Question Number 218676 by Spillover last updated on 14/Apr/25
Answered by mr W last updated on 14/Apr/25
$$\mathrm{cos}\:\theta=\frac{{R}−{r}}{{R}+{r}}=\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\sqrt{\frac{{r}}{{R}+{r}}} \\ $$$${x}=\mathrm{2}{R}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\mathrm{2}{R}\sqrt{\frac{{r}}{{R}+{r}}}\:\:\checkmark \\ $$
Commented by mr W last updated on 14/Apr/25
Commented by A5T last updated on 14/Apr/25
$$\mathrm{x}=\sqrt{\mathrm{2R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\theta\right)}=\sqrt{\mathrm{2R}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{R}−\mathrm{r}}{\mathrm{R}+\mathrm{r}}\right)} \\ $$$$\Rightarrow\mathrm{x}=\sqrt{\mathrm{2R}^{\mathrm{2}} \left(\frac{\mathrm{2r}}{\mathrm{R}+\mathrm{r}}\right)}=\mathrm{2R}\sqrt{\frac{\mathrm{r}}{\mathrm{R}+\mathrm{r}}} \\ $$
Commented by Spillover last updated on 14/Apr/25
$${great} \\ $$
Commented by Spillover last updated on 14/Apr/25
$${great} \\ $$
Answered by Spillover last updated on 14/Apr/25
