2. Sample Papers
  3. SSC

SSC Sample Paper Mock Test-1 SSC CGL Tear-II Paper-1

  • question_answer
    In the given figure, POQ is a diameter and PQRS is cyclic quadrilateral. If \[\angle PSR=130{}^\circ ,\]then \[\angle RPQ=?\]

    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.
You will be redirected in 3 sec spinner