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