sin c (sin A + sin B)=sin C (sin a+sin b), and sin c (sin A-sin B)=sin C (sin a—sin b).

Dividing these two equations successively by the preceding one; we shall have

sin A+ sin B sin C sin a+ sin b
cos A+cos B ̄R—cosC· sin (a+b)

sin A-sin B sin C

cos A+ cos BR-cos C

sin a-sin b

sin(a+b) ·

And reducing these by the formulas in Articles 24. and 25., there will result

cos (a-b)

tang}(A+B)=cot¿C·cost (a+b)

tang}(A—B)=cot}C·sin}(a+b)•
sin} (a—b)

Hence, two sides a and b with the included angle C being given, the two other angles A and B may be found by the analogies,

cos(a+b): cos(a—b) : : cot‡C : tang}(A+B)

sin}(a+b): sin}(a−b) : : cot1⁄2C : tang}(A—B).

If these same analogies are applied to the polar triangle of ABC, we shall have to put 180°-A, 180°-B, 180°—α, 180°-b, 180°-c, instead of a, b, A, B, C respectively; and for the result, we shall have these two analogies,

cos1(A+B) : cos}(A—B) : : tang‡c : tang}(a+b)

sin}(A+B) : sin1(A—B) : : tang}c : tang}(a—b,) by means of which, when a side c and the two adjacent angles A and B are given, we are enabled to find the two other sides a and b. These four proportions are known by the name of Napier's Analogies.

SOLUTION OF SPHERICAL TRIANGLES IN GENERAL.

The solution of spherical triangles includes six general cases, which we shall now explain in succession.

FIRST CASE.

LXXIII. The three sides a, b, c being given, any angle. for example the angle A opposite the side a, will be found by the formula:

a+b -C
2

sin

a+c-b
2

sin b sin c

At-67.

LXXIV. Given two sides a and b with the angle A opposite to one of them, to find the third side c, and the other two angles B and C.

First. The angle B is found from the equation sin B÷ sin A sin b

sin a

Secondly. To find the angle C, we must solve the equation, cot A sin C+ cos C cos b =cot a sin b.

For this purpose, take an auxiliary angle, such that we may

of cot A, being substituted in the equation to be solved, gives

cos b

sin &

-(cos o sin C+sin 4 cosC)=cot a sin b; whence we obtain

By this artifice, the two unknown terms of the equation are now reduced to one, from which it is easy to find the angle C.

Thirdly. The side c will be found by the equation

It might also be determined directly, by solving the equa

tion

R cos b cos c+cos A sin b sin c= -R2 cos a

Hence, by first seeking the auxiliary quantity from the equa

This second Case may have two solutions, like the analogous Case in rectilineal triangles.

THIRD CASE.

LXXV. Given two sides a and b, with the included angle C, to find the other two angles A and B and the third side c. First. The angles A and B are found by these two equations

in which the second members may be reduced to a single term, by means of an auxiliary quantity. It is simpler, however, In this case, to make use of Napier's analogies, which give

Secondly. Knowing the angles A and B, the third side c

may be computed by the equation sin c=sin a

for determining c directly, we have the equation

R2 cos c=sin a sin b cos C+R cos a cos b.

Assume the auxiliary quantity 4, such that sin b cos C=cos b

LXXVI. Given two angles A and B with the adjacent side c, to find the other two sides a and b with the third angle C. First. The two sides a and b are given by the formulas cot A sin B+ cos B cos c

They may, however, be computed more easily by Napier's analogies, namely,

Secondly. Knowing a and b we shall find C by the equation

; but C may be also found directly, by the

R2 cos C=cos c sin A sin B-R cos A cos B. Assume the auxiliary quantity 4, such that

This case and the preceding one offer no ambiguity.

LXXVII. Given two angles A and B, with the side a opposite one of them, to find the other two sides b and c and the third angle C.

First. The side b is found by the equation

Secondly. The side c depends on the equation.

cot a sin e-cos B cos c=cot A sin B

cos B

have

(sin c cos —cos c sin q)=cot A sin B; hence

sin

Thirdly. The angle C is found by solving the equation

cos a sin B sin C-R cos B cos C=R2 cos A.

cos a tang B
R

we shall have (sin C cos -cos C sin o)=

For this purpose, make cos a sin B:

=

R cos B cos
sin

or cot =

cos B
sin

This fifth Case, like the second, is susceptible of two solutions, as happens in like manner in the analogous Case of rectilineal triangles.

SIXTH CASE.

LXXVIII. The three angles A, B, C being given, we can find any side, for example the side opposite the angle A, by the formula

Grt-bo- sin &a=R\ √

-—cos ↓ (A+B+C) cos} (B+C—A)).

sin B sin C

It may be observed, that of these six general Cases, the last three might have been deduced from the first three, by the property of polar triangles; so that properly speaking, there are but three different cases in the general solution of spherical triangles. The first case is solved by a single