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SSC Sample Paper Mock Test-14 SSC CGL Tear-II Paper-1

  • question_answer
    In \[\Delta ABC,\] \[\angle C=90{}^\circ \] and \[CD\bot AB,\] also \[\angle A=65{}^\circ ,\] then \[\angle CBA\]is equal to

    A) \[25{}^\circ \]  

    B)  \[35{}^\circ \]

    C) \[65{}^\circ \]

    D)  \[40{}^\circ \]

    Correct Answer: A

    Solution :

    In \[\Delta BCD,\] \[\angle BCD=65{}^\circ \] and \[\angle BDC=90{}^\circ \] \[\angle CBA=180{}^\circ -(\angle BCD+\angle CDB)\] \[=180{}^\circ -(65{}^\circ +90{}^\circ )=180{}^\circ -155{}^\circ =25{}^\circ \]


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