2257. В треугольнике ABC угол C равен 90°, AB = 10, sin A = √21/5. Найдите AC.

Дано: $$\triangle ABC$$, $$\angle C = 90^\circ$$, $$AB = 10$$, $$\sin A = \frac{\sqrt{21}}{5}$$. Найти: $$AC$$. Решение: $$\sin^2 A + \cos^2 A = 1$$ $$\cos^2 A = 1 - \sin^2 A = 1 - (\frac{\sqrt{21}}{5})^2 = 1 - \frac{21}{25} = \frac{4}{25}$$ $$\cos A = \sqrt{\frac{4}{25}} = \frac{2}{5}$$ $$\cos A = \frac{AC}{AB}$$ $$AC = AB \cdot \cos A$$ $$AC = 10 \cdot \frac{2}{5} = 4$$ Ответ: $$AC = 4$$