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IVIIIlor's Edition is prepared as an introductory Text Book for MuEon's Grauimur, the authorized book for advanced clauses for Public Schools, so that what is learned by a pupil in an cloiiicu- tury text-book will not have to be unlearned when tiio adviinced book is used, a serious fault with many of the graded I'ublic School Books. iniller'H Edition contains all the recent examination Papers set for admission to Iligh Schools. IfllLLER'S NWIIVTOISfS LAISGUAGE LES80NM is autliorized by the Education Department of (Jntario, is adopted by the Schools of Montreal, is authorized by the Council of Public Instruction, Manitoba. To the President and Members of the County of Elgin Teachers Association : In accordiinco with a motion passed at the last regular meeting of the Association, appointing the undersigned a Oommittee to con- sider the respective merits of different English (}rammara, with a view to suggest the most suitable one for I'ublic Schools, we brg leave .0 report, that, after fully comparing the various editions tiiut have lioen recommended, we beliiAre tliat " Miller's Swintoii's Jjangunge Lessons*' is best adapted to tlie wants of Junior pupils and would urge its aiithori/.ntion on the Uoverumuut, and its intro- duction into our I'ublic Sclioola. St. Thomas, Nov. Suth, 18; d. A. F. BUTLER, Co. Inspector. J. McLI'lAN, 'I own Inspector. J. WILLKll. M.A., iioad Master Rt Thomas nigh School. A. STEEL I'!, B A.', N. M. CAMPBELL, Ayliiier High Sctiool. Co. of Elgin liklodul School. It was moved and second(>d that the report be received and adopted — Carried unanimously. I'i-ic3e, Olotli lilxtirn. «r>e. ADAM MILLEE & CO. THE EPOCH PRIMER OF £:iVO]L.ISI-I HISTORY. l^-'ing an Introductorj Volume to the series of Epocha of English Uistoryt bj Iho Rev. MANDELLCKEI'. HTON, M. A., lato FtUow and Tutor of Mer' toa College, Oxford; Editor of ' Epochs of English History.' Fc)>. 8vo.pp- 148, price 3ucts. cloth. ' In making history attractive to the young the Author has proved his apti- tude in a department of literature in wliich fuw distinguish themselves The narative is so sustained that those who take it up will have a desire to read it to the end.' DUMDKE AdTXBTISXB. ' This volume is intended tobe in- troductory to the Epocha of English History, and nothing could be hotter adapted for that purpose. The little book is admirably done in all respects, and ought to have the effect of sending pupils to other and fuller sources of historical knowledge.' Scotsman. 'Mr Cei;ighton'8 introduction to the Epochs of English History covers in a humored and forty pages more than 1800 years, but having regard to its extreme condensation is wdl worthy oi notice. On the whole the work is admirably done, and it will no doubt obtain a very considerable sale.' XrUlLffMVX. *An admirable little book that can scarcely fail to obtain a considerable popularity,! notwithstanding the great number of previous attempts made to relate the history of England in a very small eompass In tliis epitome the epochs become chapters, but an in- teresting account is given of such events iis are likely to be attractive, or even moderately intelligible to young leaders.* Wklbhuan. * The excellent series of little books published under the title of Epocha of English History, edited by the Rev. MANDEtiL Crkiohton, M. A., and Writ- ten by various able and eminent writers being niiw complete, the Editor has rrepared an introductory volume, cal- led tho Epoch Primer, comprising a concise summary of the whole series The special value of this historical out- line is that it gives the reader a com- prehensive view of the course of mem- orable events and epochs and enables him to see how they have each con tributed to make the British Nation what it is at the present day. LiTXRABY Would 'As all the leading features— political, social and popular — are given with much impartiality, it can hardly fai^ to become a » lool claos-bonk of great utility.' WORCKbTEB JoiTBNAIi. * Tho Rev. Mandkll CnEiOHioM has really succctded in making an admir- able resume of the whole of the prin- ciple events in English history, from tho time of the Roman Invasion down to the passing of the Irish Land Act in 1870. Interesting, intelligible and clear, it will prove of great value in tho elementary schools of the kingdom; and those advanced in years might find it very hanciy and useful for casual reference.' Nokthampton Herald. * This volume, taken with the eight small volumes containing the accounts of tho different epochs, presents what maybe regarded as the most tliorough course of elementary English History ever published Well suited for middle class schools, this series may also be studied with advantage by senior students, who will find, instead of tho mass of apparently unconnected facts which is too often presented in such works, a careful tracing-out of the real current of history, and an in- telligible account of the progress of the nation and its institutions.' Abkrdben JounwAii. * The whole scries may be safely commended to tho notice of parents and teachers anxious to find a suitab e work on English history for the r children, inasmuch as tho severjil volumes are simply and intrlligibjy written, without being overloaded witjh details, and care has been taken to bring every subject treated on within the comprehension of the young. The namby-pamby element, which is so often conspicuous in histories f o ' children is entirely absent, and the worksin question are certainly amongst the best cf tho kind yet issued. Tlie little voluiii<< now under notice, which brings the serii'S to a close, in fully equal in every respect to the preceding ones, and it will bo found exceedingly useful to every one who may hav«s to teach English hifitory,' LSAMINOTOM CoVRIlrR. ■^^^ gibum JllilUr k (to's IlTuthcmatkul Scries. ELEMENTARY TRIGONOMETRY. BY J, HAMBLIN SMITH, M.A., OF OONVIIiliB AND CAIUS COLLKaK, AND LATK LECTURKR AT ST. riSTKR'S CULLEQH, CAMliRlDaR. VEW EDlTIoy. TORONTO : ADAM MILLER & CO. 1879. Q ,/ ^ 111") Entered according to Act of Parliament of Canada, in the ofllce of the Minister of Agriculture, by Adam Miller ft Co., in the year 1878. rftUITKU AT TUB QUAHOIAN OFriOB, 4 COURT BTRBBT^ TORONTO. PREFACE he Minister I HAVE attempted in this work to explain and illustrali' tho principles of that portion of Plane Trigonometry which precedes de Moivre's Theorem. The method of explanation is similar to that adopted in my Elementary Algebra. The examples, progressive and easy, have been selected chiefly from College and University Ex- amination Papers, but I am indebted for many to the works of several German writers, especially those of Dienger, Meyer, Weiss, and Wiegand. I have carried on tho subject somoAvhat beyond the limits set by the Eegulations for the Examination of Candidates for Honours in tho previous Examination for two reasons: first, because I hope to see those limits extended, secondly, that my work may be more useful to those who are reading the subject in Schools and to Candidates in the Local Examinations. J. HAMBLIN SMITH. * '1 Cambridge, 1870. CONTENTS CHAP. PAUK I. Ov THE Measurement of Lines .... i II. On the Ratio ov the Circumference of a Circle . TO THE Diameter 5 III. On the Mkasurement of Angles . . . 10 IV. On the Mirriion of Converting the Measures of Angles from one to anothkr System of Measure- ment 18 V. On the Use of the Signs + and — to denote Con- trariety of Direction 26 VI. On the Trigonometrical Ratios .... 32 VII. On the Changks in Sign and Magnitude of the Trigonometrical Ratios of an Anglic as \v in- creases FROM o*" to 360*^ 39 VIII. On Ratios of Angles in the First Quadrant . 46 IX. On the Relations between the Trigonometrical Ratios for the same Angle 54 X. Comparison of Trigonometrical Ratios for dif- ferent Angles 65 XI. On the Solution of Trigonometrical Equations 73 XII. On the Trigono.mefrical Ratios of two Angles 83 XIII. On the Trigonomp:trical Ratios for Multiple and Sub-multiple Angles 99 XIV. On Logarithms 112 XV. On Trigonometrical and Logarithmic Tables . 125 XVI. On the Relations between ihk Sides of a Triangle and THE TriGONOMICTRICAL RAI'IOS OF THE ANGLES of THE Triangle 142 VIU CONTENTS. CHAP. PAGE XVII. On the Solution of Right-angled Triangles . 154 XVIII. On the Solution of Triangles other than Right- angled 164 XIX. Measurement of Heights and Distances . . 17 . XX. Propositions relating to the Areas of Triangles, Polygons, and Circles 187 Answers 206 Appendix 222 PAGE !LES . 154 Right- • • 164 • • 17 ANGLES, • • 187 • • 206 • t 222 Elementary Trigonometry. I. ON THE MEASUREMENT OF LINES. 1. To measure a line AB we fix upon some line as a standard of linear nR'asiirement : then if Alt contains the standard line p times, ^ is called the measure of AB, and the magnitude of AB is represented algebraically by the symbol p, (See Algebra, Art. 33.) Since the standard contains itself once, its measure is unity, and it will be represented by 1. ' :i. Two lines are commensurable when a line can be taken as the standard of measurement (or, as it is commonly called, the unit of length) such that it is contained in each an exact number of times. 3, If the measures of two lines AB, CD be p and q respec- tively, the ratio AB : CD is represented by the fraction -• (See Altjebra, Arts. 341, 342.) Examples.— i. 1. If the unit of length be an inch, by what number will 4 feet 6 inches be represented ] 2. If 7 inches be taken as the unit of length, by what number will 15 feet 2 inches be represented / [S.T.] A 4 ON THE MEASUREMENT OF LINES. 3. If 192 square inches be represented by the number 12, what is the unit of linear measurement ? 4. If 1000 square inches be represented by the number 40, what is the unit of linear measurement ? 5. If 216 cubic inches be represented by the number 8, what is the unit of linear measurement ? 6. If 2000 cubic inclies be represented by the number IG, what is the unit of linear measurement ? 7. If a yards be the unit of lenglh, what is the measure of h feet ? 8. A line referred to different units of len{:^th has measures 5 and 4 ; the first unit is 6 inches, what is the other? 9. A line referred to three ditferent units of len<,'t]i hucj neasuns 1, 36, 12 respectively; the unit in the first case is a yard, what is iL in the others ? 10. Express the ratio between 3^ inches and 3^ yards. 11. If the measure of m yards be c, what is the measure of n feet ] 4. Now suppose the measures of the sides of a right-angled triangle to be ;?, q', r respectively, the riglit angle being sub- tended by that side whose measure is r. Then since the geometrical property of such a triangle, ustablislied by Euclid i. 47, may be extended to the case in which the sides are represented by numbers, or symbols standing for numbers. /r + (/-^ ,.2 ON THE MEASUREMENT OF LINES. 3 If any two of the numerical quantities involved in this equation are given, we can determine the third. For example, if r = 5 and 7 = 3, ^- + 9 = 25, .'. p = 4. Examples.— ii. 1. The hypotenus^e being 51 yards, and one of the siles containing the right angle 24 yards ; find the other side. 2. Tl\e sides containing the right angle being 8 feet and G feet; find the hypotenuse. 3. A rectangular field measures 225 yards in length, and 120 yards in bieadth ; what will be the length of a diagonal ;\i}i across it? 4. A rectangular field is 300 yards long and 200 yar.ls broad; find the distance from corner to corner. 5. A rectangular plantation, wiioso :h width is 88 vards, con- tains 2^ acres ; find the distance from corner to corner acruss the plantation. 6. The si<le3 of a ri|^ht-angled triangle nn« rn nrithmenoal progression and tiic hypotenuse is s:u rewt ; find the otlicr sides 7. The sides of r right-angled triangle are in arithnu!'.^*al progression; show tliat they are ])roportinn)d to 3, 4, 5. 8. A ladder, whose foot rests in a given position, JM>t reaches a window on one side of a street, and when tun. .d about its foot, just reaches a window on the other side. If the two ])ositioiis of tiie huKlfr be at right angles to each otlitr, and the heights of the windows be ;}() and 27 fecit rcsju'ctively, find the width of the street and tin; lenglh of the ladder. 9. [n a right-angled isosceles triangle the hypotenuse id 12 feet ; find the lenj^th of each of the otliur sides. 10. What is the length of the diagonal of a «([uare wiio-f Bide is 5 inches } 4 ON THE MEASUREMENT OF LINES. 11. The area of a square is 390625 square ieet ; what is the diagonal ? 12. Each side of an equihiteral triangle is 13 ; find the lengtli of the perpendicular dropped from one of the angles on the opposite side. 13. If ABC he an equilateral trinnglc and the length of A J), a perpendicular on BC, he 15 ; fuid the length of AB. 14. The radius of a circle is 37 inches ; a chord is drawn ill the circle : if the length of this chord be 70 inches, find its distance irom the ceiitre. 15. The distance of a chord in a circle from the centre is 180 inches ; the diameter of the circle is 362 inches : find the length of the chord. 16. The length of a chord in a circle is 150 feet, and its distance from tiie centre is 308 feet find the diameter of the circle. 17. If ABC he an isosceles right-angled triangle, C being the right angle, show that AG : AB=l : V^. 18. If DEF be an equilateral triangle and a perpendicular DG lie d'-opped on EF, show that X'i : EV : 1)0=1 : 2 : A ' I le lerif'th of r>r(l is drawn IT. ON THE RATIO OF THE CIRCUM- FERENCE OF A CIRCLE TO THE DIAMETER. 6. It is evident that a straight line ran he compared as to its length with a circnhir arc, and that conseqmntly the ratio between such lines can he represented in the form of a fraction. 6. We must assume as an axiom that an arc is greater than the chord subtending it: that is, if ABD be part A of the circundVrence of a circle cut olf by the straiglit line AD, the length of ABD is greater than the length of AD 7. A figure enclosed by any number of straight lines is called a Polygon. 8. A regular polygon is one in which all the sides and angles are e(|ual. 9. Tlie prrimeter of a polygon is the sinn of the sides. Hence if AH be one of the sides of a regular polygon of ?i sides, the perimeter of the polygon will be n . A />. 10. Tlie circumference of a circle is greater than the peri- meter of any polygon which can be inscribed in the circle ; but as the number of sides of such a polygon is increased, the perimeter of the p(dygon approaches nearer to the circum- ference of the circle, as will appear from the following illus- tration. •^1 6 RATIO OF CIRCUMFERENCE TO DIAMETER. Let AB h^ tlie side of a regular bexagou ABDEFG in- ficribed in a circle. Then AB is equal to the radius of the circle. Eucl. iv. 15. Now the arc ACB is greater tlian AB, and the circumference of the circle is therefore larger than the ])erimeter of the hexagon. Hence the circumference is greater than six times the radius, and greater tliun three times the diameter. Now suppose G to be the middle point of the arc AB. Join AG, CB. These will be sides of a regular dodecagon, or figure of 12 sides, inscribed in the circle. Now A 0, CB are together greater than AB : but AG, CB are logcther less than the arc ACB. TIencc the perimeter of the dodecagon will be less than tlie circumference of the circle, but will ajtproxinuite more nearly tlian the perimeter of the hexagon to the circumference of the circle <i f . il RATIO OF CIRCUMFERENCE TO DIAMETER. 7 So the larger tlie number of sides of a polygon inscribed in a circle, tlie more nearly does the perimeter of the polygon approach to the circumference of the circle; and when the number of sides is infinitely large, the perimeter of the poly- gon will become ultimately equal to the circumference of the circle. 11. To show that the circumference of a circle varies as tne radius. M Let and be the centres of two circles. Let AB, ah be sides of regular polygons of n sides inscribed in the circles, P, p the perimeters of the polygons, and C, c the circumferences of the circles. Then GAB, oab are similar triangles. .'. OA : oa :: AB : ah :: n.AB : n .ab :: P : 2?. Now when n is very large, the perimeters of the polygons may be regarded as equal to the circumferences of the circles ; .". OA : oa :: C : c. Hence it follows that the circumference of a circle varies as the radius of the circle. 12. Since the circumference varies as the radius, the . circumfiTence . ,, ,, ,, . 1 1 ,1 r ratio - ,. 18 the same lor all circles, and thereloro radius ,, . circumference . ,, r n • i the ratio ,. — is the same for all circles, diameter t «i M n \n 1 1 1 1 I i 8 RATIO OF CIRCUMFERENCE TO DIAMETER. «* -rv ,. mi ^- circumference . , .11 ^i 13. Def. The ratio — ,. , -- — is denoted by the (liiuneter symbol it. 14. The value of this numerical quantity tt cannot be determined exactly, but it has been ajpproximatelij determined by various methods. If we take a piece of strin;^ wliich will exactly ^0 round a penny, and another piece which will exactly stretch across the diameter of the ])enny : if we then set off along a straight line seven lengths of tlie first string, and on another straight line by the side of the first we set o(f twenty-two lengths of the second string, we shall find that the tv.'o lines are very nearly equal. Hence 22 diameters are nearly ecpial to 7 circum- - .1 i. .1 .• circumference 22 , ferences, tliat is tlie ratio —v. = =- nearly, or in ' diameter 7 •'' 22 other words the fraction y is a rough approximation to the value of 7r ~- The fraction 1' [ gives a closer approximation. The accurate value of the ratio to 5 places of decimals is 3-14159. 15. Suppose we call the radius of a circle r : then the diameter =2r. Now circumference diameter circumference 2r = 7r; = ir; Hence and ;. circumference = 27rr. arc of semicircle = 7rr, arc of quadrant = - -. RATIO OF ClKCi'MhKREXCE TO JJIAME'TER. 9 Lion to the Examples.— iii. In the iblloving examjiles the value of y may be taken 22 as -^-. I 1. Tlie diameter of a circle is 5 feet, what is its circum- ference ? 2. The circumference of a circle is 542 ft. 6 in., what is its radius ? 3. The driving-wheel of a lucomotive-engine of diameter C feet makes 2 revolutions in a second ; find approximately the number of miles i)er hour at which tlie train is going. 4. Supposing the earth to be a perfect sphere whose cir- cumference is 25000 miles, what is its diameter] 5. The diameter of the sun is 883220 miles, what is its circumference ? 6. The circumference of the moon is (5850 nules, what is ils radius ? 7. Find tlu! length of an arc which is -- of the whole cir- ' ^ 12 cumference, if the radius is 12 ft. 6 in. 5 8. Find the length of an arc which is '= of the whole cir- cumference, if the diameter is 21 feet. 9. The circumference of a circle is 150 feet, what is the eide of a sipiare inscribed in it ? 10. The circumference of a circle is 200 feet, what is the side c^a sc^uare inscribed in it / 11. A water-wheel, whose diameter is 12 feet, makes 30 revolutions per minute. Find approximately the number of miles per h(mr traversed by a point on the circumfcren <e of the wheel. 12. A mill-sail, whose length is 21 feet, makes 15 revolu- tions per minute. How many miles per hour does the end of the sail traverse ] Si ■k\ m m .! I H|!!, III. ON THE MEASUREMENT OF ANGLES. 16. Trigonometry was originally, as the name imports, the science which furnished methods for determining the magnitude of the sides and angles of triangles, but it has been extended to the treatment of all theorcuis involving the cou- Bideration of angular magnitudes. 17. Euclid defines a plane rectilineal angle as the inclina- tion of two straight lines to each other, which meet, but are not in the same straight line. Hence the angles of which Euclid treats are less than two right angles. In Trigonometry the term angle is used in a more ex- tended sense, the magnitude of angles in this science being unlimited. 18. An angle in Trigonometry is defined in the following manner. Let IFQE be a fixed straight line, and QP a line which revolves about the fixed point Q, and which at first coincides with QE. Then whm QP is in the position represented in the figure, we sav that it has described the an file PQK. ON THE MEASUREMENT OF ANGLES. II The advantage of this definition is that it enables us to consider angles not only greater tlian two right angles, but greater than four right angles, viz. such as are described by the revolving line wlien it makes more than one complete 1 evolution. 19. In speaking of a trigonometrical angle we must take into account the positicm from which the line that has described the angle started. Suppose, for instance, that QP, starting from the position (?£■, and revolving in a direction contrary to that in which the luui'ls of a watch resolve, has come into the position indicated in the figure. It has then described an angle EQ,^ greater than two right angles. 20. The magnitudes of angles are represented by numbers ex]<ressing how many times the given angles contain a certain angle lixed upon as the unit of angular measure. "When we speak of an angle ^, we mean an angle which contains the unit of angular measurement Q times. 21. There are three modes of measuring angles, called 1. The Sexagesimal or English method, II. The Centesimal or French method, III. The Circular Measure, which we now proceed to describe in order. I. The Sexagesimal Method. 22. In this method we suppose a right angle to be divided into 90 equal parts, each of wliich parts is called a degree, cicli (irgree to be divided into GO e(pial parts, each of which is called a minute, and each minute to be divided into GO equal parts, each of which is called a srcond. Then the magnitude I v V^ ■*"! n.;. 12 ON THE MEASUREMENT OE AAGLES. of an angle is expressed by the nnmher of degrees, minutes and seconds, which it contains. Degrees, minutes and seconds are marked respectively by the symbols °, ', ": thus, to repre- sent 14 degrees, 9 minutes, 37*45 seconds, we write 14\9',37"-45. 23. We can express i .'asure of an angle (expressed in degrees, minutes and seconds) in degrees and decimal parts of a degree by the following process. Let the given angle be 39° . 5' . 33'', 60 I 33- 60 I 5-55 •0925 .'. 39° . 5' . 33" = 390925 degrees. Examples.— iv. Express as the decimal of a degree the following angles 1. 24M6'.5", 4. ry. 28", 2. 37°. 2'. 43", 5. 375°. 4', 3. 175^0M4", 6. 78°. 12'. 4". li I II. The Centesimal Method. 24. In this method we suppose a right aftgle to be divided into 100 equal parts, each of which parts is called a grade, each grade to be divided into 100 equal parts, each of which is called a minute, and each minute to be divided into 100 equal parts, each of which is called a second. Then the magnitude of an angle is expressed by the number of grades, minutes and seconds, which it contains. Grades, minutes and seconds are marked respectively by the symbols «, \ '' : thus, to represent 35 grades, 56 minutes, 84*53 seconds, we write 35*^. 56'. 84"*53. The advantage of this method is that we can write down the minutes and seconds as the decimal of a grade by in- sjiection. ON THE MEASUREMENT OF ANGLES. 13 Thus, if the given angle be 1-1«. 19' . 57", 10 BJnce 19' —\^7^ of ^ grade = '19 grades, and 57" = 57 of a grade = •0057 grades, looiK) 14«. 19" . 57" = 14-1957 grades. 25. If the imniber expressing tlie minutes or seconds lias oily one significant digit, we must ]>reti.\ a cipher to occupy the place of tens before we write down the minutes and seconds as the decimal of a grade. Thus and 25«.9\54" = 25M)9'.54" = 25-0954 grades, 36«.8'.4" = 36«.08'.04" = 300804 grades. EXAMPLES.— V. Express as decimals of a grade the following angles ; 1. 25M4'.25", 4. 15\7"-45, 2. 38« . 4' . 15", 5. 425'^ .13'. 5" 54, 3. 214*. 3'. 7", 6. 2«.2'.2"-22. 26. The Centesimal Method was introduced bv the French mathematicians in the 18th century. Tlie advantages that would have been obtained by its use were not considered suf- ficient to counterbalance the enormous labour which must have been spent on the rearrangement of the Mathematical Tables then in use. Iir. The Circular Measure. 27. In selecting a unit of angular measurement we may take any angle whose magnitude is invariable. Such an angle is that which is subtended at the centre of a circle by an arc equal to the radius of the circle, as we bhall now prove. ? ■•'. 14 aV THE MEASUREMENT OF ANCLES. I I '11! m 28. To show that the angle subtended at the centre of a circle by an arc equal to the radius of the circle is the same for all circles. Let be the centre of a circle, whose radius is r; AB tlie arc of a quadrant, and therefore AOB a right angle ; AP an arc equal to the radius AO. Then, AP = r and AB = % (Art. 15.) Now, by Euc. vi. 33, angle ^OP_arc AP angle'TO^ ~ hycTa I? an_gle_i40P _ r aright angle ~7rr 2 or. TT Hence ang ^le^(9P = ^^-°^'*^^i^i^ TT Thus tlie magnitude of the angle AOP is independent of r, and is therefore the same for all circles. ON THE MEASUREMENT OF AXGLES, 15 29. In the Circular System of measurement the unit of aiunilar measuivment mav be described as (1) The aiij^le subtended at the centre of a circle by an arc equal to the radius of the circle, or, w liich is the same tliiii;^', as we proved in Art. 28, as (2) The au^de whose magnitude is the ttIIi part of two ri'dit an«dej. 30. It is important that the bcj^inner should have a clear conception of the size of this angle, and this he will best obtain by considering it relatively to the magnitude of that angular unit which we call a degree. Now the unit of circular measure two right angles 180° r^^o -^r^-r, ^ = ^^— = rm ,> A = 57-2958 nearly. 17 ;314159 '' Now if BG be the quadrant of a circle, and if we suppose the arc DC to be divided into 90 equal parts, the right angle DAG will be divided by the radii which pass through these points into 90 equal angles, each of which is called a degree. A radius AP meeting the arc at a certain point between the 57th and 58th divisions, reckoned from B, will make with AB an angle equal in magnitude to the unit of circular measure. Hence an angle whose circular measure is 2 contains rather more than 114 degrees, and one whose circular measure is 3 contains nearly 172 degrees, or rather less than two right angles. oi c- *i •*.<?•! 2 right angles 31. Since the unit of circular measure = — ^ ^^ — TV times the unit of circular measure = 2 right angles. Hence an angle whose circular measure is tt is equal to 2 right angles, a right angle, It 2' '^TT 4 right angles. •■; ■% ty A< i6 11 1 1* nv: M. , OJV riJE MEASUREMENT OF ANGLES. 32. To s/ioi/; that the circular measure of an avgle is equal to a j'lactinv, u-hich has for its numerator the arc subtended by that anr/le at tlie centre of any circle, and for its denominator the radius of iliat circle. Let EOD be any aii<,'le. About as centre and with any radius, describe a circle cutting OE in yl, and OU in R. Make angle AOP erpial to tlie unit of circular measure. Then arc AP = radi us A (Art. 29). Now, b .• Euc. VI. 3IJ, MvA^AOn AR liu^AiiAOP AP' AR :. angle AOR-=~r-p. angle AOP AR AU . aniile AOP lire ,.-— . unit of circular measure: aadiu.s the circular measure of angle AOR arc ludlUil . i\ ON THE MEASUREMENT OF ANGLES. 17 33, The units in the three systems, when expressed in terms of one common standard, two right angles, stand thus : the unit in the Sexagesimal Method = r^- of two right angles, the unit in the Centesimal Method = — of two right angles, the unit in the Circular Method = - of two right angles. vr ^ 34. It is not usual to assign any distinguishing mark to angles estinuited hy tiie Third Mttliod, hut for the purpose of stating the relation between the three units in a clear and concise form, we shall use the symbol 1' to express the unit of circular measure. V' il ''■■'I '* . » iffg Then we express the relation between the units thus ; 1 1 1 V : r = itiO • ^UO ■ tt' % i!i ; Ej 24° IV. ON THE METHOD OF CONVERTING THE MEASURES OF ANGLES FROM ONE TO ANOTHER SYSTEM OF MEASUREMENT, •. tl 35. We proceed to explain the process for Converting THE Measures op Angles from each of the three systems of measurement described in Chap. in. to the other two. 36. To convert the measure of an angle expressed in degrees to the corresponding measure in grades. Illlit! Let the given angle contain D degrees. 1 degree = j;-T of a right angle; :, D degrees = ^- of a right angle c=g-r of 100 grades 1C0I> '-90 gradea 10/; 9 gradoa. Hence we obtain tlie fuHowing rule : If an angle be expressed in degrees, multiply the measure in degrees by 10, divide the result by 9, and you obtain the measure of the angle in urades CONVERTING MEASURES OF ANGLES. T9 Ex. How many grades are contained in the angle 24° . 51' . 45" ] 24' . 51' . 45" = 24-8625 degrees 10 248-C25 grades 27-625 '. the angle contains 27« . 62' . 50". ■ { EXAMPLES.— Vi. Find the number of grades, minutes, and seconds in the following angles : I. 27M5'.46''. 2. 157". 4'. 9". 3. 24'. 18". 4. 19°. CIS". 5. 143°. 9'. 6. 28°. 7. 10° . 25' . 48". 8. 27°. 38'. 12". 9. 300°. 15'. 58". 10. 422°. 7'. 22". 37. To convert the measure of an angle expressed in grades to the corresponding measure in degrees. Let the given angle contain G grades. 1 grade = -—- ^f ^ J'io^'^ '^"o^^ » :. grades =-^- of a right angle = - .'j of 90 degrees 9()G , = ^^^^^ degr.es ^G , = ^^ degrees. Hence we ohtain the following rule : If an angle he exjjresucd in grades, multiply the measure in grades by 9, divide the resuit by 10, and you obtain the measure of tlie angle in degrees. 20 CONVERTING MEASURES OF ANGLES. Ex. How many degrees are contained in the angle 42«.34'.56'W 42* .34' . 56"' = 42-3'^56 grades 9 10 j 381 1104 degrees 38-11104 CO minutes 6-66240 • 60 seconds 39-74400 .-. the angle contains 38° . 6' . 39"'744. EXAMPLES.— Vii. Find the number ol' degrees, minutes, and seconds in the following angles : I. 19«.45\95*'. 2. 124«.5'.8". 3- 29' . 75\ 4. 15«. 0\ I5'\ 5. 154«,7'.24". 6. 43« 7. 38«.7r.20"-3. 8. 50« . 76' 94"-3. 9- 170'.63\27'\ lO. 324«. 13\88"-7. 38. If the, number of dcyreea in an cnujle U yiven, to find Ui circular measure. Let the given angle conttiin 1) degrees. V — TT^T. of two right niiiiles of IT units of circular measure 180 1 "l80 "•.--• units of circular measure*; .'. D°= , -. units of circular measure, 1 ov' COXVERTLVG MEASURES OF A ANGLES. 21 Hence we obtain the following rule : Jf an angle be expressed in degrees, multipbj the measure in degrees hij tt, divide the result by 180, and you- obtain the circular measure of the angle. Ex. Find the circular measure of 45° . 15'; 45° . 15' — 45 25 degrees ; .'. circular measure required is 45-25 X TT _ 45257r _ 9057r _ 1 81 tt "" 180~" ~ 18000 ~ 3G00 ~ 720 ' 1^ \t' Examples.— viii. 3iida in tlie Express in circular measure the following angles t 5 9 10 G0°. 2. 22°. 30'. 3. ir.l5'. 315°. 6. 24°. 13', 7. 95°. 20'. The angles of an equilateral triangle. The au'des of an isosceles riuht-angled triangle 4. 270°. 8. 12\5'.4". 39. If the rirndnr measure of an angle be given to find the number of degrees which it contains. Let 6 be the given circular measure. 1«= - of two right angles TT w- of 180 degrees 180 .*. <9«= TT 0.180 TT degrees ; degrees. ITonce we obtain the following rule : If an angle be expressed in circular measure, multiply the measure by 180, divide the result by tt, and you obtain the mea- sure of the angle in degrees. CONVERTING MEASURES OF ANGLES, Ex. Express in degrees the angle whose circular measure IS 57r The measure in aegrees= — ^ = - =112-5 degrees. O X TT O Examples.— ix. Express in degrees, etc., the angles whose circular measures are IT TT I. 2' 2. ^. TT IT 6" 277- 5- 3. ,1 1^1 1 2 6. 2- 7. 3. 8. ^. 9- ^- 10. 3. 40. Similar rules will hold with respect to the equations for connecting the centesimal and the circular systems, 200 being put in the place of 180 : thus circular measure of an angle containing G grades = (r. „--^, number of jirades in the anule whose circular measure is ^=^ 200' 200 Examples.— X. Express in circular measure the following angles: I. 50«. 2. 25». 3. 6«.25'. 4. 250«. 5. 500«. 6. 13«.5\5". 7. 24«.15\2"-15. 8. 125«.0M3". 9. 3'. 10. 5". Examples.— xi. Express in grades, etc., the angles whose circular measures are TT r IT 2r 37r I. 3* 2. 6* 3- G* 4- 3* 5- 6- 1 1 1 3 .^ 6. ij' 7. 5" 8. 8' 9- 5' 10. 2 U ^11,.,: r measure measures ucasurea CONVERTIISfG MEASURES OF ANGLES. 23 41. We shall now give a set of miscellaneous examples to illustrat the principles explained in this and the two preced- ing chapters. Examples.— xii. {i^oU, Circular measure is not introduced till Ex. 17.) 1. If the unit of angular measurement be 5°, what is the measure of 22^° ? 2. If an angle of 42^" be represented by 10, what is the unit of measurement ? 3. An angle referred to different units has measures in the ratio 8 to 5; the smaller unit is 2°, what is the other? Express each unit in terms of the other. 4. An angle referred to different units has measures in the ratio 7 to 6; the smaller unit is 3°, what is the other? Express each unit in terms of the other. 5. If half a right angle be taken as the unit of angular measurement, what is the measure of an angle of 42° ? 6. Compare the angles 13° . 13' . 48" and 14« . 7\ 7. If T) be the number of degrees in any angle and Q the number of grades, show that G=iD+ -A 8. An equilateral triangle is divided into two triangles by a line bisecting one side ; express the angles of these two tri- angles in degrees and grades respectively. 9. If the angles of a triangle are in arithmetical progres- sion, show that one of them is 60°. 10. Reduce 39*"()12 to degrees, minutes, and seconds. 11. If there be m English minutes in an angle, find the number of French seconds in the same angle. 12. "What fraction of a right angle must be the unit, in order that an angle of 5° . 33' . 20" nuiy be represented by 5 ? 13. What must be the unit angle, if the sum of the measures of a degree and a grade is 1 ? ' ('■•'^i ^.■M m 14 CONVERTING MEASURES OE ANGLES. 14^ If there be three aii^'les in arithmetical pr()<:jression, and the ininiber of .grades in the greatest be ec^nal to the number of degrees in the sum of tlie other two, the angles are as 11 : 19 : 27. 1 80" 15. Prove that — ,-- = 1 1 5« . 47' Jiearl v. 16. The three angles of a triangle are in arithmetical pro- gression, and the number of grades in the least : the number of degrees in the greatest .: 2 : 9. Find the angles. 17. It being given that the angle subtended by an arc equal to the radius is 57°"29577, find the ratio of the circumference of a circle to its diameter. 18. Two angles of a triangle are in magnitude as 2 : 3. If the third angle be a right angle, express the angles of this triangle in each of the three systems of measurement. 19. Two straight lines drawn from the centre of a circle contain an angle subtended by an arc which is to the whole circumference as 13 : 27 ; express this angle in degrees. 20. An arc of a circle is to the whole circumference as 17 : 54; express in grades tlie angle A'/hich the arc subtends at the centre of the circle. 21. Determine in grades the tnagnitude of the angle sub- tended by an arc two feet long at the centre of a circle whose radius is 18 inches. 22. One angle of a triangle is 2 in circular measure, and another is 20" : find the numlier of grades in tlie tliird. 23. An arc of a circle, whose radius is 7 inches, sul)tend8 an angle of 15°. 39'. 7"; what angle will an arc of the same length subtend in a circle whose radius is 2 inches? 355 24. What is the circular measure of 11'. 30' if 7r = ~-^ % 25. if the numerical value of an angle measured by tli<^ circular system be \\\ ■> bow many degree.^? does it contain ? ■1 CO.VVERTmG MEASURES OF AACLES. 25 26. Tlic whole circumference of one circle is just long enough to subtend an angle of one grade at the centre of another circle : what part of the latter circumference will subtend an angle of 1° at the centre of the former circle ? 27. Taking 4 right angles as the unit, what number will represent 1°, 1*, T respectively ? 28. The earth being supposed a sjjhere of which the dia- meter is 7980 miles, find the length of 1° of the meridian. 29. If half a right angle be the unit of angular measure- ment, express the angles whose measures are ^ 4, TT, 4m + -, (i) in degrees, (2) in units of circular measure. 30. If the unit be an angle subtended at llie centre of a circle by an arc three times as large as the radius, what number will represent an angle of 45° ? 31. Express in degrees : (i) The angle of a regular hexagon. (2) The angle of a regular pentagon. 32. Express in grades : (i) The angle of a regular pentagon. (2) The angle of a regular octagon. '>)Tf. Express in circular measure : (i) The angle of an equilateral triangle. (2) The angle of a regular hexagon. 34.. Find the circular measure of the angle of a regnlat polygon of n sides. 35. The radius of a circle is 18 feet, find the length of an arc which sid)tends an a:igle of 10° at the centre. 36. T1k» angles in one regular pcilygon arc twice as many ;is those in anothtr polygon, and an angle of the former : an angle of the latter :: 3 : 2. Find the number of the sides in each. :ii!t .':;'• i., V. ON THE USE OF THE SIGNS + AND - TO DENOTE CONTRARIETY OF DIRECTION. 42. In a science which deals with the distances measured from a fixed point it is convenient to have some means of dis- tim^iiishinf^ a distance, measured in one direction from the point, from a distance, measured in a direction exactly oppo- site to the former. This contrariety of direction we can denote by prefixing the algebraic signs H- and — to the symbols denoting the lengths of the measured lines. 43. It must also be observed that magnitudes of things cannot properly be made subject to the rules and operations of Algebra, as these rules and operations have only been proved for algebraical symbols. We must therefore find some algebraical representative for any magnitude before we subject it to algebraical operations : such a representative is the measure of that magnitude with the proper sign prefixed. 44. "We explained in Chapters i, and iir. the principles of algebraical representation as applied to the measures of lines and angles, and we have now to explain the rules by which we are enabled to express contrariety of direction in the case of lines and angles by employing the signs + and - 45. Suppose that two straight roads NSy WE intersect one another at right angles at the point 0. CONTRARIETY OF DIRECTION'. 27 A traveller comes along SO with the intention of going to^. N W- 1-^S Suppose OE to represent a distance of 4 miles and 0\V to represent a distance of 4 miles, and suppose the traveller to walk at the rate of 4 miles an hour. If on coming to he makes a mistake, and turns to the left instead of the right, he will find himself at the end of an liour at W, 4 miles further from E than he was when he reached 0. So far from making progress towards his object, he has walked away from it: so far from gaining he has lost grcjund. In algebraic language we express the distinction between tlie distance he ought to have traversed and the distance he did traverse by saying that UE represents a positive quantity and U\V a. negative (quantity. 46. Availing ourselves of the advantages afforded by the use of the signs + and - to indicate the directions of lines, we make the following conventions : SS ' CONTRARIETY OF DIRECTION. (1) Let be a tixed point in any straight line BOA. B A Then, if distances measured from in the direction OA be considered positive, distances measured from in the direc- tion OB will pioperly be considered negative. Hence if OA and OB he equal, and the measure of each be m, the complete algebraical representative of OA is w, whereas that of OB is - w. The direction in which the positive distances are measured is quite indifferent; but when once it has been fixed, the nega- tive distances must lie in the contrary direction. (2) Let be a fixed point iu which two lines AB, CD cut one another at right angles. B' ,3 ■A Then, if we regard lines measured along OA and OC as positive, we shall properly regard lines measured along 0J> i and 01) as negative. This convention is extended to lines' parallel to AB or CD in the following manner : CONTRARIETY OF DIRECTION. 29 Lines parallel to CD are positive when they lie above AB^ negative Lelow A B. Lines parallel to AB are positive when they lie on the right of CD, negative when they lie on the left of CD. 47. We may now proceed to explain how the position of a point may be determined. iV.Vand WE are two lines cutting each other at right angles iu the point 0. N •I w- P' •p ••■':il The position of a point P is said to be known, when the lengths of the perpendiculars dropped from it on the lines NS and JVE are known, provided that we also know on which bide of each of the lines NS and IFE the point P lies. If tile perpendicular dropped from P to JP''E be above JFE, it is reckoned positive. If the perpendicular dropped from P to JFE be heluiv IFE\ it is reckoned negative. If the perpendicular dropped from P to NS be on the right ot NS, it is reckoned positive. If the perpendicular dropped from P to NS be on the left of NS, it is reckoned negative. •> H 30 CONTRARIETY OF DIRECTION. 48. Angles in Tni^'onometry must be considered not with respect to tlieir nia<,'nitude only, but also witli reference to their mode of generation ; that is to say, we shall liuve to con- sider whether they are traced out by the revolution of the generating line/ro?Ai riijht to left or from left to riyht. 49. Let a line OP starting from the position OE revolve about in tlie direction KNWSE-, that is, in a direction contrary to that in which the hands of a watch revolve. N \i' -E 'iii,'' Then all angles bo trarod out are considei'ed jiosUive, When OP reaches the li.ie ON it will liavi' traced out a right angle, Oir two right angles, OS three right angles, OE four right angles. If we suppose the line to revolve in the direction ESIVNK, we may i)roperly account the angles traced out by it to be neijaiive angles. For the sake of clearness we shall call OP the revolving line, uud OE the primitive line. CONTRARIETY OF DIRECTION. -Ji 50. Now suppose P to be a point in tlie revolving line OP. Let a perpendicular let fall from P meet the line FAV in the point i\/, and let this be done in each of the four quarters made by the intersection of NS and WE, as in the diagrams in the next article. Then in the first quarter PM is positive and DM is positive, second Pilf is positive and OiU is negative, thi'd PM is negative and OAf is negative, fourth Pil/is negative and DM is positive. Note. — When we say that a line PM is positive or negative, we mean that its measure has the + or - sign prefixed to indicate the direction in which it is drawn. Thus if the measure of PM be p, the complete algebraical representative of PuM will be p or -p, according as PM is above or below JVIiJ. So also if the measure of CM be q, the complete algebraical representative of OM will be g or -3, according as OM is on the right or left of NH ... : ^j . 1 1 . m ' ••■M I I. M VI. ON THE TRIGONOMETRICAL RATIOS. 61. Let the line OP revolving from tlie position OE about from right to left describe the angle EuP, which we shall call the Angle of Reference. From P let full the perpendicular PM on the line KOW. We then obtain a riglit-angled triangle POM, which we shall call the Triangle of Ueference. (1) (2) (^) V/ r/ \p M W M ^ M /" v/ 'Ch. "E w" (4) /U"\ M P- \ In fig. (1) the angle of reference i.s an acute angle. In fig. (2) it ia an obtuse angle. In fig. (3) it is greater than two right angles but less than three right angles. In fig. (4) it is greater than three right angles but less than four right angles. ON THE TR I GOXO METRICAL RATIOS. 33 Then the ratio PM . (1) y-p is defined to be the sine DM of the angle EOF, (2) (4) PM (Jp PM UM or (5) -^-^^ (G) OM PxV/ 62. It will be observed that cosecant EUP= , - secant EOP cosine tangent cosecant secant cotangent sine EOP* 1 cosine EOP^ cotangent EOP = - .-^tyd* " tangent EOP 53. The words sine, cosine, etc., are abbreviated, nnd the tri^'uiiunietrical ratios of an angle A are thus written : sin A, cos A, tan A, cosec A, sec A, cot A. 54. The defect of the cosine of an angle from unity is called the versed sine, tiius : versed siiu' EOP = l~ coh K()I\ The words versed sine are abbreviati'(l t»i raisin. 55. Tlie ])owers ot the trigononietricid ration are expressed in the I'ollowiug way : (sin A)^ is written thus, sin-'/l, (cos A)'^ is written thus, ros''i4, anil so for the otlier ratios. 1 •• j'it 51 ^1 I" 'Til m 34 ON THE TRIGONOMETRICAL RATIOS. 56. We have given the ratio-definitions in the most general form, but we shall lor tlie present confine the attention of the student to the particular cases of the Ratios of j!\.cute and Obtuse Angles, with which we are chiefly concerned in this treatise. Ratios for Acute Angles. 57. The six trigonometrical ratios are arithmetical quantities, denoting the relations existing between the sides of a right- angled triangle, which we call the Triangle of Reference, taken two by two. 58. Let us now look at the order in wl.ich the sides are taken to form the ratios which we call the sine and the cosiric of an acute au'de. Eeferring to fig. (1) of Art. 51, sine EOP = cosine EOP = Pil/ OP' OM OP' In each case the denominator of the fraction is formed by that side of the triangle of reference which is opposite to the right angle. Of the two other sides we may call PM the side opposite to the angle of reference, so as to distinguish it from OM Die side adjacent to the angle of reference. Hence in determining the ratio v;hich we call the sine of a given acute angh; we must take as tlie numerator of the fraction that side of our triangle of reference which is opposite to the given angle; and in determining the ratio which we call the cosine of a given acute angle we must take as the numerator of the fraction that side of our triangle of reference which i^5 adjacent to the given angle : tlu! denominator being in both cases that side of the triangle whioli subtends the riglU angle of tho triangle. •o' Note. — Omitting the word acute in tin's Artii le, the remarks will be applicable to all the diagrams uf Art. 51. OM THE TRIGONOMETRICAL RATIOS. 35 50. If tlie reiiiark-s j^iven in tlie prcceaint,' Article be clearly umlerstood, the student will find no difficulty in wiiting down the ratios for the sine and the cosine of a given acute angle, whatever may be the ])()sition in which the triangle of reference for that angle may .stand. Suppose Pil/ to be perpendicular to OA/, BN to be perpendicular to l^M. Then I'OM, PDN, DMN are three right-angled triangles. ' ■•..'■ill ■f'l'jl Now Also And sin POA/= PM UP ' sin 0PM =- sin I)PN= DM 7\V 8in/W.V= BmNDM= DX J)M' KM DM COS POM= ^-p. OP • r>N PD' cos 0PM = PM OP' PN cos DPN= pj.. cos J*I)X=--j. COS DMX = y-p, r. cos NDM-- py J)M' CO. When once the student has acquired facility in fixing on the lines which form the ratios called the tnne and the cos/?}?, he will be able to determine the other four ratios with- out unv trouble. ■' i . ill 36 ON THE TRIGONOMETRICAL RATIOS. Examples.— xiii. 1. Let ABG be a triangle. Draw from B a perpendicular Bl) on AG^ and let it be within the triangle. Then write the following ratios : sin BAD, cos BAD, tan BAD; Bin A BD, cot A FJD, cosec ABD ; sin BCD, sin CBD, tan BCD. 2. Let ABC be a right-angled triangle, having B as the right angle, and let the angles be denoted by the letters A, B, G, and the sides respectively opposite to them by the letters a, 6, c, A S 3 Show that a = 6 . sin yl =b . cos G= c . tan A = c . cot C, h = a . cosec A = a . sec C= c . sec A=c . cosec (7, c = a . cot A =a . tan C=/> . cos A = h . sin C. .ATofe. — These results are worthy of notice, as being of fre- quent use in a later part of the subject. 61. The trigonometrical ratios remain unchanged so long as Oie angle is the same. M V . >i •1 ON THE TRIGONOMETRICAL RATIOS. 37 Let EOB be cany angle. In OB take any points P, F, and draw PM, P'3/' at right .» angles to OE. Then, since PM, P'W are parallel, the triangles 0PM, OP'M are similar. PJl_P^M'^ Hence ^^j ~ /in/ » i.g. the value of tlie sine of EOB is the same so long as the angle is the same, and this result holds good for the other ratios. The figure represents the simplest case, where the given angle is less than a right angle, but the conclusion is true for all angles. Ratios for Obtuse Angles. 62. Suppose A CB to be an obtuse angle. Draw AD at right angles to BG produced. a 1 1 1 r '\ i •|>' I .11 V -1 * Then, regarding ^ C/i as an angle described by CA revolving round G from the position CB, em ACB = --,r,f AC COS ACB = -r,. AU Now suppose the measures of AD, AC, DC to be 7), 7, r respectively. Then tlie complete algebraical representative of^Dis +/>, of AC is +5, of 67^ is -r. for CD is measured from G'iu a direction exactly opposite to that of the primitive line CB. :. 8inylC7;=^, co8/10Z?=--. 2' (I Vis m 38 ON THE TRIGONOMETRICAL RATIOS. 63. Ill the ajiplication of Algebra to Geometry it is the practice of most writers to use the geometrical representative of a magnitude wliere the algebraical representative ought to be employed. Then suppose f and (/ to be the measures of two AV lines AB^ CD, we often find the fraction -7% where we ought in strictness to find the fraction -. This hjose method of nota- tion is, however, sometimes less cumbersome, and we shall therefore retain it at the risk of a slight want of clearness. 64. Whenever we represent the ratios of lines algehraicaUy, we must be careful to put the complete a]gel)raical representa- tive for each line. This cannot be t(jo strongly impressed on a beginner, and we therefore give another illustration of it. Let EUlVhe the primitive line and a diameter of a circle, NOS a diameter at right angles to EOJF, and POP' any other diameter. Draw PM and P'AF at right angles to EOIK Let p be the measure of PM and P'M\ r the measure of the radius. Then the ratio PM : PO is represented algebraically by --, Out P'M' '. FO by-~-^. VII. ON THE CHANGES IN SIGN AND MAGNITUDE OF THE TRIGONOME- TRICAL RATIOS OF AN ANGLE AS IT INCREASES FROM 0' TO 360\ 1 ~ I 65. Let NS, IVE bisect each other at right angles in the point 0, and let a line equal in length to OE be supposed to revolve ^n the positive direction from OE round the fixed point 0. Let r be the measure of OE, As the revolving line passes from the position OE to the positions OiV, OIV, OS, OE, the extremity traces out a circle , t •II m t-\\i ■Nl 40 CHANGES /y S/GJV AMD MAGNITUDE. If we take a siiecesHion of points in ^A'', as P, P, P, P, and from them let fall perpendiculars PM^ PM, PM, PM on the line OEy and do the same in the other (|iiadraiits, it is clear that In passing,' from E to N, PM (.'ontinuallv increases from zero to r, OM continually decreases irom r to zero. In passing from N to W, PM continually decreases from r to zero. OM continually increases from zero to r. In passing from W to 8, PM continually increases from zero to r, OM continually decreases from r to zero. In passing from -S' to E, PM continually decreases from r to zero, OM continually increases^ from zero to r. Again, Pilf '3 positive in the first and second quadrants, negative in the third and fourth. OM is positive in the first and fourth quadrants, negative in the second and third. OP is always positive, and always = r. fiG. To trace the changes in sign and magnitude of the sine of an angle as the angle increases from 0° to 300°. Let NOS, EOJF he two diameters of a circle at right angles. Let a radius OP, whose measnre is r, by revolving from OE trace out any angle EOP, and denote this angle by A, From P draw PM at right angles to EOJV. PM f: Then sin A = OP' C//AA'GES AV S/GN AND MAGNITUDE. A\ b As A increases from 0° to 90", OP revolves from OZ? to O.V, .*. PM increases from to r and is positive, OP is always = r positive; .*. sin A increases from to 1 positive. As A increases from 90" to 180°, OP revolves from ON to OW, :. PM decreases from r to and is positive, OP is always = r positive; .'. sin A decreases from 1 to positive. As A increases from 180° to 270°, OP revolves from OW to OS, :, PM increases I'rom to r and is negative, OP is always — r positive; .•. sin A increases from to 1 negative. As A increases from 270° to 360°, OP revolves from OH to OE, :. PM decreases from r to and is negative, OP is always = r positive; /. sin A decreases from 1 to negative. 'i i •t »i'. h:. r V ■; .' M .-> '■m lif ,r 4 J C//.'LV(7/:S /.V S/(7X .'LVD MACXl'ITDE. 07. To trace, the cliaufiea in the sirpi and iiwrjvitude of the cosine of an amjlc as the aiujlc increases from 0° to oG0°. Making the same coii.sLruction as in Art. 00, As A increasos from 0° to 90°, OP revolves from OE to OiV, .*. OM decreases from r to and is positive, Oi' is always = r positive; .'. cos A tlecreases from I to positive. As A increases from 00" to 180", C>i' revolves irom ON to OIF, :. OM increases from to r and is negative, OP is always = r positiv^e; .'. cos A increases from to 1 ne;j;ative. As A increases from 180° to 270°, OP revolves from OIV to OS, .*. OM decreases from r to and is negative, OP is always = r positive; .•. cos A. decreases from 1 to negative. As A increases from 270° to 300", OP revolves from OS to oi:, :. OM increases from to r and is j)ositive, OP is always = r })ositive ; .*. cos A increases from to 1 positive. 68. To trace the changes in the sign and magnitude of the tangent of an angle as the angle increases from 0° to 3G0\ Making the same construction as in Avt. GG, .V ..J Cf/ANCICS /.V'.SVGW AND MAGNITUDE. 43 As A iiicrcurius from 0" to 90", OV revolves from OK to CiV, /. VM iiicrcaw.'.s from to r jiiid is jxtnitivc, Oil/ decreases liom r to ])OHitive ; /. tan A increases from to qo positive. As A increases from 90° to 180°, OV revolves from O.V to c/r, /. Pilf decreases from r to and is jiositive, OM increases from to r n(';^ative ; .*, tan A decreases from oc to ue;,'ativc. As A increases from 180° to 270°, OV revolves from OW to 0.S', /. VM increas(!S from to r aiid is negative, Oil/ decreases from r to negative ; /. tan A increases from t(j co positive. As ^ increases fiom 270° to 300°, OV revolves from 06' to OE, :. PM decreases fn)m r to and is negative, Oil/ increases from to r positive ; /. tan A decreasea from co to negative. Note. — The pymlxd 00 is used to denote numbers which are infinitely great, and the symbol is used to denote numbers whicli are infinitely small. When we say that 7- = 00 , we mean that if any finite number r be divided by a number infinitely small, the ([uotient is a nmnber infinitely great 69. When A is less than, Init very nearly etjual to 90°^ tan A is very large and positiv^e ; and 'when A is very little greater than 90°, tan A is very large and negative. This is exjnvssed by saying that the tangent of an angle changes sign in i)assing through the value 00. To ex})lain this more clearly we give another method oi" tracing the changes in the sign and magnitude of tan Ay as A increases from 0° to 18()°. (■ 1 'ill 'I' ■• ^ a- * 44 CHANGES IN SIGN AND MAGNITUDE. Let A'0»S, EOJV ha two diameters of a circle at right nngles. Suppose a line OP revolving from the position OE to trace ort any angle EOF, and denote this angle by A. Draw EC, IV D at right angles to EOIF, and let them meet the revolving line in any points P, P. Let the measure of OE be r. D P S Then as A increG"='PS from 0° to 90", tan A = ^jj^,, EP increases from to x nnd is positive, 07? is always = r positive; .". tan A incrtises from to oo positive. As A increases from 90" to 180", JrP (U'creases from on to and is positive, O^r is always = r negative; .'. tan A decreases from oo to negative. Thus as the revolving line i)i\sses from one side of OiV to the other, tan A changes from f oo to -oo. 70. The changes of tho cosecant, secant, and cotangent phould bo traced for himself by the student for practice. CHANGES IM SIGN AND MAGNITUDE. a: 71. We now present the changes ol tlie trigonometriciil ratios in a convenient tabular form. Columns 1, 3, 5, 7, 9 give the values of the ratios for the particular values of the angle placed above the columns. Columns 2, 4, G, 8 give the signs of the ratios as the angle ]. asses from 0° to 90°, from 90'^ to 180° from 180' to 270", ami ii'om 270' to 3G()°. sin A cos A tan A cosec A sec A cot A lo' 90° 180° 270' -1 i ano" -^■ 1 + — 1 + — -- I - + 1 ■f 00 — + 00 __ 00 f 1 + 00 - -1 - 00 1 + X — -1 - X 4- 1 00 f <» — 00 + ~* 00 72. The sine cosecant Tl le and cosine are never greater than unit)', and secant are nevi-i less than unity. The tangent and cotangent have till values from zero to in Unity. 73. The trigonometrical ratios change sign in passing through the values and x and for no other values. 74. Froni the results given in the table (Ait. 71), we are led to the following conclusion, which will be more lully t'.\- plained hereafter. If the value of a trigonoiiu'trical ratio be given, we cannot fix on one angle to which it e.vclusively belongs. Thus if the given value of sin A he ^^, we know, since sin .1 Mi ]>asses through all values from to 1 as A increases from 0° to 90°, that one value of A lies between 0" and 90". Ibit since we also know that the value of sin A j)asses through all values between 1 and as A increases from 90'' to 180 , it is evident lliut there is another value of A lietween 90" and 180° for which sin A ~ I ■ }^ i-, >i :::^ f : 4 'i ■^ ■1 k--, n' h\Umi tif'l ijii'f: VIII. ON RATIOS OF ANGLES IN TH1£ FIRST QUADRANT. 75. "We liave now to trout of tlio vjiluos of the triErono- metnoiil ratios ior soniu ])arfii'iilar an^'le.s in tlio lirst (luadraiit. Thesic aii-^los, wliicli we .sliall take in tlioir ])roi)er order, as tlicy are trac:e<l out by tlie revolving lino, are 0°, 30°, 45°, G()\ 90°. 7(5. The sij^'ns of all the ratios for an an^de in the first ([uadrant are positive. 77. To find (he triijnnomdrical ratios for nn angle of 0" We have already proved in the preceding chapter that sin 0° = 0, cos 0° =-. 1 , • tan 0° = 0. cosec0° = x, secO°=l, cotO° = oo. 78. To find the trigonometrical ratios for an angle o/30°. N Lot OP revolving from the ])osition OK descriho an nn^lo KOP vi\\\i\\ to one-third of n rijL;ht an,<^le, that is nn angle of 30". ANGLES IN THE FIRST QUA V^ANT. 47 Draw the chord VMV at ri^^ht ,'ui<;h>a to 0/i', and join OV Tlien angle OrP = OPr = 90'-i'Oil/=C(r. Tims POP' ia an equilatoral trinn.^le, and OM bisects PP' 0P = 2PM. Lot tl le measure of PM be m e measure And the measure Then tl of OP is, 2m. o Then sin 30" f OM is ^1(41)1^ - m2)= ^/(:3m^ = m . ^^3. PM 01 ■ m J o m 9» COS 30° = tan 30" = 0M_m^f:i_J3, OP ~ 1^7/fc ~ 2 ' PM_rri ^ 1_ ' if So also coscc 30"- 2, sec 30°= -" ,, cot 30°= ^3. 79. To find the trigonometrical ratios for an angle of 4b'' Let OP rovnlvins frnni tin- position OE describe nn angle KOP eciunl 1o ball' a riglit angle, tb s an angle ul" 45°. Dr."\v PM at right angles to OK. TliLH ijiinee POM and OPM are logelher equal to a rigbt ingle, and POM ia hall' a ri-ht angle, 0PM is also half a right uiigle. pi|iil' i 1.11 lip!; ' • •iiji 48 ANGLES IN THE FIRST QUADRANT. Thus FOM is an isosceles triangle, and OM^PM. Let the measure of OM be m. Then the measure of PM ism. And :he measure of OP is ^(m^ + m-)^ y,f{2m') = m»j2. 1 Theu OM m 1 O = - = — OP mV2 V2' tan 45°=- =- = 1. CiVi m So also cosec 45°= ^2, sec 45°= ^2, cot 45° = 1. 80. To find the triyonometncal ratios for an anyk of GO", N Let OP revolvinf]j from the position OE describe an anpjle EOP equal to two-t'.iirds of a right angle, that is, an angle of G()°. Draw PM at right angles to OE, and join PE. Then POE is an e(|uilateral triangle, and PM bisects OJJJ, :. 0P = 20M. Let the measure of OM be m. Then the measure of OP is 2m. And the mcaoure of PM is ^f(-\,n'^ ~in'^^ x^(37?r)-rM ^3. ! $ Th( 81. We 6. 7. f ill . 1 ANGLES IN THE FIRST QUADRANT. 40 Then sin 60° 0P~ 2m " 2~' ^_^^o OM m \ 008 60=-^ = -^-^ = -, tan6(r = ™=-V-?=V3. So also 1 cosec 60° = — -, sec 60° --= 2, cot 60° = - , -. 81. To find the trigonometrical ratios for an angle of 90", We have already proved in the preceding chapter that sin 90° = 1 , cos 90° = 0, tan 90° = oo , cosec 90° = 1, sec 90° = 00 , cot 90° = 0. Examples.— xlv. If a = 0°, ^8=30°, 7 = 45°, 8 = 60°, = 90°, find the numerical values of the following expressions : I. cos a. sin y .cos 8, TT 2. sin . cos - . cosec 8. ^. sin ;= + cos ^ - sec a. 4. sin „ . cosec ^ . sec 5. "^2 6 3 2 5. (sin <? - cos + cosec /3) (cos 6 + sec -^ + cot 8 ). Also prove the following : 6. (sin 8 - sin y) (cos /S 4- cos y) = sin''' /J. sin^ ^ - sin** T 7. COt^ - -COt^ -; =——— — . 4 b . ., TT . ., TT 4 8. (sin^^-cos^)(sin3-cos3) = co8 3. fs.T.] II so ANGLES IN THE FIRST QUADRANT. Xmmy ■f] m d: iiiifc"" Niii i;l!' TT TT COS -- . COS ' «3 o r 1 /tt 7r\ 1 /tt 7r\ oTT TT lO. siir - - sin-^ - tin2--tnTi2-_. '^ " 3 TT TT COS- .J . COS- - 82. We are now ahle to give some simple examples of the practical use of Trigonometry in the measurement of lieighta and distances. 83. The values of the sines, cosines, tangents, and the other ratios have heen calculated for all angles succeeding each other at intervals of 1', and the results registered in Tables. Instruments have been invented for determining : (1) The angle which the line joining two distant objects subtends at the eye of the observer. (2) The angle which a line joining the eye of the observer and a distant object makes with the horizontal plane. If the object be ahove, the observer, the angle is called the Anijle of Elevation. If the object be below the observer, the angle is called the Angle of Depression. 84. To find the height of an object standing on a horizontal plane, the base of the object being accessible. Let PQ be a vertical column. From the base P measure a horizontal line AP. Then observe the angle of elevation QAP. Q A P We can then determine the height of the column, for QP'^AP. tan QAP. iiii ANGLES IN 'J HE EIRST QUADRANT. SI 85. To find the breadth of a river, R s Let RS he the horizontal line joining two objects on the opposite banks. From 0, a point in a vertical line with R, observe the angle of depression OSR. Then if OR be measureil, we can determine the length of RS, for tan~0-Si2* 86. To find the height of afiag-staff on the top of a towtr» A p Let RQ l)e the flag-stafT. From P the base of the tower measure a horizontal line AP Observe the angles RAP and QAP. Then we can find the length of RQ, for RO=RP-QP -= ^P . tan RAP - AP . tan QA P. ■ { • .1. u S2 ANGLES IN 771 E FIRST QUADRANT. ■ ■■^::|it I !li' tt! r ri'i; i::i^ Ml; r •' in 'in 87. To find the altitude of the sun. The altitude of the sun is measured by the angle between a horizontal line and a line passing through the centre of the sun. If AB be a stick standing at right angles to the horizontal plane QR, and QB the sliadow of the stick on tlie horizontal plane, a line joining i^A will pass through the centre of the sun. Q, B P . Then if we measure AB and QB, we shall know the altitude of the sun, for AB tan SQR = QB- Examples.— XV. 1. At a point 200 leet from a tower, and on a level with its base, the angle of elevation of its summit is found to be 60° ; what is the height of the tower] 2. Wliat is the height of a tower, whose top appears at an '■levation of 30° to an observer 140 feet irom its foot on a horizontal plane, his eye being 5 feet from the ground '? 3. Determine the altitude of the sun, when the length of a vertical stick is to the length of its shadow as v'^ : 1. 4. At 300 feet measured horizontally from the foot of a steeple the angle of elevation of the top is found to be 30^, what is the height of the steeple ? ; Ml' ANGLES IN THE FIRST QUADRANT. S3 ivel with be 60° ; ars at an Dot on a ? ength of bot of a be 30% 5. From the top of a rock 245 feet above the sea the aiiL^le of depression of a ship's hull is found to be 30° ; how fur is the ship distant? 6. P'rom the top of a hill there are observed two consecu- tive milestones, on a horizontal road, runnin.^ from the base. The anj^des of dcprcbsion are found to be 45° and 30". Find the height of the hill. 7. A flag-staff stands on a tower. I measure from the bottom of the tower a distance of 100 feet. I then find that the top of the flag-staff subtends an angle of 45°, and the top of the tower an angle of 30° at my place of observation. What is the height of the flag-staff ] - 8. From the summit of a tower, A\liose heiglit is 108 feet, the angles of depression of the top and bottom of a vertical column, standing on a level with the base of the tower, are found to be 30° and G0° ; find the height of the column. 9. A person observes the elevation of a tower to be G0°, and on receding from it 100 yards further he finds the eleva- tion to be 30° ; required the height of the tower. 10. A stick 10 feet in length is placed vertically in the ground, and the length of its shadow is 25 feet ; find the altitude of the sun, having given tan 25° = '4. 11. A spire stands on a tower in the form of a cube whose edge is 35 feet. From a point 23 feet above the level of the base of the tower, and 20 yards distant from the tower, the t'levHtion of the top of the spire is found to be 56° . 34'. Find the height of the spire, having given tan 5(5° . 34'= 1*5. 12. The length of a kite string is 250 yards, and the angle of elevation of the kite is 30° ; ihid the height of the kite. 13. The height of a housetop is 60 fet;t. A rope is stretched from it, and is inclined at an angle of 40° . 30' to the ground. Find the length of the rope, if sin 40° . 30'= '65. 14. A tower on the bank of a river is 120 feet high, and ihc. angle of elevation of the top of the tower from tlie opposite ba)ik is 20°; find the river's breadth, if tan 20° = '35. 15. The altitude of the sun is 36°. 30'; what is the hiigUi of the shadow of a man 6 feet high, if tan 36° . 30'= '745 \ !•.;( ' 'r1.V .:! ni! I 'IT iiili'i IX. ON THE RELATIONS BETWEEN THE TRIGONOMETRICAL RATIOS FOR THE SAME ANGLE. *•;>'.: ilHj, Ml!' iiiii nil! ii 88. Let EOP be any angle traced by OP revolving from the position OE, and let a perpendicular PM be dropped on OE or EO produced, thu3 : Q N\ E M A Let the angle EOP be denoted by A. Then we can prove the following relations : _ , . sin A I. tan A= i. cos A PM ^ , , PM (J'P sin A (jM OM cos A OP IL 8iuM4co8M = L „ . 2 , , , PM^OM^ For Bin^ A f cos^ A= r^ui-^ i OP' ' 0P2 0P-' c)7^"2 L M E P\. ' ll'!l' RATIOS FOR THE SAME ANGLE. 55 III. secM = 1 + tau'-^ ^. „ , OP- OM- + P.U'^ , Pil/2 , , 2 ^ OiV/-^ OM'-^ OF^_VM'^->rOM^ PM'~ Pil/2 For cosec- A — UM^ 89. We shall now give a number of easy examples by which the student may become familiar with the formulae which we have just obtained. lie must observe that these formula? hold good for all magnitudes of the angle which we have represented by the letter A^ that is, not only sin^ A + co.s2 A = \y but also sin^ 6 4- cos^ ^ = 1, and sin2a; + coR2x=l, and 8in2 45° + cos2 45° = l, and . sin'- G()« + cos2 GO'S =--1, and sin 9 TT 9 TT - 2 +COS- rt=l. And similarly for the other formula). 90. If then any angle be represented by 9, wc know from Alt. 88, (1) tan 6= A. ^ cos d (2) sin2 6? + cos'-^=l. (3) sec2^=l4-tan2^. (4) cosec2 6'=l + cot-^. And we also know from Art. 52, 1 (5) cosec = -r-ru ^ ' sin y (G) sec 0= - ->. cos V (7) cot = 1 tun {)' II !"1 « : li 56 RATIOS FOR JIIK i,AMIL ANGLE. li I 1!^ 91. Ex. 1. Show that sec 6^ -tan . sin ^ = cos 6. sec <9- tan 6* . sin 6*= -^ . ~ ^--^^ . siu 0, by loim (G) and (1), cos d COS "^ ^ ' ^ __!_ _sin*'^6^ ""cos cos <) ^l_-siri^ cos = cos 6^. .Ex. 2. Show that cot a - sec a cosec a (1 - 2 sin^ a) = tan a. cot a — sec a . coscc a (1 - 2 sin- a) cos a 1 1 sm a cos a sm a (l-2sin2a),by(7,6,5), costt_ 1 2 sin^a sin a cos a . sin a cos a . sm a cos''^ a - 1 + 2 sin''^ a cos a . sin a _ cos^ a - (sin- a + cos''^ a) + 2 sin'-^ a , cos a . sin a j J \ )* co s- g - sin^ a — cos^ a + 2 sin^ a cos a . sin a sin''* a cos a . sin a sin tt cos a = tan a. It "will be observed tliat in working these examples we commenced by expressing the other ratios in terms of the sine and co.sine, and the beginner will iind this the simplest course in most cases. Pr I. 3- 5- 7. 9- lo II. 12. 14. 15- 16. i7. 18. 20. RATIOS FOR THE SAME ANGLE. 57 Examples.— xvi. Prove the following' relations : I. COR . tan ^ = sin 0. 3. sin a . sec a = tun a. 5. (l + tan'^(?j.cos'-2^ = l. 7 9 10 tan''^ a ., =8m- a. 2. sin ^ , cot = cos 6. 4. cos a . cosec a = cot a. 6. (l + cot2 6^).ain2(?=L cosuc- a — 1 1 + tan- a tan X + cot aj = sec a; . cosec as. cos X . cosec X . tan x _ 8. cuseC a = cos2a. sin X . sec X . cot x II. cos X + sin X . tan x — sec x. -• tanrd^r^"'^- -a. (eos^^-l)(cot»e + l)=_l. 14. cot^ a — cos'"^ a = cot^ a . cos'-^ a. 1 5. sec^ a . cosec''^ a = sec^ a 4 cosec^ a. 1 6. sin" (/> -I- sin- i/j . tan- (/> = tan- </>. 1 7. cot'-^ . sin- (/i + sin- </) = !. 18. sec- ^- 1 =sin''^ (/) . sec''^^. 19. 2 versin </)- versiu"'^ (/) = sin'^</), sec 6/ - 1 20. sec 9 = versin 0. 92. We shall next show how to express the cosine, tanc^ent, and other ratios in terms ol" the sine. Since Again sin2^ + cos2 6^=l, cos- 6*= 1- sin- ^; sin cos d sin 9 tan 9 = • ;r. m lit 58 RATIOS FOR THE SAME ANGLE. Also coseo ^=-1- -7,, sin ^' sec d= - — 7,= liilil? i| -•wm'- cos Q ± v/(l - sin- (<)' cot 0= -1 =±^^i'^). tan 6/ sm u The doiiLle sign before the root-symbols is to be explained thus. For an assij^ned value of sin d we shall have more than one value of Q (Art. 74). Hence we have an ambiguity when we endeavour to find cos Q from the known value of sin 0. The double sign may generally be omitted in the examples which we shall Jieieafter give 93. AVe shall now give two examples of another method of arriving at expressions for the other ratios in terms of a par- ticular ratio. These examples should be carefully studied. (1) To ex-press the otlter trigonometrical ratios in terms 0/ the sine. Let PAM be an angle whose sine is s, a numerical quantit;;. Let Pilfbe drawn perpend icidar to AM, M Then if AP be represented by 1, PM will be represented by 8, and AM will therefore be represented by v'l - «^' RATIOS FOR THE SAME AXGLE. 50 Then denoting VAM l)y A, AM Vl-s2 , -, = T-n = ; = V 1 -S-: COS A = "xw= " 1 = Vl - s2= s]\- 8iu2 ^ ylP 1 ' tan -4 = Z^il/ « sin A AM Vi_s2 v/i_sin2;i' and similarly tlie other ratios may be expressed in terms of sin A. .! ,iv ' i ♦' (2) To express the other trigonometrical ratios in terms of the tau(jent. il Making the aanie construction aa in the preceding Article, Let tan A — t. Then if AM he rey.resented by 1, PM will he represented by t, and AP will be represented by \^1 +lK • 1 (.1 Therefore sin A== , ,: =-,-..— _ tan A AP ~ 7f n^"* VlTTiuM* CCS A AM I 1 AP 's^it^ \^1+Tan^i* u)id Hiniilaily tlie other ratios may be found. 6o RATIOS FOR THE SAME ANGLE. tmw EXAMPLKS. — XVii. 1. Express the other trigonometrical ratios in terms of the cosine.' 2. Express the other trigonometrical ratios in terms of the cosecant. 3. Express the other trigonometrical ratio.^ in terms of the secant. 4. Express the other trigonometrical ratios in terms of the cotangent. 94. If any one of the trigonometrical ratios he given, the others may le found. Thus suppose sin A = ^ . If PAM represent the angle, and PM })e perpendicular to AM, we may represent PM by 3, AP by 5, and consequeJitly AM by V^j^l) or 4. Then cos A 4 "5* tan A-- 3 )sec A - 5 "3' Bee A RATIOS FOR THE SAME ANGLE. 6l ins of the nis of the ms of the ms of the , tlie others a 95. If tan A = <, to find sin A and cos A, If PAM reproBent the angle, and PM he porpendicular to AM, we may represent PM by a, AM by 6, and consecjuenlly APhyK'an¥. Then sin A = a C03 A = ——-=:r=. Examples. — xviii. ;t; .*. ■■1' :0 licular to secinently I. Given sin a = ..: find cos a and tan a. «5 2. Given cos a = ;: : find sin a and tan a. .A'. 3. Given cosec = .^: find cos and Lan 6. 4. Given sin B-~ n-. : find cos ^ and tan 0. 5 Given tan a =4;^: find cosec a and sec a. 6. Given cos a= . ; find tan a and cosec a. 7. Given sin = a: find tan and sec 0. >^'f ' *.*ti I Kiiij fit vmk 62 RATIOS FOR THE SAME ANCLE, llftl il; ii V 8. Given cos 6—-h\ find tan and cosec ^. 9. Given sin ^ = -6; find cos and cot d. 10. Given cos ^ = '5 : find cot and cosec ^. 11. Given cosec 6^ = 2*4: find cos ^ and cot ^. 12. Given sec ^=1'03; find sin 6 and tan 0. 99 13. Given sin^^r-Y* find cos <^ and cot <^. 101 20 14. Given cos <^==77Ti- : find sin ^ and tan <^. 15. Given versin <? = r-^ : find sin 6 and sec 9. 96. "We may here give the geometrical solutions of the pro- Idem of constructing an angle, when its sine, cosine, or tangent is given. (1) Given that the sine of an angle is ,-, to construct the angle. The sine of an angle cannot he greater than unity, .*. a is not greater than 6. IVl B Draw a line AB=hy .'ind descrihe a circle with centre A and radius AB. Let BAG he a quadrant of this circle. Mark oil on AC the line AN'^tu RA TIOS FOR THE SAME ANGLE. 61 Draw 'NT*, PM at rigbt angles to AC, AB. Then PAM is the angle required : for PM AX a sin PAM= AP AP b' a (2) Given that the cosine of an angle is r-, to construct the angle. Making the same construction as bel'ore, PAN ia the angle required : for cos PAN= -ryi = r. AP \ a (3) Given that the tangent of an angle is r-, to construct thi angle. Take a line AM=b, and draw PM—a at right angles to AM (lig. in Art. 95). Join AP. Then PAM is the angle required ; for PM a tan PAM. AM~b' 97. We shall now give a set of examples similar to thoss in Ex. xvi., but presenting in some cases more dilliculty. * '1 •14! ' '1 ■ ; 11:1 '4 Examples.— xix. Prove the following relations : I. 8ini4 = 1 V(l + cot'M)* 1 2. cos A = — /7i— V r/-i\. v/(l +tanM) 3. cos X cot X 4. tan 03 . cos a: " ^J(\ - coa^ x)» 1" C4 RATIOS FOR THE SAME ANGLE. 5. 6. 7. 8. 9- 10. II. 12. 13. 14. 15- 16. 17. 18. 19. 20. 21. 2-> *«• 23- 24. 25. cos ^ = /•/(cosec- <^ - 1) cosec <^ tan,A= ^/(-^Jj*). sin- a = (1 + cos a) . versin a. tan2 a - tan2 /? - ^l^^iAz.^1^ Lull CI — Lull O — '-r, — T^i o — • COS'' p . cos*^ a .9 ,0 o sin- ^ - sin^ a C0t2 a - C0t2 y8= .-~^' -• -..-o- sin- a . sill- p sin- . tan'-^ ^ + cos2 Q . cot'-' ^ = tan2 + cot^ Q-\, 860"^ ^ + tan^ 6" = 1 + 2 sec- ^ . tan2 (9. cosec ^ (sec 6^ - 1) - cot 6* (1 - cos ^) = tan Q - sin ^. cot^ h + taii^ 6 = sec2 6 cosec^ 6-2. cot2 A — cos2 ^ = cos* A cosec^ A, tan2 £/ - sin- ^ = siii^ Q sec2 ^. sec2 Q cosec2 Q (sec - cosec ^) (1 + cot ^ + tan Q) ■■ cosoc 9 , ^cc 6 - _ _ — + = f,e<3 ^ ^ cosec ^. sec C7 cosec 6/ cosec ^ sec ^ * cos 6 (tan (9 + 2) (2 tan + 1) = 2 sec (9 + 5 sin 6?. f cos tc (2 sec ;k + tan re) (sec a; - 2 '.an «) = 2 cos x - 3 tan a;. (cosec ^- cot 6')^ = ! r\' ^ ' 1 + cos Bee Q . cot ^ - msec ^ . tnn t' „ ^ 7\ : 7i e= cosec . SCC 0, cos ^ - Bin 6^ sec ^ + cosec . tan^ ^ (1 + ^^sec'^ ^) = 2 sec^ ^. (sin + sec ^)2 + (cos + cosec ^)2 = (1 4- sec • cosec Oy. 1 + (cose c 6> . tan </))2 _ 1 + (cot . sin </>)' 1 + (cosec a . tan (ji)^ 1 + (cot a . sin c/))'-* (3-4 Bin2 yl) (1 - 3 taii2 A) -= (3 - tan'- A) (4 co82 ^ - 3). "osec Oy. 2^4-3). Examples.— XX. I. Fiiul till' complements of the followiii'^ angles: (i) 2-r.l4'.42". • (2) 43° . ii' . 57". (3) G1°.()M4". (4) 82°. 4'. 15". fs.T.l K f 'i .;il X. COMPARISON OF TRIGONOMETRICAL RATIOS FOR DIFFERENT ANGLES. 'i»' ^J -1. sin ^. cosec^ $ sec ^ - 3 tan X. 98. The Complemenl of an angle is that angle which must be added to it to make a right angle. Thus the complement of (50° is 30°, because 60° 4- 30° = 90°, and tlie complement of 14° . 30' . 15" is 75° . 23' . 45". Also the complement of 80^ is 20^, because 80« -t-20'^100', and the complement of 42« . 5' . 2>s'' is 57« . 94' . 72". U TT . TT TT TT TT And the complement of ^ is ^, because /. + , = ^. * 3 () 3 2 So generally, if a, (i, y be the measures of an angle in the three systems, TT complement of the angle- 90° - a-= 100*- fi -^ -y. Hence if the angle l>e negative (see Art. 48), and its mea- ^ure3 be - a, - /i, - y in the threi; systems, complement of the angle .= 90° - ( - a) = 100« _ ( » /^) r^ ^ _ ( _ y) «r90"4 a = 100M /^-g+y. ^4 • ' * 3 "i I ' • i 1- i'j '.f im I I*": i :?■!■■!! i|!i|| J i 66 /^^ TIOS FOR DIFFERENT ANGLES. (5) 125M5'.42". (6) 178°. 27'. 34". (7) 195°. (8) 254°. (9) -25°. (lo) -245°. 2. Find the complements of the following angles : , (i) 32'.23'.24^ . (2) 95«.3\75". (3) 4G«.0\84". (4) 2«.5\4". (5) 135«.2'.6". (6) 169". 0\ 3". (7) ^U3«. (8) 357«. ('?j *^5«. (lo) -245«. 3. What a? he complements of the following angles ? (0 4- (2) 3- (3) -5. (4) -4. (5) --4. 99. To compare the trigonometrical ratios vf an angle and its complement. Let NOS^ EOW be two diameters of a circle at right angles. Let a radius OP revolving from OE trace out the angle POE=A, Next, let the radius revolve from OE to ON and back again through an angle iV0P' = i4. iKIlt ' . 34". I : , 75". .3". nsles ? 5) - Stt igle and its ht angles. the angle ack again FATIOS FOR DIFFERENT ANGLES, $7 Then angle EOF = 90° - A. Draw PM and P'M' at right an[:,'lcs to EO. Now angle OP'M'==NOP'=^A=^POE. Hence the triangles P'OM' and 0PM are equal in all respects (Eucl. i. 26). Therefore P'M' 1/ sin (90° -^) = sin EOP'^ -^^r -=)fjj^ cos EOP=co3 A, OM' P M cos (90° - yl) = cos EOP' = ^^p = ^^j\ = sin EOP = sin ^, tan (90° - ^) = tan EOP' = ^'f!,' = ^^|^= cot EOP = cot ^. Oil/ FM And siniiLirly it may be shown that cosec (90° - A)=^ sec yl, sec (90° - ^i.) ■■= cosec ^, cot(90°-yl) = tanyl. This is a proposition of great prac' cal importance. We have only })roved it for the case in wh.ch A is less than 90°, but the conclusions hold good for all values of A. 100. The Supplement of an angle is that angle which inust be added to it to make two right angles. Thus the supplement of 60° is 120°, because 60° + 120° = 180", and the supplement of 24° . 43' . 1 7" is 155° . 16' . 43". Also the supplement of 80« is 120S because 808 + 120«=200«, and the supplement of 114*. 3' . 15'^ is 85« . 96' . 85'\ And the supplement of ^ is -^ , because - + -^ = 7r. So, generally, if a, (3, y be the measures of an angle in the lliree systems, supplement of the angle = 1 80° - a = 200* - /? = tt - y. Hence if the angle be negative and its measures bo -a, - (3j -y in the thr<je systems, supplement olLheangle = 180°-(- a) = 200'- (-iS) = 7r-(-y) = 18U' + a = 20U« + /^=-7r+y. ! , ], ' 4' • { ^1 .'; .:Vt| ;|i i If 11.^: lltJf ill '-' ti"|i' ■ ..il!! ! 68 /RATIOS FOR DIFFERENT ANGLES. Examples.— xxi. I. Find the supplements of the followin*,' angles (i) 34M2'.49". (3) 146°.0'.41''. (5) 179° 59'. 59". (7) i^45°. (9) -49°. (2) 132°. 24'. 47" (4) 28° . 15' . 4 (6) 100° . 49' . 53' (8) 437° . 3' . 4 (10) - 355°. 2. Find the supplements of the following unglea (l) 132«.32'.42' (3) 3'' . 97' . 98^ (5) 154«.3'.6". (7) 275^ (9) -35^ (2) 195« . 2' . 57'' (4) G5M2\8". (6) 174^0^4". (8) 527" . 2' . 14' (10) -3258. wm 3. AVhat are the supplements of the following angles ? (I) TT (2) TT (3) 47r (4) - TT (5) -- 37r 4. Find the diflerence between the supplement of the com- plement of an angle and the complement of its sui)pleinent. 101. To compare, the trig onu metrical ratios of an angle ami its su2)plenieHt. P/ w r/i' J,oi the angle FOPj=A. M t RATIOS FOR DIFFERENT ANGLES. 69 :t « ■ i i >l Produce EO to /F, and make the angle VOW = A. Take OF = OP, and draw PJ/, P'M' at ri<dit angles to KJV. Then the triangles iWC, P'M' (J are geonielrically eijual. (Eucl. I. 26.) Therefore sin (180°-^) -sin 7?0P --- cos (1 Sir- A) = cos EOP' = tan (180° - yl) = tan EOP = P'M' _PM OP'~OP = sin i4, CJ/' OiU CP'" OP P'M' _ PM OM' ~ - OM ~ cos = - tan And similarly the other ratios may be compared. 1U2, To ahoxo that sin (90° ■¥ A) = cos A, and cos (90" 4-/1)= -sin A. * * W M' Let the angle J^r; 7^ = ^. Draw OP' at right angles to OP, and make OP' = OP. Draw PiV/ and P'JI' at right angles to EOPF. Then, since the angles P'OM' and POilf make up a right angle, and the angles 0PM and PCil/ make up a right angle, angle PC>iVf = angle 0PM. .,1.5 Mi- m 70 A\I770.S /-OA' DIl I'EK EN T ANGLES. Also riglit lUiKlt' V'M'O^xV^A Jiuj^le VMO, and side P'0 = 8i«lc /'O, opposite e([ual angles in each ; .". the triangles V'OM\ OVM are o(piaI in all ivspects, and Tiicn Bin (9(r 4- /I) =sin £"0^'- y,f5= ^//)f = co8 /I. a.ui co8(0(r t-/i)=cos ij;(;r= ///r^ ~,™- -Hin /i. (>/*' (>/' 1 03. To show that sin ( 1 80" + /I ) ^ cos (180" + ^!) = ~ sm A, and -cos A. Let the angle EOP = A. Produce KO to Tfand P(9 to F, making OF=OP. Draw PJ/, P'Jf at right angles to EIF. Then the angle EOP' measured in the jtositive direction = 180° + .4. The triangles POM, P'OM' are geometrically equal. Now sin (lSO° + vl) = sin £'0F=^^^= "^™= -sin A, an d cos(180°f ^) = cosi:c>F = ^ = -^^-^= -cos A RATIOS FOR DIFFERENT ANGLES. 71 , ami m\ A, 104. To show that sin (- A)= - sin A, and cos {- A) = co3 A. Let tlio anglo iv'Oi'=yl. Draw I'M ut ri^,'ht angles to EJV, and protluco PM to F, making iV/r = MP. Join OP. Then the triangles POM, P'OM are geometrically equal, and tlie angle EOP\ which is numerically equal to EOP, will, if regarded as measured in a negative direction, be represented hy-^. Now sin ( -A) = sin EOP' = '- ~^, = - f ~ = - sin A. P'M OP PM OP and COS ( - ^) = cos ^ OP = y, = --- = cos ^. <• ^■' ircction A. 105. To show that sin (300" — A)=- sin A, and cos (^d()0'' — A)=cos A. y il\in;_,' the ramo construction as in the preceding Article, an^^le EOP' measured in the iiositive dJrection = 3G0°-il, P'M - PM Then sin (360' - ^) = sin EOF' = }.'' = — f ' - = - sin ii. and OP' OP cos (360°-yl) = cos JS:OF = ^-p = ^~^=co8 A, i i ■tj il 73 RA TIOS FOR DIFFERENT ANGLES. rPi Examples.— xxii. 1. Prove tlie following relations : (i) sec (180° -.4)- -sec y4, (2) cosecf ^ + ^ j = scc 9, (3) tan(18()° + ^) = tan^, (4) sec (tt + ^) = - sec /?, (5) tan {-0)=- tan ^, (6) cot (27r -6)=- cot (j. 2. State and prove the relations subsisting between the cosecants oi' ^" iind (90 ■{■By, also of (/> and 7r + ^. 3. State and prove the relations subsisting between the eecants of A° and (90 ■\-Ay, also of $ and •- - 0. 106. With reference to the trigonometrical ratios of dif- ferent angles discussed in this ('.ha})ter, it is to be observed that for an angle in the First Quadrant all the Ratios are Positive, Second all are Negative except the Sine and Cosecant, Third Tangent and Cotangent, Fourth Cosine and Secant. Also the following relations must be specially noticed : 9inyl=sin(180'-7l) = -sin(180'' + yl) = -sin(.3(iO''-/l) = -8in(-.4), cos^=-cos(180°-yl) = -coR(180° + .']) = cos (;30(r-yl)=cos(-^), tan^=-tan(180°-7l) = tan(180° + ^l) = -tan(3G0°-yI) = -tan(-.l). Examples.— xxiil. Find tl;c values of the following nitios : 1. sin 120", 2. cos 120', 3. sin 135", 4. cos 135', 5. sin IGO", 6. 003 150°, 7. sin 225°, 8. sin 2J0^ 9. tan 300", 10. cosec300', 11. sec 315°, i:. cot 330°. ;l'#V ,•.'«' > '-I • t - sec 0, = - cot e. etween tlie ctwccn the :ios of flif- servt'd that I Cosecant, Cotungcut, nd Secant. jticetl : -sin (-.4), -tan(-yl). ro9 ]3r.", sin 240', cot 330'. » XI. ON THE SOLUTION OF TRIGONO- METRICAL EQUATIONS. 107. The solution of a trigononu'trical equation is tlic of tindi /liat anuh ik h !r representing^ an iuij^uhir magnitude must stand for, in order that the e(juatiou may be true. (1) Suppose we have to find the value of 0, which wi'.l satisfy tlie equation cos 6^ +■ sec 9 = A. Our first step is to put ■>, in the place of sec 9, so that we '■ ' cos 9 may have only one function of the unknown angle in the equation, thus : cos 9 + 5 cos 6? 2* We then proceed to solve the ecpiation just as we ahouhl solve iin algebraical equation in which x occupied the place ol COB 9) thus : ScosV; 4 2-5 cos 9, 2 cos- -b cos 9 = n 5 COS''^^- ■- COS 9= -If COfl y-7=± 7, 4 4 eoH 0=^2 or 2* u I 74 OF TRIGONOMETRICAL EQUATIONS. l|i|f Now the value 2 is inadmissible, for the cosine of every angle is not greater than 1. The other value is the value of the cosine of 60°. (Art. 80.) Hence cos ^ = cos 60°. That is, one value of B which satisfies the equation is 60°. We shall explain hereafter our reason for writing the word one in italics. (2) To solve the equation 3 sin 6 = 2 cos^ $. 3sin ^ = 2(l-sin2^); .'. 3 sin 6' = 2-2 sin^^, or 2 sin2(9 + 3 sin 6 = 2, 6in2(9 + 2sin 6=^1, 4 4' 1 Bin (9 = ,-, or - 2. The value -2 is inadmissible, for the sine of an angle can- not be numerically greater than 1. The other value ^ is the value of the sine of 30". (Art. 78.) Hence, sin 6/ = sin 30°. That is, 07ie value of which satisfies the ceiuation is 30*. ^!'4'iiW'*0'^^i'0'im^iy4'-' .■".:..; ■ ■•■A * OF TRIGONOMETRICAL EQUATIONS. 75 of every (Art. 80.) is 60'. the word if,'Ie can- Art. 78.) 3 30^ Examples.— xxiv. Find a value of wliicli will satisfy the following equations: I. sin ^ + cos ^ = 0. 3. sin Q = tan 0. 5. 2 sin 6^ = tan 0. 7. sin + oos''^ $ . cosec 9. 4sec''2 6^-7tan'^^ = 3. (9-2. 2. sin ^- cos = 0, 4. cos 6 = cot 9. 6. 3sin0 = 2cos2a 8. tan 6^ = 4 - Scot (9. 4 10. cos . C05QC + f^iu 9 . dec = —p^. 11. 3sin2 0-cos'-^6^ + ( V3+l)(l-2 8in 0)=O» 12. Scos'-^ 0-8in2 f(V3+l)(l-2cos0)=O. 13. sec 9. cosec 6^ + 2 cot = 4. 14. sin + cos 0= v^^' 15. cot2 + 4cos''2(9 = 6. 16. tan + cot = 2. 17. sin -cos 0= V^- 18. sin 04- cos = 2 V2 sin cos 0. 1 9. V3 . si n 0=^/3- cos 0. 20. tan- + 4 sin'- = 3. 108. We have ahvadv stated (Art. 74) that, if the value of a tri;^fononietriciil rntio he ^'ivcn, we cannot fix on one parti- (ndar iin^'le, to which it exclusively helonj^s. This statement Mas confinued ))y many of the conclusions at which we arrived in Chap. x. For instance, since sin (180°-^) = sin A, it follows that the sines uf the angles A and 180' - A have the fiiinie value, that is, if we kufiw that sin A — -, A may have eillkr of iK() values, one of which is 30° and the other 150°. >..d i 76 OF TRIGONOMETRICAL £QUAIIONi>. Now we know that 8in (180°-^) = sin A and cotec (180° A)~lub>q.q. A^ cos (3(50° -A) = cos A unci sec (360" - A) = sec v4, tan (180' -\- A) = tan A and cot (180° + /I) = cot A. Thus for each given value of any one of the trij7"non;ctrical ratios there are two angles, and two only, between J° and 300" ior whicli tliat ratio is the same in magnitude and sign. 109. Suppobo OE to be the primitive line, and ()P tlie revolving line, aud let the angle EOP, less than 360°, be deuoted by A, the ci for wl value. 27r parti ( No^ ticn, Now suppose OP to make a complete revolution, tliot is, to start from the position im'.icated in the diagram and to revolve till it comes back to ihut position. Then the revolving line will have described an angle 300° + yl. Our triangh^ of reference will then be the same for an ang^e 360' + i4 as for an angle A. Hence, and sin (360°4->l)»-sinil, cos (360 -1-/1) = COM A. And tho same holds good for the other ratios. ^ 1.1 s. OF rKICONOMETRICAL EQL'ATIOys. 77 cosec i4, sec A^ cot A„ '^nomctrical 0° and 3U0" sign. nd 0? the ill 360°, be tlmt is, to 10 revolve ill! angle )r ail ung Hence, expressing ourselves for It v'ty in ihe fvn'>lK^ls of the circalar syptei/>, if r "!)e tlie circular ineasni>i of an anglci for which any one of the trigonometrical ratios has an assigned \alue, 27r + a will represent an angle, for which the value of that particular ratio is the same. Now let the revolving line make a secoml complete revolu" tion, then it will have described an angle 27r 4 27r f o, or, 47r + a. And so 47r -I- a will represent an angle, for which the value of the above-mentioned ratio will be the same. And, generally, if the revolving line, after having traced out the angle a, makes n revolutions, 2??7r -I- a will rei)rcscnt an angle, for which the value of any particular ratio is the same as it is for a. Now since n may have any integral value from 1 to oo , there ^vill be an infniite iinml)er of angles, for which the vab o of any one of the trigonometrical ratios is the same as it ih i'o. tiie angle a. Again, if a be the circular measure of an angle traced out l)y a line revolving in the positive direction, --(27r- a) will be the measure of an augle traced out by the lim revolving in the iirgative direction, for which tiie triangle of refen-nce will be the same as for the positive angle a. Tf the line then make n complete re olutions in the nega- tive direction, •• 2/(7r - (27r - a) will re])resent an angle, lor which the value of any particular ratio is the same us it is lor a. We can now explain the wny in ^^liich i^'neral expressions aie found for all angles, wbiih have a giN .i triuononietiical rat'o. '''1' . Nl *"i Ipl iii' 78 01 TRIGONOMETRICAL EQUATIONS. 110. To find a general expression for all angles which have a given sine. Let a be an angle whose sine is }]:ivftn. 0) First, reckoning in the positive direction, a and tt — a are angles with the same sine. (Art. 101.) Also 2n7r + a ) nnd 2;i7r+(7r~ a) ) are angles with the same sine. (Art. 109.) Secondly, reckoning in the negative direction, -(2;r-<t) and - |27r - (tt- a) |, that la, -(27r~a)and -(;r + a), are angles with the same sine. Also -2/i7r- (27r-a) ) ,^ and -2/i7r-(7r + a) ) ^ are angles with the same sine, n being any positive integer. Now tlie angles in (1) and (2) may be arranged tlius : "^ :imr ;- a, (2/j + l)7r - a, - (2/i + 2)7r + a, - (2h. f l)7r - u, nil of whiol'., and no others, are included in the i'urmula UTTl- (-1)". a, where n is zero or nny ])ositive or negative integer, which is therefore the general rxpressidU fur all angles which have a ^ivun bine. vs. which have a (1) (2) integer, us : TT-a, lUla , which is uh have a OF TRIGONOMETRICAL EQUATIONS. 111. To find a general expression for all angles which have- a given cosine. ,y Let a be an angle whose cosine is given. First, reckoning in the ])ositive direction, a and Stt - a are angles with the same cosine. (Art. 105.) Also 27i7r + a i and 2n7r + (27r-a) \ ^^"' are angles with the same cosine. (Art. 109.) Secondly, reckoning in a negative direction, - (27r — a) and - a are angles with the same cosine. Also -2/i7r-(27r- a) ^ * ,^. are angles with the same cosine, n being any positive rntegor. Now the angles in (1) and (2) may be arranged thus : 2n7r + a, (2)i + 2)7r-a, -(2w + 2)7r + a, -2n7r-a,, all of which, and no others, are included in the lurmula 2>J7r±a, which is therefore the geneial (.'Xpression for all angles which have a given cosine. J -». So OF TRTGONOMETRICAL EQUATIONS. 112. To find a yeneral expression for all angles which have a [liven tangent. me ail' •0) Let a be an aiijj;le whose lan{:,'ent is given. First, reckoning in the positive direction, a and tt + a are angles with the same tangent. (Art. 108.) Also 2w7r + a and 2w7r4 (7r4 a) are angles with the same tangent. (Art. 109.) Secondly, reckoning in tlie negative dirfsction, - (27r-a) and -(Tr-a) are angles with tl;e same tangent. Also - 2/i7r-(27r-a) and -- 2n7r— (tt - a) are angles witli the same tangent, ii hcing any positive integer. Now the angles in (1) and {i) may he arranged tlins : 2H;7 + a, (2?2 + l)7r4 a, - (2// + 2)7r 4 a, - (2/* + 1) 7r + a, all (jf wliich, and no others, are included in the formula NTT + a, wliich is therefore the general expression for all angles which have a given tau'.MMit. (2; v0-^m^9^^m«mr ch have a •0) (2; •v; Ulte^^e!', lis : [TT + a, ula llus wliiclj i OF TRIGONOMETRICAL EQUATIONS. 8i 113, We shall now explain how to express the tri;^'ono- metrical ratios of any angle in terms of the ratios of a positive angle less than a riglit angle. First, when the given angle is positive. If the angle is greater than 3G()°, suhtract from it 360" or any multiple of 360°, and the ratios for the resulting angle are the same as for the original angle. Thus we obtain an angk; less than 360°, and if this angle be greater than 180°, we may subtract 180° from it, and the ratios for the resulting angle will be the same in mngjiitude, but the signs of all but the tangent and cotangi-nt will be changed. (Art. 106.) Thus we obtain an angle less than 180°, and if this angle be greater than 1)0°, we may replace it by its supplement, and the ratios for the resulting angle will be the same in magnitude, but the signs of all but the sine and cosecant will be changed. {Art. 106.) Thus sin 675° = sin (360° + 315°) = sin ^15°= -sin 135°= -8in45\ tan 960° = tan {1-20° + 240°) = tan 240° = tan 60°. Secondly, when the angle is negative. Add 360°, or any multiple of 360°, so as to obtain a positive angle, for which the ratios will be the same as for the (U'iginal angU', and then proceed as before. If ihe given angle be less than 180°, apply the formuliie obtained from Art. 104. Ex. sin (-825°) = sin (1080° -825°) = sin 255°= -siu75°. tan ( - 135°) = - tan 135° = ~ tan 45°. Examples." XXV. Find the values of the following ratios ; I. sin 480°. 2. cos 480°. 3. sin 495'. 4. cos 495°, 5. sin 870". 6. coo b70°. fs.T.] V ■<'M 83 0/- TRIGONOMETRICAL EQUATIONS, 7- sin 945°. 8. sin 96(>\ 9- tan 1020°. it ' lO. cosec 1380°. II. sec 1395°. 12. cot 1410°. 1 " -1 ; ! Ilv >3. cos 420°. 14. sec 750°. 15- tan 945°. 1 6. sin 1200° 17. sin 1485°. 18. cos 1470°. 19. sin Ttt. 20. sec 87r. 21. cosec 930°. '' f'" 22. cot 1140'. 23- tan 1305°. 24. cosec 1740°. il 25. sin ( - 240°). 26. cot (-675°). 27. sec (-135°) 28. tan ( - 225°). 29. cosec ( - 690°). 30. cos (-120°) Examples. — xxrvi. "Write down the general value of Q which satisfies the fol- lowing equations : I. sin ^=1. 1 3. Bin^ = .^2- 5. 3 sin ^ = 2 cos2 ^. 7. tan2^ + 4sin2^ = 3. 9. tan ^ = 4 - 3 cot ^. 2. cos ^ = 1. 4. tan Q= V3. 6. 2 sin 6^ = tan ^. 8. C032 6' = sin2a 10. sec^ ^ - - sec ^ + 1 = 0. 114. The symbol sin-^ x denotes an angle whose sine is 0;, cos~^ X cosine is x, and a similar notation is used for the other ratios. 2 Hence if we know that, for instance, tan Q — i^ we may write the general value of thus : 2 0>=mr + tan~^5. '^mi^<\ft0MM'r-^<'^ ^'*' S". r •; n 1020°. .t 1410°. n 945". 3 1470°. sec 930°. sec 1740°. c(-135°). 3 (-120°). XII. ON THE TRIGONOMETRICAL RATIOS OF TWO ANGLES, 115. We now proceed to explain the trigonometrical functions of the sum and difference of two angles. These functions are the most important in the subject, and the student will find that his subsequent progress will depend much on the way in which he has read this Chapter. !S the fol- 1=0. me IS x. sine IS Xf we may 16. We shall first establish the following furmulse : Bin (^1 + Ij) = sin A . cos B + cob A . sin B, cos {A + B) = cos A . C0& B- sin A . sin J?, sin (.4 - B) = sin A . cos B — cos A . sin B, cos {A —B) = coii A . cos i> + sin A . sin B; by moans of which we can express the sine and cosiiie of the sum or difference of two angles in terms of the sines and cosines of the angles themselves. The diagrams which we shall employ are only applicable to the cases in which A and 7j are both positive and less than 90", also, when we are considering the sum of the angles, A-¥ Bi^i less than 90°, and when we are considering the difference of the angles,^ is greater than B. The formv J le are, hm, , . , r, true for all values of A and B. Particular cases i lav be proved by special constructions of the diagrams, but it is beyond the scope of this treatise to enter into detail on t'ii>^ and similar points. IMAGE EVALUATION TEST TARGET (MT-S) // h ^ y. ^.^^. ^ f/. V <^ 1.0 11= 11.25 UIM 125 ■" lU |2.2 1.4 1.6 '/] ^> Photographic ^Sciences Corporation 4. <- <^ ^^ ^ 23 WIST MAIN STRUT WIBSTIR.N.Y. MSIO (716) •72-4303 o^ ■*;V4 84 RA T/OS OF TIVO ANGLES. 1 i . 1' .; f : ■« •11 ^11 [i 117. and To show that sin (A + B) = sill A. cos B + cos A . sin B, cos (A + B) = cos A . cos B - sin A . sin B. / /D p / \ / ^ \ ^./^ . a c A R M B Let the an^lo BAH he represented by A and CAD by B. Then the angUi BAD will be represented by A + B. From P*, any piunt in AD, draw P/^ at right angles to AD and PQ at right angles to AC. From (3 draw (^.U at right angles to AB, and Q^V at right angles to PR. Th^Mi angle QPX=00'' - PQN==NQA=:QAM=A. Now sin {A + B) = sin PA R = 7iJP IP RN + NP QM+NP QM NP AP AP iP'^ AP Q^r AQ NP PQ AQ' AP^ PQ' AP ■am A . cos B + cos /I . sin B. COS (.4 + B) = cos PAR = Alt AP ^AM- MR^AM- NQ ^AM _NQ AI \r AP AP AM Aii_XQ P(J "AQ'AP PQ'AP ■a cos A . cos B - sin A . sin B. ' P |9 taken in the line boumlinv,' the angle under consideration, that ' ,f ' RATIOS OF TWO ANGLES. «5 118. To ahow that din (A ~ B) = sin A . cos B — cos A . sin B. and COS (A - B) = cos A . cos B + nin A . sin B. ' .^1' Lot the angle 11 Ad he rei)resonted by A and CAD by B, Tlicn tlio angle BAD will be represented by A - B. Froni P*, any point in AD, draw Pi' at right angles to AB and ]*Q at riglit angles to AC. From Q draw QM at right angles to AB and QN at riglit angles to RP jn-odnced. Then angle QPN=<)0' -PQX -^ CQN = BAC^A. N ow sm {A - B)=- Bin PA R--^ RP AP RN-NP QM-XP QM NP A J' AP P A? QM AQXP PQ "AQ ' AP PQ ' AP 'sinA . cos B - cos A . sin B, AH cos {A - 7?) - cos PA R = -rp A M + MR A M + XQ A M , NQ AP AP AP" AP AM AQ NQ r<> " J(J' JP^ PQ ' AP ■=co» A . cos Z^ + sin A . sin /?. P is taken to the hue boumliiig the angle under consideration, that ts, BAD, ' (| 86 RATIOS OF TWO ANGLES. Ti HI ■ !!' t 110. We shall now give some important examples of the application of the formulas which we have established. Ex. 1. To find the value of sin 75^ sin 75° = sin (45° + 30°) = 8in 45° . cos 30° + cos 45° . sin 30" -i2-'^^-j2'l (^ts. 78,79) 2 V2 2 V2 _y3+j " 2 x/2 • Ex. 2. To find the value of cos 15* cosl5° = cos(45°-30°) = cos 45° . cos 30° + sin 45° . sin 30' V2 • 2 "^ V2 ' 2 ^_V3 1 2 ^2 ^ 2 V2 ys + 1 ■ 2V2'' a result identical with the value of sin 75°, in accordance with Art. 99, for sin 75° = cos (90° - 75°) = cos 1 5'. Ex. 3. To show that sin (90" ^- A) = cos A. Assuminj,' the conclusions of Art. 117 to be true for all \*alues of A and B, sin (90° + id) = sin 90° . cos A + cos 90° . sin A = 1 . cos yl + . sin /I (Art. 71) = cos A, And similarly other relations between trigonometrical fuuc- tionn established in Chapter x. may be proved. I t* nples of the hed. n ince with for all RATIOS OF TWO AXGLES. , 87 Ex. 4f To find a value of which satisfies the equation sin 6 + cos 6=0, Multiply both sides by — ,- Then or. 8in^.-^ + cos(9.-i-^==0; .♦. sin $ . cos 45° + cos ^ . sin 45''«=0, 8in(^ + 45'') = 0; ' .'. (^ + 45°) = 0; .-. ^=-45^ Examples.— xxvii. Prove the following relations : 1. sin {A -f- B) . sin {A-B)= sin^ A - 8in2 B, 2. sin (a + /3) . sin (a — /?) = cos'^ yS - cos^ a. 3. cos (A+B) . cos (A- B) — cos^ ^ - sin- B, 4. cos (a + /3) . cos (a- /3) = cos^ /3 sin^ a. 5. 2 sin (a; + y) . cos {x - y) = sin 2x + sin 2j/. 6. 2 cos (x 4- 1/) . sin (x - 1/) «= sin 2x - sin 2y, ... r, sin (/l+i?) 7. tan^ + tanB = \ „ ' cos A . cos R 8. tan a -tan B= ^a. '^ cos a . 003 ff ■ 1 ■■1 s 4- -i ■ 't »-■ ., i'f i, ■■1 -■}' ; r .'" ' . ' ;. ■ » ■ 1; 1 ''1 .■ .' t ■J ■ i if- 1 fiino- 88 J?A 7/OS OF Tiro ANGLES. EXAMPLES.— xxviii. 4 4| • \ ■ \ ' » , r. Show that sin 15° = v^:5- 1 2V2 • pr 2. Show that cos 75° = -^7,; . 3. Show that tan 75° = 2 + V3. 4. Show that cot 75° = 2 - ^3. 1 2 5. If sin a = ^ and sin /!? = „, find the valne of sin (a f jS). 3 2 6. If cos a = ^ and cos /?= -, find the value of sin (a - {i). 7. If sin a = -5 and cos j^ = —7^, find the value of cos (a + /3). 8. If cos a =« -03 and sin /? = g, find the value of cos (a - /3). Examples, xxix. i H Apply the forniulnG established in this chapter to show the following relations between the trigonometrical functions of angles. I. cos(90°f yl)=-8in 4. 2. sin (180° + yl)= -sin^. '37r 3. coa (77+0)= - cos 0. 5. cosec ( -^ f a j = sec tt. 4. sin (<v + 0)- - cos 6, 6. tan (tt + a) = tan a. y. Bin (27r - 6)=^ ■• sin ^, 8. tan (27r - 6*) = - tan ^. 9. sec (180° - ^) « - sec ^. 10. cosec (tt - ^ «■ cosec ^* RATIOS OF TWO ANGLES. 89 ' I EXAMPLES.— XXX. Find a vjiliie of to sutislV the following; equations, hy a process piniilar to that given in Art. Ill), Ex. 4. I. sin 6^ -cos ^ = 0. 5. sin ^-i-cos 0= ^2. 2. sin 6^ + cos Q=\, 3. sin ^ -cos ^/^^/g. 4. sin f? + cos ^ = -^— . 6, sin ^-cos Q = 2 • ' h: % **i' 1-20. Collecting the formula) of Arts. 117, 118, we next arrange them thus : sin (A + 7?)= sin A . cos 7? + cos A . sin i?, sin (.1 - 7^) = sin A . cos B - cos A . sin 7?, cos {A - B) = cos yl . cos i5 f nin A . sin B, cos {A -f 7i) = co8 A , iios, B- sin ^ . sin 7?. Hence, by addition and subtraction, we obtain the following: sin (yl + 7?) + sin (yt - 7?) = 2 sin A . cos By Bin {A 4 B) - sin {A-B) = 1 cos A . f^in 5, cos {A - B) + cos (yl + 7?) = 2 cos A . cos 7J, cos {A -B)- cos (yl + 7i) = 2 sin .4 . sin B» Now let .4 + 7i-=P, and A-B^Q. TUea 2v4-P+^,and27?=P-^j /. ^ = — g-, and J5= ~2 ^ ; »• ■ '. '-1 i ■ *. -^1 * If i k i ? 90 J?/1 r/OS OF TWO ANGLES. So that the formula} may be put in this form : sin P + sin Q = 2 sin r+Q P-Q . cos i) » HI iii Bin P-sm G = 2 cos — — . sm — —^ cos (J + cos P=2 cos, — g— . cos — ^, cos Q - COS P = 2 sm — -^ . sm — ~ . 121. As these results are of very great importance, we shall repeat them separately, explaining each in words. (1) sin P+sin Q = 2 sin — _- . cos , that is, the sum of the sines of two angles is equal to twice the product of the sine of half the sum of the angles into the cosine of half their difference. Ex. sm 109 + sm 6^ = 2 sm «, • cos — 5 — = 2 sin 8^. cos 10. I (2) R]nP-8in(3 = 2cos-4~.sin^— ? that is, the difference of the sines of two angles is equal to tvnce the "product of the cosine of half the sum of the angles into the sine of half their difference. T7i„ • o 'A o 8tt + 4a . 8a — 4a h,X, sin 8a - sm 4a = 2 cos --- — . sm — ^ — iBs2 COS 6a . sin 2a. (3) cosQ + cosP:=2co8~^.cos:^^^, that is, the sum of the cosines of two angles is equal to twice the product of the cosine of half the sum of the angles into thu cosine of half their difference. ^} RATIOS OF TWO ANGLES. 91 Ex. cos {/ + cos 3 6/ = 2 cos . cos — 5 — = 2 cos 26^ . cos 9. (4) cos Q-cos P = 2 sin -— . sin —5—, tliat is, </ie difference of (he cosines of tico angles is equal to twice the prodiict of the sine of half the sum of the angles into the sine of half their difference. •A f f ance, we shall 17 V o >f o • 7a + 3a . 7a -3a hX. cos 3a - cos 7a = 2 sin — - — . sm — a- - - 2 tit = 2 sin 5a . sin 2a. 122. As the formulae at the end of Art. 120 teach us how to re])lace the Funi or difference of two sines or cosines by the product of two sines or cosines, so the formulas at the beginning of Art. 120, when read from right to left, thus : 2 sin yl . cos jB = sin (/I + 5) + sin {A - i?), 2 cos ^ . sin J?=sin (^ -f J5) - sin {A - B\ 2 cos yl . cos /^ = cos (yl - B) f cos {A + Z)), 2 sin i4 . sin J5 = cos (^ - U) - cos (^ + J5) , furnish rules for replacing the product of two sines or cosines by the sum or difference of sines or cosines. *»t For example, 1, . 8iu56>.cos3(9 = -Jsin(5^ + 3(9) + sin (5(9- 30) | = 5(sin80fsin20), 1 sin Q . sin 36^ = -_ } cos (3(9 - (9) - cos (3(; + 1 = q(cos20-co8 46>). -' L| 9» RATIOS OF TWO AXGI.ES. i:il WV \ Examples.— xxxl Prove the following relations : 1. sin G^ + sin 4/1 =2 sin 5^1 . cos yl. 2. sin 5/1 - sin 3/1 = 2 cos 4/1 . sin A, 3. cos 76^ + cos 9^ =2 cos 86'. cos ^. 4. cos 6^ -cos 5^ = 2 sin 3^. sin 2^. 5 sin a + sin 4tt = 2 sin oa cos 3a 2" 13a 3a 6. cos 5a - cos 8a = 2 sin -4.- . sin ^ 2 2 7. 2 sin hO . cos 76' = sin 126^ -sin 29. 8. 2 sin 3^ . sin 5(9 = cos 2^ - cos 8^. 9. 2 CO** a . cos 4a — cos 5a + cos 3a. 10. 2 cos a . sin 2a = sin 3a + sin a. II. sin A + sin B cos A + cos B = tan A + B Mf 'i i V 12. 13- 1 - cos A - cos 3/1 sin 3/1 - sin A mi 2/1 4- sin A cos 2/1 + COS A = tan 2A» , 3^ = tan ---. 14. cos (30' -0)- cos (30° + 61) = sin 6* 15. cos(^+6^) + cos(|-^) = cos^. 16. 8in(^ + aj-sin(- -aj = sin a. .»7- sin ^-^"Lf^^rnfttl^ COS /?— cos a HA 77 OS OF TIVO AXGLES. 93 I8. «^"--«^"/^ = tan^. cos p 4- COS a 2 sin 5^ + sin 3^_ ^ ^' cos 3^- cos 5^" cos a + cos j8 _ cot ^ (a 4- j8) cos ^ - cos a tan ^ (a - ^)' 20. 123. We can altso express the sum or difference of a sine and a cosine as the product of sines or cosines. For since cos ^ = sinr^ - 0\ (Art. 99), sin a + cos 6^ = sin a + sin ( ^^- - 6 J = 2sinl(a + |-(/).co4(a-r+e). A^'ain sin 40° + cos 60° = sin 40° + sin 30* — 2 sin 35° . cos 5*. t J -I EXAMPLES.—XXXii. Express as tlie i3roduct of sines and cosines : I. sin a- cos ^, 2. sin ( - +a j + cos( ^ - a j, 3. sin a + cos a, 5. sin 30° + cos 80°, w TT 7. sm^-fcosg, 4. sin a - cos a, 6. sin 20° - cos 80°, o . TT TT 8. sin - -cos =. 6 5 124. We now proceed to exphiin how the tan|j;ent of the sum and difference of two angles can be expressed in terms of the tangents of the angles themselves. M KA TIOS OF TWO ANGLES. \\ III !: =li: «Lii!^ , , „. sin (/I + /?) tan (^1+2?)= ;.- J sin yl . cos ^ 4- cos ^4 . sin Z? cos yl . cos B - sin ^ . siu U' and, dividing each term of the numerator and denominator by COS A . cos By Bin il . cos J5 cos yl . sin Z? _ cos A . cos B cos A . co.s B cos yl . cos B sin ^ . sin B cos ^ . cos B cos .1 . cos B __ tan yl + tan B "" i - tan A . tan i5* tanM-Z?)-'i-''^^^^ ^^^'^ ^^-co^{A-B) sin /I . cos B - ~~ COS yl . cos 7? -I- cos A . v,v.^ ... sin B cos A . cos 7? +• sin A . sin /? Bin A . cos /? cos A . sin 7i cos A . cos /y cos A . cos /? cos /I . cos />' si n^. si hi? cos A . cos Zi cos i4 . cos ZJ tan yl - tan B l+tfin^ . tanZ?' ■* Cor. We proved in Art. 79 that tan 45°= 1. Hence tan (45' + A)= -^"^^ t - ' "*-, ^ ^ 1 ~ tan 45° . tan i4 _ 1 + tan i4 ~ 1 -iaiiA' tan (45°-^) = tan 45" - tan /t 1 + tan 45° . tan A l-tap.4 1 +tan A' RATIOS OF TWO ANGLES. 95 ^nominator by 125, The results of the procedinrj article may be obtained without assuming the lonniilu) for the sine and cosine, thus : Taking the diagram of Art. 117, we have ^ ,. J,. PR NR + PN _QM+PN "AM-NQ QM PN AM'^AM 1- AM Now QM AM =tan A, and observing tliat PNQ, MQA are similar triangles, PN PQ ^ D AM~AQ NQ_NQ PN AM~PN ' AM rv=tan A . tan B\ tan (^ + 5) = tan A + tan B I - tan A tun B' 1 'y, \. ■ Af Again, taking tlie diagram of Art. 118, we have . , , m PR NR-PN "'^^-^-^^-ArrAMTm QM- PN '^AM+NQ Qhl PN AM' AM 1 + NT' AM n Now ~,-v— tan ^4, AM • i . I y i f * i ) t< VIS" If i! Ill- 96 KA7/0.S OF TWO AXGLES. and observing that PNQ, MQA are similar triangles, AQ NQ I\V ^ , , „ , , ... tan A - tan B :. tan (A -B) = -r-— .- , — ^ '^ 1 + tan A . tan B Examples.— xxxiii. Trove the folh)wintr rehitions tan a + tan {3 cot a + f'.ot f3 tan tt + tan^ _ cot a -tan f3 tan a . tan (3. tan a . tan (a + jS). tan a - tun /:? _ .'ot a + tan jS = tan a . tan {a- f3). */>^'/'a.*.„ *^-'/'_ 2sin<^ 4. tan - -Q ^ + tan 2 eoa </>+ cos i/-* 5. sin </) = sin \^ . cos (</>-»/') + cos i/- . sin (</> - i/). 6. cos (/) = 8in i/ . sin (</> 4- 1/) + cos i/' . cos ((/> + 1/). 7. (cos a + cos (3)\l - cos (a + /:^) * = (j*iu u + sin 13) . sin (a + p) a sin (a + /:^) + i« cos -— ^- sm a + sm (^' cos (I ... /^- sm a + i3 sin (a + /5) 2 siu a - sin li a - sin -. a + i8 JO. cot — ^ + cot a zi^ cos 2 sin a ^-C08 RATIOS OF / ItV A.VuLAS. 97 I 11. tan a ±^-^n./^^= 2sin/^ 2 2 coa a f cus ^* 12. cos a - COS ^_..../^ sin a + .siu /i = tun a I}, cot /:/- tan a = cos 5J • (a 4 /i) cos u . sin fi' 14. cot ^4- tan <l)~ 15. tan- a - tan''^ /i _ cos ((/) - 6^) COS (f) . sin ^* >in (a f-/j) sin (cr - /?) cos-* a . COS '13 16. 1 +tan a . tan /J = 17. 1 - tan a tan fS — COS (a-.«) cos a . cos fr COS (a f /j) cos a . cos B' 18. cot (I -I- tau /:f__ tan a -i- cot /:^ tan=^ ic - tan^ if cot a . tan 13. '9. ,-r:r 1 - tan'** a; . tan- y = tan (a; f </) . tau fx -y). cot ^-1 20. cot((y f 45")= -T^ - 1- ^ ' cot t7 + I 21, bin 6^ + cos ^= V^ • ^'" <'^^° ^ ^)« 21. cos 6/ -sin 6^== \/2 , sin i. - t^)- tan a - tan (i _ sin (a - (3) *"^' tan (iTlan /J~sin (tt + Ziy cot X - cot y sin (?/ - ;r) " cot .c + cot y sin (?/ 1- x)' 25. cos (.1 - Zi)+8in (A-hli) 2 sin (.4 + 45') . cos (B-45"*). 26. cos (J - /i^ - sin (. I + Z^) = 2 sin (45" A) . cos (45^ -h i/). 2;. cos (^ + ij') + bin (.4 ■ B) = 2 sin (45' \ A) . cos (45° + B). >.T. J " :.^ » « RATIOS OF TWO ANGLES. % w 28. COS (^ + 5) - sin (A - Z^) = 2 ain (45' - id ). cos (45^ h, 29. cos a + cos /? cos a - cos /S cot o+jS 2 tan «3' i 1 1 ... t si'' I , s|||i, 30. sec 72° - sec 36° = sec 60°. 31. sin 108° = (sin 81°-i-8iii 9°) (sin 8r-8in9"> cos 3° - cos 33° 32. sin 3° + sin 33 ;r„ = tan 15". sin 33° + sin 3° , ,„« •'"^ cos 33 + cos 3 cos 9° + sin 9° ' . ,. 34. ^o — . f,o = tan64\ •'^ cos 9 - sin 9 cos 27° -sin 27° ^ ,„, 35- ^^^ — ^— .^^.=t£UQ 18 . •^-' cos 27 + Bin 27 36. tan 50° + cot 50° =- 2 see X0\ ^mm' . i XIII. ON THE TRIGONOMETRICAL RATIOS FOR MULTIPLE AND SUB- . MULTIPLE ANGLES. 126. The angles 2A, 3A, 4A are called Multiples A A A of A, and the angles of^. 2' 3' 4 ,. are called Submultiples We shall first show how to express the Trigonometrical A Ratios oi 2A and ^ in terms of the ratios of ^4 . 127. sin 2-4 = sin (A +^) -=8in A . cos A 4- cos ^ . sin ^ — sin ^ . cos A 4- sin ^ . cos 4 ■■- 2 sin A . cos A. Cos 2^= cos (/I +^) = cos A . cos A - sin ^ . sin ^ =^008^ A -sin^ A. Now we may put 1 -sin'-'^ in tlie place of cos'^^ (Art. 90), and we then liave cos 2i4 = 1 — sin''' i4 - sin^ ^ -l-2.sin2/l. Or we may put 1 - cos'-^ A in the place of sin'* A (Art. 91)), and we then liave cos 2 A = cos''* ^ - (1 - cos* A) ■ cos '^A-l+ cos" A ^2 coo' A -1. ifl' fl lOo MULTIPLE AND SUBMULTIPLE ANGLES. tan 2^ = tan (yl + /I ) tan A +tau yl "~ 1 - tan A . tan ^ _ 2_tan A ~\-taiJ'A' , (Alt. 124) f \ /J 128. If we put A in the place of 2A, and ^ in the place of A, we have sin ^ =2sin -^ . cos c08-4=cos'^ sin- 1-2 sin'-^ 2 :2cos- - - 1, 2 i'N|- ' 2 tan tan^ =' 1 - tan2 4 Hence we can show that 1 - cos A 1 + cos il A =tan2-- a formula of great importance. 1 — cos A For ,— l-(l-2sin2|) 2 sin-' 1 + COB / ■1 l-f(2co8''^^ - l) 2 4ana cos^ 2 Ex AMPLES.— XXXi V. Prove the folJowin<' rulutions : I. sin 2^ 2_cot^ l+cot^r 'GLES. 124) in the place formula of 2* MULTIPLE AND SUBMULTIPLE ANGLES, ICI 2. ^^^_ C03 A __ A 1 + COS 2/1 • 1 + COS 3 ~^^^ 2"' 3- cosec^ + cot^ = cot- 2" 4. tan^ + cot6?=— — - sin 2(9* 5. 6in2<9= ^-^•''^-^-^ 1 + tan^^ 6. 2cosec2^ = sec/l.cG8ec^, 7. cos2^=^->":'^ l+tan2^* 8. sec2^=Asi°2(9 1 + sec 2(9* 9- l --2s inM^l^tanyt l+sin"2^ l+tair^i* 10. cot (9 -2 cot 2^ = tan ^. , 1 - cos a . a 12. sin20 = ^V-^^l^!«^*li.-*> ^ C0SUC2 1 3- cos 20 = 5 — -l:, sec-^ ^ cot-'t/)-!' V V 2scc2tt /• l6. cosa= ./f^-^^2ottl\ V V 2 sec 2a /• ;fl^i r « 9 I'' '' ^1 i .A I « T f I02 MULTIPLE AND SUB MULTIPLE ANGLES, 17. tan a = cosec 2a — cot 2a. 18. cot ;8 = cosec 2/3 + cot 2/?. ,..0 ^\ cos 2^ 1 9. tan (4o + ^ ) = - — . - -- . ^ ^ ^ l-sin2i4 20. cot (45° - /I) = sec 2^ + tan 2A. 1 + sina 1/, ^ a\* 2 1. = ,, ( I + tan - I . 1 +C()aa 2\ 2/ 1 - sin a 1 / ^ tt , V"* 22. ="(C0t-r-l). 1 - cos a 2 \ 2 / 23. tan Q = tan - + - tan fi . sec" r:. •^ 2 2 2 1 f sin ^ , ^ . ^^„ 24. ;j — ^t;;- ^ = (scc Q +. tan t^^2^ 1 - sin (9 sin 2i4 = -- -: _ 1 -tan2(4r)^ - ^) 26. sin 2^ = l+tan'^(45°-yiy tan(| + ^)^tan(|-ry ) tan(^ + ^) + tan(|-^y 129. We shall next show how to express the ratios of ZA in terms of the ratios of A. Bin 3^-= sin (2^1 + ^) i=sin 2^4 . cos A + cos 1A . sin A = (2 8in^ . cosyl). cos -4 -f (I - 2 sin^ A) .sin A s= 2 sin A . cos''^ A + sin A -2 sin^ A «2 sin yl . (1 - sin'-^ A) + sin /I - 2 sin^ A *5 2 sin A -2 sin^ ^ + sin A~2 sin^ J. ■a3sinil-4sin3^l. MULTIPLE AND SUBMULTIPLE ANGLES. 103 130. cos 34 = cos (2^ + -4) ^ cos 24 . cos 4 - sin 24 . sin 4 =■ (2 cos^ 4 -• I) cos 4 - (2 sin 4 .cos 4). sin 4 ■» 2 cos' 4 - cos 4 - 2 sin^ 4 . cos 4 «= 2 cos' 4 - cos 4 - 2 cos 4 (1 - cos^ 4) «= 2 cos' 4 - cos 4 - 2 cos 4 f 2 cos' A =4 cos' 4 -3 cos 4. f '^ 131. Tan 34 sin 34 " cos 34! 3sin4-4sin'4 • ,. .1. i • . cos** 4 4- 3_ C08^4 3 tan 4 . sec^ 4-4 tan" A ' "4-3Bec2 4" 3 tan A (1 +_tan^ 4 ) -^4 tan' 4 4-3(lVtan2 4) 3 tan 4 - tan' A ' ~"r^3 tan2 ^~ il Ex AMPLES.— XXXV. In soine of these Exanjplea the student may employ with advantage the formulao given in Art. 121. , . cos 30 -sin 36^ 1 n • 0/1 I. (i) -.- -^- 2i-= 1 - 2 sin W. ^ ' sm y + COS d . . . o^ , n 2tan(? + sec^ (2) Bin26? + cos0=.-j— — ,-^ ■ 'I n i i I iff:. Ml Mil' 104 MULTIPLE AND SUB MULTIPLE ANGLES, A A (3) sin A - tan y- + 2 sin^ - cot A. (4) cot^ tan^ cot A - cot 3i4 tan A - tan ZA = 1. (5) cos 4^ + cos 4Z>' = 2{ 1 -2 sin'^M + i?) ( 1 1-2 8in2(A - J5)| sec 10 - cos 26 (6) tan (45" + 6?) - tan (45" -6) = 2.' (7) cot2 6;-tan2^ = 8^--4^. ^" 1 - cos 4^ (8) 2 am A. cos 2.4 = sin 3^1 - sin A. "sin 20 (9) cos wtI - cos (n + 2) A - — 7 — r^\^i~= 1 = tan (n + l)A. sm (ri + 2) ^ - sin uA ^ ' (10) cos 9/1 f 3 cos 7/1 + 3 cos 5/1 + cos 3^ = 8 cos^/l . cos <zA . , . cosoc 2A — cot 2^ , „ . (11) —^- — ----. = tan2 A, ^ ' cosec 2.4 + cot 2.4 W^%<- 4V ^ I , , l-8HWt 1/, AY ^i^^ 2 cos 2.4 - 3 _ cos 3^-2 cos A , ^ ^^ 2cos2.1 + ;]~sTir3Z+2sin^-^^^'^- (14) tan (45° - ^) + tan (45° + ^) = 2 sec 2.4. (15) cos a - tan 2 sin a = cos 2a + tan ^ . sin la, ( 1 6) cot2 A ~ tan2 .4 = 4 cot 2 A . cosec 2 A . ( 1 7) cosec a . cot a - sec a . tan a = cosec^ 2a (cos' a - sin' a). ^,Q\ «^*2 4. <> 4 cos 2a '18) cot''a--tan-'a = - . ., ^ •» sin^ 2a V- MULTIPLE AND SUB MULTIPLE . XGLES. lo" os3/4.co8 6i4. I ( 1 9) co.^ec^ h - sec- 6 = 4 C(js ^h . cosec^ '2b. / X ,'>/'a'oA\ 2 cosec 2^ - sec ^ ( 20) cot^ I 4o + -T ) = -, Kt 7' ^ ^ \ 2/2 coscc 2 A + sec A (21) ain(-|"+6>)-sin('-|^-6^) = sin('^-6')-8m(-|+6?) (22) cot(|+ e) - tan (^ + ^) = 2 rot 26». , . (coscc a + see a)2 , • ^ (23) ^ -^ = 1 + sill 2a. ^ cosec^ a + set a , X tan^ _^ , . tan 2(9. tan ^ . „, (24) , — -^a-\ — B = cos2^. (25) .- — ^^-r vi=8in26'. tan 2^- tan ^ ^ ^' tan2^-ian^ (26) tan (a + /?) sin''' a - sin^ j8 sin a . cos a - sin ^ . cos li (27) 48in yl . sin (60° + A) . sin (60° - ^) = siii ZA, (28) cosec W + cot 4^ f cosec 46^ = cot B. Solve the ec] nations : ( i) sin 2^ + V3 . cos W = 1. TT (2) sin2 2^-sin5^ = sin2^ (3) sin hx . cos 3« = sin Oa; . cos 7a:. (4) 2 sin-^ 3^ + sin^ 6^ = 2. (5) cos2^+sinM=^. (6) cos ?>Q - cos 5^ = sin B. (7) sin 5^ - cos 3^ = sin i). (8) tan 2a = 3 tan a, j) 8in2^ + sin ^ = cnf5 2^H c.os^. (10) sin 7a-8intt = 3a \i\) corfec'^ 6 - sec'-* 6^ = 2 cosec'-^ ^ ^ 3. (T2) 8in6^ = sin4^-6in2^ il ^ If' ' • n- il. ; I I 1 » V I'i i lli' 106 MULTIPLE AND SUB MULTIPLE AKGLES. 132. To find the trigonometrical ratios for an angle of 18".* Let ^ = 18°. Then 2^1 = 36°, and 3/1 = 54°. Now the sine of an angle is f"^ al to the cosine of the com- plement of the angle ; ;. sin 3(5° = cos 54" ; .•. sin 2^= cos 3-^ : ' ,- 2 sin ^ . cos yl = 4 cos^ A~Z cos A, Divide by cos A. Then 2 sin A =4 cos^ A - 3, 2 sin ^ = 4(1 sin2/l)-3, 2sin^=4-4sinM -3, 4 sin^ i4 +• 2 sin i4 — 1, eiir' /I +- . sm A — -.-, 2 4 •0.1., 1 5 6in-^+2«m4+jg = -^, em yl + -. = ± -—, 4 4 and taking the upper sign, since sin 18° must be positive, 6inl8° = ^-\ 4 Hence we can find cos 18°, tan 18", and the other ratios. Examples.— xxxvi. I. Given sin 18°= '^ , find the value of the following ratios ; (i) 8in3G°. (2) cos36^ (3) sin 54^ (4) cos 54°. (5) sin 72°. (6) tan 72°. (7) sin 90°. (8) cos 90°. * A geometrical proof is given in the Ap])endix. (1° ^'GLF.S. ^ngleoflS"* e of the coin- MULTIPLE AND SUB MULTIPLE ANGLES. 107 2. Show that sin(36'' + ^) + sin(72°-^)-sin(36''-^)-sin(72° + ^) = sini4, and that sin(54° + ^) + sin(54"-/l)-sin(18° + ^)-sm(18''-^) = co3i4. 133. We now proceed to the fornuilse relating to sub- multiples of angles, and first we shall prove that Since . A = ±x/(i^^). cos ^ = 1 - 2 sin2 ^ (Art. 1 28) ; I , 'i } ; fV tv sitive, ratios. 2 sin^ o — 1 ■" cos A ; . ^ A 1 - cos A .•.8in2-2= 2 • *v/^" -ro8i4 2 • If the value of A be given, we know whether sin ^ is positive or negative, and hence we know which sign is to be taken. Thus, if A be between 0° and 360°, "- lies between 0° and 180", and .'. sin ^ is positive : but if A be between 3C()° and 720°, ~ lies between 180° and 360°, and .-. sin ^ 2 \.m is negative. following cos 54°. 5) cos 90'. 134. We shall next show that cos =±^/(.l±.|?J). Since ..A cos ^ = 2 cos'- 2 - 1 (Art. 128) ; :. - 2 cos- -^ = - 1 - COS .4 ; I08 MULTIPLE AND SUB MULTIPLE A/.'GLES. it! ^ H' .*. 2 cos^ " = 1 + cos -4 ; nA 1 + cos ^ 2 I I; I ' |i.Sr> i •• .'. cos i='±^/'^ cos If the value of A he given, we know whether cos ^ h positive or negative, and thus we know which sign is to be taken. For instance, if A lies between 0° and 180°, cos -jT is A positive : but if A lies between 180* and 360*, cos ^ is negative. 135. 'To prove that and 2 cos ^ = ± mJI + sin A ± v^l - sin A, 2 sin o = ± s/l + sin A + v'l - sin A, Since cos2 + sin'-^ 2^^* jiiili A A and 2 cos ^ • sin -^ =sin ^ ; .*. cos'^ ^ + 2 cos -^ . sin -x + sin^ „ = 1 + sin A, and a A n A , A . n A - . A cos^ - - 2 cos ^ . Sin ^ + sm^ - = 1 - sin /I. Hence, taking the square root of each side of both equa- tions, VGLES. MULTIPLE AND SUBMULTIPLE ANGLES. 109 cos y + «in - = ± ^/l + Hii A, ' 'I and cos 2 - sill ^ = + ^1 - sin -4. Therefore, by addition, 2 cos 2 = ± v'l + ^"1 ^ ± VI - «i" -^j ther cos ^ is sign is to be 180°, cos- is 50 , cos - IS and, by subtraction, 2 sin g = ± \/l + sin ^ + ^1 - sin -4. 136. If the value of A be given, we know the signs ut A A sin -g and cos-, as we explained in ibe preceding articles. A A We also know whether sin - is greater or less than cos -. Hence, if A be known, we can assign with certainly the signs wliich the root-syml)ols are to have in the intermediate equa- tions j[ ^ cos g + sin g = ± VI + sin A, (1) ^ ^ cos -^ - sin ^ = ± VI - sin A, (2) and hence we can select the proper signs in the final equa- tions. o^n" For instance, suppose A to lie between 180° and 270' A, both equtt- Then ^ lies between 90° and 135°. .A A . . Therefore sin ^ is positive and cos — is negative. Also, for an angle between 90° and 135° the sine is numeri- cally greater than the cosine. Hence we must take the positive sign in equation (1), ain' the negative sign in equation (2). •t ■ i ii^ ■l^mn n t no MULTIPLE AND SUBMULTIPLE ANGLES. Examples.— xxxvii. 1. Affix to the root-symbols the proper signs when . IS lo . 2. Affix to the root-symbols the proper signs when . is 300". ' • V5 - 1 3. If sin 378° = , determine cos 189° and sin 189°. .■i 4. If sin 19°. 29' = ^, what is the value of sin 9°. 44'. 30" / ' 5. If cos 315° = -i^, find the value of cos 157°. 30'. 137. We mentioned in Art. 114 what are called the Inverse Trigonometrical Functions, sin~^x, cos"^x, etc. We shall now give an example to illustrate the method of combining thesu functions. To prove that tan~^ ^ + tan~^ - = 45*. Let then Now that !s, tt = tan~'^ and ^ = tan~'^ ; tan a = ^ and tan /ii = .,. . / . o\ ta^i a -I- tan H tan {a + fS) = r— ;~^,, '^ 1 - tan a . tan f3 1 1 2^3 1-^.1 2 3 I 6 ""a "1; tan-\^-htan-»^ = 45*. w^^ tNGLES. MULTIPLE AND SVBMULTIPLE ANGLES. in signs when . signs when J .nd sin 189°. 9°. 44'. 30"/ °.30'. ed the Inverse We shall now nbining the&u EXAMPLES.- -xxxviii. 3 4 I., If ^=sin-^ =;, and jB = 8in-^ -, show that ^ + 2? = 901 p o 2. li i4 = tan-^ s, and B = tan"' -^, show that A + 2B = 45' 5 Show that sin-^ -j- + cot"^ 3 -= 45°, 1 1 11 4. Show that tan"^ - + tan*' _ + tan"' s + tan"' * = 45^ w D 7 b 5 . Show that cot"' ? + cot-' I = 1 35°. 4 / 6. Show tliat tan-» f 4 cot'^ I = cot-' J?. 5 J 18 7. Show that tan-* a; - tan~' y — tan"* ^Z^V- . ' ^ l+a;y 8. Show that sin"' a; + cos "'a; = 90°. 9 Show that sin-* - + sin-* - + sin"* ,- = k. 5 13 65 2 '^ '-^1 la. Show that 4 tan -* I - tan"* ~ =. - . 5 239 4 XIV. ON LOGARITHMS. I « ,f' ^ip>i> i;?8. Def. Till' Logarithm of a numlHT to a ^iven base is the index of the power to which the hase imist he laisetl to ;j;ivc the iiuinh«r. 'I'iuis il" m = a% x ia o-.illed the lo.i,^;uithiii of m to the base a. For instance, if the base of a system of h)garitliin3 be 2, 3 is the h)«;aritlini of tlie uiunber 8, because 8 — S'** . and if tlie base be 5, then 3 is tlie lo^'arithni of the number 125, because li2r) = r)^. i' I'M « i b' 139. The lo<,Mrithm of a number m to the base a is written thus, loi,'j»t, ; and so, if /u = a*, Hence it follows that 7?i = (i'"*'''", 140. Since 1 =a*, the lo[;arilhm of unity to any base is zero. Since a = a^, the h>}j;arithni of the base of any system is unity. 141. We now proceinl to describo tliut which ia culled the Common System of lo-rarithms. The base of the system is 10. OM LOGARITHMS. "3 lae must bo ' By u system of loi^'aiitlims to tli" Imsr 10, \vc mean a succei- sion of vuliH's of x' wliicli satisfv the ('(]uation fur all positive valuoa of ?/i, integral or fractional. Such a Bystcm is formofl by tlic scries of logarithms of the natural Tiumbers frcjm 1 to lOOOOO, which constitute tlie h).L,'arilhn»s rt'i^istered in our oi-dinary tables, und which are tiierefure called tabular loyarithms. 142. Now 1 = 1()0, l()-l()i, 1()()=1()2, 1000- 1U'», und so on. lleuce the logarithm of 1 is 0, of 10 is 1, of 100 is 2, of 1000 is 3, and so on. Hence for all numbers between I and 10 the logarithm is a decimal less than 1, between 10 and 100 the logarithm is a decimal between 1 and 2, between 100 and lOOO a decimal between '2 and 3, and so on. 143. The logarithms of the natural numbers from 1 to 12 stand thus in the tables : No. 1 o 3 4 r> 6 Log 0-0000000 0-30 10300 0"4771213 0'()020(;oo Or)98!)700 ()'778ir)l3 No. Log O-84r)0980 8 o-i)o3onoo 1) 0-9542425 10 IDOOOOOO 11 i -04 13927 12 1-0791812 The logarithms are calculated to seven places of decinuils. fs.T.] " t I, K •!■. 114 ON LOGARITHMS, Iff 1^ V 144. The integral parts of the logarithms of numbers higher than 10 are called the characteristics of those loga- rithms, and the decimal parts of the logarithms are called the mantissce. Thus 1 is the characteristic, •0791812 the mantissa, of the logarithm of 12. 145. The logarithms for 100 and the numbers that succeed it (and in some tables those tliat precede 100) have no cha- racteristic prefixed, because it can be supplied by the reader, being 2 for all numbers between 100 and 1000, 3 for all be- tween 1000 and 10000, and so on. Thus iu the tables '.ve shall End No. Log 100 0000000 101 ' 0043214 102 1 0086002 103 0128372 104 i 0170333 105 I 0211893 which we read thus the logarithm of 100 is 2, of 101 is 2-0043214, of 102 is 2-0086002 ; and so on. 146. Logarithms are of great use in making arithmetical computations more easy, for by means of a Table of Loga- rithms the operation of Multiplication is changed into that of Addition, ... Division Subtraction, ... Involution Multiplication, ... Evolution Division, as we shall show iu the next four Articles. 147. 2'he logarithm of a product is equal to the 8%im of the loijarithms of its factors. ON LOG A R ITU MS. II of numbers those loga- re called the that succeed ave no cha- ' the reader, \ for all bc- e tables '.ve ) on. irithmetical e of Loga- inn, iction, lication, t)ii. mm of the Let and 'hen 'm = tr n = a". log^ mn = x + y = log,/yi4-Tog^n. Hence it follows that log, niH}} = log, m + log, n + log, p, and similarly it may be shown that the theorem holds good for any number of factors. Thus the operation of Multiplication is changed into that of Addition. Suppose, for instance, we want to find the product of 240 and 357, we add the logarithms of the factors, and the sum is the logarithm of the product : thus log 246 = 2-3909351 log 357 = 2-5526682 their sum = 49436033 which is the logarithm of 87822, the product required. Note. We do not write logn,246, for so long as we are treating of logarithms to the particular base 10, we may onut the sutlix. 148. The logariihm of a quotient is equal to the logarithm oj the dividend dimiuislicd by the logarithm of the divisor. Let and Then m, = a", n = a*. m, n = «'-"; =^ log, //I- log, n. Thus the operation of Division is changed into that of Sub- traction. W I . '1\ ii6 ON LOGARITHMS. 1 .(t.s % l«Hl! Ml 'I t "■ If, for example, we are required to divide 371*49 by 52-376, we proceed thus, log 371-49 = 2-5099471 log 52-376=1-7191323 their difference = -8508148 which is the logarithm of 7-092752, the quotient required. 149. The logarithm of any jmvcr nf a numher is equal to the product of tlie logarithm, of the naiaber and the index denoting the power. Let m = a*. Then m' = a"'; :. loQ^ni'^rx ^=r.log,.m. Thus the operation of Involution is changed into Multipli- cation. Suppose, for instance, we have to find the fourth power of 13, we may proceed thus, log 13 = 1-1139434 4-4557736 which is the logarithm of 28561, the number required. 150. The logarithm of any root of a number is equal to the quotient arising from the division of the logarithm of the number b;i the number denoting the root. Let Then m = a". 77)'" = a'; X .*. log. m'' = - 1 , = - . log, m. Thus the operation of Evolution is changed into Division. ON LOGARITHMS. 117 9 by 52-376, 'quired. equal to the iex denoting Multipli- ;h power of ed. pial to the the number ivision. If, for example, we have to find the fifth root of 1G807, we proceed thus, 5 I 4-2254902, the log of 1CS07 ^450980 which is the logarithm of 7, the root required. 151. The common system of Lof^arithms has this ndvan tnge over all others for numerical calculations, that its babe is the same as the radix of the common scale of notation. Hence it is that the same mantissa serves for all numbers which have the same significant digits, and differ only in the position of the place of uuits relatively to those digits. i'or, since log 60 = log 10 + log G = 1 + log 6, log 600 = 1 og 1 00 + 1 og 6 = 2 + 1 og 6, log 6000 = log 1000 + log 6 = 3 + log 6, it is clear that if we know the logarithm of any number, as 6, we also know the logarithms of the numbers resulting from multiplying that number by the powers of 10. So again, if we know that log 1-7602 is -247783, we also know that log 17-692 is 1-247783, log 176-92 is 2-247783, lo- 1769-2 is 3-247783, log 17692 is 4-247783, log 176920 is 5-247783. 152. We must now treat of the logarithms of numbers less than unity. Since 1=1"'. -J I 111* I ■ !l Mi 118 C>A^ LOGARITHMS. tlie logarithm of a numlier between land *! lies between and - 1, between 'land '01 -land —2, between -01 and 001 -2 and -3, and so on. Hence the logarithms ot all numbers less than unity are negative. "We do not require a separate table for these logarithms, for we can deduce them from the logarithms of numbers greater than unity by the following process : log -6 =log ^^- =log 6 - log 10 =log 6 - 1, log -06 =log~ = log 6 -log 100 =log6-2, ' log -006 = log ^^ = log 6 - log 1000 = log 6 - 3. Now the logarithm of 6 is -7781513. Hence log -6 = - 1 -r -7781513, which is written 1-7781513, log -06 = - 2 + -7781513, which is written 2-7781513, log -006= -3 + -7781513, which is written 3-7781513, the characteristics only being negative and the mantissse positive. 153. Thus the same mantissro serve for the logarithms of all numbers, whether greater or kss than unity, M-liich have the same significant ui<,nts, and differ only in the position of the place of units relatively to those digits. It is best to regard the tabhi as a register of the logarithms of numbers which have one significant digit before tlie decimal |)oint. ON LOGARITHMS. 119 No Lo« For instance, when we read in the tables 144 | 15S3625, to interpret the entry thus, log 1-44 is -1583625. We then obtain tlie following rules for the characteristic to be attached in eacli case. I. If the decimal point be shifted one, two, three... ?? places to the right, prefix as a characteristic 1, 2, 3...U. II. If the decimal point be shifted one, two, three places to the left, prefix as a characteristic I, 2, 3...U. Thus and log 1-44 is -1583625, .-. log 14-4 is 1-1583625, log 144 is 21583625, log 1440 is 3-1583625, log -144 is 1-1583625, log -0144 is 2-1583625, iocr -00144 is 3-15836-25, n. 4 t J 164. In calculations with negative characteristics we follow the rules of algebra. Thus, (1) If we have to add the logarithms 3-64628 and 2'42367, we first add the mantissoe, and the result is 1 -06995, and then add the characteristics, and this result is 1. The final result is I + 1-06995, that is, -06995. (2) To subtract 5'6249372 from 3-2456973, we may arrange the numbers thus, - 3 + -2456973 - 5 + -6249372 1 + -6207601 the 1 carried on from the last sul)traction in decimal places changing - 5 into - 4, and then -4 subtracted from -3 giving 1 as a reb'iilt. Hence the resulting logarithm is I •6207601. ! lO ON LOGARITHMS. • f (3) To multiply 3-7482509 by 5. 3-7482509 5 12-7412845 the 3 carried on from the last multiplication of the decimal places heing added to - 15, and thus giving - 12 as a result. (4) To divide "14-2456730 hy 4. Increase the negative characteristic so that it may he exactly divisible by 4, making a proper compenoation, thus, 14-2456736 = iO + 2-2450736. ^, f4-2456736 T6 + 2-2456736 -.,rc^A^oA Then t = -. ■ — = 4+ -5014184 : 4 4 ' and so the result is 4-5614184. Examples.— xxxix. * 1. Add 3-1651553, 4-75{)5855, 66879746, 2-6150026. 2. Add 4-0843785, 5-0050657, 3-8905196, 3-4675284. • 3. Add 2-5324716, 3-6650657, 5-89()5196, -3156215. 4. From 2-483209 take 3-742891. 5. From 2-352678 take 5-428619. 6. From 5 349102 take 3024329. 7. Multiply 2-4590721 by 3. 8. Multiply 7-429083 by 6. 9. Multiply 9-2843017 by 7. 10. Divide 6-3725409 by 3. 11. Divide 14-432902 by 0. 12. Divide 4-53627188 by 9, 155. We shall now ex]>l;iin how a system of logarithms calculated to a hnpe a may be transformeil into nnother systeui t'f which the base is h. ON LOGARITHMS. 121 the decimal as a result. it may be ion, thus, 54; ) 150026. 675284. 6215. Let m be a number of which the lo;:'arithni in the first evstem is x and in the second y. Then and Hence m = a , m = b\ *• X log„6' 1_ loiT,, h Hence if we multiply the logarithm of an}' number in the svstem of which the base is a by , — , we shall obtain the ' log,/) logarithm of the same number in the systera of which the base is 6. This constant multiplier .- - Js called The Modulus of ^ log„ h the system of which the base is b with reference to the system of which the base is a. . I. 1 jafithnis r system 156. The common system of logarithms is used in all numerical calculatious, but there is auothei- system, wliicli we must notice, employed by tlie discoverer of logarithms, Napier, and hence called the Napierian System. The base of this system, denoted by the symbol e, is the number which is the sum of the series ^ + 2^-2.3-^2:3:1+ •••'^^^^^' of which sum the first eight digits are 2-7182818. In7. Our common logarithms are formed from the loga- rithms of the Napierian system by multiplying each of the ' »! '1% \ :li*l|| M *;: i \ it I 22 OAT LOGARITHMS. latter by a common multiplier called the litodulus of tlie Oommon System. This modulus is, in accordance with the conclusion of An. 155, , . log, 10 That is, if I and iV he the logarithms of the same number in the common and Napierian systems respectively, log, 10 Now log, 10 is 2-30258509 ; 1 IS 1 lou. M) ' 2-30238509 or -43429448, ^o<f and so the modulus of the common system is -43429448. 158. To prove that h>g„ 6 x log^ a = 1. Let Then ic = log, 6. 6 = a"; ■■a\ \ 1 , X o»' Thus log„6xlogja = ccx- = 1. «»* ;■ 159. The following are simple examples of the method of applying the principles explained in this Cliapter. Ex. 1. Given log 2 = -3010300, log 3 = -4771213 and log 7 = -8450980, find log 42. Since 42 = 2 x 3 x 7 log 42 = log 2 + log 3 + log 7 = -3010300 + -4771213+ •8450980 = 1-6232493. I ON LOGARITHMS. Modulus of the conclusion of ime number in 123 342944a he method of r. 213 B450980 Ex. 2. Given lo^r 2 = -3010300 and log 3 = -4771213, find the logarithms of 64, 81, and 96. log 64 = log 20 = 6 lo,i:c2 log 2 = -3010300 6 log 64= 1-8061800 log 81= log 3^ = 4 log 3 log 3 = -4771213 .-. log 81 = 1-9084852 log 96 = 1 og (32 X 3) = log 32 + log 3, and log 32 = log 25 = 5 log 2; ,. log 96 = 5 log 2 + log 3 = 1-5051500 + -4771213= 1-9822713. Ex. 3. Given log 5 = -6989700, find the logarithm of ;'(6-25). log (6-25) • = \ log 6-25 = ^ log ^^5 = ^ (log 625 - log 100) =i(log54-2) = i(4log5-2) = ]. (2-7958800 - 2) = -1136971. Examples.— xl. 1. Given log 2 = -3010300, find log 128, log 125, and log 2500. 2. Given log 2 = -3010300 and log 7 = -8450980, find the logarithms of 50, -005, and 196. 3. Given log 2 = -3010300, and log 3 = '4771213, find the logarithms of 6, 27, 54, and 576. 4. Givenlog2 = -3010300,log3 = '4771213,log7 = -8450980, find log CO, log -03, log 1*05, and log -0000432. ■\ \2\ ON LOGARITHMS. I ) i % ■*' 5. Given log 2 = '3010300, log 18= 1-2552725 and log 21 = 1-3222193, find log -00075 and log 31-5. 6. Given log 5 = -6989700, find the logarithms of 2, -064, and(^"y*. 7. Given log 2 = -3010300, find the logarithms of 5, -125, and (l^y. - 8. What are the logarithms of -01, 1, and 100 to the base l.")? What to the base -01? g. What is the characteristic of log 1593, (i) to base 10, (2) to base 12 ? 4* 10. Given -^^ = 8, and a: = 3//, find x and ?/. 11. Given log 4 = -01)20600, log 1-04= "0170333 i 1 (a) Find the logarithms of 2, 25, 83-2, (-62:))^°^ (6) How many digits are there in the integral part of (l-04)«'"^? 12. Given log 25 = 1-3079400, log 1-03 = -01 28372 : (a) Find the logarithms of 5, 4, 5 1 -5, (064)^ (ft) How many digita ore there in the integral part of (103)*"? 1 3. Having given log 3 - - 1771 2 1 3, log 7 = -8450980, log M = 1-04131)27 : 77 3 find the logarithms of 7623, ^- and ^--^. oOU Odu 14. Solve the ^([uationa ; (1) 4096'^ ^.|,. (4) oTV-'^a. (2) (yy=^6-2r). (5) a^.h'"'^c^-\ (3) u'.b'-^m, • (6) a'^-'-^c*"". '25 and log 31-5. :hms of 2, -064, /hms of 5, '125, 100 to the base (i) to base 10, 33: itegrul part of =(372 : i_ foo Legral part of 50080, i-i XV. ON TRIGONOMETRICAL AND LOGARITHMIC TABLES. 160. We sball ,Lrive in this Chapter a short description of the 'liibles ^vhlch luive been constructed for the purpose of facilitating trigonometrical calculations. The methods by which these tables are form"'! do not fall within the range of this treatise : we have merely tu exphiii: how they are applied to the solution of such siniplo ei.ampLj us we shall hereafter give. We shill arrange our remarks in the following order; I. On Tables of Logarithms of Numbers. II. On Tables of Trigonometrical Ratios. III. On Tables of Logarithms of Trigonometrical Ratios. L On Tahlei^ of Lor/arifhwn nf Nnmhers. 161. These tables are arranged so as to give the mantissas 1)1 the logarithms of the natural nundxn-s from 1 to 100000, that id of numbers containuig from one to five digits. We shidl now show how by aid of these tables, first, to find the logaiithm of any given number, and, secondly, how to de- termine the uuniber which concs^jonda to a given logarithm. ¥ ^^' «•• '<mk V 'ill .. 126 ON IRIGONOMETRICAL AND 162. When a number is given, to find its logarithm. When the qiven nnmhor has not more llian five digits, wu can take its logaiithni at once from the tables. When the given number lias more than five digits, we can determine its logarithm by a process which will be best ex- plained by an example. Suppose we require to find the logarithm of G27C153. We find from the tables log ()27()1 is 4-797(;809, and Hence and log()27()2 is 4-7l)7G!)88. log()27()200 is C-797C988, 1olM):27() UK) is ()-797()>s99. Thus for a difference of 100 in the numbers the difTerGnce of the logarithms is •00001)89. We then reason thus : If we have to add -0000089 to the logarithm of G27()100 to obtain the logarithm of G27(>200, what must we add to the logarithm of 0270100 to obtain the loga- rithm of 0270 153/ Assuming that the increase of the logarithm is proportional to the increase of the numljci" (which is nearly but not quite true), we shall have 100 : 53 =-0000089 : that which we have to add ; , ^ , ,11 53 X '0000089 , ,^,„ .-. number to be added = - -='000004717, and therefore, omitting the last two figures, log G27(;i 53 = 0-7970899 -I- '0000047 = 0-79709 10. If the first of the figni'es omitted be 5 or a digit higher than 5, it is usual to increase the figure immediately precedini^' it by I : thus if the number to be added had been -000004757 we should have added '0000048. ) LOG A A7 y 7JMJC TABLES. 127 ithm. five digits, we digits, we can ill be best ex- 27G153. the difTercnce 000089 to the ()27()200, what ituiu the lugu- is proportional but not quite to add ; )0004717, 47 I digit higher ely preceding Ln "-OOGOO^TO? 1G3. We took in tlie hist Artiele an integral number, but the same process will apply to numbers containing decimals. For suppose we have to find the logarithm of G27'6153 : we can find log 027615:1 as before, and the only dillerence to be made in the linal result is to change the characteristic Irom G to 2, that is, logG27-Gir)3 = 2-7!)7GUlG. So again, in accordance with Art. 153, log -00000270153 = 5-7970t)46. I. Given find 2. Given find 3. Given tiud 4. Given find 5. Given find 6. Given find Examples.— xli. log 52502 = 4-720175M, log 52503 = 4 -720 184 1, log 52502-5. log 3-0042 = -4777288, log 3-0043 = -4777433, log 300-425. log 3202-5 = 3-505489 1, log 3202-0 = 3-5055027, log 321)25013. log 23000 = 4-3740147, log 23001 =4 3740331, log 230-001. log 07502 = 4-8293100, log 07503 = 4-8293231, log 07-5021. log 73335 = 4-8053113, log 73330 = 4-8053172, log -007333533. ^ 1 ;: 12S ON TRIGONOMETRICAL AND 7. Given find 8. Given find 9. Given tind 10. Given find log 65931 ^4-8190897, 10^05932 = 4-8190902, log •000000593171. log 34-077 = 1-5324014, log 34-078 = 1-5324741, log 3407-78. log 39097 = 4-5921434, log 39098 = 4-5921545, log 390974. log 25819 = 4-4119394, log 25820 = 4-41 19502, log 2-581920. 164. JVJien a loyarithm is given, to find the number to which it corresponds. If tlie decimal part of the logaritlim i^ found exactly in the tables, we can take out the corresponding number. Thus if we have to find the number corresponding to the lo,i^arithm 28598045, we look in the tables for the mantissa •8598645, and we find it set down opposite the number 72421, hence 2-8598045 is the logarithm of 724-21. Next, suppose that the decimal ])ait of the logarithm is not found exactly in the tables, and that we have to find the number corresponding to the logarithm 3'9212074. We find from the tables log 8340-8 = 3-92 1 2077, aud log 8340-7 = 3-92 1 2025. Hence a difi'erence of '1 in the numbers gives 11 differcnco of •0000052 in the logarithms. a > LOGARITHMIC TABLES. 129 umbtr to which Tliun we reason thus : If we must aJd "1 to 8310-7 for an excL'SS of •()()(H)()r)2 above the loL,Mritlmi of 8:U()*7, wliat must we add for an excess of (:3-92 12074 -3-92 120:25) or •01)00049 I Assuming' that tlie increase of the number is proportional to tlie increase of the h»^'arithm, we have •0000052 : -0000049 = '1 : what is to be added to 83407 ; therefore we must add •0000049 x-1 49x-l •0000052 '^''"52 '''^•'•«»^; therefore number required is 8340-794. If the given logarithm be negative, as — (2'1401355), we can change it into atiotlier of wliich the characteristic only is negative, tlius 3^8598()45, and we can liud the number cor- responding to this logarithm, as before. The number re(|uired is -008340794. ( ] Examples.— xlii. 1 . Given log 1 2954 = 4^11 24039, logl2955 = 4^1124374, tiud the number whose logarithm is 4^1 12431. 2. Given log 4()2-45 = 2^6(;50()48, h.g4()2-46 = 2^f)«50742, hnd the number wliose lo-'arithm is 3()(J50C5 3. Given log 34572 = 4-5387245, log 34573-4-5387371, lind the number whose logarithm is 2-5387359. s u differenco 4. Given ■ log 39375 = 4 5952200, log 39370 = 4-5952310, fuiil the number wliose logarithm is 0-5952282. LS.T.J I: ( i ^! II 1^ a 130 C>A^ TRIGONOMETRICAL AND 5. Given log 3-71ol) = •5700640, log 3-7160 = -5700757, fimd the number whose logarithm is 3-5700702. 6. Given lo,": 06461 =4-9843518, log 96462 = 4-9843563, tind the number whose logarithm is 3-9843542. 7. Given log 25725 = 4-4103554, log 25726 = 4-4103723, find the number whose logarithm is 7-4103720. 8. Given log 60195 = 4-7795532, log 60196 = 4-7795604, find the number whose logarithm is 2-7795561. 9. Given log 10905 = 4-0376257, log 10906 = 4-0376655. lind the number whose logarithm is 3-0376371. i 'N:^ 'i ■' 10. Given log 26201 =4-4183179, log 26202 = 4-4183.344, find the number whose logarithm is 2-4183314. II. On Tables of the Trirjonometrical Ratios. 165. We have explained in earlier parts ol" this treatise how to find the vahrjs of certain trigonometrical ratios exactly or approximately. Thus we showed that sin 30°- .,, that is, -5 exactly. Again, tan 60°= ^3, that is, 1-73205 ai)proximately. Now the values of all tlie trigonometrical ratios for a regular succession of angles in t'^e first (jiiadrant have beei ■* LOGARITHMIC TABLES. 131 calculated and registered in tables. In some tables the angles succeed each other at intervals ot" 1", in others at intervals of 10", but in ordinary tables at intervals of 1', and to the last- mentioned we shall refer. ^ These tables are commonly called TaVdes of Natural Sines, Cosines, etc., so as to distinguish them from the Tables of the Lngarithms of the Sines, Cosines, etc., of which we shall here- after treat. We intend to explain, first, how we can determine the value of a ratio that lies between the ratios of two con- secutive angles given in the tables, and secondly, how to determine the angle to which a given ratio corresponds. .'■ i 1C6. To find the sine of a given awjle. Suppose we want to determine the sine of 25° . 14' . 20", having given sin 25° . 14' = -4203056 and sin 25° . 15' = -4205687. From sin 25° . 15' = -4265687 Take sin 25° . 14'-- -4263056 •0002631 is the difference for 1'. Now if we have to add -0002631 to the sine of 25°. 14' to obtain the sine of 25°. 15', what must we add to the sine of 25° . 14' to obtain the sine of 25° . 14'. 20" ? Assuming that an increase in th«^ angle is proportional to an increase in the sine, we have r : 20"= -0002(531 : that which w: have to add ; therefore we must add 20 X -0002631 60 -, or, -0000877 ; ;. sine 25° , 14' . 20"= -4263056 + -0000877 =- -4263933. lf^ I ;;. I Ml •-', J '32 OA^ TRIGONOMETRICAL AND 167. To find the cosine of a given angle. [f we have to determine the cosine of 74° , 45'. 40", having given cos 74° . 46' = -4263056 and cos 74° 45' = -4265687, we proceed thus : , cos 74°. 45' =-4265687 cos 74°. 46' = -4263056 •0002631 is the decrease corresponding to 1', oljserving that the cosine decreases as the angle increases from 0° to 90°. Hence 1': 40" = -0002631 : what we have to take from '4265687; therefore we ir»ust take away 40 X -0002631 60" = -0001754; .-. cos 74° . 45' . 40" = -4265687 - '0001754 = -4263933. Similar methods are to be taken to find the values of the other ratios, observing that the tangents and secants increase and the cotangents and cosecants decrease as the angle in- creases from 0° tu 90°. I. Given find 2. Given liiid Examples.— xliii, sin 42°. 15' =-6723668, sin 42°. 16' = -6725821, sin 42°. 15'. 16". bin 72°. 14'= -9523071, sin 72°. 15'= -9523958, sin 72°. 14'. 6". •^•PI'i LOGARITHMIC TABLES. 133 find Given sin 54" . 35' = -8149593, sin 54°. 30' =-8 15 1278, sin 54° . 35' . 45". ti 4. GiveL' find sin 87°. 26' =-9989968, sin 87°. 27' = -9990098, sin 87°. 26'. 15". spending to 1', 1 ^ Given increases from 1 ..a • •cm -4265687; 1 J Given 1 Given '54 1 »<>^ V tunes of the 'ants increase the angle in- I : Given 1 Given 1 liud 1 ,0. Given 1 find sin 43M4' = -6849711, 8in43°,15'=-6851830> sin 43° . 14' . 20". cos 41° 13' = -7522233, cos 41°. 14'= -7520316, cos 41°. 13'. 26". tan r. 22' =-0238573, tan 1°. 23' =-024 1 484, tan 1°.22'.30". cot 35° 6'= 1-422856], cot 35°, 7' = 1-42 19766, cot35°.ry 23". sin 67°. 22' = -922086.5, sin (57°. 23' =-9230984, ein 67° . 22' . 48"-5. cos 34°. 12' = -8270806, cos 34°. 13' =-82691 70, co.s34°.12'. 19"-6. \'Aa w Ml T kilt -7' r I iiiiiill- i mm \u'- i i- ^34 OA^ TRIGONOMETRICAL AND 168. To find the angle which corresponds to a given sine. Suppose tlie given sine to be •5082784. We find from the Tables sine 3(^.33' =5082901 Bine30°.32'=-508()396 •0002505 difference for T. given sine =5082784 sine 30°. 32' = -5080396 •0002388 Hence if x be the number of seconds to be added to 30° . 32', •0002505 : -0002388 = 60 ; x ; 2388x60 9552 ^^ ^ , ■■•^=-2505- = T67=^^'2^'"'^>^- .-. the required angle is 30° . 32' . 57"-2. 169. To find the angle which corresponds to a given cosino. Suppose the given cosine to be •5082784. We find from the Tables cos 59°. 27'= 5082001 cos 59°. 28' = '5080396 .' •0002505 difference for 1'. cos 59°. 27' =-5082901 given cosine = '5082784 •0000117 Hence if a; be the number of seconds to be odded to 59°. 27', •0002505 : •0000117 = 60 : x; 117x60 ^„ , '•'^ = "-2505'-==^"^^'''^''^^' .-. required angle is 59° . 27' . 2"-8. LOGARITHMIC TABLES. 135 given sine. • iven cosniQ. Examples.— xliv. I. Given sin 48° . 47' = -7522233, sin 48°. 46' =-7520316, find the angle of which the sine is -752140. 2. Given cos 2°. 34' = -9989968, cos 2°. 33' = -9990098, find the angle of which the cosine is -999000. 3. Given sin 43M4' = -6849711, sin 43M 5' =-6851830, find the angle of which the sine is -685. 4. Given cos 32° . 31'= 8432351, cos 32°. 32' =-8430787, iind the angle of which the cosine is -8432. 5. Given sin 24° 11' = -4096577, sin 24°. 12'= -4099230, find the angle of which the sine is -4097559. 6. Given sec 82° . 22' = 7-528249, sec 82°. 23' = 7-552 169, find the angle of which the secant is 7-53. i a 7. Given cos 53°. 7'= -6001876, cos 53°. 8' = -5999549, find the angle of which the cosine is -6. 8. Given cosec 25" . 3' = 2*36 1 79, cosec 25°. 4' = 2-36029, find the angle of which the cosecant is 2-361. '\ liH'i m\' 116 OA' T/C/GO/VOAflLr/^/CAL AND 9. Given sin 73° . 44' = '9599684, sin 73^ 45'= -9000499, find the angle of which the sine is -96. 10. Given ( liOi' ■ \\ tan 77°. 1 9' = 4-44338, tan 77°. 20' = 4-44942, find the angle of which the tangent is 4-4. III. On Tables of Logarithms of Trigonometrical Ratios. 170. The trigonometrical ratios, being nnmhers, have logarithms that correspond to them, and these logarithms are in practice much more useful than the numbers themselves. Now since the sines and cosines of all angles and the tan- gents of angles less than 45° are less than unity, the logarithms of these sines, cosines, and tangents are negative. In order to avoid the inconvenience of printing negative characteristics, the logarithms of all the ratios given in the tables are in- creased hy 10. The numbers thus registered are called The Tabular Logarithms of the sine, cosine, etc., and they are denoted by the symbol //, that is, L sin A denotes the tabular logarithm of the sine of A. When the value of any one of these Tabular Logarithms is given, we must take away 10 from it to obtain the true value of the logarithm in question ; thus L sin 25° is set down in the tables as 9*6259483, and the true value of the logtirithm of the sine of 25° is there- fore 9-6259483 - 10, that is, - -3740517, or we might adopt the usual logarithmic notation of Art. 152, and say log. sin 25° = 1-6259483. The Tables to which \ye refer are calculated for all angles in the first quadrant at intervals of,l'. cal Ratios. imhers, have jgaritlims are ;heinselves. and the tan- he logarithms In order to haracteristics, tables are in- re called The iiid they are the tabular logarithms is le true value 9483, 25° is there- of Art. 152, all angles I LOG A RnilMlC J A liL ES. 13- 171. To find the logarWimic sine- of an angle not exactly given in the tables. Suppose we have to find L sin G° . 32' . 37". LsinO\ 33' = 9-0571723 L sin 6°. 32' = 9 0500706 •0011017 diflerence for 1. Then if a: be the number to be added to 9 "056070 6 to give us Z<sinG°.32'.37", wehave /. x= - GO : 37 = -OOllOl7 : x ; •0011017x37 60 = •0006794, nearly; .-. L sin 6° , 32' . 37" = 9 -0560706 + -0006794 = 9 05675. 172. To find the logarithmic cosine of an angle not exactly given in (he tables. Suppose we have to find L co?^ 83° . 27' . 23". L cos 83°. 27' = 9-0571723 ii cos 83°. 28' = 9 -0560706 -0011017 difference for 1'. Then, if x be the number to be subtracted from 90571723 to give us L cos 83° . 27' . 23", we have 60 : 23 = -0011017 : x; •0011017x23 ^,,,^,-,^., 1 :, x= ,. , = -0004223 nearly ; oO .-. L cos 83° . 27' . 23" = 905717^3 - -0004223 = 9-05675. Similar methods must be taken to find the tabular loga- rithms of the other trigonometrical ratios : it being remem- bered that the tangent and secant increase, and the cotangent and cosecant decrease, as we pass from 0° to 90°. ■Mi i; j; t 'I « ' ♦ ! h^ if ^. i|if 138 find find find find find ON TRIGONOMETRICAL AND I. Given 2. Given 3. Given 4. Given 5. Given 6. Given thid 7. Given find S. Given find Examples.— xlv. Z sin 55°. 33' = 0-9162539, L sin 55°. 34'= 9-9163406, L sin 55°. 33'. 54". L sin 29°. 25' = 9-6912205, L sin 29°. 26' = 9 -69 14445, L sin 29' . 25' . 2". X cos 37°. 28'= 9-8996604, L cos 37°. 29' = 9-8995636, L cos 37° . 28' . 36". X sin 54°. 13' = 9-9091461, L sin 54°. 14' = 9-9092371, L sin 54°. 13'. 19". 7>tan 27°. 42' = 9-7201 600, L tan 27°. 43' = 9-7204759, L tun 27°* . 42' . 34". Ltan5°.13' = 8-9604728, Xtan5°.I4' = 8 9618659, L tan 5°. 13'. 23". Lcot 3°. 37'= 11-1992368, L cot 3°. 38'= 11 -1972347, L cot 3° . 37' . 50". I sin 39°. 25'=. 9 8027431, L8in39°.26' = 9-802H968, Lsin39°.25'.10". LOGARITHMIC TABLES. 139 9. Given find 10. Given find i>sin7()°.34' = 9-0745i52, Lsin7(r.:3r)' = 9-9745G97, Lsin70\34'. 17". L cos 88° . 54' = 8-2832434, i.cos88°.55' = 827G613C, Lcos88°.54M6". 173. To find the angle which corrc^jponds to a given Tabula':' Logarithmic Sine. Let the given Z sine be 8*878594. VVu liud irom the tables X sin 4\ 21' = 8-8799493 Z sin 4'\ 20' = 8 -8782854 •0()1G(J39 (liflerence for 1'. given Z sin = 8-8789540 L sin 4°. 20' = 8-8782854 0'00()CG8() Hence if x be the number of seconds to be added to 4" . 20', •001GG39 ; -O00GG8G = GO : x ; 6G8GxG0 „^ , -^^=-io^r=2^"^^'^^'^y' .*. re(iuiri'd angle is 4° . 20' . 24". 174. To find the angle which corresponds to a given Tabular Logarithmic Codne. Let the given L cosine bo 8*878954. We Ihid from the tables L cos 85°. 39' =8 8799 103 -Lcor.85°.40' = 8-878??854 •001GG39 (lifTerenco for 1'. 140 ON TRIGONOMETRICAL AND L cos 85°. 39' = 8-8799493 given Z cosine = 8-8789540 •0009953 Hence if x be the number ol seconds to be added to 85" . 39', •0016G39 : -0009953 = 60 : x\ 9953 X 60 ^^ ^ , :. X = -j-(.^3(p = '^^ '8 nearly ; . . required angle is 85' . 39' . 35""d. EXAMPLES.— Xlvi. 1. Given L sin 14° . 24' = 9-3956581, 7vsinl4° 25' = 9-3961499, find the angle ^vho.se L sin is 9-3959449, 2. Given 7. sin 54° 13' = 9-9091461, /.sin 54°. 14' = 9-9092371, find the angle whose L sin ia 9-9091760. 3. Given 7> sin 71° . 40' = 9-9773772, 7>sin7r.41' = 9S)774191, find the angle whose L sin is 9-9773897. 4. Given L cos 29° . 25' = 9-9400535, L cos 29\ 2(1' = 9 -9399823, * find the angle whose L cos is 9-9400512. 5. Given L tan 30° . 5(»' = 9-7759077, 7vtan30°.51' = 9-776l947, find the angle whose L tun is 9-7760397. 6. Given L cot 86° . 32' -8-7823199, X cot 86°. 33' = 8-7802218, lind the angle wlinso L cot ia 8-7814643. LOGARITHMIC 7ABLES to be added to 7- Given /-sin 24°. 8'=9-()1157G2, Z sin 24\ 9' = 9 -6 11 8580, find the angle whose L sin is 9-6117876. 8. Given L tan 11° . 39' = 9 -3 142468, i^tanir 40' = 9-3148851, find the angle wlio.se L tan is 9-3148011. 9. Given L cosec 46" . L3'= 10-1402787, L cosec 46° . 24'= 10-1401584, find the angle whose L cosec is 10-1402567. 141 10. Given L sec 29°. 54'= 10-0620326, L sec ^j)° . 55'= 10 062 1053, find the angle whose L sec is 100620665. I :|f ( lil '■ii ■ I ,1 XVI. ON THE RELATION?; BETWEEN THE SIDES OF A TRIANGLE AND THE TRIGONOMETRICAL RATIOS OF THE ANGLES OF THE TRIANGLE. 175. A Triangle is composed of six parts, three sides and three angles. Tliree of these parts heincj givei.; one at least of the tliree being a side, we can generally determine the otlier three parts. If only the three angles he given, we cannot determine tlm sides, because an infinite number of triangles may be con- structed with the three angles of the one equal to the three angles of the other, each to each. 176, Wo shall denote the angles of a triangle by the letters A, I>, ; the sides respectively opposite to them by the letters a, &, c. The student must remember the results established iu Art. 101, sin(180°-yl) = sini4, cos (180°-^)= -cos il. A B C D Thus, if AUB be an exterior angle of the triangle ACDy Bin AOn=--miACIl coaACli^-wsACD. The results of Exam})les xiii. 2, on page 36, are also fre- quently employed in this and the next Chapters. ■i RELATIONS BETWEEN THE STDES, ETC. 143 arts, three sides establislied in 177. To show that in any trianrjle c = a . cos B+b . cos A . Fig. ii. Let A, B be any two angles of tlie triangle ABC, and as oiu' of tlieni must be acute, let it be A. Tlien, according as B is acute or obtuse, draw CD at right angles to AB or to ^^ produced. Then, in fig. 1, c==AD + DB = ^C. cos A \-BC . cos 5 = & . cos ^ + a . cos J5 ; and in fig. 2, c=AD-DB = AG .cos A- CB .cos CBD s=6 . cos vl + (t . cos B, since cos CBD= -cos i45C. If the angle at B be a right angle, the theorem holds good, for then cos i? = cos 00° --0, and .'. a . cm 7? = 0, and we have N \ c — h. COS A, M I ^ 144 RELATIONS BETWEEN THE SIDES 178. To shoiv that in eveiij triamjU the sides are proportional to the sines of the oppodte angles. i^'ig- 1. i^'ig. '^. Let A. B be aiiv two angles of the triande ABC, and a& cue o f them must be an acute an.nle, let it be A Then, according:; as B is acute or obtuse, draw CD at right angles to AB or to AB produced. Then, in fig. 1, sin ^l = CD h ' sm i> = - - ; (I ^ip_/^=£P-^^=^^x sin B a a _n and in lig. 2, sin A = CD 1) ' CD Bin jr? = sin (180" - 7?) = sin CBD = —, ^\nAJW^CD CD sin 7) /' * a - _ a _a If the angle at B be a right angle, the theorem still holdd good, for then sin A = sinii=l sin A _a sin B h' AISTD ANGLES OF A TRIANGLE. '45 tre proportional Liw CD at ritrht rem still holdd Similarly it may be shown tliat in any triangle sin A a , sin B h sm 6 c sm c and therefore we conclude that bin "A sin B gin (7 a b c ' 179. To express the cosim of an angle of a tria^igle in terms of the sides. Fig. 1. Fig. 2. c c A D ' b A ^' E Let ^, J5 be any two angles of the triangle ABC, and as one of them must be acute, let it be A. Then, according as B is acute or obtuse, draw CD at right angles to AB or to AB produced. Now, in tig. 1, by Euclid ii. 13, AC^ = AB- + BC''^-2AB.BD, or h'^ = ,y^ + a'^-2c.BD. Now, from the right-angled triangle BCD, -un = COS B, or = cos i>, or BD = a . cos 7; : ■ BC ^ a : .-. 6'-^ = c- + a^ - 2ac . cos B. Again, in fig. 2, by Euclid ii. 12, AC- = AB'^ + BC^ + 2AB.BD, or h'==c'^ + a' + 2c.BD. * For another proof of this proposition, see Art. 221. ", t ■ 146 /DELATIONS BETWEEN THE SIDES BD Now =^= cos CBD--^ - cos (180° - CBlJ) = - cos B ; :. BD=-BG cos R or i?i)= - a. cos ^ ; imw fe2 = c2 + a2-2ac.cos£. Hence, in each case, 2ac .cos B-c^ + a^- 6^, or So also cos B- cos A and cos (7= • 2ac " 26c "' '+62-C'- a- 2a6 If j5 is a right angle, cos ^=0, and the theorem still holds true, lor then 180. To show that tan ¥■■ ■ a^-Vc a A-B a-h ,0 a + b cot 2' Since a sm A h sill B >> ft-6 sin yt -sin .5 a + 6~~siu ^ + sin B 2 cos A+B . A-B 2 sin 2 sin A + B cos A-B (Art. 120 tan A-B tan tan 2 A-B n f Since ^+i^ cot ;r- 2 90'- -• tan A-B a-b 2 a + b cot AND ANGLES OF A TRIANGLE. 147 ici Tf 'a + 6 + c Ibl. ns-=~^~ , we can prove tlie ibllowiiig results: (2) cos^ = The method of proof will he given in the next two Articles. eorem still holds 182. First, to show that Since 8171 -= /(^-bj (s-c) '2 V ' "l^ti cos ^ = 1 - 2 sin^ 2 sin*'^ 2 = 1 - cos ^ 2bc ■ ^^^^ Z^} (a-b-^c) 2bc Now if . a+b+c _ , .'. a +6~c = 26- -2c and a-6fc = 2s-2?) l'4r \ m'i ■r> It: i " 148 RELATIONS BETWEEN THE SIDES . . 2 sin . ,yl 2(s-c).2(s-6) 111- -- — i L > — . y • 2&C ^-l"^^'- AVe must take tb>. positive sign with the root-symbol, because A being an angle of a triangle must be less than A A 180°, and therefore -r less than 90°, and consequently sin ^- is positive. tl83. Next, to show that A Is . (s - a) cos 2 \ be Since cos A = 2 cos- -J) — 1 ; 2 cos'-^ ^j- = 1 + cos A = 1 + J*2 ^ c2 _ ^a 26c~' __26c + 6'' + c2-a2 "26c _ (6-fc- f g) (6 + c - fl) "26c " Now, if 6 + c + a '=-2 > 6 4 c 4 a = 2s and 6 + c-a = 2s-2rt; _ „ yl 2s . 2 (s - a) .. yl s.(s - a) A s . (s - a) AND ANGLES OF A TRIANGLE. 149 184.' From the prece'ling Articles we may at once derive two other formulso : (1) sin y4 = 2 sin -- . cos — the root-symbol, list be less than nsequently sin -- _o Cs'-by(s-c) ""V Fc Vs.(s-d) ^To' V^-(«-'*)(^-^^)(^-<^)> (2) tan 2" = sin -^^co&-^ ~'V he "^V" ^ '\ s .{s-a Examples.— xlvii. Prove the following relations when A, B, C are the angles of a triangle : I. B\n{A+B) = 8'inG, . A + B G 3. sm— g-^cos^. A+B n 5. ian-y-=cotg. 2. cos (^ + 5) = - cos G. A + B . G 4. cos ^-=mn^. 6. cot — ^ — = tan ^. 185. Many other relations may he established by the use of the important formuljB explained in Art. 120, and the set of examples just given. Thus, to show that, if ^ + ^ + C= 180°, Sin A + sin J54-sin C=4 cos ^ . cos -^ . cos a, ISO RELATIONS BETWEEN THE SIDES > ; we proceed thus, sin -4 + sin 5 = 2 sin — - — . cos — _ — (Art. liJO) _ G A-B = 2 cos ^ . cos — - — ; . ' A , ' n ' n c, ^ A-B „ . G C . . 8in A + aniB + sin 0= 2 cos - . cos — ; — h 2 sm ^ . cos - 2 2 2 2 — 2 cos7>( cos C/ A-B A + B 2V'°'-2-'^'°'~2 ) = 2 cos ?(2 cos y . cos ^) (Art. 120) A B G = 4 cos— . cos-^.cos-^. ^ ^ ^ Examples.— xlviii. I. li A, B, G he the angles of a triangle, prove the followin'' relations : (i) sin 2i4 + sin 25 + sin 20=4 sin A sin -K sin C. (2) 4sin^ .sin J5.sin C=sin (-yl + 5 + C) + sin(^-5 + (7) + sin(^+i?-C). G (3) cot -^ + cot ^ . „ 2 2 _ sm B ~B To "sin J.* cot -^- + cot ^ (4) tan A + tan B + tan 0= tan A . tan B . tan 0. (5) cot^ .cot i? + cot ^ .cot C+cot J5. cot 0=1. (6) cot -^ + cot — + cot 2 = cot -g- . cot - - . cot r . (7) 4 cos ^ . cos 5 . cos 0= - (1 + cos 2^4 + cos 2B + cos 20). (8) cos ^ + cos 5 + cos 0= 4 sin — . sin^ .sin ^ + 1. ^ ^ ^ AND ANGLES OF A TRIANGLE. 151 (9) 4 sin ^ . cos B . cos C= - sin 2A + sin 27? + sin 26* ABC (10) sin ^ +sin5-sin C=4sin — . sin-^. cos^. W 'Jl Jt (11) sin2vl -i-sin 27? -sin 20=4 sin C.cos^ .cos 5. ABC (12) cos ><4 + cos 5 -cos (7=4 cos — . cos-^. sin - - 1. Ji £1 A (n) cos^-^ + cos- 7r + cos2-=2 + 2sm-^. sm-Tr -smx. ^ J' <2, 2 2 2 2 2 (14) 6in2 — + sm2 — + sin2 - = 1 - 2 sin -^ . sin-^ . sin 5 . 2. In a right-angled triangle where C is the right angle, prove that (i) 6 + c = a. cot g- (2) 2cosec2^ .cot J5 = 6^ (3) «i^|=\/(^)- (4) ^"4=V'(^') . , cos 27? - COS 2^ . . . r, (5) --sin2^" '=^^^^-^'^^'^• (6) tan 2^ -sec 27? = J--. . oO (7) (sin A - sin 5)'-^ + (cos A + cos Bf = 4 sin^ --. (.2 (8) sec 2^ = ^-37-2- (9) ohc-a? , cos A\}p' . cos 5. (10) cot(5-^)+cot2(^ + ^^)==a 3. I«i any triangle prove the following relations : (1) sin ^- sin B sin (7 a — b . (2) &\i\{A-B)^ a^-h^ sin C c^ (3) tan A = T — '■ -rr (4) tiot A = -. cosec 7? - cot B. ^^' b-a cos (7 ^^^ a m^ H 152 /DELATIONS BETWEEN THE SIDES '•^.: h (5 (6 (7 (8 (ro (11 (12 (13: (14 (15 (16 (17 a + h + c={a + b) cos C+{a + c) coaB \ {b + c)coaA. (a + 0) sin jT = c . cos — ^ — . (a - 0) cos i^ = c , sin —5 — . tan5 a2 + j2_g2 ^ t;iirc"f72~/;^T^2- (9) c = a(cosB + smi?.ccti4}. a2 4-6'' + c^=2(a6.co3 G + ac. cos B + be . cos^). cos^ A + cos^ B + cos'-* G+ 2 coa -4 . cos B . cos C=»= i. cos i? — c< >pi ^ = ~ .2 COS'* -. c 2 COS ^ + COS ii = — .2 sin^ -a. c 2 a'-^ sin ^ + a6 . sin B + ac. sin C= (a-^ + 6^ + c^ sin A. cot -g- : cot -g- = f c - a : a + c - 0. ^A ^ B a + b + o cot „- . cot -TT = T 2 2 rt 4- 6 - c a . sin {B - G) + b. sin {G-A) + c. sin (^ - 5)=0. 4.. If a, 6, c be in arithmetical progression, show that * I > \ sin (-'-D-- 2 sin TT-. 5. If ABG be a triangle, and AD be drawn at right angles *;o BG, show that i r, 'j^sin C+c'sin B b + c J AND ANGLES OF A TRIANGLE. 153 6. The sides of a triangle being 4, 9, 12, sliow that the length of the line bisecting the angle between the two shorter sides is 2 — . 7. If sia A = '2 cos B . sin C, show that the triangle is isosceles. o Tr A D . /^ sin ^ + sin 5 , ., . 8. If cos ^ . cos B . sm C= — , ^, show that sec yl + sec B (7=90^ 9. If sin2 A = sin2 B + sin'-^ 0, show that A = 90'. Q? + 6"^ + C*^ 10. If f =c- and also sin ^4 . sin i? = sin^ C. show that the triangle is equilateral. 11. If C=120°, show that c^^a^-vah + V^. 12. If CD bisect the angle G and meet AB in D, show that tan ADC= r tan ^. a-b 2 13. If CD bisect AB, show that \4\ ■ mr^^ XVII. ON THE SOLUTION OF RIGHT- ANGLED TRIANGLES. 186. Let a, &, c be the skies of a tricangle aTid A, B, C the angles opposite to tliem. Of those six elements which present themselves in every triangle tJiree must be known in order that we may determine the others, and one of these three must be a side* 187. The three angles of every triangle are together equal to two right angles : that is, A + B + C=180\ Hence, if two of the angles be known, the third will be known also. 188. Several of the results obtained in Chap. xvi. are to be carefully remembered, and especial attention must be given to the following formuhe, established in Arts. 178, 179, 180. T rr«i n- , sin yl sin B sin G I. The Sine-ruie, — »— ^ — = . 'a 6 c IL The Cosine-rule, cos A '" ^2bc "' in. The Tangent-rule, tan — - — = — , . cot ,r. ii a + 6 2 180. We may now proceed to explain the method of solving right-angled triangles. OF RIGHT-ANGLED TRIANGLES. 155 rl A, B, C the vhich present 3vvn ill order f these three We shall denote the right angle by C. Then we may have the following data : (1) Two sides and an angle, (2) Two angles and a side. 100. First, when two sides and an angle are given as b, c, C. 'I ;hird will be When two sides of a right-angled triangle are known, Wte can deterniine the third side ; thus, in this case, a= mJc^ — U^, and so a is determined. a Next, sin yl = , from which we can find A» ' c Lastly, B=dO° -A, and so B is determined. 191. Next, when two angles and a side are givcny ew c, A, C. First, - = sin A, from which we can find a. Next, - = cos A, from which wo can find h. Lastly, 7? = 00°-..'l, from which we can find 7?. 102. If A be one of the anglcM of a triangle, and the value of sin A be given, wo cannot determine the value of A with- out previously knowing whether A is an acute or an nhtiiRfi angle. 15^ ON THE SOLUTION Thus, suppose we know that sin^ = jr, one value of -4 which satisfies this equation is 30°, but another value of A wliich satisfies the equation is 150°, for since the sine of an angle ia equal to the aine of the suj)plenient of the angle, sin 150° = sin(180°-150°) = sin30°. In a ri.i^lit-angled triangle A, when not the right angle, muHt be acute, and so in the cases we have considered no ambiguity can occur. In triangles other than right-angled we shall find only one case in which we cannot determine with certainty the value of A from the known value of sin A. 193. We shall now give some Examples of the practical application of the methods of solution described in the pre- ceding Articles. To take the first of the two cases, suppose we have the following data : 6 = 5, (^ = 13, C=90°. Then and rt= ylc^-l>i=. VHi9-25= Vl4i = I2. sin yl=^-[| = -9230769. Now, from the tables, we find sin ()7°. 22' = -9229865, sin 67°. 23' = '9230984. And hence, by the method explained ;n Art 168, we find the value of >1 to be 67° . 22' . 48"-5. 194. Again, to take the second case, suppose we l)ave given * c = 25, A-=m% C=90°. Then (I . . - = sin -4 ; OF FIGHT-ANGLED TRIANGLES *57 le of ^ which of A which ■ an angle is Dnsidered no ■*• 25 ~ 2 ' " ~ 2 • — cos A ; 6 _1 . ,25 •*• 25~"2' •'^~ ^' Also, £=180°-(^ + (7)-180"'-150'' = 30'' ve have the Examples.— xlix. Solve the triangles referred to in the following example? by the use of natural sines, cosines, etc., C being a right angle. 1. Given 6-3, c=--5, sin 53°. r = -:998593, sin 53°. 8'= •8000338. 2. Given 6 = 15, c=17, sin 28\ 4' = -4704986, sin 28°. 5' =-4707553. 3. Given 6 = 21 , c = 29, sin 43" . 3G' - -6896195, sin 43°. 37' -6898302. 4. Given 6 = 7, c = 25, cos 73°. 44'- -2801083, cos 73°. 45' --2798290. 5. Given 6 = 33, c = 65, cos 59° . 29' = -5077890, cos 59° 30' --5075384. 6. Given c = 13, ^ - 07° . 22' . 48"-5, sin 07° . 22' - sin 67°. 23' = 7. Given c = 41,/l--=77M9'.10"'6, sin 77°. 10'- sin 77°. 20' = 8. Gi ven c - 73, 7i - 48° . 53' . 1 0"'6, cos 48° . 53' - cos 48". 54' -•657371.2. 9229805, 9230984. 9755985, 9750023. 0575944, 158 ON THE SOLUTION •H lo. Given c = 89, i? = G4° . 0' . 38"'8, cos 64° = -4383711, cos 64M' = •4381097. Given a = 40, vl = 77M9'. 10"-6, tan 77M9' = 4-4433769, tan77°.20' = 4-4494186. 195. The process of solution by means of natural sinos, cosines, etc., can only be applied witli advantage to cases in which the measures of the sides are small numbers. We proceed to show how the use of logarithmic calculations assists us in the solution of triangles. 196. It must be observed that a formula is adapted to logarithmic calculation only when it consists of the product or quotient ol' two or more numbers. For instance, we derive no advantage from logarithms in finding c from the e([uation c- = a'"^ + 6-, when a and ^ are given. But if a and c be given, we can apply logarithms with advantage to tind h from the ccpiatiou i'-s = c^ - (X^ ; for instance, if a = 644 and c = 725, s= (c 4 a) (c - a) ^1369x81; ... log &2 = log (1369x81 ); .-. 2 log 6 = log 1369 + log 81 = 3-1364034 + 1-9084850 = 5-0448884; .-. log 6 -2-5224442; .-. fc = 333. OF RIGHT-ANGLED TRIANGLES. 159 )'=4-4433769, y = 4-4494 186. c calculations Li'itlims with 197. Let a = 644, c = 725, 6'=9cn We first find 6 = 333, as exi3lained in the preceding Article. Then sin A = a :. lo" sin A = log a - lo'x c. Now L sin A is the true lof^ sin A increased by 10, Art. 170, hence we put liere and in all similar cases L am A — 10 in place of log sin A. Thus Zy sin yl-10 — loga-logc, L sin A ^10 + 2-8088859 - 2-8603380 = 9-9485479. Now, from the tables, L sin 62°. 39' = 9-9485180, L sin 62°. 40' = 9-9485842. Hence, by the method of Art. 168, we may find ^ =62°. 39'. 27" nearly ; and therefore B = 2T. 20' . 33". Examples.— 1. Solve the triangles referred to in the following Examples by Logarithmic calculations, C being a right angle : 1. Given a = l()4, c=185, log a = 20170333, log c = 2-2671 71 7, log 153 = 2-1846914, log 289 = 2-4608978, log 81 = 1 -9084850, L sin 34°. 12'= 9-7498007, L sin 34°. 13' = 9-7499866. 2. Given a = 304, c = 425, log a = 2-4828736, log c = 2-6283889, log 297 = 2-4727564, log 729 = 2-8627275, log 121 = 2-0827854, L Bin 45°. 40'= 9-8544799, L sin 46°. 4r = 98546033. i ^' If I. ■.r* 1 60 C>iV T//£ SOLUTION 3. Given a = 840, c = 841, log a = 2-9242793, log c = 2-9247960, log 41 = 1-6127839, log 1081 = 3-2255677, Xsin 87°. 12' = 9-9994812, L sin 87°. 13' = 9-9094874. 4. Given a-: 336, c = 625, log a = 2-5263393, log c = 2-7958800, log 527 = 2-7218106, log 961 =2-9827234, log 289 = 2-4608978, L sin-32° . 31' -9-7304148, L sin 32° . 32' = 9-7306129. 5. G«.v<M; .1 = 1100, c = 1109, log a = 3-0413927, logt -^-0449315, log 141 = 2-1492191, log 2209 = 3-3441957, log 3 = -4771213, L sin 82° . 41' -= 9-9964493, L sin 82° . 42' = 9-9964655. 6. Given 6 = 1 95, c = 773, log 7^ = 2-2900346, log c = 2-8881795, log 748 = 28739016, log 968 = 2-9858754, log 578 = 2-7619278, L cos 75° . 23' = 9-4020048, L cos 75° . 24' = 9-4015201. 7. Given /; = 273, c = 785, log t = 2-4361626, log (; = 2-8948607, log 736 = 2-8668778, log 1058 = 3-0244857, log 2 = '3010300, L cos 69° . 38' = 9-5416126, L cos 69° . 39' = 9-5412721. 8. Given 6 = 609, c = 641, log 6 = 2*7846173, log c = 2-8068580, ■ log 1250 = 30960100, log 2 = '3010300, L cos 18° . 10' = 0-0777938, L cos 18° . 1 1' = 9-9777523. 9. Given a = 276, 6 = 493, log a = 2-4409091, log 6 = 2-6928469, L tan 29° . 14' = 9 7479 1 25, L tan 29° .15' = 9-7482089 fi™'' OF RIGin^-ANGLED TRIAXGLES. i6i 10. Givon a = 396, 6 = 403, Iol,' a = 2-5076952, log 6 = 2-6053050, L tan 44° . 29' -9 9921670, L tan 44" . 30' = 9-9924197. 198. AVe shall now give a few Problems to illustrate the practical use of the methods of solution of triangles explained in this Chapter. Examples.— li. 1. Having measured a distance of 220 feet in a direct horizontal line from the hottom of a steeple, ,e -ingle uf ele- vation of its top was found to be 46° . 30'. liet^ui d the height of the steeple. Given log 220 = 2-3424227, L tan 46° .30' 10 0227500, log 2 31835 = -365172; 2. A river XC, whose hrea<lth is 200 feet, runs at the foot of a tower C'i^, Avhich subtends an angle BAQ of 25°. 10' at the edge of the bank. Uecjuired the height of the tower, given log 5 - 6989700, L tan 25° . 10' - 9 6719628, log 9397 = 3 9729928. 3. A perf=:on on the top of a tower, -whose hoicrht is 50 feet, observes the angles of depression of two objects on the horizontal plane, which are in the same straight line 'with the towel', to be 30° and 45°. Find their distances from each other, and from the observer. 4. At 140 feet from the base of a tower, and on a level with the base, the angle of elevation ol tiie top was found to be 54°. 27'. Find the height of the tower, having given tan 54°. 27' -1-399364. [■^.T.! ^ -■*» . -l' 5, A person observes the an,L(le of elevation of a hill to he 32°. 14', and on approachiiif^ 500 yards nearer, he observes it to be 63° . 26'. Find the height of the hill, luiving given tan 32° 14 = "63, tan 63° . 26'= 1-998. 6. A tower 150 feet high throws a shadow 75 feet hnrr upon the horizontal plane on which it stands. Find the sun's altitude, having given log 2 = "3010300, Ltan 63°. 26' = 10-3009994, L tan 63°. 27' =10-3013153. 7. A tower stands by a river. A person on the opposite bank finds its elevation to be 60° : he recedes 40 yards in a direct line from the tower, and then linds tlie elevation to l)e 50° Find the breadth of the river, having given tan 50° = 119. 8. A rope is fastened to the top of a building 60 feet high. The length of tlie rope is 109 feet. Find the angle at which it is inclined to the horizon. Given sin 33°. 23' =-5502, sin 33° . 24' = -55043. il 9. A tower is 140 feet in height. At what angle must a rope be inclined to the horizon, which reaches from the top of the tower to the ground, and is 221 feet in length 1 Given f^in 39° . 5' = -63045, sin 39° . 6' = '6306758. 10. A person standing at the edge of a river observes that the top of a tower on the edge of the opposite side subtends an angle of 55° with a line drawn from his eye j)arallel lo the horizon ; receding backwards 30 feet, he then linds it to sub- Find the breadth of the river. tend an angle of 48 Given L sin 7° = 9 08589, L sin 35° = 9-75859. L sin 43° = 9-87107, log 3= -47712, log 1 0493= -02089. OF RIGHT-ANGLED TRIANGLES. i6- w 75 feet Ions Find the sun's = 10-3013153. II. Standing straight in front of the corner of a house which is 150 feet long, I observe that the length suLtends an angle who.se cosine is --., and its height subtends an an^le 3 whose sine is -— , ; determine the height. v3-4 12. Standing straight in front of one corner of a honse, I find that iti^ length subtends an angle whose tangent is % while its height subtends an angle whose tangent is - : the height of the house is 45 feet, find its length. g GO feet hich. .,.* rt > i XVIII. ON THE SOLTJTTON OF TRIANGLES OTHER THAN RIGHT-ANGLED. ^aR ' I \ 1 : •! 109. Ix the solution of trianfjlca otlior than riglit-an^^led, nsiially called Oblique-anglcd Triangles, Ave meet Avitli i'our distinct cases, the i'ulluwing being the data. (1) The three sides, a, h, c. (2) Two angles and a side, as ^, C, &. (3) Two sides and the angle between them, as a, h, C. (4) Two sides and an angle opposite one of them, as a,h, A. These cases we shall discuss in order. Case I. 200. Given the three sides a, b, c. A ^ B "We first lind A from one of the formula) cos i4 = lf- + c^-a^ 2bc' A_ /(.-6)T(s-~c) ^'"' 2-^J "s.(s-a) • *** SOLUTION OF TRIANGLES. 16- ?^'2 + c2-a2. The formula cos A = — — is not adapted to logantliiuic calculation; but if a, h, c contain Ics.^ tlian tlircii di.Lrits, wo may use it to find A by aid of the tuuie of liUtural cosines. Thus, if a^-G, 6 = 5, c=10, , 25 + 100-36 ^^ And hence we find A = 27\7'. 36". 201 When o, h, c, contain tliree or more digits, we may employ with advantage the formula tan^-^- /(iL^)(^-^) from which we have L tan -^- - 10 = J log (s - 6) + log (s - r) - log s - log (s - a) | , and then we may find tan ' or tan ' bij means of the same lofjarithms, thus L tan -^- - 10 = - 1 log {s-a)^- log (s - 0) - log s - log (s - 6) }. For example, let a = 9459-31, 6 = 8032-29, c = 8242-58. Then s= 12867-09, s-a = 3407-78, s- 6 = 8434-80, .s- = 4624-51. Zt:'i--10 = -|l g(s-6) + log(.s-c)-logs-log(s-a)| = \ \ 3-6843785 + 3-6650657-4-1094604 •■ 3-532471 6 { A = - -1462539 ; /. L tan -^ = 9-8537461. 166 SOLUTION OF TRIANGLES. Whence 4 = 35° . 31' . 47"-5, A and yl=7r.3'.35". Similarly we may find i? = 53° . 26', and 0=55" . 30'. 25". 202. When A has been found, we can find B i'lom the relation sin B _b 8inyl~</' and then C may be found from the lelation C=180°-(^+i?). 203. If we are required to find only one angle, as y4, we may take the formula ^^^4=V' (s-6)(s-c) 6c where Taking the example given in Art. 200, we have 11 1 .A 7(2 "" 2 ) /ll '"' 2 V r5~rroi = V2oo^ /. isin 2 - 10 = ^{log 11 -log 200} = i j 1-0413927 -2-3010300{; i ;./> sin 4 = 10 --0208180 = !l"S70r->l4. And hence we find from the tables -- = 13". 30'. 48* and thus we know that the value of A is 27°. 7'. 36". SOLUTION OF TRIANGLES. 167 °.30'.25". 1 B iroin the gle, as A^ we Oabe II. 204. Given two angles and a side, A, C, b. First, B^^IQO" -{A + G), from which we can find B. Next, -,='-. — ^r,, from which we can find a. ' b sinJ5 Lastly, r= • — T« from which we can find c. •^' sin 5' 205, Here we liavc. no <lifficulty, and we shall merely give an example to illustrate the method of finding a from the formula a sin A „ when b, A, B are known. Let 6 sin B 6 = 40, ^1-12°. 40', L' = 77MO'. Then log a = log 6 + L ^mA-L sin B = 1 -0020600 + 9 34099G3 - 9 9890137 e= 9540426, whence a=9 nearly. ){; 50'. 48", and ■it- 200. It is in practice an easier method to write the equa- tion thus, a = /; . sin A. . cosec i?, so as to save a subtraction of decimals. Thus, taking the same Exiinple, we have log a ^ log h + L sin A + L co.sec B - 20 = I-602()G0() 4 9 :MOn903 + 10'0100S03 - 20 -•9510426, ' i68 SOLUTION OF TRIANGLES. Cass III. 207. <7u'C7?. tv:o sides and the angle between them, a, b, C. We may find c from the formula c^ = a^ + ^2 _ 2a6 . cos G. Then from ' . -,v= - we can find A. sin C c Lastly, from B= ISO" - {A + C;, we can find B. Or we may pi-oeeed to find A and B before we find c, thns : by the formula established in Art. 180, A-Ji a-h a tan — . ---^ -. 7 • eot ;,, and from this we can find a + 6 A-B 2". Then, as we know that — - — — 00°- „, we .shall have two o(|uations by which we may d.'tcnuine A and B. Then we can find c from the ei|uution c sin {* 208. The first formida Lnvon, c^ = «- -f //- - 2a7) . cos C/, is not adapted to logarithuiic calculation. We must take then, in all cases where a and 6 are not HiiKill integers, the Ibrmula , \-B a-h . (; tmi -.... — =s — .cut ., fur rin<liti«r llio vulm s "f I ;iinl /,*. SOLUTION OF TRIANGLES. t6() Suppose then we li.ive given a=12y(j, 6 = 9-78, 6^=57°. 48'. 32", we proceed thus : (J a-6-3-18, a-r5 = 22-74, " = n\54MG". Then A — U p L tan - — ' - 10 = log {a-h)- loj^ (a + h) + L cot ^^ - 10, L tan -— -- = log 3-18 - log 2274 + L cot 28' . 54' . IG" = •502-1271 - I-35G7U05+ 1()-257D57I) = 9-4035945. Whence'^ :,'^=14M2'.40". 2 A + B 2 Also "-^r^- = 6r . 5'.44" (the complement of ^3) ; .-. yl = 75M8'.3()", J5^4G\52'.5.s". The otiier formulas of this case rcciuire no special remark. 209. Though the formula c-=^-n--\-lr~2'ih.co?iO ia not Hi.'ipte 1 tn logaritlimic calculation, \\\; cm luul c IVom it (in my ca.su where we do not recpiire the values of A and U also) by the following ^irocess : c- = (i''' + 6--2a6.co8 6* .a'^ + i--2r(/>.(l-2sin-'?) ^a^ + l -2((/> \-~U(h. sin' '/' ■1 \^o SOLUTIOy OF TRIANGLES. - Now, since the tangent of an angle ii'ny be oC any mo^^ni- tude, there is some angle (suppose 6) siic^ that (a - by -A Knowing a, &, Q we can apply logarithms to find from this equation. Then, from the equation c'-J = (a-&)'^jl + tan2 6'}, that is, c^ = (a - 6)2 . sec- 1/, we can apply logarithms to find c. An angle introduced, as in tliis case, to assist the solution of an equation, by breaking it up into two or more equations, is called a Subsidicmj Angle. Case IV. 210. Given two sides and an angle opjjosite one of them, a, b, A. -.- . = , ff-HBo which we have to dcterimnc B. 8111 A a If we can liiid B, we hnv« 0= 180" - (A + 7J), r which we can find C, , c fiin G .. , . , p , and -= . ., Irom which we cnn niMl r. a .11 A SOLUTION OF TRIANGLES. tyi ot any ma^oi- find 6 from st the solution Loic ec^uatioiis, one of themy \. B liu B. (liiJ C, 211. Th^is liie >-;olution of this case depends on the possi- !>ility of determining B from the equation t^J. (1). 8in A a ^ ' Now since we know a, &, ^, we obtain from this equation sin 5 = a known numerical quantity (2). But, as has been explained in Art. 74, we cannot deter- mine the value of B from a given value of sin B, unless we know whether B is greater or less tlian 90". The only way in which we can tell whether the greater or tlie smaller value of B which satisfies equation (2) is to be taken, is by knowing that a is (jreater than b. In that case A is also greater than B, and therefore B murt be less than 90°, otherwise A + B would not be less than 180°, which is impos- sible. 212. This, which is called 71ie Ainhiguous Case, may be ahown geometrically in the following manner. If from G we cnn draw a line CD equal to CB, to meet AB produced on the side of jB, both the triangles ABC, ABC have the given parts a,b, A. "Wq can always draw CD = CB, so long as AG is rjrcafer than BC, for then .i circle described with centre C and radius CB will cut AB ])roduced in two points both on the same side nf .1. r % 172 SOLUTIO^r OF TRIANGLES, m H 213, The following is u inuru complete discussion of TKq Amhigiioiis Case, j^ h.^mA , . , sin L= — a known iniinorical quantity. Now, provided 6 sin A be not > a, ^sin A . , , ~ IS not > 1, a nnd this nnnierical quantity is a possible value to be tiie sine of an an-'le. I. If 6 sin A < a, < 1, and there are two ancilcs AVe have now two cases. h si n A a which have this value for their sine, sujipknic-ntary to one another, and therefore one acute and one obtuse. Now when a is > h, A is > Ij, and therefore we cannot take I his obtuse value foi' /?, for then A and B would be toij;ether > two right angU's. Jbit when a is < b, A is < J], and we can take both tlie ncute and the obtuse vahu^ for 7>, anil then we shall have two corresponding values for (/, and two tbr r, and thus we gut two triangles having the g;''.'n parts the same. IT. [f 6 sin ^1^ a., sin 7'— 1, and 1> has only one value, viz. 00°. Of course, if a — /), B^A at once, without using the equation /' in A sin :'? at al a This ambiguity can be exhibiteil getinietrically as follows : Fi.:;. 2. C Fig 1. /' /: z / Bft A D X A Bj U B. X Let '.LV-yl, A0^\ jcussion of Tke Lmtity, to be Liie siue ire two angles luiiLiuy to one ore we cannot I /> would be tiike both the hall have two d thus we get ily one value, ^ the ctj^uation )' UH Ibllowe : B. X SOLUTION OF TRIANGLES. 173 Draw CD perpendicular to AX, then CD = b sin A. With centre C and radius e(|ual to a describe a circle. Now provided CD be not > a, this circle will meet AX. I. If CD < a, this circle will meet AX in two points, L\, and B.2. Now when a is > h, B^ and B.^ will fall on the opposite sid-'s of A as in fig. 1, and we have only one trian<.,de, viz. CABi, having the given angle A, and the sides CA, CB^, ei^u.-d to h and a. Bat when a is < h, B^ and B., will fall on the same side of A as in fig. 2, and we have two triangles, viz. CAB^, CAB., having the given angle A and the si(U'S 6'/>i and CB., ecjual to a, and the side AC equal to h, and in this case CJlyi is supplementary to CB.^X and therefore to its ec^ual CB^A. II. If CD = a, the circle will meet AX in one point only, viz. at Z), and CAD will be the required triangle. Of course, if h = a, B.^ will coincide with ^1 and we have only one triangle CAB^. 214. The formula) to be used in this case are simple, mid we have only to give instances of cases (I) ambiguous, (2) in which no ambiguity exists. (i) To tuke a very simple case, suppose tt = 5, 6 = 0, A = '3L)\ sin B b Then 6 a . . sm /> = - . sm .4 = . X -, = ,: = '6. sin A a ' Now from the tables we lind -0 to be the value of the sine of 3(r.r)2M2", and sine 14:r.r.48" is the sn])p1ement o' 36° . 52' . 1 2", it also has •() for the value of its ttine. Thus there is an ambiguity in the result. ■■'-■% HHI^^ Next, suppose a = 178-3, 6=145, B=4l°.l(y, a Then sin A= , sin B : .*. L&in A— log a - log h + L sin 5 = 2251 1513 - 2-1613680 + 9-8183919 = 9-9()81752. Hence A = 5^.2'. 22" or 125'* . 57' . 38". (2) Now il" we change the values of the sides a and h we shall get L sin ^=9-7286086 ; .-. yi=32°.21'.54", and the supplement of A cannot belong to the proposed tri- angle, because if A were 147° . 38'. 6", then, since B is greater than A, A and B would be together greater than 180°, which is impossible. So in this case there is no ambiguity. 215. The following are applications of the principles laid down in this chapter. EXAMPLES.— lii. I, Find A from the following data : (1) Given « = 37, 6=13, c = 40, sin 67°. 22' = -9229865, sin 67°. 23' = -9230984. (2) Given a = 101, 6 = 29, c = 120, sin 43° . 36'= -6890195, sin 43°. 37' = -6898302. (3) Given a-37, 6 = 13, c = 30, log 9 = -9542425, log 13 = 1-1 139434, L sin 56° . 18' = 99200994, L sin 56° . 19' = 99201836. (4) Given a = 409, 6 = 241, c = 600, log 723 = 2-8591383, log 360 = 2-556302"). I sin 29°.5T:-969nn947, 7v sip ?9' . 52' = 9-6972148. SOLUTION OF TRIANGLES. 175 y. )19 Jes a and h we le proposed tri- ice B is greater lan 180°, whicli i^rinciples laid '=•9220865, •9230984. )'= '6890195, '' = •689830:^. 2425, f)434, •9201836. -2-8591383, = 2-r)5(;3()2r). 0972148. 2. If a=5780, c=7639, i? = 43\ 8', lind A and C, having given log 185-9 = 2-2()928, log 13-419 = M2772, L cot 21" . 34'= 10-40312, L tan 19° . 18' . 50" =.9-54408. 3. It'^ = 41M3'.22", i] = 71M9'.5", a = 55, find 6, liav- iug given log 55 = 1-7403627, L sin 5 = 9'970i927, X sin ^ = 9-8188779, log 79-003 = 1-8979775. 4. If 5 = 84°. 47'. 38", C=4r. 10', c=145, find h, having given • log 145 = 21613080, L sin 41M0' = 9-8183919. L sin 84°. 47'. 38" = 9-9982()47, lug 219-37 = 2-3411808. 5. If a = 567-2341, 6 = 351-9872, i? = 31°. 27'. 18", fmd ^ having given log a = 2-7537623, log h = 2-5405209, iy sin i? = 9-7 175280, L sin 57°. 14' = 9-9247349, Lsin57°.15' = 9-92481G1. 6. "When (7=30°, 6=16, c = 8, is the triangle ambiguous or not % I'i + c2 - (t- 7. Simplify the expression cos A = ;f, in the case of an equilateral triangle. 26c 8. Given log 3 = -4771213 and L tan 57°. 19'. 11"= 10-1928032, show that, if one angle l)e 00°, and the two sides containing it as 19 to 1, the other two angles are 117° . 10'. 11", and 2°. 40'. 49". 9. The sides of a triangle are 2, »JG, and 1 + v'-^ : find the angles. 10. If a, 6, B had been given to solve a triangle, where b 13 less than a, and if Ci, c.^ be the two values found I'ur deter- mining the third side, prove ihnt h--hc^ . c.^ — a-. 'i. "'1 ■ * ay ij6 SOLUTION OF TRIANGLES. 11. Two sides of a triaiij^'lu are to each otlior as 9 : 7, aiidtlie included aiip;le is G4°. 12'; detoriuiiie the other angles. Given log 2 = -30103, L tan 57° . 51' = 10-2025255, Xtan 11° . 1G'==:9-299321G ; L tan IT. 17' = 9-2999SOi. 12. The sides- a, ?>, c of a triangle are as the numhers 4, 5, G. Find the angle R Given h)g 2 = •3010299, L cos 27° . 53' = 9-9 JG4040, log 5 = -G989700, L cos 27° . 54' = 9-94G3371. 13. If in a triangle ABC, i?C'= 70, ^1(7=35, and lAGB = 30°. 52'. 12", lind tlie remaining angles. Given log 3 = -4771213 and L cot 1S° . 2G'. G" = 10-4771213. 210. Up to this point we have snpplied the student with all the materials recpiired for the solution of each exam])le. Bnt as he ought to have some practice in making extracts from the tables, we shall suppose him to he in i)ossession rtf a set of tal>l(!>!, and we shall now give a series of examples hy which he may test his ability to apply the I'urmuia) i'ur the solution of Triangles. Examples.— liii. Solve the triangles for which the following parts ni-e given. 1. a = 4, 6 = 3, C'=90°. 2. & = 55, c = 73, (7=90". 3. « = 272, /; = 225, 6'- 90°. 4. 6 = 399, c = 401, 6'= 90°. 5. c=445, yi = 10°.52'.50"-4, ^=90". 6. c=G29, .4=4G°.59'.49"-7, ^'-90". 7. c = 449, 7; = 51°. 25' . 1 1"-7, -C - 90. 8. c=349, 7?-5S°.57'.G"-4, 6'= 90°. 9. a = 520, ^=fiG°.2'.52", C'=90°. 10. 6-31, .'1=S(r. 1H'.17", C'=90°. s: other as 9 : 7, 'lie other angles. = 10-2025255, = 9-2999804. as the numbers = 9-9164040, = 9-94G3271. 35, and lAGB '-10 4771213. le student with each example, laking extracts [)ossessi()n of a 'f examples by iinuiaa lor the .rts are ,<iiveru 0". !). OLCTION OF TRIANGLES. 177 Solve the triand parts are jriven : Examples.— liv, t's, not ri^dit-angled, lur which the foil OWU)o 1. a=197, 6 = 53, 2. a = 509, Z; = 221, c = 240. c' = 480. 3. a = 533, /y = 317. c- = 5 /, c' = 5I0. 4- " = 5G5, 6n=445 J, (-^noG. 5. « = 409, 6-241, r-.182. 6. 6 = 29, A = 43°.3(r. J(>"i, C-i2r 58' . 33"-R 6=149, yI=G9'\59'.2"-5 r,'=- '0\ 42". 30" 8. a =101, 6 = 29, 6'=32Mt)'. 53"-8 9. a = 401, 6 = 41, 6'=9( J . 57'.20"-l ro. « = 22l, 6=14.9, C=30°.J 0' . 35". II. «=10.0, 6 = fjl, 6'=(Jfr.59',25"-4. 12. « = 445, 6 = 83, 0=87 0/ . 00 t3- ^/ = 229, 6=109, C=]3r.24' 44" 14- ^« = 241, 6=169, (7=l()4\3'.5r'. 15- '« = 241, 6=169, r;=15°.22'.37". 16, a = 13, 6 = 37, ^ = 18\55'.28"7, find 7?. 17 a = 445, 6 = 56/ /I=44\29'.53", tindi; 18. a = 212-5, 6 = 836-4, .4 = 1 4°. 24'. 25", find i5. 19- « = 3: d-r. ), b. '«4-8, /1=4()\32M6". hnd B. •o. rt = 9459-31, 6 = M)32-29. .■l-.7r S.T, J .'U"-7, liud B. Af ^, <". .n^ "vv^ y^.w^. >, IMAGE EVALUATION TEST TARGET (MT-3) k A ^/ %.*5^ ..V < ^' .^ :/. J "^ "^ 1.0 I.I 1^128 |2.5 ui Hi 2.0 IK L25 i 1.4 II IIIIIJ£ 1.6 - 6'' V] <^ /2 ^}. ? r ^-5. 4V^ 7 Photographic Sciences Corporation <:1>^ <^ 23 WIST MAIN STRUT WIBSTU, NY. I4SS0 ( 7 1«) ■73-4303 '^ lA ^ f' A I: ^'r 19^^ - « '" • ' M^- m I XIX. MEASUREMENT OF HEIGHTS AND DISTANCES. 217. In this cl)a])ter we shall give examples of the applica- tion of Trigonometry in determining heights and distances. The problems which occur most frequently in practice, in addition to those given in Chap. VIII. , are the following : (1) To find the heighi of an object standing on a horizontal planet when the baso of the object is inaccessible. A B P Let PQ be a tower, of which the base P is inaccessible. Measure a distance AB in the same horizontal plane with /\ Observe the angles of elevation QBP and QAP. Then we can di'termine tlie height of the tower, for QP^QB alu QBP, and ^ni BQA AB.- »mQAP :. QP^^AB.BhiQBP. m\{k>BP-QAP)' Pin CMP Bhi{QBP - QAPy HEIGHTS AND DISTANCES. 179 1 practice, in (2) To find the heiyht of an ohject irhose foot is inaccesmhle, when a direct line between the observer and the base cannot be measured. Suppose PQ to ue a tower standiiij^' on tlie bank of a river, and B to be a point on the opposite bank. Supjtose tlie j^round to rise suddenly i'roni li, so that no distance in a direct line with JJP can be measured. Measure a line AB up the risin*,' <,M()und. Observe the angles QAB and QBA, Then in the triangle QAB two angles and the side AB are known, and therefore we can lind QB. Then if we observe the {ingle (JBP we may determine QF. (3) To find the distance beticecn tuv inaccessible ohjecls. C D Let A and B be the objects. Measure a line C/>>, and snjiposc A, II (\ D to be in one plane. Then if we observe the angles ACD and ADC we crn determine AC, because we know two angles and a side; in tic tri:ingleylC7). gll^ n-^i- ^Hffir { ■! ^iEM '*' ~ • 1 'Iv^, wi *' m ' ' 'il ' '™- t «f ■, ■ " li' t ■: .■.■il!*'i.)' t H ': ' ■ii^ ;. ■♦- i -■ i-^ '. - ■ - ^M I, y » V* IM': 1 80 h EIGHTS AND DISTANCES, A^'fiin, if we ol)serve tlie angles BCD, UDC we can deter- luiiie I>'C', l)ecau>;e we know two angles and a side in the tri- angle BCD. Thus knowing AC and DC and the included angle ACIi (which is the diflerence between the known angles ACD, BCD), ■we can determine AB. (4) 1 Jind the distance of a ski}) from the shore. A B Let S be the position of the ship. Measure AB, a straight line between two points on the shore. Observe the angles SAB and SB A. Then we can determine the distance of S from /i, for AS = AB. --— v.r,. '^ i:ur{VH{f~-ASB) . -r. sin >S7>/1 "" ^nxXSAB + SBA)' 218. In the first twenty-one of the examples that we sliall now give the results may be obtained without the aid of 'I'ablet*. HE re [ITS A,\D DISTANCES. iSi l1 aiiple ACn iACV,BCV), 01) the shoro. Examples.— iv. 1. Wishing to know the hei^'ht of an inacccssil.le hill I took the an;j;'K; of elevation of its toj) to he GO', I tin n measured 100 feet away from the hill uii«l found the an^'le of elevation to be 45^ What is the height of the hill { 2. Each of two ships, which are a mile apart, iinds the iingles subtended by the other ship and a fort to be respectively 35'. 14' and 42°. 12'. Find the distance of each from the fort. (liven sin35M4' = -577, sin 42M2' = -G71, sin 77\ 2G' = -!)7G. 3. Each of two ships, half a mile apart, linds the anf,de3 subtended by the other ship and a fort to be respectively 85° . 15' and 8',V . 45'. Find the distance of each from the fort. (Jiven sin 85". 15' = -91)65, sin 83°. 45'= -1)040, sin IT = -1908. 4. Find the angle which a flag- staff 5 yards long and standing on the top of a tower two hundred yards high sultteiids at a point in the horizontal plane lUO yards from the base of the tower. Given L tan 34' = 7-9952192, L tan 33' = 79822534, log 102 = 2 -0086002. 5. The angle of elevation of the top of a steeple is GO" from a point on the ground. That (jf the top of the tower on which the steeple rests is 45° from the same point. What [iroportion does the height of the steeple bear to that of the tower ? 6. On the bank of a river there is a column 200 feet high supporting a statue 30 feet high. The statue to an observer on the opposite bank subtends the same angle as a man G leet high standing at the base of the column. Find the bieadtli of the river. 7. A pole is fiKed on the top of a moujid and the angles of elevation of the top and bottom of the [tole are GO' and 30°, show that the lengt/i of the i>ole is twice the height of the mound. l82 riF.rcnrs and distances. i# I. 8. A porsDii at a distance a from a tower wliicli stands ou a liorizoiital jdaiu', oUsiirvcs tliat the aii^do ot" elevation a of itt^ highest point is tlic coiu|)lenient of tliat of a llaj;-statf ou the top of it. Show that the length of the llag-stalf is 2a. cot 2a. 9. If the distance of the person from the tower is nn- known, and if, when he lecedi a distance c, the angle of ele- Tation of the tower is half of what it was before, show that the length of the llag-stalf is c . cosec a . cos 2a. 10. Two spectators at two given stations observe at Ihe same time the altitude of a kite, and find it to subtend the same angle a at each place. The angle which the line join- ing one statit)n and the kite subtends at the other station is /i, and the distance between the two stations is a: liud the height of the kite. I r. Two towers stand on a horizontal jdane, and their dis- tance from each other is 120 feet. A jtcrson standing succes- sively at their bases observes that the angular elevation of one is double that of the other; but when he is half-way between them their elevations apjiear comi)lementary to each other. Show that the heights of the towers are 90 and 40 feet respec- tively. 12. /i, 5 are two inaccessible points in a horizontal jdane, and C, D are two stations, at each of which All is observed to subtend the angle :30°. AD subtends at C 19M5', and AG subtends at D 40° . 45'. Show that AB=-^j^., 13. The length of a road in which the ascent is 1 foot in 5, from the foot of a hill to the top is I5 miles. What will be the length of a zigzag road in which the ascent is one foo*; in 12 f 14. Two objects, A and B, were observed to be at the same instant in a line inclined at an angle 15° to the east of a shij)'a course, which was at the time due rrorth. The ship',5 course was then altered, and after sailing 5 miles in a N.W. direction, the same objects were observed to bear E. and N.E. respectively. Rerpiiretl the distance of A from B. V* n EIGHTS AAr/) DISTANCES. i«3 lul their dis- 15. Tlu! elevation of a tower at a ])l;i('e A due south of it is 30"; and at a ])lace /i, due west of A, and at a distance a from it, the elevation is 18' ; show tliat the height ol the Lower is re 16. A circular rini,' is placed in a vertical plane thrmij^h the Kun's centre, on the top of a vertical stalT whose iiei;^dit is eij^ht times its radius; and the extremity of the shadow of tlie ring is observed to be at a distance from the foot of the stall eijual to the stall's height. Determine the altitude of the sua. 17. The hypotenuse c of a rii^ht-angled trian;^de ABC ia trisected in the points 7), K\ prove that if CD, CE be joined, the sum of the scjuares of the sides of the triangle ODE 2 2 18. A perscm stands in the diagonal produced of the square base of St. Mary's Church tower, at a distance a from it, and observes the angles of elevation of the two outer corners of the top of the tower to be each 30°, and of the other 45'* Show that the breadth of the tower is a s/{'S± ^b). to. A towei* standing on a horizontal plane ia surrounded by a moat which is just as wi(le as the tower is high. A person on the top of another tower whose height is a and whose distance from the moat is c, observes that the first tower subtends an angle of 45°. Show that the height of the first tower is a-c' 20. A and B are two points 100 feet apart, and G is a point equally distant from A and B ; what must be the distance of G from A and B tliat the angle ACB may be 150' ? 2t. A headland G bore dtie north of a ship at A : and after the ship had sailed 10 miles due east iu H the headhind bore N.W. Required the distance of the headhind from A and B. 184 HEIGHTS AND DISTANCES. 22. The aspect ot" a uall 18 feet lii-^^h is dne poiitli, and the len.ath of tlie sliadow cast on the norih bide at noon is 16 feet. Find the huu'« ollitude. 23. At a distance of 200 yartls from the foot of a church tower, tlie an,iL,de of elevation of the to]) of the toNver was ob- served to he 30', and of the top of the spire of the tower 32°. Find the height of the tower and of the spire. 24. The distances of three objects, vl7>^, in the same liorizontal ])hine, are ylZ> = 3 niik's, /J6*=r.S niiie, ylC'=2 n ih's; from a station I) in iJA i)rodiice(l the angle ADB= 17° . 47' . 20" is observed : lind the distance of I) from B. 25. In an oblitpio triangle A,B,C, given IACB= 139° . 58', i ABC =22". IS', i>'6'= 840-5 yards, lind by how much AB dilfcrs from a mile. '.■*. 26. In an oblique-angled triangle ABC, given ylZ> = 2700 ft., iA = 50° . 20', and ^ L' - 1 10° . l:i', lind BC. To detei-mine the height of the top G of h mountain, a basj AB of 2700 feet was measured in the liorizontal jilane, tlie angle subtended by Ci> at yl. was observed to be 50°. 20', the angle subtended by ylC'at B was observed to be 110°. 12', and the angle of elevation of C from B was observed to be 10°.7' ; find the height of the mountain. 27. A llag-staff 20 feet high stands on a wall 40 feet high. At a point E on a level with the bottom of the wall the iiag- stalf subtends an angle of 10°. Find the distance of E from the wall. 28. From the top of a hill the angles of de])ression of two consecutive mile-stones on a straight level road are found to be 12° . 13' and 2° . 45'. Find the'height of the hill. 29. From the top of a tower by the sea-side 150 ft. high, it was found that the angle of de])ression of a ship's Inill was 36°. 18'. Find the distance of the ship from the foot of the tower. IJElCr/TS AXD DfSTANCES. 185 7 iimch AB 30. Given a = 0:3S3'53, 6 = 3157-76 and C'=37\2(;', lind the other parts of the triangle. 31. From each of two ships a niilo apart the ani:;le wliich is subtemk'd by tlie other sliip and a hcacou on shure is observed : these an,i4K's are 55" and 02' . 30'. Determine the distances of the ships from the 'jeacon. 32. From tine lov.'cr window of a house the an^le of ele- vation of a chnrcli tower is ol)served to he 45'', and from a window 20 feet above the Ibrnier 4l)\ How far is the- hoiiso Irom the church I 33. A line AB in lenc^th 400 yards is measured close by the side of a riviT, and a point C close; to tlie bank en liie other ?ide is observed from A and 11. Thr aiii;ie CAB is 50", and CBA C5°, find tin; periiendicuUu luvadlh of the river. 34. Two railways intersect at an angle of 35*. 20': from the point of intersi'ctioii two trains start toL,a'ther, one at the rate of 30 miles an hour : find the rate of the other train, so t1p.it after 2^ hours the trains may be, 50 Uiiles apart. Sliow that there are two velocities that will satisfy this condition, and calculate approximately either of them. 35. A base line of 600 yards was measured in a straight line close to the bank of a river, and at each end of the line the angles were observed between the other end and a tree close to the edge of the river on the opjiosile side (»f it : these angles were found to be 52" . 14' and 08" . 32'. Find the breadth of the river. 36. The angle of elevation of a tower 100 feet high, due north of an obseiver, was 50°; what will be its angle of eleva- tion after the ob.erver has walked due east 300 feet ] 37. A flag-staff, 12 feet high, on the top of a tower, sub- tends an angle of 48'. 20" to an observer at tin; distance of 100 yards from the foot of the tower: retj[uired thu height of the tower. l^*'^}^ c86 HEIGHTS AND DISTANCES. 38. At tlie foot of a hill a visible objort has an elevation of 29". 12'. 40", and when the observer has walked 300 yards U]) the hill away from the object, he finds himself on a level with it. The slope of the hill being 10°, and the ])laces (»f observation in a vertical plane with the object, find the dis- tance of the object from the first place of observation. \i% 39. yl7j, AG are two railroads inclined at an angle of 50° . 20'; a locomotive engine starts from A along AB at the rate of 30 miles an hour : after an interval of one hour, another locomotive engine starts from A along AG at the rate of 45 miles an hour: find the distance of the engines from each other, three hours after the first started. 40. A church tower stands on the bank of a river M'hich is 150 feet wide, and on the top of the tower is a spire 30 feet high. To an observer on the oi>posite bank of the river the spire subtends the same angle that a pole six feet high sub' tends jdaced u])right from the ground at the base of the tower. Show that the approximate height of the tower is 285 feet. las an elevation liked :3()0 yards iiself on a level ul the i)laces of ct, find the dia- vation. at an angle of long A]j Jit the le liour, another the rate of 45 jines from each I river Mliicli U a Bpire 30 feet f the river the feet high sub- se of the tower, is 285 feet. XX. PROPOSITIONS RELATING TO THE AREAS OF TRIANGLES, POLYGONS, AND CIRCLES. 219. Expressions for the area of a triangU, Tiie area of a triangle is equal to half the rectangle con- t.imed by one of the sides and the perpendicular drawn to meet that side from the opposite angle. Let ABG be the triangle, and as one of the angles A, B must be acute, let it be A. Draw a perpendicular from 6 to meet ^i^ or ^7? produced in D. Then, area of triangle Anc= l.AB.CD 1 . ^^.^C.sin^ g r6 . sin ^ J >< ». iS8 ARF.AS or 1 RIANCLES, tljiit is, the nrcji of a trian!;'!*' is ('(lual to half the proiluct of two sides and the sine of the aiiylc between them. 2 Also, since t^hi A = . ^h . {s -a) {s -b) {s ~ <•), hy Art. IS I, I 2 area oi trijin-K' AnC= cb.,. ^h . {s-a) {s-b) {s-c) = \^s (."? - a) {8 - b) {8 - c), whicli j,'ivc.s an oxi)r('asi()n for the area in tornia of the .-ides. For this exprest^ion the symbol used is S. 220. 7o find the area of a nyular ixdijijoa in terms of it.^ sidc^ ■It t ■ ,4. ■ •' ,.f.,'. ,r.. I' I"" 'il (■ i Lot EA, ATi, BF ha three consecutive sides of a rofrular polygon of n sidis, and let each of them =a. Bisect the angles EAB, ABF by the lines OA^ OB meeting in 0. Draw OR at right angles to AB. Now and 1 Ann ^^^^ nn-de AOR= ~~ , angle yl 07? = "'- (ImicI. i. 15, Cor.); .. ande AOR n the product of two POLYGOXS /I, YD CIRCLES. I So Ilenco area of polygon = 7i times urea of trian,L;le AUB ^ =n. AH . AR.i:ui AUU na- ^ IT = .' . e.ot . 4 n 221. To fivil the rmlius of a circle described ahout a triamjU in leniis of the sides of the triangle. in terms of its 8 of a regular 1, OB lueetiii!' D--- Lot ho the centre of tlie circle depcrihed ahout the triangle /(//C, and A' its radius. Tlirougli U draw the diameter CI) and join ]JD. Then CUD, being the angle in a semicircle, is a right angle. And i>/)C== angle CAB in the same segment =^. Now thul is, ct -,. = .sin A 2U .'. a = 211 . sin A j 2sin^' o .-. li = But, by Art. 184, sin vl = " . ;S ; DC 11' ■■,l'. lii 1^^ 190 AKEAS OF TRIANGLES, A'^o/e. Sinco -y, = 8iii yl, and similarly - ., = Bin /?> and -y, = 8iii (7, wc derive another jtroof of the Theorem sin A _8in /?_Rin (/ a ~ h ~ c ' 222. To f\)nl the r(((h'iiii of a circle imcribcil in a triangle in (trms of the sides of a iriamjle. Let l)e the centre of the inscrihi'd eircl(>, nnd r its rndiua. Then-S = areaof /l/;a rsftrea of 7j()(7 + arca of COA +area ot AOB 01]_.]}C ^^Ji'^.OJ.AU ra rh re m >r . n 4- ?) + •rs ; ■••'•-.• L'orem / in a triangle in Tid r its nidiua. AOB POL VGOXS AND CIRCLES. 191 223. To find (he radii of (he circle, cscrihcd, tluit is vMd. touch one of the radcs of a triamjle and the other side, produced. Let 1,0 11,0 cc-ntro of tho ..scrii.,.! circle that toucLos ti.e ^Klr 7iG ami the other sides pioduco.!, ami let the radiua of this circle be fj. Then qnndrilaleral /I /i6'C= triangle /IOC+ triangle yl 05 und quadrilateral yJ7jV;C-lrinii-](! ylLV+ triangle BOO :. Aoc+A on = A no + una, . AG. OK An . 0F_ lie. 01) 2 "^ 2 ~ ^ 2 ""' 6ri cr^ „ . ar 2 2 6 + c-a •• 2 +~2'~^'*'2 * 2 .r, = S; .'.(a- a)rj=;S; • • r 1 ^^ • *:JI' S ;l^: ■ 'VI i' '. .1 V i I in: AREAS OF TRIANGLES, Similjuly it niny 1)0 sliown (li.-it if r.„ 7*3 are the radii of the circles touching ylCiind Ah rcspeclivcly, »'9 = ,9 f^' 8_ s - c i 224. To fi.)\<l the arm of a rnjiihtr poJijgnn inscribed in a circle. Let he the centre of (he circle, r the rndius of the ciflIo /t // 11 Ride of the polygon. Join UA, on. Tlien area of polygon = 11 tinioa nrea of triangle AOB i '1^' = n .^^AO . OB . m\ AOn (AH. 21<X 1 . Stt n.-^.r .r . sin - 2 n t?r2 . Stt I — . Rin . POL YCOXS AND CIRCLES. ! the radii of the M inscribed in a 8 of the ciri-Ie cAOn y(Ait. 219} 193 ^^^225. To Jind the area of a regular polygon desaihcd about a Let be the centre of the circh^, r the radius, AB a «i.le of the circnmscrihiiicr poly^'on. Draw the radius cyrJ at ri<rhi, ,,,,,^x^^ ^^ ^^^^ A R B Theu area of poly^^^m^n times area of trinjigle AOB ^n .OR, All '=^11. OR, OR tan AUR *=« r.r. tan - n *=nr- . tan-. n 220. T'o find the area of a circle. T'ddnjjr tlie fi«,nire and notation of the precedini^ Article, id of circumscribing polygon = >i times area <d triangle AOIi ^n.~.An.OR ^loh'-sU.AB ■"^0/,^ X ])erinieter of i)o]ygon. Kow if the nnmher of Rides of the polygon be indefinitely iiicrrasod and tlie lengtli ol rach side indefinitely diminished, urea 194 AREAS OF TRIANGLES, ■ .< s; the perimeter of the polygon coincides with tlie circumference of the circle, and the area of the polygon is the same as the area of the circle ; .'. area of circle = ^ OR x circumference of circle = ^rx2Trr it 227. To find the area of a quadrilateral which can he inscribed in a circle in terms of its sides. '•iMW Let ABCD be the quadrilateral. Join AG. Let AB^a, BO=b, CD = c, DA=d. Then, area of figure = area of a ^45(7+ area of a ADG s= ^ a& . sin ^ -H ^ ccZ . sin 7), Now the angles at B and D are supplementary (Eucl. Ill, /. sinB = 8ini)(Art. 101); .-. area of figure = ^ («& + cd) . sin B. We have now to express sin B in terms of the sides of the figure. Now, in aABC, AC'^ = a'^ + h^-2ah .cos B, imd, in aADU, AC'^=:c^ + d^-2cd . cob 1), e circumference lie same as the of circle t can he inscribed 1 itary (Eucl. ill. POLVGOATS AND CIRCLES, Hence, observing that cos 1^= -cos i? (Art. 101), a2 + &2 _ 2ah . cos i? = c2 + d? f 2cd . cos B ; .*. COSi>: 2 {ab + cd) ' ' 19: |2(a6 + (70{2 AivLb-\-cdf ' .'. (area of figure)2=-- . (ab -f- cd)- . sin^ B 4 4(a6 + c(Z)"'^ = j^ . K2a6 + 2c(Z)2 - (aH 6''' - c2 - rf2y2 1 = ^. j(2a& + 2(:(iH-a2 + 62-c2-tZ2)(2a6 + 2c't«-a2-62 + c2 + (^Aj 16 ,.\{a-vhf-{c-df\\{c^-df~{a-bf\ = , /. . { (tt + 6 ^- c'- d!) (61 + 6 - c + (i) (c + d + a - 6) (c -I- c? - a + : ; : ,he sides of the and if « = a -!- & + c + r? 2 ' (area of figurc)2= ^-^ . j (2.9 - 2r?)' (2.s' - 2c) (2s - 2&) (2^' - ^a) ( « (s - (i) (s ~ c) (.9 - h) {s-a)\ :. area of figure = V j (s - «) (*• - h) (s - c) (s -d)\. "I 196 AREAS OF TRIANGLES, r j 1i-7 •" i. ! ' 228. On the Dip of the Horizon. Suppose the earth to be represented by the circle ABC, witli centre 0. Let EB be a tangent from the eye of an observer, looking from a lieiglit AE, to tlie eartli's surlace at 7>', and let EAG bo a straight line through the earth's centre. ■ >« If we draw a horizontal line EH, the angle HEB is called " The Dip of the Horv.on:' Then, since EA is very small as compared with AO, and therefore the arc AB very small as compared with the circum- ference of the earth, i AOB is a very small angle. Hence at spots of small elevation the Dip of tjie Horizon, which is equal to i_AOB, is very small. The distance -^f fhe horizon at sea may be approximately found by the following rule : Three times the heirjld of the jilace of ohervation in feet is equal to twice the squai'e of the distance seen in miles. This rule may be proved thus : / Let AE bean object whoso height in feet is/, —ttatt?. niilea, J o jf 5280 £B a tangent to the earth's surface whose length in miles is m. AC the diameter of the ^arth— c2— 8000 miles nenrly. jircle ABC, witli bserver, looking ajid lot EAC bo HEB is called with AO, and ith the circuni- :le. of tjie Horizon y approximately I in feet is equal '=5280""''^'' jose length in ailes nearly. POLYGONS AND CIRCLES. 197 Then sq. on i?^ = rect. CE, EA (End. in. 30); V 5280/5260 ■=5280^'^'^^ 8000/ 3/ = '5280"'''^'''^>^=2 "'^'^y* 229. To show that if 6 he the circular measure of a 'positive angle less than a right angle, si)i 0, 0, and tan are in ascending order of magnitude. Let Q be the centre rtf a circle, QE a radius cutting the chord PF at right angles, TT a tangent to the circle at E, T Let tho circular measure of the angle EQP be B* PM Then sin B= QP' PE ^ QP' tnti d-= TE _TK_ Qfrqp' Vow afssumincf that PE is greater than PM hut 1pm than TE, PM, PE, TE are in ascending order of nuignitucU Therefore sin 0, 0, tan are in ascending order of mnguitudo. m \W:V . ■■■* If... ^ ' ! ■ " .' ' • M R ■;v ■•J ■■■^ ' 'T:;) >* 198 AREAS OF TRIANGLES, The assumption which we here make that PE is inter- metliate in magnitude between Fhl and IE requires some explanation Suppose PP' to be a side of a regular polygon of n sides inscribed in the circle. Then TT' will be the side of a regular polygon of n sides described a,bout the circle, and QE will bisect PP' and TT, Now the perimeter of the inscribed polygon is always Zess than the circumference of the circle, as we explained in Art. 10, and we might show by a similar process that the perimeter of the circumscribed polygon is always greater than the cir- cumference of the circle. Now Pif =the 2?ith part of the inscribed polygon, TE= circumscribed PE= circumference; .'. PE is in magnitude intermediate between PM and TE. 230. To show that when 6 is indejinitely diminished, sin = 1. Since sin 0, 0, tan are in ascending order of magnitude, sin 0, 0, -~-n are in ascending order of magnitude, ' cos Divide each by sin 0, then , e 1 '* sin 0^ cos are in ascending order of magnitude ; a 1 •. -. — 7. lies between 1 and — x. sm 00a ^ t PE is inter- requirtis some gon of n sides rgon of n sides ^F and TT. I is always less plained in Art. .t the perimeter r than the cir- ygon, ' f PM and TE. inisliedy f magnitude, f magnitude. agnitude ; POL YGONS AND CJRCLES. 199 Now when 6^ = 0, and therefore therefore when 6=0, cos 6 = 1, 1 cos 6 6 sin 6 1; =1. Ex AMPLES.— Ivl. 1. The angle included between two sides of a triangle wh. .e lengths are 10 inches and 12 inches is 60': find the area of the triangle. 2. Two sides of a triangle are 40 and 60 feet, and they contain an angle of 30° : find the area of the triangle. 3. What is the area of a triangle whose base is 4 feet, and altitude 1| yards ] . 4. What is the area of a triangle whose sides are 5, 6, 5 inches respectively ? 5. If a = 625, 6 = 505, c=904, what is the measure of the area of the triangle ] "^ 6. If a = 409, 6 = 169, c = 510, what is the measure of the area of the triangle ? 7. If a = 577, 6 = 73, c = 520, what is the measure of the area of the triangle 1 8. In a right-angled triangle area = s. (s-c), C being the right angle. 9. If a = 52-53, 6 = 48-76, c = 44-98, log 14G-27 = 2'1G51553, log 5G-31 = 1-7505855, log 48 75 = 1 0879746, log 41-21 = 1-6150026, log 2 = -3010300, log 1-0169487 = -0072990, find the measure of area. ■ifi! 'nruslcav )) li' ,r..4- 200 AREAS OF TRIANGLES, 10. The sides of a triangle are in arithmetical progression, and its area is to that of an equilateral triangle of the same perimeter as 3 : 5. Show that its largest angle is 120°. 11. In the rectangular sheet of paper ABCD, the angular point A is turned <lo\vn so as to lie in the side CD, while the crease of the . passes through the angular point B, show that the area ^i tiie part turned down is 1 AB^ -^.-^\AB-^/{AB^--BO-% 1 2. Show that the area of a triangle _a2.sin J5. sin (7 ~~2sm{B + C) ' 13. In any triangle the area a-* .A . B . G = sin n .sm -— .sm ^ . ... ^.^ — ^i 2 2 2\sinvl siu-S sm C '(; + 62 c + 2 )• * I 14. If the radius of the inscribed circle be equal to half that of the circumscribed circle, the triangle is equilateral. 15. In any triangle ,, . A . B-G (o-c) cos -- = a. sm — g— . 16. if the points of contact of a circle inscribed in a triangle with the sides be joined, show that the area of the triangle so formed 17. The diagonals of a quadrilateral are in length a, h respectively, and intersect at an angle A. Show tliut the area of the quadrilateral = -ao am A, 18. The area of any triangle a^ - h"^ sin A . sin B """2" • tiin{A-BJ' I progression, of tlie same i 120°. the angular l\ wliile the )int B, show qnal to half liluteral. icrihed in a area of the length a, 6 lat the area POL YGONS AND CIRCLES. 301 19. In an isosceles right-angled triangle, show tiiat the radius of one of the equal escribed circles is equal to the radius of the circuniscriljed circle. 20. In any right-angled triangle, C being the right angle, cot(i?-^) + cut2(yl+?) = 0. 21. The area of anv triangle 2nt/jc A A B G ■■ --,— - . cos — . cos — . cos 7-. a+b+c 2 2 2 22. In any isosceles triangle, C being the vertical angle, A area x 32 cos* — = sin 2^1 ''2a + c)\ 23. The length of a perpendicular from A to BO ~ 6+c 24. Taking the notation adopted in this Chapter, prove the following relations : (I) r= a ,B (J cot 2 + cot ^ (2) r: 272 . sin ^ . sin - B . sin -^ G 2 2 cos- A a (3) r,^ ,,, (; ■ tan — + tan ^ A A (4) ri = 4/2 . sm g cos - . cos ^. (5) r*i + r2 + r3 = i2(3 + C08yl -f-cos^ + coaC). (6) ij! + r = 7? (cos ^ + cos B 4 cos C). ^i 11. ^ l{ -» H ' ' i|> .; % ■ > i-'i •' ,-*■' ')! ,n •'. ' 202 AREAS OF TRIANGLES, 25. Show that the area of a irgular polygon inscribed in a circle is a mean proportional between the areas of an inscribed and a circuinscribed regular polygon of hall" the number ol sides. 26. The distances between the centre of the inscribed and those of the escribed circles of a triangle ABG are 4i2 . sin ^ , 42? sin --, 4i2 sin ^, AAA E being the radius of the cii'cumscribing circle. M , -l^^''' ' 27. The points at which the line3 bisecting the angles A, 7>, (7 of a triangle cut the opposite sides are joined. Show that the area of the triangle so formed bears to that of the triangle A BG the ratio „ . A . B . C 2 sm - sin -^ sin ^ A iu A B-a U-A A-B' cos ~g— . COS — ^ .COS Q — 28. If rj, r^, r^ be the radii of the escribed circles, and s the semi-perimeter of the triangle, show that e2_ s^=rjr.;i + r^r^ + rsri 29. If Af B, C, D, be a quadrilateral capable of being inscribed in a circle, show that AG . sin A=B1) . sir B. 30. Show that the distances from the centre of the inscribed circle to the centres of the escribed circles are respectively equal to a h c T B' d' Cos X- cos Tj cos s 2 2 2 inscribed in u f an inscribed le number ol inscribed and e POL YGONS AND CIRCLES, 203 ig the angles mied. Show that of the leSj and & the ble of being sir B. the inscribed respectively 31. r is the radius of a circle inscribed in a triangle 2iBG\ show that A B G a = r cos — cosec — cosec s. i Ji 2t 32. If i?, r be the radii of the circles describod about and inscribed in the triangle ABG, and s the senii-perinieler of the triangle, prove that 33. ABG is a triangle inscribed in a circle, and a point P is taken on the arc BG : show that FA . si.i A^PB.ainB f PG . sin G. 34. Given the distances rfj, d,> '^3 fi'oni tlie angles of the point at which the sides of a plane triangle subtend ef^ual angles, find the sides and area. 35. If the lengths of three lines drawn from any point within a square to three of its angular points be a, b, c, find a side of the sijuare. 36. The areas of all triangles described about the same circle are as their perimeters. 37. If a, h, c be the sides of a triangle, and a, (3, y the perpendiculars upon them from the opposite angles, show that Py ay a/S a"" h'^ c^' 38. A person standing on the sea-shore can just see the top of a mountain, whose height he knows to be 1284*80 yards. After ascending vertically to the height of 3 miles in a balloon, he observes the angle of depression of the mountain's summit to be 2° . 15'. Find the earth's radius. Given log 3 = -4771213, L cot 2M5' = 11-4057168, log -73 = 1-863229, log 7986-4 = 3-9023533, 102 76-3551 = 1-8828381. 204 AREAS OF TRIANGLES, 39. If an e([iii].iteral triangle have its angular points in three parallel straight lines, of which the middle one is perpendicularly distant from the outside ones by a and 6, show that its side =2 V(^H— )• 40. In a triangle ABC, AC=2BC. If CD, CE respectively hisect the angle G and the exterior angle formed by producing AC, prove that the triangles CBD, ACD, ABC, CDE have their areas as 1 : 2 : 3 : 4. 41. If R, r be the radii of the circles described about and in the triangle ABC, the area of the triangle =lir (sin ^ + sin J5 + sin G). 42. In the ambiguous case prove that the circles circum- scribing the triangles will have the same radius. If the data be a = 605, 6 = 5G4, 5 = 50°. 15', find the radius of the circum- scribing circle. 43. The angles of a quadrilatgrnl inscribed in a circle taken in order, when multiplied by 1, 2, 2, 3, respectively, are in Arithmetical Progression ; find their values. '"1" 44. If from any point P in a circle lines are drawn to the extremities A, B, and to the point of contact C of a side of the circumscribing square, show that (1 + co t PCM )2 ^ q-fcot PC7i)3 cotPP^ ~'~ coiPAB' ' 45. If R, r, r„, rft, r^ are the radii of the circumscribed, inscribed and escribed circles respectively of a triangle, and C one of the angles, then 4i2 cos G=r + ra + U-r^ 46. If T be the radius of the circle inscribed "between the base of n right-angled trinngle and the other two sides pro- duced, and 7"' the radius of the circle inscribed between the altitude of the triangle and the other two sides produced, show that the area of the trians:le = r.r'. oints in three rpenclicularly at its side respectively ^y producing ^E have their d about and 'cles circum- If the data the circum- in a circle )ectivelv, are rawn to the of a side of POLYGONS AND CIRCLES. 205 "cumscribed, riangle, and between the o sides pro- iL'tween tlio !3 produced, 47. From the top of the peak of Tciierifte the dip of tlie horizon is found to be 1°. 58' . 10". If the radius of the earth be 4000 mi^es, what is the height of the mountain ? 48. "What is the dip of tlu- horizon irom the top of a mountain 1^ miles high, the radius of the earth being 4000 miles ? 49. A lamp on the top of a pole 32 feet high is just seen by a man 6 feet in height, at a dittancQ of 10 miles; find the earth's radius. 50. A ship, of which the height from the water to the summit of the top-mast is 90 feet, is sailing diioctly toward? an observer at the rate of 10 miles an hour. Fmm tlie time of its hrst a])pearance in the othng till its arrival at the stiition of the observer is 1 hour 12 minutes. Find approximately the earth's radius. 51. If the diameter of the earth be 7012 miles, what is the dip of the sea- horizon as seen from a mountain 3 miles in height \ 52. The angle subtended at the sun by the earth's radius being 8"'8(J8 and the earth's radius being 4000 miles, show that the distance of the sun from the earth is approxinuvtely 93000000 miles 53. If the distance cf the moon fiom the earth be 241118 miles, sliow that if tlie earth's radius is 4000 miles it sub- tends an angle of 57'. 1" 5 nearly at the moon. 54. The tops of two veitical rods on the earth's suvfaco, each of which is 10 feet hiuh, cease to be visible from e;ich other when 8 miles distant. Prove that the earth's vadium is neiirly 4224 miles. v 55. What is the limit of deviation in order that a circular target of 4 feet diameter may be struck at a distance d 200 yards ? 56. Explain how it is that a Bhilling can be placed befure the eye so as to hide tin; nnmn. A isr S W E R ». i. (Page 1.) I. 54. 2. 26 5. 3 inches. 6. 5 inches. 7, 9. 1 inch and 3 inches. 10. 13 : 504. 3. 4 inches. 6 3a' 4. 5 inches. 8 7i inches. II. en 3m' ii. (Page 3.) I. 45 yds. 2. 10 ft 3. 255 yds. 4. 360'5...yd. 5. 163-25 yds. nearly. 6. 12 It., 16 it. 8. 63 It., 45 ft. 9. 6v/2ft. 10. 5^2 inches. 11. 625 V2ft. 12. 13 x^ 2 • 15. 38 inches. 13. 10 V3. 16. 634 ft. 14. 12 inches. 1. 15^ ft. ill. (Pago 9.) 2. 86-3068i ft. 3. 25-714285 miles. 4. 7054— miles. 5. 2775834^ miles. 6, 1089^1 miles. ^ 11 •» 7 22 7. 6 ft. 6= in. 8. 47^ ft. ,6 ., ,0. ?-^?-^2ft. II. 12^ miles. 525 ^/2 „, 9- -22 ft- 12. 22- miles. I. 24-2G805. 4. '09i. ANSWERS. iv. (Page 12.) 2. 37-04527. 5. 375-06. 207 3. 175-0038. 6. 78-201. 4. 5 inches. 8 7^ inches. 11. en 3w' 360-5... yd. (53 ft., 45 ft. (525 V2ft. 12 inches. 285 miles. 089^? miles. f)2r) ^^ 22 ^'^• 22^ miles. I. 25-1425. 4. -150745. V. (Page 13.) 2. 38-0415. 5. 425-130554. 3. 214-0307. 6. 2-020222. Vi. (Page 19.) I. 30« 29M9-75... 2. 174^52M2-0fi2. 3. 45. 4. 21Mr.6G"-6. 5. 159«.5\55-5. 6. 31Mr.iri. 7. lP.58\88''-8. 8. 30«.70\74"-674. 9. 333«.62\ 90-1234567890. 10. 469'. 2'. 53-086419753. Vii. (Page 20.) I. ir..30'.48"-78. 2. lir.38'.44"-592. 3. 26°. 46'. 30". 4. 13\30'.4"-86. '5. 138\39'. 54"-5T(;. 6. 38'. 42'. 7. 3r.50'.26"-9772. 8. 45°. 41'. 32"-9532. 9. 153°.34'.9"-948. 10. 291° .43'. 29"-9388. Vlll. (Page 21.) I. 3' 77r 4* 3* 2. 6. TT TT 8- 3. id" 1453^ 1437r 10800' 7' 270* .« T TT TT 8. 2' 27l!);r 40500' 10. ' 1» !• 2' 4' 4 1 HI W: ■ » ■ t ^;"v; '■ • j: ^:-t : ■ . * ,1 i 208 ANSWERS. I. 90°. 2. :X. (Pago 22.) 60°. 3. 45°. 4. 30°. 5- 120°. .» 6. 90 , — degrees. 30 , 9. — degreua. TV 60 , 7. — degrees. TV ° 120 , 10. degrees, TT X. (Page 22.) 8. — degrees. It I. '^ 2 ^ 4* ^' 8* TT 3- 32- 6. •0(ir)2r)257r. 7. •1207510757r. 9- •OOOIStt. 10. •OO000257r. Xi. (Page 22.) i. 6G-6 L'radea. 2. 40«. 57r 4. -^- 5- ^ 8. -CSOOOCSjr. 4. 133*3 grades. 40 , 7. - - grades. TT ° 4G0 , 10. grades. TT 5. 120«. 8. - * grades. TT 3. 33-3 grades. . 200 , 6. -f^ — gradt'8. uTT 120 9. grade.s. TT I. 4-5. A 3^° ^ 7 xii. (Pago 23.) 2. 4'iio degrees. 14 5 15* 8. 90', GO*, 30° ; 100«, ^^X, 33 ^^^ ^ 3^" ^ ^ 6. 70 : G7. 10. 3r)°.6'.38"'8.^ )°. 5. 120° 45 TT degrees. Stt 5- -r. . •C25U0G57r. 33-3 grades. 200 3^ g^'^^^^'s. 120 grades. ^A'SfV£A'S. 209 rr. 5000?/i '27 • 12. 61 ^3. 1 ^\ 16. 20°, eo^ 100°. '7. 3-14159... :l. ,8. 3G^ 54°, 20. 125-925 grades 19. l/'3!. lGOO;r-3600 ^^- 9r ~~ grades. 90°; 40«, GO^ lOQe. '^ •'^~ TT ' 5' 10' 2* 21 ^"^^ 1 23- 54°.4G'.54"-5. 24- -1775. 25. 207r degrees. 26 - ^ fV ^ 7 1 i 144000^^ Pa^t. .7. ^^ J^^ _I^ 29- G7^^-, 180°, 45;r°, (^.180 4-15)°; 28. — ^- miles. 37r 8 30. 12' 32. 120«, 150«. 35- T feet. 36. 8 and 4. Xiii. (Pago 3G.) 31- -0°, 108°. 34. TT . n BD AD BD AD BD AU 7 AB' AB' AD' AB^ AB' AD' AD' BO' BJ^ CD DB BO' DG' ^r 5 8 5' 8' 5' ^iv. (Pago 40.) ^- :r-7^- y/2' 5- 3^2+ V3. vJ- , x/3 4. v/3. 70 : G7. 36°.C'.38"W XV. (Pago 52.) I. 346-4101 ...feet. 2. 85 •829037 ... feet. 4. IT3-?05...feet. 3- CO" T] 5. 424-352...feetfrum foot of rock. 5," t '»;'4 'I 2IO AN^iyE/^S. ..! fj i' 6. 1 "366 . . . miles. 7. 42*265 . . . feet above the tower. 8. V2 feet. 9. 50^3. yd. 10. 25°. II. 104"25 . feet above the tower. 12. 125ycL 13. 92j-2feet. 14. 342ffeet. 15. 8053. ..feet. XVii. (Page 60.) I. sm A= vl-cos^^, tanyl = -. — , sec^ = ~^ — 3, cosec A •■ 2, . . cos A . s^ =, . cot ^ = -: . V 1 - cos^ A mJI- cos'-^ a 8111-4 = 1 . Vcosec^ A — \ ■ -jy COS^= ;.- , cosec A cosec A . ^ 1 J, cosec -4 tan^= . — , sec^=— pr^— . V cosec^ A -I vcosec'-^ ^ ~ 1 cot -4 = x/cosec^ ^ - 1. . . Vsec^^-l J I \ . I — S-. 3. sm^ = --^^^-^ — , cosvl=^^^, tan ^ = VsecM - 1, cosec .4= f, ; , cot A = —.-rr^= — . V sec''^ ^ - 1 Vsec^ A-\ . - A 1 4. 6111 il 1 . coiA ^ A I Vl + cotM Vl+cotM cotA' cosec A= fjl + cot^ ^, i . \/T+cora 1, sec -4= 7— i — • I. n/5 3 ' V5* /2 1_ ^' V3' '^2* ^* 6' 4* XViii. (Page 61.) 3 3 , N^ _3 ^' "4 ' V7* V(a^ + M ) ^(a* + ¥) a 1 • N/(i-a2)' ^(i":::^)- er. 8. 72 feet. 30ve the tower. ;. 8 -053... feet. sec A = A os'-^ a' cos 1' \Q,A Vsec^ - 1, l-l tan -4 = WA cotyl' 4 ' sjf 1 7(1 -asy ANSIVERS. 211 8 V(l-_62) l_^^ lo. 9 2V14'2V14* II. - 4 4 5' 3' V40_3 V403 22 31 ' '30"' ^' 101' 26* 20 20 ^' 101' 99* 15 -^ 13 5- 13' 12* XX. (Pago G5.) I. (i)65'.45M8". (.)4C°.57'.3". (3)25-.59' 46". (4) 7°.55'.45". (5) -(35M5'.42'0. (6) -(88°. -.'. 34") (7) -105°. (8) -104°. (9)115°. (.o) 335°. 2. (I)67^76^7G^ (2) 4^96^25^ (3) os^oomg- (4)97^94^96^ (5) -(35^2^5"). (6) -(G9«.0'.3") (7) -143'. (8)-257« (9) 135'. (,o) 345'. TT 3. (Oi- (.)|. (3)--. (,)?._ 10* (5) 57r 4* xxi. (Page 68.) I. (I) 145».4r 11". (2) 47°.35M3". (3) 33°.59Mf)". (4) 15r.44'.56". (5) 1". (6) 79°. 10'. 7". (7) -65°. (8) -(257°. 3'. 4"). (9) 229°. (ro) 535°. 2. (i) G7«:67'.58". (2) 4'. 97'. 43". (3) l9f;^2^2'^ (4) 134«.8?.92". (5) 45«.9G\94''. (6) 25'.99\9G'\ (7) -75«. (8) -(327'. 2M4"). (9) 235*. (ro) 525'. '■ W I (^) I- (3) |. (4) ^J. (5) 'I 4- ?r. 212 I. 2 • V3. ANSIVERS. xxiii. (Piige 72.) 2. ~2* 3- Ji' 6. V3 2 7- 1 'V2 lO 2 V3' II. V2. 4- 8. V2- y3 2 • 12. - V3. XXiv. (Page 75.) I. -45'. 2. 45^ 3. 0°. 4. 90'. 5. 0°()r60°. 6. 30°. 7. 30°. 8. 45'. 9. 30°. 10. 60° or 30°. 11. 30°orG0°. 12. 30° 01 60°. 13. 45°. 14. 45'. 15. 30°. 16. 45°. 17. 135°. 18. 45°. 19. 30°or90°. 20. 45°. X. 2 • 5- o- 9. - V3. I3- ^''- 72* 21. -2. 25. -V- XXV. (Page 81.) 2. g. 6. -^^. 10. TT 14. 18. V3- x/3 v/3 22. --77.- V3' 26. 1. J V2- 1 va- il. V2. 15. 1, 19. 0. 23. 1. 27. ~ s!% 4. - 8. 12. V2- "2 V - J3. 16. - sJ^ 20. 24. V3* 28. -1. 29. 30. - s V2- - V3. 4. 90». 8. 45*. I. 30° or 60°. 16. 45°. )\ 20. 45°. 4. - 8. V2- "2 * 2. - J3. 16. v/3 20. 1. 4- 2 x/3- 28. -1. ANSIVERS. at 5 XXvi. (Page 82.) I. nr+(-l)' TT 2. 2 /ITT. 3- n7r+(-l)''.' TT 4" TT TT 7. W7r + (-l)-.J. 9. ?i7r + 2. 8. 2ji7r±^ or 2u7r+'-'^ 4 - 4- 10. 2?i7r± TT 5. 7. nA5 4-4^2 9 V3j-1 "2^2 • Xxviii. (Page 88.) 2U ' '• 8 y3+ v^SOO GO I. 45*. 5. 46°. XXX. (Page 80.) 2. 0°. 3. 105'. 4. 60°or30°or-3(r, 6. 60°. xxxii. (Page 93.) 2cos|(a + |-^)sin^(a-^ + /^) 2. 2 ".sin (!+«) cos TT 3. 2 sm ^ . C03 \<^- g)> 4. 2 cos -,- sin 6. 2 cos 15°. sin 5° 5. 2 sin 20°. cos 10* 7;r TT 7. 2«m--cos^-. 8. 2C08-- sin -^. |.!i i ^' ■' . / ; ■ i f ,.» 214 y^A^i-jr^i^i-. 2 ' XXXV. (Page 105.) (r) ^=45 °or-15 °. (2) (9 = 30° or 45°. (3)0: = :0° or 74'. (4) 6?= 15 ° or 30°. (5) .4 = 30° or 150°. (6) e= = 0° or 7i°. (7) ^ = 15 ° or 30°. (8) a = 0°or30°. (9) e= = 180° or 30" 1 (10) a = ()" or 15° or 60°. (11) (9 = |. xxxvi. (Page 106.) (12) ^= = 0° or 30°. I. ^V(io- 2V5). 2. i(l+V5). 3. 3(1 + ^5). 4. iV(10- 2V5). 5. iv(104-2V5). V(10 + 2V5) ^- " V5'-- 1 ^ ' 7. 1. 8. a xxxvii. (Page 110.] 1 A . ^ I. cos 2- + sm "2 = + Vl + sin ^ ; yl . A sin- = - COSg- + -^1-sin^, ^ . A 2. cos 2- + sm -■ = - Vl + sin -4 ; A COSg- . A 3in 2 = = -11 - VI -sin A. • * 3. COS 189° V5- V5+ n/3+ ^/5 . sin 189° = i| ^5-^5- V3+*/5| . 2- V2 ^ 2x/3 • - V2 + V2 • AA^SlVEJiS. 215 =0''or7i*. = 180° or 30=" '=0'or30°. (1 + V5). I. 1-2187180, 4. 4-740378. 7. 5-3790163. 10. 2-1241803. xxxix. (Page 120.) 2. 7-7074922. 5. 2-924059. 8. 40-57809a II. 3-738827. 3. 2-4036784. 6. 3-724833. 9. 62-9905319. 12. 1-61514132. xl. (Page 123.) 1. 2-1072100; 20969100; 33979400. 2. 1-0989700; 3-0989700; 2-2922560. 3. -7781513; 1-4313639; 1-7323930; 2-7604226. 4. 1-7781513; 24771213; -0211893; 5-6354839. 5. 4-8750613; 1-4983105. 6. -3010300; 2~8061800; '2916000. 7. -6989700; 1-0969100; 3-3910733. 8. -2,0,2 ; 1,0, -1.. 9. (I) 3. (3)2. xo. x=|,,=| 11. (a) -3010300; 1-3979400; 1-9201233; r-997958a (6) 103. 12. (a) -6989700; -6020600; 1-7118072: 1-9880618 (6) 8. 13. 3-8821260; 1-4093694; 3-7455326. U- (i) x = -, (4) a;= (2) a; =2. ^^^ log a + log 6' .. . i^g_c m I0.1,' a + 2 log 6' (r) 3.= 4 log 6 +log c __ 2 log c + log 6-3 log a' (6) x= ^^S^ log a + m, log 6 -f 3 log g' .«• ;. ' ■m 216 ANSWERS. Xli. (Pago 127.) I. 4-7201 79D. 2. 2-4777360 3- ir)05-.X)74. 4- 2 -37401 Go. 5 18293173 6. 3-8653132. 7 G-8190943. 8. 3 5324716 9- 6-5921478. 10. -4119438. xliL (Page 129.) I. 12954-8. 2. 4624-5095. 3. 345-7291. 4. 3L''756-9 5 3715-953. 6. •C09646153 7- •00000025725982. 8 601-95403. 9- 1090-5286. 10, 26201818. Xliii. (Page 132.) I. •6724242. 2. -9523159. 3- -8150856 4- •9990000. 5 •6850417. 6. •7521403 7- •0240028. 8 1-4225190. 9- •9230769. 10. •8270272. Xliv. (Page 135.) I. 48°. 46'. 34'. 2. 2°. 33'. 45". 3. 43M4'.8"-18. 4. 32°.31'.13"-5. 5. 24Ml'.22"-2. 6. 82°. 22'. 12"-8. 7. 53°.7'.48"'4. 8. 25° . 3' . 27"-2. 9. 73° . 44' . 23" -2. 10. 77°.19'.10"-5. Xlv. (Page 138.) I. 9-9163319. 2. 9-6912280. 3- 9-8996023. 4. 9-9091749. 5- 9-7203429. 6. 8-961006a 7. 11-1975684. 8. 9-8027687. 9- 9-9745378. 10. 8-2814755. ir)05-^74. 3-8653132. 6-5921478. 345-7291. •C09G46153. 601-95403. J. -8150856. 3. -7521403. }. -9230769. 3' 14' . 8"-18. 2°.22M2"-8. r.44'.23"-2. 9-8996023. 8 -96 1006a 9-9745378. ANSWERS. 2'7 Xlvi. (Pago 140.) I. 14°. 24'. 35". 2. 54M3M9". 3. 7r./10'.18". 4. 29°. 25'. 2". 5. 30\50'.27"-6. 6. 86° . .^2' . 24"-5. 7. 24°. 8'. 45". 8. ir.39'.52". 9. 4G\23'.li". 10. 29° . 54' . 29"-6. Xlix. (Pago 157.) 1. a=4, ^ = 53°.7'.48"-4, i? = 36°. 52'.ir-6. 2. a = 8, ^ = 28°.4'.20"-9, Z? = G1°.55'.39"-1. 3. a = 20, ^ = 43°.36'.10"-1, ^ = 46°.23'. 49"9. 4. a=24, J = 73°.44'.23"-3, i?=lGM5'.36"-7. 5. a =56, ^ = 59°.29'.23"-2, J5 = 30°.30'.3G"-8. 6. a = 12, 6 = 5, l?=22°.37'.ll"-5. 7. a = 40, 6 = 9, ^=12\40'.49"-4. 8. a = 48, 6 = 55, ^=41°.6'.43"-5, 9. a = 39, 6 = 80, ^ = 25''.59'.21"-2. 10. 6=9, c=41, 5=12°.40'.49"-4. 1. (Page 159.) 1. 6=153, ^ = 34°.12'.19"-6, ^ = 55''.47'. 40''-4. 2. 6 = 297, ^ = 45°.40'.2"-3. i?=44M9'. 57"7. 3. 6=41, ^ = 87°.12'.20"-3, i? = 2\47'.39"-7. 4. 6 = 527, ^ = 32°.31'.13"-5, i? = 57°. 28'.46"5, 5. 6=141, ^ = 82°. 41'. 44", ^ = 7° . 18'. 16". 6. a = 748, ^ = 75° . 23' . 18"-5, £ = 14° . 36' . 41"'6. W-''- .n'^ m^ ^. .-t - ■*■'. ■f y«'i' 218 ANSIVERS. 7. a=V36, yl = 69°.38'.5G"-3, ^ = 20°. 21'. 3"-7. 8. a = 200, ^ = 18". 10' . 50", J5 = 7 1° . 49' . 10". 9. c = 565, ^ = 29M4' . 30"-3, i? = 60° . 45' . 29"-7. la c = 565, ^ = 44° . 29' . 53", i? = 45° . 30' . 7". li. (Page 161.) I. 231-835 feet. 2. 93-97 feet. 3. 36-6.. .feet; 70-7.. .feet; 100 feet. 4. 196 feet nearl3\ 5. 460 yds. nearly. 6. 63°. 26'. 6". 7. 88 yds. nearly. 8. 33°.23'.55"'7» 9. 39° . 5' . 47"-9. 10. 104-93 feet. II. 45 feet* 12. 150 feet. lii. (Page 171) I. (l) 67'.22'.48"-5. > (2) 43".36'. 10"-1. (3) 112\37'.ll"-5» (4) 29°.51'.46"-L 2. A - 49° . 7' . 10", (7= 87° . 44' . 50". 3- 6 = 79-063. 4. 6 = 219-37. 5- 67°. 14'. 21" or 122°. 45'. 39". 6. N for ^ = 90°. 7. 003^ = ^^. 9. 45*, 60', 75' II. G9° . 10' . 10" and 46° . 37' . 50". 1 2. 50° . 46' . 16" 13. ^ = 116°. 33'. 54", 5 = 26°. 33'. 54". 1 liii, (ragol7C.) t. c«6, ^ = 53°.7'.48"-4, i.' = 36\ 52'. ir-6. 2, rt = 48, /I -41° . 6'. 43"-5, 7?-48° . 53'. 16"-6. t t ANSIVERS. 219 l'.3"-7. . 10". '.29"-7. 7". et. 36 feet nearl}'. 8 yds. nearly. I. 104-93 feet. '.10"-1. '.46"-h 45", 60^ 75*^ 05%46M6" •6. 3. c = 353, ^ = 50°.24'.8"1, ^=39'. 35'. 51"-9. 4. a= 40, ^ = 5° . 43' . 29"-3, J3 = 84° . 16' . 30"-7. 5. a = 84, 6 = 437, 5 = 79°. 7'. 9"-6. 6. a=460, 6 = 429, i3 = 43°.0'.10"-3. 7. a=280, 6 = 351, ^ = 38° . 34' . 48"-3. 8. a=180, 6 = 299, ^ = 3r.2'.53"'6. 9. 6 = 231, c = 569, 7? = 23°. 57'. 8". 10. a=480, c = 481, i? = 3°.41'.43''. liv. (Page 177.) 1. il = 31*.53'.26"-8, 5 = 8°.10'.16"-4, C=l39\56M6"-A. 2. ^ = 84°.32'.50"-5, i? = 25°.36'.30"-7, C=69°. 50'. 38"-8. 3. ^ = 76°. 18'. 52", J5 = 35°.18'.0"-9, C=68°. 23'. 7"'l. 4. -4 = 62°.51'.32"-9, i? = 44° . 29' . 53", 0=72°. 38'. 34"' h 5. ^=^150°. 8'. 14", 5=17°.3'.4r-6, C= 12° . 48' . 4"'5. 6. rt-101, c=120, i9=ll°.25'.16"-3. 7. a = 221, c = 222, i3'-39°. 18'. 27"-5. 8. c=78, yl-.13G°.23'.49"-9, jB=ir . 25'. 16"-3. 9. c=408, ^l = 77M9'.10"-6, 5 = 5°.43'. 29"-2. (O. c = 120, yl = 110'.0'.57"'5, 7? = 39°. 18'.27"-5. 11. c = 1 02, ^ = 79° . 36' . 40", B = 33° . 23' . 54"-6. 12. c = 450, ^ = 8r.27M6', Z?=10°.37'. 44". 13. 0-312, i4=x33°. 23'. 64"-G, B=15Mr.21"'4. 14. c-332-97, ^-45°.46'.16"'5, J5 = 30° . 0' . 52'''5. 220 ANSWERS. 15. c = 90, ^ = 134\45'.3C"-6, 7] = 29\ 51'.46"-4. 16. ^ = 67°.22'.48"-l or 112°. 37'. ll"-9. 17. 1?= G2° . 51' . 32"-9 or 117" . 8' . 27"-l. 18. i?= 78° . 19' . 24" or 101° , 40' . 36", 19. i?= 75° . 18' . 28"-2 or 104° . 41' . 31"'8. 20. 25=53°. 26'. 0"-6. r-^*: tiii Iv. (Page 181.) I. 23G-G02...feet 2. 1210 yds. and 1040'5 yds. 3. 4596 yds. nearly and 4584-48 yds. 4. 33' . 42". 6. 107 feet nearly. a 10. = sin a. sec /3. A 5. x/3 : 1. 13. 4 miles. 14. 5 (3- v/3) niilcs. 16. tan~i.-*. 20. 51*76.. .feet. 21. 10 and 14- 14... miles. 22. 48° . 22', 23. 1 15-47 yds. akd 9-503 yds. 24. 6'71307... miles. 25. 15 yds. 26. 6236-549 ft. ; 1095-47 ft. 27. 85*28 ft. or 28*14 ft. 28. 108-64 yds. nearly. 29. 204*2 ft. 30. yl = 116°.13'.20",i? = 26°.20'.40",c = 4325*26. 31. 1 mile and *923497 mile. ^ 32. 124*3 feet. 33. 306*4178 yds. 34. 34*42284 or 14-524 miles an hour. 35. 513-7045 yds. 36. 17° . 47' . 50". 37. 134 yds. 38. 169*4392 yds. 39. 76-5455 miles. 4 4 r-4. )40-5 yds. 33'. 42". ^ sin a. sec fj, 4 tan ~ 1^. 03 yda. t. or 28-14 ft. ANSWERS. 22 1 Ivi. (Pago 199.) I. 30 ^^3 s(j, in. 2. GOO sq. ft. 4. 12 sq. in. 5. 151872. 7. 12480. 9. lOlG-9487. 34. «= VW + ^3' + f7./?3), area = -^- (Jid, + ^if/3 + cUk). 3. 7^ sq. ft 6. SOGOO. 35 '1 38. 3092-835 miles. 42. 3GG-785 Gtt Ttt IItt IOtt 43* Y7' 17' 17"' Tf ' 4^^- ^ '^^ niiles. 48. r.2G', 49. 4017-79... miles. 50. 4224 miles. 51. 2M3'.5U". 55. tan-i-OOa. )-26. t'3 feet. ilcs an hour. 7°. 47'. 50". 69-4302 yds. I' ' 'i ¥: K: APPEISTDIX. 1. To/nd the trigonometrical ratios for art angle of 18". Take the figure and constniction used by Eucl. iv. 10 in describing an isosceles triangle having each of the angles at the base double of the third angle. Hence I BAD^^ of 2 rt. I'^^of 1S0° = 36°. Bisect BAD by AIiJ, which will bisect BV at right angles. Let AB = m, AC^n, and .-. BD^^n. Then, since rect. AB, 7^6'= sq. on AG, m (ni ~ n) = n^ ; n^ rm^ .'. m'^-mn + 4 ' .. m = —~- n; . n 2 in Jtj + 1 APPENDIX. 223 n J 0/ 18'. Nowsin 18° = sin7?^ff=:?^=£^l £. /. sin 18°=- ,-^- . . . multiplying numerator and denominator by V5 - 1, sin 18° = . ^^^~ J: So the other ratios may be found. 2. To explain geometrically ichy in determining sin ^ or cos ~^from cos A, we get two, but from sin A Mr different values. PAM i ^T ""1 """f 'I^"'' '"'^*"" '' ^"°^^'"' '-^^^^^ describe PAAI, the least angle which has the given cosine. UakeP'AAl = PAM. cl. IV. 10 in lie angles at lit andes. Then A must be an angle whose bounding lines are AM and either^P or AP^, that is, it must be either the an f ffAP or the angle MAP', or some angle ionned by addhu (or^knig) a multiple of ibur right angles to (or ij.;^'':^ Q^Z J^T ''''^ ''''''' "^'^ "^'^ -' Vro^n.. QA, Then ^ must liave AM and one of the four QA, R'A, RA. QA for Its bounding lines. iH-J';,; I !^ ■; Mr"' . If'.!' 224. APPENDIX, Now the ratios of all tlioae angles are of the same magnitude and can only differ in sign, there being a pair of angles which nave each ratio + and a pair — . / :. in determining from the cosine we get two values. Next, let A be an anp;le whose sine is known, and describe MAF the hiast angle which has the given sine. Make MAF equal to the BUi)plement of MAP, i. t Then A must be either MAT or MAP', or some angle formed by adding (or taking) a multiple of four right angles to (or from) either of these. Bisect these angles as before Then ^ must be an angle which has MA and one of the four QA, RA, Q'A^ B'A for its bounding lines. Now those which have either QA or RA as one of their boundaries have ratios equal in magnitude but opposite in sign. So also for tho.se having Q'A or R'A as onu of tiieir boundaries ; but the magnitudes of the ratios of the former Sets diilcr from tliose of the latter. .*. the rati'^s o*f ^ mny be oither of two sets of magnitudes and of either sign, md th'>refore hava four dilFerent values Line magnitude : angles which values. • [1, and describe r 3 • • — M • or some angle ur right angles and one of the as one of their hut opposite in as one of their »s of the former 8 of maj];nitudes erent values