Answer
The magnitude of the equilibrant is 2638.7 lb and it is directed at an angle of $167.2^{\circ}$ from the force of 1480 lb
Work Step by Step
Let $a = 1240~lb$
Let $b = 1480~lb$
Let angle $C = 180^{\circ}-28.2^{\circ} = 151.8^{\circ}$
We can use the law of cosines to find $c$, the magnitude of the equilibrant:
$c = \sqrt{a^2+b^2-2ab~cos~\theta}$
$c = \sqrt{(1240~lb)^2+(1480~lb)^2-(2)(1240~lb)(1480~lb)~cos~151.8^{\circ}}$
$c = \sqrt{6962736.19~lb^2}$
$c = 2638.7~lb$
The magnitude of the equilibrium is 2638.7 lb
We can use the law of sines to find the angle $A$ between the $b$ and $c$:
$\frac{c}{sin~C} = \frac{a}{sin~A}$
$sin~A = \frac{a~sin~C}{c}$
$A = arcsin(\frac{a~sin~C}{c})$
$A = arcsin(\frac{1240~lb~sin~151.8^{\circ}}{2638.7~lb})$
$A = arcsin(0.222)$
$A = 12.8^{\circ}$
The angle between the equilibrium and the force 1480 lb is $180^{\circ}-A = 180^{\circ}-12.8^{\circ} = 167.2^{\circ}$
The magnitude of the equilibrium is 2638.7 lb and it is directed at an angle of $167.2^{\circ}$ from the force of 1480 lb