Ex. 2. In a spherical triangle ABC there are given A=103° 59′ 57′′, B=46° 18′ 7′′, and a=42° 8′ 48′′; required the remaining parts.

There will but one triangle, since sin B<sin A.

Having given the three sides of a spherical trangle to find the angles.

For this case we use equations (3.).

'sins sin (s—a)

cos A=RV

(a+b+c)=s

sin b sin c

Ex. 1. In an oblique angled spherical triangle there are given a 56° 40', b=83° 13′ and c-114° 30'; required the

angles.

=127° 11' 30"

The addition of twice the logarithm of radius, or 20, to the numerator of the quantity under the radical just cancels the 20 which is to be subtracted on account of the arithmetical complements, to that the 20, in both cases, may be omitted.

Applying the same formulas to the angles B and C, we find,

B= 62° 55′ 46′′
C=125° 19′ 02′′.

Ex. 2. In a spherical triangle there are given a 40° 18′ 29′′, b=67° 14′ 28′′, and c=89° 47′ 6": required the three angles.

CASE IV.

Having given the three angles of a spherical triangle, to find the three sides.

For this case we employ equations (7.)

Costa RV

cos(S-B)cos(S-C)

sin B sin C

Ex. 1. In a spherical triangle ABC there are given A=48° 30', B=125° 20', and C=62° 54'; required the sides.

For this case we employ the two first of Napier's Analogies:

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

sin (a+b) sin (a-b) :: cot C: tang (A-B). Having found the half sum and the half difference of the angles A and B, the angles themselves become known; for, the greater angle is equal to the half sum plus the half difference, and the lesser is equal to the half sum minus the half diffe

rence.

The greater angle is then to be placed opposite the greater side. The remaining side of the triangle can then be found by Case II.

Ex. 1. In a spherical triangle ABC, there are given a= 68° 46' 2", b=37° 10′, and c=39° 23'; to find the remaining parts. (a+b)=52° 58′ 1′′, (a—b)=15° 48′ 1′′, C=19° 41′ 30′′.

(a+b) 52° 58' 1" log. ar.-comp. 0.220210 (a-b) 15° 48′ 1′′

C 19° 41′ 30′′

To tang (A+B) 77° 22′ 25′′

9.983271

10.446254

10.649735

(a+b) 52° 58' 1" log. ar.-comp. 0.097840

As sin

Is to sin

(a-b) 15° 48′ 1′′

Totang (A-B) 43° 37′ 21′′ `

So is cot C 19° 41′ 30′′

9.435016

10.446254

9.979110

Hence, A=77° 22′ 25′′ +43° 37′ 21′′-120° 59′ 46′′

B-77° 22′ 25′′-43° 37′ 21′′ 33° 45′ 04′′

side c

=

= 43° 37′ 37′′.

Ex. 2. In a spherical triangle ABC, there are given b=83° 19′ 42′′, c=23° 27′ 46′′, the contained angle A=20° 39′ 48′′; to find the remaining parts.

In a spherical triangle, having given two angles and the included side to find the remaining parts.

For this case we employ the second of Napier's Analogies. cos (A+B): cos (A-B) :: tang c: tang (a+b) sin (A+B) : sin † (A—B) : : tang c: tang (a—b). From which a and b are found as in the last case. The remaining angle can then be found by Case I.

Ex. 1. In a spherical triangle ABC, there are given A=81° 38′ 20′′, B=70° 9′ 38′′, c=59° 16′ 23′′; to find the remaining parts.

(A+B)=75° 53′ 59′′, (A-B)=5° 44′ 21′′, c=29° 38′ 11′′. (A+B) 75° 53′ 59′′ log. ar.-comp. 0.613287 (A-B) 5° 44′ 21′′

As cos To cos

9.997818

9.755051

10.366156

So is tang

29° 38' 11"

As sin

To sin

So is tang

*c

29° 38' 11"

(A+B) 75° 53′ 59′′ log. ar.-comp. 0.013286 (A-B) 5° 14′ 21′′

9.000000

9.755051

8.768337

Hence a=66° 42′ 52′′ +3° 21′ 25′′ 70° 04′ 17′′

b—66° 42′ 52′′′—3° 21′ 25′′—63° 21′ 27′′

angle C

=64° 46′ 33′′.

Ex. 2. In a spherical triangle ABC, there are given A=34° 15′ 3′′, B=42° 15′ 13′′, and c=76° 35′ 36′′; to find the remain

ing parts.

a=40° 0′ 10′′

Ans.

b=50° 10' 30"

C=58° 23′ 41′′.