question_answer

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                                   [SSC (CGL) 2013]

A) \[50{}^\circ \]                          

B) \[55{}^\circ \]

C) \[60{}^\circ \]  

D) \[65{}^\circ \]

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 \[\Delta ORQ,\]\[\angle OQR+\angle ORQ\]

\[180{}^\circ -110{}^\circ =70{}^\circ \]

\[\therefore \]      \[\angle OQR=\angle ORQ=35{}^\circ \]

\[\Rightarrow \]   \[\angle PRQ=35{}^\circ +25{}^\circ =60{}^\circ \]