Question

84 Балла! Відрізок ДС – перпендикуляр до площини прямокутного трикутника АВС (< АСВ=90°), ДС = 9 см, АС =15 см, ВС =20 см.. Відрізок ДЕ – перпендикуляр,опущенній із точки Д на пряму АВ. Знайдіть кут між прямою ДЕ та площиною(АВС).

Answer 1

Відповідь:

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Answer 2

To find the angle between the line DE and the plane ABC, we need to use the properties of perpendicular lines and right triangles. Let's break down the problem step by step.

Given Information:

- Length of segment DS: 9 cm - Length of segment AC: 15 cm - Length of segment BC: 20 cm

Finding the Length of Segment DE:

To find the length of segment DE, we can use the Pythagorean theorem in triangle ABC. Since triangle ABC is a right triangle with angle ACB measuring 90 degrees, we can use the theorem to find the length of segment DE.

Using the Pythagorean theorem, we have: AC^2 + BC^2 = AB^2

Substituting the given values, we get: 15^2 + 20^2 = AB^2 225 + 400 = AB^2 625 = AB^2 AB = √625 AB = 25 cm

Now, we can find the length of segment DE by subtracting the length of segment DS from AB: DE = AB - DS DE = 25 cm - 9 cm DE = 16 cm

Finding the Angle between Line DE and Plane ABC:

To find the angle between line DE and plane ABC, we need to consider the relationship between the line and the plane. The angle between a line and a plane is equal to the complement of the angle between the line and the normal vector of the plane.

In this case, the line DE is perpendicular to the plane ABC, so the angle between DE and the normal vector of the plane is 90 degrees. Therefore, the angle between DE and the plane ABC is 90 degrees.

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