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Все рёбра треугольной пирамиды равны. Найти угол наклона:
а) Бокового ребра к плоскости основы.
б) боковой грани к площине основы/
Объяснение:
АВСМ -пирамида, пусть ребро равно х.
a)Угол наклона бокового ребра к плоскости основания это ∠МАО.
Т.к АВ=ВС=АС, то высота проецируется в центр основания О , точку пересечения медиан.Тогда АО=2/3*АН, где АН медиана, ВН=х/2 .
Из ΔАВН-прямоугольного, АН=√(х²-х²/4)=(х√3)/2. Тогда АО=( х√3)/3.
ΔАОМ-прямоугольный, cos∠МАО=АО/АМ , cos∠МАО=( х√3)/3:х=√3/3,
∠МАО=arccos(√3/3) .
ОМ=√(х²-( х√3)/3)² )=(х√6)/3
б)В равностороннем ΔАВС , медиана АН является высотой . Тогда МН⊥ВС по т. о трех перпендикулярах и ∠АНМ-линейный угол между боковой гранью и плоскостью основания.
ОН=1/3*АН , ОН=(х√3)/6.
ΔОНМ-прямоугольный ,tg∠AHM=MO/OH , tg∠AHM=2√2 , ∠AHM=arctg(2√2).
Так как продолжение боковых сторон пересекаются под прямым углом, то треугольники АЕД и ВЕС прямоугольные и подобны по острому углу, так как угол ЕАС = ЕВС как соответственные углы при пересечении параллельных АД и ВС секущей АЕ.
Коэффициент подобия треугольников равен: К = ВС / АД = 8 / 12 = 2/3.
Длина отрезка ВЕ = ВС * Cos28.
ВЕ / АЕ = 2/3.
АЕ = 3 * ВЕ / 2 = 3 * 8 * Cos28 / 2 = 12 * Cos28.
Тогда АВ = АЕ – ВЕ = 12 * Cos28 – 8 * Cos28 = 4 * Cos28.
ответ: Длина стороны АВ равна 4 * Cos28.
Объяснение: