4. В треугольнике АВС угол C равен 90°, ВС = 25, cos A = 12/13 Найдите АС.

\(cos A = \frac{AC}{AB}\)
\(\frac{12}{13} = \frac{AC}{AB}\) => \(AB = \frac{13AC}{12}\)
\(AC^2 + BC^2 = AB^2\)
\(AC^2 + 25^2 = (\frac{13AC}{12})^2\)
\(AC^2 + 625 = \frac{169AC^2}{144}\)
\(144AC^2 + 144 \cdot 625 = 169AC^2\)
\(25AC^2 = 144 \cdot 625\)
\(AC^2 = \frac{144 \cdot 625}{25} = 144 \cdot 25\)
\(AC = \sqrt{144 \cdot 25} = 12 \cdot \frac{25}{1} = \frac{325}{12}\)