Split up each force into horizontal and vertical components.
• 300 N at N30°E :
(300 N) (cos(30°) i + sin(30°) j)
• 400 N at N60°E :
(400 N) (cos(60°) i + sin(60°) j)
• 500 N at N80°E :
(500 N) (cos(80°) i + sin(80°) j)
The resultant force is the sum of these forces,
∑ F = (300 cos(30°) + 400 cos(60°) + 500 cos(80°)) i
… … … + (300 sin(30°) + 400 sin(60°) + 500 sin(80°)) j N
∑ F ≈ (546.632 i + 988.814 j) N
so ∑ F has a magnitude of approximately 1129.85 N and points in the direction of approximately N61.0655°E.
