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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