7. В треугольнике АВС угол C равен 90°, СН — высота, АВ = 100, sin ∠A = 4/5. Найдите длину отрезка АН.

\(sin A = \frac{CH}{AC} = \frac{4}{5}\) => \(AC = \frac{5CH}{4}\)
\(sin A = \frac{BC}{AB} = \frac{4}{5}\) => \(BC = \frac{4AB}{5} = \frac{4 \cdot 100}{5} = 80\)
\(AC^2 + BC^2 = AB^2\)
\((\frac{5CH}{4})^2 + 80^2 = 100^2\)
\(\frac{25CH^2}{16} = 10000 - 6400 = 3600\)
\(CH^2 = \frac{3600 \cdot 16}{25}\) => \(CH = \sqrt{\frac{3600 \cdot 16}{25}} = \frac{60 \cdot 4}{5} = 48\)
\(AH = AC \cdot cos A\)
\(AC = \sqrt{AB^2 - BC^2} = \sqrt{100^2 - 80^2} = \sqrt{10000 - 6400} = \sqrt{3600} = 60\)
\(cos A = \sqrt{1 - sin^2A} = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}\)
\(AH = 60 \cdot \frac{3}{5} = 36\)