Question

In a triangle $ABC$ if $\tan \left(\frac{A}{2}\right)=\frac{5}{6}$,$\tan \left(\frac{C}{2}\right)=\frac{2}{5}$ then,

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

Answer 1

The correct option is D

$a,b$ and $c$ are in $AP$

Explanation for correct option:

Given $\tan \left(\frac{A}{2}\right)=\frac{5}{6}$ and $\tan \left(\frac{C}{2}\right)=\frac{2}{5}$

We know, from half angle formula,

$$\begin{gathered} \tan \left(\frac{A}{2}\right)=\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{s\left(s-a\right)}}\left(1\right)\left[\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}\mathrm{are}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{triangle}\right] \\ \tan \left(\frac{C}{2}\right)=\sqrt{\frac{\left(s-b\right)\left(s-a\right)}{s\left(s-c\right)}}\left(2\right) \end{gathered}$$

Multiply both the equations we get,

$$\begin{gathered} \tan \left(\frac{A}{2}\right)\tan \left(\frac{C}{2}\right)=\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{s\left(s-a\right)}}\times \sqrt{\frac{\left(s-a\right)\left(s-b\right)}{s\left(s-c\right)}} \\ \Rightarrow \frac{5}{6}\times \frac{2}{5}=\sqrt{\frac{\left(s-b\right)\left(s-b\right)}{s^2}} \\ \Rightarrow \frac{2}{6}=\frac{s-b}{s} \\ \Rightarrow s=3s-3b \\ \Rightarrow 2s=3b \\ \Rightarrow 2\left(\frac{a+b+c}{2}\right)=3b\left[\mathrm{s}=\mathrm{semiperimeter}\right] \\ \Rightarrow a+b+c=3b \\ \Rightarrow a+c=2b \end{gathered}$$

The obtained relation is the condition of an $AP$ series, hence $a,b,c$ are in $AP$.

Therefore, the correct answer is option (D).