Question

Найдите величину двугранного угла при основании правильной четырехугольной пирамиды, если ее боковые ребра наклонены к плоскости основания по углом 30'. (30' - 30 градусов). Обязательно: 1) Полное решение, с объяснением. 2) Рисунок.

Answer 1

Дано:
на картинке

Решение:
Так как пирамида правильная и SO перпендикулярно ABCD, то SOA - прямоугольный треугольник. В прямоугольном треугольнике катет, лежащий против угла в 30 градусов равен половине гипотенузы. Значит SO=SA/2.

Обозначим SA=2a, тогда SO=a. По теореме Пифагора найдем ОА:
$OA= \sqrt{SA^2-SO^2}= \sqrt{(2a)^2-a^2}= a \sqrt{3}$

Так как в основании лежат квадрат, то он имеет равные взаимно перпендикулярные диагонали, которые точкой пересечений делятся пополам. Значит, треугольник АВО - прямоугольный и АО=ВО.
По теореме Пифагора находит АВ из прямоугольного треугольника АВО:
$AB= \sqrt{AO^2+BO^2}= \sqrt{(a \sqrt{3} )^2+(a \sqrt{3} )^2}= a\sqrt{6}$

Так как точка Н - середина АВ, то НВ=НА=АВ/2
Из прямоугольного треугольника OНВ находим OН по теореме Пифагора:
$OH= \sqrt{BO^2-HB^2} = \sqrt{AO^2-HB^2} = \\\ =\sqrt{(a \sqrt{3}) ^2-( \frac{a \sqrt{6} }{2})^2} =a\sqrt{( \sqrt{3}) ^2-( \frac{ \sqrt{6} }{2})^2} =a\sqrt{3- \frac{6 }{4}} =a\sqrt{ \frac{6 }{4}} = \frac{a \sqrt{6} }{2}$

Из прямоугольного треугольника SOH:
$tgSOH= \frac{SO}{OH} =a: \frac{a \sqrt{6} }{2} = \frac{2}{ \sqrt{6} } =\frac{2\cdot \sqrt{6}}{ \sqrt{6}\cdot \sqrt{6} } =\frac{2\sqrt{6}}{6 } =\frac{\sqrt{6}}{3 } \\\ \Rightarrow SOH=\mathrm{arctg} \frac{\sqrt{6}}{3 }$

Ответ: $\mathrm{arctg} \frac{\sqrt{6}}{3 }$

Answer 2

Посмотрите в приложении

Answer 3

Calculation of the Angle

To find the measure of the dihedral angle at the base of a regular quadrilateral pyramid with its lateral edges inclined to the base plane at an angle of 30 degrees, we can use trigonometry.

Let's denote the dihedral angle as x.

Since the lateral edges are inclined to the base plane at an angle of 30 degrees, we can consider the right triangle formed by one of the lateral edges, the height of the pyramid, and the slant height of the pyramid.

In this right triangle, the angle opposite the height is 30 degrees, and the angle opposite the slant height is 90 degrees.

Using trigonometric ratios, we can relate the angles and sides of the right triangle:

- The sine of the angle opposite the height is equal to the ratio of the height to the slant height: sin(30) = height / slant height. - The cosine of the angle opposite the height is equal to the ratio of the base length to the slant height: cos(30) = base length / slant height.

Since the base of the pyramid is a regular quadrilateral, the base length is equal to the side length of the quadrilateral.

Let's denote the side length of the quadrilateral as s.

From the above trigonometric relations, we can express the height and base length in terms of the side length and slant height:

- height = slant height * sin(30) = slant height * 0.5. - base length = slant height * cos(30) = slant height * (√3 / 2).

Now, let's consider the right triangle formed by one of the lateral edges, the height, and the base length.

In this right triangle, the angle opposite the height is 90 degrees, and the angle opposite the base length is the dihedral angle x.

Using trigonometric ratios, we can relate the angles and sides of the right triangle:

- The tangent of the dihedral angle x is equal to the ratio of the height to the base length: tan(x) = height / base length.

Substituting the expressions for height and base length, we have:

- tan(x) = (slant height * 0.5) / (slant height * (√3 / 2)) = 0.5 / (√3 / 2) = 1 / √3 = √3 / 3.

To find the value of x, we can take the inverse tangent (arctan) of both sides:

- x = arctan(√3 / 3) ≈ 35.26 degrees.

Therefore, the measure of the dihedral angle at the base of the regular quadrilateral pyramid is approximately 35.26 degrees.

Diagram

Here is a diagram illustrating the regular quadrilateral pyramid with its lateral edges inclined to the base plane at an angle of 30 degrees:

          A
         /|\
        / | \
       /  |  \
      /   |   \
     /    |    \
    /     |     \
   /______|______\
  B       C       D

In the diagram, A, B, C, and D represent the vertices of the base of the pyramid, and the lines connecting the vertices represent the lateral edges. The angle between the lateral edge AB and the base plane is 30 degrees.

Please note that the diagram is not to scale and is for illustrative purposes only.

I hope this explanation and diagram help you understand how to calculate the dihedral angle at the base of a regular quadrilateral pyramid with inclined lateral edges. Let me know if you have any further questions!