Question

Let $r$ and $R$ be the inradius and the circumradius of a $△ABC$. Let $\theta$ be the angle between the line joining the incentre and the circumcentre of the $△ABC$ and $BC$. Then $\theta$ is equal to

• A. ${\tan }^{-1}\left(\frac{\cos B+\cos C-1}{\sin C-\sin B}\right)$
• B. ${\tan }^{-1}\left(\frac{\sin C-\sin B}{\cos B+\cos C-1}\right)$
• C. ${\tan }^{-1}\left(\frac{r-R\cos A}{r\cot \frac{B}{2}-R\sin A}\right)$
• D. ${\tan }^{-1}\left(\frac{r\cos A-R}{r\sin A-R\cot \frac{B}{2}}\right)$

Answer 1

Answer: (A), (C)

${\tan }^{-1}\left(\frac{r-R\cos A}{r\cot \frac{B}{2}-R\sin A}\right)$


Let $I$ be the incentre.
$O$ be the circumcentre and $OL\parallel BC$
$\angle IOL=\theta ,IM=r,OC=R,\angle NOC=A$

Now,
$\tan \theta =\frac{IL}{OL}=\frac{IM-LM}{BM-BN}=\frac{IM-ON}{BM-NC}=\frac{r-R\cos A}{r\cot \frac{B}{2}-R\sin A}=\frac{4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}-R\cos A}{4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\cot \frac{B}{2}-R\sin A}=\frac{\cos A+\cos B+\cos C-1-\cos A}{\sin A+\sin C-\sin B-\sin A}$

$\Rightarrow \tan \theta =\frac{\cos B+\cos C-1}{\sin C-\sin B}$

$\Rightarrow \theta ={\tan }^{-1}\left(\frac{\cos B+\cos C-1}{\sin C-\sin B}\right)$