Question Number 175449 by cherokeesay last updated on 30/Aug/22
Commented by som(math1967) last updated on 31/Aug/22
$${is}\:{r}=\mathrm{7}−\mathrm{3}\sqrt{\mathrm{5}}\approx\mathrm{0}.\mathrm{29}\:? \\ $$$$\frac{{S}_{\mathrm{2}} }{{S}_{\mathrm{1}} }\:=\frac{\pi\left(.\mathrm{29}\right)^{\mathrm{2}} }{\pi}×\mathrm{100}=\mathrm{8}.\mathrm{41\%}\:? \\ $$
Answered by mr W last updated on 03/Sep/22
Commented by mr W last updated on 03/Sep/22
$$\mathrm{tan}\:\mathrm{2}\theta=\frac{\mathrm{2}}{\mathrm{1}}=\mathrm{2} \\ $$$$\frac{\mathrm{2tan}\:\theta}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\theta}=\mathrm{2} \\ $$$$\mathrm{tan}\:\theta=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}+\mathrm{1}}=\frac{\mathrm{1}}{\varphi} \\ $$$${CD}=\frac{{r}}{\mathrm{tan}\:\theta}=\varphi{r} \\ $$$${BF}={DE}=\mathrm{2}−\varphi{r} \\ $$$${AF}=\mathrm{2}−{r} \\ $$$${AB}=\mathrm{2}+{r} \\ $$$$\left(\mathrm{2}+{r}\right)^{\mathrm{2}} =\left(\mathrm{2}−{r}\right)^{\mathrm{2}} +\left(\mathrm{2}−\varphi{r}\right)^{\mathrm{2}} \\ $$$$\mathrm{8}{r}=\mathrm{4}−\mathrm{4}\varphi{r}+\varphi^{\mathrm{2}} {r}^{\mathrm{2}} \\ $$$$\mathrm{0}=\mathrm{4}−\mathrm{4}\left(\varphi+\mathrm{2}\right){r}+\varphi^{\mathrm{2}} {r}^{\mathrm{2}} \\ $$$${r}=\frac{\mathrm{2}\left(\varphi+\mathrm{2}\right)−\mathrm{4}\sqrt{\varphi+\mathrm{1}}}{\varphi^{\mathrm{2}} }=\frac{\mathrm{2}\left(\mathrm{2}−\varphi\right)}{\varphi+\mathrm{1}}=\mathrm{7}−\mathrm{3}\sqrt{\mathrm{5}} \\ $$
Commented by Tawa11 last updated on 15/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$