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