Question

The vertices of an acute angled triangle are $A(x_1,x_1\tan \alpha ),B(x_2,x_2\tan \beta )$ and $C(x_3,x_2\tan \gamma ).$ If origin is the circumcentre of $△ABC$ and $H(a,b)$ be its orthocentre, then $\frac{b}{a}$ equals to
(where $x_1,x_2,x_3$ are positive)

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

Given vertices of the triangle are $A(x_1,x_1\tan \alpha ),B(x_2,x_2\tan \beta )$ and $C(x_3,x_2\tan \gamma )$

Assuming $S,G,$ and $H$ be circumcente, centroid and orthocentre of $△ABC$ respectively, then
$S\equiv (0,0)G\equiv (\frac{x_1+x_2+x_3}{3},\frac{x_1\tan \alpha +x_2\tan \beta +x_3\tan \gamma }{3})H\equiv (a,b)$

We know that, centroid divide circumcentre and orthocentre in the ratio $1:2$
$\frac{x_1+x_2+x_3}{3}=\frac{(0\times 2)+(a\times 1)}{3}\Rightarrow a=x_1+x_2+x_3\frac{x_1\tan \alpha +x_2\tan \beta +x_3\tan \gamma }{3}=\frac{(2\times 0)+(1\times b)}{3}\Rightarrow b=x_1\tan \alpha +x_2\tan \beta +x_3\tan \gamma$

As $S$ is the circumcentre, so
$AS=BS=CS=r\sqrt{(x_1-0{)}^2+(x_1\tan \alpha -0{)}^2}=r\Rightarrow x_1\sec \alpha =r\Rightarrow x_1=r\cos \alpha$
Similarly,
$x_2=r\cos \beta x_3=r\cos \gamma$

Now,
$\frac{b}{a}=\frac{x_1\tan \alpha +x_2\tan \beta +x_3\tan \gamma }{x_1+x_2+x_3}$
Putting $x_1=r\cos \alpha ,x_2=r\cos \beta ,x_3=r\cos \gamma$, we get
$\therefore \frac{b}{a}=\frac{\sin \alpha +\sin \beta +\sin \gamma }{\cos \alpha +\cos \beta +\cos \gamma }$