#### Answer

The angles of the triangle are as follows:
$A = 48.8^{\circ}, B = 71.5^{\circ},$ and $C = 59.7^{\circ}$
The lengths of the sides are as follows:
$a = 3.73~mi, b = 4.70~mi,$ and $c = 4.28~mi$

#### Work Step by Step

We can use the law of cosines to find $c$:
$c^2 = a^2+b^2-2ab~cos~C$
$c = \sqrt{a^2+b^2-2ab~cos~C}$
$c = \sqrt{(3.73~mi)^2+(4.70~mi)^2-(2)(3.73~mi)(4.70~mi)~cos~59.7^{\circ}}$
$c = \sqrt{18.313~mi^2}$
$c = 4.28~mi$
We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{(3.73)^2+(4.28)^2-(4.70)^2}{(2)(3.73)(4.28)})$
$B = arccos(0.3176)$
$B = 71.5^{\circ}$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-71.5^{\circ}-59.7^{\circ}$
$A = 48.8^{\circ}$