Imagine to have a right angle in 3d space, like a large capital "L" floating in space.

Averaged from all observation directions, what is the average projected/visible angle of the "L" shape?

(Of course, the legs of the "L" are supposed to be infinitely thin and long.)

Usual disclaimer: this is not homework; it is a question that arose in discussions. Is the average different from a right angle? What is the simplest way to answer the question?

Edit: as MvG mentioned, the average of the absolute angle is meant.

  • 1
    $\begingroup$ I assume you mean the absolute value of the angle, because of you were to consider oriented angles, then for reasons of symmetry the positive and the negative contributions would cancel out and leave you with a zero average. Have you tried a Monte Carlo simulation to get a rough idea of what the answer should approximately be? $\endgroup$
    – MvG
    Apr 23 at 8:25
  • $\begingroup$ Yes, I mean the absolute angle. No, I have not yet tried a simulation. A good idea! $\endgroup$
    – KlausK
    Apr 23 at 8:51

1 Answer 1


We can easily prove that the average apparent angle is a right angle.

Extent one of the rays backwards past the vertex so that you have two adjacent right angles. When viewed (or more formally, projected onto any plane), the sum of the apparent angles is then always 180° because the extended ray becomes a straight line in any view. So the average apparent measure for either right angle alone is half of 180°, thus back to 90°.

This proof applies to any dimensionality of space sufficient to define angles.

  • $\begingroup$ Great answer. Thank you! As a modification, if the angle is 60 degrees instead of 90 (thus one third of 180) can I reason in the same way? (I promise not to ask about more angles :-) $\endgroup$
    – KlausK
    Apr 23 at 9:52
  • 1
    $\begingroup$ Works for any divisor of 180°, then by addition for any rational multiple of 180°, then by continuity for any real angle. $\endgroup$ Apr 23 at 10:00
  • $\begingroup$ Wonderful. The solution is beautiful. $\endgroup$
    – KlausK
    Apr 23 at 10:07

Not the answer you're looking for? Browse other questions tagged .