Question 44

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waecmaths question: 

In the diagram, $\left| SR \right|=\left| QR \right|,\text{ }\angle SRP={{65}^{{}^\circ }},$$\text{and }\angle RPQ={{48}^{{}^\circ }}$ find $\angle PRQ$

Option A: 

65o

Option B: 

45o

Option C: 

25o

Option D: 

19o

waecmaths solution: 

$\begin{align}  & \angle QSR=\angle QPR={{48}^{\circ }}\text{ }\!\!\{\!\!\text{ }\angle \text{ on the same segment }\!\!\}\!\!\text{ } \\ & \angle RQS=\angle QSR={{48}^{\circ }}\text{  }\!\!\{\!\!\text{ base }\angle \text{s of issoceles }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & \angle RQS+\angle QSR+\angle QRS={{180}^{\circ }}\text{  }\!\!\{\!\!\text{ }\angle s\text{ in a }\vartriangle \} \\ & {{48}^{\circ }}+{{48}^{\circ }}+\angle QRS={{180}^{\circ }} \\ & \angle QRS={{84}^{\circ }} \\ & \angle PRQ=\angle QRS-\angle PRS={{84}^{\circ }}-{{65}^{\circ }}={{19}^{\circ }} \\\end{align}$

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