Question

In triangle ABC, angle A = 71º, angle B = 42º, and side c = 15. Solve the triangle. Round angles and sides to the nearest tenth, if necessary?

Answer 1

`C = 67^circ`

`a = {15 \ sin 71^circ}/{67^circ}`

`b = {15 \ sin 42^circ}/{sin 67^circ}`

I let you do your own calculating.

Explanation:

The third angle is

`C = 180^circ - 71^circ - 42^circ = 67^circ`

The Law of Sines gives the sides:

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

`a = c/sin C \ sinA = {15 \ sin 71^circ}/{67^circ}`

`b = c/sin C \ sin B = {15 \ sin 42^circ}/{sin 67^circ}`