# Deducing angle in equilateral triangle by the formula $\phi_2 = \alpha - \phi_1$

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 $$90^\circ$$.

Furthermore, thanks to David K's excellent answer in this question, we can find the angle $$19.5^\circ$$ by using "alternate angles":

I've now been trying to figure out how the angle $$40.5^\circ$$ was found. From there, we can use Snell's law to find the angle $$77.1^\circ$$.

According to this website, the angle $$40.5$$ is found by $$\phi_2 = \alpha - \phi_1$$ in the following image:

This computation gives us $$\phi_2 = 60^\circ - 19.5^\circ = 40.5^\circ$$, as required.

However, the website does not tell us where the formula $$\phi_2 = \alpha - \phi_1$$ came from - it just states that it is found by geometry.

I would greatly appreciate it if people could please take the time to explain where $$\phi_2 = \alpha - \phi_1$$ came from; in other words, what it is called, how it was derived, the geometric reasoning behind it, etc.

$$\phi_1+x=90,\>\>\>\>\>\phi_2+y=90$$
$$x+y+\alpha=180$$
Eliminate $$x$$ and $$y$$ to get $$\phi_1+\phi_2=\alpha$$.