В выпуклом четырёхугольнике \(\displaystyle ABCD\) биссектрисы углов \(\displaystyle A\) и \(\displaystyle B\) пересекаются в точке \(\displaystyle O{\small.}\)
Найдите угол \(\displaystyle AOB{\small,}\) если \(\displaystyle \angle C=80^{\circ}{\small,}\) \(\displaystyle \angle D=70^{\circ}{\small.}\)
\(\displaystyle \angle AOB=\) \(\displaystyle ^{\circ}\)
Обозначим на рисунке:
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Рассмотрим треугольник \(\displaystyle AOB{\small.}\)
 | Сумма внутренних углов треугольника равна \(\displaystyle 180^{\circ}{\small.}\) То есть \(\displaystyle \color{blue}{\alpha}+\color{blue}{\beta}+\color{red}{x}=180^{\circ}{\small.} \) Значит, \(\displaystyle \color{red}{x}=180^{\circ}-(\color{blue}{\alpha}+\color{blue}{\beta}){\small.} \) |
Найдем сумму \(\displaystyle (\color{blue}{\alpha}+\color{blue}{\beta}){\small.} \)
В выпуклом четырёхугольнике \(\displaystyle ABCD{\small:}\) \(\displaystyle \angle A+\angle B+\angle C+\angle D=360^{\circ}{\small.} \) То есть \(\displaystyle 2\color{blue}{\alpha}+2\color{blue}{\beta}+\color{green}{80^{\circ}}+\color{green}{70^{\circ}}=360^{\circ}{\small.} \) Получаем: \(\displaystyle 2(\color{blue}{\alpha}+\color{blue}{\beta})=360^{\circ}-\color{green}{80^{\circ}}-\color{green}{70^{\circ}}{\small;} \) \(\displaystyle 2(\color{blue}{\alpha}+\color{blue}{\beta})=210^{\circ}{\small;} \) \(\displaystyle \color{blue}{\alpha}+\color{blue}{\beta}=105^{\circ}{\small.} \) |  |
Найдем значение \(\displaystyle \color{red}{x}{\small:}\)
\(\displaystyle \color{red}{x}=180^{\circ}-(\color{blue}{\alpha}+\color{blue}{\beta}){\small;} \)
\(\displaystyle \color{red}{x}=180^{\circ}-105^{\circ}=75^{\circ}{\small.} \)
Значит,
\(\displaystyle \angle AOB=75^{\circ}{\small.}\)
Ответ: \(\displaystyle 75^{\circ}{\small.}\)