C

is 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 tri

ab

a

angles, 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 sides opposite by the small letters a, b, and c.

FORMULAS

USED IN SOLVING RIGHT-ANGLED SPHERICAL TRIANGLES.

72. Let CAB be a sperical 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 sides opposite by the letters a, b, and c, recollecting that B and C may change places, provided that b and c change places at the same time.

B

P

a

From B, draw

Draw OA, OB, and OC, each equal to 1. BP perpendicular to OA, and from P draw PQ perpendicular to OC; then join the points Q 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 spherical angle C.

We have, from the figure,

PB = sin c,

OP = cos c,

QB

= sin a, OQ = cos a.

From the right-angled triangles OQP and QPB, we have

From the right-angled triangle QPB, we have

but, from the right-angled triangle PQO, we have

QP = OQ tan QOP = cos a tan b;

substituting for QP and QB their values, we have

cos C =

cos a tan b
sin a

= cot a tan b.

· (3.)

From the right-angled triangle OQP, we have

but, from the right-angled triangle QPB, we have

substituting for QP and OP their values, we have

If, in (2), we change c and C into b and B, we have

If, in (3), we change b and C into c and B, we have

If, in (4), we change b, c, and C, into c, b, and B, we

have

Multiplying (4) by (7), member by member, we have

sin b sin c = tan b tan c cot B cot C.

Dividing both members by tan b tan c, we have

cos b cos ccot B cot C;

and substituting for cos b cos c, its value, cos a, taken from (1), we have

Formula (6) may be written under the form

(8.)

Substituting for cos a, its value, cos b cos c, taken from (1), and reducing, we have

Again, substituting for sin c, its value, sin a sin C, taken

from (2), and reducing, we have

Changing B, b, and C, in (9), into C, c, and B, we have

These ten formulas are sufficient for the solution of any right-angled spherical triangle whatever. For the purpose of classifying them under two general rules, and for convenience in remembering them, these formulas are usually put under other forms by the use of

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.

90-C

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 two others by an intervening part. In the first case, the one lying between the two other parts is called the middle part, and the two others, adjacent parts. In the second case, the one separated from both the other parts, is called the middle part, and the two others, 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; if c is the middle part, b and 90° B are adjacent parts (the right angle not being considered), and 90°C and 90° a 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 two other 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.