Question

The circum-radius and the in-radius of a triangle $ABC$ are $10$ and $3$ units respectively, then $acotA+bcotB+ccotC$ is equal to?

• A. $13$
• B. $26$
• C. $39$
• D. none of these

Answer 1

The correct option is B $26$

Given $R=10$ and $r=3$

We know $R=\frac{a}{2\sin A}$

$10=\frac{a}{2\sin A}$=$20=\frac{a}{2\sin A}$

=$a\cot A=20\cos A$ --------equ($1$) ...$\cot A=\frac{\cos A}{\sin A}$

Similary for other angle and sides

$b\cot B=20\cos B$ --------equ($2$)

$c\cot C=20\cos C$ --------equ($1$)

add all 3 equation we get,

$a\cot A+b\cot B+c\cot C=20(\cos A+\cos B+\cos C)$ ------equ($4$)

For any triangle $\cos A+\cos B+\cos C=\frac{R+r}{R}$

$a\cot A+b\cot B+c\cot C=20\times \left(\frac{13}{10}\right)$

Hence, $a\cot A+b\cot B+c\cot C=26$