WAEC 2014 · Paper 2 · Q8✱

(a)

1. Take $x$ to the right and $y$ upwards.
2. The 10 N force is $30^\circ$ from the $y$-axis, the $8\sqrt2$ N force is $45^\circ$ from the $y$-axis, and the 20 N force is $60^\circ$ below the negative $x$-axis.
3. Vertical components: $Y = 10\cos30^\circ + 8\sqrt2\cos45^\circ - 20\sin60^\circ$.
4. That is $Y = 5\sqrt3 + 8 - 10\sqrt3$
   $= 8 - 5\sqrt3$
   $\approx -0.660\text{ N}$.
5. Horizontal components: $X = 8\sqrt2\sin45^\circ - 10\sin30^\circ - 20\cos60^\circ$.
6. That is $X = 8 - 5 - 10 = -7\text{ N}$.
7. Both components are negative, so the resultant points down and to the left.
8. $\tan\theta = \dfrac{0.660}{7}$
   $\approx 0.0943$, so $\theta \approx 5.39^\circ$ below the negative $x$-axis.
9. As a bearing, this is $270^\circ - 5.39^\circ \approx 264.6^\circ$, about $265^\circ$ (S$84.6^\circ$W).