Do External Angles of Polygons Always Sum to 360 Degrees?Date: 02/13/2007 at 23:12:17 From: Cole Subject: External angles of polygons Please explain how a polygon with 16 right angles can have a total external angle of 360 degrees? I know the rule that says that the sum of all external angles of all polygons = 360 degrees, but I don't understand how it can work on all polygons when some polygons can have an infinite number of right angles which would make the sum of all external angles much more than 360 degrees. Date: 02/14/2007 at 12:46:40 From: Doctor Peterson Subject: Re: External angles of polygons Hi, Cole. A polygon can't have an _infinite_ number of angles, but I think I know what you mean; it can have as many right angles as you want, and thus can have more than enough to invalidate the claim. The trick is that in order for that to happen, it has to be a concave polygon, with some of the angles bending outward rather than inward as usual (for a convex polygon). And such reversed angles have to be counted as negative. Let's take a simple example: A-----B | | | | | C-----D | | | | F-----------E Here we have six right angles, which, as you suggest, add up to 6*90 = 540 degrees rather than 360. But look at the actual external angles (I'm going clockwise starting at A): |90 A-----B--- | |90 | | | C-----D--- | |-90 |90 90| | ---F-----------E 90| One angle bends the opposite way from the others, and is taken as negative. Then the total angle is 90-90+90+90+90+90 = 360. The claim you are challenging is true for all convex polygons without reservation; when applied to concave polygons, you have to interpret it in this way, with signed angles. See this page for another example: Sum of Interior and Exterior Angles http://mathforum.org/library/drmath/view/62228.html If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/