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Question Number 140545 by I want to learn more last updated on 09/May/21

Commented by mr W last updated on 09/May/21

$$AL=AE=3$$

Answered by mr W last updated on 09/May/21

Commented by mr W last updated on 09/May/21

$$AB=\frac{3}{\cos \alpha }$$$$AF=3\cos \alpha$$$$R=\frac{AB}{2}=\frac{3}{2\cos \alpha }$$$$CF=R\cos 2\alpha$$$${(R-r)}^2-r^2={(r+R\cos 2\alpha )}^2$$$$R(R-2r)={(r+R\cos 2\alpha )}^2$$$$r^2+2Rr(1+\cos 2\alpha )-R^2{\sin }^{2}2\alpha =0$$$$r^2+4Rr{\cos }^2\alpha -R^2{\sin }^{2}2\alpha =0$$$$r=-2R{\cos }^2\alpha +R\sqrt{{4\cos }^4\alpha +{\sin }^{2}2\alpha }$$$$r=-2R{\cos }^2\alpha +2R\cos \alpha$$$$r=2R\cos \alpha (-\cos \alpha +1)$$$$r=3(-\cos \alpha +1)$$$$AL=AF+r=3\cos \alpha +3(-\cos \alpha +1)=3$$

Commented by I want to learn more last updated on 09/May/21

$$\mathrm{I}\mathrm{really}\mathrm{appreciate}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

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