Question 35

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Jambmaths question: 

In the diagram above, PQR is a straight line and PS is a straight line and PS is a tangent to the circle QRS with $\left| PS \right|=\left| SR \right|$ and $\angle SPR={{40}^{o}}$.Find $\angle PSQ$

Option A: 

40o

Option B: 

30o

Option C: 

20o

Option D: 

10o

Jamb Maths Solution: 

$\begin{align}  & \angle PRS=\angle SPR={{40}^{o}}\text{   }\!\!\{\!\!\text{ Base angle of an isoscele }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & \angle QSR={{90}^{o}}\text{                }\!\!\{\!\!\text{ angle subtended in a semicircle }\!\!\}\!\!\text{ } \\ & \text{Consider }\vartriangle QRS \\ & \angle SQR+\angle QSP+\angle QRS={{180}^{o}}\text{  }\!\!\{\!\!\text{ Sum of angles in a triangle }\!\!\}\!\!\text{ } \\ & \angle SQR+{{90}^{o}}+{{40}^{o}}={{180}^{o}} \\ & \angle SQR={{50}^{o}} \\ & \angle PQR+\angle SQR={{180}^{o}}\text{  }\!\!\{\!\!\text{ Sum of angles on a straight line }\!\!\}\!\!\text{ } \\ & \angle PQR+{{50}^{o}}={{180}^{o}} \\ & \angle PQR={{130}^{o}} \\ & \text{considering }\vartriangle PQS \\ & {{40}^{o}}+{{130}^{o}}+\angle PSQ={{180}^{o}} \\ & \angle PSQ={{10}^{o}} \\\end{align}$ $\begin{align}  & \mathbf{Alternative}\text{ }\mathbf{method} \\ & \angle PRS=\angle SPR={{40}^{o}}\text{   }\!\!\{\!\!\text{ Base angle of an issoscele }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & \text{consider }\Delta PRS \\ & \angle SPR+\angle PSR+\angle SRP={{180}^{o}}\text{    }\!\!\{\!\!\text{ sum of angles in a }\Delta \} \\ & {{40}^{o}}+\angle PSR+{{40}^{o}}={{180}^{o}} \\ & \angle PSR={{100}^{o}} \\ & \text{Also, since }QR\text{ is a diameter} \\ & \angle QSR={{90}^{o}}\text{     }\!\!\{\!\!\text{ Angle subtended in a semicircle }\!\!\}\!\!\text{ } \\ & \angle PSR=\angle PSQ+\angle QSR \\ & {{100}^{o}}=\angle PSQ+{{90}^{o}} \\ & \angle PSQ={{10}^{o}} \\\end{align}$

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