What is the angle between any two faces on a tetrahedron?

Question

A tetrahedron has edges of 5cm each. What is the angle between any two faces? 

 

Please explain how to answer this question. Thanks :)

Answer 1

     The angle between any 2 edges is 60 degrees, but the angle between any 2 faces I got to be 70.529 degrees.See if my logic makes sense....thatis if you can understand what I'm describing.

 

Lets start with the base, which is an equilateral triangle, call it ABC. If you drew the 3 altitudes, they'd meet at point D. The altitudes are also medians. Using 30-60-90 triangles, you can say that the length of the altitude is 2.5sqrt3. If you recall, there is a theorem that says that the medians of a triangle meet at a point called the centroid, our point D. The centroid divides the median into 2 segments, on of which is twice the other. The longer segment has the centroid and the vertex of the triangle as endpoints. 

 

So lets suppose the median was 12. if we take 12 and divide it by 3, we get segments which would be 4 and 8.

Our altitude/median is 2.5sqrt3. If we divide it by 3 we get (2.5sqrt3)/3.

 

I hope you're still with me. If you draw a segment perpendicular to your triangle ABC coming from the centroid, It should hit the top vertex of the tetrahedron and this would be the height of the tetrahedron. You should hopefully remember doing this with pyramids. Point D to the edge of the base is (2.5sqrt3)/3, the height of the tetrahedron we don't know, but the altitude of one of the slanted faces is 2.5sqrt3. These 3 segments form a right triangle so we can use trigonometry. Cos ø = [ (2.5sqrt3)/3 ]   /    2.5sqrt3  . This simplifies to cos ø = 1/3 which when solved is 70.52877937 degrees.

Answer 2

reference book says arcos(1/3) = 70.53 degrees. Kudos to Ira S for doing the geometry.