#### Answer

The angles of the triangle are as follows:
$A = 44.9^{\circ}, B = 106.8^{\circ},$ and $C = 28.3^{\circ}$
The lengths of the sides are as follows:
$a = 4.21~in, b = 5.71~in,$ and $c = 2.83~in$

#### Work Step by Step

We can use the law of cosines to find $c$:
$c^2 = a^2+b^2-2ab~cos~C$
$c = \sqrt{a^2+b^2-2ab~cos~C}$
$c = \sqrt{(4.21~in)^2+(5.71~in)^2-(2)(4.21~in)(5.71~in)~cos~28.3^{\circ}}$
$c = \sqrt{7.996~in^2}$
$c = 2.83~in$
We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{(4.21)^2+(2.83)^2-(5.71)^2}{(2)(4.21)(2.83)})$
$B = arccos(-0.288)$
$B = 106.8^{\circ}$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-106.8^{\circ}-28.3^{\circ}$
$A = 44.9^{\circ}$