Question

The radius of the circle passing through the centre of in-circle of $△ABC$ and through the end points of $BC$ is given by

• A. $\left(\frac{a}{2}\right)\cos A$
• B. $\left(\frac{a}{2}\right)\sec \left(\frac{A}{2}\right)$
• C. $\left(\frac{a}{2}\right)\sin A$
• D. $a\sec \left(\frac{A}{2}\right)$

Answer 1

Answer: (B)

$\left(\frac{a}{2}\right)\sec \left(\frac{A}{2}\right)$

$\therefore \angle BIC={180}^0-(\frac{B}{2}+\frac{C}{2})$
$={180}^0-({90}^0-\frac{A}{2})$
$={90}^0+\frac{A}{2}$
$\therefore \angle BOC=2\angle BIC$
$=2({90}^0+\frac{A}{2})={180}^0+A$
Then, $\angle BOD={90}^0+\frac{A}{2}$
$\therefore \sin ({90}^0+\frac{A}{2})$
$=\frac{\frac{a}{2}}{R}=\cos \left(\frac{A}{2}\right)$
$\therefore R=\frac{a}{2}\sec \left(\frac{A}{2}\right)$