In a △ABC,Σ(b+c)cos⁡A=

In a $△ A B C, Σ (b + c) \cos A =$

A
$a + b + c$
B
$a + b - c$
C
$a - b + c$
D
none of these

Solution:

Using projection formulae, we have

$\sum (b + c) \cos A$

$= (b + c) \cos A + (c + a) \cos B + (a + b) \cos C$

$\begin{array}{l} = (b \cos C + c \cos B) + (c \cos A + a \cos C) + (a \cos B + b \cos A) \\ = a + b + c \end{array}$

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