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МатематикаГДЗ по геометрии 7 класс Мерзляк Задание 70
Задание 70
\[Рисунок\ в\ учебнике.\]
\[Дано:\ \ \]
\[\angle AOC = \angle COD = \angle DOF;\]
\[OB - биссектриса\ \angle AOC;\]
\[OE - биссектриса\ \angle DOF;\]
\[\angle BOE = 72{^\circ}.\]
\[Найти:\]
\[\angle AOF.\]
\[Решение.\]
\[1)\ \angle AOF:\]
\[\angle AOF = \angle AOC + \angle COD + \angle DOF\]
\[\angle AOF = \angle AOC + \angle AOC + \angle AOC\]
\[\angle AOF = 3\angle AOC\ \ \ \]
\[\angle AOC = \frac{1}{3}\angle AOF.\]
\[2)\ \angle\text{AOC}:\]
\[\angle AOC = \angle AOB + \angle BOC\]
\[\frac{1}{3}\angle AOF = \angle BOC + \angle BOC\]
\[\frac{1}{3}\angle AOF = 2\angle BOC\ \ \ \]
\[\angle BOC = \frac{1}{6}\angle AOF.\]
\[3)\ \angle\text{DOF}:\]
\[\angle DOF = \angle DOE + \angle EOF\]
\[\frac{1}{3}\angle AOF = \angle DOE + \angle DOE\]
\[\frac{1}{3}\angle AOF = 2\angle DOE\ \ \]
\[\angle DOE = \frac{1}{6}\angle AOF.\]
\[4)\ \angle BOE:\]
\[\angle BOE = \angle BOC + \angle COD + \angle DOE\]
\[72{^\circ} = \frac{1}{6}\angle AOF + \frac{1}{3}\angle AOF + \frac{1}{6}\angle AOF\]
\[432{^\circ} = \angle AOF + 2\angle AOF + \angle AOF\]
\[4\angle AOF = 432{^\circ}\ \ \ \]
\[\angle AOF = 108{^\circ}.\]
\[Ответ:\ \ 108{^\circ}.\]
