Correct Answer: Option A

Explanation

Given a force F, the horizontal component = \(F\cos \theta\)

R = \(-50\cos 30 + 80\cos 45\)

= \(-50(\frac{\sqrt{3}}{2}) + 80(\frac{\sqrt{2}}{2})\)

= \( -25\sqrt{3} + 40\sqrt{2} = -43.30 + 56.67 \)

= \(13.37N \approxeq 13N\)

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