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

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(A) Let $\frac{s - a}{11} = \frac{s - b}{12} = \frac{s - c}{13} = k$.Then $s - a = 11k$, $s - b = 12k$, $s - c = 13k$.Adding these, $(s - a) + (s - b)

Vedclass · Accepted Answer

$1155$

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(A) Let $\frac{s - a}{11} = \frac{s - b}{12} = \frac{s - c}{13} = k$.Then $s - a = 11k$, $s - b = 12k$, $s - c = 13k$.Adding these, $(s - a) + (s - b) + (s - c) = 36k \implies 3s - (a + b + c) = 36k$.Since $a + b + c = 2s$, we have $3s - 2s = 36k \implies s = 36k$.Then $a = s - 11k = 25k$, $b = s - 12k = 24k$, $c = s - 13k = 23k$.The formula for $	an^2 \frac{A}{2}$ is $\frac{(s - b)(s - c)}{s(s - a)}$.Substituting the values: $	an^2 \frac{A}{2} = \frac{(12k)(13k)}{(36k)(11k)} = \frac{156k^2}{396k^2} = \frac{156}{396} = \frac{13}{33}$.Given $\lambda 	an^2 \frac{A}{2} = 455$, we have $\lambda \left( \frac{13}{33} ight) = 455$.$\lambda = \frac{455 	imes 33}{13} = 35 	imes 33 = 1155$.

In a triangle $ABC$, with usual notations, if $\frac{s - a}{11} = \frac{s - b}{12} = \frac{s - c}{13}$ and $\lambda \tan^2 \frac{A}{2} = 455$, then $\lambda = $

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