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ПавелКрут
ПавелКрут
29.05.2022 07:57 •  Геометрия

Координаталық жазықтықта төбелері: а(-4; 1); в(-2; 5); с(5; 3) нүктелері болатын авс үшбұрышын салыңдар.авс үшбұрышына абсциссалар осіне қатысты симметриялы а1в1с1 үшбұрышын салыңдар.​

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Ответ:
flaming1337
flaming1337
09.06.2022 19:56

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).


Усі ребра трикутної піраміди рівні. Знайти кут нахилу: а) бічного ребра до площини основи. б) бічної
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Ответ:
1972jkmxbr
1972jkmxbr
06.01.2021 03:08

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