Question

Vertex $A$ of the acute angled triangle $ABC$ is equidistant from its circumcentre $O$ and orthocentre $H$ , then the possible value of $\angle A$ is

• A. $30^0$
• B. $60^0$
• C. $75^0$
• D. $90^0$

Answer 1

Answer: (B)

Let, the position of circumcentre $(O)$ be origin, position vector of orthocentre $(H)$ is $\overrightarrow{x}$ and position vectors of $A,B,C$ are $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ respectively
Centroid of $\Delta ABC$ $\left(G\right)=\frac{1}{3}\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)$

$\frac{\overrightarrow{x}+\overrightarrow{0}}{3}=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}\Rightarrow \overrightarrow{x}=\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$
$\left|\overrightarrow{\mathrm{OA}}\right|=\left|\overrightarrow{\mathrm{HA}}\right|$ (given)
$\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}+\overrightarrow{c}\right|\left(\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=R\right)$
${\left|\overrightarrow{a}\right|}^2={\left|\overrightarrow{b}\right|}^2+{\left|\overrightarrow{c}\right|}^2+2\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|cos\theta$
$\Rightarrow R^2=R^2+R^2+2R^{2}cos\theta$
$\Rightarrow cos\theta =-\frac{1}{2}\Rightarrow \theta =120^o$

From diagram, $\angle A=60^o$