JAMB Mathematics 2018 Paper

Concept:Use the right-angled triangle formed by the northward and eastward movements to find the angle, then express it as a bearing.Explanation:The man starts at R, which is 300 m due north of P, and walks due east to Q.So triangle $PRQ$ is right-angled at R, with $PR = 300$ m and $PQ = 600$ m as the hypotenuse.Let $\theta = \angle RPQ$, the angle between $PR$ and $PQ$ at point P.From the right triangle,$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{PR}{PQ} = \frac{300}{600} = 0.5$Thus,$\theta = \cos^{-1}(0.5) = 60^\circ$Since $PR$ points due north, the bearing of Q from P is $060^\circ$.The bearing of P from Q is the opposite direction, so add $180^\circ$:$060^\circ + 180^\circ = 240^\circ$Answer:D. $240^{\circ}$