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Question Number 187482 by ajfour last updated on 17/Feb/23

Commented by ajfour last updated on 17/Feb/23

$$If{cone}^{\prime }s\frac{R}{h}=m,greenball$$$$radius=a.Findbtheradiiof$$$$upperballs\left(equal\right),allincontact$$$$thewayshown.$$

Answered by a.lgnaoui last updated on 18/Feb/23

$$\sin \theta =\frac{a}{c}c={OO}_1=\frac{OH}{h}$$$$\tan \theta =\frac{R}{h}=m\Rightarrow c=\frac{am}{\sqrt{1+m^2}}$$$$Cone\sin \beta =\frac{R\frac{\sqrt{3}}{2}}{R-r\sin \alpha }=\cos \alpha$$$$\sqrt{1+m^2}=\frac{R\sqrt{3}}{2R-2r\frac{m}{\sqrt{1+m^2}}}$$$$2R\sqrt{1+m^2}-2mr=R\sqrt{3}$$$$$$$$\mathit{r}=\frac{\mathit{R}(2\sqrt{1+{\mathit{m}}^2}-\sqrt{3})}{2\mathit{m}}.$$

Commented by a.lgnaoui last updated on 18/Feb/23

$$r=b$$

Commented by a.lgnaoui last updated on 18/Feb/23

Answered by mr W last updated on 18/Feb/23

Commented by mr W last updated on 19/Feb/23

$$\tan \theta =\frac{R}{H}=m$$$$A^{\prime }B=\frac{2\sqrt{3}b}{3}$$$$OA=\frac{a}{\sin \theta }$$$${AA}^{\prime }=\sqrt{{(a+b)}^2-{\left(\frac{2\sqrt{3}b}{3}\right)}^2}=\sqrt{a^2+2ab-\frac{b^2}{3}}$$$${OA}^{\prime }=\sqrt{a^2+2ab-\frac{b^2}{3}}+\frac{a}{\sin \theta }$$$$OC=\frac{a}{\tan \theta }$$$${CC}^{\prime }=\sqrt{{(a+b)}^2-{(b-a)}^2}=2\sqrt{ab}$$$${OC}^{\prime }=2\sqrt{ab}+\frac{a}{\tan \theta }$$$${(\sqrt{a^2+2ab-\frac{b^2}{3}}+\frac{a}{\sin \theta })}^2+{\left(\frac{2\sqrt{3}b}{3}\right)}^2={(2\sqrt{ab}+\frac{a}{\tan \theta })}^2+b^2$$$$\sqrt{a^2+2ab-\frac{b^2}{3}}=(b-a)\sin \theta +2\sqrt{ab}\cos \theta$$$$b^2(\frac{4}{3}-{\cos }^2\theta )-a^2{\cos }^2\theta -2ab(2-{3\cos }^2\theta )+2(b-a)\sqrt{ab}\sin 2\theta =0$$$$let\lambda =\frac{b}{a}$$$$\Rightarrow (5-3\cos 2\theta ){\lambda }^2-6(1-3\cos 2\theta )\lambda +12\sin 2\theta (\lambda -1)\sqrt{\lambda }-3(1+\cos 2\theta )=0$$$$$$$$examples:$$$$\theta =30°\Rightarrow \frac{b}{a}\approx 0.8974$$$$\theta ={\sin }^{-1}\frac{1}{\sqrt{3}}\approx 35.264°\Rightarrow \frac{b}{a}=1$$$$\theta =60°\Rightarrow \frac{b}{a}\approx 1.6445$$

Commented by a.lgnaoui last updated on 18/Feb/23

$$coupetransversale$$

Commented by ajfour last updated on 19/Feb/23

Thanks Sir , correct answers.

Answered by ajfour last updated on 18/Feb/23

Commented by ajfour last updated on 18/Feb/23

$$SeesolutionalsoinQ.187535$$

Commented by a.lgnaoui last updated on 18/Feb/23

$$Mercipourexplicationpar$$$$graphe.$$

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