With usual notation in a Delta ABC,left( frac | vedclass

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

Question

(C) We know that $\frac{1}{r_1} + \frac{1}{r_2} = \frac{s-a}{\Delta} + \frac{s-b}{\Delta} = \frac{2s - a - b}{\Delta} = \frac{c}{\Delta}$.Similarly,$\

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(C) We know that $\frac{1}{r_1} + \frac{1}{r_2} = \frac{s-a}{\Delta} + \frac{s-b}{\Delta} = \frac{2s - a - b}{\Delta} = \frac{c}{\Delta}$.Similarly,$\frac{1}{r_2} + \frac{1}{r_3} = \frac{a}{\Delta}$ and $\frac{1}{r_3} + \frac{1}{r_1} = \frac{b}{\Delta}$.Therefore,the $LHS = \left( \frac{c}{\Delta} \right) \left( \frac{a}{\Delta} \right) \left( \frac{b}{\Delta} \right) = \frac{abc}{\Delta^3}$.Using the relation $\Delta = \frac{abc}{4R}$,we have $\Delta^3 = \frac{a^3 b^3 c^3}{64 R^3}$.Substituting this into the $LHS$,we get $\frac{abc}{\frac{a^3 b^3 c^3}{64 R^3}} = \frac{64 R^3}{a^2 b^2 c^2}$.Comparing this with $\frac{K R^3}{a^2 b^2 c^2}$,we find $K = 64$.

With usual notation in a $\Delta ABC$,$\left( \frac{1}{r_1} + \frac{1}{r_2} \right) \left( \frac{1}{r_2} + \frac{1}{r_3} \right) \left( \frac{1}{r_3} + \frac{1}{r_1} \right) = \frac{K R^3}{a^2 b^2 c^2}$ where $K$ has the value equal to

Similar Questions

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

The sides of a triangle inscribed in a given circle subtend angles $\alpha, \beta, \gamma$ at the center. The minimum value of the $A.M.$ of $\cos (\alpha + \frac{\pi}{2})$,$\cos (\beta + \frac{\pi}{2})$ and $\cos (\gamma + \frac{\pi}{2})$ is equal to

In a triangle $ABC$,if $\cos A \cos B + \sin A \sin B \sin C = 1$,then $\sin A + \sin B + \sin C = $

In a $\triangle ABC$,the expression $\frac{a}{\tan A} + \frac{b}{\tan B} + \frac{c}{\tan C}$ is equal to:

In $\triangle ABC$,if $\cos A \cos B + \sin A \sin B \sin C = 1$ and $C = \frac{\pi}{2}$,then $A : B =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module