Question 17

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waecmaths question: 

In the diagram PR is a diameter of the circle centre O. RS is a tangent at R and $\angle QPR={{58}^{\circ }}$, find $\angle QRS$

Option A: 

112o

Option B: 

116o

Option C: 

122o

Option D: 

148o

waecmaths solution: 

$\begin{align}  & \angle PQR={{90}^{\circ }}\text{    }\!\!\{\!\!\text{ angle in a semicircle }\!\!\}\!\!\text{ } \\ & \angle QRP={{180}^{\circ }}-(\angle QPR+\angle PQR)\text{    }\!\!\{\!\!\text{ sum of }\angle s\text{ in a }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & \angle QRP={{180}^{\circ }}-({{90}^{\circ }}+{{58}^{\circ }})={{32}^{\circ }} \\ & \angle PRS={{90}^{\circ }}\text{   }\!\!\{\!\!\text{ tangent to a circle }\!\!\}\!\!\text{ } \\ & \angle QRS=\angle PRS+\angle QRP \\ & \angle QRS={{90}^{\circ }}+{{32}^{\circ }}={{122}^{\circ }} \\\end{align}$

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