#### Answer

The angles of the triangle are as follows:
$A = 67.3^{\circ}, B = 53.2^{\circ},$ and $C = 59.5^{\circ}$
The lengths of the sides are as follows:
$a = 43.7~km, b = 37.9~km,$ and $c = 40.8~km$

#### Work Step by Step

We can use the law of cosines to find $a$:
$a^2 = b^2+c^2-2bc~cos~A$
$a = \sqrt{b^2+c^2-2bc~cos~A}$
$a = \sqrt{(37.9~km)^2+(40.8~km)^2-(2)(37.9~km)(40.8~km)~cos~67.3^{\circ}}$
$a = \sqrt{1907.58}$
$a = 43.7~km$
We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{(43.7)^2+(40.8)^2-(37.9)^2}{(2)(43.7)(40.8)})$
$B = arccos(0.5995)$
$B = 53.2^{\circ}$
We can find angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-67.3^{\circ}-53.2^{\circ}$
$C = 59.5^{\circ}$