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):

enter image description here

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

  • $\begingroup$ You already know all the angles of the projection since it is equilateral, and also the median BD in that case is also the angle bisector (as well as being perpendicular to AC). $\endgroup$
    – coffeemath
    Apr 4, 2014 at 9:05
  • 1
    $\begingroup$ There is no such situation. A projection that leaves $\triangle ABC$ equilateral would place $D$ at the triangle's center, not along its edge. $\endgroup$
    – Blue
    Apr 4, 2014 at 9:58
  • $\begingroup$ Please excuse me, my math English is poor. I meant an isosceles triangle ABC (AB = BC ≠ AC), not equilateral, of course. I've updated the text and the image in the question. $\endgroup$ Apr 4, 2014 at 12:56

1 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°
  • $\begingroup$ Do atg and actg mean the same thing as “arctan”? Could you add some explanations? Formula only answers are not really helpful. $\endgroup$
    – egreg
    Apr 4, 2014 at 13:59
  • $\begingroup$ @egreg — edited $\endgroup$ Apr 4, 2014 at 14:08

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