Question

In the triangle $ABC,\tan \frac{A}{2}=5/6,\tan \frac{C}{2}=2/5$ , 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 AP

Answer 1

The correct option is A $a,c,b$ are in AP

$\begin{array}{cc}\text{Solution - In a}△ABC\text{,}& \\ \text{it is given that,}\tan A& =\frac{5}{1},\tan \frac{C}{2}=\frac{2}{5}\end{array}$

Using the identity we get

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

Multiply equation (1) & (2)

$\begin{array}{c}\Rightarrow \mathrm{Tan}\frac{A}{2}\times \mathrm{Tan}\frac{C}{2}=\sqrt{\frac{(S-b)(S-c)}{S(S-a)}}\times \sqrt{\frac{(S-a)(S-b)}{S(S-C)}}\\ \Rightarrow \frac{5}{6}\times \frac{2}{5}=\frac{S-b}{S}\\ \Rightarrow \frac{1}{3}=\frac{a+b+c-2b}{a+b+c}\\ \Rightarrow a+b+c=3a+3c-3b\\ \Rightarrow 4b=2a+2c\\ \Rightarrow b=\frac{a+c}{2}\\ \Rightarrow a,b,c\text{are in}AP\\ \Rightarrow \text{hence,}\left(A\right)\text{is the correct option.}\end{array}$