N-J^^^J^ >^w^K^^^(lfe%^j!yvv;,,:; ^Vij\J^ LlliKARV OF THi: L.MVKKSITY OF CALlFOKNFX. C.IHT OP MISS ROSE WHrriNG. %eceived September, i8g6. Accession No.^S^id * Class y\ V^^y^^^V^^^**^*^* r. y/yMy:^y ^^?^^teB?!^*i^ '^^^"^^^%i^-^;- w»^W^vw^,u^,r^^^^^^^ ^mm^^mmmi^^ j'i'i^Y v^.. w; uu Vw^^w"^^ wV^v^v ^^^Uc '^^ •. , w' w ti ▼ ^ 'i^;^.iy^-^^'v*^w "'^vuw^yyiJ'iSS 9M(«^i V "VV A^^S/S3i v^V^^^ ■^zt^z^z. vvvw, ^vtii^ ^v¥¥wi^wr; y*?ir«^w^»^. ^;srf^ "ys^r ^kittU^ ,^V w^ ^^^V^VyV;.;./,^^^ fUA.^^^V^^C V V V Y V ▼ ^..^.'Vu'V.VvwJ^. VW/.^-^^V^i Vy ^^M^t fe^" J'^^.-.-W'^w^t/uwLCM ^■^^^1J>;««¥^' Jw^i^^^^^^V A^iV'^WW:^ VWl>*W :«Wi^^- ^VV Vl,/ JVVVVVV; '''*-vv:.?«igx;s;:r;w|«^:' v^'vyvvw'' ,w-^^*^^r'vvv/***vv^'w^. '^^:w^; M.'^;ji.vvs^<^^«.'>^;^ '^VVyv^^^ W^^^''""' y\j^y JJ^^.r^'^^^' :^.'rr!:'!'^*v%v»vw/,^^^^^^^^ ^^*****^«*'^^ ^^V.;^'.^^t "m^ ^J^-^JO^^^.Vvy^^' vvvvvW^^ Fy^VWwSBwS vwvvwv„s^.';jt,^r;y,yy;;H^ Digitized by the Internet Arciiive in 2007 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementsofplanesOOthomrich ELEMENTS PLANE AND SPHERICAL TRIGONOMETRY, >\1TU THE FIRST PRINCIPLES ANALYTIC GEOMETRY. BY JAMES THOMSON, LL.D. PftOFESSOK OF MATHEMATICS IX THE UNIVERSITY OF IJI.ASOOW. FOURTH EDITION, IVIITBRSITr VARIOUS ADDITIONS AND IMPROVEMENTS. LONDON: SIMMS AND MaNTYRE, ALDINE CHAMBERS, PATERNOSTER-ROW AND DONEGALL-STREET, BELFAST. 1844. IS- !,3j2-^' ADVERTISEMENT. The first edition of this work was intended chiefly as a Text-Book for the use of the students in the Belfast Institution, when the Author was Professor of Matiiematics in that establishment ; and it was therefore written, not as a regular and complete treatise on Trigonometry, but as an outline to be filled up, and illustrated orally in the Lectures. It was received, however, by other readers in a more favourable manner than the Author could have anticipated from its nature and form ; and, in consequence of this, he has been induced, in the subsequent editions, to make various alterations, and, it is hoped, im- provements. The investigations, though still concise, are given at such length as to be easily understood by readers of ordinary talents and attainments ; and various interesting additions are introduced, some from the best recent works on the subject, and others that have occun-ed to the Author himself. Of the latter kind are the improvements in the numerical resolution of tri- angles, established in Nos. 57, 58, 73, and 74, and exemplified in Nos. 62, 63, and 110 ; which, with the modes of operation previously known, seem to render the subject as simple and easy as can be desired. In the present edition, also, a scholium of considerable interest will be found at the end of the third section : and the last four pages of the ninth section contain some curious propositions in Spherical Geometry, most of which the Author believes to be new. It has been everywhere the aim of the Author, to comprise in a smaJl com- pass much useful and interesting matter ; and, whatever may be the imper- fections of the work, he trusts that the person who shall make himself well acquainted with what it contains, will find it easy to acquii'e a knowledge of all that is yet known in Trigonometry, and to apply it in Astronomy, and other branches of science. In the present edition, the sines, tangents, &c. are defined as mere num- bers or ratios. This mode of representing them has been in use for some time in the University of Cambridge ; and it is attended with considerable advantage, particularly in the application of Trigonometry in Natural Phi- losophy. Should any persons prefer the common mode, they may have recourse to the Note at the end of the volume. Glasgow College, July, 1844. CONTENTS. rnge Section I. — General Principles 1 Formulas regarding a Single Angle Nos. 11, 16, 17, 18, 27, 29, 32—39, 48 Two Angles Nos. 22—26, 28, 30 Tliree Angles No. 31 Particular Angles Nos. 40— 47 ^'E.C'ilo:^ 11.— Resolutwn of Plane Triangles 16 Investigation of Principles Nos. 54—59 Examples of Computation Nos. 60 — 63 Measurement of Heights and Distances Nos. 66—68 Exercises in Plane Trigonometry 25 Section III. — Theory of Splierical Trigonometry 25 Investigation of General Formulas Nos. 70 — 83 Rightangled Triangles Nos. 84, 85 Segments of the Base and Vertical Angle made by a Perpendicular Nos. 86 — 91 Extension of the Theory of Spherical Trigonometry 38 Sec tion IV. — Resolution of Spherical Triangles 40 Napier's Rules for Rightangled Triangles 45 Examples of Computation Nos. 110—115 Exercises in Spherical Trigonometry 50 Section V. — Miscellaneous Investigations 51 Symmetrical Formulas Nos. 124 — 137 Inscribed and Circumscribed Circles Nos. 139 — 147 jirea of a Spherical Triangle Nos. 148—154 Spherical Loci Nos. 155, 158 Methods of resolving certain Rightangled Triangles Nos. 157—161 Comparison of Plane and Spherical Triangles Nos. 1 62 — 1 64 Section VI. — Astronomical and Geographical Problems and Exercises . 67 ^-EGTioy Yll.— Dialling 70 Section VIII. — Multiple Arcs 85 Section IX. — Miscellaneous Propositions 91 Propositions respecting Plane Triangles Nos. 205—210 Astronomical Problems Nos. 211—216 Summation of Trigonometrical Series Nos. 217—2 1 9 Auxiliary Arcs No. 220 Solution of Quadratic Equations No. 221 Cubic Equations Nos. 222—224 Newton's Series' for the Sine and Cosine of an Arc No. 225 Trigonometrical Surveys, Legendre's Theorem, &c Nos. 226—228 Propositions iu Spherical Geometry Nos. 229—239 Section X. — Questions for Exercise • 1<*9 Section XI.— Analytic Geometry H-l^ First Principles Nos. 240-251 Applications of the First Principles Nos. 252—254 Transformation of Co-ordinates No. 22o Interchange of Rectangular and Polar Co-ordinates Nos. 256—258 NoxK 12<5 ELEMENTS OF TRIGONOMETRY.* I.— GENERAL PRINCIPLES. 1. Each of the four parts into which a circle is divided bv two dia-i meters intersecting each other at right angles, is called a quadrant. If one of the four right angles be divided into 90 equal parts by radii of the circle, each of the parts is called a degree. The parts into which these radii divide the arc of the quadrant are (Euc. III. 27) all equal, and they are therefore called degrees of the circle. A sixtieth part of a degree is called a minute; and a sixtieth of a minute, a second.i Degrees, minutes, and seconds are denoted by the charac- ters, 0, ', ". 2. It is proved by writers on geometry, that the circumferences of different circles are proportional to their radii ; and that, in the same circle, angles at the centre are proportional to the arcs on which they stand. Hence, if from the vertex of any angle A (Jig. 1) as centre, two circumferences BCD and B'C'D', be described, cutting one of the lines forming the angle in B and B', and the other in C and C, the ratio of the arc, BC> to its radius, AB, is equal to that of the other arc B'C to its radius AB'. This readily follows from the prin- ciples stated above: for, since, by No. 1, and Euc. I. 13, cor., all the angles about A amount to 360*^, we have, by the second of those principles, BC A , B'C A , BC B'C =777:7—, and -f^7777iT/ = 77777rr; whence BCD—atlOo' BCD' 3600' "»v...v.v. j^qj)— g/^/jy * Trigonometry, in its primitive meaning, is that branch of mathematical science, which determines certain sides or angles of a triangle from others that are known. It is of two kinds, Plane and Spherical: the fonner treating of triangles described on a plane ; and the latter, of those on the surface of a sphere. The principles of trigonometry, however, are now of far more gene- ral application, furnishing means of investigation in almost every branch of mathematics. f In some modern French works on mathematics, the centesimal division is adopted instead of the sexagesimal ; the right angle, and conse<][uently the qua- drant, being divided into 100 degrees; the degree, into 100 mmutes; and the minute, into 100 seconds. This division, however, is likely to fall into disuse. A 4: ELEMENTARY ILLUSTRATIONS. nery, may be supposed to revolve again and again about A. Hence A will be one right angle, when AC {fig, 2) coincides with AG two right angles, when it coincides with AF ; three, when with AH four, when having completed a revolution, it again coincides with AB five, when it falls a second time on AG ; and so on. Now, when the line AC {fig. 2 and 4) lies in the first or second right angle, BAG or GAF, the line CD, or rsinA, is on the side of BF which is towards G ; but {fig. 5 and 6) in the third and fourth right angles, FAH and HAB, it falls on the other side of BF. In the first and fourth right angles, the line AD, or rcos A, is a part of AB ; but in the second and third, it is a part of AF. In the first and third right angles, the line BE, or rtanA, is on the side of the point B which is towards G : but in the second and fourth, it is on the other side. In the first and third right angles, the line GL, or root A, lies on that side of the point G which is towards B ; but in the second and fourth, it is on the other side. Lastly, in the first and fourth right angles, tlie line AE, or rsec A, passes through C, the termina- tion of the arc BC ; while, in the other two right angles, it lies on the opposite side of the centre with regard to C : and, in the first and second right angles, the line AL, or rcosecA, passes through C; while, in the third and fourth, it lies on the opposite side of the centre. 10. It will be seen, also, tliat when, AC coinciding with AB, the angle A is nothing, the line CD is also nothing; but that, when A is a right angle, CD coincides with the radius AG, and is equal to it. In like manner, when A takes the successive values, two, three, and four right angles, CD becomes successively nothing, AH, and nothing. By dividing these, therefore (No. 6), by the radius r, and by carrying out the same principle with regard to angles still larger, we find that the sine of any angle which is nothing, or two right angles, or four right angles, or any even number of right angles, is nothing ; and that the sine of any odd number of right angles is 1. In like manner we should find, that the tangent of nothing, or of an even number of right angles, is nothing ; and that the secant of any of the same angles is 1. If, again, A be supposed to increase by the approach of AC {fig. 2) to coincidence with AG, the lines BE and AE will conti- nually increase, and may be made as large as we please. When, however, AC coincides with AG, it will be parallel to BE, and wiU therefore never meet it. In this case, BE and x\E are said to be infinite ; and therefore (No. 7), dividing them by r, we get an infi- nite quotient ; whence it appears, that the tangent and secant of a right angle are both infinite ; and the same will evidently be the case FORMULAS REGARDING A SINGLE ANGLE. 5 when A is composed of any odd number of right angles. It would appear in a similar manner, that the versed sine of nothing is nothing ; of one or three right angles, 1 ; of two right angles, 2 ; and of four right angles, nothing ; and it is easj to trace similar relations regarding the cotangent, cosecant, and coversed sine.* 11. By eliminating r, and the lines, CD, BE, &c. from the expres- sions in Nos. 7 and 8, we obtain the following trigonometrical formulas, which are of much importance : r cos A sec A = 1 (1), sin A cosec A z= 1 . . . (2), sin A *^"^^=c"3n: (^)' i cos A rtanAcot A=:lt (5), I sin2A-j-cos2A=:l (6), ■4sec2A = l-ftan2A (7), . ;Cosec2A = l-|-cot2A ...(8). To investigate the first of these, let us take the products of the members of (c) and (d) : then, r-cos Asec A = AD.AE. Now, AD. AE is equal to r^ : for, in the similar triangles, ADC, ABE (Jig. 2, 4, 5, or 6), we have AD : AC : : AB : AE ; whence, since AB and AC are each equal to r, we get (Euc. VI. 17.) AD.AE = ^2. Hence, the expression formerly found becomes r^cos Asec Azzrr^; and from this, by dividing by r^, we get (1). Formula (2) is obtained similarly from (a) and (/), by means of the similar triangles, ADC, AGL. To investigate (3), we might .find it very easily by dividing (a) by (d) ; or we may take the product of formulas (6) and (d), then * From what is pointed out above, hi connexion with what will be esta- bhshed in No. 11, it will appear that some of the values of the sines, tangents, &c. above mentioned, are positive, and some negative : and the student will have no difficulty in seeing, that if n be any number in the series, 0, 1, 2, 3, &c. sin A and tan A will be nothing, when A consists of 2n righjt angles : that cos A and cot A will be nothing, when A is 2^4- 1 right angles; tliat sin A is 1, when A is equal to 4^4-1 right angles: and to — 1, when A is 4w-}-'3 right angles : that cos A is 1 when A is 4n right angles ; and — 1, when A is 4w-f-2 right angles: that tan A and sec A are each equal to -j- os, when A is equal toin-}-! right angles ; and to — oo, when A is 4n-j- 3 right angles, &c. f Hence we have cotA=r — r. Multiply both by any quantity a; then a cot A = — -T-. From this it appears, that if a quantity be divided by the tan- gent of an arc, the quotient is the same as the product that would be obtained y multiplying the same quantity by the cotangent of the arc ; and it would be shown in a similar manner, that to divide by the cotangent is the same as to multiply by the tangent. It is evident, also, that there is the same ntutual relation between the cosine and the secant, and between the sine and the cosecant ; and, universally, between any quantity and its reciprocal. O ADDITIONAL ELEMENTARY PRINCIPLES. r2cosAtanA = AD.BE. Now, from the triangles ADC, ABE, we get AD : DC : : AB : BE : and, conseqiientlj, AD.BE=r.DC = r2 sin A, because, by (a), DC = r sin A; and, therefore, r^cosAtanA rzrr^sinA; whence (3) is obtained by dividing by r"^ and cos A. We find (4) in exactly the same manner from (a) and (/), by means of the triangles ADC, AGL, in connexion with formula {d). The easiest mode of finding (5), is to take the products of the members of (3) and (4). If we now add together the squares of the members pf (a) and (cZ), weget r2 sin2A4-r2 cos2A= AD2-f CD2. But (Euc. I. 47) AD24-CD2=AC2 = r2: and therefore r^sin^A+r^cos^A = r2 : whence we get (6) by dividing by r^. To find (7), we square the members of (6), and add to the first of the results AB2, and to the second, what is equivalent, r^ : then AB-'-j-BES or (Euc. I. 47) AE2, or, by formula (c), r2sec2A=rr2-f rHan^A ; whence we get (7) by dividing by r^. In the last place, (8) is found in a similar manner from (e), by squaring its members, adding to each AG 2, and applying to the results Euc. I. 47, and formula (/). - 12. For the purpose of extending the application of analytical for- mulas, it is often necessary to considbr the sines, tangents, &c. of angles which are greater than four right angles. Thus, we may con- sider the hue AC {fig. 2 &c.), as having revolved round A once or oftener, and having described the angle BAC besides. In this view it is evident, that in the fifth, ninth, thirteenth, &c. right angles, the sines, cosines, tangents, &c. will be the same as in the first ; in the sixth, tenth, &c. the same as in the second; the line AC always occupying the same position after the addition of four right angles. 13. To render the formulas which express the relations of sines, cosines, &c. in the first right angle, applicable in expressing the same relations in the others,* the sine is to be regarded ^s positive, when * Thus, in the first right angle, versinA=:l — cos A, which formula will Jiold true also in the second and third right angles, on the supposition, that in them the cosine is negative; and a similar illustration may be given by means of the coversed sine. " Let us lay down, then, this general principle, which is of great utility : All trigonometTical fm'mulas shoidd be fm'iTied for positive angles which do not exceed 90"; since they will serve equally for angles that are greater than 90°, and for ijegative angles, by merely making the proper changes on the signs of the trigonometrical quantities." — Cagnoli, Triijonometrie, 77. From this conventional mode of expressing difi'erence of position by the use of different signs in the analytical expressions for quanti- ties, much advantage results in extending the application of formulas both in trigonometry, and in other pai-ts of mathematics. Without this ai'tifice, in the instance already given, we should sometimes have versin A=l — cos A, and somethnes, versin A=l -j-cos A. SIGNS OF THE TIlIGONOMETllICAL QUANTITIES. / CD, the line to which it is proportional, lies on one side of AB, as towards G ; but negative, when that line lies on the other side. In like manner, the cosine is to be regarded as positive or negative, accordingly as the line AD, to which it is proportional, is a part of AB or AF. Hence, also, since (No. 11) tanA=:^ — 7, cotA = ^ ^ cos A ^0^^ 1 A 1 A A 1 -x • - — 7- =r r , sec A = -, and cosec A = . . , it is easy to trace sin A tan A cos A sm A "^ the mutations of the signs of all these quantities ; and it will appear, that the sine and cosecant are positive in the first and second, fifth and sixth, ninth and tenth, &c. right angles, and negative in the others ; that the tangent and cotangent are positive in the first, third, fifth, seventh, &c. right angles, and negative in the others ; and that the cosine and secant are positive in the first, fourth and fifth, eighth and ninth, &c.;* the tangent and cotangent changing their signs at the end of each right angle ; and the sine, cosine, secant, and cosecant, at intervals of two right angles. 14. The angle BAM (Jig. 2) lying on the opposite side of AB from the angle BAG, may be regarded as negative in relation to BAG taken as positive; and, from considering the sine, cosine, &c. of this negative angle, it will be obvious, that sin( — A)= — sin A; cos ( — A) =r cos A ; tan ( — A) = — tan A, &;c. 15. If (Jig. 2)t the angle FAN be made equal to BAG, and NO '\ be drawn perpendicular to BF, it follows (Euc. I. 26), that NO is-^l equal to GD, and AO to AD ; the two latter, however, lying in oppo- '"^ site directions. Now, if NO and AO be divided by the radius, the quotients (Nos. 6 and 8) will be the sine and cosine of the angle BAN, which (by construction and No. 4) is the supplement of BAG. Hence, it is evident, that the sines of A and its supplement are equal ; and that their cosines are also equal, but have contrary signs. * It is easy to see by an inspection of the diagram, that these signs corres- pond strictly to the positions of the lines GD, AD, BE, and GL, to which the sines, cosines, tangents, and cotangents, are proportional. That the same is the case in relation to the secant, will appear fi-om considering the circle to be described by a revolving radius commencing its motion from AB. In the first, fourth, fifth, &c. quadrants, the line AE, which is proportional to the secant, and this radius, will lie on the same side of the centre ; but in the second, third, sixth, &c. on opposite sides. A similar illustration is appli- cable with regard to the cosecant. t Everything stated here, as well as in several other cases, will hold equally in figures 4, 5, and 6 : the sole difierence in the present instance being, that in figures 5 and 6, the supplements would be negative. 8 ARCS WHICH HAVE THE SAME SINE, COSINE, ETC. If, therefore, ^ * be put to denote the quotient obtained by dividing the semicircular arc BGF by the radius AB, which quotient (No. 2) is the measure of 180°, we shall have sin A =:sin(cr — A); cos A = — cos('r — A); and consequently, by formula (3), tan A = — tan (t — A), &c. Hence, also, the sine of one angle of a triangle is equal to the sine of the sum of the other two. 16. Since (Nos. 14 and 15) sin A= =t sin (=±r A)=sin (t A); by adding to A a multiple of 2-3- (four right angles), we get (No. 12) sin A=:r±:sin(2n';r=tA)=:sin[(2n4-l)^— A} (9). In this, n is any number in the series 0, 1, 2, 3, &c., or 0, — 1, — 2, &c.: and hence any sine is the sine of an infinite number of angles. 17. In like manner, it would readily appear, that cosA = cos(2nTztA) = — cos{(2n + l)'7— A}...(10). 18. By taking n = — 1 in the concluding parts of (9) and (10), we get sin A = sin ( — cr — A), and cos A = — cos(— cr — A); or (No. 14) sin A = — sin (cr + A), and cos A = — cos (cr -f A). Dividing the former by the latter, we have, by (3), tan A = tan (t -|- A) : whence it appears, that if two angles differ by -s-, their tangents are equal. By adding, therefore, n-7r to A, and also to — A, we get, by Nos. 12 and 14, tan A = tan (n cr + A) = — tan(?ivr — A) .'...(11). Hence, also, it is plain, that (No. 13) cot A = cot (n cr -f- A) = — cot (n ^ — A). 19. It follows from Nos. 6, 7, and 8, that the sine, cosine, tan- gent, cotangent, secant, and cosecant, of an angle, are simply the ratios of the sides compared by pairs, of a righ tangled triangle, which has that angle as one of its angles. Thus, if ABC {fig. 7) be a triangle right- angled at C, we have at once, by Nos. 6 and 8, AB : AC : : 1 : sin B, and AB : BC : : 1 : cos B ; and also AC = AB sin B = AB cos A, since B is the complement of A. It thus appears, that the hypotenuse is to either leg, as 1 is to the sine of the angle opposite to that leg, or to * For the use of those who are unacquainted with Greek, the following list is given, containing the capital and small letters of the Greek alphabet, with their names, and the letters, or combinations of letters, to which they are respectively equivalent in Latin and English ; A, «, alpha, a,- B,/S or C, beta, 6.- T, y, gamma, ^ ; A, 5, delta, d ,- E, t, epsilon, e (short); Z, C or i, zeta, z ,- H, r,, eta, e (long); 0, 6 or ^, tbeta, ih,- I. /, iota, t,- K, x, kappa, k ,- A, A, lambda, / ; M, /*, mu, m; N, y, nu, n,- S, |, xi, x; O, a, omicron, o (short); 11, a- or w, pi,p; P, J, rho, r,- 2, 0- or ; , sigma, s ; T, t, tau, <,• T, y, upsilon, m or y,- *, (p, phi, /or pA,- X, Xi chi, eh; ■*", '4', psi,ps; il, u, 6mega, (long). SINES OF THE SUM AND DIFFERENCE OF TWO ARCS. 9 the cosine of the adjacent angle ; and that each leg is equal to the pro- duct of the hypotenuse and the sine of the opposite angle ^ or of the hypotenuse and the cosine of the adjacent angle. 20. We have, in like manner, by No. 7, BC : CA : : 1 : tanB, and BC : B A : : 1 : sec B : that is, one of the legs is to the other, as the radius is to the tangent of the angle opposite to the latter ; and one of the legs is to the hypotenuse as the radius is to the secant of the contained angle. o. 21. Again, let ABC {fig. 8) be any triangle, and CD its perpendi- /N cular: then (No. 19) in the rightangled triangles ACD, BCD, CD = ^^ | ACsin A=BC sinB; whence (Euc. VI. 16) AC : BC : : sin B r^^'^'"^^ sin A ; that is, in any plane triangle, the sides are proportional to the . 1 . - „ , AC sinB sines of the opposite angles. Hence, also, ^^ = - — -r. 22. One of the most important problems in trigonometry, is that in which it is required to find the sine of the sum of two angles, in terms of the sines and cosines of the angles themselves. To investi- gate this, let BAC and DAC (Jig. 9), which, for brevity, may be called 6 and^', be the two angles; and through any point C in AC, draw BCD perpendicular to AC, and meeting AB, AD in,B and D. Then (No. 19) BC = AB sin>, and CD = AD sin^; and therefore BD = AB sin & + AD sin &'; , , AB . , , AD . ,, , . whencel = — sm^-f — sin^' (gr) 13 ^ /XT oi\ -^^ sin ADC cos^' But (No. 21) and BD"" sin BAD - sin (^+^0' AD sin ABC cos d BD ~ sin BAD "" sin (^+0* By substituting these in equation (^), and multiplying by sin Q-\-&'), we get sin {&■{-&') = sin ^ cos &'-\- cos & sin &', the required formula ; or putting A for & and B for &', sin(A-f-B) = sinA cosB + cosA sinB.... (12) Hence, to find the sine of the sum of two angles, multiply the sine of each hy the cosine of the'^ other, and add the products together. From this, taking B negative, since (No. 14) sin ( — B)=: — sin B, and cos ( — B)= cos B, we get sin(A — B)=:sinA cosB — cosA sinB (13) Hence, to find the sine of the difference of two angles, from the pro- B 10 FORMULAS REGARDING TWO ANGLES. ditct of the sine of the greater and the cosine of the less, take the product of the cosine of the greater and the sine of the less. 23. By No. 3, cos(A-|-B)= sin(^ -r— -A— B>= sin J(iT— A) — B]. Now, if we take Irr — A as a single angle, it is the complement of A, and we have (No. 8) its sine = cos A, and its cosine = sin A. Hence, therefore (No. 22), cos(A-fB) = cos AcosB — sin AsinB (14)* In this, change the sign of B : then (No. 14) cos(A — B)=:cos AcosB+sin AsinB (15) 24. Take the sum and diflFerence of (12) and (13)t, and also (.f .(14) and (15): then sm(A + B) + sin(A— B) = 2sinAcosB (16) ^ sin(A-}-B)— sin(A— -B) = 2cosA sinB (17) cos(A — B)+cos(A + B) = 2cosAcosB (18) cos(A— B)— cos(A+B) = 2sinA sin B (19) 25. Let A4-B=S, and A— B = D; then A=:KS + D), and * This formula may also be investigated in the following manner : Retain- ing the same construction (fig. 9) as in No. 22, draw AE perpendicular to AD, meeting DB produced in E. Then the angle E = ^, each being the complement of E A C . Hence ( No. 1 9 ) E C = E A cos ^'. Then, because ( No. 19) BC= AB sin 0, we have ^ EB=EAcos^'— ABsin^; , EA ^ AB . ^ whence l=^prn cos^ — ^^rs sm^ (h) „,EA sinEBA sin ABC cos^ ^ AB sin E sin<^ But>:f^vrs= • T>AP = - — liT A T> = — TTT-^'f and *EB~smEAB~sinEAB~cos(^4-<^)' EB~sin EAB~cos(^ * The followmg system would be found by using as denommators the sum and difference of the cosines, and modifying them by (22) and (23): sin (A+B) _ sin|(A+B) cosB-f-co8A~~cos^(A — B) sin(A-~B) cosHA— B) ^2) cosB-~cosA~'sini(A-|-B) ' sin(A4-B) _ cosHA+B) cosB — cosA sin^(A — B) sin(A~-B) sin|(A--B) , . cosB-f cosA" cos|(A-f B) ' ^ ' t Formulas (40) and (42) are easily derived from (39) and (41), by changing B into —B. FORMULAS REGARDING TANGENTS. 13 x^A , -ON cot A cot B — 1 /^l\* ^ COt(A + B)=: -^—1 r-T- (41)* ^ ^ cot B -h cot A _. cotAcotB-j-l ,.n\ cot(A-B)= ^„,B-cotA (*2) 31. The tangent and cotangent of the sum of three angles, A, B, C, may be readily derived from the first and third of the preceding formulas, by regarding A+B as a single angle. In this way we have, first, ,* T^ . rix tan(A + B) + tanC tan(A + B + C)== — , 7a !Z\^ n > ^ ^ 1 — tan(A + B)tanC and then, by substituting for tan(A + B) its equal, according to (39), and multiplying the numerator and denominator of the result by 1 — tan A tan B, we obtain /A_i_-R_LrN— ^^"^ A + tan B + tan C — tan A tan B tan C . . „. tan ( A + B + O) _ i_tanAtanB— tanAtanC— tanBtanC"^^ By a similar process we should find, that cot A cot B cot C — c ot A — cot B — cot C ,... cot ( A + B + 0) = -^^^j^^^j^ cot A cot 4- cot B cot C — 1 "^ K It is easy to see, that formulas for the tangent and cotangent of the sum of four or more angles, might be derived in a similar manner. 32. Take B = A in (39) and (41), and B=C = A in (43) and (44); then . ^ , 2tanA ..^. tan2A== — : — ;-. (45) 1— tan^A -^^=1:^ (^«) -3A=?^|^^=S^ m _. cot^A — 3cotA ^^^^^= 3cot^A-l •-. (^^) 33. By dividing (32) by (30), we obtain 1 — cosA ^ , . ,.^v **| — ^—r— = taniA (49) ] smA ^ ^ f^ Hence, also (No. 11), cosec A— cot A = tan \ A. (50) 3 * This may be obtained by taking the reciprocals of the members of (39), and multiplymg the numerator and denominator of the second member by cot A cotB. In a. similar manner (42) might be derived from (40). 14 FORMULAS REGARDING ANGLES. 34. Double the members of (46), and perform the actual division in the second ; then (5), r 2cot2A = cotA—tanA (51) 35. In (50) change A into 90°— A; then sec A— tan A = tan (45°—^ A) (52) 36. By No. 11, cosec A + cot A = J- + ^ = i^^^^, or smA smA sin A by (31) and (30,) and by contraction, 5<[ cosecA-j-cotA = cot4 A (53) 37. In this change A into 90° — A ; then secA-f tanA = cot(45°— i A) (54) 38. Since the arcs 45° — |A and 45°+ 1 A are complements of each other, (52) and (54) will become secA— tanA = cot(45°-|-iA) (55) sec A 4- tan A = tan (45° -1-i A) (56) 39. By taking the difference of formulas (56) and (52), we obtain 2 tan A = tan (45° + ^ A) — tan (45°— I A) ; and therefore tan (45° 4-1 A) = tan (45°— i A) +2 tan A (57) 40. The sines, cosines, &c. of some angles bear remarkable relations to unity ; and therefore merit consideration. We have already seen that cosier, or cos 90°= 0. Hence if A be taken=90° in (31) and (32), we have l = 2cos245°=2sin2 45°, and therefore [ sin45°=cos45°=vT=W2 (58) Hence, also, by (3), (4), (1), and (2), tan 45°= cot 45°= 1 (59) sec45°=cosec45°=V2 (60) 41 . Again, since 60° is double of 30°, and is also its complement, we have, by (30), and by No. 8, sin 60°= 2 sin 30° cos 30°, or sin 60°= 2 sin 30° sin 60°. Hence, dividing by 2 sin 60°, we obtain ■ [ sin 30°, or cos 60°= I (61) From this, also, we derive by (6), cos30°, orsin60°=Vi=W3 (Q2) These values give, by means of (3), (4), (1), and (2), tan30°=cot60°=Vi=W3 (63) cot 30°= tan 60°= V 3 (64) FORMULAS REGARDING PARTICULAR ANGLES. 15 sec30o=cosec60°=V|-=fV3 (65) ^ cosec30o=sec60°=2 {66) _i 42. We have (37) and (38), sin 720= 2 sin 36° cos 360, ^nd 008 720=20082360—1. Hence, by (12), sinl08o= sin(360 4-72o) = sin36o cos72o-j-cos36o sin 720, or, by (38) and (37), since 72o= 360x2, sinl08o= 2sin36o 008^360— sin 360 4-2 sin 360cos236o = 4sin 360 0082360— sin36o. But, since 720-f 108o=180o, we have, by No. 15, and by (37), sin 1080= sin 720= 2 sin 360 cos 360; a^^j therefore, by equalling these values of sin 108o, we obtain 2sin360cos36o=4sin36ocos236o— siu36o. Hence, by dividing by sin 36o, and by transposition, we get 4cos2360—2oos 360=1 ; the positive root of which equation is COS360, or sin54o=i(l-f V5) (67) 43. Hence (6) sin36o, or cos54o=i.V(10— 2V5) ... (68) This is half the side of a pentagon inscribed in a circle, whose radius is 1, since an arc of 36o is one tenth part of the entire circum- ference, and (No. 6) in that circle, the sine of any angle is half the chord of twice the corresponding arc. Hence the side of the pen- tagon is iV(10— 2V5) = ^^^:y^. 44. By (38), cos72o=2oos236o— 1 ; whence, and by (67), oos72o=sinl8o=J(— 14-V5) (69) This is half the side of the inscribed decagon. The side itself, therefore, is J(V5 — 1). 45. From (69), we obtain, by means of (6), sin72o=cosl8o=JV(10-{-2V5) (70) This is the half of one of the diagonals of a regular pentagon inscribed in the same circle. 46. By taking A=60o in (17), we have, by (61), after transpo- sition, and by changing B into A, sin (600 4- A) = sin (600— A) + sin A... (71) 47 By taking (69) from (67), and doubling the remainder, we find 2sin540— 2sinl8o=l; 16 RESOLUTION OF PLANE TRIANGLES. by multiplying which by cos A, there will be obtained, by (16), sin (54°+ A) + sin (54°— A)-~sin (18<'+ A)— sin (18°— A)=cos A... (72) We should also have a similar formula by multiplying by sin A, and applying (19). 48. Two formulas giving the cosine and sine of half an angle may be thus investigated : (cos A -f sin A) 2 = cos ^ A -f- 2 cos A sin A -}- sin 2 A =l-j-sin2A, by (37), and by No. 11; and therefore cosA+sinA = V(l + sin2A). In like manner, we should find cos A — sinA= V(l — sin 2 A); and by taking half the sum and half the difference of these, we obtain cosA = iV(l + sin2A) + iV(l— sin2A) (73) sinA = iV(l+sin2A)— iV(l—sin2A) (74)* 49. Formulas might also be investigated for the secants and cose- cants of the sum and difference of angles, &c.; and we might in- vestigate many more respecting sines and tangents. As much has been done, however, as is consistent with the nature of the present publication ; and the person who shall make himself well acquainted with the mode of investigating the formulas that have been here given, will find it easy to derive others. We shall now proceed, there- fore, to the resolution of plane triangles ^ or to investigate rules and formulas for calculating certain sides or angles of plane triangles, when others are given. IL— RESOLUTION OF PLANE TRIANGLES. 50. To assist in effecting calculations in trigonometry, tables have been constructed, containing the sines, tangents, and secants of all the angles, up to 90°, which differ by small equal intervals, usually of one minute; and, in trigonometrical computations, instead of the common numbers, the logarithms not only of the numbers expressing the lengths of the sides of figures, but also those of the sines, tan- gents, and secants of angles, are almost always employed, since the invention of those remarkable numbers. The theory of logarithms, and the method of computing tables t of them, and of sines, tangents, * Let the student consider for what angles the radicals, in this formula and the last, are to be taken positive, and for what negative, t To employ those tables without being acquainted vi'ith theu' nature and RESOLUTION OF PLANE TttlANGLER. 17 and secants, are given in the Treatise on the Differential and Integral Calculus, by the Author of this work. It may suffice here to say, that the logarithms of two or more numbers, are other numbers whose sum is the logarithm of the product of the numbers to which they be- long. Hence it follows, that the difference of the logarithms of two numbers is the logarithm of the quotient obtained by dividing one of the numbers by the other ; that the logarithm of the square of a num- ber is double of the logarithm of that number ; and the logarithm of its square root, half its logarithm. 51. It is easy to see (Euc. I. 4, 8, and 26) that, with the exception mentioned in No. 53, a plane triangle is determined, when, of its sides and angles, any three, except the three angles, are given. Hence we may divide the resolution of plane triangles into three cases : I. When a side and the opposite angle, and either another side, or another angle, are given ; II. When two sides and the contained angle are given; and, III. When the three sides are given. 52. The first case is resolved on the principle, that (No. 20) the sides are proportional to the sines of the opposite angles. 53. When, in this case, two unequal sides, and the angle opposite to the less, are given, the angle opposite to the greater may (No. 15) be either that which is found in the table of sines, or its supplement, unless it be known from the nature of the problem, whether it is acute or obtuse. This will readily appear from constructing the tri- angle ABC {fig. 11) having the angle B acute, and the side AC less than AB; as it will be seen, that if from A as centre an arc be de- scribed, with AC as radius, it will cut BD in two points C and C, either of which may be taken as the extremity of AC, and the two angles ACB and AC'B are evidently supplements of each other, the triangle CAC being isosceles.* Since the sine of an angle and the sine of its supplement are equal, a similar ambiguity would always exist when the quantity to be found is a sine, were it not removed by the nature of the triangle, or by some other circumstance. construction, is not strictly scientific. The student, however, will learn then* theory and construction with greater ease, when he shall have had more expe- rience in mathematical investigations: and those who prefer the more scientific mode, may have recourse to the Treatise on the Differential and Integral Cal- culus. In all the books of tables, the method of using them is explained. * Should the computation give the angle opposite to the greater side equal to 90°, there would be no ambiguity. In this case the arc C C would touch BD. Should the value of sin C be greater than the radius the solution would be C 18 PLANE TRIGONOMETRY, SECOND CASE. 54. The method which is generally best adapted for resolving the second case, may be thus investigated. Let A, B, C be the angles of a triangle, and a, h, c* the sides respectively opposite to them. Then (No. 21) « : 6 : : sin A : sin B; whence! a — h sin A — sin B a-\-b sin A + sin B' a — h tanJ(A — B) or, by (24), (75) a^h tanl(A-f-B) and therefore r ci + h : a—h :: tani(A+B) : tan4(A — B); in which analogy the third term, as well as the first and second, is given, since (Euc. I. 32) A + BzzlSO^— C. Hence, therefore, the angles A and B will become known, half their sum and half their difference being known. 55. The angles being found, the third side may bo calculated by No. 52. In practice, however, it is generally better to use one of the following analogies -.J r cos^(A— B) : cosi(A + B) : : a + h : c (76) I sin4(A-^B) : sin4(A+B) : : a—h : c (77) impossible, the data being inconsistent with one another, and having no triangle answering to them, the arc CC neither cutting nor touching BD, * To avoid ambiguity in the use of this veiy convenient notation, A, B, C may be read angle A, angle B, avigle C ; and a, b, c, side a, side b, side c. f For a-\-b -.a — b : : sinA-|-sinB : sinA — sin B, and consequently a — 6 _ sin A — sin B a 4~ ^ ~ sill A-}- sin B* t These formulas are given in the 13th page of Thacker's Miscellany, pub- lished in 1743, and their use in resolving this case of plane triffonometi-y, was first pointed out by the late Professor Wallace of Edinburgh, in the Edin- burgh Transactions for 1823. The following geometrical proofs of formulas '15, 76, and 77, may perhaps be prefen-ed by some to the proofs aheady given. Let ABC {fig. 12) be any plane triangle having a greater than b; and h-om C as centre, with a as radius, describe a circle meeting C A produced in D and E, and BA produced in F; join DB, BE, and CF, and draw EG- parallel to AB, and meeting DB produced in Gr. Then, because DC and CE are each equal to a, DA is equal to a^b, and AE to a—b. Also (Euc. L 32), DCB = A+B, and (Euc. III. 20) DEB = |(A+B). Again (Euc. L 32), A=ACF-f- ABE, in. 31 BO the tangent of BEG, or tani(A— B); and (Euc. VL 2) DA : AE DB : BG; that is, a^b : «— ft : : tan|(A-f B) : tani(A— B), which is the same as (75). Again, DBA is the complement of ABE, or ^(A — B) ; and D the complement of BED, or i(A-f-B). But, in the triangles, ABD, ABE, PLANE TRIGONOMETRY, THIRD CASE. 19 These are easily proved in the following manner. Since (No. 21) sin A a , sin B b - — 77 r= -, and - — r^ = - , smC c smO c we have, by addition and subtraction, and by substituting (No. 15), sin(A+B) for sinC, sinA + sinB a-^-h sin A — sinB___« — b sin(A+B)- "T"' ^""^ sin (A+B) ~ "T"' or, by (33) and (35), cos.^(A — B)_«+& , sin.! (A — B)__a — b c^|(A+B)-~^' ^""^ ^i(A+BJ-T" It may be remarked, that if the latter of the two formulas just found, be divided by the former, the quotient will be (75) : and thus we have a second and an easy mode of investigating that formula. 4 5Q. In the thircl case, the three sides being given, to find one of the angles, suppose A (fig. 8) ; from C (either of the other angles) draw CD perpendicular to the opposite side; then (Euc. I. 47) a" = CD2+DB2. But CD2=:62— AD3, and DB2= (c— AD)2=rc2_ 2cx AD+AD2=c2— -26c cosA+AD^ (No. 19); and substituting these in the foregoing equation, and contracting, we obtain a' =zb^-{'G^— 2 be cos A (78) b^+c^—a^ . , , Hence we have cosA=: ^^ ; — a formula by means of 2 be -^ which A may be determined arithmetically, though in general not easily, as it is not adapted to computation by logarithms. To find a more convenient formula, we get from (78), by transposition, 26ccosA=62 + c2— a2. Subtract the members of this from 2 be, and also add them to it: then 26c(l— cosA) = a2_(52__2Jc + c2)=:a2-_(5_c)2, and 2 6c(l4.cosA)=:&2^26c+c2— a2__(5_^c)2_^2 . or, by (32) and (31), and because the difference of the squares of two quantities is equal to the product of their sum and difference, 4Z)Csin2lA=r (a — b-^c)(a-\-b — c), and 4 6c cos 21 A= (a + b -{-(}) (6-{-c— a). we have (No. 21) sin DBA : sinD : : DA : AB, and sin ABE : sin AEB : : AE : AB; that is, cosi(A— B) : cosi(A-f B) : : ar\~b : c, and sini(A— B) ; smi(A-}-B) : : a— 6 : c; which are (76) and {11). 20 PLANE TRIGONOMETRX, THIRll CASE. By putting 2« — a -f- 6 + c, these become 46csin2iA=2(5— 6). 2(s—c), and 46ccos2iA = 2s.2(s— a)* Hence, by dividing bj 45c, and extracting the square root, we get the two formulas, -4A=.y(i=^)il=^) (79) , . /s(s — a) cosiA = y-i^ (80) 57. Divide (79) bj (80); then, by (3), tan^A.^ --^— -^ (81)t In like manner we should have Divide the members of these by those of (81) ; then taniA-s-^'^'^'^taniA-ili:;^ V^^>> 58. By multiplying the values of tan^B and tan^C in the last No. by the value of tan^ A in (81), we get tan i A tan i B r= ^•=^, and tanj A tan ^ C = ^^ ... (83) s s 59. Take twice the product of (79) and (80): then (30) sinA = |Vs(s-a)(5~6)(s-c)t (84) he * For since 2s—a-\-b-{-c, by subtracting 2 6 we obtain 2s — 2b=a — 6-{-c; and similar remainders would Ibe obtained by subtracting 2 a and 2 c. f Formulas Y9, 80, and 81 were discovered by William Purser of Dubhn, probably about the year 1632, or soon after. See AVallace's Geometrical Theorems, page 1. I Multiply both members of this formula by b; then 6 sin A, or, by No. 19 {fig. 8), the perpendicular CD =- V s {s—a) (s— 6) {s—cj. If this be multiplied by ^c, we obtain for the area of the triangle, V s {s — a) {s — b) {s — c), which proves the common but important rule for finding the area, when the three sides are given. If this be divided by s, half the perimeter, we find the radius of the inscribed circle ==,^/ ~ • The third case may also be resolved by drawing the perpendicular CD {fig. 8), and thus forming the two rightangled triangles ACD, BCD : for (Euc. I. 47) AC=»— AD='=BC^— BD»; w-hence, by transposition, we have BD=»— AD='= EXAMPLES OF COMrUTATION. 21 EXAMPLES OF THE RESOLUTION OF PLANE TRIANGLES. 60. Given a=13 yards, 6=15 yards, and A=53° 8'; to find the remaining parts of the triangle. 9-90311 9-93525 1-11394 Asa : : sin A 13 15 53« 8' 67° 23', or 112° 37'; (No. 53) the do 1-11394 1-17609 9-90311 As sin A :sinC :c By taking 53° 8' 59° 29' 13 : sinB this being 9-96526 ubtful case. 14 C=14° 1-14608 15', and A and a the same as in the preceding analogy, Hence (Euc. I. 32) we find c = 4.* C=59° 29', or U'* 15'. In these operations, to find the fourth term (No. 50), the second and third terms are added together, and the first is taken from the sum. This may be done more easily by adding together the second and third terms, and the complement of the first to 10, or what re- mains after subtracting its right-hand figure from 10, and all the rest from 9, which may be done by inspection, as we proceed with the ad- dition. Thus, in the second, we have 4 and 5 are 9, and 9 are 18 ; then 1 and 9 are 10, and 2 are 12, and 8 are 20, &c. When we use this method, we must reject 10 from the result. It is still easier, however, when the quantity to be subtracted is a sine, to use instead of it the cosecant diminished by 10 in its index, and then to add all the quantities together. The reason of this is evident from the second note to No. 11. In like manner, when a cosine is to be subtracted, we may add the secant diminished by 10 ; and when a tangent or co- tangent is to be subtracted, we may add in the first case the cotan- gent, in the second, the tangent, subtracting 10, either at first, or from the final result. The following method of resolving this case is perhaps preferable to BC^'— AC^, or (Euc. II. 5, por.) (BD-f-AD) (BD— AD) = {BC+AC) (BC — AC). Now, when the perpendicular falls within the triangle, BD-|- AD= AB ; otherwise (A being obtuse) BD — AD= AB : in either case, there- fore, one of these factors is given, being equal to the base, and the other will be determined by dividing (BC -f- AC) (BC — AC) by the one which is given, or by converting the equation (by Euc. VI. 16) into an analogy having the given base for its first term. Hence the segments of the base will be known, and then each of the rightangled triangles wiU be resolved by the first case. * These results may be thus obtained by the use of natural sines and num- bers; as « : 6 : : sin A : sinB ; that is, as 13 : 15 : : '80003 : '92311 = sine of 67° 23', or 112« 37'; whence C, as before,=59° 29', or 14° 15'. Using the first of these, we have sin A : sinC : : a iC; that is, -80003 : '86148 : : 13 : 14; while, by taking CsttU** 15', we should find, in a similar manner, c=4. 22 EXAMPLES OF COMPUTATION. that which is given above. From log sin A take log a, and the remain- der, 9-90311-~M1394, is 878917; add this to log 6=M7609, and the sum 9-96526 is log sin B as before. Then, C being found, from the logarithmic sines of its values take successively the same quantity 8 "789 17, and the remainders are the logarithms of the two values of c. The reason of this is obvious, since, in the first analogy, the first and third terms are a and sin A; and in the second they are the same quantities in a reversed order. 61. Given a = 57-38 miles; 5 = 42-6 mUes, and C = 56°45'; to resolve the triangle. By (75). Then, by (76). ^sa+ft 99-98 1-99991 Ascos|(A— B) ISMS'I 9-98i31 : a—h U-Y8 1-16967 :cos|(A+B) 61^37'^ 9-67691 ::tan|(A-|-B) 6P 37'^ 10-26750 ::a + 6 99-98 1-99991 tanHA— B) 15M8'| 9-43726 Hence, by adding and subtracting the half sum and half difierence, we find 49-26 1-69251 This might also be found by either of the following analogies, (77), and No. 52 : A=76« 56', and B=46° 19'. sini(A— B) : sin^A+B) : : a— 5 : c, and sin A : sin C : : a : c. If C be a right angle, the solution is effected more easily by means of No. 20, than by the foregoing method, as we have simply a : 6 : : radius ; tan B. ^2. Given a=:679, 6=537, and c=429 ; to find the angles. Here, by adding the three sides together, we obtain 1645, the half of which, 822-5, is 5. Then, by taking from this the three sides suc- cessively, we find s — a = 143-5, s — 6 = 285-5, and s — c = 393-5.* The rest of the work, the subtraction in the first part of which may be performed in the manner pointed out in No. 60, is as foUows : 6! — a 822-5 143-5 285-5 393-5 tan I A 44° 17'^ A = 880 35' 3-91514 2-15685 2-45561 2-59494 2)19-97856 9-98928 } subt. tan I A log(s— a) log(s— 6) tan|B 26° 7'i B=52°15' log(s— — c 679*8 666-5 61*8 51*5 tan^A 6<^ U'l A= 10° 23'. 2-832381 2-753200 1-790988 1-711807 2)17-917214 8-958607 } subt, tan I A logs log(s— c)-f 20 tan|B 39M8'| B=79<'37' log(s— 6)4- 20 tan|C 45« C = 90°. 8-958607" 2*832381 11-790988' 21-711807 9*920819 add subt. 21-790088"(^ 11-790988/®^^** 10-000000 64. No easier or better solution for this case can be desired, or perhaps found, than is afforded by either of the foregoing methods ; the taking of only four logarithms from the tables being necessary in the entire operation by either of the methods. It may be resolved, however, by means of No. 5Q or 59, or of the note to No. 59 ; and the learner, for the sake of comparison and of practice, would find it useful to resolve a triangle in aU the ways here pointed out. The method in No. 59, besides being tedious, fails in determining whether the angle is acute or obtuse, and is therefore useless in practice. No. 56 affords simple and easy solutions. Q5. Rightangled triangles are resolved by the first case, except when 24 HEIGHTS AND DISTANCES. the legs are given ; and then the resolution is most easily effected by the method pointed out in No. 20. These are more easily resolved than oblique-angled triangles, as the radius may always be one of the terms. Q6. As applications of plane trigonometry, we may consider some of the simplest and most useful cases of the determination of heights and distances. The height of an accessible object AB (fg. 13), such as a tree, a spire, &c. may be found by assuming a station C on the same horizontal plane with the base B, and measuring with a line, chain, &c. the distance BC ; and with a quadrant, theodolite, &c. the angle ACB, called the angle of elevation. There will then be given the right angle B, the angle C, and the base BC, to find the height AB ; and that will be computed by means of either of the following analogies; cos C : sinC : : BC : AB, or radius : tanC : : BC : AB. 67. The data necessary for determining the height of an inacces- sible object AB (Jig. 14), may be found by measuring the distance between two objects C and D, which are on the same horizontal plane, and in the same straight line with the base of the object ; and by measuring the angles of elevation, ACB, ADB. Then, as sin CAD (=ACB— ADB, by Euc. I. 32) : sin ADB :; CD : AC; and ra- dius : sin ACB : : AC : AB, the height required. In the compu- tation, the logarithm of AC will be found by tlie first analogy, and may be used in the second without finding AC itself.* By com- bining the two analogies, we have log AB = log CD -f log sin ACB + logsinADB— logsin(ACB— ADB)— 10. When the surface on which the measurement is made is not horizontal, its inclination must be measured, and must be employed in the calculation ; and in all cases, when the angle of elevation of the summit of the object above the horizontal plane, passing through the eye, is observed, the height of the eye must be added to the final result. 68. The following is one of the most useful cases of the measure- ment of distances. Let A and B (Jig. 15) be two objects whose dis- tance is to be found, and let the base CD, in the same plane, be measured. Measure also the angles ACB, BCD, BDA, and ADC. * It follows fi-om No. 1 9, that B C = AB cot C, and BD = AB cot D. Taking the difference of these we obtain CD=AB (cot D— cot C), and consequently AB^ CD ■ cot D — cotCl' a formula which gives an easy method of computing AB by means of the table of natural tangents. EXERCISES IN PLANE TRIGONOMETRY. ZD Tlien (by case I.) in the triangle ACD, calculate AD; and in the triangle BCD, calculate BD: lastly, in the triangle ADB (case II.), calculate AB. The operation may be verified by computing AC and BC, and thence AB. The other common cases of the measurement of distances will be found in the following exercises, and will present no difficulty. EXERCISES IN PLANE TRIGONO ME TPvY. 1. Given A= 90°, 6 = 37" 52', «= 170-6 ; requu-ed h and c. Answ. h— 104-72, c= 134-68. 2. A = 90°, B = 49« 18', c=:7§9. Answ. a=1210, 6=917-3. 3. A=90«, a=157-8, 6=100. Ansiu. B = 39" 19'i, c=122'07. 4. A=90^ 6=784-3, c=940. Answ. B = 39« 51', a— 1224-2. .5. A=68° 23', B = 62« 40', a=5000. Answ. h=^111% c = 4055-9. C. A=45^ a = 64-3, 6=57. Answ. B=38° 49', c = 90-4. 7. A=45», a=57, 6 = 64-3. Ansiv. ^= 52" 64'|, c= 79-844; or B = 127° 5'|, c=ll-091. 8. A=17° 18', 6=1376, c=149. Answ. B=160° 38'i, a=1234-l. 9. «=384, 6=512, c=201. Answ. A = 41" 6', B=118'' 46'. Ex. 10. Required the breadth AB {fig. 8) of a lake, the distance fi-om A to a station C being 24-36 perches, and the angles A and C being 91*' 32' and 69" 18' respectively. Answ. AB= 69-408 perches. Ex. 11. Required the distance between two houses A and B {fig. 8), on the opposite sides of a hill ; the distance from A to a point C, from which both are visible, being 168 perches, from B to C 212 perches, and the angle ACB= 34° 48'. Answ. 121-14 perches. Ex. 12. Suppose a base AB {fi{j. 15) of 24-36 chains to be measured for ascertaining the distance between two houses C and D beyond a river, and sup- pose the angles CAD, DAB, DBC, and CBA to be 64" 38', 51" 12', 70" 44', and 49" 50', respectively : required the distance CD. Answ. 132-93 chains. III.— THEORY OF SPHERICAL TRIGONOMETRY. 69. A spherical triangle is a part of the surface of a sphere bounded by arcs of three great circles; that is, of three circles whose planes pass through the centre of the sphere. Those arcs are the sides of the triangle ; and any of its angles is the same as the inclination of the planes of the sides which contain that angle. In what follows, unless the contrary is specified, a spherical triangle will be understood as being the smaller of the two parts into which the surface of the sphere is divided by the three smaller arcs joining, by D /V Of THl^*;)(iS^ (niri^BRsiTrl 26 INVESTIGATION OF THE FUNDAMENTAL FORMtLA. pairs, three points on the surface, and which are not on the same great circle.* In such a triangle, therefore, each side is less than a semicircle. 70. To investigate the fundamental formula in spherical trigono- metry, let ABCt (fig. 16) be a spherical triangle, and S the centre of the sphere ; and let the sides opposite to the several angles A, B, C, be denoted by the corresponding small letters, a, h, c. In the planes, ASB, ASC, draw AD, AE, each perpendicular to AS, and * If through two points on the surface of a sphere, which are not diameti-i- cally opposite, a great circle be described, the points may be regarded as being joined by either the less or the greater of the two arcs into which the circle is divided at the points ; and, hence, we have one reason for the limitation in the text. Another reason is, that while this hmitation simplifies the theory, it excludes no arc or angle which it is ever necessaiy to consider in the prac- tical application of spherical trigonometiy. Besides this, if the triangle men- tioned in the text be deteimined, everything that is excluded by the hmitation flows from that triangle, without any new investigation. — See the schohum at the end of this section. The following remarks and illustrations will assist the student in under- standing the theory of spherical trigonometry. 1. If from any point in the line which is the common section of two planes, and which (Euc. XI. 3) is a straight line, two perpendiculars to that line be drawn, one in each plane, the angle of inclination of the planes is the same as the angle contained by these perpendiculars. Hence it is obvious, that a spheri- cal angle is the same as the mclination of the tangents of the containing sides. It may also be remarked, that the planes of the sides form at the centre of the sphere a triedral solid angle, the inclinations of whose planes are the same as the angles of the triangle ; and the plane angles made by the radii, or com- mon sections of the planes, are measured by the sides of the triangle. 2. Every section of a sphere by a plane is a circle. For, if the plane pass through the centre, — since, by tne definition of the sphere, all its radii ai-e equal, — ^the section is obviously a circle. But if it do not pass through the centre, a pei-pendicular to it from the centre will cut it in a point equally dis- tant (Euc. I. 47) from all points of the boundaiy of the section: and that boundaiy is therefore the circumference of a circle. A circle whose centre is not the centre of the sphere is called a small or a less circle. 3. Since the planes of all great circles pass through the centre, the common section of any two is a diameter; and hence all great circles bisect one another. 4. If a straight line be di-awn pei*pendicular to any circle of the sphere through its centre, it cuts the surface in two points called the poles of that cu-cle. Hence (Euc. I. 47) either pole of a circle is equally distant fi-om all points of its circumference ; also, if great circles be drawn through the poles, the arcs of them between the circle and either pole are (Euc. III. 28) all equal ; and, in case of a great circle, each of these is the arc of a quadi*ant. f The learner, to assist his conception, may readily make a figure of a conve- nient form in the following manner : On a piece of pasteboard, describe, with a radius of 2 or 3 inches, an arc of about 190°, marking the centre with S, and each extremity with A, and draw tangents at the extremities ; then divide the arc into three portions, AB, BC, and CA, of about 54^^, 72", and 64", respec- tively ; di-aw the radii SB and SC, and produce them to meet the tangents ah-eady drawn in D and E, and join DE. Then, let the pasteboard be cut half through in the lines SD and SE, and the parts ASD, ASE be turned up on the opposite side, till the two radii AS coincide, and the diagram will be finished; S being the centre of the sphere, and ABC the triangle. The sides here employed are chosen confonnably to the note to No. 13. INVESTIGATION OF THE FUNDAMENTAL FORMULA. 27 let them meet SB, SC produced in D and E. Now, if r be put to denote the radius of the sphere, we have (No. 1) the following expres- sions for the angles at S : BSC = -, ASC=-, and ASB=-. r T r In these, we may evidently take r =1 ; that is, we may take the radius of the sphere, as the unit to which all the other magnitudes, the arc?, sines, tangents, &c. shall be referred. On this supposition, we shall have BSC=:a, ASC = &, and ASB=:c. Hence, also, since rz=.l, we shall have (No. 7) AD = tanc, AE = tan6, SD = secc, and SE = sec h. But (78) in the triangle DAE, DE2 = EA2-f AD2_2EA.ADcosDAE, or DE^^ztan^J^-tau^c — 2tan6 tanc cos A. In a similar manner we obtain from the triangle DSE, DE^msec^^ + sec^c — 2 sec 6 secc cos a; or (7) DE2 = 2+tan264-tan2c — 2sec6 secc cosa. By equalling these values of DE^, rejecting the common quantities, transposing, and dividing by 2, we find sect secc cos « = tan 6 tanc cos A-j-l ; or (1) and (3) cosa sin6 sine cos A _ ^ cos 6 cose cos 6 cose * whence, by multiplying by cos 6 cose, we obtain cos a = sin 6 sin c cos A + cos h cos c. Hence it appears, that the cosine of one side (and it may obviously be any side) of a spherical triangle, is equal to the continual product of the cosine of the opposite angle and the sines of the other sides, to- gether with the product of the cosines of those sides, j We have, there- fore, the three following formulas, which exhibit the relation between the three sides and any of the angles, and from which all others rela- tive to spherical triangles may be derived : cosa = cosA sin6 sine + cos6 cose* (85) cos 6 = cos B sin a sine-}- cos a cose (86) ^ cos.c = cosC sin« sin6-j-cosa cos6 (87) ^ * 1. From any of these formulas we may infer, that, with the Umitation in No. 69, one side is less than the sum, and greater than the dij^erence, of the other two. Thus, (14) cos(6-f-c)=cos6 cose — sin6 sine, which is always less than >^ cos b cos c-j-cos A sin b sine, the value of cos a, since cos A cannot be — 1, every / angle being less than two right angles. But the less arc has the greater cosine, if the arc be less than a semicircle : therefore b~{-c > a. From each of these take b, then c > a — b. •^t^^>t^*y 28 AN ANGLE FOUND FROM THE THREE SIDES. 71. From the first of these, by transposing cos 6 cos c, and dividing bj sin h sin c, we obtain . cos a — cos h cose cosA=i i— ,— . — ; sm sm c a formula, which, by means of natural sines, would enable us to find an angle, when the three sides are given. 72. To obtain formulas fitted for logarithmic computation, we get from (85), by transposition, cos A sinh sinc = cos« — cos 6 cose. Subtract the members of this from sin h sin c, and likewise add them to it: then (1 — cos A) sin 6 sin c = cos 6 cosc-f sin6 sine — cos a, and (1 -|- cos A) sin 6 sin c = cos « — (cos h cos c — sin b sin c) ; or, by (32) and (31), and by (15) and (14), 2sin2iA sin 6 sin c = cos (6 — c) — cos a, and 2cos2iA sin 6 sinc = cosa — cos(6 4-c). By modifying the second members of these by (23), and halving the results, we obtain sin^lA sin6 smc=:sin-|(a — b + c) sm^(a-{-b — c), and cos^-JA sin 6 sinc = sin^(a + 6 4-c) sin^(6 + c — a). In these, put a + 6 -l-c = 25, as in No. 5(j: then, by dividing by sin b sin c, and extracting the square root, we get .in»A:=y ^'"^^-\>^'"^^-''> (88) - V sm 6 sm c ^ ^ /sin 5 sin(s— a) COS^x\ = V^ r-y- \— ' (89) ^ ^ sm sin c ^ ^ 73. Divide (88) by (89), and there wiU result, by (3), 2. Hence, we may prove that the perimeter is less tlian 360^, m- 2 tr. For let b and c {jig. IT) be produced to meet in D ; then a Dividing these by (90), we get tan|B_ 5nn(g— g)^ ^^^^ tanlC_ sin(g— a) .^^^ tan^A sin(s — h) ' tan^A sin (5 — c) 74. By multiplying the values of tan^B and tan-^ C in the last No. by the value of tan^ A, we get tonlA taniB = ?il^i=^^ and tanlA taniC = ^^^^^^.. (02) 75. Lastly, by taking twice the product of (88) and (89), we ob- tain by (30), and by some slight reductions, . , 2Vsin5sin(s — a) sin(s — 5) sin (s — c) .^^. smA:= ^ r-^i — ; yijo) sin h sm c 76. Dividing (93) by sin a, we obtain sin A 2 V sin s sin (5 — a) sin (5 — 6) sin (s — c) sin a sinasin6sino Now the second member of this equation is symmetrical in respect to «, h, and c, since they all enter into it in exactly the same manner. It is therefore equivalent also to -^-7-, or — . ; and hence, ^ sm6 smc sin A sinB sinC .^.. - — = -T-7- = ^ (94) sm« smo smc Hence, sin a : sin A : : sin 6 : sin B : : sin c : sin C ; that is, the sines of the sides of a spherical triangle are proportional to the sines of the opposite angles. Hence, also, by multiplying ex- tremes and means, we get sin A sin 6 == sin B sin « sin A sin c = sin C sin a sin B sin c = sin C sin h * In this, if a > 6, the denominator is evidently greater than the numerator, and since in the first quadrant the greater angle has the greater tangent, it follows that|Ais>|B, and consequently that A is also>B. It appeal's, therefore, that tJie greater side has the greater angle opposite to it. t This would also appear by finding, as in (No. 15), expressions for sinB and sinC, and dividing the first by sm6, and the second by sine. The for- mulas marked (94) are sometimes called the Foi^midas of the Four Sines. 30 ADDITIONAL FORMULAS. 77. From (85), (86), (87), other important formulas may be ob- tained by elimination. Thus, to eliminate sine and cose from (85) and (87), multiply the latter by cos h, and substitute the second mem- ber of the result in the former: then, by transposing cos a cos^b, the first member becomes cos a — cos a cos^^, or (6) cos a sin^b; and, after dividing by sin b, there is obtained cos a sin b = cos A sin c -|- cos € sin a cos 6. Divide all the terms of this by sm a, and for - — substitute (94) sinC ^-u x.x cot« sin 6 = cot A sinC-1-cos.C cos.6. Hence, if we call a the Jlrst side, and b the second, it appears that, C if the cotangent of the first side be multiplied into the sine of the second, the product is equal to the cotangent of the angle opposite to the first into the sine of the contained angle, together with the cosine of the con- tained angle into the cosine of the second side. t) Then, taking succes- sively, as first side and second, a and c, b and a, b and c, c and a, and c and 6, and employing this theorem, we complete the following system : «X cote? sin6:=cotA sin C -j-cosC cos 6 (95) cota sinci=cotA sinB-{-cosB cose (96) @ ^> cot& sin« = cotB sinC + cosC cos« (97) cot 6 sine =cotB sin A-j-cos A cose (98) cote sin«=cotC sin B-}- cos B cos a (99) ^y cot e sin 6 := cot C sin A + cos A cos & (1^0) These formulas exhibit the relations between a side and the oppo- site angle, and another side and angle not opposite to one another ; and they solve the problem in which two sides and the contained angle are given to find the other angles, and that in which a side and the adjacent angles are given to find the other sides, the required part in each case being found by means of its cotangent ; but they are not fitted for computation by logarithms. 78. By farther elimination, other formulas may be derived from the foregoing. Thus, to eliminate b from (95) and (97), multiply the former by sin a, and the latter by sin 6 cos C, modifying the products by (4) : then cos a sin 6 = sin a cot A sin C + sin a cos b cos C, and sin a cos 6 cos C = sin 6 cotB sinC cos C + cos a sin 6 cos^C. POLAR TRIANGLE. 31 In the former of these, substitute for sin a cos h cos C, its equal iii the latter ; transpose the last term of the resulting equation ; and then, by substituting in the first member sin^C for 1 — cos^C, and by dividing by sinC, we obtain cos a sin b sin C := cot A sin a + cot B cos C sin h. ^,., . ' ' ^A\ •!,. cos A sin a i /^r. . v In this the term cot A sm a is (4) equivalent to -. — -r — , and (94) this is equivalent to — . " . Substituting this in the last equa- tion, dividing the result by sin h, and multiplying the quotient by- sin B, we get cos a sin B sin C=:cos A + cos B cos C ; or, by transposition, cos A ^ cos a sinB sinC — cosB cosC. Hence it appears, that the (cosine of an angle is equal to the con- tinual product of the cosine of the opposite side and the sines of the other angles, ivanting the product of the cosines of those angles.^ We have, therefore, the following formulas, which exhibit the relations between the three angles and each of the sides : cos A = cos a sin B sin C — cos B cos C (101) ^ cos B = cos 6 sin A sin C — cos A cos C (1^2) { X' cosC = cosc siiiA sinB — cos A cos B (1^3) j j^ 79. On comparing these with formulas (85), (86), and (87), we observe a close resemblance, sides being merely changed for the op- posite angles, and angles for the opposite sides, and one of the terms having the contrary sign. These latter equations, indeed, are the same as would be obtained by substituting in the others, 'tt — A, cT — B, -y — C, for a, h, c; and consequently cr — a, nt — 6, cr — c, for A, B, C. Thus, (85) would become, by this substitution, cos (t — A) = cos('7r — a) sin('y — B) sin (-7 — C)-f-cos(T — B) cos('r — C); or, (No. 15) — cosA= — cosa sinB sinC — cosBx( — cosC); which, by having the signs of its terms changed, will become the same as (101). Hence, therefore, since equations (85), (86), (87), may be regarded as containing all the properties of spherical triangles, it is evident that a formula expressing any relation of the sides and angles being established, a corresponding one will be obtained by writing sides for angles, and angles for sides, and prefixing the sign minus to all the cosines ; or, which is the same, by applying the formula to a triangle which has for sides the supplements of the arcs which mea- 32 A SIDE FOUND FROM THE THREE ANGLES. sure the angles of the proposed triangle, and for angles those wWeh are measured by the supplements of the sides of the proposed one. This triangle is called the supplementary or polar triangle* 80. Equations (101), (102), (103), enable us to find, by natural sines and cosines, the sides of a triangle, when the angles are given. Thus, from the first of them we derive cos a = T i^^ . ^ — . smB smO 81. Formulas adapted to computation by logarithms may be de- rived from the same system by processes similar to those employed in Nos. 72, 73, 74, and 75. Thus, we get from (101), by transpo- sition, cos a sin B sin C = cos A -|- cos B cos C. Subtract the members of this from sin B sinC, and Hkewise add them to it: then (1 — cos a) sinB sinC = — cos A — (cosB cosC — sinB sinC), and (1 +C0S a) sin B sin C = cos A + cos B cos C -{- sin B sin C ; or, by (32) and (31), and by (14) and (15), 2sin2iasinB sinC= — cos A — cos(B4-C), and 2cos"-|a sinB sinC = cos A4-cos(B — C). * Let ABC {fig. 18) be a spherical triangle, and from A, B, and C, as polos, let arcs of great circles be described intersecting each other in A', B', and C; A'B'C is the polai* triangle. For, A being the pole of B'C, the arcs AB', AD, AE, &c. are each equal to 90". For the like reason, CB', CF, &c. are each equal to 90°; and therefore, since B'A, B'C, are each equal to 90°, B' is the pole of the arc GACE. Hence, B'E=90°; and it would be shown in a similar mannei', that C'D=90°. Now, DE-f B'D=B'E=90°, and DE-[-EC'=DC'=90°; whence, by addition, DE-f B'D-f-DE+EC, or DE-f B'C'=180°. But (page 28, note 3) DE=A; therefore B'C'=5r — A : and in a similar manner it might be shown that A'C'= tr — B, and A'B'= t — C ; and also that BC = 5r — A', AC=r T — B', and AB = *— C. Any gi'eat cu'cle divides the surface of the sphere into two equal parts ; an- other intersecting the first, divides the surface into four lunes, the opposite ones of which are equal ; and a third great circle intersecting both the former, not in the points in which they intersect each other, divides the surface into eight tri- angles, the four of which on the one hemisphere are respectively equal to the four similarly situated on the other. Hence, if the circles of which A'B', B'C, and A'C are arcs, were completed, they woiild foi-m eight triangles. Of these, however, only A'B'C, and the con-esponding triangle on the other hemisphere, have the property mentioned in the text ; each of the others having two sides the same as two angles of ABC, and two angles the same as two of its sides ; while only a side and the opposite angle of the one are supplementai-if to an angle and the opposite side of the other. The easiest mode of employing the principle above explained, which reduces the cases of spherical triangles to half their number, is, not to consider the polar triangle, but to use sides for opposite angles, and angles for opposite sides, and to prefix the sign mmus to the cosines ; or, when the halves of angles or sides occur, to use ?r — A foi* a, vr — a for A, &c. 2 ^^ sinB smC ^ ^ J A SIDE FOUND FROM THE THREE ANGLES. 33 By modifying the second members of these by (22), and halving the results we obtain Bin^ia sinB smC = — cosJ(A + B + C) cos|(B-f C — A), and cos2iasinBsinC = cosi(A— B + C)cos^(A + B— C). Hence, by putting A -[- B -{- C = 2 S, as in No. 56 ; by dividing by sin B sin C, and extracting the square root, we get . . /— cosS cos(S--A)* "1 sm^a = ^/ • Tj • n (104) X 2 V sm B sm C ^ ^ J ^ cos(S — B)cos(S— .C) sinB sinC 82. By dividing (104) by (105), we find / — cos S cos (S — A) ,iA/>x tania=^ -r^ — 7^7 y^ — -^ (10b) 2 ^ cos(S — B)cos(S — C) ^ ^ Also, by taking twice their product, we obtain 2V~^os S ^^— A) cos(S^B) cos(S— C)t .,.^. sma=— ^ -^ • ' .\ ^ . (107) smB smC * Since (No. 13) sinB and sin C are positive, the numerator of the second member of this formula must also be positive, as othei*wise the value of sin ^a would be imaginary. Now, the numerator will be positive only when cos S and cos (S — A) have contraiy signs. Of the quantities S and S — A, there- fore, one, namely S— A, the less, must (No. 13) be less than 90°; while S must be l^tween 90** and 270°. By doubling these, we find that 2 S, or the sum of the three angles is greater than two right angles, and less than six ; and that 2 S — 2 A, or B-{- C — A, that is, the excess of the sum of two angles above the third is less than two right angles. The remainder, S — A, or | (B + C — A), may be negative, since (No. 14) its cosine would still be positive; whence it appears, that one angle may he greater than the sum of the other two, which is also the case in plane triangles. Some of these conclusions might be easily derived from considering the polar ti'iangle in connexion with the notes to No. 70. t The last four formulas may be derived from (89), (88), (90), and (93), by means of the supplementary triangle. Thus, by the substitution of it — a for A, * — A for a, &c. (89) will become / sinH3^-(A4-B+C)Uini?^-(B + C-A)j . cosM^-a)=^ sin(^__B)sin(«— C) ' which, by contraction, will become the same as (104). By finding expressions, by (106), for tan|6 and tan|c, and by first dividing them, and then multiplying them, separately, by (106), we should get tan|6 cos(S — B) , tan^a~ cos(S — A) tan^c cos(S — C) ,1. tan^a~'cos(S — A) taniatan^&=^-^=|^ T {c) E 34: Napier's analogies. 83. We may now proceed to investigate four remarkable expres- sions, known by the name of Napier's analogies, from their having been discovered by Baron Napier, the inventor of logarithms. To effect this, take the members of (101) from those of (102),. and there will remain cos B — cos A= sin C (cos h sin A — cos a sin B) -}- cos C (cos B — cos A) ; or, by transposing the last term, (1 — cosC) (cosB — cos A) = sin C (cos 6 sin A — cos a sinB). By modifying this by (32) and (30), and dividing by 2sin2iC, we obtain cosB — cos A = cot J C (cos 6 sin A — cos a sinB). Dividing this successively by sin A — sinB and sin A + sin B, we get, by (27) and (28), 1/A , -ON 2. in cos 6 sin A — cos a sinB , tani(A + B) = cotJC. : — r r-.jr — , and ^^ ^ ^ smA — smB . . . -,. ^ , ^ cos h sin A — cos a sin B tani(A— B) = cotiC. . . , . -^ . ^^ ^ ^ smA-|-smB Multiply the numerators and denominators by sin c ; in the results, in both the numerators and denominators, for sin A sin c and sin B sin c, substitute (No. 76) sin a sin C and sin 6 sin C ; and divide the numerators and denominators by sin C : then X w A . T»\ ^ , ^ sin « cos 6 — cos « sin 6 , tani(A-f B) = cotJC. ; r—, , and , , . ^. , _, sin « cos h — cos a sin h tani(A— B)=cotiC. -. — 7-r , ^^ ^ -^ sin a + sm 6. Modify the numerators by (13), and to the results apply (34) and (36): then tanKA + B)=cotxC.^-^lg=|* (108) -* ^ tanKA--B) = cotiC.||^ig^> (109) taniatan|c=^-3^-^^^ {d) These formulas correspond to (91) and (92), but they are of little practical im- portance. * From this formula we may infer, that half the sum of tiuo sides of a spheri- cal triangle, and half the sum of the opposite angles, are of the same species, that iSf are either each less or each greater than 90". For (No. 69) any side of Napier's analogies. 35 By taking the difference of the members of (86) and (85), and by a process similar to the foregoing,* we should obtain w XX X 1 sin4(A— B) tan J (a — 6) = tan i c. . f;. . ^: ^^ \ ^ smi(A4-B) (110) (lll)t 3- a triangle, and consequently the difference of any two sides, being less than a semicircle, and any angle being less than two right angles, cos|(« — h) and cot^C must (No. 13) both be positive: and from this it follows that tan|(A4-B) and cos|(a-f-6) must have the same sign, which can take place only when ^(A-f-B) and \{_a-\~b) are both greater or both less than 90*^. It is also evident, that if the sum of two sides a and b be ISO*', tfie sum of tlie oppo- site angles is the same. For, in that case, cos| ( a-{-b ) = cos 90*^=0, which renders the second member, and consequently the first infinite; so that (No. 10) the first member is the tangent of 90°. * The only difference is, that sides and angles are mutually interchanged ; and that, an«r the transposition, there is l-|-cosc, instead of 1 — cosC. t The foregoing investigation, which the author believes to be new, is very direct and simple, employing only the common elementary formulas. The following, whicn is taken in substance from a work on trigonometry by Mr. Luby of Dublin, possesses much elegance, but it is more tedious from the number and variety of the reductions and preparatoiy processes which it in- volves. By (88) and (89) we have /sin (s — b) sin {s — c) s/ sin 6 sine sin {s — a) sin {s — c) sin a sine _ /sin(s — a)sin(s — b) sin a sin 6 I. sin^A II. sin|B = ^' III. BiniC = ^' 'sins sin (j IV. cosiA=^/- . ^ . V^ sm6 smc V. cos JB=y: si ns sin(g — b) sin a sine VI. cosiC= V'i24i^Hi£q:l) "V sma smo Taking the products of I. and V., of IV. and II., of IV. and V., and of I and II., and applying VI. to the first and second products, and III. to the third and fourth, we obtain VII. sin iA cos J B='i^^^. cos \ C smc VIII. cosiAsin^B: _sin(s — a) .cos|C IX. cosiAcosiB=^J^. siniC ^ smc ^ X. sin|A 8in|B = 8iu( smc •.sin|C. Take the sum and difference of VII. and VIII., and of IX. and X., and modify the first members by (12), (13), (15), and (U), and the numerators of the second members by (20) and (21), using \{a-^b^c) for s, |(a-|-6— c) for s-^, &c. and 2sin \c cos Jc for sme, and there will arise XIII.cosi(A— b; i|C sin|c XIV.cos|(A-|-B)= sin^C XL sinHAH-B)=^2ii2.cos|(^6) cos|c XII. sinHA— B)=^?^.sini(a— 6) sm^c ^^ I Now, (108) and (109) will be obtained by dividing XT. by XIV., and XII. by Xlll.; and if XIII. be divided by XIV., and XII. by XL, (110) and (111) wiU be found by multiplying the quotients by \i\w\e. .sin^(a4-^) cos ^(a-|-&) 36 RIGHTANGLED SPHERICAL TRIANGLES. 84. The formulas that have been investigated in the foregoing articles furnish the means of resolving all the elementary cases of spherical triangles. When applied to rightangled triangles, they take, in general, a simpler form, some of the terms vanishing in consequence of containing the cosine or cotangent of the right angle. Thus, if C be a right angle, we have (No. 10) cosC=:0, cotC=0, and sinC=l; and in this case (97) will become cot 6 sin a = cot B, or, bj multiplying by tan 6 (5), sin«=:tan& cotB; while, from the first and third parts of (94), we derive sin a = sine sin A. From (103), by transposition, and by dividing by sin A sinB, wo obtain cose = cot A cot B; and from (87) we derive at once cosc = cos « cos 6. From (100), also, by dividing by cos 6, we obtain cos A=i:tan6 cote; and (101) gives cos A=cosa sinB. These results may be conveniently arranged in the following form : sin«=tan6 cotB = sine sinA. (112) cose =cotA cotB = cosa cos6 (H^) cosA=tan6 cote =:cos«sinB (114) 85. From these formulas, by supplying the radius, we have the following theorems, which are sufficient for the resolution of aU the elementary cases of rightangled spherical triangles. I. The rectangle under the radius and the sine of one of the legs, is equal to the rectangle under the cotangent of the adjacent oblique angle and the tangent of the other leg, or to the rectangle under the sines of the opposite angle and the hypotenuse. II. The rectangle under the radius and the cosine of the hypote- nuse, is equal to the rectangle under the cotangents of the oblique angles, or to the rectangle under the cosines of the legs. III. The rectangle under the radius and the cosine of one of the oblique angles, is equal to the rectangle under the tangent of the ad- jacent leg and the cotangent of the hypotenuse, or to the rectangle under the cosine of the opposite leg and the sine of the other oblique angle. 86. Several useful formulas respecting oblique-angled triangles may Another investigation might be had by means of (90) and (106), and of the fbmula, tanHA±B)= ^*^^^t^*^anfB ' <«"> -"d(«)rand other in- vestigations will be found in the works on trigonometry. SEGMENTS MADE BY PERPENDICULARS. 37 be obtained by drawing their perpendiculars, and applying the for- mulas last found. Thus, let AD {fig. 20) be perpendicular to BC. Then, putting 9 to represent CD, we have BD = a — 9; and (No. 85, III.) cos C = cot 6 tan 9 ; whence (5) tan f = tan 6 cos C (115) 87. By considering 6, cos AD, and cosB=: sin (A — a>) cos AD; whence, by division, cosB^sinCA-*) cosC smO ^ 38 EXTENSION OF THE THEORY Hence, the sines of the angles^ contained by the perpendicular and the sides, are proportional to the cosines of the angles at the base. 91. Bj like processes, and by considering, first, b, «I>, AD, and c, A — , AD; and then p, O, AD, and a—)""tan6 tan^ __ tan tan (a — (p) ~tan(A — ^) Hence, the cosines of the angles, contained by the perpendicular and the sides, are reciprocally proportional to the tangents of the sides; and the tangents of those angles are proportional to the tangents of the segments of the base. SCHOLIUM. As was stated in No. 69, a spherical triangle has thus far been understood as being the smaller of the two parts into which the surface of a sphere is divided by the three smaller arcs, which join, by pairs, three points on the surface, and not on the same great circle. In strictness, however, this view of the nature of a spherical triangle, though convenient in practice, is too limited. As an arc of a gi-eat circle (No. 12) may be increased without limit, so a side of a spherical triangle may exceed, not only a semicircle, but even an entire circle of the sphere. In like manner, if the half of any great circle continue fixed, while another semicircle on the same diameter begins to revolve about that diameter fi-om coincidence, and thus to make greater and greater angles with the fixed one, as there is evidently no limit to the angular magnitude thus generated, it is plain that the spherical angle fonned by the two semi- circles, may exceed two right angles, or any assigned magnitude whatever. It is unnecessary, however, to consider sides that exceed a complete circle, as the extreme points of any such side will occupy the same positions, as when it is diminished by a circle : and, for a similar reason, it is unnecessary to con- sider any angle greater than four right angles. To enter into a minute consideration of the views thus opened up, would be unsuitable to our present limits : but the reader who wishes to prosecute the subject will find a paper by the author of this work in the London and Edinburgh Philosophical Magazine for January, 1837,.in which the subject is dis- cussed at some length. It may be proper, however, to give here a few illusti-a- tions of the subject, and a few of the results established in the paper referred to. 1. If in a spherical triangle, the angle A, and the sides b and c containing it, be given, and if we make on the surface an angle equal to A, and the arcs AC and AB equal respectively to b and c, we determine the points B and C ; and since these points may be joined by either of the parts of the great circle passing through them, there will be two triangles, each answering to the data, and differing in area by the surface of a hemisphere ; and they will be such, that if the third side of one be denoted by a, that of the other will be 2* — a This agrees exactly with formula (85), which gives the cosine of the third side OP SPHERICAL TIUGONOMETRY. 39 by means of A, h, c ; and ( 1 ) cos a is the same as cos ( 2 ?r— a). The remaining angles would be found by means of (98) and (100), and would thus have two values as they ought : and like results would be obtained by means of Napier's analogies, or any of the other modes of solving the third case. 2. If the three sides a, b, c, be given, and if we take three points A, B, C, such that BC = a, ACrr 6, and AB=c, either of the two parts into which the surface of the sphere is divided by AB, BC, and AC, will be a spherical tri- angle, having its three sides of the given magnitudes, and consequently agreeing with the conditions of the question. It is plain, also, that if A, B, and C, be the angles of one of these triangles, the angles of the other will be 2 it — A, 25r — B, and 2?i' — C. This result agrees with that which is obtained by any of the modes of solving the first case. Thus, by (85), the angle opposite to a would be determined by its cosine, and would therefore be A or 2 ) — ^logsin 4> . , . (136) 100. In the fifth case, in which two sides a and h, and an angle, A, opposite to one of them, are given, the angle B opposite to the other, is found by the following formula derived from (94) : logsin B=logsin A+log sin h — ^logsin a (137) SPHERICAL TRIANGLES — CASE V. 43 B being thus found, c and C will be determined by the following for- mulas, derived from (133) and (129) by transposition: logtanJc=loglani(a+Z>)+logcosKA+B)— logcosKA— B)(138) logcotJC=logtani(A + B)-f-logcosJ(a+6)— logcos-i(a— &) (139) It is evident that (134) and (130) would give formulas answering tlie same purpose. 101. The species of B, when not doubtful,* is known by the fol- lowing rule : When the sum of the given sides is less than 180°, the angle oppo- site to the less side is acute ; but when the sum of those sides exceeds 180°, the angle opposite to the greater side is obtuse ; and, lastly, if the sum of those sides be 180°, the sum of the opposite angles is the same. When these principles fail in determining the species of the angle, it is doubtful, t 102. The third side c might also be found by the following for- * It may be proper to remark here, that the required part, either iu plane or spherical trigonometry, is never doubtful, except when it is found by means of its sine. To understand the reason of this, it is only necessary to consider, that if the required part be determined by means of its cosine, it must (No. 13) be in the first quadrant if the cosine be positive, but in the second, if it be nega- tive : and the same holds when it is foimd by means of its tangent or co- tangent. f This rule, which is taken in substance fi-om Cagnoli, is derived from the first note to No. 83, and fi-om the note to No. 73. Thus, by the former note, if the sum of the sides be less than 180'^, the sum of the opposite angles is also less than th€ same, and, therefore, at least the less of them is less than 90°; but, by the latter note, the less is opposite to the less side, which estabUshes the first part of the inile. The proof of the second part is exactly similar, and the thu'd part follows at once from the first of the notes referred to. The nature of this case, when doubtful, will be illustrated by figures 21 and 22 ; the angles A and A' in the former being acute, and in the latter obtuse. In the former figure, if the side a be less than either h or its supplement CA', a small circle described at a distance from C equal to a would cut ABB' A' in two points B and B', and therefore there are two triangles, ABC, AB'C, either of which answers the conditions of the question. It will appear, fi'om the other figure, that a like ambiguity will exist when a is greater than either h or its supjDlement. It is also plain, that in the former figure a might be so small, and in the latter so large, that the solution would be impossible, there being no triangle answering to the data. It appears, therefore, that when the given angle is acute, tliere is ambiguity only when the opposite side is less than either the other side or its supplement; and that, when the given angle is obtuse, there is ambiguity only when the opposite side is greater than either tlie other side or its supplement. TVhen there is no ambiguity, tlie species of the required angle may be known by the p^'inciple, that the greater angle is opposite to the greater side. Some may, perhaps, prefer these principles to the rule given in the text. It may also be remarked, that in each figure, if the small circle merely touch ABB' A', the triangle will be rightangled, and there will be no ambiguity. 44 SPHERICAL TRIANGLES CASE VI. mulas, derived from (115) and (116), by changing C into A, a into c, and c into a: logtan9=logtan6+logcosA — 10 (140) logcos(c — ^)*=logcos«+logcos9 — ^logcos5 .... (141) The latter gives the difference of c and cp ; and c will be the sum or difference of

subt. sins 100" 2' 9-99331 sin(s— .a) 2 e-'zeiTG sm(s— 6) 62 U 9-94884 sin(s— c) 37 16 9.Y8213 2)22-97290 tan I A 88« 1' 53" 11-48645 A =176*' 15' 46" Then, according to (125) and (126), we have tanj A + sin(s — a) = 11-48645 + 6-76476= 18*25121; and taking from this, first, sin(5— 6)=9-94884, and, secondly, sin(5— c)=9-78213, we find taniB=8-30237, and taniC= 8-46908: whence, by the tables, we getiB = l° 8' 57", and ^0 = 1° 41' 13", the doubles of which are B=2° 17' 54", and C=3° 22' 26". In the first part of the foregoing operation, as in many other com- putations in trigonometry, the method of performing addition and subtraction at a single operation, in the manner pointed out in No. 60, may be employed with advantage. As a check on the preparatory part of the process, it may be ob- served, that, as in plane triangles, the sum of the three remainders, s — a, s — h, s — c, is equal to the half sum s. in the manner above mentioned, or from the general formulas of spheri- cal trigonometry, by taking c=90°, will resolve aU the cases of quadrantal triangles : sin A= tanB cot 6= sinC sina COsCrr — COta COt 6 = — COS A COS B cosa=: — tanBcotC=r cos A sin ^ 48 EXAMPLES OP COMPUTATION. The following are the computations of A by (122) and (123): By (122). I By (123). 8in6 sr 18' 9'78246> ' sin6 3^18' 9-78246) sine 62 46 9-94898K^"*' I sine 62 46 9-948985 ' sin(s— 6) 62 44 9*94884 sms 100 2 9-99331 sin(5— c) 37 16 9-78213 2)19-99953 sin I A 88^ 7' 9*99976 sin(s— a) 2 6*76476 2)17-02663 cos|A 88° r 53" 8*51331 A= 176*' 14' A= 176« 15' 46"* 111. Given A=:139^ 27', 3=53" 39', and C=34P 5'; to find the Here we have 8=113° 35% S— A=— 25° 51% S— 6=59° 56'^, S— C=79o 30'|; and by means of No. 81 or 82, we find a= 126° 34', 6=95° 46', and c=43° 49'. 112. Given A=42° 43', hz=4:5' 51', and c=14P 17'; to resolve the triangle. Here, by taking the sum and difference of c and b, and halving them, we get i(c+6)=93<' 34', and i(c— 6)=47° 43': and the half of A is 2P 21'^. Then, by (129) and (130), logtani(C + B)=logcotlA-f.logcosJ(c — b) — log cos ^(c 4-6), and logtani(C — B)=logcot-i-A+logsin^(c — 6) — logsini(c-|- b); whence, by performing the actual operation, we find -1(0+6)= 92° 4'i,t and ^ (C— B)=62o ll'i; and by taking the sum and dif- ference of these, we get C=154o 15'i and 6=29° 53'. 113. To find the side a, we have, by (131) and (132), logtanf=logtan6+logcosA — 10, and log cos o=logcos 6+logcos (c — (p) — log cos ^. * In this example, as A is vei*y obtuse, the use of (123) is much preferable (No. 94) to that of (122); as the "former enables us, even by tables carried to only five places of decimals, to find, by proportional parts, the result true to se- conds ; while the other does not give it, by such tables, true even to the nearest minute, there being at least two sines such that we should not know which to prefer. In what follows, the answers will generally be given true only to the nearest minute. f In the equation from which this is obtained, ^A and ^{c — b) being each less than 90 , the cotangent of the former, and the cosine of the latter, are (No. 13) both positive ; but |(c+&) being greater than 90", its cosine is nega- tive. Hence, since the product of two positive quantities is divided by a nega- tive one, the quotient, tan|(C-{~B), must be negative, and therefore (No. 13) KC-f-B) is greater than 90". For a similar reason, in No. 113, a in both methods, and (p in the second, are to be taken in the second quadrant, and are therefore the supplements of the values given by the tables. It may be stated as a general prmciple, that an odd number of negative multiphers or divisors, or of both, gives a negative result ; while in other cases the result is ppsitive. EXAMPLES OF COMPUTATION. 49 From the first of these we find f)=37° 7'^, by subtracting which from c we get c — ^ =104° 9'|. We then obtain from the second a =^102° 20'-!. We should also have from (131) and (132), log tan ^ := log tan c-j- log cos A — 10, and log cos a = log cos c + log cos (6 — N0S lET RT. Given. Answers. Given. Answers. 1. a = 56" 17' A= 62" 32' 9. B = 146" 12' a = 112" 39' b = 147 33 B = 145 6 C = 40 40 c = 43 39 c = 112 48 C = 79 33 b = 143 54 A = 60 37 2. A = B = C = 47 9 136 21 88 34 a — IQ 24: b = 164 36 c = 157 22 or a = A = 9 27 136 21 8 66 3. a r= 78 41 A = 183 15 10. A = 90 a = 91 42 b = 153 30 B = 160 39 B = 95 6 b = 95 22| C = 140 22 c = 120 60 C = 71 36 c = 71 31i 4. a = 71 45 A = 70 31 11. a =: 63 19 A = 66 40 B = 104 5 b = 102 17 b = 35 23 B = 41 32 C = 82 18 c = 86 41 C = 90 c = 60 51 5. a = 136 25 A = 123 19 12. A = 75 36 c = 31 51 c = 125 40 B = 62 6 B =r 90 b = 68 lOj C = 100 & = 46 48 a = 64 3 C = 34 38 6. b = c = B = 124 53 31 19 16 26 a = 155 35^ C = 10 19| A = 171 48| or c = b = C = 148 9 111 49| 145 22 7. a = 115 28 A = 59 39 13. a == 3 A = 36 54 b := 60 29 C = 172 43 b = 4 B = 63 10 B = 66 17 c = 172 23 c = 5 C = 90 2 or A = 120 21 14. a = 30 A = 40 39 C = 72 52 b = 40 B = 56 52 c s= 88 53 c =r 60 C = 93 41 8. A = 103 16 a t= 149 53* 15. a — 60 A = 66 62 B = 76 44 c = 164 60 b r= 80 B = 72 13 b = 30 7 C = 149 30 c =: 100 C = 107 47 * This may be found by subtraction, according to No. 104, since the sum of the given angles is 180". MISCELLANEOUS INVESTIGATIONS. 61 v.— MISCELLANEOUS INVESTIGATIONS* 116. From the three sides of the spherical triangle ABC (fig. 20), the segments of the base made by the perpendicular AD may be determined in the following manner, without calculating the angles : Putting CD = ^, we have (No. 87) cos(c«— ^ : cos 9 : : cose : cos6; whence we obtain, by composition and division, cos (a — :: tant : tanc; whence, by composition and division, cos(A — o) — cos tan 6 — tanc cos (A — a>)4-cosa) tan 5 -j- tanc* Multiply the numerator and denominator of the second member by cost cose; then, by modifying the result by (13) and (12), and the first member by (25), there is obtained tan(a)— | A)_ sin(6— c) cotiA ~sin(6 + c) *•' ^ ^ This result affords an easy means of determining the parts of the angle A; and thence, by the rightangled triangles ABD, ACD, we have another method of solving the third case. 119. It was shown in No. 88 that sin^ : sin (a — (p) : : tanB : tan C ; whence, by a process nearly the same as that employed in the last No. we obtain tan(^— l a)_ sin(B-~C ) tani« — sin(B-l-C)""^ ^ This formula affords an additional method of solving the fourth case. 120. Let AD (Jig. 24) bisect the angle A, and let CD = 6', and BD = c'. Then (94) sin 6 sin ADC , sine sinADB smo smfA smc smfA But (No. 15) sinADB = sinADC: wherefore, sin6 sine , ^, sin6 sin6' ^i-o\ -:— r-=-^ — „ and consequently -: — =-: — , ... (153) sm6 smc' ^ "^ smc sme ^ This result will be seen to be analogous to Euc. VI. 3 ; and it will furnish the means of computing the segments when the sides are given : for, by composition and division, and by (24), we have tan l(&--.c )_ tan^(6'-^') tan J (5 4- c) tan-|« 121. If the exterior angle formed by producing b through A, were bisected by a great circle cutting the base produced in D', and i SYMMETRICAL FORMULAS. 55 CD':=^h", and BT)'=c", we should have by (94), since half the ex- terior angle is the complement of ^A, sin 6 sinD' , sine sinD' , -v-r-; = - , . , and-t — 7,= j-r; whence, &iub" cosf A smc" cosJA sin6 sine , ,, sin6 sin6" /^KK\ -7——,=z'-. — ;;, and, consequently, -. — =-r—^, ... (loo) sm6" smc" ^ •^' smc smc" ^ We should also find, by a process nearly the same as in the last No. tan^(6— e) _ imj a .,^g. tani(6 + c)--tani(6''+c'0'"^ ^ 122. Let AD (Jig. 25) bisect the side BC, and let the angle BAD=B', and CAD=C', Then (94), sine sinADB , sin 6 sin ADC and sinja sinB' ' sin-Ja Divide the latter of these by the former, and since (No. 15) sin ADB= sin ADC, we get sin h sin B' sine sinC From this we obtain, as in No. 120, (157) tan^(6-e),_ tani(B-— CO tani(6 + c)-~ taniA '•• V^^^'; The learner may exercise himself in extending the formulas ob- tained thus far in this section to the triangles formed by continuing the sides of the triangle ABC till they meet ; and, by expressing the sides and angles of these triangles in terms of a, h, c, A, B, C, and -jt, he will find analogous, and, in several cases, interesting results. 123. Several formulas, remarkable for their symmetry, may be de- rived from (88), (89), (90), (93), and (104), (105), (106), (107), by finding the corresponding expressions in relation to 6, c, B, and C. In investigating these, for the sake of brevity, put Vsins sin(s — a) sin (5 — h) sin(s — c)=zn, and V— cosS cos(S— A) cos(S— B) cos(S — C)=:N. 124. Let now the continual product of the values of sin^ A, sin^ B, and sinJC (88) be taken, and there wiU result, by actual extraction of the square root, • 1 A • 1 T» • 1 />. sin(5 — a) sin(s — 6) sin(s — c) ..w^. smJA sm^B smiC = — ^ -. \ , . ^^ --.. (159) ^ ^ 't sma smo smc ^ 54 SYMMETRICAL FORMULAS. This, bj multiplying the numerator and denominator by sins, becomes sin^A sinlB siniC = -: r-^-^ r— ... (160) sm5 sma sm6 smc 125. In a similar manner, from the values of cos^ A, &c. (89) we obtain cosi A cos^B cosiC=-. — ^^^^ — (161) sma sm6 smc ^ 126. In like manner, by multiplying together tan J A, tan-JB, tan-^C (90), and contracting, we find taniAtan^B taniC=y^"K^-«) ^i°(^-^) «i°(^-<') .. (i62) Multiply the numerator and denominator by Vsins; then, taniAtaniBtanJC=: .'\ (163) 127. In like manner, from (93), we find 8n^ sin A sinB sinC = -T-^ — . „, •- (164) sm^a sm^ft sm^c ^ ^ 128. By processes exactly similar, we obtain from (104), (105), (106), and (107), • 1 -17. • 1 — NcosS ^-_^. BmJ«sin4!,smic = _^_;-g-5-^ (165) co4.cospcosio= ''°^(^-^> 7^.V-) r^'~^^ - (166) 2 2 2 sm A sm B sm C ^ "^ cos-ta cos46 cosic= s— :^ — r — . p . ^ ... (167) 2 2 2 — coslSsmAsmBsmC ^ '^ tan^atani6tanic=v — tct — tn — To — 5\ — 7^ — p^= ^^ (168) * ^ ^ ^ cos(S — A)cos(S — B)cos(S — C) N ^ ' 8N3 sina sin& smc= ... . _.p . -^ (169) sm^A sm^B sm^C ^ ^ 129. From (164) and (169) we find w=i(sin2rt sin2& sin^c sin A sinB sinC)^ ... (170) N=^(sin2A sin^B sin^C sin« sin6 sinc)^ ... (171) 130. Divide (170) by (171) ; then," n C sina sin6 sine \i sina* sin6 sine /i7o\ N \sinA sinB sinC/ sinA sinB~"sinC '""* ^ ^ * This expression and the two following, are obtained fi-om the preceding radical, on the principle (94), that / sing sin6 sine \ l_ / sm^a \ } / sin'*6 W / sin='c \^ isinAsmBsinCl ~ Vsin^A/ isin'B/ - Isin^C/ ' SYMMETRICAL FORMULAS. 00 TT 1 N w N n , N n Hence, also, -. — r-=:- — , -r— 7r = -r— r. ana-7— r^=:-: — (,i''j; sinA sma smB siii6 smO sine 131. From (93) and (107), we have sin A 2n _ sina 2N and sin a sin « sin 6 sin c' sin A sin A sin B sin C ' and, bj equalling the second member of one of these with the recip- rocal of the second member of the other, we get, by multiplying by the denominators, 4nN=:sin« sin6 sine sinA sinB sinC (174) 132. In this, substitute for sinA sinB sinC its value in (164); and, by contracting, there will arise ]Sr = -= ^.^' ■ (175) sin a smo smc By a like process we find, by means of (169), smAsmBsmC ^2 jq-2 133. From (167) by substituting . ^ ; — for . ..^ . ^ , according ^ ^ -^ ° sin 6 smc smBsinC to No. 130, and by clearing the result of fractions, we obtain — cosS cos J a cos "1 6 cosjc sin 6 sine sinArin^. But (93) sin 6 sine sinArr2w; by substituting which in the first member, and dividing by 2 n cos J a cos 16 cos-^c, we obtain, by re- storing the value of n, Vsins sin(s — a) sin(s — h) sm(s — c)__ n ~" 2cosiacos|6 cos^c 2cos^a cos^b cos|c 134. By a similar process we should obtain, from (160), . _ V— cosS c os(S~A)cos(S— B) cos(S-~C) N ..^ . "°^~ 2 sin I A sin^B sinlC -2siniA sin^B sin^C"' ^^^ 135. A curious formula, given by Cagnoli, may be thus investi- gated: Transpose the first term of the right-hand member of (85); multiply the result by cos A, and to both members of the product add sin & sine: then, the first member wiU contain sin & sine — sinS sine cos^ A; and, substituting for this its equal, sin 6 sine sin^ A, we obtain cos A cosa-|-sin6 sine sin^ArrsinS sinc+cosS cose cos A (-er) From (101), also, we find, by a similar process, cos A cosa-|-sinB sinC sin ^a^ sinB sinC — cosB cosC cos«. 5G NEW SYMMETRICAL FORMULAS. In the second term of this, substitute, according to (94), sin 6 sin A and sine sin A, for sinB sin a and sinC sin a, and there will arise cos A cosa+sin6 sine sin2A=sinB sinC — cosB cosC cosa. Hence, bj comparing this equation and equation (z), we obtain sin6 sinc-f-cosZ) cose cosA=sinB sinC — cosB cosG cosa* ... (179) 136. Bj multiplying (85) by cosB cosC, and (101) by cos& cose, and by adding the results together, we obtain, after transposition, cosa cosB cosC— cosA cosB cosC sinb sinc=:— cosA cos6 cosc+cosa cos6 cose sinB sinC (180) This formula, in common with (179) and other formulas of a similar kind, possesses the property of not being changed by the substitution of-T — A for a, rr — B for 6, &c. It is the same, therefore, in tri- angles that are supplementary to each other. 137. If we now multiply (85) by cosa sinB sinC, and (101) by cos A sin 6 sine, and subtract the latter product from the former, we shall have, after dividing by sin 6 sine, cos^a sinB sinC „ a cosa cosb cose sinB sinC , . ^^ ^ . , . cos^A=: . , . -fcosA cosB cosC. sino smc smo smc ' T A.-U' J- X /{\A\ -u x-^ ^ sin^A „ sinB sin C In this, accordmg to (94), substitute -r-r— for . , . — : then, bv sin^a sin 6 sine -^ writing 1 — sin 2 a for cos 2 a, there will arise (6) sin2A cosa cos 6 COSC sin 2 A , ^ ^ •1= r-^ f-cosA cosB cosC ; N 2 sin 2 A or, by substituting — j for -. ^ •■ , according to (172), by multiplying by n^y and, first, by transposition, and then by division, N2(l — cosa COS& cosc)=:w2(l-|-cos A cosB cosC),^ N2_ 1 + cosAcosBcosC I (181) Ji2 "" 1 — COS a cos 6 cose J From the former of these we get, by transposition, N2 — w^rrN^cosa cos6 cosc+w^cos A cosB cosC. * This formula contains all the six parts of the spherical triangle, and is re- markable for its symmetry ; the one member containing the sines or cosines of three parts, and tne other the respective sines or cosines of the parts respec- tively opposite, and the only difference being in one of the signs. Cagnoli'a investigation is founded on nearly the same principles as that given above, but is perhaps, scarcely so simple. Delambre investigates it by finding two ex- pressions for the same line — one by means of three parts of the triangle, and the other by means of the remaining parts. It is scarcely necessary to observe, that two corresponding formulas might be obtained ft-om (86), (87), and (102), (103), or from the preceding by changing the letters. Delambre gives also various other formulas containing eacn the three sides and the tlu*ee angles, INSCRIBED CIRCLE. 57 Now, by (18), cos6 cosc=icos(6+c)+icos(6— c); and multiply- ing this by cosa, and applying the same formula, we find cos a cos 6 cos c = i cos (a+ft+c) + i cos [-aJ^bJ^c) + \ cos (a- 6+c) + \ cos {a+h-c) , or cosa cos6 cosc=ico82s+^cos2{«-a)+Jcos2ls— t)+icos2(s-c). Introducing this, and the corresponding value of cos A cosB cosC, into the foregoing value of N2_n2, dividing the result by N^ andn^, and restoring the values of these quantities in the second member, we obtain cos2s-}-cos2 (s— a)-f-cos 2 (3— 6)- hcos2(8— c) J__l sins sin{s — a) sin(s — h) siii(s — c) cos2 S-[-cos2(S— A)+cos2(S— B)4-cos2(S— C) "^ —cos S cos( S—A) cos ( S— B cos ( S— C ) i ....(182) N2-2' By alternately adding and subtracting unity in the numerators of this formula, we obtain, by (31) and (32), cos 2g — sin^ (s — a) +cos^ (g — h) — sin^ (g— c) \ sins sin(s — a) sin (5 — 6) sin(s — c) 1(183) cos^S— sin2(S— A)+cos^(S— B)— sin°~(S— C) — cosScos(S— A)cos(S— B)cos(S— C) >' It is evident that this formula will admit of several variations, according to the parts of the numerators which are increased or di- minished by unity ; and it would be easy to introduce various other modifications.* 138. The foregoing formulas would furnish other expressions for n and N, and values for sin s, cos S, &c. As the investigations of such expressions, however, present no considerable difficulty, they are left for exercise to the pupil. We may now proceed, therefore, to investigate the methods of determining the inscribed and circumscribed circles ; and we shall thus obtain some curious and interesting results. 139. Let D {fig. 2^) be the pole of a circle inscribed in the spheri- cal triangle ABC, and E, F, G, the points of contact; and let DA, DE, DF, DG, be arcs of great circles. Then the angles at E, F, and G, are right angles ; and, since DF is equal to DG, and DA common to the triangles ADF, ADG, the sides AF, AG are (113) equal, as also (112) the angles DAF, DAG. In like manner, it would appear that BE=BG, and CE=CF. Hence, AF-fBE+EC is half the sum of the sides, that is, AF+a=5; and consequently AF=s — a. * The formulas in this and the preceding No. the author has never before met with. Th^y are curious, particularly those in the latter No. on account of their complete symmetry. H 58 CIRCUMSCRIBED CIRCLE. Now, if DF=r, the rightangled triangle AFD gives (112) tanDF= sinAF tanDAF, or, by (90) and No. 123, tanr= / sin(g-^) sin(g— 5) sin(g— c) _ n ^ V sins sins ^ 140. Dividing n in this formula by the value of sins found from (161), we get n n . or tan ri_ , . , ^ , ^ . .... sms cosf A cos -^ is cos-^lj sin a sin 6 smc In this, for n^ substitute its value according to (175); then, by con- tracting, there will be obtained N . 2cosiAcoslBcosiC .,„^. tanr=r-T— T-r r.^ Trq, or cotr=: ^ -^ ^ ..(185) 2siniA cos^B cosiC N ^ ^ 141. If three circles were described, each touching one of the sides externally and the other two produced, and if r' were put to denote the arc drawn from the pole to the circumference of the one touching a externally, and r" and r"' the like arcs in those touching h and c externally, it would be found, in nearly the same manner as in No. 139, that tan.' ^ /Bm^sin(.-^)sin(.-c)*^ n_ V sin (5 — a) sin(5 — a) ^ ^ tanr"=y™ilillfc=2l|!li!==l) =^^^ (187) '^ sin(s— 6) sin(s— 6) "^ ^ tanr"'=y2Slfiliit^)41f=±> =---^ (188) "V sm(s — c) sm(s — c) ^ ^ 142. By taking the continual product of the members of (184), (186), (187), and (188), we find the following remarkable symmetri- cal formula : tanr tan/ tanr" tanr"'=sins sin(s — a) sin(s — 6) sin(5 — c)=n2,(i89) 143. From D (Jig. 27), the pole of the circle described about the triangle ABC, let arcs of great circles be drawn to the several angles: let, also, the arc DE be drawn perpendicular to the side a. Then, the triangles ADB, ADC, and BDC being isosceles, the angles at their bases are equal. Hence, ABD-}-ACD=A; and therefore DBC4-DCB=:B + C — A, and DBE = ^(B-fC— A)= S— A. * This formula may be easily found from considering that a, and the continua- tions of h and c, foi-m a triangle having for sides a, w — 6, and ^ .... (194) -v — cos S cos (S — A)cos(S— B) N ^ ^ 146. Take the continual product of these and of (190); then, tanR tanR' tanR" tanR'" =:i^, or cotR cotR' cotR" cotR"'=N^... (195) * If ^ be the perpendicular from A to BO, we have sin p= sm C sin6, or (93) sin » = -: — = -r—. ^ — . Hence, from ( 1 9 1 ), by multiplying, we get sm p tan R = ^ —. ^ C08^«. In a plane triangle, this becomes (by Nos. 162 and 163) pl^^\hc\ from which, by doubhng, andputting D for 2R, we get2?D=6<;, the same as Buchd, VI. D. 60 AREA OF A SPHERICAL TRIANGLE. 147. Since the sides of the supplementary triangle are cr — A, •r — B, cr — C, by substituting these in (184) instead of a, h, c, we should obtain, in respect to the circle inscribed in that triangle, tanr- / cos(S— A) cos(S— B) cos(S— C) cot>* =y; — cosS — cosS cos(S— A) cos(S— B) cos(S— Cy Now, this value of cotr being the same as that of tanR (190), it follows that r and R are complements of each other; whence it ap- pears, that the arc drawn from the pole to the circumference of the circle inscribed in a spherical triangle, is the complement of the arc drawn from the pole to the circumference of the circle described about the polar triangle, which seems to be a new relation of these triangles. 148. To investigate the method of finding the area of a spherical triangle ABC (fg. 28), let the circle of which AB, one of the sides, is an arc, be completed, and let the continuations of AC and BC cut its circumference in D and E, and intersect each other in F, on the other hemisphere. Then, since (page 26, note 3) BCE and CEF are semicircles, BC and EF are equal; and for a similar reason, AC and DF are equal. We have also the angles F and C equal, as they are each the inclination of the planes of AC and BC. Hence the tri- angles ABC, DEF are equal. Now, the lune, bounded on the sur- face of a sphere by the halves of two great circles is evidently pro- portional to the angle at which those circles are inclined. Hence, if H denote the surface of a hemisphere, we have 180° : A : : H : H A lune ABDC=^r^o. In like manner we should find the lune TT T5 II Q BAEC = ^, and the lune CEFD=j^. Now, it is evident that the sum of these lunes is equal to the surface of the hemisphere AEDB, and twice that of the triangle ABC ; that is, — ^~i^ ' = H-f 2 ABC; whence we readily find ABC= ^ l8Qo ~ Now (Diff. and Int. Calc. No. 288), the surface of a hemisphere is equal to twice the surface of a great circle of the sphere, or to 2 err % where r is the radius of the sphere. Putting T, therefore, equal to the area of ABC, we have ^^.(A+B+C-180») jgg. '■- 180" ^ ' FORMULAS FOR THE SPHERICAL EXCESS. 61 Hence ISO^ : A+B + C — 180° :: crr^ : area of ABC; and it appears, therefore, that (in the same sphere) the area of a spherical triangle is proportional to the excess of the sum of its angles above two right angles, or to what is called its Spherical Excess.* It ia also plain, that, when the spherical excess or the sum of the three angles is known, the area can be determined from it ; and that all triangles on the same sphere which have the same spherical exc ss, are equal, however dissimilar they may be. 149. By arcs of great circles drawn from any angle of a spherical polygon to all the remote angles, we may divide it into as many tri- angles as it has sides, wanting two. Hence, if n be the number, and S the sum of its angles, we shall have the sum of the spherical exces- ses of the triangles =S — (n — 2)xl80°; and consequently, putting P to represent the area of the polygon, we have 180" : S— (n— 2)xl80o : : crr^ : P (197) 150. "When the data in a spherical triangle are not the three an- gles, the angles, or their sum, must be found by some of the methods explained in Section IV., before (196) can be applied in finding the area. Thus, if the three sides be given, the angles may be computed by Nos. 93 and 94 ; or if two sides and the contained angle be given, the half sum, and thence the sum of the remaining angles, may be found by (129). In these two cases, however, which most frequently occur, formulas have been discovered which give the spherical excess without the previous determination of the angles. Thus, putting S equal to ^(A+B-f-C), and E equal to the spherical excess, we have JE=S — 90°, sin^E=: — cosS, cosiE=sinS; and tanS=---cotiE ; and, since (108) tani(A+B)=^^?i^^'^.cotiC, we have, from (39), by taking i(A-f B) as one arc, and ^C as an- other, and by multiplying the numerator and denominator of the se- cond member by cos^{a-\-h), tanKA+B+C)=-cotiE= "°^K'^^);°t|C+cosK«+5)tan^C _ 2^ ' ' ^ 2 cos^(a-\-b) — cos -1(0 — b) By multiplying again the numerator and denominator of the second * This remarkable and beautiful theorem is ascribed to Albert Girard, a Fle- mish mathematician. In 1787, upwards of 150 years after its discovery, an important and ingenious application of it was made by General Roy, in cor- recting the angles observed m the Trigonometrical Survey of Britain; — a fact which proves, with many others, that no principle should be rejected as use- less, however long it may have been known merely as a speculative truth. 62 FORMULAS FOR THE SPHERICAL EXCESS. member of this by 2sin^C cos JC, and then by modifying the nu- merator by (31) and (32), and the denominator by (30), we obtain, by changing the signs, ot 1 E — ^Q^K^ — h)+co9^ {a-\-h)-\-{GOS^ {a — h) — cosi(a-{-5)jcosC ^ "" {GOS^(a — h) — cosi(a-f-6)} sinC From this, by modifying the denominator and the first and second terms of the numerator by (25) and (5), and dividing the denomi- nator and the rest of the numerator by their common factor, we obtain ^.^ cotia coti5-{-cosC^ COt^ E = ^ r— sni C 151. If the three sides be given, we have (177) — cos S, or (198) dni E- "^^^"^ sin(5— ci?) sin(s— 6) sin(g— c) ^ 2cos^a C0S-J6 cos^c 152. From (198) we may eliminate cos C and sin C. To do this, mul- tiply the first term of the numerator by 2 sin -J a cos 1 a. 2 sin 1 6 cos J 6, and the denominator and the remaining term of the numerator by what is equivalent, sin a sin&, and there will arise _2cos-^a. 2cos2i6-j-sina sin6 cosC cot -g Hi — : : ; : — • sma smo smO Now, in this, the first term of the numerator is (31) equivalent to (l-fcosa) (l+cos6); the second, by (87), to cose — cos a cos 6 ; and the denominator, by (93), to 2n. Substituting these, therefore, per- forming the actual multiplication, and contracting, we obtain ^.tiErr l + cosa-|-cos&-fcosc ^200) 2Vsins sin (5 — a) sin(s — h) sin(s — c) 153. The product of this and (199) is cosiE=y^ The second of these expressions is obtained by putting the nume- rator of the first under the form l+cosa+l+cosft-J-l+cosc— 2,and ia-\-GOsh+cosc _ cos^^a-{-(iOS^^h-{'(iOB^^c — 1 /oqi\ a cos^h cos4c "~ 2cosia cosi6 cos^c these t undei modifying it by (31). * This may be derived more easily, though not so conformably to a just analysis, in the following manner : By taking the reciprocals of the members of equation (c) in the note to No. 82, expanmng cos(S— C), and performing the actual division by cos C we get cot^«oot^6= — cosC — tanS sinC : and, by resolving this for — tanS, or its equal cotiE, we find (198). SPHERICAL LOCI. 63 154. By subtracting the members of (201) from unity, and divid- ing the remainders by the members of (199), we obtain, by (49), taniE= 1 — cos^^cc — cos^^h — cos^|c+2cos|ct cos^6 cos|c Vsins sin (5 — a) sm(s — h) sin (5 — c) Now, the numerator of this is the product of the two factors, cos -Jo — cos J 6 cos^c+sinJ6 sin-Jc, and — cosJa4-cos^6 cos-Jc + sini^ sinjc; or (14) and (15) cos^« — cos^ (h-\-c), and — cos ^ a -j- 00s J (6 — c); and these again are (23) equivalent to 2sin:^(a-f 64-c) sinj( — a-\-h-{-c), and 2sin^(a — h-\-c) sin^(a+6 — c); or 2 sin is sini(s — a), and 2 sin J (s — h) sin -1(5 — c). Substituting these in the numerator, modifying the denominator by (30), and dividing the numerator by the denominator, we finally obtain tan;J^E=Vtanis tanj(s — a) tan 1(5 — h) tan J (5 — c) .... (202) This beautiful formula, which gives the spherical excess by means of the three sides, was discovered by Lhuillier of Geneva. 155. If the base and area, or, which is the same, (196) if the base and the sum of the angles of a triangle be given, the locus of the ver- tex may be found in the following manner. In the circle BCB'C, (Jig. 27) take BC equal to the given base a, and BOB', CBC, each equal to a semicircle ; and make the angles BCD', CB'D' each equal to half the sum of the given angles: then a small circle described through B' and C, and having D' as its pole, will be the locus re- quired ; that is, the vertex A may be taken any where on its circum- ference.' For, drawing BAB', CAC, and AD', arcs of great circles, we have (page 26, note) the angles AB'D', AC'D' together equal to B'AC, or its equal A in the triangle ABC ; and BB'C, CC'B are respectively equal to B and C. Therefore, wherever A is taken on the circumference of the small circle, the sum of the three angles A, B, C, is equal to the sum of the angles CB'B, AB'D', BC'C, and AC'D', which, by construction, are together equal to the given sum of the angles.* * This curious proposition was discovered by Lexell, and first appeared in the Petersburgh Acts. The proof given above is more simple and easy than any other that the author has seen. It would be shown in a similar manner, that if the base B'C, and the area of the triangle AB'C, be given, the circle described about ABC is the locus of its vertex. 64 RIGHTANGLED TRIANGLES. 156. It is evident from No. 143, that if the base of a spherical tri- angle and the difference between the vertical angle and the sum of the other two be given, the locus of the vertex is the circumference of the circle described about the triangle. 157. In resolving rightangled triangles by the methods already ex- plained (Nos. 84 and 85), when the required quantity is to be found by its sine or cosine, if the feine or cosine be nearly equal to the radius, the quantity required cannot be found from the ordinary tables with much accuracy. This practical inconvenience may be obviated in different ways according to circumstances. Thus, from (112), (113), and (114), we have, by (19), (18), and (16), sina =:i[cos(Aa2c)— cos(A+c)} (203) cose =i[cos(ac/3 6)4-cos(a4-6)} (204) cosA=z|{sin(a4-B)— sin(a— B)] (205) Any of these will give the required results by addition and subtrac- tion, by means of natural sines and cosines ; and tables of these give the results in the circumstances under consideration, with more pre- cision than can be attained by means of the other tables. This inconvenience may also be obviated by first finding a part not required ; and then, by means of it and one of the given parts, the required part may be determined. Thus, if the legs were given, to find the hypotenuse, and if the legs, and consequently the hypote- nuse, were very small, instead of using the common formula (113) cos c=: cos a cos 6, we might first find A from the equation (112) sin 6 =cot A tana, and then c from the equation (114) cos A=tan6 cote; in both of which the results may be obtained with great accuracy. 158. Some of the formulas may be modified so as to give the final results very easily and accurately by means of tangents. To exem- plify this in some of the most useful instances, let A and a be given to find B: then we derive from (114) the analogy, 1 : sinB : : cosa : cos A ; whence, by composition and division, 1 — sin B _ cos a — cos A 14-sinB"" cosa4-cosA* Now, since sinB=cos(90° — B), this becomes, by (32), (31), and (25), and by extracting the square root, tan(45<'— JB)==tVtani(A-fa)tan4(A— a) ... (206) The arc thus found may be either positive or negative. Calling it, therefore =t:, we shall have B=90° d=2*. The value of B may COMPARISON OF PLANE AND SPHERICAL TRIANGLES. 65 thus be found with great accuracy, and the formula (206) is adapted to computation by logarithms. 159. We have also bj (112), 1 : sin A :: sine : sin a; and by processes exactly similar, we get tan(45o_i.).:.yjgg^:) (207) tan(45«_iA)==.ygg=g (208) 160. From (113) we have 1 : cosa : : cos6 : cose; whence, by a like process, we get tanla=:Vtani(c4-6) tani(c— 6) (209) 161. We have likewise, from (112), (113), and (114), sina=tan6 cotB, cosc=rcot A cotB, and cos A=tan6 cote. Subtracting the members of the first of these from unity, and also adding them to it, we obtain, by dividing the remainders by the sums, 1 — sin a 1 — ^tan6 cotB 1 -f sina ~ 1 -f tan 6 cot B" The first member of this may be reduced as in former instances ; and, by multiplying the numerator and denominator of the second mem- ber by cos 6 sinB, we obtain by (13) and (12), and by extracting the square root, ^„(45«-i«)=^yJ^) (210) By operations nearly similar, we obtain, from the other two formulas, tanio = y 322^1^ (211) ^ ^ COS (A — B) ^ ^ taniA=y!!?i^) (212) ^ ^ sm(c4-6) ^ ^ 162. If, while the sides of a spherical triangle retain always the same absolute length, the radius of the sphere continually increase, the triangle will become more and more nearly a plane one, and it would actually become such, if the radius were infinite. In this case, the side and its sine and tangent would coincide and become equal, while the cosine would become equal to the radius, and consequently infinite. The cotangent would also be infinite. From this it follows, that any formula respecting a spherical triangle, which contains the sines or tangents of the sides, or the sines or tangents of their sum, half sum, &c. will also hold respecting a plane triangle, if the terms I bb COMPARISON OP PLANE AND SPHERICAL TRIANGLES. sine and tangent, thus used, be omitted, so as to leave simply the side instead of its sine or tangent. In this way, the theorem in No. 76 becomes the same as that in No. 21 ; and (88), (89), (90), and (93) are reduced to (79), (80), (81), and (84). Since, also, in the plane triangle, tani(A+B)=cotiC, (109) and (151) by slight re- ductions become the same as (75). The same principle, also, reduces (91) and (92) to (82) and (83), and (1 10) and (HI) to (76) and (77). So, likewise, (117), (121), (149), (152), (153), (154), (155), (156), (157), (158), (159), (160), (161), (162), (164), and many others, all give results which are true respecting plane triangles ; and (202) be- comes, by a sHght reduction, the same as the expression for the area of a plane triangle, given in the note in page 20. 163. The formulas in (112) and (114) become immediately appli- cable to plane triangles by taking cos a equal to the radius, and by multiplying in the latter by cote. By the same means, the first for- mula in (113) gives tanA=cotB. The second, by squaring, and by (6), becomes 1 — sin^czrl— sin^a — sin^^-^-sin^a sin^h. Hence, re- jecting 1, changing the signs, and rondering the terms of the same dimensions by dividing the last term by r^, that term, in case of the plane triangle, will vanish, the radius being infinite, and the formula will be reduced to c^ =za^ -{-h^ , the same as Euclid I. 47. 164. By a like substitution, (85) would become V(l — sin2a)zr:cosA sin6 sinc+V(l — sin^^ — sin^c+sin^Z) sin^c). ^ The first member of this, by extracting the square root in series, be- comes 1 — J sin 2 a — "I- sin* a — &c.; while the second, by a like process, is changed into cosA sin6 sinc-f-1 — ^sin^h — -^sin^c+Jsin^J sin^c — -|-sin*6 — &c. Hence, by rejecting 1, dividing the terms of four dimensions by r^, multiplying by 2, and otherwise proceeding as in the last No. we get a2=62-t-c2— 26c cos A, the same as (78). ASTRONOMICAL AND GEOGHAPHICAL PROBLEMS. G7 Vr.— ASTRONOMICAL AND GEOGRAPHICAL PROBLEMS. 165. Spherical Trigonometry derived its origin ft*om the com- putations which are necessary in astronomy ; and its principal and most important applications are still furnished by the same science. Some of the most useful of these, and some of its applications in geography, are exhibited in the following problems.* * For the use of those who may be unacquainted with astronomy, or the mathematical principles of geogi-aphy, it may be proper to explain some of the terms which most frequently occur in those branches of science. The astronomer conceives an imaginary sphere of great, though indefinite magnitude, to have the same centre as the earth ; and, in all applications of spherical trigonometry, he employs, instead of the real position of any of the heavenly bodies, the point in wnicn the surface of this sphere is cut by a straight line joining the centres of the earth and the body. ( 1 ) The extremities of the axis on which this sphei'e appears to revolve, in consequence of the earth's diurnal rotation, are called the poles of the celestial sphere, or of the heavens ; and (2) the points in which the earth's surface is cut by its axis, are called the poles of the earth. (3) The great circle on the imaginary sphere which has the poles of the heavens as its poles, is called the celestial equate or the equinoctial; and (4) the corresponding great circle on the surface of the earth is called the terrestrial eaiiator. Hence it is evident, that the celestial and teiTestrial equa- tors are in the same plane. (5) In consequence of the earth's motion in its orbit, the sun appears to move eastward round the heavens, and, in the course of a year, to describe, among the fixed stars, a great circle which is named the ecliptic. (6) This circle is inclined to the equator at an angle which is called the obliquity of the ecliptic, and which, in the lapse of many centuries, varies within naiTow limits. On the first of January, 1838, the mean obliquity was 23° 27' 3*7", and at present it is diminishing at the very slow rate of 52" in a centmy ; not, however, each year by the same quantity ; and it increases and di- minishes by minute quantities at dilFerent times of the year. (Y) The points in which the equator and ecliptic intersect each other, are called the equinoctial points; — the one at which the sun appears to cross the equator northward, the vernal equinoctial point, or the Jirst point of Aries ; the other, the autumnal equinoctial point, or the first point of Libra. (8) A secondary to a great circle is another great circle perpendicular to it. (9) If a secondary "to the equator be drawn through any point of the celestial sphere, the part of it between the point and the equator, is called the declination of that point; and (10) the part of the equator extending eastward from the first point of Aries to the secondary, is called the right ascension of that point. (11) In like manner, if a secondary to the ecliptic be drawn through any point of the sphere, the part of it between'the point and the ecliptic is called the latitude of the point ; and ( 12 ) the part of the ecliptic extending eastward fi-om the first point of Aries to the secondary, is called the lx>ngitude of the point. This was formerly reckoned in si^ns of 30" each, and of which the echptic must therefore contain twelve. It is now reckoned in degrees. C8 ASTR02J0MICAL AND GEOGRAPHICAL DEFINITIONS. 166. Given the obliquity of the ecliptic, and tlie right ascension and declination of any of the heavenly bodies ; to find its latitude and longitude. In geography, (13) a secondary to the equator passing through anyplace, is called the meridian of that place ; and (14) the lotu/itude of a place is the smal- ler of the two parts into which the equator is divided by the meridian of the place and the/rsf meridian, that is, the meridian from which geographers have agreed to reckon the longitude. The British geographers assume tlie meridian of the Observatory of Greenwich as the first meridian ; but nature points out no particular meridian, in preference to another, for this pui-pose. (15) The lati- tude of a place is its distance from the equator measured on its meridian. ( 16 ) If a plane touch the earth, its intersection with the celestial sphere is called the sensible liorizon of the point of contact; and (17) a great circle parallel to this is the true or rational hmizon of the same place. (18) Of the poles of the ho- rizon of any place, that which is over the place is called the zenith, and the other the nadir. ' (19) The arc of the horizon mtercepted between the meridian and a secondary to the horizon passing through any of the heavenly bodies, is the azimuth of the body; and (20) the arc of the secondary between the body and the horizon, is its altitude. (21 ) The complement of the azimuth of a body at rising or setting, or its distance from the east point at rising, or fi'om the west point at setting, is often called its amplitude. (22) A secondary to the horizon is called a vertical or azimuth circle; and of such circles, that which is pei^pen- dicular to the meridian, and which, consequently, cuts the horizon in the east and w^est points, is called the prime vertical. Asti-onomical observations show, thit the apparent diurnal motions of the fixed stars in the circles which they appear to describe, are at all times unifonn ; and hence it follows, that the earth's rotation on its axis is likewise uniform. The sun's apparent diurnal motion is also very nearly uniform, being afl'ected only by a slight inequality arising fi'om his apparent motion in the ecliptic. If we regard his apparent motion as unifonn, or rather, if we consider him as moving with his mean or average motion, it is evident that since he appears to describe round the earth a circuit of 360° in twenty-four hours, he must describe 15" each hour, and consequently a degree in four minutes, a minute of a degree in four seconds of time, &c. Now, it is evident that the portions of the equator and of any circle parallel to it, intercepted between two circles passing through the poles, will contain the same number of degrees, each measuring the inclination of the planes of those circles. (23) Such circles, considered with respect to time, are called hmir circles ; and it follows, that the arc of the equator inter- cepted between any two of them is proportional to the time occupied by a body in passing from one of them to the othei" in its apparent diurnal revolution. Hence, also, of the arc of the equator and the corresponding time, if either be given, the other will be known, an hour and 15° being equivalent. Degrees, minutes, &c. are most easily reduced to time by multiplying by four, and taking the degrees, minutes, and seconds of the product as' minutes, seconds, and thirds of time ; and time is readily reduced to degrees, &c. by multiplying it by ten, and adding to the product half of itself ; the result, instead of hours, minutes, &c. will be degrees, minutes, &c. (24) The mean period employed by the sun in apparently moving from the meridian to the meridian again, is twenty-four hours, and is called a mean solar day. ( 25 ) The diliei-ence between the time at which the sun actually passes the meridian and the time at which he would pass it, were he to move equably in the equator, or in a parallel to it, is called the equation of time. (26) The period that is occupied by a fixed star in passing from the meridian to the meridian again, is 23 hours, 56 mi- nutes, 4'1 seconds, of solar or civil time, and is called a sidereal day. This quantity, which is the time of the earth's revolution on its axis, is always the RIGHT ASCENSION AND DECLINATION. 69 To resolve this useful problem, let EQ and KL {fig. 29) be the equator and ecliptic, P and P' their north poles, and the first point of Aries ; and let the bodj S be so situated, that its right ascension, OR, and its longitude, OG, may be each less than 90°; and its decli- nation, SR, and latitude, SG, both north. Then, in the spherical tri- angle FPS, PT is equal to LQ, the obliquity of the ecliptic, each being the complement of PL ; PS=90°— SR=90°— Jec. or the north polar distance; FS=900— SG=90°— Za^.; FPS (measured by ER) z=%0^-\'Hght asc; and PFS (measured by GL)=90°— ?on.: and the resolution will be effected by the third case of spherical trigono- metry. Thus, drawing SV perpendicular to PL, and putting

(213) sin Ion. = tsinlat. t8i>n((p-{-ol>l.) } These formulas will serve for all positions of S, if, when the lati- tude and declination are south, they, and consequently their sines, tangents, and cotangents be taken negative ; and if it be considered, that, as is evident from the diagram, when either the right ascension or longitude is between 90° and 270°, the other is between the same limits. P'SP, which is called the angle of position, is easily calcu- lated. 167. When the obliquity, and the latitude and longitude, are given, to find the declination and right ascension, we have PT, P'S, and the contained angle, and the resolution of this also is effected by the third case. Thus, putting 9:= P'V, and proceeding as in the last No. same ; while, as we have already seen, the length of the solar day varies slightly at different times of the year, being, at Christmas, thirty seconds more, and at the middle of September, twenty-one seconds less, than twenty- four hours. The number of sidereal days in the year is evidently one more than the number of solar, the sun losing one revolution in a year, in conse- quence of his apparent motion among the fixed stars. 70 RIGHT ASCENSION AND DECLINATION. we find the following formulas, which, with attention to the refmarks at the end of that No. will solve the problem in every case : tail (p = sin Ion. cotlat. ^ sindeczzzsmlat. cos(

(^14) sin E. A. z= tan dec. tan(

8" 36% * Twilight being supposed to begin and end when the sun is 18° below the horizon, we have the zenith distance 108°, the polar distance 81° 30', and the oolatitude 35° 24': and the solution will be effected by the firat case. f To illustrate the method of solving this problem, let B and w {fig. 28) be any two positions of the sun, and suppose ?f'B, wV, to be joined by arcs of great circles. Then, in the isosceles triangle BPw, compute Bu', and the adjacent angles. This is most easily effected by dividing it into rightangled triangles by an arc of a great circle bisecting the ande P, and consequently the side Bit/. Then, the three sides of the triangle WLw being given, let the angle ZB«/ be computed, and the difference between it and Vaiv will be ZBP; by means of which, and of the sides containing it, the colatitude ZP will be found by the third case. Practical modes of solving this useful problem wiU be found in the later works on navigation. The problem in which the altitudes of two known stars, taken at the same instant, in the same place, are given, to find the latitude and the time of obser- vation, is solved in the same manner, except that, unless the stars have the same declination, the triangle "BViu is not isosceles. X In solving this question, there are given, in an isosceles triangle, two sides, each equal to the complement of the latitude, and the contained angle equal to the difterence of longitude. A perpendicular to the base, fi-om the opposite angle, will bisect both ; and, hence, the resolution will be eftected by No. 85 or No. 107. The base is the distance on the arc of a great circle, and the peipen- dicular is the complement of the highest latitude. The radius of the parallel will obviously be tne cosine of the latitude, the radius of the earth being taken as unity. Hence, the distance on the parallel wiU be found by the analogy, 1 : cos lat. : : diff. Ion. : dist. on parallel. ASTRONOMIC iL AND GEOGRAPHICAL EXERCISES. 77 the arc of a great circle from Port Jackson to Cape Horn, tbeir lati- tudes being 33'^ 51' and 55° 58' S., and their difference of longitude 1400 27'^*— Answ. 72° 41'. Ex. 17. In a given latitude, and on a given day, how may the time be found at which two given stars have the same azimuth?! Ex. 18. The same being given, how may the time be found at which the stars have the same altitude or depression? J Ex. 19. Given the altitudes of two known stars, at the instant when they have the same azimuth ; to find the latitude. || Ex. 20. Given the difference of the azimuths of two known stars, at the instant when they have the same altitude; to find the latitude. Ex. 21. To find the latitude at which two given stars have always the same altitude, when two others have the same azimuth. § * Here we have two sides (the complements of the latitudes) and the con- tained angle (the difference of longitude), to find the perpendicular di'awn from that angle to the opposite side, that perpendicular being the complement of the required latitude ; and the solution may be effected by finding the remaining angles, and then the perpendicular by the rules for resolving rightangled trian- gles ; or, more easily, by finding the parts of the contained angle by (151), and thence (114) the pei-pendicular. f S and S' {Jig. 31) being the stars, find (case III.) in the triangle SPS', the angle PS'S ; and in PZS' find (case V.) the angle ZPS', which will show the difference of time between the star S' being in its present position and on the meridian. I S and S' {fig. 32) being the stars, bisect SS' in M, and join MP, MZ. Then, in the triangle SPS', compute (case III.) SS' and PSS'; and in SMP, compute (case III.) MP, SPM, and SMP. From this last angle take SMZ=: 90*^, and there will remain ZMP; and the resolution of the tnangle ZPM, by case V. will give ZPM ; the difference between which and the angle SPM will be ZPS; and this will give the required time. From this, the method of solving Ex. 20 wUl be manifest. II This problem, the method of solving which is plain from the note to Ex. 17, affords a mode of finding the latitude of a place. The instant at which the stars have the same azimuth can be ascertamed by means of a plumb-line. § To show the method of solving this question, let A, B, C, D {Jm. 31) be the given stars, and suppose gi-eat circles to be drawn joining AC, BC, BD, PA, PB, PC, PD, and ZB; suppose also ZE to be drawn pei*pendicular to AB, and consequently bisecting it. Then, since the stars are given, their right ascensions and declinations are given, and their distances asunder can be com- puted, as also the angle BCP, by No. 1G9. In the next place, the sides of the triangles ABC and BCD being given, the angles AB(f, BCD will be found by case I.; and thence CBF and FCB, their supplements, will be known. From these two angles, and the adjacent side BC, the sides BF and CF, and the remaining angle F, can be computed by case IV. Then, in the rightangled triangle ZEIi , the angle F and the side EF are given, to find ZF : fi'om which, and h-om CF, CZ will be known. The angle ZCP will also be known from BCP and BCD ; and CP being given, the colatitude ZP will be found by case III. This solution would evidently admit of several variations. 78 ASTRONOMICAL AND GEOGRAPHICAL EXERCISES. Ex. 22. Given the latitudes and longitudes of three places on the earth's surface ; to find the latitudes and longitudes of the two places equally distant from them.* Ex. 23. Given as in the last problem ; to find the latitudes and lon- gitudes of the poles of a circle touching the three great circles passing each through two of the places. Ex. 24. At a given place, to find the greatest azimuth of a given star whose declination is greater than the latitude of the place: to find also the time, on a given day, when the star will have the greatest azimuth, and when, consequently, it will appear to move perpendicu- larly to the horizon.! Ex. 25. On what days of the year is the sun on the horizons of Dubhn and Pernambuco at the same instant ; their respective lati- tudes being 53° 21' N., and S^ 13' S.; and their longitudes 6° 19' W., and 35^ 5' W.?| — Answ. The sun will set at the same instant at both on the 12th of May and tlie \st of August, and will rise at both at the same instant on the 2^th of January and the 14«A of November y the declination being 18° 6'. * There may be a similar problem respectmg stars. f The required point is that in which a vertical circle touches the parallel of the star's declination, and a cu-cle drawn from the pole to the point is perpen- dicular to that vertical circle : whence the azimuth, hour, and altitude may all be found by resolving a rightangled triangle. — This exercise will fiu-nish an ex- planation of the cm-ious ftict, that when the sun's declination is greater than the latitude of a place on the same side of the equator, the shadow of an up- right object on a norizontal plane goes backward each day dm-ing a certain period, which may be computed. In any latitude, indeed, the shadow of a pin perpendicular to a plane will move d.uring a part of each day, contrary to the usual du-ection, if the plane be so placed that the pin will be directed to a point of the meridian ftu'ther from the elevated pole than the complement of the sun's declination. X In solving this problem, it is evident that the difierence between the houi's of rising or setting at the two places must be equal to the difference in their reckoning of time, that is, to the time equivalent to their difference of longi- tude. Denoting, therefore, the latitudes of Dublin and Pernambuco by I and ?', the angles corresponding to the times from noon to rising or setting by P and F, and the difference of longitude by D, we have (note 1, page 73) cosP= —tan I tan dec. and cos P'= — tan I' tan dec. Dividing the latter by the former, and converting the quotient into an analogy, we obtain cos P' : cos P : : tan V : tanl; whence, by a process the same as in No. 118, we obtain the following analogy : sin(Z-f-Z') : — sin(Z— f) : t cot|D : tani(P-hP'). Hence, |D being half the difference of P and P', these quantities become known ; and thence the declination from the formula cos P= — tan lat. tan dec. It must be observed m the computation that the latitude of Pernambuco is to be taken negative. DIALLINa. 79 Ex. 26. Given the longitude of the ascending node* of Ceres = 80^ 54', and that of the same node of Pallas =172° 31'; given, also, the inclination of the orbit of the former to the ecHptic=10° 37'^, and that of the latter = 34° 37'; to find the mutual inclinations of the orbits of Ceres and Pallas. — Ansiv. 36° 18'. VII.— DIALLING. 175. As a farther application of the principles that have been thus far established, we may now investigate some of the more important parts of the theory of dialling ; a branch of science which, notwith- standing the very improved methods of measuring time that modern ingenuity has discovered, is still of some value, and of considerable interest. Since the sun's apparent diurnal motion is uniform, it is plain that an opaque straight line or wire, occupying, in free space, the same position as the earth's axis, would cast a shadow, the plane of which would revolve uniformly, describing an angular space of 15° in each hour ; and the same wiU hold, without sensible error, respecting any line at the earth's surface parallel to the axis, its distance from the axis being extremely small compared with the sun's distance from the earth. Hence, if HR (Jig. 30) be a great circle, parallel to the plane of a dial, such as a horizontal dial at the place whose zenith is Z, wo have only to find how HR would be divided at different times by the shadow of the axis PO, which would evidently divide the equator EQ, and consequently, the angular space about P, into parts propor- tional to the times in which they are described. To effect this, wo have, in the rightangled triangle PR5, PR = 90° — RQ = the comple- ment of the inclination of the planes of the equator and dial ; RP5, the measure in degrees of the time between the instants at which the shadow occupies the positions PQO and PLO ; and Rs will be the * The two opposite j)oints in which the orbit of a planet cuts the ecliptic, are called the nodes of its orbit; — the one in which the planet crosses the eclip- tic northward, the ascending node, — ^the other, the descending one. 80 HORIZONTAL DIAL. measure of the angle at the centre of the dial corresponding to tliat time ; and by the resolution of this triangle, we have R : sin PR (or cos RQ) :: tauRPs : tanR^. Hence, putting I to denote the inclination of the plane of the dial to that of the equator, P equal to the horary angle RP5, and H equal to the arc R5, or the hour angle on the dial, we have R : cosi : : tanP : tanll (219) 176. In case of a horizontal dial, the angle I is evidently equal to the complement of the latitude, and (219) becomes R : ^mlat. :: tanP : tanH (220) 177. In case of a vertical south or north dial, that is, a dial whose plane is perpendicular to the meridian and horizon, and which, there- fore, has its face directed exactly towards the south or north point of the horizon, I is equal to the latitude; and, therefore, (219) be- comes • R : co^lat. :: tauP : tanH .... (221) 178. To excmphfy the use of (220) in the construction of a hori- zontal dial for Belfast, by taking the latitude in that formula equal to 540 36', and P successively equal to 15°, 30°, 45°, 60°, 75^, and 900, ^e find the hour angle for one hour to be 12° 19', for two 25^ 12', for three 39° IF, for four 54° 41', for five 7P 48', and for six 90o.* Then {fig. 32) draw two parallel lines, ah, a'h', at a distance asun- der, equal to the thickness of the stile, and let them be crossed per- pendicularly by 6cc'6, tlie six o'clock hour line. Draw, also, ell, clO, c9, &c. making with ca angles respectively equal to 12*^ 19', 25° 12', &c. and draw c'l, c'2, c'3, &c. making equal angles with c'a'. The hour lines for the times before six in the morning, and after six in the evening, are found by producing 3c', 4c', 5c' , through c'; and 7c, 8c, 9c, through c. The stile is to be a firm slip of metal, or other substance, ahke thick throughout, having its back a plane inclined to the plane of the dial on the northern side, at an angle equal to the latitude, and meeting it in the points cc'. The half hours, quarters, or other minuter divisions, if the dial be on so large a scale as to ad- mit them, may be obtained with nearly sufficient accuracy by dividing the angular spaces between the hour lines into equal parts. Should * For Glasgow (latitude 55'' 52') the hour angles are 12° 30' 39*^ Z1% 55« 6'^, 72^^ 4', and 90^ EAST OR WEST l^L^ m -^ »** 8 1 much accuracy be required, however, the analogj (220),wiS^ive the angular spaces with precision by taking P in it successively equal to 7° 30', 150 + 70 30', &c. for the half hours; and equal to 3° 45', 7^ 30'4-3o 45', &c. for the quarters. It is plain that the line bounding the dial, which in Jig. 34 is a circle, may be a square, or any other figure that the maker of the dial may prefer ; and the northern side of the stile may be straight or cuiTed, or of any outline whatever. 179. A north or south erect dial would be constructed by means of (221) in a manner exactly similar. On the south erect dial it is un- necessary to put any hours, except those between six in the morning and six in the evening, as the sun can never shine on it at other times. According, also, to Ex. 11, page 75, the sun can shine on a north erect dial at Belfast, even at the longest day, only a little more than the time before seven in the morning and after five in the evening ; and, therefore, in constructing such a dial, the intermediate hours may be omitted. The angle of the stile is equal to the colatitude. 180. If the dial be a polar one, that is, if its plane be perpendicular to that of the equator, the formula (219) fails ; but the construction is more simple. Thus, in case of a vertical dial (Jig. 35) facing the cast or west, the hour lines are all parallel, and the stile is erected perpendicularly on the six o'clock hour hue, to which its back is pa- rallel. Then, if the height of the stile be assumed as radius, the dis- tances from the six o'clock line to the seven o'clock and five o'clock lines are each, according to the principles explained in No. 175, the tangent of 15° : the distances to the next lines are each the tangent of 30°, &c. Hence, if s be put to denote the height of the stile, and H' the perpendicular breadth ou the dial between the six o'clock line and any other hour line corresponding to P, the distances between the six o'clock line and the others may be computed by this analogy, R : tanP :: s : H' (222) Thus, if the height of the stile were 2 inches, we should have the dis- tance between the six o'clock hour line and the five or seven o'clock one equal to '5359 ; the distance between the same and the four or eight o'clock line equal to 1*1547, &c. These distances may be easily determined by construction, without computation, by describing from C as centre, with a radius equal to the height of the stile, an arc AB of 90°, and dividing it into six equal parts. Then, straight lines drawn from C through the several points of division will cut DE in the points 7, 8, 9, &c. The dial is to be placed so as exactly to face the east, and so that a line drawn through C, and making with CA L 82 DECLINING AND INCLINING DIALS. an angle equal to the latitude, may be horizontal ; as by this means the back of the stile will be parallel to the earth's axis. It is plain that tlie same dial would serve as a west erect one, if the numbers 1, 2, 3, &c. were substituted for 11, 10, 9, &c. 181. Having now seen the method of constructing horizontal dials, and vertical ones facing the four cardinal points of the horizon, we may consider briefly the theory of those dials whose planes occupy other positions. Of these there are two varieties, vertical declining dials, and inclining dials. A vertical declining dial has its plane perpendicular to that of the horizon, and cutting it at a distance from the east and west point called the declination of the dial ; while an inclining dial has its plane oblique to the plane of the horizon, and making with it an angle called the inclination of the dial. 182. It is plain from No. 175, that a dial which is truly constructed for any place whatever on the earth's surface, will divide the time truly at any other place on its surface, and in any position, provided its stile is perpendicular to the equator. Hence, a horizontal dial, constructed for any given place, will divide the time truly at any other, if it be so placed that its plane and its stile are parallel to their original positions ; and, conversely, a dial may be constructed on a declining or inclining plane at a given place, by finding the two places whose horizons are parallel to that plane, or, which is the same, by finding the positions of the poles of the great circle parallel to it ; as a horizontal dial for either of those places will indicate the time truly at the given place, except the difference in their reckoning of time arising from their difference of longitude, if they be not on the same meridian. 183. These principles will be illustrated by the following example: If, at Belfast, a great circle cut the horizon 33° 45' from the south towards the west, and be inclined to it at an angle of 25°, rising above it on the north-western side, it is required to find the latitude and longitude of the place in the northern hemisphere, of which this circle is the horizon. Let Z (Jig. 30) be the zenith of Belfast, and B the pole of the given circle, or the zenith of the required place. Then, since the given circle and the horizon are inclined at an angle of 25°, ZB, the distance of their poles, will likewise be 25°. It is also evident, that the angle EZB is the complement of 33° 45'; and therefore BZC is equal to 33° 45', and PZB to 123° 45'. From these, and from ZP, the co- latitude of Belfast, we find, by case III., BPZ =26° 8', PBZ = 37° 8', and PB = 52° 56'. The last of these quantities is the colatitude of INCLINING DIAL. b& the required place, and consequently its latitude is 37° 4'; also, BPZ is the difference of longitude of Belfast and the required place. Now, bj reducing this difference of longitude to time, according to note (part 23), page 68, we obtain P 44"" 32% the quantity by which the reckoning of time at the place whose zenith is B, is more advanced than the time at Belfast. Hence it appears that, were a horizontal dial constructed for latitude 37° 4', and erected at Belfast in the position described in the problem, with its stile directed to the pole, it would constantly indicate the time P 44™ 32' too far advanced ; making it appear, for instance, to be twelve o'clock, when it is only 15™ 28' past ten. The time, therefore, might be ascertained correctly at Belfast by such a dial, placed so as to be parallel to its original position, by always subtracting the constant quantity 1^ 44™ 32% This subtrac- tion, however, which would be very inconvenient, may be obviated in the following manner. 184. Draw the parallels (Jig. 36) cs, c's', making each with ah an angle of 37° 8' (= PBZ, last No.) ah being the intersection of the plane of the dial, and a vertical circle perpendicular to it : then the stile is to be erected perpendicularly on the space scc's', which is therefore called the suhstile; and, by what we saw in the last No., when the shadow falls on this space, it is 15™ 28' past ten o'clock. We have now (220) R : sin 37° 4' :: tan26° 8' : tan 16° 28', the hour angle on the dial between the substile and the twelve o'clock line. Again, R : sin 37° 4' : : tan (26° 8'— 15°) : tan 6° 46% R : sin37° 4' : : tan(26° 8'+15°) : tan27° 46', R : sin37° 4' : : tan(26° 8'— 30°) : tan(— 2° 20'), R : sin 37° 4' :: tan (26° 8'+ 30°) : tan41o 46'. These are the angles which the hour lines for eleven, one, ten, and two o'clock make respectively with the substile ; and the other hour angles are found in a similar manner. The angle of the stile is 37° 4'; and the dial is to be erected in such a manner that, if two plumb- lines be suspended, one above a and the other above h, a horizontal line passing through them may be directed to a point of the horizon 56° 15' (=90°— 33° 45') from the south towards the east, and so that the line ah may have an elevation of 25°, at the north-western side. 185. By taking ? = 37° 4', and d = 23° 28' (in the note, page 73), we get P=7'' 17™, which is half the length of the longest day at a place in the latitude of 37° 4'. Subtracting this from lO*' 15™, and also 84 EQUINOCTIAL DIAL.— MERIDIAN LINE. adding it, we get 2'> 58™ and 17'' 32™. From this it appears that, at Belfast, the sun would cease to shine on this dial on the longest day at thirty-two minutes past five in the evening ; and that, if he rose so soon, he would begin to shine on it at two minutes before three in the morning. Hence, as the sun rises that day at half-past three, it is unnecessary to mark any hours on the dial except those between that hour and half-past five or six in the evening. 186. In case of a vertical decHning dial, the process is rather more simple, ZB {fig. 30) being a quadrant. The simplest of all dials, however, is the equatorial or equinoctial one, so called from its plane being parallel to that of the equator. It will appear, from the slight- est consideration, that each hour line will make, with the one next to it, an angle of \6^\ and consequently that the positions of the hour lines^will be obtained by dividing each quadrant into six equal parts. It is manifest, also, that the stile will be a pin perpendicular to the plane of the dial ; and that, during the six summer months, the sun will shine on the upper side of the dial, and, during the rest of the year, on the other side. 187. The determination of the meridian line is necessary to enable us to give to a dial its proper position at any particular place. This may be effected by marking the position of the shadow of an object perpendicular to a horizontal plane at a particular instant, and at the same time measuring the sun's altitude by a quadrant or other instru- ment. Then, in the triangle ZBP {fig. 30) there are given the three sides to find the angle PZB, the azimuth from the north; and one of the lines drawn on the horizontal plane, making with the direction of the shadow an angle equal to the computed one, wiU be the meridian line. The same may also be effected by bisecting the angle contained by the shadows of a perpendicular object, on the same day, at the two times when it has the same altitude. If this method be employed, the observations should be made about the solstices. It may be observed, in conclusion on this subject, that, to find the true time by means of a sun dial, the time which it indicates must always be corrected for the equation of time, the table of which is given in almanacs, and various other publications.* * The foregoing section exhibits the general principles on which the construc- tion of the commoner kinds of dials depends. Dials may also be described on curve surfaces ; and there are various other modes of determining time by means of shadows, several of which are ingenious and interesting. JPor information MULTIPLE ARCS. 85 VIII.— MULTIPLE ARCS. 188. Bj squaring cos.A+sinAV — 1, we obtain (cosA4-sinAV^^)^=cos2A — sin2A-|-2cos A sin AV — i; or, by the application of (38) and (37), (cosA-f-sinAV — l)-=:cos2 A-i-sin2AV — 1. Multiply by cos A -f sin A V — 1, and modify the second member of the result by (14) and (12); then, (cos A -j- sin A V — 1 )^= cos 3 A -f- sin 3 A V — 1 . By successive multiplications by cos A -|- sin A V — 1, and by means of (14) and (12), we should find that, n being any whole positive number, (cos A -f sin A V^l )"= cos n A -f sin n A V^ . . . (223) It will appear hereafter (Nos. 190 and 191) that this formula holds true when n is anif number, whole or fractional, positive or negative. 189. By taking A negative, the last formula becomes (No. 14) (cos A — siuAV — l)"=cosnA — siawAV — 1 ... (224) This result might also be obtained from cos A — sin AV — 1, by a pro- cess nearly the same as that in No. 188 ; and it will appear hereafter (Nos. 190 and 191), that this formula will also hold when n is any number, either whole or fractional. 190. If cos n A + sin n A V — 1 be multiplied by cos n A — sin n AV^l , the product is cos^nA+sin^nA; which (6) is= 1. Hence, by di- vision, we get 1 ~ — : . , — r=:cos?2A — sinnAV — 1 ; coswA+.smnAV — i or, by substituting for the denominator its equal in (223), 1 ; T~, : ;; , — ^.„=cosnA — sinwAV — 1, or (cosA-fsmAV — 1)" ' (cos A + sin A V — l)-"=cos»iA — sinnAV^. respecting these contrivances, as weU as many other pai*ticulars connected with the subject, should the student feel a wish to devote time to a study that is more curious than useful, recourse may be had to the treatises written expressly on this branch of science. 86 DE moivre's formulas. Now (No. 14), cosnA=cos( — wA), and sinnA=: — sm( — nA). The last formula, therefore, may be written (cos A + sin A V — 1 )""= cos ( — n A) -}- sin ( — n A) V^ ; a formula which proves that (223) is true when n is a negative in- teger, as well as when it is a positive one. 191. The same formula is also true when w is a rational fraction. To prove this, it follows, from (223), that p and q being any whole numbers, positive or negative, (cos- A -1- sin- A V — l)'?=cos|)A+sinpAV — 1 = (cos A 4- sin A V — 1)'' ; whence, by extraction, and by taking the second member first, (cosA+sinAV — l)'^=cos-A+sin- AV — 1; ' q q a formula which, if n be taken instead of the fractional index, is the same as (223). It would be shown, in a similar manner, that (224) is true, while n is rational, whether it is whole or fractional, positive or negative. If n be a surd, or any other irrational number, we can find a frac- tion differing from it by a quantity less than anything that can be assigned ; and hence (223) and (224) are proved to be true for any real value whatever of n. These two important formulas were dis- covered by De Moivre. 192. The formulas (223) and (224) are true without any modifi- cation, when »i is a whole number. To show how they are true, and how they give the different values of the left-hand members, when n is a fraction, we must substitute for cos A and sin A their equals, by (10) and (9), cos(2n'';r-f-A) and sin(2»iV-i-A), n' being an integer. By this means, the first members remaining unchanged, we obtain (cosA-f-sinAV — l)"=cos(2n»i'«-}-nA)-f-sin(2nw'?r-f wA)V — 1 (225) (cosA— 8inAV--i)"=cos(2nw'«'-f wA)--sin(2nw'{r-f nA)V— -i (226) These formulas are true for every value whatever of n. If it be an integer, the second members become simply cos n Art: sinn A V — 1, and the formulas are therefore reduced to (223) and (224). If n be a fraction, the left-hand members will have, by the theory of equa- tions, as many values as there are units in the denominator of n; and these values will be found by taking n' successively equal to 0, 1, 2, 3, &c. When, in these successive substitutions, n' is at length taken SINE AND COSINE OF A MULTIPLE ARC. 87 equal to a denominator of n, the product nn' will be a whole num- ber, and (No. 12) the resulting value of the second member is the same as when »i'=0 ; so that the right-hand member will thus, as it ought, have just as many values as the left. 193. To illustrate the foregoing remarks by an example, let n = J; then (225) becomes (cosA+sinAV^)5=cos(fwV4-|A)-{-sin(|nV4-f A)V^1; and by taking n' successively equal to 0, 1, 2, and 3, we find the only four values of the second, and, consequently, of the first member to be cos|A + sin|AV — 1, cos(270°+|A) + sin(2700-fJA)V^, cos(540o + |A)-i-sin(540o-ff A)V^, and cos(810°+jA)-fsin(810<'+}A)V^; or, by contraction, by means of No. 12, and formulas (12) and (14), cosJA-{-sin|AV — 1, sin|A — cosfAV — 1, — cos| A — sin|AV — 1, and — sinJA-fcosfAV — 1. Were n' taken equal to 4, 5, 6, &;c. we should have the same series of values recurring perpetually. 194. By taking half the sum and half the difference of (223) and (224), and dividing the latter by V — 1, we obtain cos»iA=i[(cosA4-sinAV^)"+(cosA— sinAV^)"? (227) sin/iA= — ={(cosA+sinAV^)"— (cosA— sinAV^)'*K228) ^v — -L 195. By expanding the parts composing the second members of these, by the binomial theorem, and contracting the results, the ima- ginary expressions disappear, and we obtain tlie following interesting formulas for the cosine and sine of a multiple arc : cosn A=:cos"A \ o cos''~^A sin^A _^ n(n-l)(n-2)(n-S) ^^^„_,^ ^.^,^_ ^ ^^^O) A « 1 A ■ A w(n — 1) (n — 2) ^ , . ... sm n A= n cos""' A sm A ^^ rp— o cos""' A sm^ A ^n (n-l)(n-2Xn-3)(n-i) ^^^„_,^ ^.^,^_^ ^230) 88 POWERS OF COSINES. When w is a whole positive number, some one of the factors n — 1, n — 2, &c. will vanish by being n — n, and the series will consist of a finite number of terms ; but if n be fractional, the series wiU be in- finite. 196. Bj taking n successively equal to 2, 3, 4, &c. in these, and substituting 1 — cos^A for sin^A in the first, and 1 — sin^A for cos ^ A in the second, we obtain the following systems of the cosines and sines of multiple arcs : cos2A= 2cos2A— 1 -\ cos3A= 4cos^A — 3 cos A | cos4A= 8cos*A — Scos^A+l \^ /'^sn cos5A= IGcos^A— 20cos^A+5cosA /" ^"' ^ cos6 A= 32cos6A— 48cos^A+18cos'-^A— 1 1 &c. &c. &C. J sin 2 A= 2 sin A cos A sin3A= 3 sin A — 4sin^A sin4A=r (4sin A — 8sin^A)cosA \ ^232") sin5A= 5sinA— 20sin3A+16sin^A ^ ^ ^ sin 6 A= (6 sin A]— 32 sin^ A -f 32 sin* A) cos A &c. &c. &c. 197. We may now investigate the method of transforming a power of the cosine or sine of an angle into an expression composed of sines or cosines of its multiples, — an important problem which is the con- verse of the one investigated in No. 195. To effect this, assume cos A 4- sin A V — 1 = ^, and cos A — sin A V — 1 = v. Adding these together, we get 2cosA=^-J-^; whence, by the bino- mial theorem, on n A J nil ^(W— 1) „ , 3 1 W(M— 1) (W — 2) „ , , , „ or, as it may be expressed, 2" cos''A=^-^n^"-^^^;-^ ^^^~^^ ^"-^^V•f ^^^""^jl^^""^ ^^"-^^»^;''-^ &c. But, by No. 190, ^v=:l, and consequently its powers, z^v^, &c. are equal to the same. These factors, therefore, wTU disappear in the foregoing development. We have also (223) -3r"z=coswA-f.sinnAV — 1, ^"-2=008 (n— 2) A4-sin(w— 2) AVUl, &c. &c. &c. POWERS OF COSINES. 89 Then, hy substituting these for their equals, and by separating tlie real and imaginary parts, we obtain 2" cos" A I \ cosnA+ncos{n—2)\+"-——^CQa{n—4)\-^^ ~ ^coi(n— 6)A+ &c. = < r 12.3 _5.{233) / + ] sinwA+nsiD(n-2)A+lllli^sin(n-4JA | "^"72!,^"~?^ siD(n-6)A+&c. I V v_ t 1'2 1.2.0 J 198. This formula is general, being, like those from which it is derived, equally applicable, whether n is integral or fractional. When w is a whole positive number, the second part of the series vanishes, and the expression becomes simply the following, which will consist of n 4- 1 terms, the first and last of which are equal, as also those equally distant from them : 2"cosnA=cosnA+nco8(n-2)A+'^^^^^cos(n— 4) A+''L^!—^^—^cos{n-6)A+ &c (S34) To illustrate this, it is plain that, in both parts of the series, the c<>efficients of all the terms following the first n-|-l terms would vanish, in consequence of containing the factor n — n. It will also be seen that the numerator of the last coefiicient will be w(« — 1) 3.2.1, and its denominator 1.2.3 (n — l)n, so that the numerator and denominator being equal, the coefliicient itself will be unity, the same as the coefficient of the first term ; while, in the coefficient of the last term but one, the numerator will be n(n — 1) 3.2, and the denominator 1.2.3 (n — 1), so that the coefficient itself becomes simply n, the same as that of the second term : and a like illustration is applicable in respect to the other terms. It is also easy to see that the arc in the last term is — wA, that in the preceding term — (n — 2) A, M 90 POWERS OF SINES. 2cos«A=:cos2A-f 1 4cos^A = COS 3 A+ 3cos A 8 cos*A = cos 4 A-|- 4 cos 2 A + 3 16cos^A = cos5 A+5cos3 A+ lOcos A 32cos6A = cos6A-h6cos4A+15cos2A-i-10 &c. &c. &c. (235) 200. Take the difference of the formulas at the beginning of No. 197, and multiply by — V^ ; then, 2sinA=::(^ — v) ( — V^). Expanding this by the binomial theorem, and proceeding as in No. 197, we find 2" sin'' A cosnA — ncos(n— 2)A + -^— — ^co»(n— 4) A— &c. < sinnA— nsin(n— 2)A-^ __isin(n— 4) A— &c. > V- _>X(-V-ir.-(236) 201. We may now consider this formula when w is a whole positive number,* and we shall find that it takes different forms according to the form of n. It would appear, as in No. 198, that when n is even, the second part of the series vanishes, the coefficient of V — 1 be- coming nothing. In this case, also, if n be 4, 8, 12, &g. or, in gene- ral, if it be of the form 4 m, m being a whole number, the multiplier ( — IV — l)** will become simply 1; but, if n be 2, 6, 10, &c. or 4m -1-2, the multiplier wiU become — 1. On this supposition, there- fore (236) will be reduced to the following, in which the upper sign is to be used when n^4m; and the lower, when n = 4m-j-2: 2''sin"A=i±= I cosjiA— wcos(n — 2)A-f- ^ ?T^ ' cqs(m — 4) A — &c.\ . (237) 202. If n be an odd number, it would appear, in nearly the same manner, that the first part of (236) would vanish, and that the formula would be converted into the following, in which the upper quently the number of terms even, the second members will be found simply by substituting n in the general formula (234), and taking all the terms in which the arc is positive ; and, in addition to this, when 7i is even, the last term, which is a number, is to be halved. The same is to be observed in respect to (237) and (238). * When n is a, fraction, (233) and (236) give the several roots or values of 2" cos" A and 2" sin"A ; but the examination of this part of the theory is not suf- ficiently elementary to be given here. The subject of this Section has been lately examined by the French mathematicians, Poisson and Poinsot, and freed from errors which had escaped the observation of former writers. TANGENT AND COTANaENT OF A MULTIPLE ARC. 91 sign is to be used when n = 4m-|-l, and the lower whenn=: 4m4-3: 2"sin''A==t: | sinnA— n 8in(n— 2) A-f-^^^^=i-^ sin(w.-4) A—&c.} (238) 203. By taking n equal to 2, 3, 4, &c. we obtain from the last two formulas the following: 2sin2A =— C0S2A+1 4 sin^A =: — sin3 A4- 3 sin A 8sin*A= cos4A — 4cos2A-f-3 V (239^ 16sin^A= sin5A— 5sin3A+;10sinA ^ ' ^ ^ 32sin6A=r— cos6A+6cos4A— i5cos2A+10 &c. &c. &c. These formulas might also be obtained, very simply and easily, from formulas (235), by substituting 90° — A for A. 204. By dividing (230) by (229), and again (229) by (230), and then by dividing the numerators and denominators of the second members respectively by cos" A and sin" A, we get , . n{n — l){n — 2), „. , o tanwA= J- r^ 7-'^^\w liYT— ^x ■' (240) ^_^jn-l)^^^,^_^H(n-l)(n-2)(n-3)^^^,^__^ &e. cot"A- '^^''7^^ cot"-2A+ &c. cotnA= h^ (241) ncot"-^A-^^-:^](f-^) cot«-A+ &c. I IX.— MISCELLANEOUS PROPOSITIONS.* 205. Prove that the sum of the tangents of the three angles of a plane triangle is equal to their product. This follows at once from Euc. I. 32., and from (43); as, in case of a plane triangle, the first member of this equation vanishes, which * This Section will be found to contain, in small compass, much useful mat- ter that could not be given at greater length, without swelling this work beyond its intended limits. By studying it with care, the student wiU not only be pre- sented with useful exercise on the preceding part of the work, but he will find himself enabled to apply with success, the principles already established, in such cases as may turn up in the farther prosecution of his studies. 92 PROBLEMS RESPECTING PLANE TRIANGLES. can take place only when the numerator of the second member also vanishes; that is, when tanA+tanB4-tanC=tanA tanB tanC. It is evident, also, that, if the sum of three arcs or angles be any multiple of 180^, the product of their tangents is equal to their sum. We should find, in like manner, from (44), that the sum of the co- tangents of three arcs is equal to their product, when the sum of the arcs is (n-\-^)'7r, n being any whole number. 206. Given the base, the vertical angle, and the sum of the sides of a plane triangle ; to find the sides. Since (Euc. I. 32) half the sum of two angles of a triangle is the complement of half the remaining angle, we have, from (76), c : a-\''b :: sin-|C : cosJ(A — B); whence the remaining angles will be known, and thence the remain- ing sides. We should have, in like manner, from (77), c : a — 5 :: cos^C : sin-|(A — B); an analogy which solves the problem in which the base, the vertical angle, and the difference of the sides, are given. The same formulas may also be adapted to solve the problems in which the base, the difference of the angles at the base, and either the sum or difference of the sides are given; and they solve, without modification, the problems in which there are given two angles and the sum or dif- ference of the opposite sides. 207. Given the base of a plane triangle, one of the angles at the base, and the difference of the other sides ; to resolve the triangle. Here, taking c as base, and taking half the difference, and half the sum of it and a — 6, we find s — a and s — h ; and if A be the given angle, we have (82) s — h : s — a :: tanJA : tan^B. 208. Given the base of a plane triangle, one of the angles at the base, and the sum of the other sides ; to resolve the triangle. This will be solved by means of formula (83). 209. Given the angles and the perimeter of a plane triangle ; to re- solve it. By dividing the value of sin^C, according to (79), by the product of the values of cos J A and cos^B, according to (80), and converting the product into an analogy, we obtain cos^Acos^B : sinJC :: s : c. ASTRONOMICAL PROBLEMS. 93 210. Given two sides of a plane triangle, and the difference of the opposite angles ; to resolve the triangle. By (75), a—h : a+h :: tani(A— B) : tanJ(A+B). ^ 211. Given the times at which the sun sets and is west on the same day, at a particular place ; to find the latitude of the place, and the sun's declination. To solve this, taking formulas 1 and 6 in the note, page 73, divide the first of them by the second, and also find their product : then, if the signs be changed, the square roots of the results will be the tan- gents of the latitude and declination. 212. Given the time of the sun's setting, and his altitude at six o'clock on the same day, and at the same place ; to find the latitude and the declination. This question will be easily solved by dividing equation 3 by equa- tion 1 in the same note ; as the sum and difference of equation 3 and the quotient will become, by (14), (15), (31), and (32), ,. 2sma cos-^F , ,, ,. 2 sin a sin ^iP cos (I + d)= ^— , and cos (I co d)= — -^- ^ _, ^ ^ ' ^ cosP ^ cosF 213. Given the azimuth at setting, and the altitude at six o'clock ; to find as before. By equalling the values of sind in the second and third of the same equations, we find, by an easy process, 2 sin a sm2l. cosZ.' 214. Given the azimuths of the sun at setting and at six o'clock ; to find as before. Multiply the second of the same equations by the fourth, divide the result by cos?, and multiply by sind: then, by writing 1 — cos^d for sin^d, there will arise 1 — cos^dzzcosZ cotZ' cos<^; from which the value of cosd will be found by the resolution of a quadratic. 215. Given the sun's meridian altitude, and his altitude at six o'clock, at the same place, and on the same day ; to find the latitude and declination. Let a" and a be the respective altitudes : then it will be seen from fig. 30, that Ew = Hw + EZ--ZH, or d=za"+l—W; whence sindzz 94 SUMMATION OF TRIGONOMETRICAL SERIES'. — CQs(a"-\-l). By multiplying this bj sin?, and comparing the pro- duct with equation 3, note, page 73, we obtain siul cos (a'' -}-?))= — sina, or (17) sin (a" •{- 21) — sina^'^ — 2sin«. From this a"-\-2l, and consequently I itself, will be known. 216. Given the sun's declination, and the interval between the times at which he is west and sets ; to find the latitude. Let T be the given time ; then P— F=T, and (15) cosP cosF+sinP sinF=cosT. We have also, by taking the product of equations 1 and 6, in the note, page 73, cosP cosP'= — tan^c?. From the double of this take the former, and there will remain cos P cos P'— sin P sin F or (14) cos(P4- F) = — cotT— 2tan2(?. Having thus the sum and difference of P and P', we can find these angles; and, from either of them and the declination, the latitude can be found by the note already referred to. 217. Required the sum of n terms of the two series', sin. 9 4" sin .2 ^-|- sin .3^-}- -{-sinn ... (243) loga?"=log^'+21ogcotig— 20. J The latter of these cases is impossible, when the data are such as to render sin^ greater than the radius; that is, when 4ac is greater than hK These methods of solution may be employed with advantage when the coefficients are large numbers, or fractions with large numerators or denominators. 222. Cubic equations may also be solved on similar principles. Thus, let the proposed equation be x^-\-ax=zh. By substituting y-\-z for X, this is changed into y^-]rz^ + ^yz{y-\-z)-\-a{y-\'Z)=zh; which, by taking Zyzz=. — a, becomes y^-\-z^=zh. From the square of this, take four times the cube of yz, found from the equation ^yz=z — a, and the square root of the remainder will be the value of y^ — z^. Then, taking the cube roots of half the sum and half the difference of this, and of y^-\-z^z=ihf and adding them together, we find Now, be positive, put ^=yj = *3^ ^ 9> which gives -J 6= cot ^^ ^ ; then, 0^=^ 5^6(14- sec^)]+^5i6(l—sec^)J; or, by substituting its equal for ^6, and by an easy reduction, «;= Vi a X ( v^ cot i ^— ./C^ tan I ^) . 100 C*UBIC EQUATIONS. — IRREDUCIBLE CASE. Put -y/tanj^zztanp, and consequentlj -i^cotj^izcotg; then (51) _^ 2ar __ a;— 2cot2gV^a;*wheretanp=-3yVia, and tan|=^rHan^p..(244) If in this case h be negative, the value of x will be the negative of the foregoing ; but every thing else will be the same as above. 223. If a be negative, this method fails. In that case, assume sin 2^=--—, and tan^ as before. Then, by a process similar to the foregoing, and by using at the conclusion the formula found by ad- ding together (50) and (53), we obtain 2ar d;=2cosec2^V^a; where sm^r2tan^^.(245) If in this case h be negative, the value of x, as in the last No. will be the negative of the foregoing ; but the method of computation will be the same. 224. Even this method also fails, if 4a^ be greater than 27 6 S as cos^ would be imaginary, and the equation would belong to what has been called the irreducible case. To investigate the method of solu- tion in this case, we have the equation x^ — ax = h, where the co- efficient of X is negative, and b either positive or negative. Now, using ^z instekd of A in the second of formulas (231), supplying a radius r, and dividing by 4, we obtain cos^^-er — ^r^cos,^zz=ir^coBz, Com- paring this with the proposed equation, we have a:=cos^^, a=|r2 h=ir^C0Bz. The second of these gives r=2^/}a; while, by using this value for q T r, we find, from the third, cos^=' — ; and it is obvious, that if the radius were equal to the value of r above mentioned, we should have a;=cos^^. To find the value of x, therefore, to the radius 1, we di- vide the value of cos-s^, and multiply that of x, by the foregoing value of r, and we thus obtain COS^: /3 36 , co3izx2^/ia ^ -r-, or cos-2r= =, and xzz — ^ ^—. 4ta — , jj „ Now, since (No. 17) cos ^ is the same as cos (....(246) cos(4-^+60°)x2Vi« , 36r r ^ ' xzz ^ — ' ^^^ 2-; where cos -8^= = \ *• 2a^/ia J When h is negative, the signs of these roots are to be changed. 225. To find series' expressing the sine and cosine of an arc in terms of the arc itself, divide both members of (229) and (230) by cos" A, and the terms in the second members, exclusive of the coef- ficients, will become 1, tan* A, tan* A, &c.; and tan A, tan^A, tan^A, &c. In the results thus obtained, let nA=x, and consequently nzzj; A. then retaining the denominators of the left-hand members unchanged, we shall have, instead of their numerators, cos a; and sin a;,- while the right-hand members will become , a;(^— A)tan2A , a:(a;— A)(a;— 2A)(a;— 3A)tan*A „ , ^- 1.2 A^ + 1.2.3.4A^ ^^-'^^^ a;tan A x(x — A) (x — 2 A) tan^A . . -A ^ l-Ols + *"• tan A. Let, now, A=0; then, — -r--=l*, cos"A=l (No. 10); and, A dis- appearing in the coefficients, we shall have cosx=l-f^2 + j|^-j^|l3^+ &c. ... (247) '^^=*-m+rm3-'*^ <248) * That the ratio of a diminishing arc and its tangent tends to become that of equality as its limit, will appear as follows. The area of the triangle ACH {pg. 1) is equal to ^CA.AH; and the area of the sector ACF is equal to I CA.AF, as may be shown by resolving it into an infinite number of infi- nitely small sectors by radii, which sectors may be regarded as triangles. Hence, CAH : CAF : : ^CA.AH : ^CA.AF : : AH : AF. But CAH :^ CAF; therefore, AH >AF; that is, tanA>A, or, by dividing by A, — T— > 1. Now, since a straight line is the shortest distance between two 102 legendre's theorem. These series' were discovered bj Newton; and they enable us to compute the sine and cosine of any arc, the length of which is given, to the radius 1. 226. Given, in a survey, the angular elevations of two points above the plane of the horizon, and their angular distance asunder ; to find the horizontal angle, or angular distance of the projections of the given points on the plane of the horizon by perpendiculars. This angle is evidently the same as the vertical angle of a spherical triangle, of which the measured distance is the base, and the comple- ments of the elevations the side's, as would appear by producing the perpendiculars till they meet in the zenith. The calculation, there- fore, is effected by the first case of spherical trigonometry. When the elevations are small, the observed angle may be reduced to the horizontal one by an easy approximation ; which, with several others of a similar kind, may be investigated most easily by means of the differential calculus. 227. The following interesting theorem was discovered by Le- gendre : 1/ there he a spherical triangle, the sides of which are very small compared with the radius of the sphere, it is very nearly equivalent to a plane triangle which has its sides equal to the sides of the pro- posed triangle, and its angles equal to the angles of the same dimin- ished respectively by one third of the spherical excess.) (See No. 148.) To prove this, we have, from (85), cos A sin& sinc=cosa — cos 6 cose. This, by applying (247) and (248) to all the sines and cosines con- tained in it, except cos A, and by rejecting all quantities rising above four dimensions, will become 5ccosA{l — i(h^'{-c^) = l—La^ + -^^a'^l-\-i(b^-\-c^)—i^(b*+c')^ih^c^ =^(b'-\-c^^a')—^(b'+c'—a')—ib^cK points, and since the sine of an arc is half the chord of twice the same arc, we have sin A < A, or — j — < 1; and, therefore, dividing by cos A, we get A — T — < -. Hence, — 7 — is of a magnitude intermediate between 1 and A cosA ' A * r , the latter of which (No. 10) tends continually to 1 as its limit, when A cosA' ' ' •' is diminished indefinitely ; and hence, when the arc is infinitely small, the ratio of it and its tangent will differ in an infinitely smaU degree fi-om that of equality ; so that, when the arc is in a vanishing state, it and its tangent are to be regarded as equal. LEGENDRE S THEOREM. 103 By contracting this, dividing by the coefficient of cos A, and perform- ing the actual division by 1 — ^(i>^'\-c^) we obtain ""^^= Wc -^ 246^ • Now, if A' be the angle of a rectilineal triangle, the sides of which are equal in length to the arcs, a, h, c, the first part of this is equiva- lent (78) to cos A'; and, by taking the square of this value of cos A' « , , I , . 1 1 . 6csin2A' from unity, we find the second part to be equal to ; so that . ., 6c sin 2 A' cos A= cos A ^ . b If A=:xV-f-a;, we get cos A = cos A' cos a; — sinA'sina?; which, by applying (247) and (248), and rejecting the second and higher powers of 07, becomes cos A=: cos A' — x sin A'. Equalling this and the for- 1 • , • A / i. &c sin A' , mer value of cos A, and rejectmg cos A, we get x= — ^ — ; and, therefore, A=A'+^ = A'+i^ii^'. Now, according to the note, page 20, ^hc sin A' is the area of the rec- tilineal triangle whose sides are a, h, c; and this will not differ sen- sibly from the area of the spherical triangle, which, by No. 148, is proportional to the spherical excess. Putting, therefore, the excess equal to E, we have AzrA'+^^E, and it might be shown, in a simi- lar manner, that B=B'+^E, and C=:C'+^E. By the introduction of the radius r, the proof would be rendered apparently more strict. The final conclusion, however, is the same ; and the proof is somewhat more simple, as given above.* * By means of this theorem, a spherical triangle, whose sides do not exceed a degree or two, may be resolved as a plane one by taking from each of its angles one third of the spherical excess. Hence, if the angles and one side be deteniiined by measurement, the excess will be known ; and the other sides will be computed by the first case of plane trigonometry. This theorem, as well as those in Nos. 226 and 228, is useful in geodetical operations, that is, in surveys of kingdoms or other large portions of the earth's surface, in measuring the lengths of degrees, or in determining the relative positions of noted places. It would be inconsistent with the nature of this work to give minute details respecting this important branch of practical science. At the same time, it may be interesting to the student to have some idea of the general nature of such operations. 104 TRIGONOMETRICAL SURVEYS. 228. If, in a spherical triangle, two sides h and c and the contained angle A be given, and if A' be put to denote the angle contained by the chords of h and c, then cos A'=cosA cos ^6 cos^c+sin-Jfe sin^c. For (38) cosa=l— .2sin2ia=l_p2^ cos6=l— j;b'2, and cosc= 1 — i^"S Ar, k', and k'\ being the chords of «, 6, and c; also (37) sin6=2sinJ6cosi6=A;'cosi6, and sinc=F cos-|c. The substi- tution of these in (85) gives, after rejecting 1 in each member, trans- posing — 1^"'2 and — ^k"^, and dividing by k'k", k'^ 4- k"^ k^ OTJTJ, =cos A cosiJ cos^c + i^'^'. But (78) the first member of this is equivalent to cos A'; and, by our assumption, ikfk"^&m^h sinic; whence, cosA'=cosA cosJ6 cos^c + smi6 siu^c... (249) If, by this formula, all the three angles of the triangle formed by the chords, be computed, their sum should be 180°; and this affords a test of the accuracy both of the observations and of the computa- Let us suppose, then, that the earth is an exact sphere, or rather that the ob- servations made on its surface, varied as it is, are reduced, by No. 226, to what they vrould be on a sphere. Then, if signals, such as spires, towers, poles erect- ed for the purpose, or other objects, be assumed at as great a distance asunder as will admit of distinct and accurate observations, with telescopes of considerable power attached to the instruments used in measuring the angles, the country wiU be divided into a series of primary triangles ; and if any side of any one of these be measured, the remaining sides of all of them may be computed by Legendre's theorem. By means exactly similar, each of these triangles is resolved into a number of others, called secondary triangles; and thus the positions of towns, and other remarkable objects, are determined. The length of the hase^ or line measured, which is an arc of a great circle, must be determined with extreme accui*acy ; as an error in measuring it would affect the entire survey. For checking the measurements and the computa- tions, it is proper to measure some other line at a considerable distance from the first ; as the comparison of its measured and computed lengths will be a test of the accuracy of the intermediate operations. Such a line is called a hose of verification. The measurement of a base is one of the principal diffi- culties in the survey, chiefly on account of the inequalities of the surface, and the variations in the lengths of the measuring instrument, arising from the change of temperature. On this account, the base is assumed on as flat a portion of country as can be obtained ; and the chain, or other measuring instrument is constructed with extreme care. The degree of accuracy ob- tained by these means, is very remarkable. Thus, in the Trigonometrical Sur- vey of England, the difference between the measured and computed lengths of a base of verification on Romney Marsh, nearly five miles and a half in length, was only about two feet ; though the series of triangles, connecting it with the original base on Hounslow Heath, extended over a space of eighty miles. PROPOSITIONS IN SPHERICAL GEOMETRY. 105 tion. Delambre has given another formula instead of the foregoing, and has computed tables to facilitate the calculations. 229. Let AD=^ {fig. 37) be an arc of a great circle drawn from the vertex, to any point in the base, of the spherical triangle ABC, and let the segments of the base CD, DB be respectively denoted by h', c'. Then, by (85), and (No. 15), cos6=cosADC sin6' sin^+cos6' cosq (250) and cosc= — cos ADC sine' sin^+cosc' cos^' (251) These two equations contain six quantities, 6, c, h', c', q, and ADB. If, however, any one of them be given, or if one of them be a function of one or two others, — that is, if it depend upon them, so that it may be determined by means of them, the number of inde- pendent quantities will be reduced to five. By eliminating one of these between the two equations, an equation will be obtained con- taining only four quantities, and the resolution of it will give any one of them in terms of the other three. It is only in a few cases, how- ever, that formulas of elegance or value wiU be thus obtained. The following are some of the more interesting. 230. Let h'=zc'=ia. Then, by adding (250) and (251), we get cos 6+ cos c= 2 cos la cos^; or (22) cosi(64-c) cos-J(6 — c)=cos-Ja cosg^ (252) Hence cos-Ja : cos^(h-\-c) :: cos^(b — c) ; cos AD; whence we can find AD, the line bisecting the base, when the three sides are given. Square both members of (252) : then by (6), &c. »in^^(64-c)-f-sin2^(6 — c)— sin^|(6-|-c) sin^^(6 — c) =am^a-{-sm^q — sin^a sin^q. By dividing the last terms of these equals by r^, &c.; as in No. 163, we have, in a plane triangle, J(6-f-c)2+i(6— c)2=ia24-^2; or, by doubling, &c. 62-fc2=i«2^2^2. — a well known theorem. (Euc. II. A.) 231. Let, again, h=:c: then, by taking the difference of (250) and (251), and by transposition, we get cos A DC sin q ( sin h' + sin c') z=z cos q (cos c' — cos h') . Divide by cosq and sin 6' -f sine'; then, by (28), cos ADC tan(^=tani(6'— c') (253) This formula might also be obtained by No. 107, from the right O UFI7BESITr] 106 PROPOSITIONS IN SPHERICAL GEOMETRY. angled triangle contained bj the perpendicular of the isosceles tri- angle hj q, and by the part of the base intercepted between them. 232. If ^ = 4^, we have, by (250) and (251), cos 6 = cos ADC sin 6', and cosc = — cos ADC sine". TT cose — cos 6 sin i' 4- sine' ,,-,^. j ^n^x Hence ; ^= — . ,, — -. — ,; or, (25) and (24), cose 4- cos 6 sin 6' — sine' ^ ^ ^ ^ tanj-(6— e) _ tan^c^ coti(6 + c)~ tanj(2>— O ^ ^ This formula affords the means of finding the segments of the base made by either of the quadrantal arcs drawn from the vertex. By taking ADC = 90°, and dividing the difference of (250) and (251) by their sum, we should get formula (149). 233. Putting now the angle BAD = B', and CAD = C', we have by (101) and No. 15, cosB=:cos^ siuB' sin ADC + cos B' cos ADC "^ (2^K\ cos C = cos g sin C sin ADC — cos C cos ADC / ' " Like (250) and (251), these formulas contain six quantities, and will give different formulas according to the values or relations of those quantities. 234. Thus, let B'=C = iA. Then, by taking the first from the second, and by (23), sinKB + C)sini(B — C) =— cos-JAcosADC .... (256) Hence, cosJA : sini(B+C) :: sini(B — C) : — cosADC, or cos ADB. We have thus the means of computing the angles which the arc bisecting the vertical angle makes with the base. 235. Let us now take B=C ; and, by subtracting, transposing, &c., we get cos^ sinADC(sinB'— sinC')=— cosADC(cosC'+cosB'). Hence, by resolving this for cos q, and by (29), cosgrr—cotADC coti(B'— C) ... (257) This might be obtained by means of No. 107, from the rightangled triangle mentioned in No. 231. 236. By taking ^=^cr, and dividing the difference of formulas (255) by their sum, we get, by (25 ), tan^ (B + C) _ coti(B'— CQ . cotl(B— C)~ tan^A ":' '^- ^ By taking ADC=90^ we should get formula (150). 237. Let DBCD' and DB'C'D' {fig. 38) be the halves of two great PROPOSITIONS IN SPHERICAL GEOMETRY. 107 circles iutei-secting each other in D and D'; and through any point A let great circles be drawn cutting them in B, B', C, C: join also AD. Then, putting AB=c, AB'=c', AC=:h, and AC'=h\ we have, by (95), cote sinADzzcotADB sin DAB+ cos DAB cos AD, cote' sin AD=cot ADB' sin DAB -|- cos DAB cos AD, cot6 sinAD=cotADB sinDAC + cosDAC cos AD, cot6'sinAD=cotADB'sinDAC4-cosDAC cos AD. Divide the diflPerence of the first and second of these by the difference of the third and fourth : then cote' — cote sin DAB r9^()\ eot6'— cot6~sinDAC ^ ^ Multiply the numerators of this by sine sine', and the denominators by sin 6 sin 6': then, by (13), sin(e — c') sin DAB sine sine' sm(b — b') sinDAC sin6 sin6'' Now, by the formulas of the four sines, in the triangles DAB, DAC, sin DAB sine=siu ADB sinDB, and sinDAC sin6=sin ADB sin DC: and by substituting these in the last, and contracting, we obtain, since CD'=:-t— DC, sin(e — e') sin DB sine' sin DB sine' sin (6 — 6') sin DC sin 6' sin CD' sin 6' (260) Hence, when DB=rCD', we get the curious and interesting for- mula, sin(c — c') sine' . ^. sinBB' sinAB' ^^„, , -T-yr 77<=^-775 that IS . ^^, Z=-. r-p^ .... (261) sin(6 — b') sin6' sinCC sin AC ^ "^ From this we have the analogy, sinAB' : sinB'B : sin AC : sinCC; which, when the radius of the sphere is infinite, becomes AB' : B'B :: AC : CC, the same as Euclid, VI. 2. Formulas of some interest would be derived from (260), and (261), by taking h and e each =90°; by taking b'±:c'; by taking b — 6'= c — e', &c. 238. Let BEC (Jigs. 39 and 40) be a circle (great or small) on the 108 PROPOSITIONS IN SPHERICAL GEOMETRY. surface of a sphere, and through any point A on the surface, draw great circles cutting it in B, C, D, and E ; then tanJAC tan^ABrrtaniAE tan^AD (262) To prove this, take P the pole of BEC, and join PA, PC, PE ; draw also PF perpendicular to BC, and PG to DE. Then, by (116), cosFC _ cosPC cos GE _ cos PE _ cos PC cos F A "" cos PA' cos G A "~ cos PA "~ cos PA* Equalling the first members of these, we get, by composition and di- vision, and by (25), the expression given above. This formula corresponds to Euc. III. 35 and 36. It gives those propositions, in fact, by taking the radius of the sphere infinite, quadrupling the results, and making one of the circles drawn from the external point touch the given circle. 239. Let BAC {fig. 20) be a right angle, and let AD be perpen- dicular to BC. Then (No. 107) sinAD=tanBD cot BAD, and sinAD=tanDC cot CAD. By taking the product of these, since BAD=i'r — CAD, we get sin2AD=tanBDtanDC (263) Again, (No. 107), cosC=tan AC cotBC, and cosC=tanDC cot AC. Hence, by equalling the second members, and multiplying by tan AC tanBC, we get tan2AC=tanBC tanDC (264) When the radius of the sphere is infinite, these two formulas be- come the same as the corollary to Euclid, VI. 8. QUESTIONS FOR EXERCISE. - 109 X—QUESTIONS FOR EXERCISE. Ex. 1...4. Prove the truth of the following formulas: (1) sin^A— sin^Brircos^B— cos2A = sm(A-|-B) siu(A— B); (2) cos^'A— sin2B=cos=^B— sin2A=cos(A-f-B) cos(A— B); /ON. 3A . 'R sin(A+B)s m(A— B) , (3) tan^A— tan2B= — .,/ ^d ' ^ ' cos'A cos"^ B /.x .2-n .2A sin(A+B)sin(A— B) Ex. 5... 18. Prove also the following (5) sin 9"=-iV(34-V5)— iV(5— V5); (6) sin81«=iV(34- V5)+iV(5— V5); (7) sm22° 30'=| V(2— V2) ; (8) siner 30'=^V(2 + V2); (9)sinl5°=|V(2— ;V3); (10)siii75°=JV(24-V3); (17) sin 3«=i(— 1+V5)V(2 + V3)— iV(10 + 2V5)V(2— V3); (18) 8in87°=i(— l-hV5)V(2— V3)-l-iV(10-f 2V5)V(2-fV3). (11) tan 15"== 2— V3; (12) tan75°=2-|-V3; (13) tanl8«=V(l— IV5) (U) tan72°=V(5 + 2V5): (15) tan36°=^V(5— 2V5) (16) tan54''=V(l+3V5) Ex. 19. In a plane triangle, prove the following formulas: 1. tanj A tan^B taniC= 2. versin,A=^^^ f^ -; DC ^ bj-c _ cos^ (B— C) _ l + tanlBtan^C , ' a4-& + c""2cos^B cos-JC" 2 b + c _ cos^(B— C) l + cotiBco tjC ^- ^,4-c— «~2siniBsiniC'" 2 r being the radius of the inscribed circle, and s half the perimeter, Ex. 20. In a plane triangle, if C=60°, c^ = a^ + b^ — ab ; but if C=120o, c'-=:a^ + b^ + ab. Required the proof. Ex. 21. Prove that, in a plane triangle, the straight line drawn from . 22. If equilateral triangles be described externally on the three sides of any plane triangle, the square of the straight line joining the centres of any two of these triangles is equal to ^(a^ + ^^ + c^)-!- |AV3, where A is the area of the given triangle. Required the proof. 110 QUESTIONS FOR EXERCISE. Ex. 23. Prove that, in a spherical triangle, tan4(a-f^) : tani(a— 6) :: tani(A-|-B) : tani(A— B); and show from this how the problems may be solved, in which two sides and the sum or difference of the opposite angles, or two angles and the sum or difference of the opposite sides, are given. Ex. 24. In a spherical triangle, if the sum of the three sides be 180°, the sine of half any angle is a mean proportional between the cotangents of the containing sides ; but if the sum of the three an- gles be 360°, the cosine of half any side is a mean proportional be- tween the cotangents of the adjacent angles. Required the proofs. Ex. 25... 36. Required the method of resolving a right-angled spherical triangle, from any of the following data: — (25, 26) The hypotenuse and the sum or difference of the legs; (27, 28) The hypotenuse and the sum or difference of the adjacent angles; (29, 30) An angle and the sum or difference of the opposite leg and the hypotenuse; (31, 32) An angle and the sum or difference of the adjacent leg and the hypotenuse ; (33, 34) A leg and the sum or dif- ference of the other leg and the hypotenuse ; (35, 36) . A leg and the sum or difference of the adjacent angle and the other leg.* Ex. 37... 40. Given the base and the vertical angle of a spherical triangle; given, also (37, <38) the sum or difference of the other sides ; and (39, 40) the sum or difference of the other angles ; to re- solve the triangle.! Ex. 41. Given the altitudes of the sun at six o'clock and when east or west, on the same day, and at the same place ; to find the lati- tude and declination. Ex. 42... 51. Required the method of finding the latitude and de- clination from any of the following data, according to the notation adopted in the note, page 73 : (42) «, Z'; (44) a', 7J\ (46) Z, P; (48) a', P; (50) P, Z'; (43) a', Z; (45) a', F; (47) Z', F; (49) a, F; (51) F, Z. Ex. 52... 56. Given the sun's meridian altitude, and {^2) his alti- tude when west ; (53) the time when west ; (54) his azimuth at six * Solutions of all these, and several similar questions may be obtained fi'oin (203) to (212), inclusive. f These questions will be solved by means of XI, XII, XIII, XIV, in the note, page 35. Questions similar to these may be solved by means of (108), (109), (110), and (111); and also of (149), (158), and the mtermediate for- mulas. QUESTIONS FOR EXERCISE. Ill o'clock; (55) the time of rising or setting ; and (56) the amplitude : to find, in each case, the latitude and declination. Ex. 57. Given the interval between the times at which the sun rises and is east, at a place whose latitude is known ; to find his de- clination. Ex. 58. Given the angle contained hy two hour lines (such as the three and four o'clock ones) on a horizontal dial ; to find the latitude. Ex. 59. In what latitude will the angle contained by the five and six o'clock lines on a horizontal dial, be double of the angle con- tained by the twelve and one o'clock lines? — Answ. 44° (V-i. Ex. 60. In what latitude are the hour lines for ten and five o'clock, on a south vertical dial, perpendicular to each other? — Answ. 47° 3'-^. Ex. 61. To find the latitude at which, on a given day, the hour angle on a horizontal dial, at the time when the sun is east or west, will be of a given magnitude. Ex. 62. On a horizontal dial for latitude 54P 36', what two horn- angles differ by 15°, while the corresponding times differ by an hour ? Answ.— The times are T" 41 J™ and 3'' 41^"^. Ex. 63. In latitude 54° 36', for what time are the hour angles, on a horizontal, and on a south vertical dial, complements of each other? AnsxD. 3'* 42- l^ Ex. 64. Required the declination of a star, which in latitude 54° 36' N. would shine on a north vertical dial during half the time of its continuance above the horizon. — Answ. 27° 23' N. Ex. ^5. Investigate the following formulas ; ra denoting the num- ber of terms in the second member before the last or fractional term cosnA , ,>. , o\ A . / K\ A _, cos(n — 2m)A -r- — r =cos n— 1 A— cos n— 3 A +COS n— 5 A— . . . rfc — \ -^ 2cosA ^ ' ^ 2cosA cosnA . , ,,. . , o\A • / K\A , cos(n — 2m) A ■— — -=— sin w — 1)A — sin(n — 3) A — sin(n — 5) A — ..-\ —. — j-!— 2smA 2smA sinnA . . ,, . . . ox A . • / -\A _. sin(n— 2m)A -: -=:sin(n — 1)A — sin(n — 3)A-|-sin(w — o)A — ...z±z — ; 2cosA ^ ' ^ 2cosA sin»iA , TV. , / o\ A . / K\ A I . sin(n — 2m) A 77-^— r=rcos(»2 — l)A+cos(n — 3)A-|-cos(n — 5)A-f-...H \^. — -r-!—- 2sinA / • \ / I I 2sinA Ex. ^^. Prove that the sums of n terms of the series', sinmo-f- sin(m + r)9-f-sin(m-4-2r)^ + sin(m-f- 3r)f) -|- &c., and cosm^ + cos(m4- r)(|ji -f-cos(m-|- 2r)^ -\- cos(m + 3r)^ •\- &c., are, respectively, sin-Lnr^sin )m-|-l(n — 1)»*}^ , sin-|nr^cosJm-f.^(w — l)»*?f ; :j -, and. - 7 -, - • { li^ QUESTIONS FOR EXERCISE. Ex. 67. Prove that ^ sin ^ + sin394-sin5^4-sin7^-|- ), and2/''z=:aa;"+i (^); where a and h are to be found in terms of a/, y\ x", and y": and, by any of the common methods of elimination, we should find a = ^, — —„ and 6 = — ^S tt-'- the substitution of which in the gene- X — X X — X ° ral equation (265) gives y-x'—x"''^ x'—x" ' the equation required. ' The following method, however, is rather more simple : Subtract (cf) from (p) ; the remainder will give as be- fore, a = -, — ^. Take the diiference of (jp) and {2^h^, substitute X —X in it this value of a, and there will result y — y':=(x — x') , , -; or, when cleared of fractions, (y-y') (^'-x") = (x^x') {y'^y'T (268) * As an example of the use of this, suppose the co-ordinates to be rectangu- lar, and ^^=2, y=r4:, y= — 2, and y"=l. Substitute these in (268); then (v — 4)(2-f 2)— (a;— -2) (4 — 1) = 0, or, by contracting, &c. ^— |a;=2|, tne equation of the line to be drawn. Hence, when y=zO,we have x— — 3J ; 118 INTERSECTION 01* TWO STRAIGHT LINES. the equation required, which might be easily reduced to the form found above. If the line were required to pass through only one point, ol\ y', we should have, bj subtracting (^) from (265), y—y'=a{x—x'y (269) This, which is the equation required, is indeterminate, a being un- known ; and consequently the line may have an indefinite number of positions, as is also evident from the nature of the problem. 247. li y^iax-^h, and y=:a'x-\- h', be the equations of two lines, the co-ordinates x' and y' of their point of intersection, will be de- termined by taking in each equation x equal to a/, and y equal to y', and then determining the values of x' and y' from the two equations. We should thus find ^/_ and 2/'= 7 (270) a — a' ^ a — a' ^ ^ 248. To find, for rectangular co-ordinates, the equation of a straight line passing through a given point x' , y' (L, jig. 42) and making a given angle & (HPK) with a given straight line (HGP) whose equation is yz^ax-^h : let the required equation be 2/ = «'^' + ?>'• and when a?=:0, 3/=2|. To trace the required line, therefore, draw [fvg. 42) AB and OC at right angles to each other, and in AO take OH = 3 J, tliat is, equal to three times the Tine assumed as the Hneal measure, and a third of the same, and in OC take 0G-=2|; the straight fine passing through G- and H will be the locus required. The coefiicient — | is the tangent of 143° 8', the angle which HG- makes with HA; and by means of this also we might be assisted in determining the position of the required locus. Exercises. Required the equations of the straight lines passing through points whose co-ordinates are as follows : 1. y= 2, y= 3 ; a!"=l, f= 2. 3. af= 0, y= 4 ; x"z=z 3, y'= 1. 2. y=2,y=— 1; x"= — l,y"=—l. 4. a^=l,y= — 2; cr"=— 3,y'=4. * In like manner, if y" be another point, we have y—y"=a{x—x"); and dividing (269) by this, we get^^^= ^~^^, , another form of (268). t Exercises. Show which of the lines represented by the following pairs of equations, intersect, and which do not : and, when they do intersect, find the co-ordinates of their points of intersection : (y^x-i-1, fy=2x--.3, ^ r3i/=2^-.l, \2/=2a;+l. '\y=3x—2. '\2x=3y., „ r22/=^+l, 4 (2y=xi-2, (2x=y, '\ y=^2x+l \6.v = 3a;— 1. ' \3x=—2y. EQUATION OF PERPENDICUJ I i ^^ ,-< V',/il9 For the point x', y', this will become y'=ia'x'-\-h'\ whence, by sub- traction, y—y'^a'ix—x!) (/>) Now (No. 245) a = tan^, and a'=tanLKO=tan(^— ^'). Hence, a — tan^' , by (40), and by substituting a for tan^, we get « =rT — T^'' this changes (/>) into y-y'=: ^T^^''\X ^-^T (271) From this, which is the required equation for any value of &', we derive the equation of a line passing through a given point x', y' , and parallel to a given line, y=:ax-\-h, simply by taking ^'=0, and we thus get y-y'=a(x^x') (272) From the same equation, by multiplying the numerator and deno- minator of the fraction in the second member by cos^', and then taking &'=^'r, we find, for the equation of a straight line passing through a given point, x', y' , and perpendicular to a given straight line y=zax-\-h, y-y'=-\{x-x) (273) To find the equation of a straight line passing through the origin, and making a given angle &' with a given straight line, y=:ax-\-hy take a/, y', each =0 in (271), and the required equation is found to be a — tan^' ^nnA\ V = i : — ,,.x (274) ^ l + atan^' ^ ^ By the same means we find from (273) the equation of a straight line passing through the origin, and cutting y=:ax-\-h perpendicu- larly, to be y = -l (275) If ^'=^=0, the line is parallel to the axis AB. In that case (271) becomes simply y — y'—Q, a being =0; and x will be indeterminate. Again, from (271), x — x'=—^^ — j, {y — y'). In this, change a — tan Q * By changing od, 'if, into a?", y, and dividing (2'71) by the result, we get • =^ =■• ^„ another form of (268), the same as in the note to (269). y — y X — «r 120 LENGTH OF A STRAfGHT LINE. ^' into —(900—0: then tan ^'=— cot ^= ; and LPK being then parallel to the axis OC, we have for its equation x — a?'=:0, y being indeterminate.* 249. If it be required to draw a line through x', y\ making a given angle &' with the axis AB, we have, bj (269) and No. 245, 2/— 2/'= tan ^'(^—^') (^76) 250. For rectangular co-ordinates, the length of a Hue joining two points, X, y, and x', y\ is (Euc. I. 47) V5(^ — ^'Y-{-{y — y'Y]- If x', y', one of the points, be the origin, this will become simply ^/{x'^■\^y^), or a;V(l + «^), because (No. 244) y=.ax. 251. To find the length of a perpendicular drawn from a given point x' ,y\tQ 2, given line whose equation is y—ax^h (;)) By (273) the equation of the perpendicular to that line from the point x'y y' is From (p) take y'=y'+ax' — ax'; then y—y'=a(x—x') + h—y'+ ax' Equalling the second members of this and (q), and transposing, we get (« + -)(^ — ^') = 2/' — ^^' — ^' or, multiplying by a, and dividing by 1 + a-, "^""^ ~TT^ (2/'— ao?'- &). If we divide this by — a, and compare the result with (q), we obtain 2/— 2/'= — ^-p^ (S/'— aa;'— 6) By taking the sum of the squares of these values of a? — x' and y — y\ * Exercises. — 1. What is the equation of a straight Hne drawn through a point whose co-ordinates are afz=4i andy= — 1, and making an angle of 45**, with a straight line whose equation is ^-}-2/=l ? 2. Find the equation of a straight line passing through the origin of the co- ordinates, and perpendicular to the line x — 2/=l. 3. What is the equation of a straight line passing through the point whose co-ordinates are x'=5, and y=0, and parallel to the line a;-|-2/+l = 0? INTERSECTION OF THE PERPENDICULARS OF A TRIANGLE. 121 and extracting the square root, we get, after a slight modification, the required length equal to "^ y, ^. . 252. As an application of these principles, let it be required to discover whether the three perpendiculars AD, BE, CF, (^jig. 43) drawn from the three angles of a triangle ABC to the opposite sides, intersect in the same point. To facilitate this investigation, let BC be assumed as the axis of the abscissas, and a perpendicular to it through B, as the axis of the ordinates : then, assuming BC = cd' , and the co-ordinates of h.-=ix\ y\ we have those of B = 0, 0, and of * It may be convenient, for the sake of reference, to bring together the re- sults thus far obtained, which may be regarded as the elementary principles of analytic geometry. We have, therefore, the following table : 1 . The equation of a straight line passing through the origin of the co-ordi- nates \^yz=ax. (No. 244.) 2. The coefficient a= . ?"^ — r; or, if w=:90^ a=tan^. (No. 245). sm(w — ^) ' ' ' 3. Another form of the equation of a straight line is, by No. 245, ysin(w — ^)=^sin^-|-^ sin(w — ^); or, if w=90®, 3/=z=^tan^-f-&. 4. The equation of a line passing through the points, .r', y, and x", y'\ is (3/-y)(^— 1-")=(-^— ^')(y-y'). (No.246.) 6. The equation of a Hue passing through one point sf, y, is 3/ — yf=a{x — .r'). .6. The co-ordinates x', y\ of the intersection of two lines y=ax-\-h, and , , ,, , h — y , , ah' — dh t/= aJx + 6', are a;'= ^, and y= -. ( No. 247. ) I. sf, y, being a given point, and y — ax-\-hVi\Q equation of a given line for rectangular co-ordinates, the equation of a line making with that hne a ^ ^, . , a — tan^ , given angled', IS t/—y=^-p^^^^(a:—:r'). (No. 248.) 8. The equation of a Une from the point x', y, pei-pendicular to the line y=ax-\-h, w being =90^ ist/— y=— l(^_a!'). (No. 248.) 9. The equation of a line parallel ioy^axA-h, and passing through the point x',y, w being =90«,' IS i/—y=a(a?_;r'). (No. 248.) 10. The equation of a Hne through or', y\ parallel to the axis AB {fvg. 33), is y— y=0 ; and that of one parallel to CD is x—:d=^. (No. 248). II. The equation of a line passing through o(i, y, and making a given angle &' with the axis AB, is ?/—y= tan /(.'«;— a;'). (No. 249.) 12. When w=90« the length of a line joining the points a/, y, and x", y, is V[(^-^')='4-(y-y')^j. (No. 250.) ^ ^ 13. When w=90°, the length of a pei-pendicular from the point ^, y, to the Hne„=a. + 6,is^^^^. (No. 251.) Q 122 INTERSECTION OF LINES BISECTING THE SIDES, ETC. Cz=x", 0. Then (No. 251, note 4) the equation of AB is (y—y') x zz:(x — x')y', or by contraction, yx'=:xy'; whence y = ^: also 00 x' (same note, 8) the equation of CF is ?/= f^x — .r"). In a similar manner we should find the equation of BE to be v = — j— x. Now, V (by No. 251, note 6) we find from these equations of BE and CF, that the abscissa of the point of intersection of these lines is := x'\ and this being the abscissa of the point A, it follows that these perpendi- culars intersect one another in the remaining one AD. 253. As a second example, let AD, BE, CF (^fig. 44) be drawn from the angles, and bisecting the sides. Then, retaining the same notation, we have obviously the co-ordinates of D = Jx", 0; of E = \{x'-\-x"),\y'', and of F=:Jx', \y'\ whence (No. 251, note 4) wo find the equation of AD to be yicd — \x!')-=.{x — \xl')y'\ of BE, y{x'^x"^r=:xy'\ and of CF, y{\x'—3(^')=.\y\x—x"). Now by taking x and y in the first of these equal to x and y in the second, we readily find y=.\y'\ and xz=.\{x'-\-x")\ and also, by treating the second and third equations in the same manner, we still find y-=.\y'\ and x-=i\{3c!-^x")'. whence it appears that the three lines pass through the same point, since the co-ordinates of the intersection of AD and BE are the same as those of the intersection of BE and CF. It is also obvious, since y = ^y', that DO = ^DA, being the point of intersection ; and consequently the lines divide each other in the ratio of 1 : 2. 254. As another example, let it be required to determine whether the three perpendiculars drawn through D, E, and F, (Jig. 45) the points of bisection of the sides BC, AC, and AB, pass through the same point. Here, retaining the same notation, we have (No. 251, note 4) the equations of the three sides BC, BA, and AC, y=0, x'y=xy', and y(x' — x")z=(x — x'')y'; and (same note, 8) from these equations, and from the co-ordinates of D, E, and F, we find the equations of the perpendiculars through D, F, and E, to be x—ix''=0, (2/— i2/02/'=— (^— i^0^'» and (2/~iy)2/'=(^"-^'') (^-i^-i^n- Then, by taking x the same in the first and second, and also in the first and third, and by taking y also the same, we find the abscissas of the intersections of the perpendicular through D with those through TRANSFORMATION OF CO-ORDINATES. 123 F and E, each equal to ^x", and the ordinates of the same each equal to i^-27'<^"^ -^)' ^' ^ 2y ' whence it appears that the intersections coincide, or that the per- pendiculars all pass through the same point. Let this point be ; and, if we take the sum of the squares of BD (^x") and DO nr-\'X''—x'x"\ „ , T^Qo (y'^ + x'^^-^x'x y + x''^ ^ Xow, if the three sides be put = a, h, c, we have x"=za, x'- + y''-=zc\ and(x"—x'y-{-y'^ = b\ or x"^^2x"x'-{-x'^^y'^z=ih\ ora'-—2ax'+c^z=bK Hence, (c2_^y/)2^^2^/2 c*—2ac^x'-\-a^x'^-\-a^y''' — 42^/2 — 42^/2 c*—2ac^x'+a^x' ^ + a^(c'-—x'^)_ c*—2ac^x'+a^c^ — 4 yr2 — ^ _ 42,^2 (c^—2ax'+a^)c^ _h'-c\ T)C cjLuc b ^ and consequently BO = 2^, = ^ = ^, where A denotes the area of the triangle. This expression bemg ahke related to the three sides, it is evident that the same would be found for the hue drawn from O to either of the other angles. Hence is the centre of the circumscribed circle ; and it appears that the radius of that circle may be found bj dividing the continual product of the three sides by four times the area. 255. The equation of a line referred to one system of co-ordinates being given, it is often of use to find its equation in relation to an- other system. To effect this, let y=ax-\-b (Jig. 46) be the equa- tion of a line in relation to the axes OB, OC, x and y being the co- ordinates of any point P in the line, and the angle BOG being =w ; * This agrees with what is found in the note to No, 1 14, 124 INTERCHANGE OF RECTANGULAR AND POLAR CO-ORDINATES. and let it be required to find its equation in relation to the axes O'B^ O'C^ which are so situated with respect to OB, OC, that OE'' being parallel to OC, OB'^^x" and WO'=y'; and that O'S being parallel to OB, the angle B'O'S is =&, and C'0'S = ^'. Then, re- presenting O^E', and E'P, the co-ordinates of P in relation to the new axes, bj X and Y, and drawing E'R parallel to OB, and E'Q to OC, we have EP or 2/ = ES-fSR-hRP=E''0'-l-QE'+RP (p) and a:=OE''+0'Q + E'R (q) But 0'E"=2/', and OE"=y: and we have also in the triangles O'QE', E'RP, sinO'QE' : sinQO'E' : : O'E' : QE', or sinw : sin^ : : X : QE'; sinO'QE' : sinO'E'Q : : O'E' : O'Q, or sinw : sin(w-~^) : : X : O'Q; sinE'RP : sin RET : : E'P : RP, or sinw : sin^' : : Y : RP; sinE'RP : sinE'PR : : E'P ; E'R, or sinw : sin(w— ^0 : : Y : E'R. By means of these analogies we find the values of QE', O'Q, &c.: by substituting these again, along with the values of O'E" and OE", in (p) and (q), we obtain the values of y and x : and lastly, by substitu- ting these values of y and x in the equation yz=ax-\-b, we get, for the required equation, , , Xsin^-I-Ysin^' , , Xsin(w — ^ + Ysiu(w— ^ , , /o^^x y' ^ J :3: ax' -\^ a. ^^ ~ ^- ^ + b..(277) ^ smw sinw • v / This general equation becomes simplified in particular cases. Thus, if w = 90°, the denominators disappear. If 0' be in OB, y'= ; if in OC, 0:^=0; and if 0' and coincide, x^ and y' are each equal to 0. Also, if either of the new axes coincide with the corresponding original one, or be parallel to it, we shall have the angled, or w — ^ = 0. 25Q. If the co-ordinates be rectangular, and P (Jig. 41) a point whose co-ordinates are x and y, we have OP, or, (see note, page 100) v=V(^2_^2/0; OE = OPcosPOE, or x=zvcos B y S. /B C D\ JO /2. 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