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Question Number 87285 by ajfour last updated on 03/Apr/20

Commented by ajfour last updated on 03/Apr/20

$$IfregionsAandBhaveequal$$$$areas,determineR/r.$$

Answered by mr W last updated on 03/Apr/20

$$\sin \frac{\alpha }{2}=\frac{r}{R}=\lambda$$$$\Rightarrow \alpha =2{\sin }^{-1}\lambda$$$$\sin \alpha =2\sin \frac{\alpha }{2}\sqrt{1-{\sin }^2\frac{\alpha }{2}}=2\lambda \sqrt{1-{\lambda }^2}$$$$\frac{R^2}{2}(2{\sin }^{-1}\lambda -2\lambda \sqrt{1-{\lambda }^2})=\frac{\pi r^2}{4}$$$$\Rightarrow {\sin }^{-1}\lambda =\lambda (\frac{\pi \lambda }{4}+\sqrt{1-{\lambda }^2})$$$$\Rightarrow \lambda =\frac{r}{R}\approx 0.84945$$

Commented by ajfour last updated on 03/Apr/20

$$ThanksSir,Igotthesame,$$$$\frac{R}{r}\approx 1.1772265$$

Commented by Ar Brandon last updated on 03/Apr/20

$$PleasemayIknowhowyoumanaged\mathit{\lambda }?Iarrived$$$$at{\sin }^{-1}\lambda =\lambda (\frac{\pi \lambda }{4}+\sqrt{1-{\lambda }^2})but{didn}^{\prime }tknowwhattodo$$$$with\lambda .Howdidyoudothat?$$

Commented by ajfour last updated on 03/Apr/20

$$calculatorSir.$$

Commented by mr W last updated on 03/Apr/20

$$itisnotpossibletosolvetheequation$$$$exactly.usuallyiusetheappGrapher$$$$togettheapproximation.$$

Commented by Ar Brandon last updated on 03/Apr/20

$$Oh!Isee.AndIwasstrugglingtogettheanswerdirectly.$$$$haha!$$

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