I might post the detailed mathematics here , but I think I ' ve found that a d 10 inscribed in a sphere would have two right angles on each face , with an angle just under 52 degrees at the apex , and just over 128 degrees for the remaining angle triangle 0 , 0 , 1 , 2 sqrtsqrt 51 % 2 F 5 , 0 , 2 sqrt 55 % 2 F 5 , sqrt 1 % 2 F 5 % 2 B 1 % 2 Fsqrt 5 , sqrt 1 % 2 B 3 % 2 Fsqrt 5 , 12 % 2 Fsqrt 5 This dotproduct calculation , using the relevant two sides of this triangle , shows that the d 10 does in fact have right angles on its faces dotproduct 0 , 0 , 1 sqrt 1 % 2 F 5 % 2 B 1 % 2 Fsqrt 5 , sqrt 1 % 2 B 3 % 2 Fsqrt 5 , 12 % 2 Fsqrt 5 , 2 sqrtsqrt 51 % 2 F 5 , 0 , 2 sqrt 55 % 2 F 5 sqrt 1 % 2 F 5 % 2 B 1 % 2 Fsqrt 5 , sqrt 1 % 2 B 3 % 2 Fsqrt 5 , 12 % 2 Fsqrt 5 The key is that if the d 10 is centered at the origin , and oriented so that the apexes are on the zaxis and so that one of the top faces is bisected by the xzplane , the azimuthal angle of the vertex of that face that is opposite to the apex will be arccos 12 sqrt 5 , or about 96 . 06 degrees . I might also figure out the angles between faces , or even actually plot the d 10 itself . I suspect , however , that an actual d 10 wouldn ' t have these proportions , because once the die lands with one apex down , it would be too unlikely for it to bounce and land with the other apex down that is , it wouldn ' t be fair . It is also curious to figure out the proportions of a d 10 that has the property that an inscribed sphere would be tangent to all faces , at their geometric centers .