In triangle ABC,if frac{s-a}{11}=frac{s-b}{12 | vedclass

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

Question

(A) Given that,$\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}=k$. $s-a=11k$,$s-b=12k$,$s-c=13k$. Adding these,we get $3s-(a+b+c) = 36k$. Since $a+b+c=2

Vedclass · Accepted Answer

$\frac{290}{429}$

Vedclass · Answer

(A) Given that,$\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}=k$. $s-a=11k$,$s-b=12k$,$s-c=13k$. Adding these,we get $3s-(a+b+c) = 36k$. Since $a+b+c=2s$,we have $3s-2s=36k$,so $s=36k$. Using the formula $	an^2\left(\frac{A}{2}ight) = \frac{(s-b)(s-c)}{s(s-a)}$ and $	an^2\left(\frac{C}{2}ight) = \frac{(s-a)(s-b)}{s(s-c)}$,$	an^2\left(\frac{A}{2}ight)+	an^2\left(\frac{C}{2}ight) = \frac{(12k)(13k)}{(36k)(11k)} + \frac{(11k)(12k)}{(36k)(13k)}$. $= \frac{12 	imes 13}{36 	imes 11} + \frac{11 	imes 12}{36 	imes 13} = \frac{1}{3} \left( \frac{13}{11} + \frac{11}{13} ight)$. $= \frac{1}{3} \left( \frac{169+121}{143} ight) = \frac{1}{3} 	imes \frac{290}{143} = \frac{290}{429}$.

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

Similar Questions

In a triangle $ABC$, if $r_1=2 r_2=3 r_3$, then $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=$

If in a triangle $ABC$,$a, b, c$ are sides and angle $A$ is given,and $c \sin A < a < c$,and $b_1$ and $b_2$ are two possible values of $b$,then:

In $\triangle ABC$ with usual notations, if $1 + \tan(\frac{A}{2}) \tan(\frac{B}{2}) = \frac{k}{s}$ (where $s$ is the semi-perimeter), then the value of $k$ is...

In a $\Delta ABC$,if $C = 90^{\circ}$,then $\frac{a^{2} + b^{2}}{a^{2} - b^{2}} \sin(A - B)$ is equal to:

In $\triangle ABC$, with usual notation, if $\cot A, \cot B, \cot C$ are in arithmetic progression, then

Vedclass Test Series

Exam Paper Generator

Online Exam Module