This has probably already been answered but I can't find any solution using only high-school trigonometry. I'm not a native English speaker, so I probably lack the proper wording to find what I was looking for. Anyway:

In a plan $P$, I have two secant lines $d_1$ and $d_2$ making an angle of $\alpha$. I have a second plan $P'$ containing $d_2$ and making an angle of $\beta$ with $P$ (see image below).

orthogonal projection of angles

Using only $\alpha$, $\beta$ and the basic trigonometric functions, I would like to express the angle $\alpha '$ made by the projection of $d_1$ and $d_2$ onto the $P'$ plan.

I came myself to the conclusion that $\alpha ' = atan(tan \beta . cos \alpha)$, but I'm not confident at all in the path I followed to reach that. It appears I was wrong. See answer below.

  • $\begingroup$ Is $d_2$ always along the “hinge,” then? $\endgroup$
    – amd
    Jul 1, 2019 at 22:29
  • $\begingroup$ Yes, $d_2$ is always on the intersection of the planes. $\endgroup$ Jul 2, 2019 at 5:29

1 Answer 1


enter image description here

Filling in a bit more of your diagram, we are projecting the angle $\alpha$ orthogonally onto the angle $alpha'$. Since it is an orthogonal projection, the image of $C$ is a point $D$ that is the foot of the perpendicular from $C$ to the plane of $\alpha'.$

Moreover, $CD$ lies in a plane perpendicular to both of the first two planes. So have right triangles $\triangle ABC,$ $\triangle ABD,$ and $\triangle BDC$ with right angles $\angle ABC,$ $\angle ABD,$ and $\angle BDC$ respectively.

Taking the length of $AB$ as $1$ for simplicity, we have $$BC = \tan\alpha$$ (due to $\triangle ABC$) and therefore $$BD = \tan\alpha \cos\beta$$ (due to $\triangle BDC$). Looking at $\triangle ABD,$ we find that $$\tan\alpha' = \tan\alpha \cos\beta.$$

  • $\begingroup$ Thank you very much David. It helped me a lot in understanding how to solve such problems. One question though: you said "$BE = \tan\alpha \cos\beta$". Isn't that $BD$ instead? $\endgroup$ Jul 2, 2019 at 16:48
  • 1
    $\begingroup$ Yes, that was a typo. Well spotted! $\endgroup$
    – David K
    Jul 2, 2019 at 17:48

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