Answer
$F_R=393lb$
$\theta=353^{\circ}$
Work Step by Step
We are asked to determine the magnitude of the force, $F_R$, and its direction, measured counterclockwise
from the positive x axis.
We can treat the forces like a triangle and use the law of cosines:
$c^2=a^2+b^2−2a∗b∗\cos C$
$c=\sqrt{a^2+b^2−2a∗b∗\cos C}$
$c=\sqrt{(250lb)^2+(375lb)^2−2∗250lb∗375lb∗\cos(75^{\circ})}$
$c=393lb$
$F_R=393lb$
Now we can use the law of sines to find the angle between $F$ and the x-axis.
$\sin\alpha/250lb=\sin75^{\circ}/393lb$
$\alpha=\arcsin(\sin75^{\circ}/393lb∗250lb)$
$\alpha=37.89^{\circ}$
This is measured from $F_2$. We need to subtract 45 from 365 and then add our angle to get the measurement from the x-axis
$\theta=360^{\circ}-45^{\circ}+37.89^{\circ}=353^{\circ}$
