Equations (6) and (7) and their permuted values give as If we convert equations (1) and (2) of (F) into proportions, we have 14 cos. 1 (a+b): cos. (a - b) :: cot. C: tan. (A + B), } which proportions are known as Napier's first and second Analogies. They are used in solving a triangle when two sides and the included angle are given. $61. To express the tangent of the sum and difference of two sides of a spherical triangle, in terms of the angles opposite to them, and the third side of the triangle. Let A, B, C, a, b, c be the angles and sides of a spherical triangle, A', B', C', a', b', c', the corresponding parts of the polar triangle; then by (F), we have tan. +[(180°—a)+(180°-6)] cos. [(180°-A)-(180°-B)] cos. [(180°-A)+(180°-B)] Xcot. † (180°—c). This becomes tan. (a+b) = cos. (AB) x tan. c. (1.) Similarly we have Therefore, sin. § (a' — b′) tan. (A' — B') = sin. (a' + b') x cot. C'. Equations (1) and (2), together with their permuted values, The first and second conditions of (G) being converted into proportions, gives Napier's third and fourth Analogies, as fol lows: cos. (A + B): cos. (A – B) : : tan. c: tan. (a+b), sin. (A + B) : sin. (A – B) :: tan. c : tan. 1⁄2 (a — b). These analogies are employed in the solution of a triangle when two angles and the interjacent side are given. §62. To express the cotangent of an angle of a spherical triangle, in terms of the side opposite, one of the other sides, and the angle included between these two sides. Equation (2) of § 60, when cleared of fractions, becomes cos. A sin. c = cos. a sin. b-cos. b sin. a cos. C. (1.) By (E) we have sin. A sin. c = sin. C sin. a. (2.) Dividing (1) by (2), we find cot. A cot. a sin. b cosec. C—cos. b cot. C. (3.) (4.) By interchanging B and C, b and c, (3) will give cot. A = cot. a sin. c cosec. B — cos. c cot. B. Proceeding in like manner for the other angles, we shall ob tain the following group: cot. A = cot. a sin. b cosec. C— cos. b cot. C = cot. a sin. c cosec. B-cos. c cot. B; cot. B=cot. b sin. c cosec. A cos. c cot. A (H.) cot. b sin. a cosec. C — cos. a cot. C; cot. C — cot. c sin. a cosec. B — cos, a cot. B =cot. c sin. b cosec. A — cos. b cot. A. These six equations are of such a nature that they do not yield any new relations by the application of the principle of the Polar Triangle (§ 56). §63. By aid of the eight groups of formulas designated by (A), (B), (C), (D), (E), (F), (G), (H), we shall be enabled to solve all the cases of spherical triangles, whether right-angled or oblique-angled. We shall, in the next chapter, proceed to apply these formulas. §64. Before, however, passing to the solution of spherical triangles, we will deduce some other general formulas of spherics, which will frequently be found useful in Astronomy. Equation (6) of § 55, becomes, by permuting, cos. c cos. a + sin. c sin. a cos. B cos. b. (1.) Multiplying (6) of § 55, by cos. c, it becomes cos. c cos. α = cos. b cos.2 c + sin. b sin. c cos. c cos. A. Subtracting (2) from (1) and substituting sin.2c for 1 - cos.2 c, (2.) we have sin. c sin. a cos. B = cos. b sin.2 c sin. b sin. c cos. c cos. A ; which, divided by sin. c, becomes sin. a cos. B = cos. b sin.c - sin. b cos. c cos. A. (3.) Interchanging B, C, and b, c, (3) becomes sin. a cos. C =cos. c sin. b — sin. c cos. ¿ cos. A. Equations (3) and (4) with their permutations, give sin. a cos. B =cos. b sin.c (4.) sin. b cos. C = cos. c sin. a sin.c cos.a cos. B; (I.) If we apply the principle of the Polar Triangle to equations (I) they become sin. A cos. b = cos. B sin. C + sin. B cos. C cos. (J.) CHAPTER VI. SOLUTION OF SPHERICAL RIGHT TRIANGLES. $65. We shall confine ourselves to such triangles as have only one right angle; those that have two or three right angles will be considered hereafter. (See $74.) A spherical triangle consists of 6 parts, 3 sides and 3 angles; and any three of these being known the others may be found. In the present case, one of the angles being a right angle, it follows that any other two parts being known the other three may be found. The number of combinations of 5 things taken 3 and 3 at a 5 × 4x 3 time, is 10; therefore, ten different cases present 1×2×3 themselves in the solution of right-angled spherical triangles. $66. The solution of these ten cases may all be comprised in two rules first given by Napier, and known under the name of Napier's Rules for Circular Parts. OF THE CIRCULAR PARTS. The right angle is not taken into consideration. The two sides, the complements of the two angles, and the complement of the hypotenuse constitute the five circular parts. Let ABC be a spherical triangle, rightangled at C. If we arrange the five circulating parts upon the circumference of its circumscribed circle, we shall readily discover which are adjacent, and which are opposite, when either of the five parts is chosen as the middle part. NAPIER'S RULES. 90°-€ 90°-B A 90°- A B AS I. The sine of the middle part is equal to the product of the tangents of the adjacent parts. II. The sine of the middle part is equal to the product of the cosines of the opposite parts. If now we take in turn each of the five parts as the middle part, and apply these Rules, we shall obtain ten formulas, as follows. First. Let a be the middle part, then will b and (90°-B) be the adjacent parts, and (90°— c) and (90° – A) the opposite parts. (1.) RULE I. sin. α = tan. b tan. (90°- B) = tan. b cot. B. RULE II. sin. a cos. (90°-c) cos. (90°-A) = sin. c sin. A. (2.) Secondly. Let b be the middle part, then will a and (90° — A) be the adjacent parts, and (90°— c) and (90°— B) the opposite parts. (3.) RULE I. sin. b = tan. a tan. (90° - A) = tan. a cot. A. |