Предмет: Геометрия, автор: bulinalera14

Помогите пожалуйста решить одну задачу с объяснением.
Очень нужно, даю 60 баллов и лучший ответ, тому кто ответит первым.

Приложения:

Ответы

Автор ответа: zmeura1204

3

Ответ:
КА=40ед.

Решение:
∆АЕК- прямоугольный треугольник АК- касательная;
∆ЕКВ- прямоугольный треугольник
ВК- касательная
∆АЕК=∆ЕКВ, по гипотенузе и катету.
ЕК- биссектрисса угла ∠АКВ.
∠АКЕ=120°:2=60°
∠АЕК=180°-∠ЕАК-∠АКЕ=180°-90°-60°=
=30°
АК- катет против угла 30°
АК=ЕК/2=80/2=40ед

Приложения:

bulinalera14: Спасибо огромное!!
Просто от всей души!

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