Question Number 222747 by fantastic last updated on 06/Jul/25
Commented by fantastic last updated on 06/Jul/25
$${r}\:{in}\:{terms}\:{of}\:{R}\:{and}\:\alpha \\ $$
Answered by mr W last updated on 06/Jul/25
Commented by mr W last updated on 07/Jul/25
$$\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }={r}\:\mathrm{cot}\:\frac{\alpha}{\mathrm{2}}−{R} \\ $$$${R}^{\mathrm{2}} −\mathrm{2}{Rr}={r}^{\mathrm{2}} \mathrm{cot}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}−\mathrm{2}{rR}\:\mathrm{cot}\:\frac{\alpha}{\mathrm{2}}+{R}^{\mathrm{2}} \\ $$$${r}=\frac{\mathrm{2}{R}\left(\mathrm{cot}\:\frac{\alpha}{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{cot}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}} \\ $$$$\:\:=\mathrm{2}{R}\left(\mathrm{1}−\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\right)\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\:\checkmark \\ $$
Commented by fantastic last updated on 06/Jul/25
$${thanks}\:{I}\:{got}\:{the}\:{same} \\ $$