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A quadrantal spherical triangle together with Napier's circle for use in his mnemonics. A quadrantal spherical triangle is defined to be a spherical triangle in which one of the sides subtends an angle of π /2 radians at the centre of the sphere: on the unit sphere the side has length π /2.
The snub square antiprism (J 85) can be seen as a square antiprism with a chain of equilateral triangles inserted around the middle. The sphenocorona ( J 86 ) and the sphenomegacorona ( J 88 ) are other Johnson solids that, like the square antiprism, consist of two squares and an even number of equilateral triangles.
Finite spherical symmetry groups are also called point groups in three dimensions. There are five fundamental symmetry classes which have triangular fundamental domains: dihedral, cyclic, tetrahedral, octahedral, and icosahedral symmetry. This article lists the groups by Schoenflies notation, Coxeter notation, [1] orbifold notation, [2] and order.
Because these are double angles, each of q, r, and s represents two applications of the rotation implied by an edge of the spherical triangle. From the definitions, it follows that srq = uw −1 wv −1 vu −1 = 1, which tells us that the composition of these rotations is the identity transformation. In particular, rq = s −1 gives us
An area formula for spherical triangles analogous to the formula for planar triangles. Given a fixed base , an arc of a great circle on a sphere, and two apex points and on the same side of great circle , Lexell's theorem holds that the surface area of the spherical triangle is equal to that of if and only if lies on the small-circle arc , where and are the points antipodal to and , respectively.
The conjugacy definition would also allow a mirror image of the structure, but this is not needed, the structure itself is achiral. For example, if a symmetry group contains a 3-fold axis of rotation, it contains rotations in two opposite directions. (The structure is chiral for 11 pairs of space groups with a screw axis.)
As can be seen from Fig. 1, these problems involve solving the triangle NAB given one angle, α 1 for the direct problem and λ 12 = λ 2 − λ 1 for the inverse problem, and its two adjacent sides. For a sphere the solutions to these problems are simple exercises in spherical trigonometry , whose solution is given by formulas for solving a ...
Consider the projective (spherical) triangle at the point ; the vertices of this projective triangle are the three lines that join with the other three vertices of the tetrahedron. The edges will have spherical lengths α i , j , α i , k , α i , l {\displaystyle \alpha _{i,j},\alpha _{i,k},\alpha _{i,l}} and the respective opposite spherical ...