Answer
The angles of the triangle are as follows:
$A = 64.6^{\circ}, B = 74.8^{\circ},$ and $C = 40.6^{\circ}$
The lengths of the sides are as follows:
$a = 8.92~in, b = 9.53~in,$ and $c = 6.43~in$
Work Step by Step
We can use the law of cosines to find $b$:
$b^2 = a^2+c^2-2ac~cos~B$
$b = \sqrt{a^2+c^2-2ac~cos~B}$
$b = \sqrt{(8.92~in)^2+(6.43~in)^2-(2)(8.92~in)(6.43~in)~cos~74.8^{\circ}}$
$b = \sqrt{90.835~in^2}$
$b = 9.53~in$
We can use the law of cosines to find $C$:
$c^2 = a^2+b^2-2ab~cos~C$
$2ab~cos~C = a^2+b^2-c^2$
$cos~C = \frac{a^2+b^2-c^2}{2ab}$
$C = arccos(\frac{a^2+b^2-c^2}{2ab})$
$C = arccos(\frac{8.92^2+9.53^2-6.43^2}{(2)(8.92)(9.53)})$
$C = arccos(0.759)$
$C = 40.6^{\circ}$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-74.8^{\circ}-40.6^{\circ}$
$A = 64.6^{\circ}$
