Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}$, then ${\tan }^2\left(\frac{A}{2}\right)$ is equal to

• A. $\frac{143}{432}$
• B. $\frac{13}{33}$
• C. $\frac{11}{39}$
• D. $\frac{12}{37}$

Answer 1

The correct option is B $\frac{13}{33}$

Let $\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}=p$

so, we can write , $s-a=11p\Rightarrow a=s-11p......\left(1\right)s-b=12p\Rightarrow b=s-12p......\left(2\right)s-c=13p\Rightarrow c=s-13p......\left(3\right)$

We know that ,

$a+b+c=2s\Rightarrow s-11p+s-12p+s-13p=2s\Rightarrow s=36p$

So, from(1) $a=36p-11p=25p$

Similarly, $b=24p,c=23p$

We know that ${\tan }^2\frac{A}{2}=\frac{(s-b)(s-c)}{s(s-a)}=\frac{12p\times 13p}{36p\times 11p}=\frac{13}{33}$