Answer
The magnitude of the equilibrant is 1759.2 lb and it is directed at an angle of $167.0^{\circ}$ from the force of 960 lb
Work Step by Step
Let $a = 840~lb$
Let $b = 960~lb$
Let angle $C = 180^{\circ}-24.5^{\circ} = 155.5^{\circ}$
We can use the law of cosines to find $c$, the magnitude of the equilibrant:
$c = \sqrt{a^2+b^2-2ab~cos~\theta}$
$c = \sqrt{(840~lb)^2+(960~lb)^2-(2)(840~lb)(960~lb)~cos~155.5^{\circ}}$
$c = \sqrt{3094785.5~lb^2}$
$c = 1759.2~lb$
The magnitude of the equilibrant is 1759.2 lb
We can use the law of sines to find the angle $A$ between the $b$ and $c$:
$\frac{c}{sin~C} = \frac{a}{sin~A}$
$sin~A = \frac{a~sin~C}{c}$
$A = arcsin(\frac{a~sin~C}{c})$
$A = arcsin(\frac{840~lb~sin~151.8^{\circ}}{1759.2~lb})$
$A = arcsin(0.225638)$
$A = 13.0^{\circ}$
The angle between the equilibrant and the force of 960 lb is $180^{\circ}-A = 180^{\circ}-13.0^{\circ} = 167.0^{\circ}$
The magnitude of the equilibrant is 1759.2 lb and it is directed at an angle of $167.0^{\circ}$ from the force of 960 lb
