I was just wondering if anyone had any idea how to solve this problem:
What is the angle between lateral faces of a rectangular-based pyramid with length a, width b, and height h, in terms of a, b and h?
Any responses are much appreciated.
[Math] Angles between lateral faces of any rectangular-based pyramid
anglegeometrytrigonometry
Related Solutions
The lengths of the sides of the regular tetrahedron are irrelevant.
Hint: Find the vectors from the tetrahedron's center to the center of two faces. Then calculate the angle between them.
The answer comes from spherical trigonometry, but don't let the "spherical" part throw you.
Let $a$, $b$, $c$ be the "face-corner" angles (bounded by pairs of edges), and let $A$, $B$, $C$ be the "dihedral angles" (bounded by pairs of faces), with $a$ opposite $A$, $b$ opposite $B$, and $c$ opposite $C$.
The two Spherical Laws of Cosines relates these angles thusly:
$$\begin{align} \cos c &= \phantom{-}\cos a\cos b+\sin a \sin b \cos C \tag{1}\\[4pt] \cos C &= -\cos A\cos B+\sin A\sin B \cos c \tag{2} \end{align}$$
Note: If the plane containing face-angle $c$ is perpendicular to the edge along dihedral angle $C$, then $a=b=90^\circ$ so that $(1)$ reduces to $\cos c=\cos C$, so that $c=C$. This is consistent with OP's parenthetical about those angles matching in this situation.
FYI: There's also the Spherical Law of Sines:
$$\frac{\sin a}{\sin A}=\frac{\sin b}{\sin B}=\frac{\sin c}{\sin C} \tag{3}$$
Best Answer
Set up a coordinate system such that pyramid vertex is $V=(0,0,h)$ and base vertices are $A=(b/2,a/2,0)$, $B=(-b/2,a/2,0)$, $C=(-b/2,-a/2,0)$, $D=(b/2,-a/2,0)$.
A vector $N_1$ perpendicular to face $VAB$ (and directed outwards) can be found by $$ N_1=(A-V)\times(B-V)= \left({b\over2},{a\over2},-h\right)\times\left(-{b\over2},{a\over2},-h\right) =\left(0,hb,{ab\over2}\right). $$ Analogously, one can find a vector $N_2$ perpendicular to face $VAD$ and directed inwards: $$ N_2=-(D-V)\times(A-V)= -\left({b\over2},-{a\over2},-h\right)\times\left({b\over2},{a\over2},-h\right) =-\left(ha,0,{ab\over2}\right). $$ Angle $\theta$ between those faces is also the angle between $N_1$ and $N_2$, so it can be found by $$ \cos\theta={N_1\cdot N_2\over|N_1|\cdot |N_2|}= {-1\over\sqrt{1+4h^2/a^2}\sqrt{1+4h^2/b^2}}. $$