if cos@sina.sin b. (cos C)2,

and cos = cos(a + b).

(W. Ch. viii. prop. 13; Leg. Geom. Note x; L. 177, 263-6.) (24.) Let R, R2 be the circular radii of the inscribed, and circumscribing circles,

(25.)

Solution of right-angled spherical triangles.
Let A, B, be the oblique angles, C, the right angle;
a, b, the sides, c, the hypothenuse.

Naper's Rules. The circular parts are

90°-A, 90°-B, a, b, 90°-c;

any one of which being called the middle part, m, the two adjacent to м on each side of it, 41, 429 and the two remaining or opposite parts, 01, 025

These are all the forms essentially different; four more analagous to [3], [4], [5], [6], may be obtained by changing A, a, into B, b, and vice versâ.

These rules may be applied to a quadrantal triangle, in which c=90°, if

A, B, (90°C), 90°-a, 90° - b,

be taken as the circular parts.

Any angle, and the side opposite are either both > 90°, or both < 90°.

An oblique angle cannot be less, if acute, nor greater, if obtuse, than the opposite side.

The sides are both > 90°, or both < 90°, if the hypothenuse <90°; and one side > 90°, and the other < 90°, if the hypothenuse > 90°.

A side is > or < hypothenuse, according as it is> or < 90°.
A+B and a + b are both>, or both <, 180°.

A + B is always > 90°, and A~B always < 90°.
(W. Chap. x; L. 202—9.)

The results obtained in [2], [4], are doubtful; the ambiguity may in some cases be removed by attending to the conditions stated in page 118.

=

When a small angle is to be determined from its cosine, or an angle nearly 90° from its sine, a small error in the sine or cosine gives a large one in the angle; in such cases some of the following formulæ will give more accurate results.

(26.) Solution of oblique-angled spherical triangles. First Case: given a, b, c.

[1] sin A=

2

sin b.sin c

. sin s. sin (s — a). sin (s — b). sin (s — c) |* ;

log Sin 4 = log 2+ log r-log Sin b-log Sin c

+ {log Sins + log Sin (s — a) + log Sin (s—b) + log Sin (s — c)}.

log Sin4={log Sin (sb) + log Sin (sc)

log Tana={log Sin (s—b) + log Sin (s—c)

-log Sin slog Sin (sa)}+ logr.

[5] Assume cos 0=cos b. cos c,

log Cos-log Cos b + log Cos c-logr; then

log Cos A = log 2 + log Sin 1 (0 + a) + log Sin 1 (0 — a)

-log Sin b-log Sin c + logr.

A-90°, and -a have the same sign.

log Tan = log Cos b + log Cos c-log Sin a; then

log Cos A = log Cos (a + p) + 3 log r

― log Cos - log Sin b―log Sin c.
Q