Question-141291

Question Number 141291 by mr W last updated on 17/May/21

Commented by mr W last updated on 17/May/21

OB=(√((R−r)^2 −r^2 ))=(√(R^2 −2Rr)) AB=R+(√(R^2 −2Rr))=(r/(tan 22.5°)) BC=R−(√(R^2 −2Rr))=1 R−1=(√(R^2 −2Rr)) R^2 −2R+1=R^2 −2Rr 1=2(1−r)R 2R=(1/(1−r)) 2R=1+(r/(tan 22.5°)) (1/(1−r))=1+(r/(tan 22.5°)) (r/(tan 22.5°))=(1/(tan 22.5°))−1 AE=AB=(1/(tan 22.5°))−1=1+(√2)−1=(√2)

$${OB}=\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }=\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rr}} \\ $$$${AB}={R}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rr}}=\frac{{r}}{\mathrm{tan}\:\mathrm{22}.\mathrm{5}°} \\ $$$${BC}={R}−\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rr}}=\mathrm{1} \\ $$$${R}−\mathrm{1}=\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rr}} \\ $$$${R}^{\mathrm{2}} −\mathrm{2}{R}+\mathrm{1}={R}^{\mathrm{2}} −\mathrm{2}{Rr} \\ $$$$\mathrm{1}=\mathrm{2}\left(\mathrm{1}−{r}\right){R} \\ $$$$\mathrm{2}{R}=\frac{\mathrm{1}}{\mathrm{1}−{r}} \\ $$$$\mathrm{2}{R}=\mathrm{1}+\frac{{r}}{\mathrm{tan}\:\mathrm{22}.\mathrm{5}°} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{r}}=\mathrm{1}+\frac{{r}}{\mathrm{tan}\:\mathrm{22}.\mathrm{5}°} \\ $$$$\frac{{r}}{\mathrm{tan}\:\mathrm{22}.\mathrm{5}°}=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{22}.\mathrm{5}°}−\mathrm{1} \\ $$$${AE}={AB}=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{22}.\mathrm{5}°}−\mathrm{1}=\mathrm{1}+\sqrt{\mathrm{2}}−\mathrm{1}=\sqrt{\mathrm{2}} \\ $$

Commented by mnjuly1970 last updated on 17/May/21

$${grateful}\:{mr}\:\:{W}\:…{thanks}\:{alot} \\ $$

Commented by BHOOPENDRA last updated on 17/May/21

$${great}\:{sir} \\ $$

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Question-141291

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