Mathematics - Straight Lines Question with Solution | TestHub

Let ABC be the triangle in which $\mathrm{AB}=\mathrm{AC}$. Let $I,P$ respectively be the incentre and the orthocentre of the triangle. $AI=r\mathrm{cosec}\frac{A}{2},AP=2R\cos A$. Since $P$ lies on the in-circle, $AI=AP+r$ : $r\mathrm{cosec}\frac{A}{2}=2R\cos A+r$. Using $r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$ with $B=C=\frac{\pi -A}{2}$ and ${\sin }^2\frac{\pi -A}{4}=\frac{1-\sin \frac{A}{2}}{2}$ : $2\left(1-\sin \frac{A}{2}\right)=2\cos A+2\sin \frac{A}{2}\left(1-\sin \frac{A}{2}\right)\Rightarrow 6{\sin }^2\frac{A}{2}-4\sin \frac{A}{2}=0\Rightarrow \sin \frac{A}{2}=\frac{2}{3}$. Base angle $B=\frac{\pi -A}{2}$, so $\cos B=\sin \frac{A}{2}=\frac{2}{3}$ (A true) and $\tan B=\frac{\sqrt{5}}{2}$ (C true). $\cos A=1-2\cdot \frac{4}{9}=\frac{1}{9}>0$, so all angles are acute (B true). $\frac{r}{R}=4\sin \frac{A}{2}{\sin }^2\frac{\pi -A}{4}=\frac{8}{3}\cdot \frac{1}{6}=\frac{4}{9}$, so $S{I}^2=R^2-2Rr\Rightarrow \frac{SI}{R}=\sqrt{1-\frac{8}{9}}=\frac{1}{3}$ (D true).