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Question Number 218676 by Spillover last updated on 14/Apr/25

Answered by mr W last updated on 14/Apr/25

$$\cos \theta =\frac{R-r}{R+r}=1-2{\sin }^2\frac{\theta }{2}$$$$\Rightarrow \sin \frac{\theta }{2}=\sqrt{\frac{r}{R+r}}$$$$x=2R\sin \frac{\theta }{2}=2R\sqrt{\frac{r}{R+r}}✓$$

Commented by mr W last updated on 14/Apr/25

Commented by A5T last updated on 14/Apr/25

$$\mathrm{x}=\sqrt{{2\mathrm{R}}^2(1-\cos \theta )}=\sqrt{{2\mathrm{R}}^2(1-\frac{\mathrm{R}-\mathrm{r}}{\mathrm{R}+\mathrm{r}})}$$$$\Rightarrow \mathrm{x}=\sqrt{{2\mathrm{R}}^2\left(\frac{2\mathrm{r}}{\mathrm{R}+\mathrm{r}}\right)}=2\mathrm{R}\sqrt{\frac{\mathrm{r}}{\mathrm{R}+\mathrm{r}}}$$

Commented by Spillover last updated on 14/Apr/25

$$great$$

Commented by Spillover last updated on 14/Apr/25

$$great$$

Answered by Spillover last updated on 14/Apr/25

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