Question 28

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Jambmaths question: 

 In the diagram above, O is the centre of the circle, POM is a diameter and $\angle MNQ={{42}^{\text{o}}}$. Calculate $\angle QMP$  

Option A: 

42o

Option B: 

48o

Option C: 

132o

Option D: 

138o

Jamb Maths Solution: 

$\begin{align}  & \angle MNQ={{42}^{\text{o}}} \\ & \angle MPQ=\angle MNQ={{42}^{o}}\text{     }\!\!\{\!\!\text{ }\angle \text{s in the segment }\!\!\}\!\!\text{ } \\ & \angle MQP={{90}^{o}}\text{                    }\!\!\{\!\!\text{ }\angle \text{s in a semicircle }\!\!\}\!\!\text{ } \\ & \angle MPQ+\angle MQP+QMP={{180}^{o}}\text{   }\!\!\{\!\!\text{ Sum of }\angle \text{s in a }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & {{42}^{o}}+{{90}^{o}}+\angle QMP={{180}^{o}} \\ & \angle QMP={{48}^{o}} \\\end{align}$

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