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 a $\triangle ABC$,the altitude $AD$ and the median $AE$ divide $\angle A$ into three equal parts. If $BC=28$,then the nearest integer to $AB+AC$ is

In a triangle,if the ex-radii $r_1, r_2, r_3$ are in the ratio $1: 2: 3$,then its sides are in the ratio

If the angles of a triangle are in the ratio $1:2:3$,then their corresponding sides are in the ratio

In a $\triangle ABC$,if $a=5, b=6, c=7$,then the length of the median drawn from $B$ is

If in $\triangle ABC$,with usual notations,$a \cdot \cos^2 \frac{C}{2} + c \cdot \cos^2 \frac{A}{2} = \frac{3b}{2}$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module