Suppose these circles to be invariable whilst another great circle DFCB revolves about the pole F, and let cn be at right angles to the great circle mnofsd; then the three triangles ABC, CGF, and EDF will be variable, viz. I. In the right-angled triangle ABC, the L A will be a fixed quantity, and the other parts will be variable; viz. Bm will be the increment of ab; no the increment of BC; co the increment of Ac; and as the increment of the arc in which measures the ¿c. II. In the right-angled triangle cGF, the side FG will be a fixed quantity, and the other parts will be variable; viz. co will be the decrement of cg; no the decrement of FC; Bm, the decrement of the LCFG; and Ds the increment of the LC. III. In the right-angled triangle FDE, the hypothenuse EF will be a fixed quantity, and the other parts will be variable; viz. sd =no, will be the increment of FD=BC; sd the decrement of ED; and Bm, the decrement of the ZEFD= L CFG (N. 135.) PROPOSITION 11. (Plate IV. Fig. 3.) (Y) In any right-angled spherical triangle ABC, right-angled at B, suppose one of the angles as a to remain constant, it is required to find the ratios of the fluxions of the other parts. 1. In the triangles FBM, Fcn, having the same acute , at F, sine FB : sine fc::tang Bm : tang cn (M. 167.) But FB=90°, Fc is the complement of BC, and Bm and cn, being very small arcs, have the same ratio to each other as their tangents. COS BC ... rad : cos BC:: Bm : CN= rad 2. In the triangles DCI, con, where the 2 DCI may be supposed equal to the con. Tang di : sine cı::tang cn : sine no (M.167.) But di is the measure of the Lc, ci=90', and the tangent of cn and the sine of no, have the same ratio to each other as Вт. the arcs, COS BC COS BC o :: tang 2.c: rad::cn (= BM) : no= Bm. rad tang LC Again, sine di : sine dc:: sine cn : sine co (M. 167.) But d=LC, Dc=90°, and the very small arcs cn and co, have the same ratio to each other as their sines. COS BC .:: sine Lc: rad:: cn Bm) : co= COS BC Bm. 3. In the triangles BFM, DFs, where F=FD extremely near. sine FB : sine Bm::sine Fs ; sine ds. (M. 167.) But FB=90°, FS=FD=BC, and the small arcs are as their sines. sine BC and ds COS BC COS BC BC rad But no will represent the fluxion of Bc; co the fluxion of ac; Ds the fluxion of di= Lc; and Bm the fluxion of AB. (B. 126.) Therefore, sine BC X AB; AC= X AB; LCR X AB. tang Lc sine Lc rad (Z) Hence, in any right-angled spherical triangle ABC, right-angled at B; by denoting the sides opposite to the angles A, B, C, by a, b, c, respectively, we derive by substitution and reduction, the following general equations, which include all the varieties that can possibly happen wherein the LB is 90° and one of the other angles (viz. A) constant. A B *1. a= sine c COS a cot a X c. tang C COS a rad cot a II. = хс X c. COS C sine a III. C= tang C ха — cot a rad X c. sine a . å sinec x. IV. := tang C (A) Any of the foregoing equations may be turned into proportions, or varied in the expression, by reference to pages 104, 105, &c. * Simpson's Fluxions, page 280. COS Thus, from the fourth set of equations c x cos a =) x sin 2 c. that is, ' : c::cos a : sin c. o sinc sin 6 COS C COS BC COS BC sin BC вт. . PROPOSITION III. (Plate IV. Fig. 3.) (B) In any right-angled spherical triangle CGF, right-angled at G, suppose one of the sides as FG to remain constant, it is required to find the fluxions of the other parts. By the foregoing proposition no = BM ; CO tang 4c Bm; and ds= But no is the decrement of rad FC; Bm the decrement of the LF; co the decrement of cg; and Ds the increment of the 2.c, also fc is the complement of BC. Therefore, x-_#;-caz X-L7; and LC= tang 2c sin LE sin Lc x-; also _ i = tang 2c X CG = rad sin cf rad sin LC sin CF sin CF CF = COS CF X CF sin CF COS CF (C) Hence, if the fixed side to be represented by c, the hypothenuse Fc by b, the side ca by a, and the angles opposite to these sides by C, B, and a respectively (as at Z. 346.) we derive the following general equations, or formulæ, which comprehend all the different cases that can possibly happen wherein the angle B is 90° and one of its adjacent sides (viz, c.) constant. tang C sine c rad sinec Х å. rad tang b sine 1 rad III. a = sine c * Vince's Trigonometry, 2d edition, page 139; Traité de Trigonométrie, par M. Cagnoli, art. 677, page 329. IV. Ő tang 6 Ха. tang C sine 6 COS C x A = xa rad (D) The preceding formulæ may be varied in the expression by reference to pages 104, 105, &c. or they may be turned into proportions thus, from the first set of equations A x sine 6 = sine cxa, that is, à: ::sine 6 : sine c.* PROPOSITION IV. (Plate IV. Fig. 3.) (E) In any right-angled spherical triangle FDE, right-angled at D, suppose the hypothenuse EF to remain constant, it is required to find the fluxions of the other parts. COS BC COS BC By Prop. 2d, no= - Bm; CO = Bm; and sine LC tang Lc sine BC DS = Вт. . rad But no is the increment of BC, or of its=FD; Bm is the measure of the BFm, or of its equal sed, and consequently it is the decrement of the L EFD; also BC = FD, and the Lc is measured by the arc di which is the complement of ED. COS FD Hence FD= Х cot ED Again, co is the decrement of cG, and co is the complement of gi the measure of the 2 E, therefore co is the increment of the E. COS FD Hence L E = COS ED LF; also ds is the decrement of ED, sine FD Consequently- ED= Х LF, and by reduction rad rad X LE = sine fd (F) Hence, if the hypothenuse EF be represented by b, the side Fd by C, ED by a, and their opposite angles by B, C, and a X ED. * Vince's Trigonometry, 2d edition, page 140. COS a XCE Х COS C COS C Х cos a xc. COS a respectively, (as at Z. 346.) we derive the following general formulæ, which include all the varieties that can possibly happen, wherein the angle B is 90°, and the hypothenuse (viz. b) constant. cot a rad 1. À å. sinec tang c rad II. c = хат sine a COS C cot c sine a X Ха xc. rad COS a Хcc X C. c. cot c (G) These formulæ may be varied in the expression or turned into proportions, &c. as before observed. Thus from the third set of equations cot c rad? rado and cotas xa. Hence a: c::tang c : tang a.* III. C= tang c tang a tangc Авт. . PROPOSITION V. (Plate IV. Fig. 4.) (H) In any oblique-angled spherical triangle ABC, suppose the angle A and its adjacent side AB to remain constant t, it is required to find the fluxions of the other parts. Let Bc, by its revolution about B, be changed to Bm, then AC will become am, and the LABC will be reduced to the L Now, mc will be the decrement of Ac, the CBm will be the decrement of the LB, and if cn be drawn perpendicular to Bn, mn will be the decrement of Bc; produce en and BC so that Bp and so may be quadrants, then op will be the measure of the Z CBM. Hence, I. cn : op::sine Bc : sine bo, and in the small straight-lined triangle cnm right-angled at n mn: cn::sine mcn : cos L mcn. (Y. 34.) .::mn : op::sine BC x sine Zmcn : sine Bo x cos Lmcn. * Vince's Trigonometry, Prop. 49th, page 140, 2d edition. + Traité de Trigonométrie, par Cagnoli, page 311. La Lande's Astror |