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Question Number 218734 by Spillover last updated on 14/Apr/25

Answered by mr W last updated on 15/Apr/25

Commented by mr W last updated on 15/Apr/25

$$a^2=R^2+{(2R-a)}^2$$$$\Rightarrow a=\frac{5R}{4}$$$$\cos \theta =-\frac{a^2+{(a-r)}^2-{(R+r)}^2}{2a(a-r)}=\frac{2R-a}{a}$$$$6r=4a-R=5R-R=4R$$$$\Rightarrow r=\frac{2R}{3}$$$$\Rightarrow \frac{R}{r}=\frac{3}{2}✓$$

Commented by Spillover last updated on 15/Apr/25

$$correct.thanks$$

Answered by Spillover last updated on 15/Apr/25

Answered by Spillover last updated on 15/Apr/25

Answered by Spillover last updated on 15/Apr/25

Answered by Spillover last updated on 15/Apr/25

Answered by Spillover last updated on 15/Apr/25

Answered by Spillover last updated on 15/Apr/25

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