waecmaths question:
In the diagram PR is a diameter of the circle centre O. RS is a tangent at R and $\angle QPR={{58}^{\circ }}$, find $\angle QRS$
Option A:
112o
Option B:
116o
Option C:
122o
Option D:
148o
waecmaths solution:
$\begin{align} & \angle PQR={{90}^{\circ }}\text{ }\!\!\{\!\!\text{ angle in a semicircle }\!\!\}\!\!\text{ } \\ & \angle QRP={{180}^{\circ }}-(\angle QPR+\angle PQR)\text{ }\!\!\{\!\!\text{ sum of }\angle s\text{ in a }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & \angle QRP={{180}^{\circ }}-({{90}^{\circ }}+{{58}^{\circ }})={{32}^{\circ }} \\ & \angle PRS={{90}^{\circ }}\text{ }\!\!\{\!\!\text{ tangent to a circle }\!\!\}\!\!\text{ } \\ & \angle QRS=\angle PRS+\angle QRP \\ & \angle QRS={{90}^{\circ }}+{{32}^{\circ }}={{122}^{\circ }} \\\end{align}$
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