Solve each triangle. round each side length to the nearest tenth and angle measures to the nearest degree. a= 14, c= 20, b= 38 degrees.

Okay after scratching my head and some scrap paper :P  I was really surprised to see that I don't remember ever running into this before...

Using the Law of Cosines we find the third side.

c=√(196+400-560cos38)≈12.4384  (we can round further for the final answer)

Now using the Law of Sines for the other angles.

sin38/12.4384=sinA/14=sinB/20

A=arcsin(14sin38/12.4384)≈43.8645°

BUT what I didn't realize before is that B was obtuse which means:

B=180-arcsin(20sin38/12.4384)≈98.135°

So solved for sides and angles (with rounding asked for)

(14, 44°), (12.4, 38°), and (20, 98°)

A= r2 × 3.14 11×11 ↓ ↓ 121 × 3.14 = 379.94 → 379.9