Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively, if $\tan \frac{A}{2}=\frac{5}{6}$ and $\tan \frac{B}{2}=\frac{20}{37}$

then which of the following is true?

• A. $2a=b+c$
• B. $a>b>c$
• C. $2c=a+b$
• D. none of these

Answer 1

The correct option is B $a>b>c$

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

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

$=\frac{\frac{6}{5}.\frac{37}{20}-1}{\frac{6}{5}+\frac{37}{20}}=\frac{222-200}{120+185}=\frac{122}{305}=\frac{2}{5}$

Now $\tan \frac{A}{2}\tan \frac{C}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$

$\Rightarrow \frac{5}{6}.\frac{2}{5}=\frac{s-b}{s}\Rightarrow 3(s-b)=s\Rightarrow 2s=3b$

$\Rightarrow a+b+c=3b\Rightarrow a+c=2b$

Thus $a,b,c$ are in $A.P.$

Thus the correct order is $a>b>c$