Question

In a $△ABC,\tan \frac{A}{2}=\frac{5}{6},\tan \frac{C}{2}=\frac{2}{5}$, then :

• A. $a,c,b$ are in $AP$
• B. $a,b,c$ are in $AP$
• C. $b,a,c$ are in $AP$
• D. $a,b,c$ are in GP

Answer 1

Answer: (B)

Key Idea : $\tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$
$\because \tan \frac{A}{2}=\frac{5}{6}$ and $\tan \frac{C}{2}=\frac{2}{5}$
Now, $\tan \frac{A}{2}\tan \frac{C}{2}=\frac{5}{6}\times \frac{2}{5}$
$\Rightarrow \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\cdot \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}=\frac{1}{3}$
$\Rightarrow \frac{s-b}{s}=\frac{1}{3}$
$\Rightarrow 3s-3b=s$
$\Rightarrow 2s=3b$
$\Rightarrow a+b+c=3b$
$\Rightarrow a+c=2b$
$\Rightarrow a,b,c$ are in $AP$