Question Number 166676 by ajfour last updated on 24/Feb/22
Answered by ajfour last updated on 24/Feb/22
$${R}\mathrm{sin}\:\theta+\mathrm{2}{r}={R} \\ $$$$\mathrm{sin}\:\theta=\frac{{r}}{{R}−{r}} \\ $$$$\Rightarrow\:\:{Rr}=\left({R}−\mathrm{2}{r}\right)\left({R}−{r}\right) \\ $$$$\Rightarrow\:\:{R}^{\mathrm{2}} −\mathrm{4}{rR}+\mathrm{2}{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\frac{{R}}{{r}}=\mathrm{2}+\sqrt{\mathrm{2}} \\ $$