Answer
There is one triangle and the solution is ${{C}_{1}}\left( \text{or }C \right)\approx 37{}^\circ,\ B\approx 7{}^\circ,\ \text{ and }\ b\approx 10.1$.
Work Step by Step
We will use the law of sines to find C:
$\begin{align}
& \frac{a}{\sin A}=\frac{c}{\sin C} \\
& \frac{57.5}{\sin 136{}^\circ }=\frac{49.8}{\sin C} \\
& \sin C=\frac{49.8\sin 136{}^\circ }{57.5} \\
& \approx 0.6016
\end{align}$
There are two angles between 0° and 180° for which $\sin C=0.6016$ is possible:
${{C}_{1}}\approx 37{}^\circ $
Since we know that the sine function is positive in the second quadrant, the second angle is
$\begin{align}
& {{C}_{2}}\approx 180{}^\circ -37{}^\circ \\
& \approx 143{}^\circ
\end{align}$
Here, ${{C}_{2}}$ is impossible, as $136{}^\circ +143{}^\circ =279{}^\circ $.
Now, find $B$ using ${{C}_{1}}$ and $A=136{}^\circ $.
Using the angle sum property, we get
$\begin{align}
& B=180{}^\circ -{{C}_{1}}-A \\
& \approx 180{}^\circ -37{}^\circ -136{}^\circ \\
& \approx 7{}^\circ
\end{align}$
Using the law of sines, we will find side b:
$\begin{align}
& \frac{b}{\sin B}=\frac{a}{\sin A} \\
& \frac{b}{\sin 7{}^\circ }=\frac{57.5}{\sin 136{}^\circ } \\
& b=\frac{57.5\sin 7{}^\circ }{\sin 136{}^\circ } \\
& \approx 10.1
\end{align}$
