Question

if $x,y$ and $z$ are the distances of in centre from the vertices of the triangle $ABC$ respectively then $\frac{abc}{xyz}$ is equal to-

• A. $\coprod \tan \frac{A}{2}$
• B. $\sum \cot \frac{A}{2}$
• C. $\sum \tan \frac{A}{2}$
• D. $\frac{s^2}{\Delta }$

Answer 1

The correct option is A $\coprod \tan \frac{A}{2}$

Since,

$x=r\cos ec\frac{A}{2}y=r\cos ec\frac{B}{2}z=r\cos ec\frac{c}{2}$

$Alsoa=r\left(\cot \frac{B}{2}+\cot \frac{C}{2}\right)$

$Now,\frac{a}{x}=\left(\cot \frac{B}{2}+\cot \frac{C}{2}\right)\cdot \sin \frac{A}{2}\frac{a}{x}=\frac{\sin \frac{A}{2}\cdot \cos \frac{A}{2}}{\sin \frac{B}{2}\cdot \sin \frac{C}{2}}$

$Similarly$

$\frac{b}{y}=\frac{\sin \frac{B}{2}\cdot \cos \frac{B}{2}}{\sin \frac{A}{2}\cdot \sin \frac{C}{2}}\frac{c}{z}=\frac{\sin \frac{C}{2}\cdot \cos \frac{C}{2}}{\sin \frac{A}{2}\cdot \sin \frac{B}{2}}\frac{abc}{xyz}=\tan \frac{A}{2}\cdot \tan \frac{B}{2}\cdot \tan \frac{C}{2}$

$=\coprod \tan \frac{A}{2}$

Hence, this is the answer.