Now in the limit when is infinite, the terms involving disappear; hence in the limit, This is true of the cos. A'cos. B' cos. C' sin. B' sin. C' Cos. a' = sides and angles of every polar triangle. Now it appears from Prop. xi., p. 259, that every triangle may be regarded as the polar triangle of some other; hence the above formula is perfectly general, and is true of every triangle, and we have Con. 3.-Formulas similar to (3) and (4) are, of course, true of Cos. B and Cos. C, and of Cos. b and Cos. c. (3.) To express the Formulas of the last Article in a Form adapted for Logarithmic sin. b sin. c 2 s, then b + c — a = 2 (s — a) a− b + c = 2 (s — b), and C= 2 (s c). these formulas are analogous to the formulas on p. 324 of plane trigonometry, which can be shown to be the limits of these in the same manner as in Cor. (1). Art. (1), and Cor. (1) Art. (2). It is to be observed that, since any two sides of a triangle are greater than the third, s a, s — b, s - c, are positive; and since all the sides of a triangle are less than four right-angles, s is less than two right-angles, and à fortiori, s — -a, s- b, 8- c, are each less than two right-angles: so that, sin. s., sin. (s — a), sin. (s — b), sin. (sc), are each positive, also b and c are each less than 180°; so that sin. b and sin. c are always positive. Hence, (5), (6), (7), (8), are always positive, and .. the Cor: If we consider the case of the polar triangle A = 180° - · a':. A' b 180° B' c 180° — c'. a = 180° .. cos. S where 2 S A+ B+ C. It is to be observed that A + B + C must be less than six right-angles, and greater than two right-angles; .. S> 90° < 270°. is always negative. Also S—A being equal to ¦. (B+C—A) = } { (180° — (b'′ + c'—a') } must be -B) < 90°, since b′ + c′ > a' .. cos. (S — A) is always positive, and similarly cos. (S and cos. (S-C) are always positive, and .. (9) (10) (11) (12) though in appearance negative, are really always positive, and .. give us real values for sin. a a Cos. 2 2' tan. and sin. a. 2' sin. sin. Similarly tan. a COS. 1 (a—1) cos. 1 (28—a—b) sin. ssin. (s—c) These formulas are clearly analogous to (36) on p. 325 Plane Trigonometry. Cor: If we take the polar triangle, since tan. and tan. tan. Hence, remembering that the formulas for the polar triangle are perfectly general, we have (5.) To prove the Formula Cotan. A sin. C = Cotan. a sin. b This formula is used in certain propositions: e.g. it is employed in the astronomical problem of finding the aberration in declination. There are as many as six different cases of right-angled spherical triangles, as will appear from the following considerations : Let A B C be the triangle, having a right angle at C. Then using the ordinary notation, all possible cases are the following: (1.) Given the base and perpendicular, i.e., given a and b. (2.) Given the hypothenuse and another side, i.e, given c and a, or c and b. (3.) Given the base or perpendicular, and an adjacent angle, i.e., given a and B, or b and A. (4.) Given the base or perpendicular and an opposite angle, i.e., given a and A, or b and B. (5.) Given the hypothenuse and an angle, i.e., given c and A, or c and B. (6.) Given the two angles, i.e., A and B. Fig. 3. B If these cases be compared with those on p. 361, for plane triangles, it will be seen that the third case of plane triangles diverges into two cases, viz., the third and fourth of spherical triangles, while the sixth case is peculiar to spherical triangles. Both of these differences are due to the circumstance, that in the spherical triangle, A + B + C is not known, whereas in the plane triangle A + B + C 180° 6. To investigate the Formulas on which the Solution of Right-angled Spherical Triangles depends. |