# Angles enclosed in parallel lines

Imagine that you are going for a walk, from west to east, in the direction of the arrow along the top line. You divert to take the zigzag path, adding up the clockwise angles of diversion each time you turn a corner, and end up still going east with the arrow on the bottom line. In degrees, the total angle turned is $$0$$ :$$0=60+(180-a)-90+(180-b)-120-c.$$So $$a+b+c=210$$.

If we draw a line perpendicular to the parallel lines on the left of the zigzag, we enclose an octagon whose angles should sum to $$1080^circ$$. Subtracting the given angles inside the octagon – two $$90^circ$$ angles on the perpendicular, then $$120^circ,270^circ,300^circ$$ – gives $$a+b+c=210^circ$$.

I got it by the following way:
$$90^{circ}+180^{circ}-alpha+120^{circ}-gamma+180^{circ}-beta=360^{circ},$$
which gives $$alpha+beta+gamma=210^{circ}.$$