sin (AB-AC): cot (BAD+CAD):'tan (BAD-CAD). Now, when AD is within the triangle, BAD + CAD= BAC, and therefore sin (AB+ AC): sin (AB-AC):: cot BAC: tan (BAD-CAD).

But if AD be without the triangle, BAD-CAD-BAC, and therefore sin (AB + AC) sin (AB-AC):: cot (BAD+CAD): tan BAC; or because cot (BAD+CAD): tan BAC:: cot BAC : tan (BAD4CAD), sin (AB+AC); sin (AB—AC): : cot BAC tan (BAD + CAD). Wherefore, &c. Q. ED.

LEMMA.

The sum of the tangents of any two arches, is to the difference of their tangents, as the sine of the sum of the arches, to the sine of their dif ference.

Let A and B be two arches, tan A+tan B: tan A-tan B:: sin (A+B): sin (A-B).

For, by § 6. page 340, sin A x cos B+cos Axsin Bsin (A+B), and therefore dividing all by cos A cos B, sin A sin. B

cos A

+

sin (A+B)

sin A

that is because

cos B cos Ax cos B' tan A, tan A+ tan B

=

cos A

sin (A+B)

cos Axcos B'.

2

In the same manner it is proved that tan A-tan B= sin (A-B)

Therefore tan A+tan B; tan A—tan B : : cos AX COS outre

COS

Β ́

sin (A+B): sin (A-B). Q. E. D.

The sine of half the sum of any two angles of a spherical triangle is to the sine of half their difference, as the tangent of half the side adjacent to these angles is to the tangent of half the difference of the sides opposite to them and the cosine of half the sum of the same angles is to the cosine of half their difference, as the tangent of half the side adjacent to them, to the tangent of half the sum of the sides opposite.

Let C+B=2S, C-B-2D, the base BC=2B, and the difference of the segments of the base, or BD-CD =2X. Then, because (30.) sin (C+B): sin (C—B) :: tan BC: tan (BD-CD), sin 2S: sin 2 D:: tan B: tan X. Now, sin 2S sin (S+S) = 2 sin S x cos S, (Sect. III. cor. Pl. Tr.) In the same manner, sin 2D= 2 sin Dxcos D. Therefore sin Sx cos S: sin D x cos D:: tan B tan X.

AM

A gain, in the spherical triangle ABC it has been proved, that sin C+ sin B: sin C-sin B:: sin AB+sin AC: sin A B-sin AC, and since sin C+ sin B-2sin (C+B) × cos (C-B,) (Sect. III. 7. Pl. Tr.) = 2 sin S x cos D;

and sin C-sin B2 cos(C+B) x sin

(CB): 2 cos Sx sin D. Therefore 2 sin S x cos D: 2 cos S x sin D sin AB + sin AC: sin AB-sin AC. But (3. Pl. Tr.) sin AB+ sin AC sin AB-sin AC tan (AB+AC); tan ! (AB—AC):: tan : tan4, Σ being equal to (AB+ AC) and a to (AB-AC). Therefore sin Sx cos D: cos Sxsin D:: tan : tan 4. Since

by multiplying equals by equals,

(sin D)2 x cos Sx cos D ___ (sin D)2

(sin S)2 x cos S× cos D (sin S)2.

tan

(BD-DC) tan (AB+AC), that is,

= tan

(AB-AC)

tan BC

But (29.)

tan X tan

tan X

=

and therefore,

tan Extan Al

tan A

tan B'

tan A sin D

tan B sin S'

which is the first part of the proptan (AB-AC);

sin (C+B): sin (C-B):: tan BC: tan (AB-AC);

tan E or cos S: cos D:: tan B: tan E, that is cos (C+B):

tan B'

2

cos (C-B): tan BC: tan (AB+AC); which is the second part of the proposition. Therefore, &c. Q. E. D.

COR. 1. By applying this proposition to the triangle supplemental to ABC (11.), and by considering, that the sine of one half of the sum or half the difference of the sup plements of two arches, is the same with the sine of half the sum or half the difference of the arches themselves; and that the same is true of the cosines, and of the tangents of half the sum or half the difference of the supplements of two arches; but that the tangent of half the supplement of an arch is the same with the cotangent of half the arch itself; it will follow, that the sine of half the sum of any two sides of a spherical triangle, is to the sine of half their difference, as the cotangent of half the angle contained between them, to the tangent of half the difference of the angles opposite to them; and also that the cosine of half the sum of these sides is to the cosine of half their difference, as the cotangent of half the angle contained between them, to the tangent of half the sum of the angles opposite to them.

COR. 2. If therefore A, B, C be the three angles of a spherical triangle, a, b, c the sides opposite to them.

In a right angled spherical triangle, of the three sides and three angles, any two being given, besides the right angle, to find the other three.

This problem has sixteen cases, the solutions of which are contained in the following table, where ABC is any spherical triangle right angled at A.

R: sin BC:: sin B: sin AC, (19) 1
R: cos B:: tan BC: tan AB,(21) 2
R: cos BC:: tan B: cot C, (20). 3

R; sinAC:: tan C: tan AB, (18).
cos C: R:: tan AC: tan BC, (21). 5
Rcos AC: sin C: cos B, (23). 6

tan B: tan AC::R: sin AB, (18). 7
sin B: sin AC:: R: sin BC, (19). 8
cos AC: cos B::R:sin C, (23). 9

cos AC: cos BC::R: cos AB, (22). 10 sin BC: sin AC:: R: sin B, (19). 11 tan BC: tan AC:: R: cos C, (21), 12

R: cos AB:: cos AC: cos BC, (22). 13 sin AB: R:: tan AC: tan B, (18). 14 sin AC: R:: tan AB: tan C, (18).14

sin B: cos C:: R: cos AB, (23). 15 sin C cos B:: R: cos AC, (23). 15 tan B: cot C:: R: cos BC, (20). 16