WAEC Further Mathematics 2013 Paper

Concept:Split each force into its horizontal component using $F\cos\theta$, then add the components with the correct signs.Explanation:Take the rightward direction as positive and the leftward direction as negative.The $50$ N force pushes to the left at $30^\circ$ to the horizontal.Its horizontal component is $-50\cos30^\circ$.Since $\cos30^\circ = \frac{\sqrt3}{2}$, this becomes $-50 \times \frac{\sqrt3}{2} = -25\sqrt3$.The $80$ N force pushes to the right at $45^\circ$ to the horizontal.Its horizontal component is $80\cos45^\circ$.Since $\cos45^\circ = \frac{\sqrt2}{2}$, this becomes $80 \times \frac{\sqrt2}{2} = 40\sqrt2$.The resultant horizontal component is $R = -25\sqrt3 + 40\sqrt2$.Substituting $\sqrt3 \approx 1.732$ and $\sqrt2 \approx 1.414$ gives $R = -43.30 + 56.57 = 13.27$ N.To the nearest whole number, $R \approx 13$ N.Answer:$13$ N, which corresponds to option A.