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?

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Answer 1 · 2018-04-26T20:11:07

$C = 67^{\circ}$ $C = 67^circ$

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

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

I let you do your own calculating.

The third angle is

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

The Law of Sines gives the sides:

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

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

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