# Orthogonal projection of angles

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).

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.

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

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 $$\tan\alpha' = \tan\alpha \cos\beta.$$
• 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? Jul 2, 2019 at 16:48