# Given a non-right triangle with vertex C = 91 &#x2218;<!-- ∘ --> </msup> , <

Given a non-right triangle with vertex , 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 $c$

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

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

but I'm getting marked wrong for both answers b) and c). What am I doing wrong?
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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.
$\mathrm{\angle }A\left(\approx {38.96}^{\circ }\right)+\mathrm{\angle }{B}^{\circ }+{91}^{\circ }={180}^{\circ }\to \mathrm{\angle }B\approx {50.04}^{\circ }$
###### Not exactly what you’re looking for?
xonycutieoxl1
On the last answer you are using 51.9, which is the length of side c, instead of the correct value of 91 degrees. As for calculating the angle A, I am guessing that it is marked wrong because of an insufficient explanation. You indeed show that $sin\left(A\right)=0.63$, but this does not always imply that $A=38.96$ degrees.