<s>[INST] Sides of non-right triangle 
Given a non-right triangle with vertex $C = 91^{\circ}, \
\overline{AC} =39.9 \ cm, \ \overline{BC} = 32.6 \ cm$, find 
a) $\overline{AB}$ 
b) $m\measuredangle A$
c) $m\measuredangle B$

I was able to find the length of $\overline{AB}$, which I labeled $c$
\begin{align*} c=& \ \sqrt{32.6^2+39.9^2-2(32.6*39.9)\cos(91^\circ)}
\\ c =& \ 51.9 \ cm
\end{align*}
but I'm having a hard time finding the angle measures of $A$ and $B$,
here's my attempt: 
a)
\begin{align*} \frac{32.6}{\sin A} =& \frac{51.9}{\sin 91^\circ} \\ \\ \sin A =& \frac{32.6 \sin 91^\circ}{51.9} \\
\end{align*}
So given some theta A, we get the vertical length $0.63$ radians. To find A, I used
\begin{align*} \arcsin (0.63) \approx 0.68 \ \text{rad}
\end{align*}
So I figured I just needed to convert this to degree measure, getting me  $$m\measuredangle A \approx 38.96^\circ$$
b) using the fact that the sum of angle measures of a triangle is $180^\circ$, I tried to solve for angle $B$ using 
\begin{align*} 38.96^\circ + 51.9^\circ + m\measuredangle B \approx& \ 180^\circ \\ m\measuredangle B \approx& \ 89.14^\circ
\end{align*}
but I'm getting marked wrong for both answers b) and c). What am I doing wrong?
 [/INST] Side length $c \approx 51.9$ is ok.
The triangle side lengths are close to the right triangle of sides $ 30,40,50$  in proportion.
$$ \angle A ( \approx 38.96^{\circ})+ \angle B^{\circ} + 91 ^{\circ} = 180^{\circ}\rightarrow \angle B\approx 50.04 ^{\circ} $$
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