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Question Number 116466 by mr W last updated on 04/Oct/20

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

$$R=radiusofbigcircles$$$$r=radiusofsmallcircles$$$$let\lambda =\frac{r}{R}$$$$\cos \alpha =\frac{R}{R+r}=\frac{1}{1+\lambda }$$$$\sin \frac{\beta }{2}=\frac{r}{R+r}=\frac{\lambda }{1+\lambda }$$$$\alpha +\frac{\beta }{2}=30°$$$$\sin (\alpha +\frac{\beta }{2})=\frac{1}{2}$$$$\sin \alpha \cos \frac{\beta }{2}+\cos \alpha \sin \frac{\beta }{2}=\frac{1}{2}$$$$\frac{\sqrt{{(1+\lambda )}^2-1}}{1+\lambda }\times \frac{\sqrt{{(1+\lambda )}^2-{\lambda }^2}}{1+\lambda }+\frac{1}{1+\lambda }\times \frac{\lambda }{1+\lambda }=\frac{1}{2}$$$$\frac{\sqrt{\lambda (2+\lambda )(2\lambda +1)}}{{(1+\lambda )}^2}+\frac{\lambda }{{(1+\lambda )}^2}=\frac{1}{2}$$$$2\sqrt{\lambda (2\lambda +1)(\lambda +2)}={(1+\lambda )}^2-2\lambda$$$$4\lambda (2\lambda +1)(\lambda +2)={({\lambda }^2+1)}^2$$$$8{\lambda }^3+20{\lambda }^2+8\lambda ={\lambda }^4+2{\lambda }^2+1$$$$\Rightarrow {\lambda }^4-8{\lambda }^3-18{\lambda }^2-8\lambda +1=0$$$$\Rightarrow {(\lambda +1)}^2({\lambda }^2-10\lambda +1)=0$$$$\Rightarrow \lambda =\frac{r}{R}=5-2\sqrt{6}\approx 0.101$$

Commented by I want to learn more last updated on 04/Oct/20

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{Happy}\mathrm{for}\mathrm{this}.$$

Commented by I want to learn more last updated on 04/Oct/20

$$\mathrm{sir},\mathrm{do}\mathrm{you}\mathrm{write}\mathrm{a}\mathrm{maths}\mathrm{book}??$$

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

$$no$$

Commented by I want to learn more last updated on 04/Oct/20

$$\mathrm{Okay}\mathrm{sir}$$

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