Answer
The missing quantities for the triangle are $A={{109}^{{}^\circ }},a=60.1,b=33.7$.
Work Step by Step
Using the angle sum property of triangle, we get,
$\begin{align}
& A+B+C={{180}^{{}^\circ }} \\
& A+{{32}^{{}^\circ }}+{{39}^{{}^\circ }}={{180}^{{}^\circ }} \\
& A+{{71}^{{}^\circ }}={{180}^{{}^\circ }} \\
& A={{109}^{{}^\circ }}
\end{align}$
Using the law of sines, we get,
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
For a,
$\begin{align}
& \frac{\sin {{109}^{{}^\circ }}}{a}=\frac{\sin {{39}^{{}^\circ }}}{40} \\
& a=\frac{40\times \sin {{109}^{{}^\circ }}}{\sin {{39}^{{}^\circ }}} \\
& a=60.09 \\
& a\approx 60.1
\end{align}$
For b,
$\begin{align}
& \frac{\sin {{32}^{{}^\circ }}}{b}=\frac{\sin {{39}^{{}^\circ }}}{40} \\
& b=\frac{40\times \sin {{32}^{{}^\circ }}}{\sin {{39}^{{}^\circ }}} \\
& b=33.68 \\
& b\approx 33.7
\end{align}$
Thus, for the given triangle, $A={{109}^{{}^\circ }},a=60.1,b=33.7$.
