#### Answer

The angles of the triangle are as follows:
$A = 60^{\circ}, B = 40.9^{\circ},$ and $C = 79.1^{\circ}$
The lengths of the sides are as follows:
$a = 39.7~m, b = 30~m,$ and $c = 45~m$

#### Work Step by Step

$A = 60^{\circ}$
$b = 30~m$
$c = 45~m$
We can use the law of cosines to find $a$:
$a^2 = b^2+c^2-2bc~cos~A$
$a = \sqrt{b^2+c^2-2bc~cos~A}$
$a = \sqrt{(30~m)^2+(45~m)^2-(2)(30~m)(45~m)~cos~60^{\circ}}$
$a = \sqrt{1575~m^2}$
$a = 39.7~m$
We can use the law of sines to find angle $B$:
$\frac{a}{sin~A} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~A}{a}$
$B = arcsin(\frac{b~sin~A}{a})$
$B = arcsin(\frac{30~sin~60^{\circ}}{39.7})$
$B = arcsin(0.6544)$
$B = 40.9^{\circ}$
We can find angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-60^{\circ}-40.9^{\circ}$
$C = 79.1^{\circ}$