Question

Given an isosceles triangle with equal sides of length $b$, base angle $\alpha <\frac{\pi }{4}$ and $R,r$ are the radii of circumcircle and incircle and $O,I$ are the centres of circumcircle and incircle respectively. Then:

• A. $R=\frac{1}{2}b\text{cosec}\alpha$
• B. $\Delta =2b^2\sin 2\alpha$
• C. $r=\frac{b\sin 2\alpha }{2(1+\cos \alpha )}$
• D. $OI=\left|\frac{b\cos \left(\frac{3\alpha }{2}\right)}{2\sin \alpha \cos \left(\frac{\alpha }{2}\right)}\right|$

Answer 1

Answer: (A), (C), (D)

$OI=\left|\frac{b\cos \left(\frac{3\alpha }{2}\right)}{2\sin \alpha \cos \left(\frac{\alpha }{2}\right)}\right|$

In triangle $OAB,\frac{R}{\sin \frac{A}{2}}=\frac{b}{\sin 2\alpha }$$\Rightarrow R=\frac{b}{2\sin \alpha }=\frac{b}{2}\text{cosec}\alpha (\because A=\pi -2\alpha )$
$\Delta =\frac{1}{2}{b}^2\sin A=\frac{1}{2}{b}^2\sin ({180}^{\circ }-2\alpha )=\frac{1}{2}{b}^2\sin 2\alpha$
Also, we have $r=4R\sin \frac{\alpha }{2}\sin \frac{\alpha }{2}\sin ({90}^{\circ }-\alpha )$
$\Rightarrow r=\frac{2b}{\sin \alpha }{\sin }^2\frac{\alpha }{2}\cos \alpha$
$\Rightarrow r=\frac{b(1-\cos \alpha )\sin 2\alpha }{2(1-{\cos }^2\alpha )}$
$\Rightarrow r=\frac{b\sin 2\alpha }{2(1+\cos \alpha )}$
Now, we have $OI=\sqrt{R^2-2Rr}=R\sqrt{(1-\frac{2r}{R})}$
$=R\sqrt{1-4\cos \alpha +4{\cos }^2\alpha }=R(2\cos \alpha -1)$
$=R\{2(2{\cos }^2\frac{\alpha }{2}-1)-1\}=R(4{\cos }^2\frac{\alpha }{2}-3)$
$\Rightarrow OI=\frac{R(4{\cos }^3\frac{\alpha }{2}-3\cos \frac{\alpha }{2})}{\cos \frac{\alpha }{2}}=\frac{R\cos \frac{3\alpha }{2}}{\cos \frac{\alpha }{2}}$
$\Rightarrow OI=\left|\frac{b\cos \left(\frac{3\alpha }{2}\right)}{2\sin \alpha \cos \left(\frac{\alpha }{2}\right)}\right|$