(a) Each of these triangles has three right angles, for an angle sum of 270degrees.(b) The total angle sum is 8·270 = 2160 degrees, which is 720 degrees more than thetotal angle sum of eight ordinary plane triangles, 8·180 = 1440 degrees. The difference of720 degrees is the total angle defect of a polyhedron, as expected.6.Repeat problem 5 for either the tetrahedron or the dodecahedron. (B+S 4.6.33 or 34)Solution for the tetrahedron.(a) Each triangle has three angles. Three of these anglesmeet at the center of each face, so each angle is 360/3 = 120 degrees.The angle sum ofeach triangle is 360 degrees. (b) The sum of the angles of all four triangles is 4·360 = 1440degrees, which is 720 degrees more than the 720-degree total for four plane triangles.Solution for the dodecahedron.(a) Each triangle has three angles. Five of these anglesmeet at the center of each face, so each angle is 360/5 = 72 degrees. The angle sum of eachtriangle is 216 degrees. (b) The sum of the angles of all twenty triangles is 20·216 = 4320degrees, which is 720 degrees more than the 3600-degree total for twenty plane triangles.3
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Polyhedron, 120 degrees, Art Gallery Theorem, 72 degrees, Angle sum