 sin ^ (1 - sin2^) + sin J - 2 sin^ A = 2 sin ^ - 2 sin=*^ + sin J^ - 2 sin^A = 3sinA-4iSin^A. The second formula may be proved in a similar manner. ^ , ^ , X o . 3 tan A - tan^ A Example, Prove that tan 3A = — :j — — — ^ — • „, , ,^ , ,, tan2u4 + tan^ tan3^ = tan(2^ + ^)=^^^^^2lTt^^ 2tan^ 1 - t an'-^^ _ 2 tan A + tan A - tan^.4 " , 2 tan A ^ ~ 1 - tan^^ - 2 tan'-^-i 1 - - — r— 0-; • tan A 1 - tan-^ 3 tan A - tan^^ ■" 1-3 tan2^ • EXAMPLES. XXXI. Prove the following statements: _ sin 3il ^ n 4 . -, rt cos 3^ _ « . ^ 1. —. — 7=2cos2^ + l. 2. - —r=2 cos 2^-1. sm A cos A ^ 3 sin ^ - sin 3^ , , . . , _ . cot^ ^ - 3 cot ii 3. ;r^ — o , =tan3.4. 4. cotdA = ~- — -r- — — - . cos3^ + 3cos^ 3cot2^-l _ sin 3^ - sin ^ ^ , ^ sin 3^ - cos 3^ „ . ^ , ^ 5. TTi 7 = tan^. 6. — ^ — : — = 2 sin 2^-1. cos dA + cos A sm ^ 4- cos A _ sin 3^ + cos 3^ c^ - ^ ^ ^ cos ^ - sm ^ ^' tan34-tan^'^cot^-cot3l'^^^*^^'^* / 3sin^-sin 3^y /sec 2^^-1X3 • \3 cos A + cos 3AJ "" \sec 2^ + ij * 10- ^^"=(^-^-^)^- CHAPTER XL On Logarithms. IIL In Algebra it is explained i (i) that the multiplication of different powers of the same quantity is effected bj adding the indices of those powers; (ii) that division is effected by subtracting the indices; (iii) that involution and evolution are respectively effected by the multiplication and division of the indices. Example 1. Let m = a^i n = a^, then nixn = a^xa*=a^'^^ (i ) , m-r-n = a^-i-a'' = a^^^ (ii), . } (iii)- Example 2. Given that 347 = lO^'S^ossus * and 461 = 102-6637009^ prove t7iai 347x461 = 105 20-10304. We have 347 x 461 = 1025403295 ^ 102G637009 — 102*5403295 + 2-6637009 ==105-2040304, Q.E.D. EXAMPLES. XXXIL 1. If m = a*, n = a*, express in terms of a, h and it, (i) m^xn^. (ii) m^y-n^. (iii) '^m^ x n^, (iv) {"i/m^xn^Y. 2. If 458=102-6560982 and 650=102-8i29i34, find the indices of the powers of 10 which are equal to (i) 453x650. (ii) (453)4. (iii) 6503x4532. (iv) ^453. (v) V453x<^650. (vi) ^453 x (650)3. (yii) V453 x 650. 3. Express in powers of 2 the numbers, 8, 32, J, ^\, '125, 128. 4. Express in powers of 3 the numbers, 9, 81, |, -^V* '^> tV- * The number 347 lies between 100 and 1000, 1. e. between 10^ and 103. Hence, if there is a power of 10 which is equal to 347, its index must be greater than 2 and less than 3, i. e. equal to 2 + a fraction. LOGARITHMS. 77 112. Suppose that some convenient number (such as 10) having been chosen, we are given a list of the indices of the powers of that number, which are equivalent to every whole number from 1 up to 100000. Such a list could be used to shorten Arithmetical calculations. Example 1. Multiply 3759 by 4781 and divide the result by 2690. Looking in our list we should find 3759 = 103-5750723^ 4781 = 103-6795187^ 2690 = 103 4"97523. Therefore 3759 x 4781 ~ 2690 = 103-5750723 x 103-6795187 ^ 103-4297523 _ 103-6750723+3-6795187-3-4297523 _ 1038248387, The list will give us that 103 8248387 = 6680-9. Therefore the answer correct to five significant figures is 6680*9. Example 2, Simplify 36 x 210^-4/17601. The list gives 2 = 10-3oio3m 3 = 10-4771213 and 17601= 104-2455373, Thus 36 X 2'^<>-^^rmi= (10-4771213)6 X (10-3010300)10^(104-2455373j^ _- 102-8627278 ^ 103-0103000 _i_ 101-4151791 ~ 102-8627278+3-0103000-l-4151791= 104-4578487, And from our list we find 1044578487 = 28697, nearly. EXAMPLES. XXXm. Given that 2 = lO-soiosoo^ 3 = 10-477i2i3 and 7=10-845098o^ find the indices of the powers of 10 equivalent to the following numbers. 1. 22, 32, 23, 2x3, 24, 72. 2. 14, 16, 18, 24, 27, 42. 3. 10, 5, 15, 25, 30, 35. 4. 36, 40, 48, 50, 200, 1000. 5. 310 x7l0-^ 220, 212x3^-0-7-711. 6. -^2rx v^l8, ^/-iyxl^x /^34ir2io; 7. Find approximately the numerical value of v 42, having given that 101623249 = 1.4532 nearly. 8 Find approximately the numerical value of 4^(42)4 x \/(42)3, having given that 10338177 =2408-6. 9. Find the value (i) of ^6x^/7x^9, (ii) of Jy2x3"^x7", having given that 10-6615067=4.5868 and 10-o285094^. 93546. 10. Find the value of (67-21)1 x (49-62)i x (3-971)"^, having given that 67-21 = 1018274339^ 49*62 = 101-695G568^ 3-971 = 10'S988999 and 10-59713ip = 3-9549. 11. Find the area of a square field whose side is 640-12 feet, having given that 640-12 = 1028062614 and that 105-6125228=40975.3. 12. Find the edge of a solid cube which contains 42601 cubic inches, having given 42601 = 1046294198 and 101-^^399=34.925. 13. Find the edge of a solid cube which contains 34*701 cubic inches, ha\ing given that 34-701 = 1015403420^ and 10-5134473 = 3.2617. 14. Find the volume of the cube the length of one of whose edges is 47-931 yds.; given 47-931 = 1016806165^ 105-o4i8495 = 110115. 78 TRIGONOMETRY, 113. The powers of any other number than 10 might be used in the manner explained above, but 10 is the most convenient number, as will presently appear. 114. This method, in which the indices of the powers of a certain fixed number (such as 10) are made use of, is called the Method of Logarithms. Indices thus used are called logarithms. The fixed number whose powers are used is called the "base. Hence we have the following definition : DEF. The logarithm of a number to a given base is the index of that power of the base, which is equal to the given number. If I be the logarithm of the number n to the base a, then a^=n, 115. The notation used is logji = I. Here, logji is an abbreviation for the words * the loga- rithm of the number n to the base a.' And this means, as we have explained above, *the index of that power of a which is equal to the number nJ Example 1. What is the logarithm of a^^ to the base a? That is, what is the index of the power of a which is a - ? The index is f ; therefore f is the required logarithm, or logaa^=|. Example 2. What is the logarithm of 32 to the base 2? That is, what is the index of the power of 2 which is equal to 32? Now 32 = 25, .'. the required index is 5 ; orlog232 = 5. The use of Logarithms is based upon the following propositions : — I. The logarithm of the product of two numbers is equal to the logarithm of one of the numbers + the logarithm of the other. For, let log^m=x and log^n = y, then, m = a^, n = ay, loga (m X 7i) = loga [a^ X aV) = log^ (a^+J') = x + y = log^^ m + log^ n, II. The logarithm of the quotient of two numbers is the logarithm of the dividend - the logarithm of the divisor. For, log, (^j =logJ^\=: log, {a^-y) = x-y [as above] = log^77z-log„n. LOGARITHMS, 7\) III. The logarithm of a number raised to a power k is k times the logarithm of the number. For, loga (771*) = loga { (a*)*} = log, (a*^) =:kx—k times log^ m. Examples. Given logjo 2 = -3010300, log^o 3 = -4771213, logjQ 7 = -8450980, Jind the values of the following : (i) logio 6 = logio (2x3) = logio 2 + logjo 3 = -3010300 + -4771213 = -7781513. [by I.] (ii) logio J = logio 7 - logio 3 = -8450980 - -4771213 = -3679767. [by IL] (iii) logio 35 = 5 ti^es logio3 = 5 x -3010300 = 1-5051500. [by III.] .... ^/Sx4: . /3x^ti 1 .. dx\^ _, ^^^, (iv) logio Y -7- "" ^°^io \T') ^ ^ ^^ ^^^10 ~7~ ^^ '^ = I of (log 3 + log 4 - log 7) = i of {-4771213 + twice -30103 - -8450980} = i of -2340833 = -0780278. [by I. and II.] ( v) logio 5 = logio Y- = logio 10 - logio 2 = 1- -3010300 = -6989700. EXAMPLES. XXXIV. 1. Find the logarithms to the base a of a^, a"^*' ^Ja, ^ a^, "^ * 2. Find the logarithms to the base 2 of 8, 64, J, -125, -015625, 4/64. 3. Find the logarithms to the base 3 of 9, 81, ^, ^Y> *i> -gV 4. Find the logarithms to base 4 of 8, ^16, Jl>, Ay-015625. 5. Find the value of log28, log2-5, log3 243, log5(-04), logjolOOO, logi^-OOl. 6. Find the value of log^a'^", logi,^})^, log82, log27 3, logjoolO. If logio2 = -30103, logio 3 = -4771213, logio 7 = '845098, find the values of 7. Iogio6, logio 42, logio 16. 8. logio 49, logjo 36, logio 63. 9. Iogio200, logio 600, logio 70. 10. logio 5, logio 3-3, logio 50. 11. Iogio35, logiol50, logio-2. 12. logio3-5, logio7-29, logio-081. 13. Given logio2, logio3, logio 7, find the value (i) of 4/6 x 4/7 x ^9. (ii)ofjy2x3'^x7^^ [•6615067 =logio4-5868; - -0285094 =logio-93646]. 14. Prove that (i) log {4/2 x 4/7-J-4/9} = ^ log 2 + ^ log 7 - 1 log 3, (ii)log{]i^2x3~^x7^^}=iVlog2-ilog3 + ^rlog7. 80 TRIGONOMETRY. Common Logarithms. 116. That System of Logarithms whose base is 10, is called the Common System of Logarithms. In speaking of logarithms hereafter, common logarithms are referred to unless the contrary is expressly stated. We shall assume that a power of 10 can be found which is practically equivalent to any number. 117. The indices of these powers of 10, i.e, the Com- mon Logarithms, are in general incommensurable numbers. Their value for every whole number, from 1 to 100000, has been calculated to 7 significant figures. Thus any cal- culation made with the aid of logarithms is as exact as the most carefully observed measurement. 118. Now, the greater the index of any power of 10, the greater will be the numerical value of that power ', and the less the index, the less will be the numerical value of the power. Hence, if one number be less than another, the loga- rithm of the first will be less than the logarithm of the second. But the student should notice that logarithms (or indices) are not proportional to the corresponding numbers. Example, 1000 is less than 10000; and the logarithm to base 10 of the first is 3 and of the second is 4. But 1000, 10000, 3, 4 are not in proportion, 119. We know from Algebra that 1= 10*, 10 = 10' and that '1= J^ =10-' 100 = 10^ -01= tJ^ =10-* 1000 = 10^ -001 = toVtt = 10 10000 = 10* -0001=^^^ = 10-* loiru" 'TO 0(5'' and so on. Hence, the logarithm of 1 is 0. The (common) logarithm of any number greater than 1 is positive. The logarithm of any positive number less than 1 is negative. ON LOGARlTiniS. 81 120. We observe also that the logarithm of any number between 1 and 10 is a positive decimal fraction ; that the logarithm of any number between 10 and 100, i. e. between 10^ and 10°, is of the form 1 + a decimal fraction ; that the logarithm of any number between 1000 and lOpOO, L e. between 10^ and 10*, is of the form 3 + a decimal fraction ; and so on. 121. "We observe also that the logarithm of any number between 1 and '1, i.e, between 10" and 10~\ can be written in the form — 1 + a decimal fraction ; that the logarithm of any number between '1 and -01, i.e. between 10"^ and 10"^, can be written in the form — 2 -i- a decimal fraction ; and so on. Example 1. How many digits are contained in the integral part of the number whose logarithm is 3 '67192? The number is IO^'^tiss ^nd this is greater than 10^, i.e. greater than 1000, and it is less than 10*, i. e. less than 10000. Therefore the number lies between 1000 and 10000, and therefore the integral part of it contains 4 figures. Example 2. Given that 3 = lO'^'^i^is^ find the number of the digits in the integral part of 320. We have 3 = 10-477i2i3^ ,«. 320_ n 0'*^^213\20_ 1Q9-5424260 Therefore there are 10 digits in the integral part of 3"^'; for it is greater than 10^ and less than lO^^*. Example 3. Supposing that the decimal part of the logarithm is to be kept positive, find the integral part of the logarithm of -000123.1:. This number is greater than -0001 i.e. than 10~* and less than •001, i. e. than 10*3. Therefore its logarithm lies between — 3 and - 4, and therefore it is -4 + a fraction; the integral part is therefore -4. L. T. B. 6 82 TRIGONOMETRY. EXAMPLES. XXXV. Note. The decimal part of a logarithm is to be kept positive, 1. Write down the integral part of the common logarithms ol 17601, 361-1, 4-01, 723000, 29. 2. Write down the integral part of the common logarithms oi •04, -0000612, -7963, -001201. (See Note above.) 3. Write down the integral part of the common logarithms oi 7963, -1, 2-61, 79-6341, 1-0006, '00000079. 4. How many digits are there in the integral part of the numbers whose common logarithms are respectively 3-461, -3020300, 5-4712301, 2-6710100? 5. Give the position of the first significant figure in the numbers whose logarithms are - 2 + -4612310, - 1 + -2793400, - 6 + -1763241. 6. Give the position of the first significant figure in the numbers whose common logarithms are 4-2990713, -3040695, 2-5860244, -3 + -1760913, -1 + -3180633, -4980347. 7. Given that 2 = 10 3010300^ fin^i tjie number of digits in the in- tegral part of 810, 212, 1520^ 2100. 8. Given that log 7 ='8450980, find the number of digits in the integral part of 1^\ 49^, 343^", (^7^)20^ (4-9)i2, (3 -43)10. 9. Find the position of the first significant figure in 5^2, [iy\ (-V)2o, (-02)^ (-49)«. 10. Find the position of the first significant figure in the numerical value of 20^, (-02)7, (-007)2, (3-43)^7, (-0343)8, (.0343)tt^. 122. Prop. To prove that when two numbers expressed in the decimal notation have the same digits {so that they differ only in the position of the decimal 2>oint)^ their loga- rithms to the base 10 differ only by an integer. The decimal point in a number is moved by multiplying or dividing the number by some integral power of 10. Let the numbers be m and n\ then m = nxW when k is a whole number (positive or negative); then log m = log (^ X 10*) = log ri + log 10* = log 7i 4- k. That is log m and log n diflfer by an integer, q. e. d. Example i. log 1779-2=log {(1-6792) x lO^} = log 1-6792 + log lO^ = log 1-6792 + 3. ON LOGAEITHMS. S3 Example ii. Given that log 1-7692 = -247770, ^^ ^„ ,^ -t;Jl find (i) log 16792, (ii) log -0016792, (iii) log 167-92. ^>^?^ " Here log 16723 = log (1-6792 xiO^) = 4-247776, . '^ f^^ log -0016792 = log (1-6792 x 10-3) = - 3 + -247776, l "1 \/\ >* log 167-92 = log (1-6792 x 10^) = 2-247776. . ' 123. It is convenient to keep the decimal part of com- mon logarithms always positive, because then the decimal part of the logarithms of any numbers expressed by the same digits will be always the same. 124. The decimal part of a logarithm is called the mantissa. 125. The integral part is called the characteristic. 126. The characteristic of a logarithm can be always obtained by the following rule, which is evident from page 81. EXILE. The characteristic of the logarithm of a number greater than unity is one less than the number of integral figures in that number. The characteristic of a number less than unity is nega- tive, and (w^hen the number is expressed as a decimal,) is one more than the number of cyphers between the decimal point and the first significant figure to the right of the decimal point. 127. When the characteristic is negative, as for ex- ample in the logarithm -3 + -1760913, the logarithm is abbreviated thus, 3-1760913. Example 1. The characteristics of 36741, 36-741, -0036741, 3-6741 and -36741 are respectively 4, 1, - 3, 0, and - 1. Example 2. Given that the mantissa of the logarithm of 36741 is 5651510, we can at once write down the logarithm of any number whose digits are 36741. Thus log 3674100 =6-5651510, log 36741 =4-5651510, log 367-41 =2-5651510, log -36741 =1-5651510, log -00036741 = 4-5651510, and so on. 84 TRIGONOMETRY. 128. In any set of tables of common logarithms the student will find the mantissa only corresponding to any set of digits. It would obviously be superfluous to give the charac- teristic, 129. It is most important to remember to keep the mantissa always positive. Example. Find the fifth root of -OOOGSOGl. Here log^o -OOOGSOGl =4-8133207, .-. logio (-00065061)* = i(I-8133207) = i ( - 4 + -8133207) = 1 ( _ 5 + 1^8133207) = - 1 + -3626641 = 1 -3626641, and 1-3626641 = log -23050, : .'. the fifth root of -00065062 = -23050 nearly. I EXAMPLES. XXXVI. 1. Write down the logarithms of 776*43, 7*7643, -00077643 and 776430. (The table gives opposite the numbers 77643, the figures 8901023.) ^ 2. Given that log'^o 59082 = 4*7714552, write down the logarithms of 5908200, 5*9082, -00059082, 59P;82 and 5908*2. 3. Find the fourth root of -0059082, having given that log 5*9082 = -7714552; 4 -4428638 = logio 27724. 4. Find the product of -00059082 and -027724, having given that •21431= log 16380 (of. Question 3). 5. Find the 10th root of -077643 (cf. Question 1), having given that •8890102 = log 7-7448. 6. Find the product of (-27724)2 and -077643. (See Questions 1 and 3; 7758288= log 59680.) MISCELLANEOUS EXAMPLES. XXXVIL 1. Find logs 8, logg 1, logg 2, logy 1, loggg 128. 2. Show that the logarithms of all except eight of the numbers from 1 to 30 inclusive, can be calculated in terms of log 2, log 3 and log?. 3. Show that the logarithms of the numbers 1 to 10 inclusive may- be found in terms of the logarithms of 8, 14, 21. 4. The mantissa of the log of 85762 is 9332949. Find the log of v^-0085762. Find how many figures there are in the integral part of (85762) ^^ (i) 2^^x34^=7^ (ii) 32^=128x7*-*, (iii) 12^=49, (iv) 28^=21-^-3^. Given logjo 7, find log7 490. Given logjo 3, find logg 270. Given logjo 2, find logg 10. ON LOGARITHMS. 85 5. Find the product of 47*609, 47G-09, -47609, -000047609, having given that log 4-7609 = -6776891 and -7107564 = log 5-1375. 6. What are the characteristics of the logarithm of 3742 to the bases 3, 6, 10 and 12 respectively? 7. Having given that log 2 = -3010300, log 3 = -4771213 and log 7 = -8450980, solve the following equations : 8. 9. 10. 11. Given logg 9 = V 15', stands for the tabular logarithm of sin 3r 15', and is equal to {log (sin 31' 15') + 10}. The words logarithmic sine are used as abbreviation for tabular logarithm of the sine. Thus in the tables we find L sin Sl^ 15'= 9-7149776. Therefore log (sin SP 15') = 9-7149776 - 10 = i-7149776. USE OF LOGARITUMIG TABLES. 97 Example 1. Find sin 31® 6' 25". The Tables give sin SO^ 6'= -5165333 (i), Bin 310 7'= -5167824 (ii)^ Let sin 310 7' 25" = -5165333 + d (iiij. The difference between the first two angles is 60". The difference between the first and third angle is 25". The differences between the corresponding sines are -0002491 and d. By the Bule these four differences are in proportion. Therefore 60" : 35"= -0002491 ; ^, .-. (Z= -0002491 X If = -0001038. Hence from (iii) sin 31^7' 25" =-5165333 + -0001038= -5166371. Example 2. Find the angle whose logarithmic cosine is 9*7858083. The table gives 9-7857611 = 1, cos 520 22' (i), 9-7859249 = Lcos520 2r (ii). The cosine diminishes as the angle increases. Hence correspond- ing to an increase in the angle there is a diminution of the cosine. Hence, let 9-7858083 = 1* cos (500 22' -D) (iii). Subtracting the first tabular logarithm from the second the differ- ence is -0001638. Subtracting the first tabular logarithm from the third, the differ- ence is -0000472. Subtracting the first angle from the second, the difference is - 60". Subtracting the first angle from the third, the difference is - D. By the Kule these four differences are in proportion. Therefore -0001638 : -0000472 =- 60" : -D. .-. D = 60"xt\V^ = 17-3". Hence 9 '7858083 = L cos (520 22' - 17") = iLcos520 21'43", EXAMPLES. XLI. 1. Find sin 420 21' 30" tiaving given that sin 42^ 21' = -6736577 sin 42^22'= -6738727. 2. Find cos 470 38' 30" aaving given that cos 47^ 38' = -6738727 cos 470 39'= -6736577. 3. Find cos 210 27' 45" [laving given that cos 210 27'= -9307370 cos 210 28'= -9306306. L. T. B. 7 98 TLIGONOMETRY. XLL 4. Find the angle whose «ine is '6666666 having given that -6665325 = sin il^ 48' •6667493 = sin 410 49'. 5. Find the angle whose cosine is -3333333 having given that -3332584 = cos 70^ 32' •3335326 = cos 700 31'. 6. Find the angle whose cosine is -25 having given that -2498167 = cos 75^ 32' •2500984 = cos 750 31'. 7. Find Z sin 45016' 30" • having given that L sin 45^ 16' = 9-8514969 L sin 45017'= 9-8516220. 8. Find L tan 270 13' 45" having given that L tan 270 13' = 9*7112148 i tan 270 14'= 9-7115254. 9. Find X cot 360 18' 20" having given that L cot 36o 18' = 10-1339650 L cot 360 19/^ 10-1337003. 10. Find the angle whose Logarithmic tangent is 9'8464028 having given that 9-8463018=1. tan 35o 4' 9-8465705=Ltan350 5'. 11. Find the angle whose Logarithmic cosine is 9-9448230 having given that 9-9447802 = L cos 28o 17' 9-9448541 =Z cos 280 16'. 12. Find the angle whose Logarithmic cosecant is 10-4274623 having given that 10-4273638 = L cosec 210 57' 10-4276774 = 1, cosec 210 56'. 146. Problems in which each of the lines involvei contains an exact number of feet, and each angle an exac number of degrees, do not occur in practical work. As from time to time the skill of observers and of ir strum ent- makers has increased, so also has the number c significant figures by which observations have been recordec Thus the want was felt of some method by which th labour involved in the multiplication and division of Ion numerical quantities could be avoided. In the year 1614 Scotch mathematician, John Napier, Baron of Merchistor proposed his method of * Logarithms ' ; i.e. the method rejDresenting numbers by indices; * which, by reducing t a few days the labour of many months, doubles, as i were, the life of an astronomer, besides freeing him fror the errors and disgust inseparable from long calculations Laplace, ON THE USE OF LOGARITHMIC TABLES, 99 147. We shall now give a few examples of the practical use of logarithms. Example 1. The sides containing the right angle in a right- angled triangle ABC contain 3456*4 ft. and 4543*5 ft. respectively; find the angles of the triangle, and the length of the hypotenuse. Let a, 6, c be the lengths of the sides of the triangle opposite the angles A^ B, C respectively. See figure, p. 25. Then a = 3456*4 feet, & = 4543-6 feet. ^ , a 3456*4 *"^^ = 5=4543:5- In the Tables we find log 3456*4 = 3*5386240. log 4543*5 = 3-6573905. .-. log ^= log a -log 6. = 3*5386240 -3-6573905. .-. log tan ^ = 1-8812335. .*. L tan ^ = 9-8812335. In the Tables we find 9*8810522 = X tan 37n5'. 9*8813144 =i tan 37n6'. Whence we find by the Eule of Proportional Differences 9*8812335 =L tan 37^ 15' 42". .-. .1 = 37015' 42". ' Also B = (900 _ j)^ .«. 5 = 520 44' iq'^^ and - = cosec A = cosec 370 15' 42", a ' .'. log c = log a + log cosec 370 15' 42" = log a + 1, cosec 370 15'42"-10 = 3*5386240 + 10-2179174 -10 =3-7565414 =log 5708-8, ••. the hypotenuse contains 5708*8 feet. Thus we have found the angles and the third side of the triangle. 7—2 100 TRIGONOMETBY, 148. There are some formulse which are seldom used ii practical work, because they are not adapted to logarithm! calculation. They are those in which powers of quantitie are connected by the signs + or — . Example. In the above example we might have found the lengtl of the hypotenuse by means of the formula c^ = a^ + bK But we should have had to go through the process of calculatin: by multiplication the values of a'^ and b^. For this reason, a formula which consists entirely o factors is always preferred to one which consists of terms when any of those terms contain any power of the quantitie involved. If in the above example the lengths of the hypotenuse c and of on side a were given, then the formula U^^c^ - a^ = {c - a) (c + a) will giv the length of 6. For log b'^ = log { (c - a) (c + a)} , or, 21og6 = log(c-a)+log(c + a). And the values of (c + a) and (c — a) are easily written down fror the given values of c and a, EXAMPLES. XLII. In the following questions A,B, G are the angles of a right-angle triangle of which C is a right angle, and a, 6, c are the lengths of th sides opposite those angles respectively. 1. Given that a = 1046-7 yards, c = 1856-2 yards, a^W, find A, log 1046-7 = 3-0198222, log 1856-2 = 3-2686248, i sin 34019' = 9-7510991, i sin 340 20'= 9-7512842. 2. Given that a = 843-2 feet, C=90^ and A =34^ 15', find c, log 848-2 = 2-9259306, L cosec 34^ 15' = 10-2496421, log 1-4982 = -17557. 3. Given that a = 4845 yards, 6 = 4742 yards, and C = 90, find A. log 4845 = 3-6852938, log 4742 = 3-6759615, L tan 450 36' = 10 -0090965, L tan 46^ 37' = 10-0093492. 4. Given that c = 8762 feet, {7= 90, and ^ = 37^ 10', find a and b, log 8762 = 3-9426032, L sin 37^ 10' = 9 -7811344, L cos 370 10' = 9-9013938, log 5-2934 = -72373, log 6-9823 = -843997. 5. Given that b = 1694-2 chains, G = 90^, and ^ = I80 47', find a. log 1694-2 = 3-2289647, L cot 18^ 47' = 10-4683893, log 5-7620 = -76057. 6. Given that a = 1072 chains, c = 4849 chains, and G = 90^ find \ log 5921 = 3-7723951, log 3777 = 3-5771470, log 4-729 = -67477. PJRACTICAL EXAMPLES. 101 7. Given that 6=841 feet, c=3762 feet, and (7= W, find a. log 4603 = 3 -6630410, log 2921 = 3-4655316, log 3-6668= -56428. 8. Given that a = 7694-5 chains, 6 = 8471 chains, (7=90^ find A and c. log 7694-5 = 3-8861804, log 8471 = 3-9279347, L tan 420 15'= 995824, L cosec 420 15'= 10-1723937, log 11444 = -05857. MISCELLANEOUS EXAMPLES. XLIIL 1. A balloon is at a height of 2500 feet above a plain and its angle of elevation at a point in the plain is 40" 35'. How far is the balloon from the point of observation ? L cosec 40<> 35' = 10 -18672. 2. A tower standing on a horizontal plain subtends an angle of 37" 19' at a point in the plain distant 369-5 feet from the foot of the tower. Find the height of the tower. L tan 37" 19'=9-88210. 3. The shadow of a tower on a horizontal plain in the sunUght is observed to be 176-2 feet and the elevation of the sun at that moment is 330 12'. Find the height of the tower. L tan = 9-81583. 4. From the top of a tower 163-5 feet high by the side of a river the angle of depression of a post on the opposite bank of the river is 290 47'. Find tne distance of the post from the foot of the tower. L cot 390 47' = 10-67952. 5. Given a=673, 6 = 416 chains, C=900, find A and B. L tan 580 17' =10-20900. 6. Given a =576, c = 873 chains, C=900, find 6 and A. L sin 4P 17' = 9-81940, Lcos4P 17' =9 '87590. 7. From the top of a light-house 112-5 feet high, the angles of depression of two ships, when the line joining the ships points to the foot of the light-house, are 27" 18' and 20^ 36' respectively. Find the distance between the ships. L cot 27^18' = 10-28723, L cot 20^ 36' = 10-42496. 8. From the top of a cliff the angles of depression of the top and bottom of a light-house 97-25 feet high are observed to be 23" 17' and 240 19' respectively. How much higher is the cliff than the light-house ? Ltan23n7'=9-63379, L tan 24" 19' =6 -65501. 9. Find the distance in space travelled in an hour, in consequence of the earth's rotation, by St Paul's Cathedral. (Latitude of London = 51*25', earth's diameter = 7914 miles.) X cos 5P 25' = 9 -79494. 10. The angle of elevation of a balloon from a station due south of it is 47*18', and from another station due west of the former and distant 671 feet from it the elevation is 41*14'. Find the height of the balloon. cot 47* 18' = -92277, cot 41* 14' = 1-14095. CHAPTER XITL On the Relations between the Sides and Angles OF A Triangle. 149. The three sides and the three angles of any triangle, are called its six parts. By the letters A, B, C we shall indicate geometrically, the three angular points of the triangle ABC', algebraically, the three angles at those angular points re- spectively. A By the letters a, h, c we shall indicate the measures of the sides BC, CA, AB opposite the angles A, B, C respec- tively. 150. I. We know that, A + B + C= 180'. [Euc. i. 32.] 151. Also if A be an angle of a triangle, then A may have any value between 0' and 180'. Hence, (i) sin A must be positive (and less than 1), (ii) cos A may be positive or negative (but must be numerically less than 1), (iii) tan A may have any value whatever, positive or negative. 102* APPENDIX. In some Examinations, as for instance tliat of the L^iil stage, Mathematics, of the South Kensington Science and Art Department, Chapters ix. and x. of this book (the A, B ', S, T] and 2 A formulae) are not required. As, however, the student is required to solve Triangles by the aid of Logarithms he must use [see Arts. 158, 159, 161, 162] the two following propositions. The proofs here given are deduced from Euclid iii. Prop. I. To prove that cos ^ = 2 cos2 1 J: - 1 = 1 - 2 sin2 ^A. Let ROP be the angle A ; with as centre and any radius OR describe the semicircle RFL\ join PL^ PR, and draw PM perpendicular to LOR. j. Then POM^OLP+OPL^^WLP, I .-. OLP = iPOM=^A. LM-LO 2LM ^ OP iOP JNow, cos^ — yYp - OP = 2 OP ^. ^-1 = 2 cos OLP. cos OLP- :2cos2i^-l (i) -2(l-sin2l^)-l = l-2sin2iil (ii). Note, sin A MP = 2 . MP LP _^ MP LP 0P~'"ZP ' 20P" " ' LP ' LR = 2 sin OLP . cos OLP = 2 sin \A . cos I A, [Sea.,4rt 161.J 103* APPENDIX. Prop. II. To prov^ that in any triangle , B-C h-c A \yr tan — - — = , cot — . TC^ 2 6 + c 2 \ Let ABC be a triangle of which the angle B is greater than C, Make the angle BCD = B and produce BA to i>. In the triangle A CD inscribe the circle LMN, centre /, touching the sides in Z, M, N', join IL, IM, IN, I A, IC. Then ICM= i LCM =1{DCB- ACB) = l{B - C), IAM=^DAC =^(180'' - C AB) = {90° -^A), CM= CL = CD-Ln = BD-WD = BN= BA + AM-, .-. CM=l{CM+ BA + AM) = 1{AC + AB) = J(5 + c), and AM = AC - CM=h- l{h + c) = l{h - c). B-C Hence ^^^ 2 ttinlCM tan/Cif cot A 2 tan (90° - -\A) tan lAM = IM CM ^ im" AM AM CM~ 1(6 + c) h-c b + c' Q.E.D. < SIDES AND ANGLES OF A TRIANGLE, 103 152. Also, if we are given the value of (i) sin A, there are two angles, each less than 180**, which have the given positive value for their sine. (ii) cos A, or (iii) tan A, then there is only one value of A, which value can be found from the Tables. ^ 153. i + | + ^ = 90^ Therefore ^ is less than 90", A and its Trigonometrical Ratios are all. positive. Also, -^ is known, when the value of any one of its ratios is given. Similar remarks of course apply to the angles B and G, Exaviple 1. To prove sin (^ + J5)=sin (7. [Art. 96.] ^ + 5 + 0=1800; .-. A+B = 1S0^-G, and .-. sin (^ +i5) = sin (1800 -C) = sin C. [p. 61.] Example 2. To prove sin — ^ — = cos - . Now =90^ .*. = 90^ - H- » and .-. sin^— = sinr900 -^^=cos^. [Art. 94.] EXAIUPLES. XLIV. Find A from each of the six following equations, A being an angle of a triangle. 1. cos^ = J. 2. cos^=-J. 3. BmA = i. 4. tan^=-l. 5. j2 8mA=:h 6. tanA^-JS. Prove the following statements, A, Bj G being the angles of a triangle. 7. sin(^ + B + C) = 0. 8. cos(^+J5 + C)=-l. 9. Binj(^ + B + 0) = 1. 10. cosi{A + B + C)=0, 11, tan(^ + J5)=-tanC. 12. cot J (J5+.C) = tanp. 13. cos(^+^)=-cosC. 14. cos(^ + 5-C)=-cos2a 15. tan A-cotB=cos G . sec A . cosec B, - - sin ^ - sin P , G . A-B 16. -^ — : : — ^ = tan — . tan — ^r— . "■ 6m^4-sm5 2 2 sin 3 B -sin 3(7 _^ 3jl ^^' cos3C-cos3^~*^'' 2 • 104 TRIGONOMETRY, 154. II. To prove a = ^JJosjC^ 4- c cos^S^ From A, any one of the angular points, draw AD per- pendicular to BO, or to £C produced if necessary. There will be three cases. Fig. i. when both B and are acute angles ; Fig. ii. when one of them (^) is obtuse ; Fig. iii. when one of them (B) is a right angle. Then, CD rig. i. 7TT= ^os ^^^ i ^^' CD = b cos 0, T)T> and —7-j- = cos A BD ; or, DB = c cos B, AB ,\ a= CD + DB = b cos C + c cos B. CD Fig. ii. -p^ = cos ACD ; or, CD = h cos C', = cos ABD ; or, BD = c cos (1 80' - i?), .-. a=CD-BD = hcosC-ccoB{180'-B) = h cos (7 4- c cos ^. Fig. iii. a= CB = b cos C = 6 cos (7 + c cos B. [For, cos ^ = cos 90' = 0.] Similarly it may be proved that, h = c cos A + a cos C ', c = a cos B + h cos ^. /* £2^ ^l SIDES AND ANGLES OF A TRIANGLE. 105 /^ 155. in. To prove that in any triangle, the sides are ' proportional to the sines of the opposite angles ; or, 2^o prove 7 ^ that sin A sin B sin (7 ' From u4, any one of the angular points, draw AD per- pendicular to BCy or to £0 produced if necessary. Then, A D Fig. i. AI) = bsmC', for, -j^ = sin C [Def.] ; also AD = csinB ; for, —-p- = sin D. ^ AB ,•. b sin(7 = c sini?; b c or, - — - = -.— ^. sm^ sinO Pig. 11. ^D = 5 sin (7, and AD=c sin ABD = c sin (180" -i?). .•. AD = csLQ j?; .'. b sin C = c sin B ; 6 ^ c ' sinjS ~sin(7' Fig. ill. AB = AC .sin C ; or, c = bsinG ; .-. -T^ = -J-^ . [For sin ^ = sin 90" = 1.1 smG suiij '- -^ Similarly it may be proved that a b sin A sinB' a b c Bin A sm B sm (7 / 106 )J^ r TRIGONOMETRY. \ y 1 56. IV. To prove that a^ = V + c^- 2bc cos A, / Take one of the angles A, Then of the other two, one must be acute. Let £ be an acute angle. From C draw OF perpendicular to £A, or to BA produced if necessary. ^ There will be three figures according as A is less, greater k than, or equal to a right angle. Then, I h If [Euc. II. 13] ^ Fig. i. J]C' = CA' + AB' ~2.BA,FA; or, a' = b' + c'- 2c, FA i- = h' + c'- 2cb cos A. [For FA = h. cos A.]^^ Fig. ii. BC =CA' + AB'+2.BA.AF; [Euc. ii. 12] Ci or, a^ = b^-hc^+ 2cb cos FAC = b' + c'-2bccosA, [For FAC = ISO' -A.] /' Fig. iii. BC = CA' + AB' ; [Euc. i. 47] "" or, a' = b^' + c^- 2bc cos A. [For cos ^ = cos 90^ = 0.] Similarly it may be proved that b^ = c^ + a^- 2ca cos B, and that c^ = a^ 4- 5^ — 2ab cos C. v; 157. V. Hence, cos^ = ;^158 5^ + c^-a^ 25c ~ VI. cos^ = c' + a'-b' 2ca cos (7== a'-^b'-c' ^ ^ :7^o prove that sin^ — = ^^ ^-^ -^ Since cos A = l b' + c 2 sm^ - . 2 ' 46c [Art. 109] and cos A = 2bc [Art. 157] ^j^ SIDES AND ANGLES OF A TRIANGLE. 107 _ 2bc-{b' + c''-a') ^ a' -{b' -210 + 6') 2bc " 2bc _ a^-{h- cf ^ {a-{b-c)}{a + {b-c) ] 2hc 2bc . ^A (a + c-b)(a-hb'-c) .-. sm^-=^ it "• ^•^•''• iro rr J.^ . 2^ {a + b + c){ b + c-a) 159. To prove that cos 7: = ,7 -. ^ ^ 2 46c Since cos ^ = 2 cos' ~ - 1 j [Art. 109] .•• 2 cos^ -=l+cos^ = l + ^j^ — ; [Art. 157] ^A (b + cY -a^ (b + c + a)(b + c-a) ••• ^^^'2 = —ibv- = Wo • ^•"•^• S< ICO. VIL Now let s stand for ^"^^"^^ ^ so that (a + b + c) = 2s. Then, {b + c-a) = (b + c-\-a-2a) = {2s'-2a) = 2{s- a), and (c + a - 6) = (c + a + 6 - 26) = (2s - 26) = 2 (5 - 6), and (a + b-c) = {a + b + c-2c)= (2s - 2c) = 2 (5 - c). Then the result of Arts. 158, 159 may be written . ,A 2(s-6)2(s-c) . A /(s-b){s-c) sin^ — = — ^^ i^ ^ : or, sni — = / ^^ ^-^ ^ '2 ~V " 2 Wo ^ "^' ^^^^2'-V fc ' J 2 ^ 2s 2 (s — a) A Is is - a) and cos^ - = -V ^ : or, cos - = . / -^ ^, 2 46c ^ ' 2 V 6c and so on. . A rr . , A ''''2 V(s-6)(s-c) Hence, tan -^ = —7 = ^ ^7-^ -^ -^ . ' 2 A Js(s-a) 108^ ' TBIGONOMETBY. /■ Example. Write down the corresponding formulas for / Sin - , for cos — , and for tan - . IGl. VIII. Again, / ^ ^, A A sin A = 2 sin— . cos y ; [Art. 109] V be \l ho jr ,1 2 / .■ '^. ''^^ The letter aS' usually stands for Js {s -a)(s — h) (s - c), so that the above may be written = -y- . ' a ri ^. ., , sin^ 2S sin (7 Similarly, _^=_=__. ^ 162. IX. To prove that . . cot — =tan — ^ — • ^ ^ h+c 2 2 Since — — ^ = -; — -^ , let each of these fractions = d. sm B sm C7 Then 6 = (/ sin B, and c = cZ sin (7. h — c_d sin B - d sin (7 sin ^ - sin (7 ' ' 6 + c d sin i^ + c^ sin (7 sin B + sin (7 ^ . ^-6' ^+(7 ^ B-G 2 sm — ^r — . cos — -^ — tan — ^r — ^ . B+G B-G~ , B+G 2 sm — - — . cos — H — tan — ^ — tan 2 B-G ^A cot -2 , B-G tan- Tsince tan ^^ = tan ^90'' - ^ V h-c A 2 A ^ B-G hVc' "^^-2 = T"- cot ^ =tan-^. q.e.d. cot-^ SIDES AND ANGLES OF A TRIANGLE. 109 Similarly, c-a B ^ G-A a-h C ^ A-B . cot ■jr= tan — ^r — , ^ . cot ^ = tan — ^r— . . c+a 2 2 ' a+h 2 2 1 63. The student is advised to make himself thoroughly familiar with the following formulae : a = h cos C + c cos B (ii), isrj~is:§"sr(7'-='^J = 2^ ^'''^' cos^ = — - — - (V), ^ M. ^ /sis -a) I '¥ = V-6r- J cos- ^in A = J- J s {s - a) {s — h) {s - c) = ^ (viii), 5_(7 5-c A *^^-2-=rrc-'"*2 • " ^^^)- EXAMPLES. XLV. In any triangle ABC prove the following statements: sin ^ + 2 sin B _ sin G _ sin^ A-m. sin^ B _ sin^ C 3, a coaA + hcoQB-c cos (7= 2c cos ^ . cosJB. 4. (a + &) sm - = c cos . 5. (&-c) cos — = asin — ^. 6. asin(J5-C) + 6sin(C-^) + csin(^-J!5) = 0. _ a-& _cosB-cos^ & + C _cos5 + cos (7 c ~ 1 + cos C * a ~ 1 - cos A ft /; — '—^ — -—n ^^ sin (7 + c2 sin P 6 + c 10. a + h + c = {h + c)co8 A + {c + a) cos B + {a + h) cos (7. 11. h + c-a=(b + c) co3A-{c-a) cos B + {a — b) cos G, -rt . . a sin (7 .« tanB a2 + Z>2_c*^ 12. tan^ = - — . 13. - — -= » .o . o » t - a cos (7 tan C7 a^ - Z>2 + c^ no TRIGONOMETRY. 14. a (Z>2 + c2) cos ^ + 6 (c2 + a2) coa B + c (a^ + S^) cos 0= Sahe. 15. a cos (it + J5 + C) - & cos (E + ^ ) - c cos {A + C) = 0. _ - cos A cos B cos C a^ + &2 + ^2 lb. h — 5 V = 7z—. • a c zauc 17. h cos^ - + c cos^ - = «. 18. tan 77 . tan - 2 ' 2 • *"• " 2 2 t + c + a* 19. tan -(6 +c-a) = tan- (c + a- 6). 20. c2=(a + &)2sin2^ + (a-6)2cos2?. MISCELLANEOUS EXAMPLES. XLVL 1. Simplify the formul89 in the case of an equilateral triangle. 2. The sides of a triangle are as 2 : ^6 : 1 4-/^/3, find the angles. 3. The sides of a triangle are as 4, 2;^2, 2(^3-1), find the angles. 4. Given O=1200, 0=^19, a = 2, find &. 5. Given A = W, h = 4:Jl,c = ^ ^7, find a. 6. Given A = 45^, B = 60^ and a = 2, find c. 7. The sides of a triangle are as 7 : 8 : 13, find the greatest angle. 8. The sides of a triangle are 1, 2, ^7, find the greatest angle. 9. The sides of a triangle are as a : & : ^(a^ + ab + b'^), find the greatest angle. 10. When a : & : c as 3 : 4 : 5, find the greatest and least angles ; given cos 36^ 52' = '8. 11. If a = 5 miles, 6=6 miles, c = 10 miles, find the greatest angle, [cos 490 33' =-65.] 12. Ifa = 4, & = 5, c = 8, find C; given that cos 54^54' = -575. 13. a : b~J^ : 1, and 0=30^ find the other angles. 14. Given C = 18^, a = ^/S + 1, c = ;y/5 - 1, find the other angles. 15. If & = 3, C=120o, c=J13, find a and the sines of the other angles. 16. Given A = 1050, B = 45^, c =^2, solve the triangle. 17. Given B = 75\ C=300, c=^8, solve the triangle. 18. Given J5 = 45", c=;^75, 6=^50, solve the triangle. ON THE SOLUTION OF TRIANGLES . Ill 19. Given ^ = 300, c = 150, 6 = 50^^3, show that of the two triangles which satisfy the data one will be isosceles and the other light-angled. Find the third side in the greatest of these triangles. 20. Are there two triangles in which B = 30<>, c = 150, & = 75 ? 21. If the angles adjacent to the base of a triangle are 224<> and 112^^, show that the perpendicular altitude will be half the base. 22. If a = 2, 6 = 4 - 2 ^3, c= ^6 (^^3 - 1), solve the triangle. 23. liA=9^,B = 450, b = V6, find c. 24. Given B = 15^, 6 =;^3 - 1, c =;^/3 + 1, solve the triangle. 25. Given sin^='25, a=5, 6 = 2*5, find A, Draw a figure to explain the result. 26. Given C=15^ c=4, a = 4+/^/48, solve the triangle. 27. Two sides of a triangle are 3 JQ yards and 3 {^S + 1) yards, and the included angle 45^, solve the triangle. 28. If C = 30^ 6 = 100, c = 45, is the triangle ambiguous ? 29. Prove that if ^ = 450 and B = GO^ then 2c = a (1 + ^3) . 30. The cosines of two of the angles of a triangle are J and |, find the ratio of the sides. CHAPTER XIY. On the Solution op Triangles. 164. The problem known as the Solution of Triangles may be stated thus : When a sufficient number of the i^arts of a triangle are given, to find the magnitude of each of the other parts, o 165. "When three parts of a Triangle (one of which must be a side) are given, the other parts can in general be determined. There are four cases. ^^^r Given tliree sides. [Compare Euc. i. 8.] 11. Given one side and two angles. [Euc. l 2Q,] - m. Given two sides and the angle between them. [Euc. I. 4.] IV. Given two sides and the angle opposite one of them. * [Compare Euc. VL 7.] 112 TRIGONOMETRY. Case L 166. Given three sides, a, h, c, [Euc. i. 8; vi. 5.] We find two of the angles from the formulse tai4= /5HMZZ) 2 V s{s-a) tan|= / {^-c)(s-a) 2 V s{s-b) The third angle C=180-^-i?. 167. In practical work we proceed as follows : logtan4 = log /(EMZI); ^ V 8{s-a) or, X tan — ~ 1 = J {log (s - 5) + log (s - c) - log s - log (5 - a)}. Similarly, Z tan — - 10 = J {log {s-c) + log {s-a)- log s - log {s ~ 5)}. 168. Either of the formulse sin 4 = / (^' " ^ ) (^ - ^) 2 V 5c cos 2 = / ?J^ ^ may also be used as above. V he ^ . A A The sin -^ and the cos -^ formulse are either of them as A convenient as the tan — formulse, when one of the angles only is to be found. If all the angles are to be found the tangent formula is convenient, because we can find the L tangents of two half angles from the same/oi^r logs, viz. log s, log {s — a\ log (s - h\ log (s - c). To find the L sines of two half angles we require the six logarithms, viz. log (s-a), log {s — b), log {s - c), log a, log b, log c. ON THE SOLUTION OF TRIANGLES. 113 Example. Given a = 275 -35,. 6 = 189-28, c = 301-47 chains, find A andB. p-yc ^^. 10 n U '*-< .c^ A \'¥^(^-:°)' - ^ ^ L" - ^^^ u %) , i;v ■^ ■r» •{., c^ .. .1 >;L.-^w J7 ^ ^0, 112 ■^tt-€.. TniGONOMETET. L /. y/// 1 \ / (;. -ViV (?,^/V^ ?6- - :- . //^" ^.^^id /./76/ •tl i 'V) , 1 ■" '^..S-J«^i - - 1 if . ^ r 1 r^-n t 1 V — y ^?a-^^ \y ' 'X^ ^..^UU-^ ON THE SOLUTION OF TRIANGLES. 113 Example. Given a = 275 -35,. 6 = 189-28, c = 301-47 chains, find A andB. Here, s=383-05, «-a = 107-70, «-6 = 193-77, a-c = 81-58. Then X tan ^ = 10 + i {log 193-77 + log 81-58 - log 383-05 - log 107-70} = 10 + i { 2-2872865 + 1 -9115837 - 2-5832555 - 2 -0322157 } = 9-7916995 [from the Tables], whence ~ = 310 45' 28-5"; .% A = 630 30' 57". Again, L tan^ = 10 + J {log 81-58 + log 107-70 -log 383 05 -log 193-77} = 9-5366287 =L tan 18^ 59' 9-8" ; .•. :B = 370 58'20"; (7=180<>-^--B = 780 30'43". 169. This Case may also be solved bj the formula h' + c'-a' cos A = —J . 26c But this formula is not adapted for logarithmic calculation, and therefore is seldom used in practice. It may sometimes be used with advantage, when the given lengths of a, b, c each contain less than three digits. Example. Find the greatest angle of the triangle whose sides are 13, 14, 15. Let a =15, 6 = 14, c = 13. Then the greatest angle is A. 142 + 132-152 _ ~ 2x14x13 ~2x =cos67<> 23', nearly. „ . 142 + 132-152 140 . „,.,. Now, cos A = —^ — —, — -r— = ,r — zr-. — ^ = A = '384615 ' 2x14x13 2x14x13 ^^ [By the Table of natural cosines. ] .•. the greatest angle =67<* 23'. yi EXAMPLES. XLVIL 1. If a = 352-25, 6 = 513-27, c=482-68 yards, find the angle J, having given log 674-10=2-8287243, log 321-85 = 2-5076535, log 160-83 = 2-2063401, log 191-42 = 2-2819873, L tan 200 38'=9-5758104, L tan 200 39' = 9*5761934. L. T. B. 8 t. 114 TBIGONOMETRY, 2. Find the two largest angles of the triangle whose sides are 484, 370, 522 chains, having given that log 6-91 = -8394780, log 3-15 = -4983106, log 2-07 = -3159703, log 1-69 = -2278867, L tan 36H6' 6'' = 9-8734581, L tan 3P 23' 9"= 9-7853745. 3. If a=5238, 6 = 5662, c = 9384 yards, find the angles A and B, having given log 1-0142= -0061236, log 4-904= 6905505, log 4-48= -6512780, log 7 -58 = -8796692, L tan 140 33' = 9-4168099, L tan 150 57' = 9-4500641, L tan 140 39' = 9-4173265, L tan 15^ 58' = 9-4565420. 4. If a = 4090, 6 = 3850, c = 3811 yards, find A, having given log 5-8755= -7600448, log 3-85 = -5854607, log 1-7855 = -2517599, log 3-811 = -5810389, L cos 320 15' = 9 -9272306, L cos 320 16' = 9-9271509. 5. Find the greatest angle in a triangle whose sides are 7 feet, 8 feet, and 9 feet, having given log 3 = -4771213, L cos 360 42' = 9-9040529, log 1-4 = -146128, diff. for 60"= -0000942. 6. Find the smallest angle of the triangle whose sides are 8 feet, 10 feet, and 12 feet, having given that lo^ 2 = -30103, X sin 200 42' = 9-5483585, diff. for 60" = -0003342. 7. If a : 6 : c=4 : 5 : 6, find C, having given log 2 = -3010300, log 3 = -4771213, X cos 410 25' =9-8750142, diff. for 60"= -0001115. 8. The sides of a triangle are 2, Ay6, and 1 + \/3> ^^^ t^® angles. 9. The sides of a triangle are 2, \/2, and \J^ - 1, find the angles. Case n. 170. Given one side and two angles, as a, B, C. [Euc. I. 26 ; YI. 4.] First, ^ = 180'' - ^ - (7 ; which determines A, h a . a . sin ^ ^^^*> :A:rD = 7^:n » or, 6 = and, sin ^ sin ^ ' ' sin J. ' c a a. sin C sin (7 sin J. ' ' sin A These determine h and c. ON THE SOLUTION OF TRIANGLES. 115 171. In practical work we proceed as follows : a . sin B Since b - sin A ' a . sin B • *. W h = loof "" "^ smA .*. log b = loga + log (sin ^) + 10 - (10 + log sin A)^ or, log b = log a + LsinB -L sin 4- Similarly, log c = log a 4- Z sin G — L sin 4. Example. Given that c = 1764-3 feet, (7=180 27', and B= 66^39', find ft. From the Tables we find log 1764-3 = 3-2465724. L sin 18« 27' = 9-5003421, L sin 66^ 39' = 9-9628904 ; .-. log h = 3-2465724 + 9-9628904 - 9-5003421 = 3.-7091207 = log 5118-2 ; ' .-. & = 5118-2 feet. ^i- - ^ ' EXAMPLES. XLVm. 1. If ^ =53^24', 5 = 66^27', c = 338'65 yards, find C and a, having given that L sin 53024'=9-9046168, log 3-3865 = -5297511, i sin 600 9'=9-9381851, log 3-1346 = -4961821, log 3-1347= -4961960. 2. If ^ = 480, B = 540, and c = 38 inches, find a and &, having given that log 38 = 1-5797836, log 2-88704= -4604527, log 3-14295 = -4973368, L sin 54^ = 9-9079576, L sin 780 ^ 9 -9904044, L sin 480 = 9-8710735. 3. Find c, having given that a = 1000 yards, ^ = 500, 0=660, and that L sin 500=9-8842540, L sin 660=9-9607302, log 1-19255= -0764762. 4. Find &, having given that 5 =320 15', (7=21047'20", a = 34feet. log 3-4 = -531479, L sin 320 15' = 9-727228, log 2-241 = -350442, Lsin540 2'=9-908141, log 2-242= -350636, I,sin640 3'=9-908233. 5. Find a, &, C, having given ^ = 7204', ^=41056'18", c = 24 feet. log 2-4 =-3802112, Lsin7204' =9-9783702, logl-755 =-2442771, L sin 41056'10"= 9-8249725, log 1-756 =-2445245, L sin 410 56'20"= 9-8249959, log 2-4995 = -3978531, i sin 650 59' =9-9606739, log 2 -4996 = -3978701, Lsin66o =9-9607302. 8—2 116 TRIGONOMETRY. Case III. 172. Given two sides and the included angle, as hyC,A. [Euc. I. 4; VI. 6.] First, i5 + C = 180° - ^. Thus (B + C) is determined. Next, tan — ^r — = ^ cot -x- . ' 2 6 + c 2 Thus {B—C) is determined. And i? and C can be found when the values of (B + C) and {B - C) are known. _ ., a h 6. sin .4 Lastly, - — J = -; — =T, or a = — ; — jj- . •" sin ^ sm i^ sm i^ Whence a is determined. 173. In practical work we proceed as follows : B^G h-c A Since tan — ^r— = ^ cot -^ , 2 h + c 2 .-. log(tan^) + 10 = log(6-c)-log(& + c) + logf cot-^ j + 10, B — G A or, L tan — ^r— = log {& - c) - log (6 + c) + Z cot — . . , . 5 . sin ^ Also, smce a = — ; — 77— » sinjB .'. log a = log 5 + Z sin ^ — Z sin ^, as in Case II. Example. Given 6 = 456*12 chains, c = 296-86 chains, and^ = 74020', find the other angles. Here, 6 -c = 159*26, 6 + c= 752-98. From the Table we find log 169-26 = 2-2021067, and log 752-98 = 2-8767834, i cot 370 10' = 10-1202593; ... L tan ?^=: 2-2021067 -2-8767834 + 10-1202593 = 9-4455826 = L tan 15^ 35' 18". .% B-C=3P10'36", and5 + C=1800-74020'. Thus ^ + 0=105040'; A 2B = 136050'36"; 2C7=74029'24", or, i? = 68025'18"; or, C=37n4'42". ON THE SOLUTION OF TRIANGLES, 117 174. The formula a^ = 6V c^ - 2hc cos A may be used in simple cases. Example. If 6=35 feet, c = 21 feet, and^ = 500, fin^ a, given that CcULj2 ^32C . A ^ -t hi"- ^6 ^ •- . »- -^T, / . ^/^-/i^.v/^ -^ g o. V*- ^7 /<^6 - '^'Z :.'Kc- ^ /'-if .^v., '"fe^ /^ -^ .^ fi/0^- 116 TRIGONOMETRY. Case III. 172. Given two sides and the included angle, as 5, c, A. [Euc. I. 4; YI. 6.] ON THE SOLUTION OF TRIANGLES. 117 1 74. The formula a' = b^ + c'~ 2bc cos A may be used in simple cases. Example, If 6=35 feet, c=21 feet, and-4=60<^, find a, given that cos 500 = -643. Here a^=S5^ + 2P-2 x 35 x 21 x cos500; .-. ^2 = 5'' + 32 - 2 X 5 X 3 X cos SO®, =25 + 9- 30 X -643, =14-71. 1 = 3-82 nearly; or, a=26-74 = about 26| feet. EXAMPLES. XLIX. 1. Find B and C, having given that ^ = 40^, 6 = 131, c = 72j log 5-9 =-7708520, Xcot200 =10-4389341, log 2-03 = -3074960, Ltan38036'= 9-9021604, i tan 380 37'= 9-9024195. 2. Find A and B, having given that a = 35 feet, 6 =21 feet, C= 500. log 2= -301030, i tan 28nr =9-729020, L tan 650= 10-331327, l tan 280 12'= 9*72923. 3. If 6 = 19 chains, c = 20 chains, ^1 = 600, find B and C, having given that log 3-9= -591065, L tan 20 32' =8-645853, i cot 300=10-238561, L tan 20 33' = 8-648704. 4. Giventhata=376-375chains,6 = 251-765chains,and(7=78026', find ^ and ^. L cot 390 13' = 10-0882755, log 1-2461 = -0955529, Ltan 13039'=: 9*3853370, log 6-2814 = -7980565, Xtanl30 40'= 9*3858876. 5. If a = 135, 6 = 105, G = 600, find A , having given that log 2 = -3010300, Ltanl2012'=9*3348711, log 3 = -4771213, itan 12013^=9*3354823. 6. If a = 21 chains, 6 = 20 chains, C=60o, find c. 7. Find c in the triangle of Example 5. 8. In a triangle the ratio of two sides is 5 : 3 and the included angle is 760 30'. Find the other angles. log 2 = -3010300, L cot 350 15' = 10-1507464, X tan 19028' 50"= 9-5486864. 118 TBIGONOMETRY. Case IV. 175. Given two sides and the angle opposite one of them, as b, c, B. [Omitted in Euc. I. ; Euc. YI. 7.] T-,. . . c h . ^ csinB ± irst, since -: — -^ = - — ^ ; .'. sm (7 = — ; — . sm G smB b C must be found from this equation. When C is known, A = lSO'-B-C, - h sin A and, a = —. — ^ , Which solves the triangle. 176. We remark, however, that the angle C, found from the trigonometrical equation sinC = a given quantity^ where G is an angle of a triangle, has two values, one less than 90°, and one greater than 90'. [Art. 152.] The question arises, Are both these values admissible? This may be decided as follows : If B is not less than 90°, C must be less than 90°; and the smaller value for G only is admissible. If B is less than 90° we proceed thus. 1. If 6 is less than c sin B^ then sin G, which = — ^ — , is greater than 1. This is impossible. Therefore if 6 is less than c sin B, there is no solution whatever. 2. If b is equal to c sin^, then sin (7= 1, and therefore (7= 90°; and there is only one value of (7, viz. 90°. 3. If b is greater than c sin B, and less than c, then B is less than G, and G may be obtuse or acute. In this case G may have either of the values found from the equa- tion sin G = — ^ — . Hence there are two solutions, and the triangle is said to be ambiguous. 4. If b is equal to or greater than c, then B is equal to or greater than (7, so that G must be an acute angle ; and the smaller value for G only is admissible. ON THE SOLUTION OF TBI ANGLES, 119 177. The same results may be obtained geometri- cally. Construction. Draw AB = c ; make the angle ABD - the given angle JB \ with centre A and radius = h describe a circle ; draw AD perpendicular to BD, f'g-ll-y] k C^r h >^ V. D „^''' "% Then ^i) = c sin ^. 1. If & is less than csin-S, i.e. less than AB^ the circle will not cut BB at all, and the construction fails. (Fig. i.) 2. If h is equal to AB^ the circle will touch the line BB in the point D, and the required triangle is the right-angled triangle ABB. (Fig. ii.) 3. If h is greater than AB and less than AB, i.e. than c, the circle will cut the line BB in two points Cj , C^ each on the same side of B. And we get two triangles ABG^y ABC 2 each satisfying the given condition. (Fig. iii.) 4. If h is equal to c, the circle cuts BB in B and in one other point C ; if 6 is greater than c the circle cuts BB in two points, but on opposite sides of B. In either case there is only one triangle satisfying the given condition. (Fig. iv.) 120 TRIGONOMETRY, 178. We may also obtain the same results algebrai- cally, from the formula 6^ = c* + a^ - 2c a cos £. In this b, c, £ are given, a is unknown. Write x for a and we get the cLuadratic ec[uation x^ - 2c cos B,x = h^- c\ Whence, ic^ - 2c cos ^ . cc + c^ cos' ^ = 6' - c' + c^ cos^ B .-. x = ccosB^Jb'-'c'sm'B. Let a^ , a^ be the two values of x thus obtained, then «j = c cos j5 + ^6^-c'sin^i5| ^2 = c cos ^ - Jb^ - c' sin^ B) * Which of these two solutions is admissible may be decided as follows : 1. When b is less than csin^, then (5^ - c' sin^ j5) is negative, so that a^ , a^ are impossible quantities. 2. When b is equal to c sin ^, then (b^ - c^ sin" B) = 0, and a^ = a^ ; thus the two solutions become one. 3. When b is greater than c sin j5, then the two values ttj, a^ are different and positive unless Jb'' — c^ sin^ i? is > c cos i?, ' i. e. unless b^ - c' sin' B > c' cos' B, i. e. unless 6' > c'. 4. When 6 is equal to c, then a^ = 0; if 6 is greater than c, then (X^ is negative and is therefore inadmissible. In either of these cases a^ is the only available solution. 179. We give two examples. In the first there are two solutions, in the second there is only one. Example 1. Find A and C, having given that 6 = 379*41 chains, c=. 483-74 chains, and j5 = 34nr. L sin C=logc + Lsin J5-log6 = 2-6846120 + 9- 7496148 - 2-5791088 = 9-8551180 = 1/ sin 45^ 45'; .-. C=45H5', or, 1800- 45045' =134015'. Since h is less than c, each of these values is admissible. When C7= 45045', then ^ = lOOO 4'. When C = 1340 15', then ^ = lio 34'. ON THE SOLUTION OF TRIANGLES. 121 Example 2. Find A and C, when 6 = 483*74 chains, c =379-14 chains, and 5 = 340 11'. isin C =log c + L sin B -log b = 2-5791088 + 9-7496148 - 2-6846120 = 9-6441116 = L sin 2609'; .-. C = 2609', or, ISO^- 26^9' =1530 51'. Since 6 is greater than c, C must be less than 90®, and the larger value for G is inadmissible. [It is also clear that (153051' + 34m') is > 1800]. .-. C = 260 9', .1 = 119040'. EXAMPLES. L. 1. If J? = 400, J = 140-5 feet, a = 170-6 feet, find A and G. log 1 -405 = -1476763, L sin 400 = 9*8080675, log 1-706= -2319790, L sin 510 18' = 9-8923342, i sin 510 19' =9-8924354. 2. Find B and (7, having given that A - 50o, b - 119 chains, a= 97 chains, and that log 1-19 = -075547, isin 500 =9-884254, log 9-7 =-986772, L sin 700 =9.972986, L sin 700 r = 9-973032. 3. Find B, C, and c, having given that A = 50o, 6 = 97, a = 119 (see Example 2). log 1-553= -191169, i sin 380 38' 24" = 9-795479, L sin 880 37' 24" = 9-999876. 4. Find A, having given that a = 24, c = 25, C=650 59', and that log 2-5 = -3979400, L sin'650 59' = 9-9606739, log 2-4= -3802112, L sin OP 16' = 9-9429335, i sin 610 17' = 9-9430028. 5. If a =25, c = 24, and 0= 650 59', find A, B and the greater value of b. log 1-755 = -2442771, L sin 720 4' = 9-9783702, log 1-756 = -2445245, L sin 720 5' = 9-9784111, L sin 410 56' 10" = 9-8249725, L sin 410 56' 26" = 9-8249959 (see Example 4.) 6. Supposing the data for the solution of a triangle to be as in the three following cases (a), (j8), (7), point out whether the solution will be ambiguous or not, and find the third side in the obtuse-angled triangle in the ambiguous case : (a) ^ = 300, a = 125 feet, c = 250 feet, (/3) ^ = 300, a =200 feet, c = 250 feet, (7) ^ = 300, a = 200 feet, c = 125 feet. log 2 = -3010300, L sin 380 41' = 9-7958800, log 6-0389 = -7809578, Xsin 80 41' = 9-1789001, log 6-0390= -7809050. 122 TRIGONOMETRY. 1 80. In the following Examples the student must find the necessary logarithms etc. from the Tables. MISCELLANEOUS EXAMPLES. LL 1. Find ^ when a = 374-5, 6 = 676-2, c = 759 '3 feet. 2. Find B when a=4001, 6 = 9760, c = 7942 yards. 3. Find G when a= 8761-2, 6 = 7643, c = 4693-8 chains. 4. Find B when ^ = 86o 19 ', 6 = 4930, c = 5471 chains. 5. Find G when 5 = 320 58', c = 1873-5, a = 764-2 chains. 6. Find c when G = 108^ 27', a = 36541, b = 89170 feet. 7. Find c when jB = 740 10', ^=62^ 45', 6 = 3720 yards. 8. Find 6 when B = 100^ 19', (7=440 59', a=1000 chains. 9. Find a when B = 123^ 7' 20", C= 15^ 9', c = 9964 yards. Find the other two angles in the six following triangles. 10. G= 1000 37 ', 6 = 1450, c = 6374 chains. 11. (7=52010', 6 = 643, c = 872 chains. 12. ^ = 7602' 30", 6 = 1000, a =2000 chains. 13. C=54023', 6 = 873-4, c = 75 2 -8 feet. 14. C7=180 21', 6 = 674-5, c = 269-7 chains. 15. A = 290 11' 43", 6 = 7934, a = 4379 feet. 16. The difference between the angles at the base of a triangle is 17^48', and the sides subtending those angles are 105-25 feet and 76-75 feet ; find the third angle. 17. If 6 : c = 4 : 5, a = 1000 yards and A = B7^ 19', find 6. The student will find some Examples of Solution of Triangles without the aid of logarithms, in an Appendix. CHAPTER XY. On the Measurement of Heights and Distances. 181. We have said (Art. 58) that the measurement, with scientific accuracy, of a line of any considerable length involves a long and difficult process. On the other hand, sometimes it is required to find the direction of a line that it may point to an object which is not visible from the point from which the line is drawn. As, for example, when a tunnel has to be constructed. MEASUREMENTS OF HEIGHTS AND DISTANCES. 123 By the aid of the Solution of Triangles we can find the length of the distance between points which are inaccessible ; we can calculate the magnitude of angles which cannot be practically observed; we can find the relative heights of distant and inaccessible points. The method on which the Trigonometrical Survey of a country is conducted affords the following illustration. 182. To find the distance between two distant objects. ^^A /^<\ B Two convenient positions A and B, on a level plain as far apart as possible, having been selected, the distance between A and B is measured with the greatest possible care. This line AB is called the base line. (In the survey of England, the base line is on Salisbury Plain, and is about 36,578 feet long.) Next, the two distant objects, P and Q (church spires, for instance), visible from A and B, are chosen. The angles PAB, PBA are observed. Then by Case II. Chapter xiv, the lengths of the lines PA, PB are cal- culated. Again, the angles QAB, QBA are observed; and by Case II. the lengths of QA and QB are calculated. Thus the lengths of PA and QA are found. The angle PAQ is observed; and then by Case III. the length of PQ is calculated. 124 TRIGONOMETRY. 183. Thus the distance between two points P and Q lias been found. The points F and Q are not necessarily accessible; the only condition being that F and Q must be visible from both A and F, 184. In practice, the points F and Q will generally be accessible, and then the line FQ, whose length has been cal- culated, may be used as a new base to find other distances. 185. To find the height of a distant object above the point of observation. P M Let B be the point of observation; F the distant object. From B measure a base line BA of any convenient length, in any convenient direction; observe the angles FAB, FBA, and by Case II. calculate the length of BF, Next observe at B the * angle of elevation ' of P; that is, the angle which the line BF makes with the horizontal line BM, J/ being the point in which the vertical line through F cuts the horizontal plane through B. Then FM, which is the vertical height of F above B can be calculated, for FM= BF . sin MBF, Example 1. The distance hetiveen a church spire A and a milestone B is known to he llQiS feet ; C is a distant spire. 'The angle CAB is 94^ 54', and the angle CBA is 66^ 39'. Find the distance of G from A. ABC is a triangle, and we know one side c and two angles (A and R), and therefore it can be solved by Case II. The angle ACB = ISO^ - 94^ 54' - 660 39' = 180 27'. Therefore the triangle is the same as that solved on page 115. There- fore .40=5118-2 feet. PRACTICAL EXAMPLES. 125 Example 2. If the spire C in the last Example stands on a hill, and the angle of elevation of its highest point is observed at A to be 49 19'; find how much higher is than A. The required height x = AG, Bin i^ 19' and ^C is 6118-2 feet, .-. loga; = log(^C.8in4n9') = log 5118-2 + L sin 4n9' - 10 t = 3-7091173 + 8-8766150 - 10 = 2-5857323 = log 385-24. Therefore a; =385 ft. 3 in. nearly. EXAMPLES. LH. (Exercises x. and xliii. consist of easy Examples on this subject.) 1. Two straight roads inclined to one another at an angle of 60^, lead from a town A to two villages B and ; jB on one road distant 30 miles from A^ and C on the other road distant 15 miles from A. Find the distance from B to C, Ans. 25*98 m. 2. Two ships leave harbour together, one sailing N.E. at the rate of 7^ miles an hour and the other sailing North at the rate of 10 miles an hour. Prove that the distance between the ships after an hour and a half is 10 -6 miles. 3. A and B are two consecutive milestones on a straight road and (7 is a distant spire. The angles ABC and BAC are observed to be 120^ and 45<^ respectively. Show that the distance of the spire from A is 3-346 miles. 4. If the spire C in the last question stands on a hill, and its angle of elevation at A is 15^, show that it is '616 of a mile higher than il. ^^^i> 5. If in Question (3) there is another spire D such that the angles DBA and DAB are 45^ and 90^ respectively and the angle DAC is 45<^; prove that the distance from C to D is 2| miles very nearly. 6. A and B are two consecutive milestones on a straight road, and G is the chimney of a house visible from both A and B. The angles CAB and CBA are observed to be 36^ 18' and 120^ 27' re- spectively. Show that C is 2639*5 yards from J5, log 1760 =3-2455127 L sin 36n8'= 9*7723314 log 2639*5 = 3-42152 L cosec 23^ 15' = 10*4036846. 7. A and B are two points on opposite sides of a mountain, and is a place visible from both A and B. It is ascertained that C is distant 1794 feet and 3140 feet from A and B respectively and the angle ACB is 58^ 17'. Show that the angle which the line pointing from A to B makes with ^C is 86<> 55' 49", log 1346 = 3*1290451 L cot 29^ 8' 30" = 10*2537194 log 4934 = 3*6931991 L tan 2Q^ 4' 19"= 9*6895654, ri^R.-Ar^ 126 TEIGONOMETRY. LII. 8. A and B are two hill-tops 34920 feet apart, and C is the top of a distant hill. The angles CAB and CBA are observed to be 61^ 53' and 76^ 49' respectively. Prove that the distance from A to C is 51515 feet, log 34920 = 4-5430742 L sin 760 49'= 9-9884008 log 51515 = 4-71193 L cosec 410 18' = 10-1804552. 9. From two stations A and B on shore, 3742 yards apart, a ship G is observed at sea. The angles BAG^ ABG are simultaneously observed to be 72^^ 34' and 81*^ 41' respectively. Prove that the dis- tance from A to the ship is 8522*7 yards, log 3742 = 3-5731038 L sin 810 41' = 9 '9954087 log 8522-7 = 3-9005774 L cosec 25^ 45'= 10-3620649. 10. The distance between two mountain peaks is known to be 4970 yards, and the angle of elevation of one of them when seen from the other is 9^ 14'. How much higher is the first than the second? Sin 9^ 14'= -1604555. Ans. 797*5 yards. 11. Two straight railways intersect at an angle of 60^. From their point of intersection two trains start, one on each line, one at the rate of 40 miles an hour. Find the rate of the second train that at the end of an hour they may be 35 miles apart. Ans, Either 25 or 15 miles an hour. (Art. 264.) 12. A and B are two positions on opposite sides of a mountain ; C is a point visible from A and B ; AG aud BG are 10 miles and 8 miles respectively, and the angle BGA is 60*^. Prove that the distance between A and B is 9*165 miles. 13. In the last question, if the angle of elevation of C at ^ is S^, and at B is 2^ 48' 24'' : show that the height of A above B is one mile very nearly. sin 80= -1391731 sin 2^ 48' 24"= -0489664. 14. Show that the angles which a tunnel going through the mountain from A to B^ in Questions 12 and 13, would make (i) with the horizon, (ii) with the line joining A and C, are respectively 60 16' and 49^ 6' 24". sin 60 16'=-1091; tan IQo 53' 36"=-192450. 15. A and B are consecutive milestones on a straight road; C is the top of a distant mountain. At A the angle GAB is observed to be 380 19/, at B the angle GBA is observed to be 132^42', and the angle of elevation of C at jB is 10015'. Show that the top of the mountain is 1243* 5 yards higher than B. L sin 380 19'= 9-7923968 log 1760 = 3-2455127 L cosec 80 59' = 10 -8064659 log 1243-5 = 3-09465 jL sin 100 15'= 9-2502822. 16. A base line AB, 1000 feet long, is measured along the straight bank of a river; G is an object on the opposite bank; the angles BAG and CBA are observed to be 65o 37' and 530 4' respectively. PRACTICAL EXAMPLES, 127 Prove that the perpendicular breadth of the river at C is 829-87 feet; having given L sin 650 37' = 9.9594248, L sin 530 4' = 9*9027289 L cosec 6P 19' = 10-0568J89, log 8-2987= -91901. MISCELLANEOUS EXAMPLES. LHI. 1. A man walking along a straight road at the rate of three miles an hour sees, in front of him at an elevation of 60^ a balloon which is travelling horizontally in the same direction at the rate of six miles an hour ; ten minutes after he observes that the elevation is 300. Prove that the height of the balloon above the road is 440/^3 yards. 2, A person standing at a point A, due south of a tower built on a horizontal plain, observes the altitude of the tower to be 6OO. He then walks to a point B due west from A and observes the altitude to be 450, and then at the point C in AB produced he observes the altitude to be 30o. Prove that AB = BC. 3, The angle of elevation of a balloon, which is ascending uni- formly and vertically, when it is one mile high is observed to be 350 20 ; 20 minutes later the elevation is observed to be 55^ 40'. How fast is the balloon moving? Am, 3 (sin 20^ 20') (sec 55® 40^ (cosec 35® 20') miles per hour. 4. The angular elevation of a tower at a place A due south of it is 300 . and at a place B due west of A , and at a distance a from it, the elevation is I80 ; show that the height of the tower is a{2 + 2J5}'K 5, The angular elevation of the top of a steeple at a place due south of it is 450, and at another place due west of the former station and distant a feet from it the elevation is 15o ; show that the height of the steeple is | (3* - 3"*) feet. 6. A tower stands at the foot of an inclined plane whose in- clination to the horizon is 90; a line is measured up the incline from the foot of the tower of 100 feet in length. At the upper extremity of this line the tower subtends an angle of 540. Find the height of the tower. Ans. 114-4 ft. 7, The altitude of a certain rock is observed to be 47 0, and after walking 1000 feet towards the rock, up a slope inclined at an angle of 320 to the horizon, the observer finds that the altitude is 770. Prove that the vertical height of the rock above the first point of observation is 1034 ft. Sin 470= -73135. 8. At the top of a chimney 150 feet high standing at one corner of a triangular yard, the angle subtended by the adjacent sides of the yard are 30o and 450 respectively ; while that subtended by the opposite side is 30o. Show that the lengths of the sides are 150 ft. 86-6 ft. and lOG ft. respectively. CHAPTER XVI. On Triangles and Circles. 186. To find the Area of a Triangle. The area of the triangle ABC is denoted by A. J / H. K. B C a W . Through A draw UK parallel to BC^ and through ABC draw lines AD^ BK, CH perpendicular to BC. The area of the triangle ABC is half that of the rectan- gular parallelogram BCHK [Euc. i. 41]. BC.CH BC.DA Therefore A = ' 2 a . b sin C : 2 .(i). But sin C = — • ^s (s - «) (s - J) (s - c) ; .-. A=^s{s-a)(s- b) {s - c) =S (ii). TRIANGLES AND CIRCLES, 129 ^ 187. To find the Radius of the Circumscribing Circle. Let a circle AA'CB be described about the triangle ABC. Let R stand for its radius. Let be its centre. Join BOy and produce it to cut the circumference in A\ Join A!C, Then, Fig. i. the angles BAC^ BA'C in the same segment are equal ; Fig. ii. the angles BAC^ BA'C are supplementary; also the angle BCA' in a semicircle is a right angle. Finr. I. CB Therefore —7-77 =^ sin CA 'B = sin CAB = sin ^ , A!B ' a . . 2^ = sin^; .-. 2i? = sin J[ 188. Similarly, it may be proved that 1R h c ; and that 2i? = -r Hence, siujS a h c sin A sin B sin C sin C ' -27?. Thus d, the value of each of these fractions, is tlie diameter of the circumscribing circle. T.. T Fi f\ 12^ TRIGONOMETRY, 189. To find the radius of the Inscribed Circle. f B F, Let i), E, F be the points in which the circle inscribed in the triangle ABC touches the sides. Let / be the centre of the circle ; let r be its radius. Then ID = IE = IF = r. The area of the triangle ABC ~ area of IBC + area of ICA + area of lAB. And the area of the triangle IBC = |/i> . BC =\r .a, .-. area oi ABC = \ID . BC + \IE . CA + \IF , AB ~\ra ^ \rb ■\- \rc ) or, ' A = |r (a + 5 + c) = |r . 25 = rs. ^ = ^ = '1 190. A circle which touches one of the sides of a triangle and the other two sides produced is called an Escribed Circle of the triangle. 191. To find the radius of an Escribed Circle. Let an escribed circle touch the side BC and the sides AC, AB produced in the points D^, E^, F^ respectively. Let /j be its centre, r^ its radius. Then The area of the triangle ABC = area of ABI^C — area of I^BC, = area of I^CA + area of I^AB ~ area of I^BCy TRIANGLES AND CIRCLES. 131 or A = J7,^, . CA + ^I^F^ . AB - ^I^D^. DC = 1^1 (^ + c - 0^) = Jr^ (2s - 2a) = r^ (s - a). * s — a s — a* 192. Similarly if r^ and r^ be the radii of the other two escribed circles of the triangle ABO, then S S EXAMPLES. LIV. (1) Find the area of the triangle ABC when (i) a = 4, 6 = 10 feet, C=300. (ii) 6 = 5, c = 20 inches, .4 = 600. (iii) c = 66i a = 15 yards, 6 = 17^ 14' [sin 17" 14' = -29626]. (iv) o = 13, 6 = 14, c = 15 chains. (v) a = 10, the perpendicular from A on BC=20 feet, (vi) a= 625, 6 = 505, c = 904 yards. (2) Find the Kadii of the Inscribed an^d each of the Escribed Circles of the triangle ABC when a = 13, 6 = 14, c = 15 feet. (3) Show that the triangles in which (i) a = 2, A = 60^ ; (ii) 6 = § . \/3, ^ = 30^ can be inscribed in the same circle. (4) Prove that ■K = v|^ ; find R in the triangle of (2). (5) Prove that if a series of triangles of equal perimeter are described about the same circle, they are equal in area. (6) If ^=600, a = V3, 6 = V2, prove that the area = i (3 + V3). (7) Prove that each of the following expressions represents the area of the triangle ABC : (i) ^. (ii) 2E2sin^.sin5.sinC. (iii) rs. (iv) i?r (sin ^ + Bin 5 + sin C). (v) ^a^ sin J^ . sin C cosec A . (vi) ra cosec I A cos ^ i^ cos ^ C. (vii) (rrjT^r-^y, (viii) ^ (a^ - 6^) sin ^ . sin 2? . cosec (A-B). Prove the following statements : (8) If rt, 6, c are in A. P., then ac = 6rR. (9) The area of the greatest triangle, two of whose sides are 50 and 60 feet, is 1500 sq. feet. (10) If the altitude of an isosceles triangle is equal to the base, R is five-eighths of the base. 9—2 EXAMPLES FOR EXERCISE. LV. 1, Define the terms sine, cotangent ; and prove that if A be any angle, sin^ ^ + cos^ ^ = 1, If tan^=|, find sin^ and cos^i. 2, Find the sine, cosine and tangent of 30^ In a triangle ABC the angle (7 is a right angle, the angle A is 60^ and the length of the perpendicular let fall from C on AB is 20 feet; Und the length of AB. 3. Prove geometrically that cos (ISO*^ - -4) = - cos A, Find 4 if 2 sin ^ =tan A, 4. Prove (1) sin (A + B), sin (^ - J5) = sin^^ - sin^ B ; . . sin A + sinB _ tan ^{A-hB) ^ ^ sin A - sin ^ ~ tani(^-i^) * 5, Prove that cos2^ - cos A cos (600 + a) + sin^ (30<> - ^) = f. 6. Find the greatest side of the triangle of which one side is 2183 feet and the adjacent angles are 78^ 14' and 71^ 24'. log 2183 = 3-3390537, log 42274 = 4-6260733, L sin 7b« 14' = 9-9907766, log 42275 = 4-6260836. L Bin 300 22' =9-7037486, 7. Express the other trigonometrical ratios in terms of the cosine. 8. Prova sin (180 + ^) = - sin 2I; tan(90 + ^)=-cot^. 9. Write down the sines of all the angles which are multiples of 300 and less than BQO\ ,^ T^ ^ „ . l-cos2^ 10. Prove tan2 A = ,— ; ^-. . •*'^' l + cos2^ 11. If tan ^ + sec ^ = 2, prove that sin ^ = f , when A is less than 900. If sin^ = f, prove that tan J +sec^ = 3, when A is less than 90®. 12. The length of the greatest side of a triangle is 1035-43 feet, and the three angles are 44®, 66^, and 70®. Solve the triangle, having given L sin 440 = 9-8417713, - L sin 660 = 9*9607302, L sin 700 = 9-9729858, jog 1035-43 = 3-0151212, log 765432 = 5-8839067, log 10066 = 4-0028656. EXAMPLES FOR EXERCISE. LV. 133 13. Express the other trigonometrical ratios in terms of the cotangent. 14. Prove that cos {ISQO - ^) = - cos .4 ; cosec (180° + ^) = - cosec A. 15. "Write down the tangents of all the angles which are multiples of 30° and less than 3G00. 16. If tan ^ + sec ^ = 3, prove that sin ^i =|, when A is less than 900. If sin ^ =f , prove that tan ^ + sec -4 = 2, when A is less than 90^ 17. Find the sines of the three angles of the triangle whose sides are 193, 194, and 195 feet. 18. Investigate the following formulie : 3-4 (1) cos -^ = (2 cos -4 - 1) cos J4 ; (2) cos 6 - cos {d + d) = sin 6 sin 5 (1 + cot 6 tan Jo). 19. Define the secant of an angle. Prove the formula — ^-r + r-: = 1. sec^^ cosec^-d If sin ^ = J, find sec ^. 20. Find the logarithms of ^^(32) and of -03125 to the base {/2. 21. Express the sine, cosine, and tangent of each of the angles 1962^, 2376<^, 2844^, in terms of the trigonometrical functions of angles lying between and 45°. 22. Prove the formula to express the cosine of the sum of two angles in terms of the sines and cosines of those angles. Express cos 5a in terms of cos a. 23. Eind solutions of the equations (i) sec 6 cosec 6-cotd=,J3; (ii) sin 2^ - sin ^ = cos 2^ + cos ^. 24. A ring 10 inches in diameter is suspended from a point 1 foot above its centre by six equal strings attached to its circumference at equal intervals ; find the cosine of the angle between two consecutive strings. 25 » Define 1^. Assuming that ^^ is the circular measure of two right angles, express the angle A^ in circular measure. Find the number of degrees in the angle whose circular measure is-1. 26. Find the trigonometrical ratios of the angle whose cosine is^. 134 TRIGONOMETRY. 27. Prove that (1) cos (1800 + ^) = cos (1800-^); (2) tan (900 + ^j ^ cot (1800 -A). X 3^ 28. Prove sin x (2 cos a; - 1) = 2 sin - cos -^ , 29. Express logj^ 5-832, logio^(35) and logio'3048 in terms of logio2, logio3, logio?. 30. If the angle opposite the side a be 60^, and if 6, c be the remaining sides of the triangle, prove that (a + 6 + c) (6 + c-a) = 36c. 31. Assuming ^f- to be the circular measure of two right angles, express in degrees the angle whose circular measure is Q, Pind the number of degrees in an angle whose circular measure is \, 32. Shew from the definitions of the trigonometrical function that sin2^ + cot2^ + cos'^^ = cosec2^. tan A + sec ^ + 1 sec A-\-\ Prove that tan ^ + sec ^ - 1 tan A 3aj 33. Prove sin x (2 cos a; + 1) = 2 cos - sin - 34. Find the logarithms of /^(27) and -637 to the base 4/3. 35. If (sin ^ + sin ^ + sin C) (sin A -\- sin E - sin (7) = 3 sin A sin I?, and ^+jB + C7 = 1800, prove that (7=60o. 36. Given A = 180, B = 144o, and 6 = 1, solve the triangle. 37. Give the trigonometrical definition of an angle. "What angle does the minute-hand of a clock describe between twelve o'clock and 20 minutes to four? 38. Express the cosine and the tangent of an angle in terms of the sine. The angle A is greater than 90^ but less than 180^, and sin A = \, Eind cos A, 39. Eind all the values of B between and 27r for which cos ^ + cos 2^ = 0. 40. If in a triangle a cos A=.h cos J5, the triangle will be either isosceles or right-angled. 41. The sides are 1 foot and sj^ feet respectively, and the angle opposite to the shorter side is 30^ ; solve the triangle. 42. The sides of a triangle are 2, 3, 4. Eind the greatest angle, having given log 2= -3010300, log 3= -4771213, i tan 520 . 15/^ 10-1111004, L tan 520 . 14/ = 10-1108395. EXAMPLES FOR EXERCISE. LV. 135 43. Distinguish between Euclid's definition of an angle and the trigonometrical definition. What angle does the minute-hand of a clock describe between half- past four and a quarter-past six? 44. Express the sine and the cosine of an angle in terms of the tangent. The angle A is greater than ISO** but legs than 270<^, and tan^ = J. Find sin A, A^ -r^ /•%.«/ 2 cot^ 45. Prove (i) sm 2^ = ^^^^-^^^ . (ii) Show that if ^ + 2? + C= 90^ sin 2^ + sin 2Z?-}-sin 2C=4cos^ cosBcosC 46. "Find, all the values of between and 2ir for which sin ^ + sin 2^ = 0. 47. If in a triangle &cos^ = acos J5, show that the triangle is isosceles. 48. The sides are 1 foot and s/^ feet respectively, and the angle opposite to the shorter side is 30®; solve the triangle. 49. Express in degrees, minutes, etc. (1) the angle whose circular measure is -^tt ; (2) the angle whose circular measure is 5. K the angle subtended at the centre of a circle by the side of a regular heptagon be the unit of angular measurement, by what number is an angle of 45® represented ? 50. Prove that (sin 30® + cos 30®) (sin 120® + cos 120®) = sin 30®. 51. Prove the formulaB : (1) cos2(a + /3)-sin2a = cosj9cos(2a+iS); (2) 1 + cot a cot la = cosec a cot ^a. 52. Eind solutions of the equations: (1) 5tan'a;-sec2a; = ll; (2) sin 5(?- sin 30=^/2. cos 4^. 53. Two sides of a triangle are 10 feet and 15 feet in length, and the angle between them is 30®. What is its area? 54. Given that sin 40®29'=0-6492268, sin 40® 30' = 0-6494480, find the angle whose sine is 0-6493000. 55. Express in circular measure (1) 10', (2) ^ of a right angle. If the angle subtended at the centre of a circle by the side of a regular pentagon be the unit of angular measurement, by what number is a right angle represented? 136 TEIGONOMETBY. 56. If sec a = 7, find tan a and cosec a. 57. Prove the formulae : (1) cos2 (a - /S) - sin2 (a + /5) = cos 2a cos 2^; (2) l + tanatan^a = seca. 58. Find solutions of the equations: (1) 6tan2a; + sec2a; = 7; (2) cos 6^ + cos 3^=^72 .cos 4^. 59. The lengths of the sides of a triangle are 3 feet, 6 feet, and 6 feet. What is its area? 60. Given that sin 380 25'= 0-6213757, sin 380 26' = 0-6216036, find the angle whose sine is (0-6215000). 61. Which is greater, 76s or l-2«? [Art. 32.] 62. Determine geometrically cos 30^ and cos 45<*. If sin A be the arithmetic mean between sin B and cos B, then cos2^ = cos2(B-f450). 63. Establish the following relations : (1) tanM-sinM = tanM sinM; (2) cot A - cot 2A = cosec '2,A ; ^ '' sm2a; + sm2?/ ^ ^' 64. Express logio^(28), logio3-888, logio'1742 in terms of log^oS, logioS, logio7. 65. Prove that sin (A+B)=. sin A cos B + cos A sin B, and deduce the expression for cos {A + B), Show that sin A cos (B + 0) - sin B cos {A-\-G) = sin {A - B) cos (7. 66. One side of a triangular lawn is 102 feet long, its inclinations to the other sides being 70<^ 30', 78^ 10' respectively. Determine the other sides and the area. L sin 700 30' = 9-974, log 102 = 2-009, L sin 78n0'=9-990, logl85 = 2-267, i sin 310 20' = 9-716, log 192 = 2283, log2 = -301, log 9234 = 3-965. 67, Which is greater, 126^ or the angle whose circular measure is 2-3? 68. Establish the following relations: (1) cot2^ - cos2^ = cotM cosM ; (2) tan ^ + cot 2i4 = cosec 2^ ; ,„. cos [x - 3?/) - cos i^x-y) ^ . , . EXAMPLES FOR EXERCISE, LV. 137 69. Given log^o 2 = -3010300, logjo 9 = -9542425 ; find without using tables, logjoS, logjoO, logjo -0216 and logjo ;^( -375). 70. Prove that sin 30^ + sin 120o = ^2 cos 16<>. 71. Establish the identities: (1) l + co8^ + sin^ = /v/2(l + cos^) (1 + sin^); cosec^ul (2) cosec 2 A = — . - ; 2 ^GOsec'^A - 1 ,_. . 27r . 47r . Gtt . . ir . Stt . Srr (3) sm — + siny-siny = 4sm-siny siny . 72. The sides of a triangular lawn are 102, 185, and 192 feet in length, the smallest angle being approximately 31<^20'. Pind its other angles and its area. log 102 = 2 009, L sin 310 20' = 9-716, log 185 = 2-267, i sin 700 30' = 9-974, log 192 = 2 -283, L sin 78« 10' = 9-990, log 2 = -301, log 9234 = 3-965. 73. If the circumference of a circle be divided into five parts in arithmetical progression, the greatest part being six times the least, express in radians the angle each subtends at the centre. 74. Define the sine of an angle, wording your definition so as to include angles of any magnitude. Prove that sin (90^ + ^) = cos ^, and cos (90® + A) = - sin A, and by means of these deduce the formulae sin (1800 + ^) = - sin ui, cos (1800 + ^) = - cos J. 75. Prove the formulae : (1) cot2^ = cosec2^-l; (2) cot^^ + cot^A = cosec^^ - cosecM. Verify (2) when ^ = 300. 76. Evaluate to 4 significant figures by the aid of the table of 7 -8^1 logarithms 1^^ x ^(-008931). 77i If sin B be the geometric mean between sin A and cos Aj then cos25 = 2cos2(^ + 45**). 78. The lengths of two of the sides of a triangle are 1 foot and ^2 feet respectively, the angle opposite the shorter side is 30^. Prove that there are two triangles which satisfy these conditions ; find their angles, and show that their areas are in the ratio ^JS + 1 ; ;^3 — 1. 79. If the circumference of a circle be divided into six parts in arithmetical progression, the greatest being six times the least, express in radians the angle each subtends at the centre. 138 TRIGONOMETRY, 80. Define tlie tangent of an angle, wording your definition so as to include angles of any magnitude. Prove that tan (90^ + ^) = - cot J, and by means of this formula deduce the formula tan (180^ •\-A) = tan A. 81. Compute by means of tables the value of 6*12 ,,., ,^ 82. Prove that cos (A + B) = cos ^ cos B - sin^ sin P, and deduce the expression for sin (A+B), Show that cos A cos (5 + C) - cos B cos (^ + (7) = sin {A - B) sin G, 83. EstabUsh the identities: (1) 1 + cos^ -sin^ = Ay2(l + cos^) (1-sin^); sec^^ (2) sec2^ = jr — ; ^ ' 2-sec^^ /o\ 27r 47r , Gtt . ir Stt Stt ^ ^ (3) cos — + cos — + cos — + 4 cos =- cos — cos — - + 1 = 0. 84. Two adjacent sides of a parallelogram 5 in. and 3 in. lon;^ respectively, include an angle of 60^. Find the lengths of the two diagonals and the area of the figure. 85. Investigate the following formulae : ^A (1) sin-7r-=(l + 2cos^) sinj^; (2) sin (^ + 5) - sin ^ = cos ^ sin 5 (1 - tan 6 tan J 5). 86. Prove that (1) sin W -f sin 50^= sin 70^ ; (2) ^^3 + tan 400 + tan 80^ = ^/3 tan 40^ tan 80^ ; (3) if^ + J5 + C=180o, sin ^ - sin B cos (7 _ sin J5 - sin ^ cos G cos 5 "" cos^ * 87. Prove by means of the logarithmic table that -1-= 1-846 nearjy. 73-T 88. The length of one side of a triangle is 1006*62 feet and the adjacent angles are 44^^ and 70^. Solve the triangle, having given L sin 440 = 9-8417713, L sin 70^ = 9-9729858, L sin 660 = 9-9607302, log 1006-62 = 3-0028656, log 7654321 = 6-8839067, log 103543 = 5-0151212. EXAMPLES FOR EXERCISE. LV. 139 89. Find the length of the arc of a circle whose radius is 8 feet which subtends at the centre an angle of 50^, having given 7r = 3-1416. 90. Prove that sin ^ = - sin (A - ISO^). Find the sines of 30^ and 20100. 91. Given that the integral part of (3-1622)iooooo contains fifty thousand digits, find log^o 81622 to five places of decimals. 92. Prove that (1) cos2 A + cos2 B -2 cos A cos B cos (A+B) = sin^ {A + B) ; (2) cos2 A + sin2 A cos 2B = cos2 B + sin^ B cos 2A. 93. Prove that in any triangle a2 cos 2B + 62 cos 2A = a'^ + h^~ 4a& sin A sin B. 94. If a = 123, 5 = 29<> 17', C=1350, find c, having given log 123 =2-0899051, log 2 = -3010300, log 3211 = 4-5066403, diff . for 1 = 1352. X sin 150 43'= 9-4327777. 95. Define the unit of circular measure, and prove that it is an invariable angle. If an arc of 12 feet subtend at the centre of a circle an angle of 50®, what is the radius of the circle, v being equal to 3-1416? 96. Express the cosine and cotangent in terms of the cosecant. If cot A + cosec A = 5f find cos A. 97. Given that the integral part of (3-981)^ooooo contains sixty thousand digits, calculate log^Q 39810 correct to 5 places of decimals. 98. Prove that (1) Biii^A + sin^B + 2smAsmBcos{A+B) = sm'^{A + B); (2) sin2 A - cos2 A cos 2B = sin^ J5 - cos2 B cos 2 A, 99. On the birth of an infant £1500 is invested so that it may accumulate at Compound Interest (3 per cent, per annum payable half-yearly) during the child's minority ; calculate by logarithms the amount at the end of 21 years. 100. Prove that in any triangle cos 2^ cos 2B _1 1 a^ 6- ~" a- b'^ * ANSWERS TO THE EXAMPLES. I. 1. 80. 2. 10. 3. 16. 4. 109^1. 5. 5 acres. 6. — J — . 7. -k- y^s. 8. A shilling and a three-penny piece. u o II. 1. 10 ft. 2. 80 yds. 3. 20 ft. 4. 50 ft. 5. 90 ft. a 20^f nearly. 7. 5a feet. 8. 12a yards. 10. ^ a yards. 12. — ^afeet. 13. 1:^2. 14. \/8lft. 15. 2V9^2^"Pffc, III. 1. 3fyds. 2. 25fft. 3. 150fin. 4. 3x\ft. 5. T/yft. 6. 560. 7. 15i nearly. 8. 33600. 9. 32f. 10. 7 ft. 11. 553f, 13-8in. 12. 339fft. 13. 443 in. 14. 235 in. 15. 203 in. 16. 1886 in. V. 1. -09175 of a right angle = 9^7' 50\ 2. -0675 „ =68 75\ 3. 1-07875 „ =107^87^50". 4. -180429612345679 =18^4^29", etc. 5. 1-467 „ =146s7r77-7"- 6. -54 „ =54g 44^44 -r. 7. in4'15". 8. 70 52' 30". 9. 1530 24' 29 -34". 10. 21036'8-1". 11. 16n2'37-26". 12. 31030'. VI. I. (1) 2 right angles or 180^. (2) f of a right angle. 2 6 (3) - T^g^^ angles. (4) - right angles. (5) 2 right angles. TT TT A Off (6) "2 ^8^* angles. (7) — right angles. (8) '002 of a right angle. (9) 20 right angles. '''' 3. v±; Q. v^/ 180* ^"^ "• ^'^ 180' II. (1) TT. (2) 27r. (3) J. (4) ^. (5) t|^. (6) 1". (7) "" (8) J-. (9) ^. III.(1)|. (2)f. (3)^. (4)2^. (5)20^0- (6)2»- (7)2^0- (S)l-. (9)5.. IV. (l)i. {2)Y. (3)1. (4)|^. (5)ff. (6)3!^. ANSWERS. 141 VII. 1. f. 2. DO. 3. 4|. 4. nsjft. 5. s^jft. 6. 838000 miles. 7. -J radian = 6^^ degrees. 8. 21|i degrees. 9. 51^". 10. about 34 yds. H. 1 : 3-1416. 12. 3-1416. 13. 3-1416. 14. 400:1. 15. -0000484.... 16. 49^im. 17. |i. e. a right angle, 19. 473 : 489. 20. (i) A-1, (ii) ft = ^. 21. 380, iqo. 22. j^.' 23. (i) 1200, 133-3», y , (ii) 1350, 150«, ^, (iii) 1560, 173 'S^ ^. 24. (i) 3f, (ii) ~. 25. W' 26. a right angle. 27. -g^ • ^^ 9a + 106, ^- ISOOtt o/^ n i/» 28. -10^ ^^g^^^^- 29. i7r.-^i800- 30. 9orlG. VIII. 1. (i) DA, BD. (ii) DB, AD. (iii) DA, CD. (iv) DC, Ji). ^ ,., Di^ ,.., DC ,..., CD ,. . DA , , DB 2. W 2^ • (") CA' (^^^ Zd- (^^) :bI- (") Id- ... Da , ... CD . .... Dl /. ^ ^^ , ^ DA ^^'^ AG' (^"^Cl- (^"^^CD- ^'^^BA' W Cl- DJ5 B^ CD CB DB BA ^' ^'^ GB* CA' ^''^OB' CA' ^^''^ CD* CB' ,. . DB BQ , , ^D AB , ., DD DC ('^) Id' Ic- W Id' Ic* ^''^ ad* ab' . ,,DA ,.., BA AC ,..., DC ,. , AB *• (^) Dl • (^^^ H °" DC • (^^) DC • (^^) Id • ^ . ^D AB I .s BD , ... DD BA AE <^^ Id "^ Ic- ("^) DC • ("^) CD ' °^ CD' "^ cl • , .... DA ,. , BA AG , , DC , ., DB BG (vm) — . (IX) ^ or — . (X) _ . (xi) ^ or - • (^) ID- 5. sin^ = f, cos -4 =4, tan^ = f; sinD = f, cosD=|, tanD = J. 7. Of the smaller angle, the sine = Y^^, cosine = ^§, tangent = ^'V. Of the larger angle, the sine =15, cosine = 3^, tangent =-i^. v/3 1 8. Of the smaller angle, the sine = ^, cosine =-^ , tangent = -7r. Of the larger angle, the sine=i\/3> cosine = J, tangent = \/3. U2 TRIGONOMETRY, 10. Bc=^S; sinA=l\/S, cos^ = i, ta.nA=J3. 12. .4C=V2; sin^ = ^J, sinS = -^. X. 1. 179 ft. 2. 346 ft. 3. 86-6 ft. 4. 138-5 ft. 5. 7ift. 6. 600, 173 ft. 7^ 63-17 yds. 8. 277-3 ft. 9. 192-8 ft. 10. 78 ft. 11. 34-15 ft. 12. 73 2ft; 13. 86-6 ft. 14. -866 miles = 1524 yds. 15. -173-2 yds. 17. 373 ft. 18. 3733 ft. 19. W6 miles = 6465 ft. 20. ^;— . 21. 300. ou 22. About 523-6 miles. XL 27. 2cos2 6'-l, l-2sin2^. 28. (1-2 cos2^)2, (2 sin2^-l)2. 2cos2(9-l l-2sin2(9 J^^' cos*^ ' (l-sin2^)2' 30. l-3cos2f = .. Vl + tanM Vl + tanM tan^» ecu J3. /^ A T KOiXJ. ^ tan^ XIII. 1. s, s. 2v/2 2. 3 1 ' 2vr 3. 1, !. 4. -^- ^. 5 Vis- ^ 2 » *• 6. 3 • '• 7. 8. a 1 . 9. • Va^-l a 1 t' \/a' + l ' Va2+1 11, , ;»^(l + fc2)=l. Jc^'b-^' XIV. 2. sec d increases continuously from 1 to oo . 3. sin A diminishes continuously from 1 to 0. 4. cot d diminishes continuously from oo to 0. XV. 1. 430. 2. 300. 3^ 450. 4^ 600. 5^ 300. 6. 300. 7. 300. 3^ 00, or 450. 9. 900, or 600. ^q^ (jqo, 11. 450. 12. 450. 13. 900, or 450. i^ 450, 15^ 450, 16. 450. 17. 300. 18. 300. XVI. 3. The value 3 is inadmissible. 4. l(2±x/2). 5. I, or i. 6. I, or |. 7. The value - 1(7 Vs) is inadmissible, 9, 1-sinM. 10. l-3sin2^ + 3sin^^. 1 - 2 cos^^ + 2_cos^ 1-sin^ •^•*- c'os^'^ * ^^' 1 + sin^* 14. cosec decreases continuously from 00 to 1. 15. cot increases continuously from to go • 16. = iTr, = ^T7r. 144 TRIGONOMETRY. XVII. 1. +6. 2. 0. 3. +2. 4. +3. 5. +10. 6. 0. 7. +7. 8. +7. XIX. 1. The second. 2. The fourth. 3. The second. 4. The third. 5. The fourth. 6. The first. 7. The second. 8. The first. 9. The first. 10. The fourth. H. The fourth. 12. The first, if n be even, the third, if n be odd. XX. 1. +, +, +. 2. + , - • » ~ • 3. — , — 1 +• 4. -, +, -. 5. ~, +: > "" • 6. -, -, +. 7. +, -, -. 8. 10. +, +, +. 11. + , +, + , -. , +. 9. +, +, +. 12. -, +, -. XXI. 1. +h ~^,- 3. +^, -i, -v/3. 1 ^/3' 2. 4. ^v/2' V2' ^' + 2 ' ^3- 11 9- +*'+X'+^- 6. 8. 10. -^|, +i +n/3. .-1- -^ +1 11. -^, -i, +V3. 12. -^-, +i, -x/3. 11 14. f, 'h -V3. 15. -i, --g-. +;y3- 16. 300, 1500^ _ 2100, - 3300. 17. 450,1350, -2250, - 3150. 18. 600,1200, -240, -3000. 19^ _300, -1500, 2100, 3300. 20. 200, 1600, 3800, 5200. 21. f TT, ItT, ^^TT, J^TT. 26. No. 22. fT, i^T, yir, -V-ir. 25. The tan. XXIII. 1. 600. 5. 115. 6. 4100. 2. - 7. - -lOO". -JT. 3. 00. 4. -2600. 8. |r. XXIX. 1. sin {e + 4>) + sin {0 - «/>). 3. sin (2a + 3^) + sin (2a - 3^3). 5. sin 8^ -sin 2^. 6. cos ^ + cos 2^ 2. cos (a - j3) + cos (a + /3). 4. cos 2a-)- cos 2/3. 7. 1 (cos 3^ -cos 5^). 8. i (sin 4^- sin ^). 9. sin 600 + sin 40o. ANSWERS, 145 10. Hsin000-sin30«). H. 2 cos 3^ cos 2^. 12. -cos 4^ sin 2^. 13, 4cos2|sin2^. XXXII. 1. (i) a2^+3*. (ii) a*^-«*. (iii)a8'^8. (iy) a^^T^ 2. (i) 5-4690116. (ii) 10-6243928. (iii) 13-7509366. (iv) -8853661. (v) 1-7968680. (vi) 8-9699598. (vii) 2-7345058. 3. 2=*, 25, 2-1, 2-^ 2-3, 27. 4. 32, 34^ 3-1^ 3-3^ 3-2^ 3-4. XXXIII. 1. -60206, -9542426, -90309, '7781513, 1-20412, 1-690196. 2. 1*146128, 1-20412, 1-2552726, 1-3802113, 1-4313639, 1-6232493. 3. 1, '69897, 1-1760913, 1*39794, 1*4771213, 1-5440680. 4. 1-5563026, 1-60206, 1-6812413, 1*69897, 2-30103, 3. 5. 7-201593,3-858708. 6. '7545579,2-989843. 7. 1*4532. 8. 2408-6. 9. (i) 4-5868. (ii) -93646. 10. 3-9549. 11. 40975-3 sq.ft. 12. 34-925 in. 13. 3-2617 in. 14. 110115 cub. yds. XXXIV. 1. 3, V-, h h -f- 2. 3, 6, -1, .3, -6, 2. 3. 2,4, -1, -3, -2, -4. 4. 5. 3, -1,5, -2,3, -3. 6. 7. -7781513, 1-6232493, 1-20412. 8. 1-6901960, 1-5563026, 1-7993406. 9. 2-30103, 2-7781513, 1-845098. 10. '60897, 6228787, 1-69897. 11. 1*544068,2-1760913, -1 + -30103. 12. -5440680, -8627278, -2 + -9084852. XXXV. 1. 4,2,0,5,1. 2. -2,-5,-1,-3. 3. 3, -1, 0,1,0, -7. 4. 4,1,6,3. 5. the second decimal place, the first dec. pi., the sixth dec. pi. 6. ten thousands, units, hundreds, third dec. pi., first dec. pi., units. 7. 10, 4, 25, 31. 8. 9, 11, 85, 4, 9, 6. 9, units, fourth dec. pi., thousands, seventh dec. pi., second dec. pi. 10. tenth integral pi., twelfth dec. pL, fifth dec. pi., units, twelfth dec. pi., first dec. pi. XXXVI. 1. 2-8901023, -8901023, 4-8901023,5-8901023. 2. 6-7714552, -7714552, 4-7714552, 2-7714552, 3*7714552. 3. -27724... 4. -00001638... 5. *77448... 6. -005968... L. T. B. 10 2. 3,6, -1. S,J. -h -1. 1.1. h h h 146 8. 11. 12. 19. XXXVII. •51375. (i) x= (iii) X 1 2 + TRIGONOMETRY. 1. 3, 0, J, 0,|. 2 log 7 4. 6. 1-8121177, 55. 7, 4, 3, 3. log2 + 41og3* 2 log 7 21og2+log3* 0, logio7' 1 9. i + logio 3 ... _ 71og2 + 41oor7 ^"^ ^~ 21og3 + log7 • ^'""^ '^~81og2 + 3(log3 + log7)' 10. n .L o » 3a 2 & 3a + 2 6c 6+1' 26 + 2' 6 + 1' 6 + 1' 26 + 2' 6 + 1* 63 - 31 = 32. 13. (a" - aio) integers. 14. 1-9485 nearly. 2-53855. 20. 4-59999. 21. 167 years. 2. XXXVIII. 1. -8839066. 4. 6-8086920. 5. '5710750. 8. 2492837. 9. -000439658, XXXIX. 1. 7-669. 2. 3-809. 5. 12-03. 6. -04023. 7. 8287. , 2-7513738. 3. 4-9413333. 5. 3-70404. 7, 45740-26. 10. 5-689158. 3. 47-32. 4. 55460. 8. 1165. 9. -3107. 10. -6731. 11. 1096. 12. 823-6. 13. 2-624. 14. 22-51. 15. 23-28. 16. 2801. 17. -'8243. 18. XL. 1. £48. 2. £-477^9s. 64c?. 3. 23-4. 4 5. £73-07. 6. 140 years. 7. £1869. 8. 36-9 years, 1407. 17-7. 9. £5066 alouU 11, -0679 miles per hour. XLI. 1. -6737652. 4. 410 48' 37". 7. 9-8515594. 10. 350 4' 23". XLII. 1. 340 19' 31-8". 10. About 67,100,000 pence. 12. 1-24 yds. 2. -6737652. 3. -9306572. 5. 700 31' 43-6". 6. 750 31' 21". 8. 9-7114477. 9. 10-1338768. 11. 280 16' 27-5". 12. 210 56' 41". 2. 1498-2 ft. 3. 450 36' 56". 5293*4 ft., 6982-3 ft. 5. 576-2 chains. 6. 4729 chains. 3666-8 feet. 8. 42015', 11444 chains. XLIII. 1. 3843 ft. 2. 281-7 ft. 3. 115 ft. 286 ft. 5. 580 17', 310 42'. 6. 656 chains, 41017'. 7. 81ft. 8. 1942 ft. 9. 646-7 miles. 10. XLIV. 5. 450. 600. 1200." 2. 1200. 3. 300. 1000 ft. 4. 1350. ANSWERS. 147 XL VI. 1. cos^ = ^, cosii = ^v/3. 2. 45«, 600,750. 3. 1350,300,150. 4. 3. 5. 14. 6. l + ^/3. 7. ISO*. 8.' 1200. 9^ 1200. 10. 900,360 52'. H. 1300 27'. 12. 1250 6'. 13. 1200. 14. ^ = 540 or 1260, B = 1080 or 360. 15. a = l. 16. C=300, a=;^3 + l, 6 = 2. 17. ^ = 750, a = 6=^3 + l. 18. C = 000orl200. 19. 100^3. 20. No. 22. J=105o, C=600, B = X50. 23. ls/^U^ + ^)' 24. ^=900 or 600, c = 750 or IO50, a = 2 ^2 or ^^6. 25. 300 or 1500. 26. ^=450 or 1350, i? = 300 or 1200, &= ^2 (1 + ^/3) or <^6(l+;^3). 27. 600, 750^ Q y^s. 28. It is impossible. 30. 15 : 8^3 : 4^^5 + 6. XLVII. 1. 410 1(5/ 51.5'', 2. 730 32' 12", 620 46' 18". 3. 290 17' 16", 310 55' 31". 4^ 640 31' 58". 5^ 730, 23' 54-4". 6. 410 24' 34-6". 7^ 820 49' 9". 3^ 750, 6OO, 450. 9. 1350, 300, 100. XLVIII. 1.313-46 yds. 2. 28-87 in., 31-43 in. 3. 1192-55 yds. 4. 22 -415 ft. 5. 24-995 = 25 ft. nearly, 17*559 ft., 650 59' 42". XLIX. 1. 1080 36' 30", 310 23' 30". 2. 930 11' 49", 360 48' 11". 3. 67" 27' 25 -4", 620 32' 34-6". 4. 640 26' 47", 370, 7', 13". 6. 720 12' 59". 6. 20-5 chains. 7.122-7. 8. 71o 13' 50", 320 16' 10". L. 1. ^ = 51018'21",(7=880 41'39"; or^ = 128041'39",C=ll018'21". 2, B = 700 0' 56", C= 590 59' 4" ; or B = IO90 59' 4", C= 200 0' 56". 3. P = 38038'24", C=9102r36", c = 155-3. 4. 61016'10". 5. ^ = 720 4'48", ^ = 410 56' 12"; or ^ = 1070 55' 12", JS = 60 5'48", & = 17-56. 6. /3 is ambiguous; 60-3893 ft. LI. The angles are given correct to the nearest second. 1. 280 35' 39". 2. 1040 44' 39". 3. 320 20' 48". 4. 430 40'. 5. 1280 23' 13". 6. 106531ft. 7. 3437-6 yds. 8. 1728-2 chains. 9. 25376 yds. 10. ^ = 660 27' 48", 5 = 120 55' 12". n^ ^ = 92012' 53", ^ = 350377". 12. i5 = 290 1' 40", C = 740 55' 50". 13. J5 = 700 35' 24" ; or IO90 24' 36". 14' B = 51056'17"; or 1280 3'43". 15. J5 = 6206'10"; or 1170 53'50". 16. Very nearly 900. 17. 1319-6 yds. LV- L sin^ = f, cos^ = |. 2. -^^^3 ft. = 46-19.. .ft. 3. ^=nx 1800; or, 713600 ±600. 5, 4227-47 feet. 9. 300, 600, 900^ 1200, etc. have for sine i, J^, 1, ^f, J, 0, -J, - Vf"l. -Jh - i respectively. 148 TRIGONOMETRY. 12. The other sides are 765-4321 ft. ; 1006-6 ft. 15. 30^ 600, 900^ etc. have for tan^^S, J3, oo , -^3, -J^3, 0, ^ ;^3, GO , - ^3, -^sj^ respectively. -_ 168 168 32592 ,^ . ,, ,« «^ ^ 17- m'l95'r93xl95- 19- ^^^^ = SV2. 20. (i) ^; (ii) -15. 21. +sinl8^ -cos ISO, -tanlS^; -sin300, -cos860, +tan360; -sin36o, +COS360, -tanSe^. 22. cos5a = 16cos5a-20cos=*a + 5cosa. 23. (i) 0, TiTT, ^tt; (ii) cos^ = J, or, sin (^ - 450) = — : . 24. T2Vv/651 = -981... 25. fV^^ radians; 5-729650. 26. sine, I ; tan, |- ; cot, | ; cosec, f ; sec, f . 29. (i) 6 logio 3 + 3 logjo 2 - 3 ; (ii) ^{log lo 7 + 1 - logjo 2} ; (iii) 31og3o7 + 31ogio2-21ogio3-2. 31. -«A«-^deg.; 19-09854». 34. f; -9. 36. C7=18o, a=c = 2^J(10-2V5). 37. -13200. 38. -1^2. 39. Tr\h-^\ fTT. 41. 1 foot, 1200, 300. or 2 feet, 600, 90o. 42. 1040 28 39". 43. -6300. 44. -J^5. 46. 0; tt; |7r; |7r. 48. i {N/6iv/2} and 150, 1350; or, 105o, 450. 49. 9^ 2860. 28'. 41-16"; |. 52. (i) nir ± Jtt. (ii) JwTT ± ^TT, or WTT + ( - 1)»^ Jr. 53. 374 sq. ft. 54. 400 . 29' . 19-85". 55. ttdV^'t; tV^; f. 56. tan a = 4^3, cosec a = ^^^^3. 58. (i) nir^lir. (ii) Jn7ri|^7r; or, 27i7r±j7r. 59. 2^14 sq.ft. 60. 380.25'. 32-725". 61. 1-2 radians = 76-39416s. 64. (i) 1-Iogio5 + Jlogio7; l-41ogio5 + 51ogio3; 2-5 logio 5-2 logio 3 + 2 log^o 7. 66. 192 ft., 185 ft. and 9234 sq. ft. 67. 2-3 radians = 131- 7799260. 69. -6989700; '7781513; 2-3344538; 1*9148063. 72. 780 10', 700 30', 9234 sq. ft. 73. /^t; ^tt; ^Itt; ifx; IfTT. 76. 116-6. 78. 1350, 150 ; or 450, 1050. 79. ^r ; f^ir ; ^tt ; ^\ir ; |f tt ; ^lir, 81. 32-92... 84. 7 ft.; v^l9 ft.; \«V3 sq.ft. 88. 1035-43 ft.; 765-4321 ft.; 660. 89. 6-981 feet. 90. 4; -J. 91. 4-49999. 94. 3210-793. 95. 13-751 ft. 96. }J. 97. 4-59999. 99. £2803 nearly. CAMBRIDGE : PRINTED BY C. J. CLAY, M.A. AND SONS, AT THE UNIVERSITY PRESS. V /■ i 'in UNIVERSITY OF CALIFORNIA LIBRARY THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW f^^f^ 8 1919 )ni -1^ -« JAN 11 1922 SEf- U im fWfll? 4 inr-rf-ff -'^x\ R£C'D' LD 9k~.'./^' • ^> Ai^-* In/k.