Question

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

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

Answer 1

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

Given: $\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}$
Let the ratio be equal to $k,$ then
$s-a=11k,\cdots \left(i\right)s-b=12k,\cdots \left(ii\right)s-c=13k\cdots \left(iii\right)$
Adding $\left(i\right),\left(ii\right),\left(iii\right)$
$\Rightarrow s=36k$
$\therefore {\tan }^2\frac{A}{2}=\frac{(s-b)(s-c)}{s(s-a)}=\frac{12k\times 13k}{36k\times 11k}$
$\therefore {\tan }^2\frac{A}{2}=\frac{13}{33}$