Math Problem Statement

For a regular tetrahedron with edge length 1, find a) the angles the lateral edges make with the base; b) the angles the lateral faces make with the base4 .

Solution

Given a regular tetrahedron with edge length 1, we need to calculate:

Part a) The angles the lateral edges make with the base.

To compute this, we can use geometric properties of a regular tetrahedron:

  1. A regular tetrahedron has four equilateral triangular faces, and each of its edges has the same length.
  2. Consider one triangular face as the base, and let’s examine the angles formed between the lateral edges and this base.

Step 1: Determine the coordinates of the vertices

We can place the tetrahedron’s vertices in 3D space as follows:

  • Place the first vertex at the origin A(0,0,0)A(0, 0, 0).
  • The second vertex can be at B(1,0,0)B(1, 0, 0).
  • The third vertex at C(12,32,0)C\left(\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right) (this forms an equilateral triangle base in the xyxy-plane).
  • The fourth vertex DD lies above the plane of the base, directly along the perpendicular from the centroid of the base triangle. Its coordinates can be calculated using symmetry and the Pythagorean theorem. The zz-coordinate of DD is z=23z = \frac{\sqrt{2}}{\sqrt{3}}.

Step 2: Calculate the angle between the lateral edge and the base

For simplicity, we calculate the angle between the edge ADAD and the plane containing ABCABC. The angle between a vector and a plane is given by:

θ=arccos(projection of the vector onto the normal of the planelength of the vector)\theta = \arccos \left( \frac{ \text{projection of the vector onto the normal of the plane}}{\text{length of the vector}} \right)

After solving, the angle between a lateral edge and the base comes out to be:

θ=arccos(13)54.74\theta = \arccos \left( \frac{1}{\sqrt{3}} \right) \approx 54.74^\circ

Part b) The angles the lateral faces make with the base.

The angle between two planes can be found using the normals of the planes. Since all the faces are equilateral triangles, we can consider the dihedral angle between one of the lateral faces and the base face.

The formula for the dihedral angle ϕ\phi between two faces of a regular tetrahedron is:

ϕ=arccos(13)70.53\phi = \arccos \left( \frac{1}{3} \right) \approx 70.53^\circ

Thus, the angle between any lateral face and the base is approximately 70.5370.53^\circ.


Would you like more details on the geometric steps involved, or clarification on any specific part?

Related Questions:

  1. How can we compute the volume of a regular tetrahedron?
  2. What is the height of the regular tetrahedron with edge length 1?
  3. How do you find the centroid of a tetrahedron?
  4. What are the properties of the circumscribed sphere of a tetrahedron?
  5. How do the angles change if the tetrahedron is not regular?

Tip:

When dealing with polyhedra, it's often helpful to work with symmetry and coordinate geometry for precise calculations.

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Math Problem Analysis

Mathematical Concepts

Geometry
3D Coordinate Geometry
Tetrahedron
Angles in Polyhedra

Formulas

Angle between a vector and a plane: θ = arccos( (projection of the vector onto the normal of the plane) / length of the vector )
Dihedral angle formula: φ = arccos(1/3)

Theorems

Geometric properties of a regular tetrahedron
Pythagorean Theorem for height calculation in 3D

Suitable Grade Level

Grades 10-12