Putting s = a + b + c, we shall have, {s—a=} (b.tc—a), }s—b=}(a+c—b), and }s—c=}(a+b−c); 10. From equations (4) and (6) we obtain, sin (sc) sin (s - b) sin (s) sin (s — a) tan A = 11 We may deduce the value of the side of a trian gle in terms of the three angles by applying equations (5), to the polar triangle. Thus, if a', b', c', A', B', C', repre sent the sides and angles of the polar triangle, we shall have (B. IX., P. 6), hence, omitting the ', since the equations are applicable to any triangle, we shall have, cos (A + B − C) cps 1⁄2 (B + C − A), cos (A + 0 B) cos (B + C − A), S=A+B+C, we shall have (8) } S~ A = (C + B- A), S- B = } (A + C - B), } S − C = }(A + B − C') ; cos (SC) cos ( S − B) and, cos c sin A sin O 12. All the formulas necessary for the solution of spheri cal triangles, may be deduced from equations marked (1) If we substitute for cos b in the third equation, its value taken from the second, and substitute for cos2 a its value 1 - sin2 a, and then divide by the common factor, sin U we shall have, cos c sin a = sin c cos a cos B + sin b cos C. Therefore, cot c sin a = cos a cos B+cot C sin B. Hence we may write the three symmetrical equations, cot a sin b = cos b cos C+ cot A sin C, cot b sin c = cos c cos A+ cot B sin A, (10) cot c sin a = cos a cos B+ cot C sin B. That is: In every spherical triangle, the cotangent of one o the sides into the sine of a second side, is equal to the cosine of the second side into the cosine of the included angle, plus the cotangent of the angle opposite the first side into the sine of the included angle. NAPIER'S ANALOGIES. 13. lf from the first and third of equations (1), cos e be eliminated, there will result, after a little reduction, cos A sin c = cos a sin b cos C sin a cos b. From the second and third of equations (1), we get, cos B sin c = cos b sin a cos sin b cos a. - Hence, by adding these two equations, and reducing, we shall have, sin c (cos A+ cos B) = (1 — cos C) sin (a + b). and, sin c (sin Asin B) = sin C (sin a — sin b). Dividing these two equations, successively, by the preced ing, member by member, we shall have, reducing these by the formulas (Plane Trig., Arts. 85, 86), Hence, two sides, a and b, with the included angle being given, the two other angles A and B may be found by the proportions, cos (a+b): cos (ab): cot sin (a + b) sin (a - b) :: cot C: tang (A + B), C: tang (A - We may apply the same proportions to the triangle, polar to ABC, by putting by means of which, when a side c and the two adjacent angles A and B are given, we are enabled to find the two other sides a and b. These four proportions are known by the name of Napier's Analogies. 14. In the case in which there are given two sides and an angle opposite one of them, there will in general be two solutions corresponding to the two results in Case II., of rectilineal triangles. It is also plain, that this ambiguity will extend itself to the corresponding case of the polar triangle, that is, to the case in which there are given two angles and a side opposite one of them. In every case we shall avoid all false solutions by recollecting, 1st. That every angle, and every side of a spherical triangle is less than 180°. 2d. That the greater angle lies opposite the greater side, and the least angle opposite the least side, and reciprocally. NAPIER'S CIRCULAR PARTS. 15. Besides the analogies of Napier already demonstrat ed, that Geometer invented rules for the solution of all the cases of right-angled spherical triangles. In every right-angled spher ical triangle BAC, there are six parts: three sides and three angles. If we omit the consideration of the right angle, which is always known, there are five remaining parts, two of which must be given before the others can be determined. B Comp.a Сотр. Comp. being a circular part, is supposed not to separate the cir cular parts c and b, so that these parts are considered as lying adjacent to each other. If any two parts of the triangle are given, their cor responding circular parts are also known, and these, together with a required part, will make three parts under consideration. Now, these three parts will all lie together, or one of them will be separated from both of the others. For example, if B and c were given, and a required, the three parts considered would lie together. But, if B and C were given, and b required, the parts would not lie together; for B would be separated from comp. by the part comp. a, and from b by the part c. In either case, comp. B is the middle part. Hence, when there are three of the circular parts under consideration, the middle part is that one of them to which both of the others are adjacent, or from which both of them are separated. In the former case, the parts are said to be adjacent, and in the latter case, the parts are said to be opposite. This being premised, we are now to prove the following theorems for the solution of right-angled spherical triangles, which, it must be remembered, apply to the circular parts, as already defined. 1st. Radius into the sine of the middle part is equal to the rectangle of the tangents of the adjacent parts. 2d. Radius into the sine of the middle part is equal to the rectangle of the cosines of the opposite parts. These theorems are proved by assuming each of the five circular parts, in succession, as the middle part, and by taking the extremes first opposite, and then adjacent. Having thus fixed the three parts which are to be consid |