Answer
There are two triangles possible and the solution are ${{B}_{1}}\approx 56{}^\circ,{{C}_{1}}\approx 112{}^\circ,{{c}_{1}}\approx 31.2$ and
${{B}_{2}}\approx 124{}^\circ,{{C}_{2}}\approx 44{}^\circ,{{c}_{2}}\approx 23.4$.
Work Step by Step
Using the ratio $\frac{a}{\sin A}$,or $\frac{7}{\sin 12{}^\circ }$,
We will use the law of sines to find B.
$\begin{align}
& \frac{a}{\sin A}=\frac{b}{\sin B} \\
& \frac{7}{\sin 12{}^\circ }=\frac{28}{\sin B} \\
& \sin B=\frac{28\sin 12{}^\circ }{7} \\
& \sin B\approx 0.8316
\end{align}$
There are two angles possible for this value:
$\begin{align}
& {{B}_{1}}=56{}^\circ \\
& {{B}_{2}}=180{}^\circ -56{}^\circ \\
& =124{}^\circ
\end{align}$
This implies there are two triangles.
In the first triangle,
$\begin{align}
& A=12{}^\circ \\
& {{C}_{1}}=180{}^\circ -{{B}_{1}}-A \\
& =180{}^\circ -56{}^\circ -12{}^\circ \\
& =112{}^\circ
\end{align}$
In the second triangle,
$\begin{align}
& A=12{}^\circ \\
& {{C}_{2}}=180{}^\circ -{{B}_{2}}-A \\
& =180{}^\circ -124{}^\circ -12{}^\circ \\
& =44{}^\circ
\end{align}$
Using the law of sines we will find the side ${{c}_{1}}$ and ${{c}_{2}}$.
For ${{c}_{1}}$ consider,
$\begin{align}
& \frac{{{c}_{1}}}{\sin {{C}_{1}}}=\frac{a}{\sin A} \\
& \frac{{{c}_{1}}}{\sin 112{}^\circ }=\frac{7}{\sin 12{}^\circ } \\
& {{c}_{1}}=\frac{7\sin 112{}^\circ }{\sin 12{}^\circ } \\
& {{c}_{1}}\approx 31.2
\end{align}$
For ${{c}_{2}}$ consider,
$\begin{align}
& \frac{{{c}_{2}}}{\sin {{C}_{2}}}=\frac{a}{\sin A} \\
& \frac{{{c}_{2}}}{\sin 44{}^\circ }=\frac{7}{\sin 12{}^\circ } \\
& {{c}_{2}}=\frac{7\sin 44{}^\circ }{\sin 12{}^\circ } \\
& {{c}_{2}}\approx 23.4
\end{align}$
In case of the first triangle,
${{B}_{1}}\approx 56{}^\circ,{{C}_{1}}\approx 112{}^\circ,{{c}_{1}}\approx 31.2$
In case of the second triangle,
${{B}_{2}}\approx 124{}^\circ,{{C}_{2}}\approx 44{}^\circ,{{c}_{2}}\approx 23.4$
Hence, there are two triangles possible and the solutions are ${{B}_{1}}\approx 56{}^\circ,{{C}_{1}}\approx 112{}^\circ,{{c}_{1}}\approx 31.2$ and
${{B}_{2}}\approx 124{}^\circ,{{C}_{2}}\approx 44{}^\circ,{{c}_{2}}\approx 23.4$.
