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Question Number 154088 by iloveisrael last updated on 14/Sep/21

Commented by Tawa11 last updated on 14/Sep/21

$$\mathrm{Weldone}\mathrm{sir}$$

Answered by mr W last updated on 14/Sep/21

$$\tan 3\alpha =2\tan 2\alpha$$$$\frac{\tan \alpha (3-{\tan }^2\alpha )}{1-3{\tan }^2\alpha }=2\times \frac{2\tan \alpha }{1-{\tan }^2\alpha }$$$$\frac{3-{\tan }^2\alpha }{1-3{\tan }^2\alpha }=\frac{4}{1-{\tan }^2\alpha }$$$${\tan }^4\alpha +8{\tan }^2\alpha -1=0$$$$\Rightarrow {\tan }^2\alpha =\sqrt{17}-4$$$$\Rightarrow \tan \alpha =\sqrt{\sqrt{17}-4}$$$$\tan 2\alpha =\frac{2\sqrt{\sqrt{17}-4}}{1-(\sqrt{17}-4)}=\frac{2\sqrt{\sqrt{17}-4}}{5-\sqrt{17}}$$$$x=BE\times \tan 2\alpha =12\times \sin 2\alpha \times \tan 2\alpha$$$$=12\times \frac{2\sqrt{\sqrt{17}-4}}{5-\sqrt{17}}\times \frac{2\sqrt{\sqrt{17}-4}}{\sqrt{26-6\sqrt{17}}}$$$$=\frac{24}{4}$$$$=6$$

Commented by mr W last updated on 14/Sep/21

$$ifoundamistake.yes,itis6.$$

Commented by iloveisrael last updated on 14/Sep/21

$$not6?$$

Commented by iloveisrael last updated on 14/Sep/21

$$yes$$

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