What is the sine of C if triangle ABC has b = 3, c = 7, and angle B = 15°?

Solution:

Sine Law should be used to solve this problem.

It can be represented using the triangle.

a/sin A = b/sin B = c/sin C

The values given are b = 3, c = 7, and angle B = 15°

Substituting the values in the equation

3/sin 15° = 7/sin C

By cross multiplication

3 sin C = 7 sin 15°

3 sin C = 7 × 0.2588190451

So we get, 3 sin C = 1.811733316

sin C = 1.811733316 ÷ 3

sin C = 0.6039111053

Therefore, the sine of C if triangle ABC has b = 3, c = 7, and angle B = 15° is 0.6039111053.

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What is the sine of C if triangle ABC has b = 3, c = 7, and angle B = 15°?

Summary:

The sine of C if triangle ABC has b = 3, c = 7, and angle B = 15° is 0.6039111053.