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  4. In a triangle A B C, tan (A/2)=(5/6), tan (C/2)=(2/5), then :

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Q. In a $\triangle A B C, \tan \frac{A}{2}=\frac{5}{6}, \tan \frac{C}{2}=\frac{2}{5}$, then :

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Solution:

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, $\tan \frac{A}{2} \tan \frac{C}{2}=\frac{5}{6} \times \frac{2}{5}$
$\Rightarrow \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \cdot \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}=\frac{1}{3}$
$\Rightarrow \frac{s-b}{s}=\frac{1}{3}$
$\Rightarrow 3 s-3 b=s$
$\Rightarrow 2 s=3 b $
$\Rightarrow a + b + c = 3 b$
$\Rightarrow a + c = 2 b$
$\Rightarrow a, b, c$ are in $A P$