Answer
$F=10.8kN$
$\phi=3.16^{\circ}$
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 $F_R$.
$c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$
$c=\sqrt{8^2+6^2-2*8*6*\cos(100^{\circ})}$
$c=10.8$
$F=10.8kN$
Using this result, we can apply the law of sines to find the angle between the top rope and $F_R$, ($\theta$)
$\sin\theta / 6 = \sin 100^{\circ} / 10.8$
Solving for $\theta$, we obtain:
$\theta=33.16^{\circ}$
We will need to subtract $30^{\circ}$ from this answer to get the angle to $F_R$ as measured from the x-axis.
$\phi=33.16^{\circ}-30^{\circ}=3.16^{\circ}$
