Answer
There is one triangle that is possible with the given parts.
$A = 96.80^{\circ}, B = 37.77^{\circ},$ and $C = 45.43^{\circ}$
$a = 5.818~ft, b = 3.589~ft,$ and $c = 4.174~ft$
Work Step by Step
We can use the law of sines to find the angle $B$:
$\frac{a}{sin~A} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~A}{a}$
$sin~B = \frac{(3.589~ft)~sin~(96.80^{\circ})}{5.818~ft}$
$B = arcsin(0.6125)$
$B = 37.77^{\circ}$
We can find angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-96.80^{\circ}-37.77^{\circ}$
$C = 45.43^{\circ}$
We can find the length of side $c$:
$\frac{a}{sin~A} = \frac{c}{sin~C}$
$c = \frac{a~sin~C}{sin~A}$
$c = \frac{(5.818~ft)~sin~(45.43^{\circ})}{sin~96.80^{\circ}}$
$c = 4.174~ft$
Note that we can also find another angle for B.
$B = 180-37.77^{\circ} = 142.23^{\circ}$
However, we can not form a triangle with this angle B and angle A since these two angles sum to more than $180^{\circ}$
