№3. Найти AE.

В прямоугольном треугольнике ABC угол A равен 30°, а катет BC равен 7. Угол BCE = 60°. $$\tan(30°) = \frac{BC}{AC}$$ $$\frac{\sqrt{3}}{3} = \frac{7}{AC}$$ $$AC = \frac{7 * 3}{\sqrt{3}} = \frac{21}{\sqrt{3}} = \frac{21\sqrt{3}}{3} = 7\sqrt{3}$$ $$\tan(60°) = \frac{BE}{EC}$$ $$\sqrt{3} = \frac{BE}{EC}$$ $$\frac{AC}{AE + EC} = AC$$ $$AE = AC - EC$$ $$AE = 7\sqrt{3} - EC$$ $$EC = \frac{BE}{\sqrt{3}}$$ $$\frac{BE}{AB} = \sin(30°)$$ $$AB = \sqrt{(7\sqrt{3})^2 + 7^2} = \sqrt{147 + 49} = \sqrt{196} = 14$$ $$\frac{BE}{14} = \frac{1}{2}$$ $$BE = 7$$ $$EC = \frac{7}{\sqrt{3}} = \frac{7\sqrt{3}}{3}$$ $$AE = 7\sqrt{3} - \frac{7\sqrt{3}}{3} = \frac{21\sqrt{3} - 7\sqrt{3}}{3} = \frac{14\sqrt{3}}{3}$$ Ответ: $$AE = \frac{14\sqrt{3}}{3}$$