Question Number 140545 by I want to learn more last updated on 09/May/21
Commented by mr W last updated on 09/May/21
$${AL}={AE}=\mathrm{3} \\ $$
Answered by mr W last updated on 09/May/21
Commented by mr W last updated on 09/May/21
$${AB}=\frac{\mathrm{3}}{\mathrm{cos}\:\alpha} \\ $$$${AF}=\mathrm{3}\:\mathrm{cos}\:\alpha \\ $$$${R}=\frac{{AB}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2cos}\:\alpha} \\ $$$${CF}={R}\mathrm{cos}\:\mathrm{2}\alpha \\ $$$$\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} =\left({r}+{R}\mathrm{cos}\:\mathrm{2}\alpha\right)^{\mathrm{2}} \\ $$$${R}\left({R}−\mathrm{2}{r}\right)=\left({r}+{R}\mathrm{cos}\:\mathrm{2}\alpha\right)^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} +\mathrm{2}{Rr}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\alpha\right)−{R}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\alpha=\mathrm{0} \\ $$$${r}^{\mathrm{2}} +\mathrm{4}{Rr}\mathrm{cos}^{\mathrm{2}} \alpha−{R}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\alpha=\mathrm{0} \\ $$$${r}=−\mathrm{2}{R}\mathrm{cos}^{\mathrm{2}} \:\alpha+{R}\sqrt{\mathrm{4cos}^{\mathrm{4}} \:\alpha+\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\alpha} \\ $$$${r}=−\mathrm{2}{R}\mathrm{cos}^{\mathrm{2}} \:\alpha+\mathrm{2}{R}\mathrm{cos}\:\alpha \\ $$$${r}=\mathrm{2}{R}\mathrm{cos}\:\alpha\left(−\mathrm{cos}\:\alpha+\mathrm{1}\right) \\ $$$${r}=\mathrm{3}\left(−\mathrm{cos}\:\alpha+\mathrm{1}\right) \\ $$$${AL}={AF}+{r}=\mathrm{3}\:\mathrm{cos}\:\alpha+\mathrm{3}\left(−\mathrm{cos}\:\alpha+\mathrm{1}\right)=\mathrm{3} \\ $$
Commented by I want to learn more last updated on 09/May/21
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$