\[Схематический\ рисунок.\]
\[Дано:\]
\[O - центр\ опис.\ окружности;\]
\[AD - диаметр;\]
\[\angle ABC = 108{^\circ};\]
\[\angle BCD = 132{^\circ}.\]
\[Найти:\]
\[\angle BAD;\ \angle ADC;\]
\[\angle CAD;\ \angle BDA.\]
\[Решение\]
\[\angle BAD + \angle BCD = 180{^\circ}\]
\[\angle BAD + 132{^\circ} = 180{^\circ}\]
\[\angle BAD = 48{^\circ}.\]
\[\angle ADC + \angle ABC = 180{^\circ}\]
\[\angle ADC + 108{^\circ} = 180{^\circ}\]
\[\angle ADC = 72{^\circ}.\]
\[\cup ABC = 2\angle ADC = 144{^\circ};\]
\[\cup CD = 180{^\circ} - \cup ABC = 36{^\circ};\]
\[\angle CAD = \frac{1}{2} \cup CD = 18{^\circ};\]
\[\cup DCB = 2\angle BAD = 96{^\circ};\]
\[\cup AB = 180{^\circ} - \cup DCB = 84{^\circ};\]
\[\angle BDA = \frac{1}{2} \cup AB = 42{^\circ}.\]
\[Ответ:\ \ 48{^\circ};\ 72{^\circ};\ 18{^\circ};\ 42{^\circ}.\]