In a △ABC,Σ(b+c)cos⁡A= 

In a ABC,Σ(b+c)cosA= 

  1. A

    a+b+c

  2. B

    a+bc

  3. C

    ab+c

  4. D

    none of these 

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    Solution:

    Using projection formulae, we have  

    (b+c)cosA

    =(b+c)cosA+(c+a)cosB+(a+b)cosC

    =(bcosC+ccosB)+(ccosA+acosC)+(acosB+bcosA)=a+b+c

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