Answer
The third side of the triangle is $4.4$ units long and the other two angles are $45{}^\circ $ and $96{}^\circ $.
Work Step by Step
We use the cosine law to find out the side of the triangle:
$\begin{align}
& {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A \\
& {{a}^{2}}={{5}^{2}}+{{7}^{2}}-2.5.7\cos 39{}^\circ \\
& =19.5998
\end{align}$
The value of the third side is as given below:
$\begin{align}
& a=\sqrt{19.5998} \\
& \approx 4.4
\end{align}$
Then, to calculate angles, apply the law of sines
$\begin{align}
& \frac{a}{\sin A}=\frac{b}{\sin B} \\
& \frac{4.4}{\sin 39{}^\circ }=\frac{5}{\sin B} \\
& \sin B=\frac{5\sin 39{}^\circ }{4.4} \\
& B\approx 45{}^\circ
\end{align}$
Now, the third angle would be $180-39-45=96{}^\circ $
Thus, the third side of the tringle is 4.4 units long and the other two angles are $45{}^\circ $ and $96{}^\circ $.
