Question

In a $∆ABC$ , $(b+c)\cos A+(a+c)\cos B+(a+b)\cos C$ is equal to :

• A. $0$
• B. $abc$
• C. $a+b+c$
• D. $1$

Answer 1

The correct option is C

$a+b+c$

Determine the value of $(b+c)\cos A+(a+c)\cos B+(a+b)\cos C$.

Expanding the given equation we get,

$\Rightarrow b\cos A+c\cos A+a\cos B+c\cos B+a\cos C+b\cos C...\left(1\right)$

From the projection rule, we know that :

$$\begin{gathered} a=b\cos C+c\cos B, \\ b=c\cos A+a\cos C, \\ c=a\cos B+b\cos A. \end{gathered}$$

Therefore rearranging the equation $\left(1\right)$ we have,

$$\begin{gathered} \Rightarrow (b\cos C+c\cos B)+(c\cos A+a\cos C)+(acosB+b\cos A) \\ \Rightarrow a+b+c\text{[By projection rule]} \end{gathered}$$

Hence, option (C) is correct.