#### Answer

The angles are $A=29.7^{\circ}, B=49.1^{\circ}$, and $C=101.2^{\circ}$
The angles are $A=29.7^{\circ}, B=130.9^{\circ}$, and $C=19.4^{\circ}$

#### Work Step by Step

We can use the law of sines to find the angle $B$:
$\frac{a}{sin~A} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~A}{a}$
$sin~B = \frac{(41.5~ft)~sin~(29.7^{\circ})}{27.2~ft}$
$B = arcsin(0.7559)$
$B = 49.1^{\circ}$
We can find angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-29.7^{\circ}-49.1^{\circ}$
$C = 101.2^{\circ}$
The angles are $A=29.7^{\circ}, B=49.1^{\circ}$, and $C=101.2^{\circ}$
Note that we can also construct another triangle.
$B = 180-49.1^{\circ} = 130.9^{\circ}$
We can find angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-29.7^{\circ}-130.9^{\circ}$
$C = 19.4^{\circ}$
The angles are $A=29.7^{\circ}, B=130.9^{\circ}$, and $C=19.4^{\circ}$