In a triangle ABC,if 4a = b + c,then tan frac | vedclass

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

Question

(B) We know that $	an \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$ and $	an \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$.Multiplying these,we

Vedclass · Accepted Answer

$\frac{3}{5}$

Vedclass · Answer

(B) We know that $	an \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$ and $	an \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$.Multiplying these,we get $	an \frac{B}{2} 	an \frac{C}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)} \cdot \frac{(s-a)(s-b)}{s(s-c)}} = \sqrt{\frac{(s-a)^2}{s^2}} = \frac{s-a}{s}$.Substituting $s = \frac{a+b+c}{2}$,we have $\frac{s-a}{s} = \frac{\frac{a+b+c}{2} - a}{\frac{a+b+c}{2}} = \frac{b+c-a}{b+c+a}$.Given $b+c = 4a$,we substitute this into the expression:$\frac{4a-a}{4a+a} = \frac{3a}{5a} = \frac{3}{5}$.

In a $\triangle ABC$,if $4a = b + c$,then $\tan \frac{B}{2} \tan \frac{C}{2} =$

Similar Questions

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

The sum of the radii of the inscribed and circumscribed circles for an $n$-sided regular polygon of side $a$ is:

In a triangle $ABC$ with usual notations,find the value of $\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2}$.

With usual notations in $\Delta ABC$,$a=3$,$c=2$ and $\sin C=\frac{2}{3}$,then $\angle A=$

In a triangle $ABC$,with usual notations,if $\frac{2 \cos A}{a} + \frac{\cos B}{b} + \frac{2 \cos C}{c} = \frac{a}{bc} + \frac{b}{ca}$,then $\angle A = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module