Question

In triangle ABC, (b+c) cos A+(c+a)cos B+(a+b)cos C is equal to

• A. $0$
• B. $1$
• C. $a+b+c$
• D. $2(a+b+c)$

Answer 1

Answer: (C)

$a+b+c$

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

$\Rightarrow$ $b\cos A+c\cos A+c\cos B+a\cos B+a\cos C+b\cos C$

$\Rightarrow$ $(b\cos C+c\cos B)+(c\cos A+a\cos C)+(a\cos B+b\cos A)$ ----( 1 )

Using projection formula,

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

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

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

Substituting above values in ( 1 ) we get,

$\Rightarrow$ $a+b+c$

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