Question Number 175149 by cherokeesay last updated on 20/Aug/22
Answered by som(math1967) last updated on 21/Aug/22
Commented by som(math1967) last updated on 21/Aug/22
$${AB}=\mathrm{2}{R}\:\:,{BC}={R} \\ $$$$\therefore\:{AC}=\sqrt{\mathrm{4}{R}^{\mathrm{2}} +{R}^{\mathrm{2}} }=\sqrt{\mathrm{5}}{R} \\ $$$${AE}={AB}−{BE}=\mathrm{2}{R}−{r} \\ $$$$\:\bigtriangleup\boldsymbol{{ABC}}\sim\bigtriangleup\boldsymbol{{ADE}} \\ $$$$\therefore\:\frac{\mathrm{2}{R}−{r}}{\:\sqrt{\mathrm{5}}{R}}=\frac{{r}}{{R}} \\ $$$$\Rightarrow\mathrm{2}{R}−{r}=\sqrt{\mathrm{5}}{r} \\ $$$$\Rightarrow\mathrm{2}{R}=\left(\sqrt{\mathrm{5}}+\mathrm{1}\right){r} \\ $$$$\:\therefore\frac{{R}}{{r}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}=\Phi \\ $$
Commented by cherokeesay last updated on 21/Aug/22
$${Nice} \\ $$$${thank}\:{you}\:! \\ $$
Commented by Tawa11 last updated on 21/Aug/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$