Question

If $r_1,r_2,$ and $r_3$ be the radii of encircles $ABC$, then $\frac{\sum r_1}{\sqrt{\sum r_1{r}_2}}$is equal to:

• A. $\sum \cot \frac{A}{2}$
• B. $\sum \cot \frac{A}{2}\cot \frac{B}{2}$
• C. $\sum \tan \frac{A}{2}$
• D. $\parallel \tan \frac{A}{2}$

Answer 1

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

Since,

$\frac{r_1+r_2+r_3}{r_1{r}_2+r_2{r}_3+r_3{r}_1}=\frac{\frac{\Delta }{s-a}+\frac{\Delta }{s-b}+\frac{\Delta }{s-c}}{\Delta \sqrt{\frac{1}{\left(s-a\right)\left(s-b\right)}+\frac{1}{\left(s-b\right)\left(s-c\right)}+\frac{1}{\left(s-c\right)\left(s-a\right)}}}=\frac{\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}}{\sqrt{\frac{\left(s-c\right)+\left(s-a\right)+\left(s-b\right)}{\left(s-a\right)\left(s-b\right)\left(s-c\right)}}}=\frac{\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}}{\frac{s}{\Delta }}=\frac{\frac{\Delta }{s-a}+\frac{\Delta }{s-b}+\frac{\Delta }{s-c}}{s}=\frac{\Delta }{s\left(s-a\right)}+\frac{\Delta }{s\left(s-b\right)}+\frac{\Delta }{s\left(s-c\right)}=\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}=\sum \tan \frac{A}{2}$

Hence, the option $C$ is the correct answer.