the same as that included between the planes AOC and AOB; and the side a is the measure of the plane angle BOC, O 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 B A 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, and let O be the centre of the sphere on which it is situated. Denote the angles of the triangle by the letters A, B, and C, and the opposite sides by the letters a, b, and c, recollecting that B and C may change places, provided that band change places at the same time. B a P 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. We have, from the figure, PB = sin c, OP = cos c, QB = sin a, OQ: = COS α. From the right-angled triangles OQP and QPB, we have, OQ = OP_cos AOC; or, cos a = cos c cos b · (1.) sin c = sin a sin C · (2.) = PB QB sin PQB; or, If, in (2), we change c and C, into b and B, we have, sin b = sin ɑ sin B · a ・ ・ (5.) If, in (3), we change b and C, into c and B, we have, cos B = tan (90°-a) tan c (6.) 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') cos a, taken Formula (6) may be written under the form, (8.) Changing B, b, and C, in (9), into C, c, and B, we have, cos C = cos c sin B. (10.) These ten formulas are sufficient for the solution of any right-angled spherical triangle whatever. NAPIER'S CIRCUIAR 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. If these parts be arranged in their order, as shown in the figure, we see that each part is adjacent to two of the others, and that it is separated from each of two remaining parts by an intervening part. If any part be taken as a middle part, those which are adjacent to it are called adjacent parts, and those which are separated from it, are called opposite parts. Thus, 90° B, and 90° C, are adjacent parts to 90° - a; and c and bare opposite parts; and so on, for each of the other parts. 74. Formulas (1), (2), (5), (9), and (10), of Art. 72, may be written as follows: 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. Formulas (8), (7), (4), (6), and (3), of Art. 72, 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 each 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 same 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, |