#### Answer

The angles of the triangle are as follows:
$A = 28.6^{\circ}, B = 53.3^{\circ},$ and $C = 98.1^{\circ}$
The lengths of the sides are as follows:
$a = 28~ft, b = 47~ft,$ and $c = 58~ft$

#### Work Step by Step

We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{28^2+58^2-47^2}{(2)(28)(58)})$
$B = arccos(0.597)$
$B = 53.3^{\circ}$
We can use the law of cosines to find $C$:
$c^2 = a^2+b^2-2ab~cos~C$
$2ab~cos~C = a^2+b^2-c^2$
$cos~C = \frac{a^2+b^2-c^2}{2ab}$
$C = arccos(\frac{a^2+b^2-c^2}{2ab})$
$C = arccos(\frac{28^2+47^2-58^2}{(2)(28)(47)})$
$C = arccos(-0.14)$
$C = 98.1^{\circ}$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-53.3^{\circ}-98.1^{\circ}$
$A = 28.6^{\circ}$