#### Answer

The angles of the triangle are as follows:
$A = 42.0^{\circ}, B = 35.9^{\circ},$ and $C = 102.1^{\circ}$
The lengths of the sides are as follows:
$a = 42.9~m, b = 37.6~m,$ and $c = 62.7~m$

#### Work Step by Step

We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{42.9^2+62.7^2-37.6^2}{(2)(42.9)(62.7)})$
$B = arccos(0.81)$
$B = 35.9^{\circ}$
We can use the law of cosines to find $C$:
$c^2 = a^2+b^2-2ab~cos~C$
$2ab~cos~C = a^2+b^2-c^2$
$cos~C = \frac{a^2+b^2-c^2}{2ab}$
$C = arccos(\frac{a^2+b^2-c^2}{2ab})$
$C = arccos(\frac{42.9^2+37.6^2-62.7^2}{(2)(42.9)(37.6)})$
$C = arccos(-0.21)$
$C = 102.1^{\circ}$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-35.9^{\circ}-102.1^{\circ}$
$A = 42.0^{\circ}$