Question

The vertices of a triangle are $A(x_1,x_1\tan {\theta }_1),B(x_2,x_2\tan {\theta }_2)$ & $C(x_3,x_3\tan {\theta }_3)$. If the circumcentre $O$ of the $△ABC$ is at the origin & $H(\overline{x},\overline{y})$ be its orthocentre, then $\frac{\overline{x}}{\overline{y}}$=

• A. $\frac{\cos {\theta }_1-\cos {\theta }_2-\cos {\theta }_3}{\sin {\theta }_1+\sin {\theta }_2+\sin {\theta }_3}$
• B. $\frac{\cos {\theta }_1+\cos {\theta }_2+\cos {\theta }_3}{\sin {\theta }_1+\sin {\theta }_2+\sin {\theta }_3}$
• C. $\frac{{\cos }^2{\theta }_1+{\cos }^2{\theta }_2+{\cos }^2{\theta }_3}{{\sin }^2{\theta }_1+{\sin }^2{\theta }_2+{\sin }^2{\theta }_3}$
• D. $\frac{\cos {\theta }_1-\cos {\theta }_2+\cos {\theta }_3}{\sin {\theta }_1-\sin {\theta }_2+\sin {\theta }_3}$

Answer 1

The correct option is B $\frac{\cos {\theta }_1+\cos {\theta }_2+\cos {\theta }_3}{\sin {\theta }_1+\sin {\theta }_2+\sin {\theta }_3}$

Circumcentre is origin

$\therefore O{A}^2=O{B}^2=O{C}^2$

$x_1^2+x_1^2{\tan }^2{\theta }_1=x_2^2+x_2^2{\tan }^2{\theta }_2$

$=x_3^2+x_3^2{\tan }^2{\theta }_3=r^2$

$x_1=r\cos {\theta }_1,x_2=r\cos {\theta }_2,x_3=r\cos {\theta }_3$

$\therefore$ Co-ordinate of vertices of the triangle become

$A(r\cos {\theta }_1,r\sin {\theta }_1),B(r\cos {\theta }_2,r\sin {\theta }_2),C(r\cos {\theta }_3,r\sin {\theta }_3)$

$x^{\prime }=\frac{\sum r\cos {\theta }_1}{3},y^{\prime }=\frac{\sum r\sin {\theta }_1}{3}$
Now, $x^{\prime }=\frac{\overline{x}}{3}$

$\overline{x}=r(\cos {\theta }_1+\cos {\theta }_2+\cos {\theta }_3)$

$\overline{y}=r(\sin {\theta }_1+\sin {\theta }_2+\sin {\theta }_3)$

$\therefore \frac{\overline{x}}{\overline{y}}=\frac{\cos {\theta }_1+\cos {\theta }_2+\cos {\theta }_3}{\sin {\theta }_1+\sin {\theta }_2+\sin {\theta }_3}$