#### Answer

The angle opposite the side of length 45 feet is $26.4^{\circ}$
The angle opposite the side of length 60 feet is $36.3^{\circ}$

#### Work Step by Step

Let $a = 45~ft$, let $b = 60~ft$, and let $c = 90~ft$.
We can use the law of cosines to find $A$, which is the angle which subtends the side $a$:
$a^2 = b^2+c^2-2bc~cos~A$
$2bc~cos~A = b^2+c^2-a^2$
$cos~A = \frac{b^2+c^2-a^2}{2bc}$
$A = arccos(\frac{b^2+c^2-a^2}{2bc})$
$A = arccos(\frac{60^2+90^2-45^2}{(2)(60)(90)})$
$A = arccos(0.89583)$
$A = 26.4^{\circ}$
The angle opposite the side of length 45 feet is $26.4^{\circ}$
We can use the law of cosines to find $B$, which is the angle which subtends the side $b$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{45^2+90^2-60^2}{(2)(45)(90)})$
$B = arccos(0.8056)$
$B = 36.3^{\circ}$
The angle opposite the side of length 60 feet is $36.3^{\circ}$