Question 8

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Jambmaths question: 

In the diagram POQ is a diameter of the circle. PQRS. If $\angle PSR={{145}^{o}}$. Find xo           

 

Option A: 

45o

Option B: 

25o

Option C: 

55o   

Option D: 

35o

Jamb Maths Solution: 

$\begin{align}  & \text{consider }\Delta PSQ \\ & \angle PSQ={{90}^{o}}\text{    }\!\!\{\!\!\text{ angle in semicircle }\!\!\}\!\!\text{ } \\ & \angle QSR={{145}^{o}}-{{90}^{o}}={{55}^{o}} \\ & \text{Join point }P\text{ and }Q\text{ together} \\ & \text{Consider }\Delta QRS\text{ and }\Delta RQP\ \text{which has }QR\text{ has its base} \\ & \angle RPQ=\angle QSR={{55}^{o}}\text{    }\!\!\{\!\!\text{ Angle on the same segment and base }\!\!\}\!\!\text{ } \\ & \angle RPO=x={{55}^{o}} \\\end{align}$

Jamb Maths Topic: 
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