Question

In an acute angle $△ABC,2\cos A=\frac{\sin B}{\sin C}$​ and $2^{{\tan }^{2}B}$ is a solution of the equation $x^2-9x+8=0,$ then the distance between incentre to the circumcentre of $△ABC$ is

• A. $0$
• B. $2\sqrt{3}$
• C. $1$
• D. $1+\sqrt{3}$

Answer 1

The correct option is A $0$

$\because x^2-9x+8=0$
$\therefore (x-1)(x-8)=0$
$\therefore x=1,8$
$2^{{\tan }^{2}B}=1,8$
$\Rightarrow 2^{{\tan }^{2}B}=2^0,2^3$
$\Rightarrow {\tan }^{2}B=0,3$
$\Rightarrow {\tan }^{2}B\ne 0$
$\therefore {\tan }^{2}B=3$
$\Rightarrow \tan B=\pm \sqrt{3}$
$\therefore \angle B={60}^{\circ },{120}^{\circ }\left(\text{rejected}\right)$
Also, given $2\cos A=\frac{\sin B}{\sin C}$
$\Rightarrow 2\cos A\sin C=\sin B$
Using compound angle formula, we have
$\Rightarrow \sin (A+C)-\sin (A-C)=\sin B$
$\Rightarrow \sin (\pi -B)-\sin (A-C)=\sin B$
$\Rightarrow \sin B-\sin (A-C)=\sin B$
$\Rightarrow \sin (A-C)=0$
$\Rightarrow A-C=0orA=C$
Hence, $A+B+C={180}^0$
$A+{60}^0+A={180}^0$
$\therefore 2A={180}^0-{60}^0={120}^0$
Hence, $A=C=B={60}^0$
Hence the given triangle is equilateral.
So Incentre and circumcentre will coincide.