Question 20

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waecmaths question: 

In the diagram \[\left| PR \right|=\left| QR \right|\] and $\left| PR \right|=\left| RS \right|=\left| SP \right|$ , calculate the size of $\angle QRS$

Option A: 

150o

Option B: 

120o

Option C: 

90o

Option D: 

60o

waecmaths solution: 

\[\begin{align}  & \text{In }\vartriangle PRS \\ & y+y+y={{180}^{\circ }}\text{    }\!\!\{\!\!\text{ }\angle \text{s in a equilateral }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & y={{60}^{\circ }} \\ & \angle PSR+\angle PQR={{180}^{\circ }}\text{   }\!\!\{\!\!\text{ opp }\angle s\text{ of a cyclic quad }\!\!\}\!\!\text{ } \\ & \text{6}{{\text{0}}^{\circ }}+\angle PQR={{180}^{\circ }} \\ & \angle PQR={{120}^{\circ }} \\ & In\text{ }\vartriangle PQR \\ & \angle QPR=\angle QRP\text{   }\!\!\{\!\!\text{ Base }\angle s\text{ of Iss}\text{. }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & x+x+{{120}^{\circ }}={{180}^{\circ }} \\ & x={{30}^{\circ }} \\ & \angle QRS=x+y={{30}^{\circ }}+{{60}^{\circ }}={{90}^{\circ }} \\\end{align}\]

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