I have a question about how angles behave under projection. Consider 2 lines making an angle of $\theta$ that are rotated out of the plane (say by an angle $\alpha$). Now, the projected angle $\theta_p$ (the angle between the red lines below) is larger than the initial $\theta$, and reaches a maximum value of $\pi$ for rotation $\alpha=\pi/2$.
Can someone please give me an equation giving the projected angle $\theta_p$ for a given $\alpha$? And also tell me how to derive it. I am guessing it involves basic trigonometry, but that only relates the sides under rotation. How do I then go on to relate the angles?
EXAMPLE:
This is what the projection looks like for $\theta=\pi/4$ and rotation angle $\alpha=\pi/4$. The blue lines are the initial lines (lying in the x-y plane say), and the red lines are the rotated lines:
Side view:
ADDENDUM TO ANSWER BELOW:
The accepted response below is perfectly correct. Let me just complement with some more details. Here is the full intended picture:
Error alert: I mislabeled $\alpha$ in the above picture. It should be $\angle AOC = \alpha$, not $\angle BOD$.
Now, by symmetry, we only need to consider half of the triangle, hence the apex angle is $\theta/2$. Also, AO is the perpendicular bisector to the base, so $\angle AOB = \pi/2$.
Triangle COD is the projection of triangle AOB. Thus, CO = AO $\cos \alpha$.
- The crucial fact we need to employ is that the base length AB is unchanged under the rotation $\alpha$. This is because it is parallel to the axis of rotation.
The rest follows is simple, see below.
Check limits: For $\alpha=0$, $\theta_p = \theta$. For $\alpha=\pi/2$, $\theta_p=\pi$. Perfect!
Consider $AOB$ to be a half-angle $\theta/2$ projected onto plane $DCO$ (all angles at $C$ are right). $\angle AOC=\alpha$.
If $OA=1$, then $AB=\tan(\theta/2)$ and $CO=\cos\alpha$. Since $AB=CD$, then in triangle $COD$, $\tan\angle COD = CD/CO = \tan(\theta/2)/\cos\alpha$. Finally, $$ \tan\frac{\theta_p}2 = \frac{1}{\cos\alpha}\tan\frac\theta2 $$