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Question Number 183631 by a.lgnaoui last updated on 27/Dec/22

$$determiner\mathit{r}?$$$$AB=6AE=5$$

Commented by a.lgnaoui last updated on 27/Dec/22

Answered by mr W last updated on 28/Dec/22

Commented by mr W last updated on 28/Dec/22

$${BE}^2=5^2+6^2-2\times 5\times 6\cos 30°$$$$\Rightarrow BE=\sqrt{61-30\sqrt{3}}$$$$R=\frac{\sqrt{61-30\sqrt{3}}}{2\sin 30°}=\sqrt{61-30\sqrt{3}}$$$$\cos \alpha =\frac{6}{2\sqrt{61-30\sqrt{3}}}=\frac{3}{\sqrt{61-30\sqrt{3}}}$$$$\Rightarrow \sin \alpha =\frac{\sqrt{52-30\sqrt{3}}}{\sqrt{61-30\sqrt{3}}}$$$$AF=\frac{r}{\sin 15°}=\frac{4r}{\sqrt{6}-\sqrt{2}}=(\sqrt{6}+\sqrt{2})r$$$$\cos (\alpha +15°)=\frac{R^2+r^2{(\sqrt{6}+\sqrt{2})}^2-{(R-r)}^2}{2Rr(\sqrt{6}+\sqrt{2})}$$$$\cos \alpha \cos 15°-\sin \alpha \sin 15°=\frac{(7+4\sqrt{3})r+2R}{2R(\sqrt{6}+\sqrt{2})}$$$$\frac{3}{\sqrt{61-30\sqrt{3}}}\times \frac{\sqrt{6}+\sqrt{2}}{4}-\frac{\sqrt{52-30\sqrt{3}}}{\sqrt{61-30\sqrt{3}}}\times \frac{\sqrt{6}-\sqrt{2}}{4}=\frac{(7+4\sqrt{3})r+2R}{2R(\sqrt{6}+\sqrt{2})}$$$$3(2+\sqrt{3})-\sqrt{52-30\sqrt{3}}=\frac{(7+4\sqrt{3})r}{2}+R$$$$\Rightarrow r=2(7-4\sqrt{3})(6+3\sqrt{3}-\sqrt{52-30\sqrt{3}}-\sqrt{61-30\sqrt{3}})\approx 1.1478$$

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