Answer
The missing values are $B=3{}^\circ,C=15{}^\circ \text{ and }a\approx 59$.
Work Step by Step
For any triangle,
The law of sines states that:
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\operatorname{sinC}}$
The law of cosines states that:
$\begin{align}
& {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2\cdot b\cdot c\cdot \cos A \\
& {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2\cdot a\cdot c\cdot \cos B \\
& {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2\cdot a\cdot b\cdot \cos C \\
\end{align}$
Use the law of cosines to obtain:
$\begin{align}
& {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2\cdot b\cdot c\cdot \cos A \\
& {{a}^{2}}={{11.2}^{2}}+{{48.2}^{2}}-2\left( 11.2 \right)\left( 48.2 \right)\cos 162{}^\circ \\
& {{a}^{2}}\approx 347.5 \\
& a\approx \sqrt{347.5}
\end{align}$
This gives $a\approx 59$.
Using the law of sines we get,
$\begin{align}
& \frac{a}{\sin A}=\frac{b}{\sin B} \\
& \frac{59}{\sin 162{}^\circ }=\frac{11.2}{\sin B} \\
& 59\cdot \sin B=11.2\left( \sin 162{}^\circ \right) \\
& \sin B=\frac{11.2\left( \sin 162{}^\circ \right)}{59}
\end{align}$
We will simplify it further to obtain the value of angle B as follows:
$\begin{align}
& \sin B=0.05 \\
& B={{\sin }^{-1}}\left( 0.05 \right) \\
& B\approx 3{}^\circ \\
\end{align}$
Using the angle sum property we will obtain the measure of angle C.
$\begin{align}
& A+B+C=180{}^\circ \\
& 162{}^\circ +3{}^\circ +C=180{}^\circ \\
& C=180{}^\circ -165{}^\circ \\
& C=15{}^\circ
\end{align}$
So, $B=3{}^\circ,C=15{}^\circ \text{ and }a\approx 59.0$.
Hence, $B=3{}^\circ,C=15{}^\circ \text{ and }a\approx 59$.
