Answer
$B=58^{\circ}20'$
$c=68.4\,km$
$b=58.2\,km$
Work Step by Step
The sum of angles in a triangle is $180^{\circ}$.
$\implies$ Angle $B=179^{\circ}60'-(90^{\circ}00'+31^{\circ}40')=58^{\circ}20'$
$\sin A=\sin 31^{\circ}40'=\sin 31.6667^{\circ}=\frac{\text{opposite side}}{\text{hypotenuse}}=\frac{35.9\,km}{c}$
$\implies c=\frac{35.9\,km}{\sin31.6667^{\circ}}=68.4\,km$
$\tan A=\tan 31.6667^{\circ}=\frac{\text{opposite side}}{\text{adjacent side}}=\frac{35.9\,km}{b}$
$\implies b=\frac{35.9\,km}{\tan 31.6667^{\circ}}=58.2\,km$
