In a triangle ABC,with usual notations,if tan | vedclass

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(C) Given $	an \left(\frac{A}{2}ight) = \frac{5}{6}$ and $	an \left(\frac{C}{2}ight) = \frac{2}{5}$.Using the formula $	an \left(\frac{B}{2}i

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$a, b, c$ are in $A$.$P$.

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(C) Given $	an \left(\frac{A}{2}ight) = \frac{5}{6}$ and $	an \left(\frac{C}{2}ight) = \frac{2}{5}$.Using the formula $	an \left(\frac{B}{2}ight) = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$,we know that $	an \left(\frac{A}{2}ight) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $	an \left(\frac{C}{2}ight) = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$.Let $x = \sqrt{s-a}, y = \sqrt{s-b}, z = \sqrt{s-c}$. Then $	an \left(\frac{A}{2}ight) = \frac{y}{x} \cdot \frac{z}{\sqrt{s}} = \frac{5}{6}$ and $	an \left(\frac{C}{2}ight) = \frac{x}{z} \cdot \frac{y}{\sqrt{s}} = \frac{2}{5}$.Multiplying these,$	an \left(\frac{A}{2}ight) 	an \left(\frac{C}{2}ight) = \frac{y^2}{s} = \frac{5}{6} 	imes \frac{2}{5} = \frac{1}{3}$.Since $s = \frac{a+b+c}{2}$,we have $\frac{s-b}{s} = \frac{1}{3} \implies 3(s-b) = s \implies 3s - 3b = s \implies 2s = 3b$.Substituting $2s = a+b+c$,we get $a+b+c = 3b$,which simplifies to $a+c = 2b$.This condition implies that $a, b, c$ are in $A$.$P$.

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

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