Adeoye moves a distance of \(4.0 km\) from a point, A on a bearing of \(N 30^{\circ} E\) to a point \(B\) and then a distance of \(3.0 km\) on a bearing of \(S 60^{\circ} E\) to a point, C, Calculate Adeoye's resultant displacement from point, A
A. \(10 km\), \(N 60^{\circ} E\) B. \(5 km , N 67^{\circ} E\) C. \(3 km , S 30^{\circ} E\) D. \(4 km , S 60^{\circ} E\)
Correct Answer: B
Explanation
We obtain from the diagram a right angled triangle, so we use Pythagoras theorem i.e. \begin{aligned} &R^{2}=4^{2}+3^{2}=16+9 \\ &R=\sqrt{25}=5 km \end{aligned} to find the angles \begin{aligned} &\sin \theta=\frac{\text { opp }}{\text { hyp }}=\frac{3}{5} \\ &\theta=\sin ^{-1}\left(\frac{3}{5}\right)=36.870 \end{aligned} so that its bearing is \(36.87^{\circ}+30^{\circ}=66.87^{\circ} \approx 67^{\circ}\) i.e. we finally have \(5 km , N 67^{\circ} E\)
