My friend finished making dodecahedrons, and her next project is to make regular icosahedrons (20-sided dice). Again she wants to know: what is the dihedral angle between any two adjacent faces?
Perhaps this can be solved without the use of spherical trig? ;P
An icosahedron has five triangular faces meeting at each vertex. Erect a small sphere centered upon one vertex. The five faces intersect the sphere in five arcs of great circles forming a spherical regular pentagon. As each side of the pentagon is the arc subtended by one angle of a plane equilateral triangle, each is 60 degrees. The angles of the spherical pentagon are the dihedral angles of the original icosahedron.
Connect the center of this spherical pentagon (on the sphere) to each of the pentagon’s corners and to each of the centers of the five sides, forming 10 congruent spherical right triangles (5 of each handedness).
The angle of each right triangle that is located at the center of the pentagon is 360/10 degrees, or 36 degrees. The angle at the midpoint of each side is 90° by symmetry. The leg opposite the 36-degree angle is half the length of the side of the pentagon, and therefore 30 degrees.
In spherical trigonometry there is a version of the law of cosines that relates three angles and one side:
cos A = -cos B cos C + sin B sin C cos a, where capital letters refer to angles and lower case to sides, and side a is opposite angle A.
As the angle and side pair we know are 36 degrees and 30 degrees, and we know the angle adjacent to the 30-degree side is 90 degrees, we get
cos(36) = - cos(90) cos C + sin(90) sin C cos(30)
and as cos(90)=0, we can solve
sin C = cos(36)/cos(30)
Angle C is half the angle of the spherical pentagon, that is, half the dihedral angle sought, and therefore acute. The dihedral angle is twice this, and therefore 2 arcsin(cos(36)/cos(30)) = 138.1896851042214… degrees.
Posted by Charlie
on 2003-04-17 03:59:45