In the given figure PQ is a diameter of a circle with centre O and PT is a tangent at P. QT meets the circle at R. If ∠POR=72o, ∠PTR is
A
118o
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B
36o
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C
54o
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D
26o
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Solution
The correct option is B54o Given−PQisadiameterofacirclewithcentreO.PTisatangenttothecirclefromTatP.QTintersectsthecircleatR.∠POR=72o.Tofindout−∠PTR=?Solution−ThediameterPQortheradiusOPmeetsthetangentPTatP.∴PQ⊥PTsincetheradiusthroughthepointofcontactofatangenttoacircleisperpendiculartothetangent.∴∠QPT=90o.AgaintheminorarcPRsubtends∠PORtothecentreOand∠PQRtothecicumference.∴∠PQR=12×∠POR=12×72o=36o.SoinΔPQT∠QTP=180o−(∠PQR+∠QPT)(byanglesumpropertyoftriangles)=180o−(36o+90o)=54o.Ans−OptionC.