Maciej Kosinski wrote: > > How to prove that the angule of the regular tetrahedron is 109 > degrees (or was it 109,5 degrees) in NH3 molecule? > One of my ideas is to use inscribed and outscribed > spheres which will give the answer: 2*arctan(r/R), where r is > radius of the inscribed sphere and R is the radius of the out > scribed sphere, but I don't know the relation of the r and R. > Please help.
OK, each face of the tetrahedron is an equilateral triangle. Let's consider the side (edge of tetrahedron) to have length 1.
B e e be e e b e e e b e e e b e e e b e e F e f b e e f e f b e e f d d C D d d a e e a A e e e e e e e e
In the diagram, showing one face, the edges are identified as "e". The center of the bottom face (not shown) is C. The perpendicular bisector of an edge of the bottom face is "a", and point A is the midpoint of the edge (not drawn to scale). "d" is an angle bisector of the bottom face, intersecting line "a" at point C. (I know it's crooked in my drawing; please imagine a straight line there.) "b" is an altitude of the tetrahedron. A, B, C, D, and F are points. Each of the edges "e" has length 1.
BD = 1
The altitude of the triangle with a side of 1 is sin(60 deg.) = distance AB (line segment not shown here; the diagram gets messy fast as one adds lines) AB = sqrt(3)/2
AD = 1/2 AC = length of line segment "a" = AD*tan(30 deg.) = (1/2)*tan(30 deg.) = (1/2)*(1/sqrt(3)) = sqrt(3)/6 DC = length of line segment "d" = (1/2)/cos(30 deg.) = (1/2)*(2/sqrt(3)) = sqrt(3)/3
The angle that I think you want is the angle as measured at the center of the tetrahedron to (any) two vertices.
We need a point on line "b" equidistant from B and D.