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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

$$\cos \alpha =\frac{2}{6}=\frac{1}{3}\Rightarrow \cot \alpha =\frac{1}{2\sqrt{2}}$$$$\beta =\frac{\pi -\alpha }{2}\Rightarrow \cot \beta =\tan \frac{\alpha }{2}=\frac{\frac{2\sqrt{2}}{3}}{1+\frac{1}{3}}=\frac{1}{\sqrt{2}}$$$$AD=\frac{x}{\sin \beta }$$$$\frac{ED}{\sin (\alpha -\beta )}=\frac{AD}{\sin \alpha }$$$$\frac{4}{\sin (\alpha -\beta )}=\frac{x}{\sin \beta \sin \alpha }$$$$x=\frac{4\sin \alpha \sin \beta }{\sin (\alpha -\beta )}=\frac{4}{\cot \beta -\cot \alpha }$$$$=\frac{4}{\frac{1}{\sqrt{2}}-\frac{1}{2\sqrt{2}}}=8\sqrt{2}✓$$

Commented by Tawa11 last updated on 21/Jun/24

$$\mathrm{weldone}\mathrm{sir}$$

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