A) \[40{}^\circ \]
B) \[50{}^\circ \]
C) \[60{}^\circ \]
D) \[70{}^\circ \]
Correct Answer: A
Solution :
PQRS is cyclic quadrilateral. \[\therefore \] \[\angle PSR+\angle PQR=180{}^\circ \] \[\Rightarrow \] \[130{}^\circ +\angle PQR=180{}^\circ \] \[\Rightarrow \] \[\angle PQR=50{}^\circ .\] Also, \[\angle PRQ=90{}^\circ \](angle in a semi-circle) In\[\Delta PQR,\]we have \[\angle PQR+\angle PRQ+\angle RPQ=180{}^\circ \]\[\Rightarrow \]\[50{}^\circ +90{}^\circ \angle RPQ=180{}^\circ \] \[\Rightarrow \] \[\angle RPQ=40{}^\circ .\]You need to login to perform this action.
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