Answer
There is one triangle and the solution is ${{B}_{1}}\left( \text{or }B \right)\approx 31{}^\circ,\ C\approx 99{}^\circ,\ \text{ and }\ c\approx 38.7$.
Work Step by Step
At first we will use the law of sines to find B:
$\begin{align}
& \frac{a}{\sin A}=\frac{b}{\sin B} \\
& \frac{30}{\sin 50{}^\circ }=\frac{20}{\sin B} \\
& \sin B=\frac{20\sin 50{}^\circ }{30} \\
& \approx 0.5107
\end{align}$
There are two angles between 0° and 180° for which $\sin B=0.5107$ is possible:
${{B}_{1}}\approx 31{}^\circ $
Since we know that the sine function is positive in the second quadrant, the second angle is
$\begin{align}
& {{B}_{2}}\approx 180{}^\circ -31{}^\circ \\
& \approx 149{}^\circ
\end{align}$
Here, ${{B}_{2}}$ is impossible, as $50{}^\circ +149{}^\circ =199{}^\circ $.
Now, we will find $C$ using ${{B}_{1}}$ and $A=50{}^\circ $.
Using the angle sum property, we will find C:
$\begin{align}
& C=180{}^\circ -{{B}_{1}}-A \\
& \approx 180{}^\circ -31{}^\circ -50{}^\circ \\
& \approx 99{}^\circ
\end{align}$
Using the law of sines, we get side c:
$\begin{align}
& \frac{c}{\sin C}=\frac{a}{\sin A} \\
& \frac{c}{\sin 99{}^\circ }=\frac{30}{\sin 50{}^\circ } \\
& c=\frac{30\sin 99{}^\circ }{\sin 50{}^\circ } \\
& \approx 38.7
\end{align}$
