Write each force in i,j component form and sum them up to get the resultant force.
• (34.3 N, 62.0°) :
F₁ = (34.3 N) (cos(62.0°) i + sin(62.0°) j) ≈ (16.1 i + 30.3 j) N
• (56.2 N, 125°) :
F₂ = (56.2 N) (cos(125°) i + sin(125°) j) ≈ (-32.2 i + 46.0 j) N
• (45.4 N, 251°) :
F₃ = (45.4 N) (cos(251°) i + sin(251°) j) ≈ (-14.8 i - 42.9 j) N
Then the resultant is approximately
∑ F = F₁ + F₂ + F₃ ≈ (-30.9 i + 33.4 j) N
Its magnitude is
|∑ F| = √((-30.9)² + 33.4²) N ≈ 45.5 N
Let θ be the resultant's direction, so that
tan(θ) ≈ 33.4/(-30.9)
Taking the inverse tangent of both sides gives
θ ≈ arctan(33.4/(-30.9)) ≈ -47.2°
but this would mean the resultant points out into the fourth quadrant. However, we found that the i-component is negative and the j-component is positive, so the resultant should actually point out into the second quadrant.
To get the correct direction for the resultant, we use the fact that for any angle θ,
tan(θ) = tan(θ + 180°)
Then
θ ≈ 180° + arctan(33.4/(-30.9)) ≈ 133°
