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I'm trying to review exterior angles after many years. It's my understanding that the sum of a polygon's exterior angles must equal 360°. How would you find the exterior angles in this polygon?

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  • $\begingroup$ $90°$$ \text{ }$ $\endgroup$
    – user93957
    Nov 13, 2013 at 18:55

2 Answers 2

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You've got most of it already.

An algorithm for drawing the exterior angles would be to visit each vertex in turn (clockwise, say). As you hit each vertex and change direction, you extend and label the exterior angles as you have with the solid red lines. The extensions are parallel to the "new" direction, just as you have them.

Now, the one with the question mark shoots inside the polygon. So, I'd keep the horizontal dashed line, get rid of the vertical dashed line, and put "-90 degrees" above the horizontal line.

Add all of them up now, and you get 360 degrees.

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For a consistent convention of positive counterclockwise angular rotation...measured from previous tangent direction of a polygon to present direction,

it is $270^0$ turn at the re-entrant vertex marked $?$

Five regular right angles shown make up $450^0.$

The total is $720^0$ or two full rotations.

For a polygon with arbitrary number of re-entrant vertices the anticlockwise sum should be an integer multiple of $360^0.$

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