Question Number 43338 by ajfour last updated on 09/Sep/18
Commented by ajfour last updated on 10/Sep/18
$${Find}\:\boldsymbol{{r}}\:{in}\:{terms}\:{of}\:\boldsymbol{{R}} \\ $$$$\left({in}\:{the}\:{case}\:{of}\:{circles}\right). \\ $$
Answered by ajfour last updated on 10/Sep/18
$$ \\ $$$$\left({R}−{r}\right)\mathrm{cos}\:\frac{\pi}{\mathrm{6}}\:=\:\frac{{R}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:{r}\:=\:{R}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:. \\ $$