Answer
The magnitude is about 528N. It's 120 degrees clockwise from force A
Work Step by Step
1) Finding x-component of the force
$(\rightarrow +) \sum \overrightarrow{F}_{x} =0 $
$395N-\cos(75^{\circ})485N + \overrightarrow{F}_{x}=0$
$\overrightarrow{F}_{x}= -395N+\cos(75^{\circ})485N$
$\overrightarrow{F}_{x}\approx-252N$
2)Finding the y-component
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$\sin(75^{\circ})475N + \overrightarrow{F}_{y}=0$
$\overrightarrow{F}_{y}=-\sin(75^{\circ})475N$
$\overrightarrow{F}_{y}\approx-459N$
3) Finding the magnitude of \overrightarrow{F}
$\Vert\overrightarrow{F}\Vert=\sqrt {F^2_x+F^2_y}$
$\Vert\overrightarrow{F}\Vert\approx\sqrt {(-252N)^2+(-459N)^2}$
$\Vert\overrightarrow{F}\Vert\approx528N$
3) Finding the direction
$\arctan{F_y/F_x}\approx60^{\circ}$
$180^{\circ}+60^{\circ}=120^{\circ}$
120 degrees clockwise from force A
