Finding out the missing angle

Question

Path of a light ray passing through a $90, 60, 30$ prism(I am not allowed to post images yet)

I found this picture in my Physics book, where it was said that $\angle FED$ is $30°$. I cannot figure out how, is it possible to prove it geometrically?

Also $\angle CEF$ and the angle at $D$ is a right angle.

Answer 1

Here, $\angle CDG = 90° \therefore \angle CDE=90-60=30°$.

Since sum of all angles in a triangle is $180°$, in $\triangle CED$, $\angle CED=180-60=120°$.

$$\because\angle CEF=90°$$ $$\angle FED=\angle CED-\angle CEF = (120-90)°= 30°$$

Answer 2

$\angle EDC=90^\circ-60^\circ=30^\circ$ and $\angle FEC=90^\circ$. Also, the angles of $\Delta CED$ must add up to $180^\circ$ (and hence, $\angle CED=120^\circ$).

Does that help?