Answer
There are two triangles and the solutions are ${{B}_{1}}\approx 27{}^\circ,\ {{C}_{1}}\approx 133{}^\circ,\ \text{ and }\ {{c}_{1}}\approx 64.2$, and
${{B}_{2}}\approx 153{}^\circ,\ {{C}_{2}}\approx 7{}^\circ,\ \text{ and }\ {{c}_{2}}\approx 10.7$.
Work Step by Step
We will use the law of sines to find B:
$\begin{align}
& \frac{a}{\sin A}=\frac{b}{\sin B} \\
& \frac{30}{\sin 20{}^\circ }=\frac{40}{\sin B} \\
& \sin B=\frac{40\sin 20{}^\circ }{30} \\
& \approx 0.4560
\end{align}$
There are two angles between 0° and 180° for which $\sin B\approx 0.4560$ is possible:
${{B}_{1}}\approx 27{}^\circ $
As we know that the sine function is positive in the second quadrant, the second angle is
$\begin{align}
& {{B}_{2}}\approx 180{}^\circ -27{}^\circ \\
& \approx 153{}^\circ
\end{align}$
So, there are two triangles.
For ${{B}_{1}}\approx 27{}^\circ $,
$\begin{align}
& {{C}_{1}}=180{}^\circ -{{B}_{1}}-A \\
& =180{}^\circ -27{}^\circ -20{}^\circ \\
& =133{}^\circ
\end{align}$
For ${{c}_{1}}$,
$\begin{align}
& \frac{{{c}_{1}}}{\sin {{C}_{1}}}=\frac{a}{\sin A} \\
& \frac{{{c}_{1}}}{\sin 133{}^\circ }=\frac{30}{\sin 20{}^\circ } \\
& {{c}_{1}}=\frac{30\sin 133{}^\circ }{\sin 20{}^\circ } \\
& \approx 64.2
\end{align}$
For ${{B}_{2}}\approx 153{}^\circ $,
$\begin{align}
& {{C}_{2}}=180{}^\circ -{{B}_{2}}-A \\
& =180{}^\circ -153{}^\circ -20{}^\circ \\
& =7{}^\circ
\end{align}$
For ${{c}_{2}}$,
$\begin{align}
& \frac{{{c}_{2}}}{\sin {{C}_{2}}}=\frac{a}{\sin A} \\
& \frac{{{c}_{2}}}{\sin 7{}^\circ }=\frac{30}{\sin 20{}^\circ } \\
& {{c}_{2}}=\frac{30\sin 7{}^\circ }{\sin 20{}^\circ } \\
& \approx 10.7
\end{align}$
In case of the first triangle, the solution is
${{B}_{1}}\approx 27{}^\circ,\ {{C}_{1}}\approx 133{}^\circ,\ \text{ and }\ {{c}_{1}}\approx 64.2$
In the other triangle, the solution is
${{B}_{2}}\approx 153{}^\circ,\ {{C}_{2}}\approx 7{}^\circ,\ \text{ and }\ {{c}_{2}}\approx 10.7$
