Предмет: Геометрия, автор: ayananurbay07

35 БАЛЛОВ ABCDE - это правильный пятиугольник, а ABF - равносторонний треугольник. Каков угол AFE?​

Приложения:

Данил12437: Если тебе не трудно, отметь решение лучшим)))

Ответы

Автор ответа: Данил12437
1

Відповідь: 72 градуса

Пояснення: Равные хорды стягивают равные дуги. Пять равных хорд, являющихся сторонами пентагона, делят окружность на пять равных дуг, т.е.

∪AB=∪BC=∪CD=∪DE=∪AE =360°/5=72°

AD, BE - диагонали пентагона, пересекающиеся в точке F.

Вписанный угол равен половине угловой меры дуги, на которую опирается.

∠AEB=∪AB/2 =72°/2

∠DAE=∪DE/2 =72°/2

Сумма углов треугольника равна 180°. Отсюда:

△AFE: ∠AFE= 180°-∠AEB-∠DAE =180°-72° =108°

Мерой угла между пересекающимися прямыми считается мера меньшего из образованных углов.

∠AFB=180°-108=72;

Ответ:∠AFE =72°

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