Given a trirectangular tetrahedron in which the base corners are A, B, and C, what is the angle between the base and any side, measured from edge AB, BC, or CA?
Edited 10/28/2013:
I am not nearly as facile with mathematics as others on this forum. I am actually trying to fabricate an object that must fill the negative space in the corner of a carport where two walls meet a flat ceiling. Given that the negative space conforms to the dimensions of a trirectangular tetrahedron, with as an example, 90° angles at the top of all three vertices of the sides; legs of identical length; and an equilateral triangle for the base, I would imagine the angle I refer to above, i.e. the angle created where any of the sides meets the base on the interior of the tetrohedron, should be the same number, no matter where the measurement is taken along AB, BC, or CA. I'm not seeking a formula here, but rather an actual number measured in degrees. I built a model from cardboard but it was not substantial, nor accurate enough to produce an accurate, uniform measurement between the base and all three sides. The angle would appear to be somewhere around 54° but I'm not sure. I lack the geometric/trigonometric expertise to define the angle to a certainty. I would imagine this is child's play for most of you, and so, I apologize for my lack of expertise and correct nomenclature and would greatly appreciate an answer in lay terms, in the form of an exact angle. Thank you one and all for your patience and kindness.
Best Answer
You can use tetrahedral Laws of Cosines to get these angles.
Let $W$, $X$, $Y$, $Z$ be the areas of faces opposite respective vertices $O$, $A$, $B$, $C$. Write "$\angle PQ$" for the dihedral angle along edge $\overline{PQ}$. Then we have what I call the "First" Law of Cosines: $$\begin{align}W^2 &= X^2 + Y^2 + Z^2 - 2 Y Z \cos \angle OA - 2 Z X \cos \angle OB - 2 X Y \cos \angle OC & (1)\end{align}$$ as well as this "Second" one: $$\begin{align} Y^2 + Z^2 - 2 Y Z \cos \angle OA &= W^2 + X^2 - 2 W X \cos \angle BC \\[4pt] Z^2 + X^2 - 2 Z X \cos \angle OB &= W^2 + Y^2 - 2 W Y \cos \angle CA &(2) \\[4pt] X^2 + Y^2 - 2 X Y \cos \angle OC &= W^2 + Z^2 - 2 W Z \cos \angle AB \end{align}$$
In a tri-rectangular (or "right-corner") tetrahedron, with mutually-orthogonal edges $\overline{OA}$, $\overline{OB}$, $\overline{OC}$, the corresponding dihedral angles are right angles whose cosines vanish. The First Law of Cosines reduces to the tetrahedral Pythagorean Theorem:
$$W^2 = X^2 + Y^2 + Z^2 \qquad (1^\star)$$
while the aspects of the Second Law simplify to $$X = W \cos \angle BC \qquad Y = W \cos \angle CA \qquad Z = W \cos \angle AB \qquad (2^\star)$$
Faces $X$, $Y$, $Z$ being right triangles, the individual areas are given by $$X = \frac{1}{2}|\overline{OB}||\overline{OC}| \qquad Y = \frac{1}{2}|\overline{OC}||\overline{OA}| \qquad Z = \frac{1}{2}|\overline{OA}||\overline{OB}|$$ with $W$ computed via $(1^\star)$.
Another approach to $(2^\star)$ ...
If you look at the tetrahedron with, say, edge $\overline{BC}$ directed at your eye, then $\overline{BC}$ collapses to a point (call it $D$) and the tetrahedron collapses to a right triangle $\triangle OAD$ with right angle at $O$. Leg $\overline{OD}$ and hypotenuse $\overline{AD}$ are projections of altitudes from $O$ and $A$ dropped to $\overline{BC}$ in the tetrahedron. Since $$X = \frac{1}{2} |\overline{BC}| |\overline{OD}| \qquad W = \frac{1}{2} |\overline{BC}||\overline{AD}|$$ we have $$\cos \angle BC = \cos D = \frac{|\overline{OD}|}{|\overline{AD}|} = \frac{2X/|\overline{BC}|}{2W/|\overline{BC}|} = \frac{X}{W}$$
Edit. The updated question asks specifically about a right-corner tetrahedron $OABC$ where $|\overline{OA}| = |\overline{OB}| = |\overline{OC}|$. Writing "$a$" for that common length, we have $$X = Y = Z = \frac{1}{2}a^2 \qquad \qquad W^2 = 3 X^2 = \frac{3}{4}a^4 \quad\to\quad W = \frac{\sqrt{3}}{2} a^2$$ so that $$\cos \angle BC = \cos \angle CA = \cos \angle AB = \frac{X}{W} = \frac{\frac{1}{2}a^2}{\frac{\sqrt{3}}{2}a^2} = \frac{1}{\sqrt{3}}$$ whence $$\angle BC = \angle CA = \angle AB = \operatorname{acos}\frac{1}{\sqrt{3}} \approx 54.7356^\circ$$