Question-218734

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 ⇒a=((5R)/4) cos θ=−((a^2 +(a−r)^2 −(R+r)^2 )/(2a(a−r)))=((2R−a)/a) 6r=4a−R=5R−R=4R ⇒r=((2R)/3) ⇒(R/r)=(3/2) ✓

$${a}^{\mathrm{2}} ={R}^{\mathrm{2}} +\left(\mathrm{2}{R}−{a}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{5}{R}}{\mathrm{4}} \\ $$$$\mathrm{cos}\:\theta=−\frac{{a}^{\mathrm{2}} +\left({a}−{r}\right)^{\mathrm{2}} −\left({R}+{r}\right)^{\mathrm{2}} }{\mathrm{2}{a}\left({a}−{r}\right)}=\frac{\mathrm{2}{R}−{a}}{{a}} \\ $$$$\mathrm{6}{r}=\mathrm{4}{a}−{R}=\mathrm{5}{R}−{R}=\mathrm{4}{R} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}{R}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{{R}}{{r}}=\frac{\mathrm{3}}{\mathrm{2}}\:\checkmark \\ $$

Commented by Spillover last updated on 15/Apr/25