Answer
The angles of the triangle are as follows:
$A = 73.8^{\circ}, B = 53.1^{\circ},$ and $C = 53.1^{\circ}$
The sides of the triangle are as follows:
$a = 12, b = 10,$ and $c = 10$
Work Step by Step
Let $a = 12,$ let $b = 10,$ and let $c=10$
We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{12^2+10^2-10^2}{(2)(12)(10)})$
$B = arccos(0.6)$
$B = 53.1^{\circ}$
By symmetry, angle $C = 53.1^{\circ}$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-53.1^{\circ}-53.1^{\circ}$
$A = 73.8^{\circ}$
