back to solid geometry

How to find the angle between two faces in a regular tetrahedron?

Question:

In the figure shown, P-ABC is a regular tetrahedron with side length a. Find the degree measure of the angle between the plane PAB and the plane ABC.

How to find the angle between two faces in a regular tetrahedron?

Solution:

Draw a line from the vertex P perpendicular to the plane ABC, this line intersects the plane ABC at the point O. Because P-ABC is a regular tetrahedron, so the point O is in the center of the equilateral triangle ABC.

Draw a line from the vertex P perpendicular to the line segment AB at point D. Because P-ABC is a regular tetrahedron, so triangle PAB is an equilateral triangle. PD is the distance from the vertex P to the side AB. From the property of an equilateral triangle, PD = (square root of 3) a/2 in which a is the length of the side of the equilateral triangle PAB.

How to find the angle between two faces in a regular tetrahedron?

Connect OD, because line segment PO perpendicular to plane ABC and because line segment OD lies on the plane ABC, so, line segment PO perpendicular to the line segment OD. Because PD perpendicular to AB and because OD is the projection of PD on the plane ABC, so, line segment OD perpendicular to line segment AB. Because P-ABC is a regular tetrahedron, so, triangle ABC is an equilateral triangle. OD is the distance from the center of the equilateral triangle to a side. From the property of an equilateral triangle, the distance OD = (square root of 3) a/6 in which a is the side of the equilateral triangle ABC. So, the angle between the plane PAB and the plane ABC is the angle between the line segment PD and the line segment OD. Let alfa be the angle PDO, because line segment PO perpendicular to line segment OD, so triangle POD is a right triangle.

In right triangle POD,
cos alfa = OD/PD
= (square root of 3) a/6 ÷ (square root of 3) a/2
= (square root of 3) a/6 × 2/(square root of 3)a
= 2/6 = 1/3 = 0.3333
alfa = arc cos 0.3333 = 70.53o

Therefore, the angle between the plane PAB and the plane ABC is 70.53 degrees. Please watch the video for more details.

Because P-ABC is a regular tetrahedron, so, the angle between the plane PAB and the plane ABC is the angle between ant two faces. That is, in a regular tetrahedron, the angle between any two faces are 70.53 degrees.