Question Number 214603 by ajfour last updated on 13/Dec/24
Answered by TonyCWX08 last updated on 13/Dec/24
$${Where}\:{is}\:{R}? \\ $$$${Where}\:{is}\:{r}? \\ $$
Commented by ajfour last updated on 13/Dec/24
$${sorry}\:{wrong}\:{question} \\ $$
Answered by mr W last updated on 13/Dec/24
Commented by mr W last updated on 13/Dec/24
$$\left({R}−{r}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} +\left(\mathrm{2}{r}\right)^{\mathrm{2}} −\mathrm{4}{Rr}\:\mathrm{cos}\:\mathrm{30}° \\ $$$$\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right){R}=\mathrm{3}{r} \\ $$$$\Rightarrow\frac{{R}}{{r}}=\frac{\mathrm{3}}{\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}=\frac{\mathrm{3}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{4}}\approx\mathrm{2}.\mathrm{049} \\ $$
Answered by A5T last updated on 13/Dec/24
$$\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }={R}−\sqrt{\left(\mathrm{2}{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\Rightarrow\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}={R}−{r}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{R}^{\mathrm{2}} −\mathrm{2}{rR}={R}^{\mathrm{2}} +\mathrm{3}{r}^{\mathrm{2}} −\mathrm{2}{Rr}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}{Rr}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)=\mathrm{3}{r}^{\mathrm{2}} \Rightarrow\frac{{R}}{{r}}=\frac{\mathrm{3}}{\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}=\frac{\mathrm{3}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{4}} \\ $$
Commented by ajfour last updated on 13/Dec/24
$${Thank}\:{you}!\:{Prompt}\:{ways}! \\ $$
Commented by ajfour last updated on 13/Dec/24
https://youtu.be/z1RwkyYqV34?si=luatRZ1-3shsgmnX