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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

$$AC=ytan\left(3\alpha \right);x=ytan\alpha ;z=\frac{y}{cos\alpha };c=y(tan3\alpha -tan\alpha )$$$$EC=\frac{y}{cos3\alpha }$$$$\frac{sin(90+\alpha )}{\frac{y}{cos\alpha }}=\frac{sin\left(2\alpha \right)}{8}\Rightarrow \frac{sin(90+\alpha )cos\alpha }{y}=\frac{2sin\alpha cos\alpha }{8}$$$$\Rightarrow \frac{sin(90+\alpha )}{y}=\frac{sin\alpha }{4}\Rightarrow tan\alpha =\frac{4}{y}=\frac{x}{y}\Rightarrow x=4$$

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