Answer
The angles of the triangle are as follows:
$A = 50.1^{\circ}, B = 85.1^{\circ},$ and $C = 44.8^{\circ}$
The lengths of the sides are as follows:
$a = 324~m, b = 421~m,$ and $c = 298~m$
Work Step by Step
We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{324^2+298^2-421^2}{(2)(324)(298)})$
$B = arccos(0.0856)$
$B = 85.1^{\circ}$
We can use the law of cosines to find $C$:
$c^2 = a^2+b^2-2ab~cos~C$
$2ab~cos~C = a^2+b^2-c^2$
$cos~C = \frac{a^2+b^2-c^2}{2ab}$
$C = arccos(\frac{a^2+b^2-c^2}{2ab})$
$C = arccos(\frac{324^2+421^2-298^2}{(2)(324)(421)})$
$C = arccos(0.709)$
$C = 44.8^{\circ}$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-85.1^{\circ}-44.8^{\circ}$
$A = 50.1^{\circ}$
