Problem 12 (3 marks). 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 base


To solve this problem for a regular tetrahedron with edge length 1, we'll break it down into two parts as asked in the question: 

### Part a) Finding the angles the lateral edges make with the base
A regular tetrahedron is a three-dimensional shape with four equilateral triangular faces. Let's label the tetrahedron vertices as \( A \), \( B \), \( C \), and \( D \), with \( ABC \) as the base and \( D \) as the apex. The lateral edges are the edges from vertex \( D \) to the vertices on the base (i.e., edges \( DA \), \( DB \), and \( DC \)).

We need to calculate the angle between one of the lateral edges (e.g., \( DA \)) and the base plane \( ABC \). This is essentially the angle between the edge \( DA \) and the plane containing \( A \), \( B \), and \( C \).

#### Step 1: Calculate the height of the tetrahedron
The height of the regular tetrahedron is the perpendicular distance from the vertex \( D \) to the base plane \( ABC \). For a regular tetrahedron with edge length \( 1 \), the formula for the height \( h \) is:

\[
h = \frac{\sqrt{2}}{\sqrt{3}} \cdot \text{edge length} = \frac{\sqrt{2}}{\sqrt{3}} \cdot 1 = \frac{\sqrt{2}}{\sqrt{3}}
\]

#### Step 2: Calculate the distance from the center of the base to a vertex
Since the base \( ABC \) is an equilateral triangle, the distance from the centroid of the triangle (which is also the projection of \( D \) onto the plane) to any vertex of the base triangle is:

\[
r = \frac{\text{edge length}}{\sqrt{3}} = \frac{1}{\sqrt{3}}
\]

#### Step 3: Apply trigonometry to find the angle
The angle \( \theta \) between the lateral edge and the base can be found using basic trigonometry. If we consider the right triangle formed by the height of the tetrahedron, the centroid of the base, and one of the base vertices, we have:

\[
\tan(\theta) = \frac{r}{h} = \frac{1/\sqrt{3}}{\sqrt{2}/\sqrt{3}} = \frac{1}{\sqrt{2}}
\]

So:

\[
\theta = \arctan\left(\frac{1}{\sqrt{2}}\right)
\]

The exact value of \( \theta \) is:

\[
\theta = 35.26^\circ
\]

Thus, the angles that the lateral edges make with the base are \( 35.26^\circ \).

### Part b) Finding the angles the lateral faces make with the base
We now need to calculate the angle between one of the lateral faces (e.g., face \( DAB \)) and the base plane \( ABC \). This is the dihedral angle between these planes.

#### Step 1: Recall the formula for the dihedral angle
For a regular tetrahedron, the dihedral angle between any two faces is given by:

\[
\phi = \arccos\left(\frac{1}{3}\right)
\]

#### Step 2: Calculate the angle
The dihedral angle between the lateral faces and the base plane is:

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

Thus, the angles that the lateral faces make with the base are \( 70.53^\circ \).

### Final Answer:
a) The angles the lateral edges make with the base are \( 35.26^\circ \).  
b) The angles the lateral faces make with the base are \( 70.53^\circ \).

Would you like any further details or clarifications?

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Here are 5 related questions:
1. How is the height of a tetrahedron derived from the edge length?
2. What is the general formula for calculating dihedral angles in polyhedra?
3. How does the centroid of an equilateral triangle relate to the geometry of a tetrahedron?
4. What are the properties of a regular tetrahedron's faces and edges?
5. How do trigonometric functions apply to three-dimensional geometry?

**Tip**: When solving geometric problems involving polyhedra, visualizing or sketching the shape and using symmetry can often simplify complex calculations.

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 base.

Explore the geometric properties of a regular tetrahedron with edge length 1. Learn how to calculate the angles between the lateral edges and the base (35.26°) and the dihedral angles between the lateral faces and the base (70.53°). This solution involves 3D geometry and trigonometry, suitable for Grades 10-12 students.

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