#### Answer

$F_{R} = 1.96 KN$
$\theta = 4.12 °$

#### Work Step by Step

Projecting the three forces on the x axis:
$F_{R}x = 900 + 750 \times \cos 45 ° + 650 \times 4/5 = 1950.33N$
Projecting the three forces on the y axis:
$F_{R}y = 750 \times \sin 45 ° - 650 \times 3/5 = 140.33N$
The magnitude is obtained:
$F_{R} = \sqrt {F_{R}y^{2}+F_{R}x^{2}} = 1955N \approx 1.96KN$
The direction of the force:
$\theta = \tan^{-1} (\frac{F_{R}y}{F_{R}x}) = 4.12°$