\[Схематический\ рисунок.\]
\[Дано:\]
\[\angle ABC = 90{^\circ};\]
\[\angle ACB = 32{^\circ};\]
\[AC = 12\ см.\]
\[Найти:\]
\[\cup AB;\ \ \]
\[\cup BC;\ \ \]
\[\cup AC;\ \]
\[\text{R.}\]
\[Решение.\]
\[1)\ \mathrm{\Delta}ABC - прямоугольный:\]
\[\angle ACB + \angle CAB = 90{^\circ}\]
\[32{^\circ} + \angle CAB = 90{^\circ}\]
\[\angle CAB = 58{^\circ}.\]
\[2)\ \cup AC = 2\angle ABC = 180{^\circ};\]
\[\cup AB = 2\angle ACB = 64{^\circ};\]
\[\cup BC = 2\angle CAB = 116{^\circ}.\]
\[AC - диаметр:\]
\[R = \frac{1}{2}AC = 6\ см.\]
\[Ответ:\ \ 64{^\circ};\ 116{^\circ};\ 180{^\circ};\ 6\ см.\]