This is going to be so obvious, but I’ve forgotten:
The sum of the interior angles of a triangle is 180º. Any other polygon, the sum is 360º.
Why?
My guess is that a triangle actually has four sides and four angles, with one of the “angles” being degenerate and measuring 180º. Am I on the right track?
I think I speak for most of the boards here when I say:
Whatchu talkin’ about, Willis?
More seriously, the sum of the interior angles of “any polygon” is not always 360 degrees, only in squares and other four-sided figures. Take a pentagon, for example; in a regular figure there are five internal angles of what – 108 degrees? I forget – anyhow, five angles larger than 90 degrees, so obviously the sum of all of them will be more than 360.
Or are you talking about something else and I’m the one not getting it?
I believe the formula for interior angles of a polygon is:
(n-2)*180= sum of interior angles, where “n” is the number of sides of the polygon.
I’m not sure that you’re on a track at all. Or maybe I just don’t understand the question.
I believe the formula for interior angles of a polygon is:
(n-2)*180= sum of interior angles, where “n” is the number of sides of the polygon.
I’m not sure that you’re on a track at all. Or maybe I just don’t understand the question.
I think you meant quadrilateral when you said polygon. A triangle has 180 degrees. To add one more side, you can just put a vertex at some point on one of the sides; Then you have a forth vertex with 180 degrees. Every time you add a side, you add 180 degrees. So if n is the number of sides, the number of degrees will be 180 + (n-3)*180 or (n-2)*180. A regular pentagon, for example, will have (5-2)*180/5 = 108 degrees at each vertex
Planely, I am assuming Euclidian geometry here.
I was going to post the formula too, but got here too late. I did want to say, DrMatrix, that you slay me.
My guess is that Mjollnir confused interior and exterior angles.
The exterior angles of any polygon will sum to 360 degrees. (This is observably true. If I start at one vertex and walk along the perimeter of the figure, at each turn corner I will turn an angle equal to the exterior angle at that vertex. When I’ve completed walking all the way around, I’ll have gone in a complete circle and thus have gone through 360 degrees.)
So a triangle has exterior angles summing to 360, like all the other polygons.
As already explained, the sum of the interior angles of polygons varies with the polygon. 180 for a triangle, 360 for a quadrilateral, etc. The above fact can be used to derive the formula for regular polygons (since the angles are all equal, and interior angles will be supplementary to the exterior angles).
Could this be what caused the confusion?
Actually, panamajack, that’s not right. The sum of the exterior angles will be:
(n+2) * 180
For example, on a square, the exterior angles are each 270 degrees (360-90). The sum is 1080.
I don’t think anyone has given an explanation of how the angles of a triangle add up to 180[sup]o[/sup], so here’s one. It’s easy to see if you draw the right picture. Picture a triangle, now pick a vertex and draw a line through that vertex, parallel to the side of the triangle opposite that vertex. Obviously the three angles at this vertex (the angle of the triangle, and the two on either side of it) add up to 180[sup]o[/sup], and it’s easy to see these angles are equal to the angles of the triangle.
To show that the interior angles of an n-gon add up to 180(n-2), just notice that it’s easy to divide an n-gon into n-2 triangles.
panamajack is correct. The exterior angles are the angles from the extension of the side you are on to the next side.
______ ........
| |A
| |
| |
|______|
If that was a square (trust me, it’s a square - I’d draw a picture, but I don’t know if I can post it here?) then the exterior angle is the angle between the dotted line and the adjacent side. (180 - the interior angle). This is the angle you turn when walking around the square.
The sum of these exterior angles is always 360 degrees for any closed shape, so if you walked around the shape, you’d end up in the same position facing the same way after completing a full revolution.
This was a good example of petitio principii (and you know who you are).
And DrMatrix, that was a good point, although not affine one.
Okay, maybe it’s been too long since I took geometry. The way I was defining exterior angle was a little different. I was thinking this more along the “lines” of this:
/
/ A
/____
with A (say 60[sup]o[/sup]) being the interior angle and the exterior angle being the reciprocal angle (300[sup]o[/sup]) being the exterior angle. The sum of these would then follow my formula.
Using your definition, which I think is the correct one, don’t you have more exterior angles than sides of a polygon? Modifying your diagram a little:
: :
:C B:
.... ______ ....
D| |A
| |
E| |H
....|______|....
:F G:
: :
Wouldn’t all eight be considered exterior angles? If so, than a polygon has n*2 exterior angles whose sum is 720[sup]o[/sup].
In other words, if you walk in a circle, you walk 360[sup]o[/sup].
Quoth Cabbage:
Unless, of course, you’re Riemann, in which case it’s not easy at all to see that. Which is why, of course, the sum of the angles of a triangle isn’t 180[sup]o[/sup] in non-Euclidean geometries.
JeffB, you only put one exterior angle at each vertex. It doesn’t matter which one you choose, of course, but they’re usually drawn either all clockwise or all counterclockwise.