Question

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

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

Answer 1

Answer: (A)

A.P.

In $△ABC$

$\tan \frac{C}{2}=\tan (90-\frac{A+B}{2})$

$=\cot \left(\frac{A+B}{2}\right)$

$=\frac{1-\tan \frac{A}{2}\tan \frac{B}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}}$

$=\frac{1-\frac{5}{6}\times \frac{20}{37}}{\frac{5}{6}+\frac{20}{37}}$

$=\frac{\frac{61}{111}}{\frac{305}{222}}$

$\tan \frac{C}{2}=\frac{2}{5}$$\sin A=\frac{2\tan A/2}{1+{\tan }^{2}A/2}$

$=\frac{2\left(5/6\right)}{1+\frac{25}{36}}$

$=\frac{10/6}{61/366}=\frac{60}{61}$

$\sin B=\frac{2\times \frac{20}{37}}{1+\frac{400}{1369}}$

$=\frac{40\times 37}{1769}=\frac{1480}{1769}$

$\sin C=\frac{2\left(2/5\right)}{1+4/25}=\frac{20}{29}$

$\therefore 2\sin B=\sin A+\sin C$

$\sin A,\sin B$ & $\sin C$ are in A.P.

So, according to sine rule sides are in A.P.