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Question Number 166063 by cortano1 last updated on 12/Feb/22

Commented by blackmamba last updated on 12/Feb/22

$$\tan \alpha =\frac{2}{6-2\sqrt{3}}=\frac{1}{3-\sqrt{3}}=\frac{3+\sqrt{3}}{6}$$$$\tan 2\alpha =\frac{2(15+7\sqrt{3})}{13}$$$$\tan (90°-2\alpha )=\frac{1}{\tan 2\alpha }=\frac{15-7\sqrt{3}}{12}$$$$m=\frac{15-7\sqrt{3}}{12}$$

Commented by MaxiMaths last updated on 12/Feb/22

$$\mathrm{cool}...$$

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