Question

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

• A. $R=\frac{1}{2}b\csc \alpha$
• B. $△=2b^2\sin 2\alpha$
• 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)

The correct options are
A $R=\frac{1}{2}b\csc \alpha$
C $r=\frac{b\sin 2\alpha }{2(1+\cos \alpha )}$
D $OI=\left|\frac{b\cos \left(\frac{3\alpha }{2}\right)}{2\sin \alpha \cos \left(\frac{\alpha }{2}\right)}\right|$

$R=\frac{b}{2\sin \alpha }=\frac{b}{2}\csc \alpha$
$△=\frac{1}{2}{b}^2\sin (180-2\alpha )=\frac{1}{2}{b}^2\sin 2\alpha$
and $r=4R\sin \frac{\alpha }{2}\sin \frac{\alpha }{2}\sin (90-\alpha )$
Substitute $R=\frac{1}{2}{b}^2\sin 2\alpha$ in above equation
$r=4\left(\frac{b}{2\sin \alpha }\right){\sin }^2\frac{\alpha }{2}\cos \alpha$
Multiply and divide both sides by $\sin \alpha$ we get
$=4\left(\frac{b}{2\sin \alpha }\right){\sin }^2\frac{\alpha }{2}\cos \alpha \times \frac{\sin \alpha }{\sin \alpha }$
$=\frac{b}{\sin \alpha }{\sin }^2\frac{\alpha }{2}\frac{2\sin \alpha \cos \alpha }{\sin alpha}$
$=\frac{b}{\sin \alpha }{\sin }^2\frac{\alpha }{2}\frac{\sin 2\alpha }{\sin \alpha }$
$=\frac{b{\sin }^2\frac{\alpha }{2}\sin 2\alpha }{{\sin }^2\alpha }$
$=\frac{b(1-\cos \alpha )\sin 2\alpha }{2(1-{\cos }^2\alpha )}$
$=\frac{b(1-\cos \alpha )\sin 2\alpha }{2(1-\cos \alpha )(1+\cos \alpha )}$
$=\frac{b\sin 2\alpha }{2(1+\cos \alpha )}$
and $OI=\sqrt{(R^2-2Rr)}$
$=R\sqrt{(1-\frac{2r}{R})}$
$=R\sqrt{1-4\cos \alpha +4{\cos }^2\alpha }$
$=R\sqrt{{\left(2\cos \alpha -1\right)}^2}$
$=R(2\cos \alpha -1)$
$=R[2(2{\cos }^2\frac{\alpha }{2}-1)-1]$
$=R[4{\cos }^2\frac{\alpha }{2}-3]$
Multiply and divide by $\cos \frac{\alpha }{2}$ we get
$=\frac{R[4{\cos }^3\frac{\alpha }{2}-3\cos \frac{\alpha }{2}]}{\cos \frac{\alpha }{2}}$
$=\frac{R\cos \frac{3\alpha }{2}}{\cos \frac{\alpha }{2}}$
$=\left|\frac{\frac{b}{2\sin \alpha }\cos \frac{3\alpha }{2}}{\cos \frac{\alpha }{2}}\right|$
$=\left|\frac{b\cos \frac{3\alpha }{2}}{2\sin \alpha \cos \frac{\alpha }{2}}\right|$