question_answer

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                                                                                                                       [SSC (CGL) 2013]

A)  \[50{}^\circ \]

B)  \[55{}^\circ \]  

C)  \[60{}^\circ \]

D)  \[65{}^\circ \]

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]

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

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

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