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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2002

  • question_answer
    In triangle\[ABC,\text{ }a({{b}^{2}}+{{c}^{2}})cos\text{ }A+b({{c}^{2}}+{{a}^{2}})\] \[cos\text{ }B+c({{a}^{2}}+{{b}^{2}})cos\text{ }C\] is equal to:

    A)  \[abc\]                                

    B)         \[2\,abc\]

    C)         \[3\,abc\]                                          

    D)         \[4\,abc\]

    E)         0

    Correct Answer: C

    Solution :

     \[({{b}^{2}}+{{c}^{2}})a\cos A+({{c}^{2}}+{{a}^{2}})b\cos B+\] \[({{a}^{2}}+{{b}^{2}})c\cos C\] \[=ab(b\cos A+a\cos B)+\] \[bc(b\cos C+c\cos B)+ca(a\cos C+c\cos A)\] \[=abc++abc+abc=3abc\]            


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