The net force on the cart has to be calculated by adding the two forces that are acting on the cart. As force is a vector quantity, we need to perform vector addition.
The first force of 200 N pulls the cart towards the south.
The second force of 150 N, pulls the cart in a direction 25 degrees west of south. The component of the force towards the south is 150*cos 25 = 135.94 N. The component towards the west is 150*sin 25 = 63.39 N.
The total force towards the south is 335.94 N and the force towards the west is 63.39 N.
This gives the magnitude of the net force as sqrt(335.94^2 + 63.39^2) = 341.86 N acting in the direction arc tan(63.39/335.94) = 10.68 degrees west of south.
The net force of 341.86 N acts in a direction 10.68 degrees west of south.
To sole the problem we
let F1 = 200N, South
F2 = 150N, 25º West of South
R the Net Force
R² = Fx² + Fy²
where the x-axis along the East-West and the y-axis along the Meridian line
Assuming that component force towards the N & E as positive and component force towards S & W as negative, then
Fx1 = 0
Fy1 = -200N
Fx2 = -150sin25º
Fx2 = -63.39N
Fy2 = -150cos25º
Fy2 = -135.95N
Fx = Fx1 + Fx2
Fx = -63.39 N
Fy = Fy1 + Fy2
Fy = -200 + (-135.95)
Fy = -335.95 N
R² = (-63.38)² + (-335.95)²
R = 341.87N
The direction (θ) of the net force can be computed by
tan θ = |Fx/Fy|
tan θ = |-63.39/-335.95|
Note (-) Fx means the direction is West and (-) Fy means the direction is South, then
θ = Arctan|-63.38/-335.95|
θ = 10º30'17.5" South-West
Therefore the net force is 341.87N at 10º30'17.5" South-West