Angles do not add up to $180^\circ$ In $\triangle ABC$, $\alpha = 35^\circ$, side $b = 10$ and side $c = 4$.
(i) Show that the length of side $a$ is $7.10$
(ii) Find the remaining angles $\beta$ and $\gamma$
I used the cosine rule and found the $7.10$, but I got $53.89^\circ$ for angle $\beta$ and $18.85^\circ$ for angle $\gamma$, which don't add to make $180^\circ$.
 A: One issue with the law of sines is that using it to find an angle is ambiguous, since the sine function doesn't differentiate between an angle and its supplementary angle. The law of cosines doesn't have this problem. Once you found side $a$, you could use the law of cosines for angle $B$, giving $100=66.4-56.8 \cos(B)$, which reveals that $B$ is obtuse since $\cos(B)<0$. (I took $a=7.1$ as given here.)
If you instead use the law of sines to find an angle, then in general you must split into cases: using your numbers, you get that $B$ is either $53.89^\circ$ or $126.11^\circ$. When you assume that it is $53.89^\circ$ and then calculate $C$ by the angles adding to $180^\circ$, you get a contradiction in that $B$ is not the largest angle of the triangle, even though $b$ is the largest side of the triangle. So you can go this way to conclude that $B=126.11^\circ$ is correct.
A: Hint
Try sketching the triangle after finding the third side length. It will give you a good idea of what the other two angles should roughly be (or at the very least, whether they're acute or obtuse) 
Then remember that when you apply the sin rule, $\sin(x)=\sin(180^\circ-x)$ holds for $0\leq x\leq180^\circ$ so you have two possible values for each angle. Compare the two possibilities with your sketched triangle, only one will make sense on each, as one possibility is acute and the other obtuse (why?)
