Answer
(a) $F_x = -8.10~N$
$F_y = 3.00~N$
(b) The magnitude of the resultant force is 8.64 N.
Work Step by Step
(a) $F_x = F_{1x}+F_{2x}$
$F_x = (9.00~N)~cos(120^{\circ}) + (6.00~N)~cos(233.1^{\circ})$
$F_x = -8.10~N$
$F_y = F_{1y}+F_{2y}$
$F_y = (9.00~N)~sin(120^{\circ}) + (6.00~N)~sin(233.1^{\circ})$
$F_y = 3.00~N$
(b) $F = \sqrt{(-8.10~N)^2+(3.00~N)^2}$
$F = 8.64~N$
The magnitude of the resultant force is 8.64 N.
