In △ABC,s−a11=s−b12=s−c13, then tan2⁡A2=

s−a=11k,s−b=12k,s−c=13k ⇒s-a+s-b+s-c=11k+12k+13k ⇒3s-(a+b+c)=36k∴3s-2s=36k ⇒s=36k, then tan2A2=(s-b)(s-c)s(s-a)=12133611=1333

In △ABC,s−a11=s−b12=s−c13, then tan2⁡A2=

s−a=11k,s−b=12k,s−c=13k ⇒s-a+s-b+s-c=11k+12k+13k ⇒3s-(a+b+c)=36k∴3s-2s=36k ⇒s=36k, then tan2A2=(s-b)(s-c)s(s-a)=12133611=1333

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In $△ ABC, \frac{s - a}{11} = \frac{s - b}{12} = \frac{s - c}{13}, then \tan^{2} (\frac{A}{2}) =$

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$143 / 432$

$13 / 33$

$11 / 39$

$12 / 37$

answer is B.

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Detailed Solution

$s - a = 11 k, s - b = 12 k, s - c = 13 k \Rightarrow s - a + s - b + s - c = 11 k + 12 k + 13 k \Rightarrow 3 s - (a + b + c) = 36 k ∴ 3 s - 2 s = 36 k \Rightarrow s = 36 k, then \tan^{2} (\frac{A}{2}) = \frac{(s - b) (s - c)}{s (s - a)} = \frac{(12) (13)}{36 (11)} = \frac{13}{33}$