Question

In a triangle ABC, if $tan\left(A/2\right)=5/6$ and $tan\left(\beta /2\right)=20/37$,

then sides $a,b,c$ of the triangle are in?

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

Answer 1

The correct option is A A.P.

$\tan \left(\frac{A+B}{2}\right)$ $=\frac{\tan \left(\frac{A}{2}\right)+\tan \left(\frac{B}{2}\right)}{1-\tan \frac{A}{2}.\tan \frac{B}{2}}$
$=\frac{5\left(37\right)+20\left(6\right)}{6\left(37\right)-100}$
$=\frac{5}{2}$

Also, $\tan \left(\frac{C}{2}\right)=\tan ({90}^o-\frac{(A+B)}{2})$

$=\cot \left(\frac{A+B}{2}\right)=\frac{\cot \left(A/2\right)\cot \left(B/2\right)-1}{\cot \left(A/2\right)+\cot \left(B/2\right)}$
$=\frac{\frac{6}{5}.\frac{37}{20}-1}{\frac{6}{5}+\frac{37}{20}}=\frac{222-100}{120+185}=\frac{122}{305}=\frac{2}{5}$
$⟹\tan \left(\frac{C}{2}\right)=\frac{2}{5}$.
Hence
$\tan \left(\frac{A}{2}\right).\tan \left(\frac{C}{2}\right)$
$=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}\times \frac{(s-a)(s-b)}{s(s-c)}}$ $=\frac{s-b}{s}$
$⟹\frac{2}{5}.\frac{5}{6}=1-\frac{b}{s}$
$⟹\frac{b}{s}=\frac{2}{3}$
$⟹3b=2s$
$⟹3b=a+b+c$
$⟹2b=a+c$.
Hence $a,b,c$ are in A.P.