Answer
There is one triangle and the solution is ${{B}_{1}}\left( \text{or }B \right)\approx 12{}^\circ,\ C\approx 6{}^\circ,\ \text{ and }\ c\approx 2.1$.
Work Step by Step
We will use the law of sines to find B:
$\begin{align}
& \frac{a}{\sin A}=\frac{b}{\sin B} \\
& \frac{6.1}{\sin 162{}^\circ }=\frac{4}{\sin B} \\
& \sin B=\frac{4\sin 162{}^\circ }{6.1} \\
& \approx 0.2026
\end{align}$
There are two angles between 0° and 180° for which $\sin B\approx 0.2026$ is possible:
${{B}_{1}}\approx 12{}^\circ $
As we know, the sine function is positive in the second quadrant, so the second angle is
$\begin{align}
& {{B}_{2}}\approx 180{}^\circ -12{}^\circ \\
& \approx 168{}^\circ
\end{align}$
Here, ${{B}_{2}}$ is impossible, as $162{}^\circ +168{}^\circ =330{}^\circ $.
Now, find $C$ using ${{B}_{1}}$ and $A=162{}^\circ $.
Using the angle sum property of triangles:
$\begin{align}
& C=180{}^\circ -{{B}_{1}}-A \\
& =180{}^\circ -12{}^\circ -162{}^\circ \\
& =6{}^\circ
\end{align}$
Using the law of sines to find side c:
$\begin{align}
& \frac{c}{\sin C}=\frac{a}{\sin A} \\
& \frac{c}{\sin 6{}^\circ }=\frac{6.1}{\sin 162{}^\circ } \\
& c=\frac{6.1\sin 6{}^\circ }{\sin 162{}^\circ } \\
& \approx 2.1
\end{align}$
