| In the given figure, O is the centre of a circle, PQL and PRM axe the tangents at the points Q and R, respectively and S is a point on the circle such that \[\angle SQL=50{}^\circ \]and \[\angle SRM=60{}^\circ .\]Then, \[\angle QSR\]is equal to |
 |
A) \[40{}^\circ \]
B) \[50{}^\circ \]
C) \[60{}^\circ \]
D) \[70{}^\circ \]
Correct Answer: D
Solution :
| Since, \[PQL\]is a tangent and \[OQ\]is a radius, so \[\angle OQL=90{}^\circ \] |
| \[\angle OQS=(90{}^\circ -50{}^\circ )=40{}^\circ \] |
| Now, \[OQ=OS\] |
| \[\Rightarrow \] \[\angle OSQ=\angle OQS=40{}^\circ \] |
| Similarly, \[\angle ORQ=(90{}^\circ -60{}^\circ )=30{}^\circ \] |
| and \[OR=OS\] |
| \[\Rightarrow \] \[\angle OSR=\angle ORS=30{}^\circ \] |
| \[\Rightarrow \]\[\angle QSR=\angle OSQ+\angle OSR\] |
| \[=(40{}^\circ +30{}^\circ )=70{}^\circ \] |
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