Answer
$\begin{aligned} & F=1302 \mathrm{~N} \\ & \theta=84.5^{\circ} \mathrm{} \\ & x=1.36 \mathrm{~m} \text { (to the right) }\end{aligned}$
Work Step by Step
$
\begin{aligned}
& \pm F_{R x}=\Sigma F_x ; \quad F_{R x}=450 \cos 60^{\circ}-700 \sin 30^{\circ}=-125 \mathrm{~N}\\&=125 \mathrm{~N} \leftarrow \\
& +\uparrow F_{R y}=\Sigma F_y ; \quad F_{R y}=-450 \sin 60^{\circ}-700 \cos 30^{\circ}-300=-1296 \mathrm{~N}=1296 \mathrm{~N} \\
& F=\sqrt{(-125)^2+(-1296)^2}=1302 \mathrm{~N} \\
& \theta=\tan ^{-1}\left(\frac{1296}{125}\right)=84.5^{\circ} P \\
& ↺+M_{R B}=\Sigma M_B ; \quad 1296(x)\\&=-450 \sin 60^{\circ}(4)+700 \cos 30^{\circ}(3)+1500 \\
& x=136 \mathrm{~m} \text { (to the right) }
\end{aligned}
$
