\[\boxed{\mathbf{223.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[Дано:\ \]
\[\mathrm{\Delta}\text{ABC.}\]
\[Найти:\ \]
\[\angle C - ?\]
\[Решение.\]
\[По\ теореме\ о\ сумме\ углов\ в\ \]
\[треугольнике:\]
\[\angle A + \angle B + \angle C = 180{^\circ}\ .\]
\[\textbf{а)}\ \angle A = 65{^\circ};\ \angle B = 57{^\circ}:\]
\[\angle C = 180{^\circ} - \angle A - \angle B =\]
\[= 180{^\circ} - 65{^\circ} - 57{^\circ} =\]
\[= 180{^\circ} - 122{^\circ} = 58{^\circ}.\]
\[\textbf{б)}\ \angle A = 24{^\circ};\ \angle B = 130{^\circ}:\]
\[\angle C = 180{^\circ} - \angle A - \angle B =\]
\[= 180{^\circ} - 24{^\circ} - 130{^\circ} =\]
\[= 180{^\circ} - 157{^\circ} = 26{^\circ}.\]
\[\textbf{в)}\ \angle A = \alpha;\ \angle B = 2\alpha:\]
\[\angle C = 180{^\circ} - \angle A - \angle B =\]
\[= 180{^\circ} - \alpha - 2\alpha = 180{^\circ} - 3\alpha.\]
\[\textbf{г)}\ \angle A = 60{^\circ} + \alpha;\ \angle B = 60{^\circ} - \alpha:\]
\[\angle C = 180{^\circ} - \angle A - \angle B =\]
\[= 180{^\circ} - 60{^\circ} - \alpha - 60{^\circ} + \alpha =\]
\[= 180{^\circ} - 120{^\circ} = 60{^\circ}.\]
\[Ответ:а)\ 58{^\circ};б)\ 26{^\circ};\]
\[\textbf{в)}\ 180{^\circ} - 3\alpha;г)\ 60{^\circ}.\]