Answer
$F=980lb$
$\phi=19.4^{\circ}$
Work Step by Step
We are asked to determine the magnitude of the resultant force, $F_R$, and its direction, measured counterclockwise
from the positive x axis.
We can apply the law of cosines to find the magnitude of $F_R$.
$c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$
$c=\sqrt{800^2+500^2-2*800*500*\cos(95^{\circ})}$
$c=980lb$
$F=980lb$
Using this result, we can apply the law of sines to find the angle between the top rope and $F_R$, ($\theta$)
$\sin\theta / 500lb = \sin 95^{\circ} / 980lb$
Solving for $\theta$, we obtain:
$\theta=30.6^{\circ}$
We will need to subtract this answer from the given angle in figure, (50^{\circ}), to get the angle to $F_R$ as measured from the x-axis.
$\phi=50^{\circ}-30.6^{\circ}=19.4^{\circ}$
