#### Answer

$F_{R} = 983N$
$\theta = 21.8°$

#### Work Step by Step

Projecting both forces on the x axis:
$F_{R}x = 400 \times \cos(30°) + 800 \times \sin (45°) = 912.10N $
Projecting both forces on the y axis:
$F_{R}y = 400 \times \sin(30°) + 800 \times \cos (45°) = 365.69N $
The magnitude is obtained:
$F_{R} = \sqrt {912.1^{2} + 365.69^{2}} = 983N$
The direction about the x axis is:
$\theta = \tan^{-1} (\frac{F_{R}y}{F_{R}x}) = 21.84° \approx 21.8°$