Answer
The required values are $C=90{}^\circ,a\approx 8.0$, and $b\approx 8.9$.
Work Step by Step
At first we will find C.
Properties of a triangle:
Sum of three angles is $A+B+C=180{}^\circ $
$\begin{align}
& A+B+C=180{}^\circ \\
& 42{}^\circ +48{}^\circ +C=180{}^\circ \\
& C=180{}^\circ -90{}^\circ \\
& C=90{}^\circ
\end{align}$
Now, to find the remaining sides we will use the ratio $\frac{c}{\sin C}$ or $\frac{12}{\sin 90{}^\circ }$,
Now, we will use the law of Sines to find a.
$\begin{align}
& \frac{a}{\sin A}=\frac{c}{\sin C} \\
& \frac{a}{\sin 42{}^\circ }=\frac{12}{\sin 90{}^\circ } \\
& a=\frac{12\sin 42{}^\circ }{\sin 90{}^\circ } \\
& a=8.0
\end{align}$
. Again use the law of sines to find b
$\begin{align}
& \frac{b}{\sin B}=\frac{c}{\sin C} \\
& \frac{b}{\sin 48{}^\circ }=\frac{12}{\sin 90{}^\circ } \\
& b=\frac{12\sin 48{}^\circ }{\sin 90{}^\circ } \\
& b=8.9
\end{align}$
The solution is $C=90{}^\circ,a\approx 8.0$, and $b\approx 8.9$.
