Answer
There is one triangle and the solution is ${{B}_{1}}\left( \text{or }B \right)\approx 29{}^\circ,\ C\approx 111{}^\circ,\ \text{ and }\ c\approx 29.0$.
Work Step by Step
We will use the law of sines to find B:
$\begin{align}
& \frac{a}{\sin A}=\frac{b}{\sin B} \\
& \frac{20}{\sin 40{}^\circ }=\frac{15}{\sin B} \\
& \sin B=\frac{15\sin 40{}^\circ }{20} \\
& \approx 0.4821
\end{align}$
There are two angles between 0° and 180° for which $\sin B=0.4821$ is possible:
${{B}_{1}}\approx 29{}^\circ $
As we know, the sine function is positive in the second quadrant, so the second angle is
$\begin{align}
& {{B}_{2}}\approx 180{}^\circ -29{}^\circ \\
& \approx 151{}^\circ
\end{align}$
Here, ${{B}_{2}}$ is impossible, as $40{}^\circ +151{}^\circ =191{}^\circ $.
Now, find $C$ using ${{B}_{1}}$ and $A=40{}^\circ $.
By the angle sum property,
$\begin{align}
& C=180{}^\circ -{{B}_{1}}-A \\
& \approx 180{}^\circ -29{}^\circ -40{}^\circ \\
& \approx 111{}^\circ
\end{align}$
Using the law of sines we will find side c:
$\begin{align}
& \frac{c}{\sin C}=\frac{a}{\sin A} \\
& \frac{c}{\sin 111{}^\circ }=\frac{20}{\sin 40{}^\circ } \\
& c=\frac{20\sin 111{}^\circ }{\sin 40{}^\circ } \\
& \approx 29.0
\end{align}$
