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Manipal Engineering Manipal Engineering Solved Paper-2011

  • question_answer
    In\[\Delta \,ABC,\,\,\frac{b-c}{{{r}_{1}}}+\frac{c-a}{{{r}_{2}}}+\frac{a-b}{{{r}_{3}}}\]is equal to

    A) \[0\]                                     

    B) \[abc\]

    C) \[a+b+c\]           

    D)        \[ab+bc+ca\]

    Correct Answer: A

    Solution :

    Now,     \[\frac{b-c}{{{r}_{1}}}+\frac{c-a}{{{r}_{2}}}+\frac{a-b}{{{r}_{3}}}\] \[=\frac{[(b-c)(s-a)+(c-a)(s-b)+(a-b)(s-c)]}{\Delta }\] \[=0\]


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