Question

The vector of a triangle are $A\left(x_1,x_1\tan \alpha \right),B\left(x_2,x_2\tan \beta \right)\&C\left(x_3,x_3\tan \gamma \right)$. If the circumcentre of $\Delta ABC$ coincides with the origin and $H(a,b)$ be its orthocentre, then $\frac{a}{b}$ is equal to

• A. $\frac{\cos \alpha +\cos \beta +\cos \gamma }{\cos \alpha cos\beta cos\gamma }$
• B. $\frac{\sin \alpha +\sin \beta +\sin \gamma }{\sin \alpha \sin \beta \sin \gamma }$
• C. $\frac{\tan \alpha +\tan \beta +\tan \gamma }{\tan \alpha \tan \beta \tan \gamma }$
• D. $\frac{\cos \alpha +\cos \beta +\cos \gamma }{\sin \alpha +\sin \beta +\sin \gamma }$

Answer 1

The correct option is D $\frac{\cos \alpha +\cos \beta +\cos \gamma }{\sin \alpha +\sin \beta +\sin \gamma }$

Given vertices of triangle are,

$A(x_1,x_{1}tan\alpha ),B(x_2,x_{2}tan\beta ),C(x_3,x_{2}tan\gamma )$

Orthocenter $H(a,b)$

and circumcenter is (0,0)

Let G be the centroid of given triangle

Cordinates of G are=$(\frac{x_1+x_2+x_3}{3},\frac{x_{1}tan\alpha +x_{2}tan\beta +x_{3}tan\gamma }{3})-\left(1\right)$

As we know centroid Divides the line joing the orthocenter and circumcenter in ratio 1:2

Therefore cordinates of centroid =$(\frac{2a}{3},\frac{2b}{3})-\left(2\right)$

From eqs. (1) and (2)

H(a,b)=$(\frac{3}{2}\frac{x_1+x_2+x_3}{3},\frac{3}{2}\frac{x_{1}tan\alpha +x_{2}tan\beta +x_{3}tan\gamma }{3})$

This nothin but the circle whose radius is given by,

$r=\sqrt{x_1^2+x_1^{2}ta{n}^2\alpha }$

$\to r=\sqrt{x_1^{2}se{c}^2\alpha }$

$\to x_1=\frac{r}{sec\alpha }$

Similarly $x_2=\frac{r}{sec\beta }$ and

$x_3=\frac{r}{sec\gamma }$

$\therefore \frac{a}{b}=\frac{r(cos\alpha +cos\beta +cos\gamma )}{r(sin\alpha +sin\beta +sin\gamma )}$

$\therefore \frac{a}{b}=\frac{(cos\alpha +cos\beta +cos\gamma )}{(sin\alpha +sin\beta +sin\gamma )}$