the same as that included between the planes AOC and AOB; and the side a is the measure of the plane angle BOC, 0 being the centre of the sphere, and OB the radius, equal to 1. 71. Spherical triangles, like plane triangles, are divided into two classes, right-angled spherical A B triangles, and oblique-angled spherical triangles. Each class will be considered in turn. We shall, as before, denote the angles by the capital letters A, B, and C, and the opposite sides by the small, letters a, b, and c. FORMULAS USED IN SOLVING RIGHT-ANGLED SPHERICAL TRIANGLES. 72. Let CAB be a spherical triangle, right-angled at A, Draw OA, OB, and OC, each of which will be equal to 1. From B, draw BP perpendicular to OA, and from P draw PQ perpendicular to OC; then join the points and B, by the line QB. The line QB will be perpendicular to OC (B. VI., P. VI.), and the angle PQB will be equal to the inclination of the planes OCB and OCA; that is, it will be equal to the angle C. From the right-angled triangles OQP and QPB, we have, If, in (2), we change c and C, into b and B, we QP the fraction by OQ, and QB (90° — a), we have, tan (90° a) tan b. (3.) fraction QP OP' by PR, and (90° C), we have, or, sin b tan c tan (90° C). (4.) have, sin b sin a sin B (5.) If, in (3), we change b and C, into c and B, we If, in (4), we change b, c, and C, into c, b, and Multiplying (4) by (7), member by member, we have, sin b sin c tan b tan c tan (90°—B) tan (90°— C). Dividing both members by tan b tan c, we have, cos b cos c = tan (90°-B) tan (90° — C'); and substituting for cos b cos c, its value, from (1), we have, cos a = tan (90°—B) tan (90°— C') Formula (6) may be written under the form, Changing B, b, and C, in (9), into C, c, and B, we These ten formulas are sufficient for the solution of any right-angled spherical triangle whatever. NAPIER'S CIRCULAR PARTS. 73. The two sides about the right angle, the complements of their opposite angles, and the complement of the hypothenuse, are called Napier's Circular Parts. b 0-06 90-a 90-B B If we take any three of the five parts, as shown in the figure, they will either be adjacent to each other, or one of them will be separated from each of the other two, by an intervening part. In the first case, the one lying between the other two parts, is called the middle part, and the other two, adjacent parts. In the second case, the one separated from both the other parts, is called the middle part, and the other two, opposite parts. Thus, if 90°-a, is the middle part, 90° B, and 90° C, are adjacent parts; and b and c, are opposite parts; and similarly, for each of the other parts, taken as a middle part. 74. Let us now consider, in succession, each of the five parts as a middle part, when the other two parts are opposite. Beginning with the hypothenuse, we have, from formulas (1), (2), (5), (9), and (10), Art. 72, Comparing these formulas with the figure, we see that. The sine of the middle part is equal to the rectangle of the cosines of the opposite parts. Let us now take the same middle parts, and the other parts adjacent. Formulas (8), (7), (4), (6), and (3), Art. 72, give Comparing these formulas with the figure, we see that, The sine of the middle part is equal to the rectangle of the tangents of the adjacent parts. These two rules are called Napier's rules for Circular Parts, and they are sufficient to solve any right-angled spherical triangle. 75. In applying Napier's rules for circular parts, the part sought will be determined by its sine. Now, the same sine corresponds to two different arcs, supplements of cach other; it is, therefore, necessary to discover such relations between the given and required parts, as will serve to point out which of the two arcs is to be taken. Two parts of a spherical triangle are said to be of the ame species, when they are both less than 90°, or both greater than 90°; and of different species, when one is less and the other greater than 90°. From Formulas (9) and (10), Art. 72, we have, |