Предмет: Геометрия, автор: polinapakhalina

Треугольник авс угол в 60 градусов сторона ав 6 см вс 10 см найти площадь

Ответы

Автор ответа: John339

0

Ответ:

15√3 см^2.

Объяснение:

$S=\frac{1}{2} AB*BC*sinB$ .

Вычисляем:

$S=\frac{1}{2} *6*10*\frac{\sqrt{3} }{2} =15\sqrt{3}$ (см^2).

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