In a ∆ABC , (b+c)cosA+(a+c)cosB+(a+b)cosC is equal to :
0
abc
a+b+c
1
Determine the value of (b+c)cosA+(a+c)cosB+(a+b)cosC.
Expanding the given equation we get,
⇒bcosA+ccosA+acosB+ccosB+acosC+bcosC...(1)
From the projection rule, we know that :
a=bcosC+ccosB,b=ccosA+acosC,c=acosB+bcosA.
Therefore rearranging the equation (1) we have,
⇒(bcosC+ccosB)+(ccosA+acosC)+(acosB+bcosA)⇒a+b+c[Byprojectionrule]
Hence, option (C) is correct.
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