Question Number 87369 by ajfour last updated on 04/Apr/20
Commented by ajfour last updated on 04/Apr/20
$${I}\:{think},\:\frac{{R}}{{r}}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:\:\:\:{only},\:{Sir}. \\ $$
Commented by ajfour last updated on 04/Apr/20
$${Find}\:{R}/{r}\:. \\ $$
Commented by Prithwish Sen 1 last updated on 04/Apr/20
$$\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}} \\ $$
Commented by Prithwish Sen 1 last updated on 04/Apr/20
$$\mathrm{yes}\:.\:\mathrm{thank}\:\mathrm{you}. \\ $$
Answered by mr W last updated on 04/Apr/20
Commented by mr W last updated on 04/Apr/20
$$\beta+\gamma=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\beta=\frac{{R}−{r}}{{R}+{r}} \\ $$$$\mathrm{cos}\:\gamma=\frac{{R}−{r}}{{R}+{r}}=\mathrm{cos}\:\beta \\ $$$$\Rightarrow\beta=\gamma=\frac{\pi}{\mathrm{4}} \\ $$$$\frac{{R}−{r}}{{R}+{r}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\frac{\frac{{R}}{{r}}−\mathrm{1}}{\frac{{R}}{{r}}+\mathrm{1}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{R}}{{r}}=\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$
Commented by Prithwish Sen 1 last updated on 04/Apr/20
$$\mathrm{wow}!\:\mathrm{excellent}\:\mathrm{sir}. \\ $$
Commented by ajfour last updated on 04/Apr/20
$${Thank}\:{you}\:{both}\:{mrW}\:{Sir}\:\&{Sen}\:{Sir}. \\ $$