(a) The magnitude of force vector B at the given resultant force is 120 N.
(b) The direction of force vector A is 142.46°
Resultant of the forces
The resultant forces is calculated as follows;
Acosθ + Bcosθ = 0 --- (1)
Asinθ + Bsinθ = 150 ---- (2)
Magnitude of B
Acosθ + Bcos60 = 0
-60 + Bsin60 = 0 (Acosθ = x-component of A = Ax)
Bcos60 = 60
B = 60/cos(60)
B = 120 N
Direction of A
Asinθ + Bsinθ = 150
Ay + Bsinθ = 150
Ay + 120sin60 = 150
Ay = 150 - 120sin60
Ay = 46.1 N
[tex]tan \theta = \frac{A_y}{A_x} \\\\\theta = tan^{-1} (\frac{A_y}{A_x})\\\\\theta = tan^{-1} (\frac{46.1}{-60})\\\\\theta = -37.53^0\\\\\theta = -37.53^0 + 180 = 142.46^0[/tex]
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