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. $6 0^{0}$
• C. $75^{0}$
• D. $9 0^{0}$

Answer 1

Answer: (B)

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

$\frac{\overset{ \rightarrow }{x} + \overset{ \rightarrow }{0}}{3}=\frac{\overset{ \rightarrow }{a} + \overset{ \rightarrow }{b} + \overset{ \rightarrow }{c}}{3}\Rightarrow \overset{ \rightarrow }{x}=\overset{ \rightarrow }{a}+\overset{ \rightarrow }{b}+\overset{ \rightarrow }{c}$
$\left|\overset{ \rightarrow }{O A}\right|=\left|\overset{ \rightarrow }{H A}\right|$ (given)
$\left|\overset{ \rightarrow }{a}\right|=\left|\overset{ \rightarrow }{b} + \overset{ \rightarrow }{c}\right| \, \left(\left|\overset{ \rightarrow }{a}\right| = \left|\overset{ \rightarrow }{b}\right| = \left|\overset{ \rightarrow }{c}\right| = R\right)$
$\left|\overset{ \rightarrow }{a}\right|^{2}=\left|\overset{ \rightarrow }{b}\right|^{2}+\left|\overset{ \rightarrow }{c}\right|^{2}+2\left|\overset{ \rightarrow }{b}\right|\left|\overset{ \rightarrow }{c}\right|cos \theta $
$\Rightarrow R^{2}=R^{2}+R^{2}+2R^{2}cos \theta $
$\Rightarrow cos \theta =-\frac{1}{2}\Rightarrow \theta =12 0^{o}$

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