In a triangle ABC, if tan frac{A}{2} : tan fr | vedclass

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(D) Let $	an \frac{A}{2} = 15k$, $	an \frac{B}{2} = 10k$, and $	an \frac{C}{2} = 6k$. Using the formula $	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}

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$4$

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(D) Let $	an \frac{A}{2} = 15k$, $	an \frac{B}{2} = 10k$, and $	an \frac{C}{2} = 6k$. Using the formula $	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$, we have: $\frac{	an(A/2)}{	an(B/2)} = \sqrt{\frac{(s-b)^2}{(s-a)^2}} = \frac{s-b}{s-a} = \frac{15}{10} = \frac{3}{2}$. This gives $2s - 2b = 3s - 3a$, so $s = 3a - 2b$. Similarly, $\frac{	an(B/2)}{	an(C/2)} = \sqrt{\frac{(s-c)^2}{(s-b)^2}} = \frac{s-c}{s-b} = \frac{10}{6} = \frac{5}{3}$. This gives $3s - 3c = 5s - 5b$, so $2s = 5b - 3c$. Substituting $s = \frac{a+b+c}{2}$, we get $a+b+c = 5b - 3c$, which simplifies to $a = 4b - 4c$. Therefore, $\frac{a}{b-c} = 4$.

In a triangle $ABC$, if $\tan \frac{A}{2} : \tan \frac{B}{2} : \tan \frac{C}{2} = 15 : 10 : 6$, then $\frac{a}{b-c} =$

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