# Deducing angles in an equilateral triangle with normals to surfaces

I have the following equilateral triangle:

In the following picture of the same equilateral triangle, the dotted lines are the normals to the surfaces:

Since this is an equilateral triangle, we know that the three angles are $$60^\circ$$. And the since the dashed lines are normal to the surfaces of the triangle, we know that the angle they make with the triangle surface is $$90^\circ$$. Using these two facts and some geometry, we can deduce that the angle between the left-most red arrow and the dashed line is $$30^\circ$$, as shown in the above image.

As you can see, I understand the process here, but I'm missing the geometry knowledge that allows one to deduce that the angle between the left-most red arrow and the dashed line is $$30^\circ$$. What is the geometrical reasoning here? Is it similarity of triangles? I would greatly appreciate it if people could please take the time to explain this.

• There seems to be a missing piece of information here. We have to know the exact direction of the red arrow first. Is it parallel to the bottom edge of the triangle? – David K Nov 4 '19 at 3:39
• @DavidK The problem begins as follows: "A beam of light is incident upon an equilateral prism with index of refraction of $1.5$, pictured below." As you can see, it is a physics problem, but it does not give any information with regards to what you're describing. Even so, it seems to me that the picture is implicitly implying that the arrow is parallel to the bottom edge? I mean, if it wasn't parallel to the bottom edge, then the angle it makes with the normal (dashed) line wouldn't be $30^\circ$, right? – The Pointer Nov 4 '19 at 3:43
• Well, shame on that author then. Just making something that looks like it might be parallel is not the same as declaring or marking it parallel. But it does seem practically certain that they meant it to be parallel. – David K Nov 4 '19 at 3:47
• @DavidK Yes, sorry about that. I'm obviously even more lost than you are, so I'm just trying to learn from what I've been provided with, haha. – The Pointer Nov 4 '19 at 3:48
• @ThePointer What is the source of your question, could you please give a reference from where you got this question? – Rene Duchamp Nov 13 '19 at 15:55

The particular angles involved in your question are called alternate angles, but to be honest I had to look that up. What I remember is just that in the figure all the angles labeled $$1$$ are equal to each other and all the angles labeled $$2$$ are equal to each other.
Now in the problem you're looking at, the first red arrow and the bottom edge of the triangle are parts of two parallel lines, and the left edge of the triangle is part of another line that crosses both of those parallel lines. If you extended all these lines you'd have a lot of angles equal to each other, but even without the other parts of the lines we can still see two angles that are equal, and since the angle inside the triangle is $$60^\circ$$ the other is also $$60^\circ$$. These two angles are marked $$60$$ in the figure below.
Now the angle between the normal and the edge of the triangle is a right angle, $$90$$ degrees. The red arrow divides this angle in two parts, one of which is $$60$$ degrees, so the other part is $$30$$ degrees.