If in the Delta ABC,AB = 2BC,then find the ra | vedclass

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(D) Given $AB = c$ and $BC = a$. Since $AB = 2BC$,we have $c = 2a$.Using the Napier's Analogy,we know that $	an \frac{C - A}{2} = \frac{c - a}{c + a}

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$1:3$

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(D) Given $AB = c$ and $BC = a$. Since $AB = 2BC$,we have $c = 2a$.Using the Napier's Analogy,we know that $	an \frac{C - A}{2} = \frac{c - a}{c + a} \cot \frac{B}{2}$.Rearranging this,we get $\frac{	an (B/2)}{\cot ((C - A)/2)} = 	an \frac{B}{2} 	an \frac{C - A}{2}$.Substituting the expression for $	an \frac{C - A}{2}$,we get $	an \frac{B}{2} \left( \frac{c - a}{c + a} \cot \frac{B}{2} ight) = \frac{c - a}{c + a}$.Substituting $c = 2a$,we get $\frac{2a - a}{2a + a} = \frac{a}{3a} = \frac{1}{3}$.Thus,the ratio is $1:3$.

If in the $\Delta ABC$,$AB = 2BC$,then find the ratio $\tan \frac{B}{2} : \cot \left( \frac{C - A}{2} \right)$.

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