Answer
The solution is $C=107{}^\circ,b=24.8,c=36.1$
Work Step by Step
The provided angles and sides of the triangle are: $A=32{}^\circ,B=41{}^\circ,a=20$
As the sum of all the angles of the triangle is $180{}^\circ $ therefore, the third angle of the provided triangle is $107{}^\circ $, which is calculated as below:
$\begin{align}
& 180{}^\circ -\left( A+B \right)=180-\left( 32{}^\circ +41{}^\circ \right) \\
& =107{}^\circ
\end{align}$
Using the law of sines we will find the side b of the triangle. That is,
$\begin{align}
& \frac{\sin A}{a}=\frac{\sin B}{b} \\
& \frac{\sin 32{}^\circ }{20}=\frac{\sin 41{}^\circ }{b}
\end{align}$
This implies that,
$\begin{align}
& b=\sin 41{}^\circ \times \frac{20}{\sin 32{}^\circ } \\
& =24.8
\end{align}$
Now, by using the law of sines we will find the third side c of the triangle. That is,
$\begin{align}
& \frac{\sin A}{a}=\frac{\sin C}{c} \\
& \frac{\sin 32{}^\circ }{20}=\frac{\sin 107{}^\circ }{c}
\end{align}$
Therefore,
$\begin{align}
& c=\sin 107{}^\circ \times \frac{20}{\sin 32{}^\circ } \\
& =36.1
\end{align}$
The required angle and sides of the provided triangle are $C=107{}^\circ,b=24.8,\ \text{ and }\ c=36.1$
