Question 27


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Jambmaths question: 

In the diagram above, PQR is a circle O. If $\angle \mathbf{QRP}$is xo, Find $\angle \mathbf{QRP}$

Option A: 


Option B: 

(90 – x)o

Option C: 

(90 + x)o

Option D: 

(180 – x)o

Jamb Maths Solution: 

$\begin{align}  & \angle PQR={{90}^{\circ }}\text{     }\!\!\{\!\!\text{ Angle subtend in a semicirle }\!\!\}\!\!\text{ } \\ & \angle PQR+\angle RPQ+\angle QRP={{180}^{\circ }}\text{   }\!\!\{\!\!\text{ sum of angles in }\Delta \text{ }\!\!\}\!\!\text{ } \\ & {{90}^{\circ }}+{{x}^{\circ }}+\angle QRP={{180}^{\circ }} \\ & \angle QRP=90-{{x}^{\circ }} \\\end{align}$

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