If,in a triangle ABC,tan frac{A}{2} = frac{5} | vedclass

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(B) We know that $	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $	an \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$.Multiplying these,we

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$2b = a + c$

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(B) We know that $	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $	an \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$.Multiplying these,we get $	an \frac{A}{2} 	an \frac{C}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)} \cdot \frac{(s-a)(s-b)}{s(s-c)}} = \sqrt{\frac{(s-b)^2}{s^2}} = \frac{s-b}{s}$.Given $	an \frac{A}{2} = \frac{5}{6}$ and $	an \frac{C}{2} = \frac{2}{5}$,we have $\frac{s-b}{s} = \frac{5}{6} \cdot \frac{2}{5} = \frac{1}{3}$.Thus,$3(s-b) = s$ $\Rightarrow 3s - 3b = s$ $\Rightarrow 2s = 3b$.Since $2s = a + b + c$,we have $a + b + c = 3b$,which simplifies to $a + c = 2b$.

If,in a $\triangle ABC$,$\tan \frac{A}{2} = \frac{5}{6}$ and $\tan \frac{C}{2} = \frac{2}{5}$,then $a, b, c$ are such that :

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