Given a non-right triangle with vertex $C={91}^{\circ},\text{}\overline{AC}=39.9\text{}cm,\text{}\overline{BC}=32.6\text{}cm$, 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 $c$

$\begin{array}{rl}c=& \text{}\sqrt{{32.6}^{2}+{39.9}^{2}-2(32.6\ast 39.9)\mathrm{cos}({91}^{\circ})}\\ c=& \text{}51.9\text{}cm\end{array}$

but I'm having a hard time finding the angle measures of $A$ and $B$, here's my attempt:

a) $\begin{array}{rl}\frac{32.6}{\mathrm{sin}A}=& \frac{51.9}{\mathrm{sin}{91}^{\circ}}\\ \\ \mathrm{sin}A=& \frac{32.6\mathrm{sin}{91}^{\circ}}{51.9}\end{array}$

So given some theta A, we get the vertical length 0.63 radians. To find A, I used

$\begin{array}{r}\mathrm{arcsin}(0.63)\approx 0.68\text{}\text{rad}\end{array}$

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

$\begin{array}{rl}{38.96}^{\circ}+{51.9}^{\circ}+m\measuredangle B\approx & \text{}{180}^{\circ}\\ m\measuredangle B\approx & \text{}{89.14}^{\circ}\end{array}$

but I'm getting marked wrong for both answers b) and c). What am I doing wrong?

a) $\overline{AB}$

b) $m\measuredangle A$

c) $m\measuredangle B$

I was able to find the length of $\overline{AB}$, which I labeled $c$

$\begin{array}{rl}c=& \text{}\sqrt{{32.6}^{2}+{39.9}^{2}-2(32.6\ast 39.9)\mathrm{cos}({91}^{\circ})}\\ c=& \text{}51.9\text{}cm\end{array}$

but I'm having a hard time finding the angle measures of $A$ and $B$, here's my attempt:

a) $\begin{array}{rl}\frac{32.6}{\mathrm{sin}A}=& \frac{51.9}{\mathrm{sin}{91}^{\circ}}\\ \\ \mathrm{sin}A=& \frac{32.6\mathrm{sin}{91}^{\circ}}{51.9}\end{array}$

So given some theta A, we get the vertical length 0.63 radians. To find A, I used

$\begin{array}{r}\mathrm{arcsin}(0.63)\approx 0.68\text{}\text{rad}\end{array}$

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

$\begin{array}{rl}{38.96}^{\circ}+{51.9}^{\circ}+m\measuredangle B\approx & \text{}{180}^{\circ}\\ m\measuredangle B\approx & \text{}{89.14}^{\circ}\end{array}$

but I'm getting marked wrong for both answers b) and c). What am I doing wrong?