Question

In triangle $ABC$, if $\tan \left(\frac{A}{2}\right)=\frac{5}{6}$ and $\tan \left(\frac{B}{2}\right)=\frac{20}{37}$, then the sides $a,b$ and $c$ are in

• A. A.P.
• B. G.P.
• C. H.P.
• D. none of these

Answer 1

The correct option is D none of these

Let $\Delta$ be the area of the given triangle, then

$tan\frac{A}{2}=\frac{(s-b)(s-c)}{\Delta }=\frac{5}{6}$ ... (i)

Similarly,

$tan\frac{B}{2}=\frac{(s-c)(s-a)}{\Delta }=\frac{20}{37}$ ...(ii)

Dividing the equation (i) by (ii) , we get

$\frac{s-b}{s-a}=\frac{5}{6}\times \frac{37}{20}$

substituting $s=\frac{a+b+c}{2}$, we get

$\frac{a+c-b}{b+c-a}=\frac{37}{24}$

$24a+24c-24b=37b+37c-37a$

On solving, we have

$61a-61b=13c$

It can be easily observed that a, b and c are not in A.P, G.P and H.P

Hence, the correct option is 'D'.