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Category: Geometry

Question-218812

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Question Number 218812 by Spillover last updated on 15/Apr/25 Answered by A5T last updated on 16/Apr/25 $$\sqrt{\left(\mathrm{R}−\mathrm{x}\right)^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }=\sqrt{\mathrm{x}^{\mathrm{2}} −\left(\mathrm{x}−\mathrm{h}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\left(\mathrm{R}−\mathrm{2x}\right)\mathrm{R}=\mathrm{h}\left(\mathrm{2x}−\mathrm{h}\right) \\ $$$$\Rightarrow\mathrm{R}^{\mathrm{2}}…

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Question-218676

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Question Number 218676 by Spillover last updated on 14/Apr/25 Answered by mr W last updated on 14/Apr/25 $$\mathrm{cos}\:\theta=\frac{{R}−{r}}{{R}+{r}}=\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\sqrt{\frac{{r}}{{R}+{r}}} \\ $$$${x}=\mathrm{2}{R}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\mathrm{2}{R}\sqrt{\frac{{r}}{{R}+{r}}}\:\:\checkmark \\ $$…

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