Answer
The angles of the triangle are as follows:
$A = 81.8^{\circ}, B = 37.4^{\circ},$ and $C = 60.8^{\circ}$
The lengths of the sides are as follows:
$a = 9.3~cm, b = 5.7~cm,$ and $c = 8.2~cm$
Work Step by Step
We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{9.3^2+8.2^2-5.7^2}{(2)(9.3)(8.2)})$
$B = arccos(0.7949)$
$B = 37.4^{\circ}$
We can use the law of cosines to find $C$:
$c^2 = a^2+b^2-2ab~cos~C$
$2ab~cos~C = a^2+b^2-c^2$
$cos~C = \frac{a^2+b^2-c^2}{2ab}$
$C = arccos(\frac{a^2+b^2-c^2}{2ab})$
$C = arccos(\frac{9.3^2+5.7^2-8.2^2}{(2)(9.3)(5.7)})$
$C = arccos(0.488)$
$C = 60.8^{\circ}$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-37.4^{\circ}-60.8^{\circ}$
$A = 81.8^{\circ}$
