\[Схематический\ рисунок.\]
\[Дано:\ \ \]
\[\angle ABC = 160{^\circ};\]
\[BK\bot AB;\]
\[BM\bot BC.\]
\[Найти:\]
\[\angle MBK.\]
\[Решение.\]
\[1)\ \angle MBC = \angle MBK + \angle CBK\]
\[\angle CBK = \angle MBC - \angle MBK.\]
\[2)\ \angle ABC = \angle ABK + \angle CBK\]
\[\angle ABC = \angle ABK + \angle MBC - \angle MBK\]
\[160{^\circ} = 90{^\circ} + 90{^\circ} - \angle MBK\]
\[\angle MBK = 180{^\circ} - 160{^\circ} = 20{^\circ}.\]
\[Ответ:\ \ 20{^\circ}.\]