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Question Number 218456 by Spillover last updated on 10/Apr/25

Answered by vnm last updated on 10/Apr/25

$$\mathrm{d}\mathrm{is}\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{centers}$$$$\phi \mathrm{is}\mathcal{L}\cancel{\underbrace{}}$$

Commented by Spillover last updated on 10/Apr/25

$$correct$$

Answered by mr W last updated on 10/Apr/25

Commented by mr W last updated on 10/Apr/25

$$R=\frac{2+8}{2}=5$$$${(R-r)}^2-r^2=3^2$$$$\Rightarrow r=\frac{8}{5}$$$$\cos 2\theta =\frac{r}{R-r}=\frac{8}{17}$$$$2{\cos }^2\theta -1=\frac{8}{17}$$$$\Rightarrow \cos \theta =\frac{5}{\sqrt{34}}$$$$?=x=2r\cos \theta$$$$=2\times \frac{8}{5}\times \frac{5}{\sqrt{34}}=\frac{8\sqrt{34}}{17}\approx 2.744$$

Answered by Spillover last updated on 10/Apr/25

Answered by Spillover last updated on 10/Apr/25

Answered by Spillover last updated on 10/Apr/25

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