Answer
given,
F₁ = 15 lb
F₂ = 8 lb
θ₁ = 45°
θ₂ = 25°
Assuming the question's diagram is attached below.
now,
computing the horizontal component of the forces.
F_h = F₁ cos θ₁ - F₂ cos θ₂
F_h = 15 cos 45° - 8 cos 25°
F_h = 3.36 lb
now, vertical component of the forces
F_v = F₁ sin θ₁ + F₂ sin θ₂
F_v = 15 sin 45° + 8 sin 25°
F_v = 13.98 lb
resultant force would be equal to
F = 14.38 lb
the magnitude of resultant force is equal to 14.38 lb
direction of forces
θ = 76.48°
