Answer
$a).$ (2) Second Quadrant.
$b).$ 184lbs.
Work Step by Step
a). From the figure, we expect the third force to be in the (2) second quadrant.
b). $F_{1}+F_{2}=[(150+30)x+(0-40)y]lb=(180x-40y)lb$
For the body to be at rest, the third force must be =$ (-180x+40y)lb$
Now, $\theta=tan^{-1}\frac{40}{-180}=167.5^{\circ}$ since it is in the second quadrant.
Net $F=\sqrt (180^{2}+40^{2})=184lb$
