Question 30

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waecmaths question: 

Q is 32km away from P on a bearing of 042o and R is 25km from P on a bearing of 132o. Calculate the bearing of R from Q

Option A: 

122o

Option B: 

184o

Option C: 

190o

Option D: 

226o

waecmaths solution: 

$\begin{align}  & \tan \theta =\frac{\left| PR \right|}{\left| PQ \right|}=\frac{25}{32} \\ & \theta ={{\tan }^{-1}}(\tfrac{25}{12})={{38}^{\circ }} \\ & \angle PQA=\angle BPQ={{42}^{\circ }} \\ & \angle RQA=\angle PQA-\angle PQR \\ & \angle RQA={{42}^{\circ }}-{{38}^{\circ }}={{4}^{\circ }} \\\end{align}$The bearing of R of Q is 180o­­ + 4o =184o

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