Answer
The missing values are: $C=55{}^\circ,b\approx 10.4\text{ and }c\approx 10.5$
Work Step by Step
For any triangle,
The law of sines states:
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\operatorname{sinC}}$
Using the angle sum property, we will obtain:
$\begin{align}
& A+B+C=180{}^\circ \\
& 70{}^\circ +55{}^\circ +C=180{}^\circ \\
& C=180{}^\circ -125{}^\circ \\
& C=55{}^\circ
\end{align}$
Using the law of sines we get,
$\begin{align}
& \frac{a}{\sin A}=\frac{c}{\operatorname{sinC}} \\
& \frac{12}{\sin 70{}^\circ }=\frac{c}{\sin 55{}^\circ } \\
& c\cdot \sin 70{}^\circ =12\cdot \sin 55{}^\circ \\
& c=\frac{12\cdot \sin 55{}^\circ }{\sin 70{}^\circ }
\end{align}$
This gives $c\approx 10.5$ to the nearest tenth.
Also,
$\begin{align}
& \frac{a}{\sin A}=\frac{b}{\sin B} \\
& \frac{12}{\sin 70{}^\circ }=\frac{b}{\sin 55{}^\circ } \\
& b\cdot \sin 70{}^\circ =12\cdot \sin 55{}^\circ \\
& b=\frac{12\cdot \sin 55{}^\circ }{\sin 70{}^\circ }
\end{align}$
This gives $b\approx 10.5$ to the nearest tenth.
So, $C=55{}^\circ,b\approx 10.4\text{ and }c\approx 10.5$.
Hence, $C=55{}^\circ,b\approx 10.4\text{ and }c\approx 10.5$.
