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Question Number 87369 by ajfour last updated on 04/Apr/20

Commented by ajfour last updated on 04/Apr/20

I think, (R/r)=3+2(√2) only, Sir.

$${I}\:{think},\:\frac{{R}}{{r}}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:\:\:\:{only},\:{Sir}. \\ $$

Commented by ajfour last updated on 04/Apr/20

Find R/r .

$${Find}\:{R}/{r}\:. \\ $$

Commented by Prithwish Sen 1 last updated on 04/Apr/20

3±2(√2)

$$\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}} \\ $$

Commented by Prithwish Sen 1 last updated on 04/Apr/20

yes . thank you.

$$\mathrm{yes}\:.\:\mathrm{thank}\:\mathrm{you}. \\ $$

Answered by mr W last updated on 04/Apr/20

Commented by mr W last updated on 04/Apr/20

β+γ=(π/2) cos β=((R−r)/(R+r)) cos γ=((R−r)/(R+r))=cos β ⇒β=γ=(π/4) ((R−r)/(R+r))=((√2)/2) (((R/r)−1)/((R/r)+1))=((√2)/2) ⇒(R/r)=((2+(√2))/(2−(√2)))=3+2(√2)

$$\beta+\gamma=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\beta=\frac{{R}−{r}}{{R}+{r}} \\ $$$$\mathrm{cos}\:\gamma=\frac{{R}−{r}}{{R}+{r}}=\mathrm{cos}\:\beta \\ $$$$\Rightarrow\beta=\gamma=\frac{\pi}{\mathrm{4}} \\ $$$$\frac{{R}−{r}}{{R}+{r}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\frac{\frac{{R}}{{r}}−\mathrm{1}}{\frac{{R}}{{r}}+\mathrm{1}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{R}}{{r}}=\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$

Commented by Prithwish Sen 1 last updated on 04/Apr/20

wow! excellent sir.

$$\mathrm{wow}!\:\mathrm{excellent}\:\mathrm{sir}. \\ $$

Commented by ajfour last updated on 04/Apr/20

Thank you both mrW Sir &Sen Sir.

$${Thank}\:{you}\:{both}\:{mrW}\:{Sir}\:\&{Sen}\:{Sir}. \\ $$

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