Answer
The angles of the triangle are as follows:
$A = 47.7^{\circ}, B = 74.0^{\circ},$ and $C = 58.3^{\circ}$
The lengths of the sides are as follows:
$a = 94.6~yd$, $b = 123~yd$, and $c = 109~yd$
Work Step by Step
$a = 94.6~yd$
$b = 123~yd$
$c = 109~yd$
We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{94.6^2+109^2-123^2}{(2)( 94.6)( 109)})$
$B = arccos(0.27645)$
$B = 74.0^{\circ}$
We can use the law of sines to find the angle $A$:
$\frac{b}{sin~B} = \frac{a}{sin~A}$
$sin~A = \frac{a~sin~B}{b}$
$sin~A = \frac{(94.6)~sin~(74.0^{\circ})}{123}$
$sin~A = 0.7393$
$A = arcsin(0.7393)$
$A = 47.7^{\circ}$
We can find angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-47.7^{\circ}-74.0^{\circ}$
$C = 58.3^{\circ}$
