Answer
$A=90{}^\circ,b\approx 7.9$ and $c\approx 1.4$.
Work Step by Step
First we will find the value of A
Properties of a triangle:
The sum of three angles is $A+B+C=180{}^\circ $
$\begin{align}
& A+B+C=180{}^\circ \\
& A+80{}^\circ +10{}^\circ =180{}^\circ \\
& A=180{}^\circ -90{}^\circ \\
& A=90{}^\circ
\end{align}$
Now, we will find the remaining sides using the ratio:
$\frac{a}{\sin A}$,or $\frac{8}{\sin 90{}^\circ }$,
Now, we will use the law of sines to find b:
$\begin{align}
& \frac{b}{\sin B}=\frac{a}{\sin A} \\
& \frac{b}{\sin 80{}^\circ }=\frac{8}{\sin 90{}^\circ } \\
& b=\frac{8\sin 80{}^\circ }{\sin 90{}^\circ } \\
& b\approx 7.9
\end{align}$
Using the law of sines again, we will find c.
$\begin{align}
& \frac{c}{\sin C}=\frac{a}{\sin A} \\
& \frac{c}{\sin 10{}^\circ }=\frac{8}{\sin 90{}^\circ } \\
& c=\frac{8\sin 10{}^\circ }{\sin 90{}^\circ } \\
& c\approx 1.4
\end{align}$
