Answer
There are two triangles and the solutions are ${{B}_{1}}\approx 54{}^\circ,\ {{C}_{1}}\approx 89{}^\circ,\ \text{ and }\ {{c}_{1}}\approx 19.9$, and
${{B}_{2}}\approx 126{}^\circ,\ {{C}_{2}}\approx 17{}^\circ,\ \text{ and }\ {{c}_{2}}\approx 5.8$.
Work Step by Step
We will use the law of sines to find B:
$\begin{align}
& \frac{a}{\sin A}=\frac{b}{\sin B} \\
& \frac{12}{\sin 37{}^\circ }=\frac{16.1}{\sin B} \\
& \sin B=\frac{16.1\sin 37{}^\circ }{12} \\
& \sin B\approx 0.8074
\end{align}$
There are two angles between 0° and 180° for which $\sin B\approx 0.8074$ is possible:
${{B}_{1}}=54{}^\circ $
Since the sine function is positive in the second quadrant, the second angle is
$\begin{align}
& {{B}_{2}}=180{}^\circ -54{}^\circ \\
& =126{}^\circ
\end{align}$
So, there are two triangles.
In case of the first triangle,
$\begin{align}
& {{C}_{1}}=180{}^\circ -{{B}_{1}}-A \\
& =180{}^\circ -54{}^\circ -37{}^\circ \\
& =89{}^\circ
\end{align}$
In case of the second triangle,
$\begin{align}
& {{C}_{2}}=180{}^\circ -{{B}_{2}}-A \\
& =180{}^\circ -126{}^\circ -37{}^\circ \\
& =17{}^\circ
\end{align}$
Using the law of sines we will find the sides ${{c}_{1}}$ and ${{c}_{2}}$.
For ${{c}_{1}}$,
$\begin{align}
& \frac{{{c}_{1}}}{\sin {{C}_{1}}}=\frac{a}{\sin A} \\
& \frac{{{c}_{1}}}{\sin 89{}^\circ }=\frac{12}{\sin 37{}^\circ } \\
& {{c}_{1}}=\frac{12\sin 89{}^\circ }{\sin 37{}^\circ } \\
& {{c}_{1}}\approx 19.9
\end{align}$
For ${{c}_{2}}$,
$\begin{align}
& \frac{{{c}_{2}}}{\sin {{C}_{2}}}=\frac{a}{\sin A} \\
& \frac{{{c}_{2}}}{\sin 17{}^\circ }=\frac{12}{\sin 37{}^\circ } \\
& {{c}_{2}}=\frac{12\sin 17{}^\circ }{\sin 37{}^\circ } \\
& {{c}_{2}}\approx 5.8
\end{align}$
In first triangle, the solution is
${{B}_{1}}\approx 54{}^\circ,\ {{C}_{1}}\approx 89{}^\circ,\ \text{ and }\ {{c}_{1}}\approx 19.9$
In the other triangle, the solution is
${{B}_{2}}\approx 126{}^\circ,\ {{C}_{2}}\approx 17{}^\circ,\ \text{ and }\ {{c}_{2}}\approx 5.8$
