Calculate angles of a projection of a tetrahedron

Question

ABCD is a regular tetrahedron. It is projected on a plane in such a way that the projection forms an isosceles triangle ABC (AB = BC ≠ AC) with D lying in the middle of AC (right image):

Problem: calculate the angles of the ABC triangle of the projection (right image).

Answer 1

Height of the tetrahedron = a×√⅔ = BD
AC = a → AD = a×½
From ABD:
B = 2×atan(1/(2√⅔)) ≈ 2×31.48° ≈ 62.96°
A = C = actan(2√⅔) ≈ 58.52°