What is the sum of the five outer angles in a given figure of a star?

I also want to know how!

Experiment with geometry software.

Imagine walking and turning.

reflex angle

Exterior angles: around one small triangle, angles equal sum of angles of star

Inscribe regular one in circle (assuming the sum is the same for all); angles subtended.

Move one vertex to nearby one; angle at target becomes sum, other angle drops to 0; move that line, to make triangle with same angle sum

Draw AC. The vertical angles at F are congruent, so 1 + 2 = 3 + 4. Therefore the sum of the star's angles equals sum of the angles in triangle ABC.

Previous one was one of the techniques .This is a second one-

You can find the sum of the angles easily if you first think not about 
the internal angles A, but the external angles B. Imagine you're 
taking a walk around the star, following the edges, starting in the 
middle of one side, until you get back to the same place. At each 
point you will be turning B degrees. How many times will you turn 
completely around in your travels? This total turn will be the sum of 
all the "B" angles; from that you can find the sum of the "A"s.

I'll put a post in the ground in the middle of my star, and attach a 
rope to the top. Let's see how many times the rope will wrap around:

                                 /
                                /
                             3 /
                              + X3
                             /                             /        \                     /             \  X5             /       \               1
           +----------------->+------->----------+--------
          5   \         /   start   \      Y  /   X1
                  \    /             \    /
                      \       *       \/
                     /   \         /                       /        \  /                          /        /    \                         /     /           \                   X2 /  /                   \                  +                           + 4
            /    2                        X4         /                                                                                    
I first come to point 1, then to 2, then to 3. At that point, I've 
gone around once, since I'm in the same direction from the pole as the 
place I started. Since I've walked around the pole once, I've turned 
360 degrees.

Now I continue to point 4, then point 5, and back to my starting 
point, going around the pole a second time. I've now turned a total of 
720 degrees.

At each of the five points, I turned some number of degrees X. In 
general, these might all be different, so I'll call them X1, X2, X3, 
X4, and X5. Do you see why these angles are the amount I turned? At 
point 1, I was first pointing to the right along the line I drew, but 
I turned to the right until I pointed to 2, so I turned through X 
degrees.

If in five turns I turned a total of 720 degrees, then the sum of the 
five X's must be 720 degrees.

But each X is the supplement of the acute angle at that point, or 
vertex; that is, it's 180 degrees minus the angle Y at that vertex. 
This gives me this equation:

     (180-Y1) + (180-Y2) + (180-Y3) + (180-Y4) + (180-Y5) = 720

Now you can solve this equation to find what Y1+Y2+Y3+Y4+Y5 is.