Question Number 208110 by Ismoiljon_008 last updated on 05/Jun/24
Answered by mr W last updated on 05/Jun/24
Commented by mr W last updated on 05/Jun/24
$$\mathrm{cos}\:\alpha=\frac{\mathrm{2}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\mathrm{cot}\:\alpha=\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\beta=\frac{\pi−\alpha}{\mathrm{2}}\:\Rightarrow\mathrm{cot}\:\beta=\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\frac{\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$${AD}=\frac{{x}}{\mathrm{sin}\:\beta} \\ $$$$\frac{{ED}}{\mathrm{sin}\:\left(\alpha−\beta\right)}=\frac{{AD}}{\mathrm{sin}\:\alpha} \\ $$$$\frac{\mathrm{4}}{\mathrm{sin}\:\left(\alpha−\beta\right)}=\frac{{x}}{\mathrm{sin}\:\beta\:\mathrm{sin}\:\alpha} \\ $$$${x}=\frac{\mathrm{4}\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\left(\alpha−\beta\right)}=\frac{\mathrm{4}}{\mathrm{cot}\:\beta−\mathrm{cot}\:\alpha} \\ $$$$\:\:\:=\frac{\mathrm{4}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}}=\mathrm{8}\sqrt{\mathrm{2}}\:\checkmark \\ $$
Commented by Tawa11 last updated on 21/Jun/24
$$\mathrm{weldone}\:\mathrm{sir} \\ $$