In a $\triangle ABC$ we have $\tan (A/2) = 5/6$,$\tan (C/2) = 2/5$ then we have to comment on sides of $\triangle ABC$.
MY ATTEMPT: Firstly i found the value of $\sin A$, $\sin C$ then $\sin B$ by $\sin (A+C)= \sin(\pi-B)$. Now I tried using sine rule but the value of $\sin B$ could not make it possible to reach on any result. How shall I proceed?
On
Note that $\displaystyle \tan\frac{B}{2}=\cot\left(\frac{A}{2}+\frac{C}{2}\right)=\frac{1-\frac{5}{6}\cdot\frac{2}{5}}{\frac{5}{6}+\frac{2}{5}}=\frac{20}{37}$.
Let $I$ be the incentre of the triangle and $r$ be the inradius. If the incircle touches $BC$, $CA$ and $AB$ at $D$, $E$ and $F$ respectively. Then
$$BC=BD+CD=r\cot\frac{B}{2}+r\cot\frac{C}{2}=\frac{87}{20}r$$
$$CA=CE+AE=r\cot\frac{C}{2}+r\cot\frac{A}{2}=\frac{37}{10}r$$
$$AB=AF+BF=r\cot\frac{A}{2}+r\cot\frac{B}{2}=\frac{61}{20}r$$
So, $BC:CA:AB=87:74:61$.
$\tan(A/2)=5/6$, let's use half angle tangent formulas: $$\sin A=\frac{2\tan(A/2)}{1+tan^2(A/2)}=60/61$$ Similarly, $\sin C=20/29$. Now, $\sin B=\sin(\pi-A-C)=\sin(A+C)=\sin A\cos C+\cos A\sin C$. Apply half-angle tangent formula for $\cos A$, $\cos B$. I leave that up to you. Finally, apply sine theorem to find relationships between sides.