Question

The circumradius of an isosceles triangle $ABC$ if four times as that of inradius of triangle. If $\angle A=\angle B,$ then

• A. $8{\cos }^{2}A-8\cos A+1=0$
• B. $4{\cos }^{2}A-10\cos A+1=0$
• C. ${\cos }^{2}A-\cos A-3=0$
• D. ${\cos }^{2}A-\cos A-8=0$

Answer 1

The correct option is A $8{\cos }^{2}A-8\cos A+1=0$

$4r=R\Rightarrow 4\left(4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\right)=R$
$\Rightarrow 16{\sin }^2\frac{A}{2}\sin \frac{180°-2A}{2}=1$
$\Rightarrow 16\frac{1-\cos A}{2}\cos A=1$
$\Rightarrow 8{\cos }^{2}A-8\cos A+1=0$