Question

In a $\Delta ABC,$ the value of $\frac{a\cos A+b\cos B+c\cos C}{a+b+c}$ is equal to

• A. $\frac{r}{R}$
• B. $\frac{R}{2r}$
• C. $\frac{R}{r}$
• D. $\frac{2r}{R}$

Answer 1

The correct option is A $\frac{r}{R}$

We have;

$\frac{a\cos A+b\cos B+c\cos C}{a+b+c}$

Using sine law;

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

$⟹a=2R\sin A;b=2R\sin B;c=2R\sin C$

$=\frac{2R\sin A.\cos A+2R\sin B.\cos B+2R\sin C.\cos C}{2R\sin A+2R\sin B+2R\sin C}$

$=\frac{R(\sin 2A+\sin 2B+\sin 2C)}{2R(\sin A+\sin B+\sin C)}$

we know that

$ifA+B+C=\pi \sin 2A+\sin 2B+\sin 2C=4\sin A.\sin B.\sin C\sin A+\sin B+\sin C=4\cos A/2.\cos B/2.\cos C/2$

$=\frac{4R(\sin A.\sin B.\sin C)}{2R\times 4\cos A/2.\cos B/2.\cos C/2}$

$=\frac{4R\sin A/2.\sin B/2.\sin C/2.8\cos A/2.\cos B/2.\cos C/2}{2R\times 4\cos A/2.\cos B/2.\cos C/2}$

$=\frac{r}{R}$ $[\because r=4R\sin A/2.\sin B/2.\sin C/2]$