Answer
The angles of the triangle are as follows:
$A = 33.8^{\circ}, B = 38.9^{\circ},$ and $C = 107.3^{\circ}$
The lengths of the sides are as follows:
$a = 189~yd, b = 214~yd,$ and $c = 325~yd$
Work Step by Step
We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{189^2+325^2-214^2}{(2)(189)(325)})$
$B = arccos(0.778)$
$B = 38.9^{\circ}$
We can use the law of cosines to find $C$:
$c^2 = a^2+b^2-2ab~cos~C$
$2ab~cos~C = a^2+b^2-c^2$
$cos~C = \frac{a^2+b^2-c^2}{2ab}$
$C = arccos(\frac{a^2+b^2-c^2}{2ab})$
$C = arccos(\frac{189^2+214^2-325^2}{(2)(189)(214)})$
$C = arccos(-0.298)$
$C = 107.3^{\circ}$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-38.9^{\circ}-107.3^{\circ}$
$A = 33.8^{\circ}$
