The circumradius R of an isosceles triangle A | vedclass

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Question

(A) Given $R = 4r$. In any triangle,$r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$.Since $A = B$,$C = 180^{\circ} - 2A$. Thus,$r = 4R

Vedclass · Accepted Answer

$8 \cos^2 A - 8 \cos A + 1 = 0$

Vedclass · Answer

(A) Given $R = 4r$. In any triangle,$r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$.Since $A = B$,$C = 180^{\circ} - 2A$. Thus,$r = 4R \sin^2(\frac{A}{2}) \sin(\frac{180^{\circ}-2A}{2}) = 4R \sin^2(\frac{A}{2}) \cos A$.Substituting $R = 4r$,we get $r = 4(4r) \sin^2(\frac{A}{2}) \cos A$,which simplifies to $1 = 16 \sin^2(\frac{A}{2}) \cos A$.Using the identity $2 \sin^2(\frac{A}{2}) = 1 - \cos A$,we have $1 = 8(1 - \cos A) \cos A$.$1 = 8 \cos A - 8 \cos^2 A$.Rearranging gives $8 \cos^2 A - 8 \cos A + 1 = 0$.

The circumradius $R$ of an isosceles triangle $ABC$ is four times its inradius $r$. If $A = B$,then which of the following is true?

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