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This is a really basic question, yet I can't remember my old geometry classes nor could I find an answer via google.

Given a circle "tilted" at angle a to the horizontal plane, and given angle b inside the circle, I want to calculate the vertically projected angle, c, on the horizontal plane.

It's obvious that the circle's vertical projection on the plane is an ellipse, and that cos(b) = cos(c) but I still can't work out the formula that gives me c given only the 2 first angles.

I'm assuming this can be worked out regardless of the dimensions of the circle, although temporary arbritary dimensions can be used to calculate the solution.

projected angle

Note: As an example, in the above sketchup drawing a = 70 degrees, b = 45 degrees, and (what I want to calculate mathematically) c = 18.8 degrees.

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Imagine that the circle is centered at the origin of a (titled by angle $a$) cartesian $x$-$y$ plane, with the $x$-axis along the intersection of the circle’s diameter and the ellipse’s major axis. Projection of points from the tilted plane onto the horizontal plane leaves $x$-coordinates alone and scales $y$-coordinates by a factor of $\cos a$. So the point $(\cos b,\sin b)$ (on the circle at angle $b$) is projected onto the point $(\cos b,\cos a\sin b)$ on the horizontal plane. Then $\tan c=\frac{\cos a \sin b}{\cos b}$, or $c=\arctan(\cos a \tan b)$. Note that $\arctan(\cos 70^\circ \tan 45^\circ)\approx18.882^\circ$

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  • $\begingroup$ Brilliant. Thank you! $\endgroup$ – dude Mar 18 '15 at 3:16

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