- Выразим cos A через отношение сторон:
\[\cos A = \frac{AC}{AB} = \frac{\sqrt{5}}{5}\]
- Найдем AB:
\[\frac{3}{AB} = \frac{\sqrt{5}}{5}\]
\[AB = \frac{3 \cdot 5}{\sqrt{5}} = \frac{15}{\sqrt{5}} = \frac{15\sqrt{5}}{5} = 3\sqrt{5}\]
- Применим теорему Пифагора:
\[AB^2 = AC^2 + BC^2\]
\[BC^2 = AB^2 - AC^2\]
\[BC^2 = (3\sqrt{5})^2 - 3^2 = 45 - 9 = 36\]
\[BC = \sqrt{36} = 6\]