Sorry I’ve just been having trouble with this question If you don’t mind could you include the solving steps

68.8 N 13.8°N of W

Explanation:

F₁ is 50 N 30°N of W.  The terminal angle is 150°.

F₂ is 25 N 20°S of W.  The terminal angle is -160°.

Graphically, you can add the vectors using head-to-tail method.  Move F₂ so that the tail of the vector is at the head of F₁.  The resultant vector will be from the tail of F₁ to the head of F₂.

Algebraically, find the x and y components of each vector.

F₁ₓ = 50 N cos(150°) = -43.3 N

F₁ᵧ = 50 N sin(150°) = 25 N

F₂ₓ = 25 N cos(-160°) = -23.5 N

F₂ᵧ = 25 N sin(-160°) = -8.6 N

The x and y components of the resultant vector are the sums:

Fₓ = -43.3 N + -23.5 N = -66.8 N

Fᵧ = 25 N + -8.6 N = 16.4 N

The magnitude of the resultant force is:

F = √(Fₓ² + Fᵧ²)

F = √((-66.8 N)² + (16.4 N)²)

F = 68.8 N

The direction of the resultant force is:

θ = tan⁻¹(Fᵧ / Fₓ)

θ = tan⁻¹(16.4 N / -66.8 N)

θ = 166.2°

θ = 13.8°N of W

probably not if net force is zero