Answer
$F_R = 839 N$
$Θ_R = 14.8^0$
Work Step by Step
* Summing the force components in X, we have
$F_{Rx} = 700 Sin 30° - 600 Cos 20°$
$F_{Rx} = -213.8 N$
* Summing the force components in X, we have
$F_{Ry} = 700 Cos 30° - 600 Sin 20°$
$F_{Ry} = 811.4 N$
* The magnitude of the resultant force $F_R$ is
$F_R = \sqrt {F_{Rx}^2+F_{Ry}^2}$
$F_R = \sqrt {{(-213.8)}^2+{(811.4)}^2}$
$F_R = 839N$
The directional angle $Θ$ measured counterclockwise from the positive y-axis is
$Θ = tan^{-1}\frac{F_Rx}{F_Ry}$
$Θ = tan^{-1}\frac{231.8}{811.4}$
$Θ = 14.8^0$
