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Question Number 211992 by ajfour last updated on 26/Sep/24

Commented by ajfour last updated on 26/Sep/24

$$FindrintermsofRanda.$$

Commented by ajfour last updated on 26/Sep/24

https://youtu.be/xI8mXZpX-qM?si=hInwvqv42gbBwtVx

Answered by mr W last updated on 27/Sep/24

$$\cos 2\alpha =\frac{a}{R}=1-2{\sin }^2\alpha$$$${\sin }^2\alpha =\frac{1-\frac{a}{R}}{2},{\cos }^2\alpha =\frac{1+\frac{a}{R}}{2}$$$$\tan \alpha =\frac{r}{R+\sqrt{{(R-r)}^2-r^2}}$$$${\left[\frac{r}{R+\sqrt{R(R-2r)}}\right]}^2=\frac{1-\frac{a}{R}}{1+\frac{a}{R}}$$$$sayk=\frac{r}{R},\lambda =\frac{a}{R}$$$$\Rightarrow {\left[\frac{k}{1+\sqrt{1-2k}}\right]}^2=\frac{1-\lambda }{1+\lambda }$$

Commented by mr W last updated on 27/Sep/24

$$yes,thanks.$$

Commented by ajfour last updated on 27/Sep/24

$$$$$$\Rightarrow {\left[\frac{k}{1+\sqrt{1-2k}}\right]}^2=\frac{1-\lambda }{1+\lambda }=m^2=\tan {}^2\alpha$$$$k=m+m\sqrt{1-2k}$$$$k^2+\cancel{m^2}-2km=\cancel{m^2}-2m^{2}k$$$$\Rightarrow \frac{k}{2}=\frac{r}{2R}=m-m^2$$$$r=2R\{\sqrt{\frac{1-\lambda }{1+\lambda }}-\left(\frac{1-\lambda }{1+\lambda }\right)\}$$

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