Find the height of the regular pyramid with each edge measuring l cm.
Also,
if `alpha` is angle between any edge and face not containing that edge, then prove that `cosalpha=1/sqrt3`

Explanation

In the figure we have regular pyramid.

All the edges have length l cm.
So, each face of the pyramid is an equilateral triangle.
From vertex D drop perpendicular to meet opposite face ABC at point G, which is the centre of the triangle ABC.
Here, G is also centroid of triangle ABC as it an equailateral triangle.
Clearly, `AG=2/3AM`
In triangle `AMB, AM=ABsin60^@=sqrt3/2l cm`
`:. AG=2/3xxAM=2/3xxsqrt3/2l=1/(2sqrt3)lcm`
In triangle AGD, using Pythagoras, we get
`AD^2=AG^2+GD^2`
`:. l^2=1/3l^2+GD^2`
`:. GD=sqrt(2/3)1cm`
For angle between any edge and face not containing that edge, consider edge AD and face ABC.
Angle between them is `alpha`,which is angle between AD and AM.
In triangle AGD, `cosalpha=(AG)/(AD)=(1/sqrt3l)/l=1/sqrt3`
Angle between two faces ABC and BCD is the angle between AM and DM, which is `beta`.
In triangle `DGM,cos beta=(GM)/(DM) =1/3`