Question

In a $△ABC$, if $\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}$, then ${\tan }^2\frac{A}{2}$ is equal to

• A. $\frac{11}{39}$
• B. $\frac{142}{432}$
• C. $\frac{13}{33}$
• D. $\frac{12}{37}$

Answer 1

Answer: (C)

$\frac{13}{33}$

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

$s-a+s-b+s-c=K(11+12+13)=36K$

$⟹s=36K$

${\tan }^2\frac{A}{2}=\frac{(s-b)(s-c)}{s(s-a)}=\frac{12\cdot 13}{36\cdot 11}=\frac{13}{33}$