CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    In any triangle\[ABC,\text{ }{{c}^{2}}sin\text{ }2B+{{b}^{2}}sin\text{ }2C\]is equal to

    A)  \[\frac{\Delta }{2}\]         

    B)                         \[\Delta \]

    C)  \[2\Delta \]                       

    D)         \[3\Delta \]

    E)  \[4\Delta \]

    Correct Answer: E

    Solution :

    \[{{c}^{2}}sin\text{ }2B+{{b}^{2}}sin\text{ }2C\] \[={{c}^{2}}(2sinB\,cosB)+{{b}^{2}}(2sinC\,cosC)\] \[=2{{c}^{2}}\left( \frac{2\Delta }{ac}\cos B \right)+2{{b}^{2}}\left( \frac{2\Delta }{ab}\cos C \right)\] \[=4\Delta \left( \frac{c\cos B+b\cos C}{a} \right)\] \[=4\Delta \left( \frac{a}{a} \right)=4\Delta \]


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