Answer
$F_{R} = 413N$
$\theta = 24.2°$
Work Step by Step
Projecting both forces on the x axis:
$F_{R}x = 400 \times \sin 30° + 250 \times \cos 45° = 376.78N$
Projecting on the y axis:
$F_{R}y = 400 \times \cos 30° - 250 \times \sin 45° = 169.63N$
The magnitude of the force is obtained:
$F_{R} = \sqrt {376.78^{2} + 169.63^{2}} = 413N$
The direction is obtained:
$\theta = \tan^{-1} (\frac{F_{R}y}{F_{R}x}) = 24.2°$
