Question

1) Основание прямой призмы АВСА1В1С1 является равнобедренный треугольник АВС, АВ=АС=20, ВС=30. Высота призмы равна 22. На высоте АА1 взята точка F, такая что AF : FА1 = 8:3. Найдите угол между плоскостями А1В1С1 и ВСF. Пожалуйста с решением и рисунком, умоляю

Answer 1

Ответ:

$\alpha =arctg \dfrac{16\sqrt{7}}{35}$

Объяснение:

Плоскости оснований призмы параллельны, значит плоскость (BCF) образует с плоскостями оснований равные углы.

Пусть Н - середина ВС.

СН = 0,5ВС = 0,5 ·  30 = 15

АН - медиана и высота равнобедренного треугольника АВС,

АН⊥ВС.

АН - проекция FH на (АВС), значит FH⊥BC по теореме о трех перпендикулярах, значит

∠FHA = α - линейный угол двугранного угла между плоскостями (BCF)  и (АВС) - искомый.

ΔАНС: ∠АНС = 90°, по теореме Пифагора

  АН = √(АС² - СН²) = √(20² - 15²) = √((20 - 15)(20 + 15)) = √(5 · 35) = 5√7

AF : FA₁ = 8 : 3, значит

$FA=\dfrac{8}{11}AA_1 = \dfrac{8}{11}\cdot 22=16$

ΔFHA:  ∠FAH = 90°

$tg\; \alpha =\dfrac{FA}{AH}=\dfrac{16}{5\sqrt{7}}=\dfrac{16\sqrt{7}}{35}$$tg\; \alpha =\dfrac{FA}{AH}=\dfrac{16}{5\sqrt{7}}=\dfrac{16\sqrt{7}}{35}$

$\alpha =arctg \dfrac{16\sqrt{7}}{35}$

Answer 2

Given Information:

We are given the following information about the prism ABCA1B1C1: - The base ABC is an isosceles triangle with AB = AC = 20 and BC = 30. - The height of the prism is 22. - Point F is taken on the height AA1 such that AF : FA1 = 8 : 3.

We need to find the angle between the planes A1B1C1 and BCF.

Solution:

To find the angle between the planes A1B1C1 and BCF, we first need to determine the normal vectors to these planes.

The normal vector to the plane A1B1C1 can be found by taking the cross product of the vectors A1B1 and A1C1. Similarly, the normal vector to the plane BCF can be found by taking the cross product of the vectors BC and BF.

Let's calculate these vectors step by step:

1. Calculate the vectors A1B1 and A1C1: - Vector A1B1 can be obtained by subtracting the coordinates of point A1 from point B1: A1B1 = B1 - A1. - Vector A1C1 can be obtained by subtracting the coordinates of point A1 from point C1: A1C1 = C1 - A1.

2. Calculate the vectors BC and BF: - Vector BC can be obtained by subtracting the coordinates of point B from point C: BC = C - B. - Vector BF can be obtained by subtracting the coordinates of point B from point F: BF = F - B.

3. Calculate the cross products: - The cross product of vectors A1B1 and A1C1 will give us the normal vector to the plane A1B1C1: n1 = A1B1 x A1C1. - The cross product of vectors BC and BF will give us the normal vector to the plane BCF: n2 = BC x BF.

4. Find the angle between the planes: - The angle between two planes can be found using the dot product of their normal vectors and the formula: cos(theta) = (n1 · n2) / (|n1| |n2|). - Once we have the cosine of the angle, we can find the angle itself by taking the inverse cosine: theta = arccos(cos(theta)).

Let's calculate the vectors and find the angle between the planes A1B1C1 and BCF.

Calculation:

1. Calculate the vectors A1B1 and A1C1: - A1B1 = B1 - A1 = (20, 0, 0) - (0, 0, 0) = (20, 0, 0). - A1C1 = C1 - A1 = (0, 30, 0) - (0, 0, 0) = (0, 30, 0).

2. Calculate the vectors BC and BF: - BC = C - B = (0, 30, 0) - (20, 0, 0) = (-20, 30, 0). - BF = F - B = (8/11) * (A1 - B) = (8/11) * (0, 0, 22) = (0, 0, 16).

3. Calculate the cross products: - n1 = A1B1 x A1C1 = (20, 0, 0) x (0, 30, 0) = (0, 0, 600). - n2 = BC x BF = (-20, 30, 0) x (0, 0, 16) = (480, 320, 600).

4. Find the angle between the planes: - cos(theta) = (n1 · n2) / (|n1| |n2|) = ((0, 0, 600) · (480, 320, 600)) / (|n1| |n2|) = 0.8. - theta = arccos(cos(theta)) = arccos(0.8) ≈ 37 degrees.

Therefore, the angle between the planes A1B1C1 and BCF is approximately 37 degrees.

I hope this helps! Let me know if you have any further questions.