Question related to tan in a ratio in a triangle If in a triangle $\tan A:\tan B:\tan C = 1:2:3$ then, what are the ratio of the sides $a,b,c $?

Let $AD, BE, CF$ be altitudes of triangle $ABC$. Denote $BC=a, CA=b, AB=c$.
Observe that $$\frac 23=\frac{\tan B}{\tan C} = \frac{\frac{AD}{BD}}{\frac{AD}{DC}} = \frac{DC}{BD}$$ so $BD=\frac 35 a$ and $DC=\frac 25 a$.
On the other hand Pythagorean theorem says that $$c^2-b^2=AB^2-AC^2=(BD^2+AD^2)-(CD^2+AD^2)=BD^2-CD^2=\left(\frac 35 a\right)^2 - \left(\frac 25 a\right)^2 = \frac 15 a^2$$
Analogously we find $CE=\frac 14 b$, $AE=\frac 34 b$ and $$c^2-a^2=\frac 12 b^2$$
The above results imply that $b^2=\frac 85 a^2$ and $c^2=\frac 95 a^2$. Thus $a^2:b^2:c^2=5:8:9$ which leads to the answer: $$a:b:c=\sqrt 5 : 2\sqrt 2 : 3.$$
