waecmaths question:
In the diagram, O is the centre of the circle $\angle QPS={{100}^{\circ }},\text{ }\angle PSQ={{60}^{\circ }}\text{ and }\angle QSR={{80}^{\circ }}$ Calculate $\angle SQR$
Option A:
20o
Option B:
40o
Option C:
60o
Option D:
80o
waecmaths solution:
$\begin{align} & \angle PRQ=\angle PSQ={{60}^{\circ }}\text{ }\!\!\{\!\!\text{ }\angle s\text{ on the same segment }\!\!\}\!\!\text{ } \\ & \angle ORP=\angle PRQ={{60}^{\circ }} \\ & OQ=OR\text{ }\!\!\{\!\!\text{ Radius of a circle }\!\!\}\!\!\text{ } \\ & \angle OQR=\angle ORP={{60}^{\circ }}\text{ }\!\!\{\!\!\text{ Base }\angle s\text{ of an isso}\text{. }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & \angle SQR=\angle OQR={{60}^{\circ }} \\\end{align}$
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