Question

If in $△ABC,\frac{a\cos A+b\cos B+c\cos C}{a\sin B+b\sin C+c\sin A}=\frac{a+b+c}{9R},$ then the $△ABC$ is

• A. isosceles
• B. equilateral
• C. right-angled
• D. none of these.

Answer 1

The correct option is B equilateral

Since,$\frac{a\cos A+b\cos B+c\cos C}{a\sin B+b\sin C+c\sin A}=\frac{a+b+c}{9R}$
$\Rightarrow \frac{2R\sin A\cos A+2R\sin B\cos B+2R\sin C\cos C}{a\times \frac{b}{2R}+b\times \frac{c}{2R}+c\times \frac{a}{2R}}=\frac{a+b+c}{9R}$
$\Rightarrow \frac{2R^2[\sin 2A+\sin 2B+\sin 2C]}{ab+bc+ca}=\frac{a+b+c}{9R}$ .............$\left(1\right)$
Now, $\sin 2A+\sin 2B+\sin 2C$
$=2\sin (A+B)\cos (A-B)+2\sin C\cos C$
$=2\sin (\pi -C)\cos (A-B)+2\sin C\cos C$
$=2\sin C\cos (A-B)+2\sin C\cos (\pi -(A+B))$
$=2\sin C[\cos (A-B)-\cos (A+B)]$
$=2\sin C[-2\sin \left(\frac{A-B+A+B}{2}\right)\sin \left(\frac{A-B-A-B}{2}\right)]$
$=4\sin C\sin A\sin B$
$=4\sin A\sin B\sin C$
Substituting for $\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$ in $\left(1\right)$ we get
$\Rightarrow \frac{2R^2\left[4\sin A\sin B\sin C\right]}{ab+bc+ca}=\frac{a+b+c}{9R}$
$\Rightarrow \frac{2R^2\left[4\frac{a}{2R}\frac{b}{2R}\frac{c}{2R}\right]}{ab+bc+ca}=\frac{a+b+c}{9R}$
On simplifying we get
$\Rightarrow \frac{abc}{ab+bc+ca}=\frac{a+b+c}{9}$
$\Rightarrow 9abc=(a+b+c)(ab+bc+ca)$
On Re-arranging we get
$\Rightarrow a(b^2+c^2-2bc)+b(c^2+a^2-2ca)+c(a^2+b^2-2ab)=0$
$\Rightarrow a{(b-c)}^2+b{(c-a)}^2+c{(a-b)}^2=0$
which is possible when $a-b=0,b-c=0,c-a=0$
$\therefore a=b=c$