Quantitative Aptitude Ques 1635

Question: If O be the circumcentre of a $ \Delta PQR $ and $ \angle QOR=110{}^\circ , $ $ \angle OPR=25{}^\circ , $ then the measure of $ \angle PRQ $ is

Options:

A) $ 50{}^\circ $

B) $ 55{}^\circ $

C) $ 60{}^\circ $

D) $ 65{}^\circ $

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Answer:

Correct Answer: C

Solution:

  • [c] Given, $ \angle QOR=110{}^\circ , $ $ \angle OPR=25{}^\circ $ $ \angle OPR=\angle ORP=25{}^\circ $ [ $ \because $ angles opposite of equal side are equal $ \therefore OP=OB=radius $ ] Now, in $ \Delta QOR, $ $ \angle QOR+\angle OQR+\angle ORQ=180{}^\circ $ [angles sum property]

$ \Rightarrow $ $ \angle QOR+\angle ORQ+\angle ORQ=180{}^\circ $ $ \because $ $ \angle OQR=\angle ORQ $ [angles opposite to equal side in a triangle are equal]

$ \Rightarrow $ $ 2,\angle ORQ=180{}^\circ -110{}^\circ $

$ \Rightarrow $ $ \angle ORQ=\frac{70{}^\circ }{2}=35{}^\circ $

$ \therefore $ $ \angle PRQ=\angle ORP+\angle ORQ=25{}^\circ +35{}^\circ =60{}^\circ $