Answer
$F_R = 389$
$Θ_R' = 42.71^0$
Work Step by Step
* Summing Forces In X direction
$F_{Rx} =300Sin70^0 -\frac{12}{13}(180) $
$F_{Rx} = 115.8N$
* Summing Forces In X direction
$F_{Ry} =300SCos70^0 +200+\frac{5}{13}(180) $
$F_{Ry} = 371.8N$
Magnitude:
$F_R = \sqrt {(115.8)^2+(371.8)^2}$
$F_R = 389N$
Directrion $Θ_R$
$Θ_R = tan^{-1}[\frac{371.8}{115.8}]$
$Θ_R = 72.71^0$
Direction wrt positive x' axis $Θ_R'$
$Θ_R' = 72.71^0 - 30^0$
$Θ_R' = 42.71^0$
