Split F₁ into its horizontal and vertical components:
F₁ = F₁ cos(θ) i + F₁ sin(θ) j
(boldface = vector; regular font = magnitude)
By Newton's second law, if the object is in equilibrium, then
• the net horizontal force on the block is
∑ F = F₁ cos(θ) - F₃ = 0 → F₁ cos(θ) = 70 N
• the net vertical force is
∑ F = F₁ sin(θ) + F₂ - W = 0 → F₁ sin(θ) = 65 N
Recall that cos²(θ) + sin²(θ) = 1 for any θ, so we have
(F₁ cos(θ))² + (F₁ sin(θ))² = (70 N)² + (65 N)²
F₁² = 9125 N²
F₁ = √(9125 N²) ≈ 95.5 N