Answer
There is one triangle and the solution is ${{C}_{1}}\left( \text{or }C \right)\approx 52{}^\circ,\ B\approx 65{}^\circ,\ \text{ and }\ b\approx 10.2$.
Work Step by Step
We will use the law of sines to find C:
$\begin{align}
& \frac{a}{\sin A}=\frac{c}{\sin C} \\
& \frac{10}{\sin 63{}^\circ }=\frac{8.9}{\sin C} \\
& \sin C=\frac{8.9\sin 63{}^\circ }{10} \\
& =0.7930
\end{align}$
There are two angles between 0° and 180° for which $\sin C=0.7930$ is possible:
${{C}_{1}}=52{}^\circ $
Since we know that the sine function is positive in the second quadrant, the second angle is
$\begin{align}
& {{C}_{2}}=180{}^\circ -52{}^\circ \\
& =128{}^\circ
\end{align}$
Here, ${{C}_{2}}$ is impossible, as $63{}^\circ +128{}^\circ =191{}^\circ $.
Now, we will find $B$ using ${{C}_{1}}$ and $A=63{}^\circ $.
Using the angle sum property, we get
$\begin{align}
& B=180{}^\circ -{{C}_{1}}-A \\
& =180{}^\circ -52{}^\circ -63{}^\circ \\
& =65{}^\circ
\end{align}$
Using the law of sines, we will find side b.
$\begin{align}
& \frac{b}{\sin B}=\frac{a}{\sin A} \\
& \frac{b}{\sin 65{}^\circ }=\frac{10}{\sin 63{}^\circ } \\
& b=\frac{10\sin 65{}^\circ }{\sin 63{}^\circ } \\
& \approx 10.2
\end{align}$
