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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer
    If the co-ordinate of the vertices of a triangle ABC be\[A(-1,3,2),B(2,3,5)\]and \[C(3,5,-2),\] then\[\angle A\]is equal to:

    A)  \[45{}^\circ \]                                  

    B)  \[60{}^\circ \]                  

    C)  \[90{}^\circ \]                  

    D)         \[30{}^\circ \]

    E)  \[135{}^\circ \]

    Correct Answer: C

    Solution :

    \[A(-1,3,2),B(2,3,5)\]and\[C(3,5,-2)\] \[\therefore \]  \[AB=\sqrt{{{3}^{2}}+0+{{3}^{2}}}=\sqrt{18}\]                 \[CA=\sqrt{16+4+16}=6\] and        \[BC=\sqrt{1+4+49}=\sqrt{54}\] \[\because \]     \[A{{B}^{2}}+C{{A}^{2}}=B{{C}^{2}}\] \[\therefore \]\[\Delta ABC\] is right angled triangle, right angled at A. \[\therefore \] \[\angle A=90{}^\circ \]


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