Answer
There are two possible triangles with the given parts.
One triangle has the following values:
$A = 53.23^{\circ}, B = 39.68^{\circ},$ and $C = 87.09^{\circ}$
$a = 29.81~m, b = 23.76~m,$ and $c = 37.16~m$
Another triangle has the following values:
$A = 126.77^{\circ}, B = 39.68^{\circ},$ and $C = 13.55^{\circ}$
$a = 29.81~m, b = 23.76~m,$ and $c = 8.72~m$
Work Step by Step
We can use the law of sines to find the angle $A$:
$\frac{b}{sin~B} = \frac{a}{sin~A}$
$sin~A = \frac{a~sin~B}{b}$
$sin~A = \frac{(29.81~m)~sin~(39.68^{\circ})}{23.76~m}$
$sin~A = 0.801$
$A = arcsin(0.801)$
$A = 53.23^{\circ}$
We can find angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-53.23^{\circ}-39.68^{\circ}$
$C = 87.09^{\circ}$
We can find the length of side $c$:
$\frac{b}{sin~B} = \frac{c}{sin~C}$
$c = \frac{b~sin~C}{sin~B}$
$c = \frac{(23.76~m)~sin~(87.09^{\circ})}{sin~39.68^{\circ}}$
$c = 37.16~m$
Note that we can also find another angle for A.
$A = 180-53.23^{\circ} = 126.77^{\circ}$
We can find angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-126.77^{\circ}-39.68^{\circ}$
$C = 13.55^{\circ}$
We can find the length of side $c$:
$\frac{b}{sin~B} = \frac{c}{sin~C}$
$c = \frac{b~sin~C}{sin~B}$
$c = \frac{(23.76~m)~sin~(13.55^{\circ})}{sin~39.68^{\circ}}$
$c = 8.72~m$
