Answer
$F_B = 6300 ~N$
$F_A + F_B = 8400 ~N$
Work Step by Step
The sum of the horizontal forces is 0. Therefore:
$F_{Bx} = F_{Ax}$
$F_B ~sin(32^{\circ}) = 4500 ~sin(48^{\circ})$
$F_B = \frac{4500 ~sin(48^{\circ})}{sin(32^{\circ})}$
$F_B = 6300 ~N$
Next, we find $F_A + F_B$:
$F_A + F_B = F_{Ay} + F_{By}$
$F_A + F_B = 4500 ~cos(48^{\circ}) + 6300 ~cos(32^{\circ})$
$F_A + F_B = 8400 ~N$
