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] |
\[\Rightarrow \]\[2\,\angle ORQ=180{}^\circ -110{}^\circ \] |
\[\Rightarrow \]\[\angle ORQ=\frac{70{}^\circ }{2}=35{}^\circ \] |
\[\therefore \] \[\angle PRQ=\angle ORP+\angle ORQ=25{}^\circ +35{}^\circ =60{}^\circ \] |
You need to login to perform this action.
You will be redirected in
3 sec