If s is the semi-perimeter of triangle ABC an | vedclass

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(D) Let $\frac{s-a}{4} = \frac{s-b}{5} = \frac{s-c}{6} = k$. Then $s-a = 4k$,$s-b = 5k$,and $s-c = 6k$. Adding these gives $3s - (a+b+c) = 15k$. Since

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$\frac{25}{33}$

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(D) Let $\frac{s-a}{4} = \frac{s-b}{5} = \frac{s-c}{6} = k$. Then $s-a = 4k$,$s-b = 5k$,and $s-c = 6k$. Adding these gives $3s - (a+b+c) = 15k$. Since $a+b+c = 2s$,we have $3s - 2s = 15k$,so $s = 15k$. Then $a = s - 4k = 11k$,$b = s - 5k = 10k$,and $c = s - 6k = 9k$. We know $\sin^2\left(\frac{A}{2}\right) = \frac{(s-b)(s-c)}{bc}$. Thus,$\sum \sin^2\left(\frac{A}{2}\right) = \frac{(s-b)(s-c)}{bc} + \frac{(s-c)(s-a)}{ca} + \frac{(s-a)(s-b)}{ab}$. Substituting the values: $= \frac{(5k)(6k)}{(10k)(9k)} + \frac{(6k)(4k)}{(9k)(11k)} + \frac{(4k)(5k)}{(11k)(10k)} = \frac{30}{90} + \frac{24}{99} + \frac{20}{110} = \frac{1}{3} + \frac{8}{33} + \frac{2}{11}$. $= \frac{11 + 8 + 6}{33} = \frac{25}{33}$.

If $s$ is the semi-perimeter of $\triangle ABC$ and if $\frac{s-a}{4}=\frac{s-b}{5}=\frac{s-c}{6}$,then $\sum \sin^2\left(\frac{A}{2}\right) = $

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