Вопрос школьника по предмету Геометрия
Один из острых углов прямоугольного треугольника равен 42⁰. Найдите угол между высотой и биссектрисой ,проведенными из вершины прямого угла треугольника.
(С ОБЪЯСНЕНИЕМ ПОЖАЛУЙСТА :3)
Ответ учителя по предмету Геометрия
Ответ:
3 градуса
Объяснение:
- Пусть СН — высота, СL — биссектриса в треугольнике ABC с прямым углом С
- Рассмотрим треугольник СВН: в нём угол СНВ = 90° — т.к. СН перпендикулярно АВ. Тогда угол НСВ = 180 — 90 — 42 = 48° (сумма углов треугольника равна 180°).
- угол СНL = угол НСВ — угол LСВ. Угол LCB = 90°/2 = 45° — т.к. CL — биссектриса прямого угла.
- тогда угол СНL = 48°-45° = 3°
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