В треугольнике \(\displaystyle ABC\) угол \(\displaystyle C\) равен \(\displaystyle 90^{\circ} {\small,}\) \(\displaystyle \cos \angle B=\frac{2}{5} {\small,}\) \(\displaystyle AB=10 {\small.}\) Найдите \(\displaystyle BC {\small.}\)

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