In a triangle ABC,tan frac{A}{2} = frac{5}{6} | vedclass

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(D) We know that $	an \frac{A}{2} 	an \frac{C}{2} = \frac{s-b}{s}$.Given $	an \frac{A}{2} = \frac{5}{6}$ and $	an \frac{C}{2} = \frac{2}{5}$,we ha

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$a, b, c$ and $\sin A, \sin B, \sin C$ are in $A.P.$

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(D) We know that $	an \frac{A}{2} 	an \frac{C}{2} = \frac{s-b}{s}$.Given $	an \frac{A}{2} = \frac{5}{6}$ and $	an \frac{C}{2} = \frac{2}{5}$,we have $\frac{5}{6} 	imes \frac{2}{5} = \frac{s-b}{s}$.$\frac{1}{3} = \frac{s-b}{s}$ $\Rightarrow s = 3s - 3b$ $\Rightarrow 2s = 3b$.Since $2s = a + b + c$,we have $a + b + c = 3b$,which implies $a + c = 2b$.Thus,$a, b, c$ are in $A.P.$By the Sine Rule,$a = 2R \sin A, b = 2R \sin B, c = 2R \sin C$.Since $a, b, c$ are in $A.P.$,$2R \sin A, 2R \sin B, 2R \sin C$ are in $A.P.$Therefore,$\sin A, \sin B, \sin C$ are also in $A.P.$

In a triangle $ABC$,$\tan \frac{A}{2} = \frac{5}{6}$ and $\tan \frac{C}{2} = \frac{2}{5}$,then

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