Answer
There is one triangle and the solution is ${{C}_{1}}\left( \text{or }C \right)\approx 55{}^\circ,\ B\approx 13{}^\circ,\ \text{ and }\ b\approx 10.2$.
Work Step by Step
Using the law of sines to find C:
$\begin{align}
& \frac{a}{\sin A}=\frac{c}{\sin C} \\
& \frac{42.1}{\sin 112{}^\circ }=\frac{37}{\sin C} \\
& \sin C=\frac{37\sin 112{}^\circ }{42.1} \\
& \approx 0.8149
\end{align}$
There are two angles between 0° and 180° for which $\sin C=0.8149$ is possible:
${{C}_{1}}\approx 55{}^\circ $
As we know that the sine function is positive in the second quadrant, the second angle is
$\begin{align}
& {{C}_{2}}\approx 180{}^\circ -55{}^\circ \\
& \approx 125{}^\circ
\end{align}$
Here, ${{C}_{2}}$ is impossible, as $112{}^\circ +125{}^\circ =237{}^\circ $.
Now, we will find $B$ using ${{C}_{1}}$ and $A=112{}^\circ $.
Using the angle sum property of triangles:
$\begin{align}
& B=180{}^\circ -{{C}_{1}}-A \\
& \approx 180{}^\circ -55{}^\circ -112{}^\circ \\
& \approx 13{}^\circ
\end{align}$
Using the law of sines we will find side b:
$\begin{align}
& \frac{b}{\sin B}=\frac{a}{\sin A} \\
& \frac{b}{\sin 13{}^\circ }=\frac{42.1}{\sin 112{}^\circ } \\
& b=\frac{42.1\sin 13{}^\circ }{\sin 112{}^\circ } \\
& \approx 10.2
\end{align}$
