Question Number 205892 by cortano12 last updated on 02/Apr/24
Answered by A5T last updated on 02/Apr/24
Commented by A5T last updated on 02/Apr/24
$${AC}={ytan}\left(\mathrm{3}\alpha\right);{x}={ytan}\alpha;{z}=\frac{{y}}{{cos}\alpha};{c}={y}\left({tan}\mathrm{3}\alpha−{tan}\alpha\right) \\ $$$${EC}=\frac{{y}}{{cos}\mathrm{3}\alpha} \\ $$$$\frac{{sin}\left(\mathrm{90}+\alpha\right)}{\frac{{y}}{{cos}\alpha}}=\frac{{sin}\left(\mathrm{2}\alpha\right)}{\mathrm{8}}\Rightarrow\frac{{sin}\left(\mathrm{90}+\alpha\right){cos}\alpha}{{y}}=\frac{\mathrm{2}{sin}\alpha{cos}\alpha}{\mathrm{8}} \\ $$$$\Rightarrow\frac{{sin}\left(\mathrm{90}+\alpha\right)}{{y}}=\frac{{sin}\alpha}{\mathrm{4}}\Rightarrow{tan}\alpha=\frac{\mathrm{4}}{{y}}=\frac{{x}}{{y}}\Rightarrow{x}=\mathrm{4} \\ $$