Answer
The solution is $A=28{}^\circ,B=42{}^\circ,\ \text{ and }\ C=110{}^\circ $.
Work Step by Step
The provided sides of the triangle are given below:
$a=5.0,b=7.2,c=10.1$
Using the law of cosines, we will find the side b of the triangle. That is,
${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\,\cos \,C$
Therefore,
$\begin{align}
& \cos \,C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab} \\
& \cos \,C=\frac{{{5}^{2}}+{{7.2}^{2}}-{{10.1}^{2}}}{2\times 5\times 7.2} \\
& \cos \,C=-0.3496
\end{align}$
Thus, $C=110{}^\circ $.
Now, by using the law of sines we will find the angle A of the triangle. That is,
$\begin{align}
& \frac{\sin \,A}{a}=\frac{\sin \,C}{c} \\
& \frac{\sin \,A}{5}=\frac{\sin \,110{}^\circ }{10.1}
\end{align}$
This implies that
$\begin{align}
& \sin \,A=5\times \frac{\sin \,110{}^\circ }{10.1} \\
& =0.4652
\end{align}$
This implies that $A=28{}^\circ $.
Now the third angle will be found as below:
$\begin{align}
& B=180{}^\circ -A-C \\
& =180{}^\circ -28{}^\circ -110{}^\circ \\
& B=42{}^\circ
\end{align}$
Therefore, the solution is $A=28{}^\circ,B=42{}^\circ,\ \text{ and }\ C=110{}^\circ $.
