Answer
The solution is $A=26{}^\circ,C=44{}^\circ \ \text{ and }\ b=21.6$
Work Step by Step
The given angles and sides of the triangle are: $B=110{}^\circ,a=10,c=16$
Using the law of cosines, we will find the side b of the triangle. That is,
$\begin{align}
& {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\cos B \\
& {{b}^{2}}={{10}^{2}}+{{16}^{2}}-2\left( 10 \right)\left( 16 \right)\cos 110{}^\circ \\
& {{b}^{2}}=465.4464 \\
& b=21.6
\end{align}$
Now, by using the law of sines we will find the angle A of the triangle. That is,
$\begin{align}
& \frac{\sin A}{a}=\frac{\operatorname{sinB}}{b} \\
& \frac{\sin A}{10}=\frac{\sin 110{}^\circ }{21.6}
\end{align}$
This implies that,
$\begin{align}
& \sin A=10\times \frac{\sin 110{}^\circ }{21.6} \\
& =0.4350
\end{align}$
Therefore, $A=26{}^\circ $
Now, the third angle will be found as below:
$\begin{align}
& C=180{}^\circ -A-B \\
& =180{}^\circ -26{}^\circ -110{}^\circ \\
& C=44{}^\circ
\end{align}$
Therefore, the solution is: $A=26{}^\circ,C=44{}^\circ \ \text{ and }\ b=21.6$
