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Question

In the given figure POQ is a diameter of a circle with centre O and PQRS is a cyclic quadrilateral. SQ is joined. If R=138o. Then PQS is

A
90o
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B
42o
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C
48o
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D
38o
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Solution

The correct option is C 48o
GivenPOQisthediameterofacirclewithcentreO.PQRSisaquadrilateralinscribedinthegivencircle.SQhasbeenjoined.QRS=138o.TofindoutPQS=?SolutionPQRSisacyclicquadrilateral.QPS+QRS=180o(sincethesumoftheoppositeanglesofacyclicquadrilateralis180o).QPS=180oQRS=180o138o=42o.NowPQisthediameterwhichsubtendsQPStothecircumferenceofthegivencircleatS.QPS=90osinceitistheangleinasemicircle.SoinΔPQSwehaveQSP+QPS+PQS=180o(anglesumpropertyoftriangles).PQS=180o(QSP+QPS)=180o(90o+42o)=48o.AnsOptionC.
311960_244543_ans_55682903186e4f18a19b92d0a5092090.png

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