3. Given b = 152.67 yds., and C = 50° 18′ 32′′, to find the other parts.

Ans. B 39° 41′ 28′′, c = 183.95, and a = 239.05.

4. Given c = 379.628, and C 39° 26' 16", to find B, a, and b.

Ans. B 50° 33′ 44", a = 597.613, and b = 461.55,

CASE III.

Given the two sides about the right angle, to find the re maining parts.

40. The angle at the base may be found by Formula (12), and the solution may be completed as in Case II.

EXAMPLES.

1. Given b = 26, and c = 15, to find

C, B, and e

OPERATION.

Applying logarithms to Formula (12), we have,

As in Case II., log c+ 10 log sin C = log a;

(a. c.) log sin C (29° 58' 54") 0.301271

Ans.

C 29° 58′ 54′′, B = 60° 01′ 06′′, and a = 30.017.

2. Given b = 1052 yds., and c = 347.21 yds., to find B, C, and a.

B = 71° 44′ 05′′, C = 18° 15′ 55′′, and a = 1107.82 yds.

3. Given b = 122.416, and c = 118.297, to find B,

and a.

B = 45° 58′ 50′′, C = 44° 1′ 10′′, and a = 170.225

4. Given b = 103, and c = 101,

and a.

to fird B, C

B = 45° 33′ 42′′, C = 44° 26' 18", and a = 144.256.

Given the hypothenuse and either side about the right angle, to find the remaining parts.

41. The angle at the base may be found by one of Formulas (10) and (11), and the remaining side may then be found by one of Formulas (7) and (8).

EXAMPLES.

1. Given a = 2391.76, and b = 385.7, to to find C, B, and c.

OPERATION.

Applying logarithms to Formula (11), we have,

and c = 2360.45.

Ans. B 9° 16' 49", C = 80° 43' 11",

2. Given a = 127.174 yds., and c = 125.7 yds., to find C B, and b.

Ans. B = 8° 43′ 54′′, C = 81° 16′ 6′′, and b = 19.3 yds.

3. Given a 100, and b = 60, to find B, C, and c

Ans. B = 36° 52′ 11′′, C = 53° 7′ 49′′, and c = 80.

SOLUTION OF OBLIQUE-ANGLED

TRIANGLES.

42. In the solution of oblique-angled triangles, four cases may arise. We shall discuss these cases in order.

CASE I.

Given one side and two angles, to determine the remaining

Since a and b are any two sides, and A and B the angles lying opposite to them, we have the following principle:

The sides of a plane triangle are proportional to the sines of their opposite angles.

It is to be observed that Formula (13) is true for any value of the radius. Hence, to solve a triangle, when a side and two angles are given: