#### Answer

There is one possible triangle with the given parts.
$A = 42.5^{\circ}, B = 20.6^{\circ},$ and $C = 116.9^{\circ}$
$a = 15.6~ft, b = 8.14~ft,$ and $c = 20.6~ft$

#### Work Step by Step

We can use the law of sines to find the angle $B$:
$\frac{a}{sin~A} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~A}{a}$
$sin~B = \frac{(8.14~ft)~sin~(42.5^{\circ})}{15.6~ft}$
$B = arcsin(0.3525)$
$B = 20.6^{\circ}$
We can find angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-42.5^{\circ}-20.6^{\circ}$
$C = 116.9^{\circ}$
We can find the length of side $c$:
$\frac{a}{sin~A} = \frac{c}{sin~C}$
$c = \frac{a~sin~C}{sin~A}$
$c = \frac{(15.6~ft)~sin~(116.9^{\circ})}{sin~42.5^{\circ}}$
$c = 20.6~ft$
Note that we can also find another angle for B.
$B = 180-20.6^{\circ} = 159.4^{\circ}$
However, we can not form a triangle with this angle B and angle A since these two angles sum to more than $180^{\circ}$