In a triangle ABC,if a+c=5b,then cot frac{A}{ | vedclass

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

Question

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

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) We know that $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.Multiplying these,we get $\cot \frac{A}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-a) \cdot s(s-c)}{(s-b)(s-c) \cdot (s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-b)^2}} = \frac{s}{s-b}$.Since $s = \frac{a+b+c}{2}$,we have $2s = a+b+c$.Thus,$\frac{s}{s-b} = \frac{2s}{2s-2b} = \frac{a+b+c}{a+b+c-2b} = \frac{(a+c)+b}{(a+c)-b}$.Given $a+c=5b$,substituting this into the expression gives $\frac{5b+b}{5b-b} = \frac{6b}{4b} = \frac{3}{2}$.

In a $\triangle ABC$,if $a+c=5b$,then $\cot \frac{A}{2} \cot \frac{C}{2}$ is equal to

Similar Questions

In a $\triangle ABC$,if $a=6$,$b=5$,and $c=4$,then find the value of $\cos 2A$.

In $\Delta ABC,$ if $\cot A, \cot B, \cot C$ are in $A.P.,$ then $a^2, b^2, c^2$ are in

In $\triangle ABC, (r_2+r_3) \operatorname{cosec}^2 \frac{A}{2} =$

In a triangle $ABC$,if $\tan \left(\frac{A-B}{2}\right) = \frac{1}{3} \tan \left(\frac{A+B}{2}\right)$,then $a : b =$

If $a = 9, b = 8$ and $c = x$ satisfies $3 \cos C = 2$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module