Question

Please help me solve for the following angle alpha (click to see image).

Answer 1

The correct answer is about 150*. I used the law of sines and my answer is about 29*.

When I subtract 29 from 180, 180-29= ~ 150*, I get the correct answer.

Answer 2

Right, law of sines is the way to go here:
\[\frac{\sin\alpha^\circ}{16}=\frac{\sin15^\circ}{8.53}~~\implies~~\sin\alpha^\circ\approx0.485~~\implies~~\alpha^\circ=\arcsin0.485\]
When you use the inverse sine function, you have to be mindful of its domain. Recall that \(\sin x\) starts to repeat values as \(x\) moves from the first quadrant to the second, which means \(\sin x\) is not one-to-one on this interval and so the inverse doesn't exist, but it does when \(x\) is in the first or fourth quadrant. By default, this is how your calculator computes the inverse sine. Like you said, you end up getting \(\alpha\approx29.04\).

To resolve this issue, you shift the argument of the sine function by \(\pi\) radians, or \(180^\circ\), so that \(x\) is in either the third or fourth quadrant. All this means is that you find the "congruent" angle in the third or fourth quadrant by taking this compute \(\alpha\) from \(180^\circ\).