Question

If in a triangle $ABC$,
$\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}$.
If $k{\tan }^2\left(A/2\right)=429$ then the value of $k$ is?

Answer 1

$\begin{array}{c}\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}=k\\ s-a=11ks-b=12ks-c=13k\\ s-a+s-b+s-c=36k\\ 3s-(a+b+c)=36k\{\begin{array}{c}s=\frac{a+b+c}{2}\\ 2s=a+b+c\end{array}\\ 3s-2s=36k\\ s=36k\\ \tan \frac{A}{2}=\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{\left(s-a\right)}}\\ =\sqrt{\frac{12k\times 13k}{36k\times 11k}}\\ =\sqrt{\frac{13}{33}}\\ {\tan }^2\frac{A}{2}=\frac{13}{33}\\ k{\tan }^2\frac{A}{2}=429\\ k=\frac{429\times 33}{13}\\ ={33}^2=1089Ans.\end{array}$