In the figure, PQ is a diameter of a circle with centre at O and OR⊥PQ, where R is a point on the circle. If S is another point on the circle such that ∠RPS=32∘, then ∠QRS is
A
13∘
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B
26∘
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C
45∘
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D
30∘
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Solution
The correct option is A13∘ Given−PQisthedoameterofacirclewithcentreO.R&SaretwopointsonthecircumferenceofthegivencirclesuchthatOR⊥PQ&∠RPS=32o.Tofindout−∠QRS=?Solution−WejoinRQ&RS.InΔOPROP=OR(radiiofthesamecircle).SoΔOPRisanisoscelesonewithPRasbaseand∠POR=90o(sinceOR⊥PQ).∴∠OPR=∠ORP⟹∠OPR+∠ORP=2∠OPR.So2∠OPR=180o−∠POR⟹2∠OPR=180o−90o=90o⟹∠OPR=45o.∴∠QPS=∠OPR−∠PRS=45o−32o=13o.NowthechordQSsubtends∠QPS&∠QRStothecircumferenceofthegivencircleatP&Rrespectively.∴∠QPS=∠QRS=13o(sincetheangles,subtendedbyachordofacircletoitscircumference,areequal).Ans−OptionA.