Question

The vertices of a triangle are $A(x_1,x_1\tan \alpha ),B(x_2,x_2\tan \beta )$ and $C(x_3,x_3\tan \gamma )$. If the circumcentre of triangle $ABC$ coincides with the origin and $H(a,b)$ be its orthocentre then $\frac{b}{a}=$

• 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. $\frac{\sin \alpha +\sin \beta +\sin \gamma }{\cos \alpha +\cos \beta +\cos \gamma }$

Answer 1

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

$A(x_1,x_1\tan \alpha ),B(x_2,x_2\tan \beta ),C(x_3,x_3\tan \gamma )$ are vertices of $\Delta ABC$

$\therefore$ centroid coordinates are given by:-

$(\frac{x_1+x_2+x_3}{3},\frac{x_1\tan \alpha +x_2\tan \beta +x_3\tan \gamma }{3})$

$\therefore$ circumcentre $\equiv (0,0)\to$ given

We know that centroid divide circumcentre and orthocentre in the ratio $1:2$

$\therefore \frac{x_1+x_2+x_3}{3}=\frac{0\times 2+a\times 1}{3}$

$\therefore a=x_1+x_2+x_3$

$\therefore \frac{x_1\tan \alpha +x_2\tan \beta +x_3\tan \gamma }{3}=\frac{1\times b+2/0}{3}$

$\therefore b=x_1\tan \alpha +x_2\tan \beta +x_3\tan \gamma$

$\therefore \frac{b}{a}=\frac{x_1\tan \alpha +x_2\tan \beta +x_3\tan \gamma }{x_1+x_2+x_3}$

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