Answer
There is one possible triangle with the given parts.
$A = 60.91^{\circ}, B = 30.39^{\circ},$ and $C = 88.70^{\circ}$
$a = 98.25~m, b = 56.87~m,$ and $c = 112.4~m$
Work Step by Step
We can use the law of sines to find the angle $B$:
$\frac{c}{sin~C} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~C}{c}$
$sin~B = \frac{(56.87~m)~sin~(88.70^{\circ})}{112.4~m}$
$B = arcsin(0.5058)$
$B = 30.39^{\circ}$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-30.39^{\circ}-88.70^{\circ}$
$A = 60.91^{\circ}$
We can find the length of side $a$:
$\frac{a}{sin~A} = \frac{c}{sin~C}$
$a = \frac{c~sin~A}{sin~C}$
$a = \frac{(112.4~m)~sin~(60.91^{\circ})}{sin~88.70^{\circ}}$
$a = 98.25~m$
Note that we can also find another angle for B.
$B = 180-30.39^{\circ} = 149.61^{\circ}$
However, we can not form a triangle with this angle B and angle C since these two angles sum to more than $180^{\circ}$
