This may come as a surprise, but there is more than one way to classify an exterior angle. You will want to have both of these definitions in mind when dealing with questions on the test since both instances are possible. Don’t worry! It won’t be too difficult to remember.

**Type I**: At each vertex, extend the segment, and measure the angle of the “turn”.

For any convex polygon, this kind of exterior angle is always *less than 180°* at each vertex. For this kind of exterior angle, the sum of all the exterior angles for any polygon is 360°.

**Type II**: At each vertex, take all of the angle not inside the polygon

For any convex polygon, this kind of exterior angle is always greater than 180° at each vertex. The sum of these is much bigger, and depends on the kind of polygon.

Let n be the number of vertices in a polygon (n = 3 for triangle, n = 4 for a quadrilateral, etc.)

We have to add up all the circles, and subtract the angles inside the polygon.

(sum of exterior angles) = n*(360°) – (sum of angles inside the polygon)

= n*(360°) – (n – 2)*180°

For a triangle (n = 3), sum of Type II exterior angles = 900°

For a quadrilateral (n = 4), sum = 1080°

For a pentagon (n = 5), sum = 1260°

For a hexagon (n = 6), sum = 1440°

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I did not know that the GRE/GSCE talk about exterior angles as only part of the exterior angle. I was supprised to see the formular vertices / 360 = external angles…

I came to my own conclusion on this exterior angles problem. I assumed that the whole angle from one edge round the outside to another hinged edge is the external angle.. the type II angle.

For this type II, there are easy formulas for working out the sum of angles (in degrees) where n represents the number of straight edges to a 2d polygon.

Sum of internal angle = (n-2)*180

Sum of external angles = (n+2)*180

Sum of all edges = n* 360

(or anotherway to say it is n*2pi)

If you think about 180 degrees is half a rotation (which is also pi rotation), and a polygon is a closed shape like a circle, then we can break the sums down easily.

If you imaging say 5 bendy straws, connected in a straight line to make a long straw, then as you bend the straws around so the head joins the tail. you will make the outside edge rotate a whole 360 degrees (thus +2pi) and conversly the inside edge has counter-rotated 360 degrees (thus -2pi)

So for a 17 Sided polygon we know that the total sum of all angles is 17*2*180 (17*2pi) degrees. (34Pi)

the internal angles will be (17-2)*180 degrees (15pi )

the external angles will be (17+2)*180 degrees (19pi )

15pi+19pi = 34pi

Ways to see this:

total angles = sum of Internal angles + sum of external angles = corners (or sides) * 360 degrees = corners *2pi-rotations = corners * 360 degrees = Corners * full rotations.

I hope that helps someone.

This blog post is misleading. There is only one accepted definition of “exterior angle”, and that’s the type I definition given here. The type II definition is *not* the definition of “exterior angle”.

Hey Mike I know this thread is fairly old but im considering the product to help me prepare for math which I haven’t had to do at this level for many years. When I saw the n * 360 – (n-2) * 180 I was very confused as re-learner. I had to look up that (n-2)*180 is the equation for the sum of interior angles. You may want to specify that in this kind of tutorial.

Best,

Zach

Hi Zach,

Thanks for your feedback! I’m sorry that you found the explanation confusing. We tried to make the formula as clear as possible, replacing the quantity (sum of angles inside the polygon) in the first equation with (n – 2)*180° in the next line. The sum of the angles inside a polygon is equivalent to the sum of the interior angles of a polygon.

Thanks, again, for your message. We appreciate the feedback and also hope this message will help other students who may be confused on the equation 🙂

Ah I see now I could of drawn that conclusion, but like I said this is very rusty for me so sometimes being explicit I think helps a lot with those of us that are already feeling a little flustered and may overlook things. If one has no idea bout exterior angles then they most likely dont know much about interior angles either. I appreciate the response! So far even just the free material is helping me greatly.

Best,

Zach

Hi

My son was marked wrong because he used type II ( 360° – interior angle ) His teacher said that it should of been 180° – interior angle.

Is there a more accepted way for a UK GCSE Maths student?

Stephen Geary, Dillon’s dad.

So I get that this does not hold true for a concave polygon. Can you give me a succinct explanation? Is it just the case that there isn’t an exterior angle measure for an angle greater than 180°? The sum of the interior angles of a polygon holds true whether convex or concave, so why not the exterior angle sum property?

Sir,

I’m not too clear on this, would sum of the exterior angles always remain 360 irrespective of the number of sides the polygon would have or do we need to calculate ??

Dear Aparna,

That’s perfectly correct! The number of sides is completely irrelevant. Think about it: think about if you could walk around on a path shaped like the polygon. At every vertex, you would turn by the exterior angle to align yourself with the next segment, and as a result of all these turns, you would return to your starting point facing the same direction — at that point, you would have turned a full 360 degrees.

Does all this make sense?

Mike 🙂

Sir ,

I have a doubt regarding the following statement of yours

“For any convex polygon, this kind of exterior angle is always less than 180° at each vertex. For this kind of exterior angle, the sum of all the exterior angles for any polygon is 360°.”

Is it always 360° or depends on the number of sides the polygon has ?

Dear Viggnesh,

I’m happy to respond. 🙂 Yes, for any number of sides, the sum is 360. Think about it this way. Suppose the polygon were big, and you were going to walk around its edge. At each vertex, you would turn your direction — the angle of that turn would be the exterior angle at that point. When you had walked all the way around, and returned to your starting point, you would have turned your direction a full 360 and be facing the same direction again. If you have a hard time picturing this in your head, imagine some polygons on the floor, walk around them, and pay attention to how much you turn at each vertex and how much your direction turns in a full trip around the polygon.

Does all this make sense?

Mike 🙂

Son of a gun, this is so helpful!

Dear Jacklyn,

I’m very glad you found this helpful. Best of luck to you!

Mike 🙂

Excellent…!! such subtle things can be there on real test, great help.. 🙂

I need one more help , can u post a updated one on properties of Normal distribution curve like comparison b/w two normal distri curves or what can be inferred from them and bimodel normal distribution ??

inferences like which one is having greater mean , greater SD or great population ??

Dear AArendy,

This is Mike, the GMAT expert at Magoosh. I’m happy to respond. 🙂 I’m glad you found this post helpful. As for your question, here’s a blog about the basics of the Normal Distribution:

https://magoosh.com/gre/2012/normal-distribution-on-the-gre/

Here’s a blog that discusses the subtle relationship of median & mean:

https://magoosh.com/gre/2012/gre-math-histograms/

Here’s a sophisticated GMAT blog about the properties of standard deviation:

https://magoosh.com/gmat/2012/standard-deviation-on-the-gmat/

As to your questions —

(a) the mean/median/mode of a normal distribution is the peak of the central hump, wherever that falls on the horizontal axis. If there are two normal distributions on the same graph, whichever one has a hump further to the right has a higher mean.

(b) Standard deviation is a measure of spread, of how far the data points are from each other. If a normal distribution is wide, it has a large standard deviation, and if it is skinny and narrow, it has a small standard deviation.

(c) The normal distribution, by its very nature, is a relative distribution, a distribution based on percentages. It is absolutely impossible to tell how many people are in the population, unless you are given the number of people in a particular section. Much in the same way, if I tell you, on a certain test, such-and-such grade was the 70% percentile, and such-and-such grade was the 90%, there would be no way to determine from that how many people had taken the test.

Does all this make sense?

Mike 🙂

Third link u gave me, that just rocked… best info one can get abt SD and its properties , amazing

Finally, now i can’t stop myself from buying Magoosh premium account..!!

gonna join you guys asap

Thanks Mike , great great help

Dear AArendy,

You are more than welcome, my friend. Best of luck to you.

Mike 🙂