I thought I could solve this quick. "Exterior angles of a polygon"? I remember that, they always sum to 360˚. Well, kind of. The exterior supplementary angles of a polygon always sum to 360˚. See https://www.mathsisfun.com/geometry/ext ... ygons.html to see what I'm talking about.

The angles shown here, the angles around each vertex (or point) are not the same as "exterior angles of a polygon" one may have heard about before. As Greenlight showed, they are 360˚ minus whatever the interior angle is. That seems obvious when you look at it.

Given that the ratio is just asking for the comparison of sums of angles, we could say the numerator is 180˚, because we know the sum of interior angles of any triangle is 180˚.

As for the sum of the exterior surrounding angles. We could just look at one example, and then multiply that by 3. Since we know the average angle for the interior angles of the triangle is 60˚. We could say the average exterior surrounding angle is 360˚ - 60˚ which is 300˚. Since there are 3 exterior surrounding angles, we can just add 300˚ three times to find the sum of the exterior surrounding angles. So we get 3(300˚) = 900˚.

So, we have \(180˚/900˚\) = \(\frac{18}{90}\) = \(\frac{2}{10}\) = \(\frac{1}{5}\)




- Extra

I looked into this sort of problem some more, and based on my own calculations (I'm sure someone else has discovered it too), there is a formula one can use to find the sum of the exterior surrounding angles for any polygon. Given n is the number of sides (or vertices) of a polygon, it is:

sum of exterior surrounding angles of polygon = (n+2)(180˚)

You may find this looks familiar. It looks remarkably similar to the formula for finding the sum of the interior angles of a polygon, which is:

sum of interior angles of polygon = (n-2)(180˚)

One can see, the only difference between the two formulas is that a plus symbol is used for the sum of the exterior surrounding angles.

If one knew this, one could use these formulas to solve this problem, like so.

Given: a triangle


\(\frac{sum Interior Angles}{Sum Of Exterior Surrounding Angles}\) =

\(\frac{(n-2)(180)}{(n+2)(180)}\)

. the 180 values cancel out leaving

\(\frac{(n-2)}{(n+2)}\)

. we can plug in 3 for the number of sides and we get

\(\frac{(3-2)}{(3+2)}\)

\(\frac{(1)}{(5)}\)


Why has this formula not been mentioned before? I don't know. It works though. For example, what is the sum of the surrounding exterior angles for an octagon? We could find the value for each individual interior angle, and then subtract that from 360˚ to give us the value for each surrounding exterior angle, and then multiply that by 8 to get our answer, like so:

. sum of interior angles of octagon = (n-2)(180) = (8-2)(180) = (6)(180) = 1080

. value of one interior angle of octagon = 1080˚/8 angles = 135˚ per angle

. value of one exterior surrounding angle = 360˚ - 135˚ = 225˚

. sum of all exterior surrounding angles of octagon = 225˚/angle • 8 angles = 1800˚


Now let's compare that to the formula for surrounding exterior angles for polygon

. sum of surrounding exterior angles for polygon = (n+2)(180˚) = (8+2)(180˚) = (10)(180˚) = 1800˚

I also calculated sum of exterior surrounding angles for square, pentagon, hexagon, using both methods, that's where I noticed the 180˚ step

sum exterior surrounding angles triangle = 900˚
sum exterior surrounding angles square = 1080˚
sum exterior surrounding angles pentagon = 1260˚

You can see the sum goes up by 180˚ for each extra side or angle of the polygon.

Conclusion on Extra section
Is it useful to know that the sum of exterior angles of a polygon are equal to (n+2)(180˚)? A little. If you can easily remember formulas then it doesn't seem like it would hurt to learn it. If you have difficulty memorizing formulas, then you can easily pass on it. For example, just by looking at one corner of a polygon and extrapolating from that one could solve this type of problem. At least, that's my thoughts on it.

PS I tried using   to make spaces in formula, but it didn't work, so I just used camel case.

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