Question

If in a triangle $\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}$ then $\lambda {\tan }^2\left(\frac{A}{2}\right)=455$ if $\lambda$ must be

• A. $1155$
• B. $1515$
• C. $5151$
• D. $5511$

Answer 1

The correct option is A $1155$

Given that
$\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}=\frac{s-a+s-b+s-c}{11+12+13}$
$=\frac{3s-(a+b+c)}{36}$
$=\frac{3s-2s}{36}$ where $a+b+c=2s$
$=\frac{s}{36}$
$\therefore s-a=\frac{11s}{36},s-b=\frac{12s}{36},s-c=\frac{13s}{36}$
$\therefore {\tan }^2\frac{A}{2}=\frac{(s-b)(s-c)}{s(s-a)}$
$=\frac{\frac{12s}{36}\times \frac{13s}{36}}{s\times \frac{11s}{36}}$
$=\frac{13}{33}$
$\lambda {\tan }^2\frac{A}{2}=455$ (given)
$\lambda .\frac{13}{33}=455$
$\Rightarrow \lambda =35\times 33=1155$ (on simplification)