Number of right angled triangles formed in a cube by joining three of its vertices We can form a maximum of 4 right angled triangles in a rectangle/square. Due to symmetry any right angled triangle in a cube will also be a part of a rectangle/square formed by its vertices.  
I can count 6 faces $(a\times a)$ and 6 rectangles formed by opposite edges of the cube $(a\times \sqrt 2a)$ So a total of 48 triangles. The answer given in my textbook is 72. 

The answer of your book is simply impossible. In fact a cube has $8$ vertexs. This means that also taking all of the possible non-ordered triples of vertex:
$${8\choose 3} =56<72$$
:)
