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 \[\Delta ORQ,\]\[\angle OQR+\angle ORQ\] |
\[180{}^\circ -110{}^\circ =70{}^\circ \] |
\[\therefore \] \[\angle OQR=\angle ORQ=35{}^\circ \] |
\[\Rightarrow \] \[\angle PRQ=35{}^\circ +25{}^\circ =60{}^\circ \] |
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