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VIT Engineering VIT Engineering Solved Paper-2008

  • question_answer
    In a triangle ABC, the sides b and c are the roots of the equation \[{{x}^{2}}-61x+820=0\] and\[A={{\tan }^{-1}}\left( \frac{4}{3} \right)\] then \[{{a}^{2}}\] is equal to

    A)  1098         

    B)  1096

    C)  1097         

    D)  1095

    Correct Answer: C

    Solution :

    The given equation is \[{{x}^{2}}-61x+820=0\] \[\Rightarrow \]\[{{x}^{2}}-41x-20x+820=0\] \[\Rightarrow \] \[(x-41)\,(x-20)=0\] \[\Rightarrow \] \[x=41,\,20\] Let b = 41 and c = 20 Also, \[A={{\tan }^{-1}}\left( \frac{4}{3} \right)\] \[\Rightarrow \] \[\cos A=\frac{3}{5}\] \[\therefore \]By cosine formula \[{{a}^{2}}={{b}^{2}}+{{c}^{2}}-2\,bc\,\cos A\] \[={{41}^{2}}+{{20}^{2}}-2\times 41\times 20\times \frac{3}{5}\] \[=2081-984=1097\]


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