\[Схематический\ рисунок.\]
\[Дано:\]
\[BD - высота;\]
\[\angle A = 60{^\circ};\]
\[\angle C = 45{^\circ};\]
\[AB = 10\ см.\]
\[Найти:\]
\[\text{BC.}\]
\[1)\ В\ \mathrm{\Delta}ABD:\]
\[\angle BDA = 90{^\circ};\]
\[\angle ABD = 90{^\circ} - \angle BAD = 30{^\circ};\]
\[AD = \frac{1}{2}AB = \frac{1}{2} \bullet 10 = 5\ см;\]
\[BD = \sqrt{AB^{2} - AD^{2}} =\]
\[= \sqrt{100 - 25} = \sqrt{75} = 5\sqrt{5}\ см.\]
\[2)\ В\ \mathrm{\Delta}BDC:\]
\[\angle BDC = 90{^\circ};\ \ \ \angle BCD = 45{^\circ};\]
\[\angle DBC = 90{^\circ} - \angle BCD = 45{^\circ};\]
\[CD = BD = \sqrt{75};\]
\[BC = \sqrt{CD^{2} + BD^{2}};\]
\[BC = \sqrt{75 + 75} =\]
\[= \sqrt{150} = 5\sqrt{6}\ см.\]
\[Ответ:\ \ 5\sqrt{6}\ см.\]