Question: If O is 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:
- Given, $ \angle QOR=110{}^\circ , $ $ \angle QOR=25{}^\circ $
$ \angle OPR=\angle ORP=25{}^\circ $
[since, angles opposite of radius are equal]
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 sides 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 $
$ \because $ $ \angle OPR=25{}^\circ $
$ \therefore $ $ \angle ORP=25{}^\circ $ [because OP and OR are equal]
In $ \Delta ORQ, $ $ \angle OQR+\angle ORQ $
$ 180{}^\circ -110{}^\circ =70{}^\circ $
$ \therefore $ $ \angle OQR=\angle ORQ=35{}^\circ $
$ \Rightarrow $ $ \angle PRQ=35{}^\circ +25{}^\circ =60{}^\circ $
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