Then in the right triangle A'BC, A'C 180° - b.

=

That is, the sides a and A'C of the triangle A'BC are each < 90°; and by I., A'B and the angles A' and A'BC are each 90°.

But, c 180° — A'B, A = A', and B = 180° — A'BC.

=

Whence, c is > 90°, A < 90°, and B > 90°.

In like manner, if a is > 90° and b < 90°, then c is > 90°, A> 90°, and B < 90°.

Then in the right triangle ABC', AC'=180° — b, and BC' = 180°

α.

That is, the sides AC' and BC' of the triangle ABC' are each< 90°; and by I., AB and the angles BAC' and ABC' are each< 90°.

But, A = 180° - BAC', and B = 180° - ABC'.

Whence, c is < 90°, A> 90°, and B > 90°.

Hence, in any spherical right triangle :

1. If the sides including the right angle are in the same quadrant, the hypotenuse is < 90°; if they are in different quadrants, the hypotenuse is > 90°.

2. An angle is in the same quadrant as its opposite side.

XI. SPHERICAL RIGHT TRIANGLES.

139. Let C be the right angle of the spherical right triangle ABC, and O the centre of the sphere.

At any point A' of OA draw A'B' and A'C' perpendicular to 04, and join B'C'.

Then by Art. 131, the sides a, b, and c measure the angles BOC, COA, and AOB, respectively, and the angle B'A'C' is equal to the angle A of the spherical triangle.

Since OA is perpendicular to A'B' and A'C', it is perpendicular to the plane A'B'C'.

Whence, since each of the planes A'B'C' and OBC is perpendicular to the plane OAC, their intersection B'C' is perpendicular to OAC.

Therefore B'C' is perpendicular to A'C' and OC".

In the right triangle OA'B', we have

But in the right triangles OB'C' and OC'A',

141. Since sin a = cos a tan a (Art. 20), (75) may be

142. From (74), (81), and (82), we have

143. The proofs of Art. 139 cannot be regarded as general, for in the construction of the figure we have assumed a and b, and therefore c and A (Art. 138), to be less than 90°. To prove formulæ (74) to (78) universally, it is necessary to consider two additional cases:

CASE I. other > 90°.

When one of the sides a and b is < 90°, and the

In the right triangle ABC, let a be < 90° and b > 90°. Complete the lune ABA'C; then in the triangle A'BC,

A'B 180° - c,

=

A'C 180° — b,

A' A,

A'BC = 180° - B.

But by Art. 138, c is > 90°, A < 90°, and B > 90°. Hence each element, except the right angle, of the right triangle A'BC is < 90°, and we have by Art. 139,

Putting for A'B, A'C, A', and A'BC their values, we have

and we obtain the formulæ (74) to (78) as before.

In like manner, the formulæ may be proved in the case where a is > 90° and b < 90°.

CASE II. When both a and b are > 90°.

In the right triangle ABC, let a and b be > 90°.
Complete the lune ACBC'.

By Art. 138, c is < 90°, A > 90°, and B > 90°.

Hence each element, except the right angle, of the right triangle ABC' is < 90°, and we have by Art. 139,