#### Answer

The angles of the triangle are as follows:
$A = 29.9^{\circ}, B = 56.3^{\circ},$ and $C = 93.8^{\circ}$
The lengths of the sides are as follows:
$a = 3.0~ft, b = 5.0~ft,$ and $c = 6.0~ft$

#### Work Step by Step

We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{3.0^2+6.0^2-5.0^2}{(2)(3.0)(6.0)})$
$B = arccos(\frac{5}{9})$
$B = 56.3^{\circ}$
We can use the law of cosines to find $C$:
$c^2 = a^2+b^2-2ab~cos~C$
$2ab~cos~C = a^2+b^2-c^2$
$cos~C = \frac{a^2+b^2-c^2}{2ab}$
$C = arccos(\frac{a^2+b^2-c^2}{2ab})$
$C = arccos(\frac{3.0^2+5.0^2-6.0^2}{(2)(3.0)(5.0)})$
$C = arccos(-\frac{1}{15})$
$C = 93.8^{\circ}$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-56.3^{\circ}-93.8^{\circ}$
$A = 29.9^{\circ}$