Question

The radius of the circle passing through the centre of the incircle of $\Delta ABC$ and through the end points of $BC$, is given by (circum radius of $\Delta ABC$ is $R)$

• A. $\frac{a}{2}\cos A$
• B. $2R\sin \frac{A}{2}$
• C. $a\sec \frac{A}{2}$
• D. $2R\cos \frac{A}{2}$

Answer 1

Answer: (B)

$2R\sin \frac{A}{2}$

$\angle BIC=\pi -\frac{B+C}{2}$
$=\frac{\pi }{2}+\frac{A}{2}$
$\angle BOC=2(\pi -\angle BIC)$
$=2\pi -\pi -A$
$=\pi -A$
$\therefore$ In $\Delta BOC$,

$a^2=r^2+r^2-2r^{2}cos(\pi -A)$
$\Rightarrow a^2=2r^2(1+cosA)$
$\Rightarrow r=\frac{a}{2}\sec \frac{A}{2}=R\sin A\sec \frac{A}{2}(\therefore a=2RsinA)$
$=2R\sin \frac{A}{2}$