Question-219116

Question Number 219116 by Spillover last updated on 19/Apr/25

Answered by A5T last updated on 20/Apr/25

tan30°=(r/x)⇒x=r(√3) k=sin15°=(R/((2r+2r(√3))−R)) ⇒R+Rk=kr(2+2(√3))⇒(r/R)=((1+k)/(k(2+2(√3)))) sin30°=2sin15°(√(1−sin^2 15°)) ⇒k=sin15°=((√(2−(√3)))/2)=(((√3)−1)/(2(√2))) ⇒(r/R)=((2(√2)+(√3)−1)/( 2((√3)−1)((√3)+1)))=((2(√2)+(√3)−1)/4)

$$\mathrm{tan30}°=\frac{\mathrm{r}}{\mathrm{x}}\Rightarrow\mathrm{x}=\mathrm{r}\sqrt{\mathrm{3}} \\ $$$$\mathrm{k}=\mathrm{sin15}°=\frac{\mathrm{R}}{\left(\mathrm{2r}+\mathrm{2r}\sqrt{\mathrm{3}}\right)−\mathrm{R}} \\ $$$$\Rightarrow\mathrm{R}+\mathrm{Rk}=\mathrm{kr}\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}\right)\Rightarrow\frac{\mathrm{r}}{\mathrm{R}}=\frac{\mathrm{1}+\mathrm{k}}{\mathrm{k}\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}\right)} \\ $$$$\mathrm{sin30}°=\mathrm{2sin15}°\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{15}°} \\ $$$$\Rightarrow\mathrm{k}=\mathrm{sin15}°=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\frac{\mathrm{r}}{\mathrm{R}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}−\mathrm{1}}{\:\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}=\frac{\mathrm{2}\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{4}} \\ $$

Commented by Spillover last updated on 20/Apr/25