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Question Number 164808 by Mathematification last updated on 22/Jan/22

Answered by mr W last updated on 22/Jan/22

Commented by mr W last updated on 22/Jan/22

$$1-2{\sin }^{2}15°=\cos 30°=\frac{\sqrt{3}}{2}$$$${\sin }^{2}15°=\frac{2-\sqrt{3}}{4}$$$$\sin 15°=\frac{\sqrt{2-\sqrt{3}}}{2}$$$$$$$$AB=\frac{r}{\sin 15°}$$$$OB=R-r$$$$OA=R$$$${(R-r)}^2=R^2+{\left(\frac{r}{\sin 15°}\right)}^2-2R\left(\frac{r}{\sin 15°}\right)\cos 30°$$$$\Rightarrow \frac{r}{R}=\frac{\frac{\sqrt{3}}{\sin 15°}-2}{\frac{1}{{\sin }^{2}15°}-1}$$$$=\frac{\frac{2\sqrt{3}}{\sqrt{2-\sqrt{3}}}-2}{\frac{4}{2-\sqrt{3}}-1}=\frac{2(\sqrt{3}-2+\sqrt{3(2-\sqrt{3})})}{2+\sqrt{3}}$$$$withR=\frac{1}{2}$$$$\Rightarrow r=\frac{\sqrt{3}-2+\sqrt{3(2-\sqrt{3})}}{2+\sqrt{3}}\approx 0.168$$

Commented by Tawa11 last updated on 22/Jan/22

$$\mathrm{Great}\mathrm{sir}.$$

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