Cfrgtkky

I want to find out about the angle $angle PdS$ of a square-based pyramid with four equilateral triangle faces, but I am limited by my knowledge of trigonometry and I would like to know if is even possible to solve for it. If it is possible to find out the angle of elevation/depression of a four-faced equilateral triangle pyramid, how do I apply the concept for $n$-faced equilateral triangle pyramid?

Here is what I have so far.

Let's consider the pyramid whose peak is above the unit square with vertices at $(0,0,0),(0,1,0),(1,0,0),(1,1,0).$ Since the peak is equidistant from these vertices, then it will be at the point $(1/2,1/2,z)$ for some positive $z.$ More precisely, it is at distance $1$ from the vertices, and from the origin, in particular. The distance formula then gives us $$1=left(frac12-0right)^2+left(frac12-0right)^2+(z-0)^2=frac14+frac14+z^2=frac12+z^2.$$ Consequently, $z=sqrt{frac12}=frac{sqrt{2}}2.$ Now, to find the angle of elevation of a face, we consider a right triangle with vertices at the center of the base, at the peak, and at the midpoint of one of the base's edges. The leg opposite the angle of elevation has length $frac{sqrt{2}}2,$ as we found earlier, and the adjacent leg has length $frac12,$ so the tangent of the angle of elevation is $frac{sqrt{2}/2}{1/2}=sqrt2,$ so that the angle of elevation is the arctangent of $sqrt2,$ roughly $54.74$ degrees. The angle of elevation of an edge is easier. In that case, the adjacent leg has length $frac{sqrt{2}}2$ as well, so that the angle of elevation is $45$ degrees.

We can do something similar for a pyramid whose base is a regular pentagon, though we must first determine its vertices, which is more difficult than for the square. We can also do this for a tetrahedron--a pyramid whose base is also an equilateral triangle. However, for $n=6,$ our "pyramid" will be degenerate--that is, it will have height $0,$ since a regular pentagon can be decomposed into six equilateral triangles. For $n>6,$ the base angles become too large for the equilateral triangles' edges to even meet, so we don't even get a degenerate pyramid.

Let me know if you're uncertain about any of this, and I'll try to clear it up for you. Welcome to Math.SE!

Yes, it is definitely possible! :-)

Let assume the length of all sides is $a$. Then the diagonals of the base square are of length $asqrt{2}$. Thus, in the rectangular triangle $PDS$ we have $|DP|=asqrt{2}/2$ and cathedus $|DS|=a$. Using the definition of the cosine we deduct:

$cos(angle PDS)=|DP|/|DS|=sqrt{2}/2$ which leads to $angle PDS=45°$.

However there is a much nicer solution. If you look carefully, you find that the triangle $BDP$ has exactly the same sides as $BDA$. Thus their angles are identical and the solution is already there.