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] |
| \[\Rightarrow \]\[2\,\angle ORQ=180{}^\circ -110{}^\circ \] |
| \[\Rightarrow \]\[\angle ORQ=\frac{70{}^\circ }{2}=35{}^\circ \] |
| \[\therefore \] \[\angle PRQ=\angle ORP+\angle ORQ=25{}^\circ +35{}^\circ =60{}^\circ \] |
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