В прямоугольном треугольнике ABC:
\[ \sin A = \frac{BC}{AB} \]
\[ \sin A = \frac{\sqrt{19}}{10} \]
и\[ AB = 4 \]
\[ \frac{\sqrt{19}}{10} = \frac{BC}{4} \]
\[ BC = 4 \cdot \frac{\sqrt{19}}{10} = \frac{2\sqrt{19}}{5} \]
\[ AC^2 + BC^2 = AB^2 \]
\[ AC^2 + \left( \frac{2\sqrt{19}}{5} \right)^2 = 4^2 \]
\[ \left( \frac{2\sqrt{19}}{5} \right)^2 = \frac{4 \cdot 19}{25} = \frac{76}{25} \]
\[ 4^2 = 16 \]
\[ AC^2 + \frac{76}{25} = 16 \]
\[ AC^2 = 16 - \frac{76}{25} = \frac{16 \cdot 25 - 76}{25} = \frac{400 - 76}{25} = \frac{324}{25} \]
\[ AC = \sqrt{\frac{324}{25}} = \frac{\sqrt{324}}{\sqrt{25}} = \frac{18}{5} \]
Ответ: 18/5