Question 41

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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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