Angles enclosed in parallel lines 
I am not sure of the best way to answer this question. We have to find $a + b + c$.
I have got the answer of $210$ by drawing in parallel lines to split the angles $a$, $90$, $b$ and $60$, but then that is quite demanding for younger students. So I was wondering if there was a quicker solution?
 A: 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 $a+b+c=210$.
A: 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 $a+b+c=210^\circ$.
A: I got it by the following way:
$$90^{\circ}+180^{\circ}-\alpha+120^{\circ}-\gamma+180^{\circ}-\beta=360^{\circ},$$
which gives $$\alpha+\beta+\gamma=210^{\circ}.$$
A: Supplemet of angle $120$ is $180 -120=60$
then
the sum of the angles facing to the left$=$the sum the angles facing to the right
$a+b+c=60+90+60=210$
A: In order that a heading be maintained irrespective of traversed distances, to depend purely on angle summation,
$$\sum{clockwise \;angles}=\sum{anti-clockwise \;angles} $$
convention clockwise <0, anticlockwise >0
$$ 60+90+60=210=a+b+c $$
How is it an Olympiad problem?
A: Draw a horizontal line at each vertex: these cut each intermediate angle in two, and use that the two acute angles in a shape of Z agree:
$$a=a_1+a_2,\ a_1=60,\ 90=a_2+b_1,\\  b=b_1+b_2,\ 60=b_2+c\,.$$

 $a+b+c=a_1+a_2+b_1+b_2+c=60+90+60$

