In a triangle ABC, with usual notations, if f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $\frac{s - a}{11} = \frac{s - b}{12} = \frac{s - c}{13} = k$.Then $s - a = 11k$, $s - b = 12k$, $s - c = 13k$.Adding these, $(s - a) + (s - b)

Vedclass · Accepted Answer

$1155$

Vedclass · Answer

(A) Let $\frac{s - a}{11} = \frac{s - b}{12} = \frac{s - c}{13} = k$.Then $s - a = 11k$, $s - b = 12k$, $s - c = 13k$.Adding these, $(s - a) + (s - b) + (s - c) = 36k \implies 3s - (a + b + c) = 36k$.Since $a + b + c = 2s$, we have $3s - 2s = 36k \implies s = 36k$.Then $a = s - 11k = 25k$, $b = s - 12k = 24k$, $c = s - 13k = 23k$.The formula for $	an^2 \frac{A}{2}$ is $\frac{(s - b)(s - c)}{s(s - a)}$.Substituting the values: $	an^2 \frac{A}{2} = \frac{(12k)(13k)}{(36k)(11k)} = \frac{156k^2}{396k^2} = \frac{156}{396} = \frac{13}{33}$.Given $\lambda 	an^2 \frac{A}{2} = 455$, we have $\lambda \left( \frac{13}{33} ight) = 455$.$\lambda = \frac{455 	imes 33}{13} = 35 	imes 33 = 1155$.

In a triangle $ABC$, with usual notations, if $\frac{s - a}{11} = \frac{s - b}{12} = \frac{s - c}{13}$ and $\lambda \tan^2 \frac{A}{2} = 455$, then $\lambda = $

Similar Questions

In $\triangle ABC$,if $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$,then $\angle C$ is equal to: (in $^{\circ}$)

In $\triangle ABC$,if $s-a : s-b : s-c = 2 : 3 : 4$,then $\cot A : \cot C =$

If $a, b, c$ are the sides of a triangle $ABC,$ then which of the following inequalities is not true?

In a $\triangle ABC$,if $(a+b+c)(b+c-a) = \lambda bc$,then which of the following is true?

In $\triangle ABC$,$m \angle B = \frac{\pi}{3}$ and $m \angle C = \frac{\pi}{4}$. Let point $D$ divide $BC$ internally in the ratio $1:3$. Then,the value of $\frac{\sin(\angle BAD)}{\sin(\angle CAD)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module