geometry
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.
Your best bet is to setup an xyz-coordinate system as shown.
After determining the co-ordinates of each point in your figure (eg B = (0, 1, 0)), write down the equation of EM and BP.
The angle between them can then be found.
http://www.nabla.hr/PC-LinePlaneIn3DSp2.htm is a website that provides examples.
Another question asks for the dihedral angle of a general tetrahedron in terms of its edge lengths. I chose to answer by giving a formulas in terms of face areas (because those relations deserve to be better-known). Most-relevant here is this version, for a tetrahedron $OABC$, ... $$W^2=X^2+Y^2+Z^2-2YZ\cos\angle OA - 2ZX\cos\angle OB-2XY\cos\angle OC \tag{$\star$}$$ where $W := |\triangle ABC|$, $X := |\triangle OBC|$, $Y := |\triangle AOC|$, $Z := |\triangle ABO|$ are face areas, and $\angle OA$, $\angle OB$, $\angle OC$ are dihedral angles along respective edges $\overline{OA}$, $\overline{OB}$, $\overline{OC}$.
In a regular tetrahedron, all face areas are equal ($W=X=Y=Z$), as are all dihedral angle measures ($\angle OA = \angle OB = \angle OC$), so $$W^2 = 3 W^2 - 6W^2\cos\angle OA \qquad\to\qquad \cos\angle OA = \frac{1}{3} \tag{$\star\star$}$$
I'll leave calculating the angle measure itself to the reader. $\square$
FYI ... The $0$-dimensional point, $1$-dimensional segment, $2$-dimensional triangle and $3$-dimensional tetrahedron have higher-dimensional counterparts, each known simply by the umbrella term "$n$-dimensional simplex" (or "$n$-simplex"). The bounding elements (counterparts of "vertices", "edges", and "faces") are called "facets", and we can write, for $n\geq 2$ ...
Fun Fact. The angle between two neighboring facets of a regular $n$-simplex is $\operatorname{arccos}\frac{1}{n}$ .
So, always remember and never forget: $\cos 60^\circ$ isn't merely "half of one"; it's the reciprocal of the equilateral triangle's dimension!
Best Answer
Yes, it is definitely possible! :-)
Let assume the length of all sides is $a$. Then the diagonals of the base square are of length $a\sqrt{2}$. Thus, in the rectangular triangle $PDS$ we have $|DP|=a\sqrt{2}/2$ 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.