Question 27

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waecmaths question: 

In the diagram, P, Q and R  are three points in a plane such that the bearing of R from Q is 110o and the bearing of Q from P is 050o. Find $\angle PQR$

Option A: 

60o 

Option B: 

70o

Option C: 

120o

Option D: 

160o

waecmaths solution: 

$\begin{align}  & \angle CQP=\angle APQ={{50}^{\circ }}\text{    }\!\!\{\!\!\text{ alternate angles }\!\!\}\!\!\text{ } \\ & \angle CQR={{180}^{\circ }}-\angle BQR\text{    }\!\!\{\!\!\text{ sum of }\angle s\text{ on a straight line }\!\!\}\!\!\text{ } \\ & \angle CQR={{180}^{\circ }}-{{110}^{\circ }}={{70}^{\circ }} \\ & \angle PQR=\angle CQP+\angle CQR={{50}^{\circ }}+{{70}^{\circ }}={{120}^{\circ }} \\\end{align}$

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