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anonymous

  • one year ago

How would you prove that the sum of the interior angles of a n-gon is 180(n-2) degrees in the case that at least one of the interior angles is greater than 180 degrees?

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  1. anonymous
    • one year ago
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    Basically how would you prove the case where its a concave polygon? You wouldn't always be able to create a line segment that can connect any two vertices

  2. ganeshie8
    • one year ago
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    like for example |dw:1436929826981:dw| ?

  3. anonymous
    • one year ago
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    Yeah, exactly.

  4. ganeshie8
    • one year ago
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    It will be easy to prove if we consider "signed" angle for sum of exterior angles of a polygon

  5. ganeshie8
    • one year ago
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    I'm thinking of using the fact that the sum of "signed" exterior angles of any polygon add up to 360

  6. ganeshie8
    • one year ago
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    count counterclockwise as positive and clockwise as negative |dw:1436930131588:dw|

  7. ganeshie8
    • one year ago
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    \[\color{blue}{a-b+c-d+e = 360}\]

  8. anonymous
    • one year ago
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    How can we then extend that to what the interior angles would be? If we had triangle constructed, I know the exterior angle is equal to the sum of the other 2 interior angles. But I don't think we can make a triangle construction, if that would even help.

  9. anonymous
    • one year ago
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    Well, at least make a triangle construction that would work in any arbitrary case, but Im not positive on that.

  10. ganeshie8
    • one year ago
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    We may use this : \[\text{interior angle} + \text{exterior angle} = 180\]

  11. ganeshie8
    • one year ago
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    |dw:1436930692495:dw|

  12. anonymous
    • one year ago
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    Alrighty, Ill mess with that idea for a minute or two.

  13. anonymous
    • one year ago
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    And then we would need some angle f at the starting point, right?

  14. ganeshie8
    • one year ago
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    sry forgot angle \(f\) : |dw:1436931109510:dw|

  15. ganeshie8
    • one year ago
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    I also don't have the proof yet, still trying..

  16. anonymous
    • one year ago
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    Ah, okay :)

  17. anonymous
    • one year ago
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    No worries

  18. ganeshie8
    • one year ago
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    If we agree that above theorem works, looks we're done!

  19. ganeshie8
    • one year ago
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    \[\text{interior angle} + \text{exterior angle} = 180\] \[\sum\limits_{1}^{n}\text{interior angle} + \text{exterior angle} = \sum\limits_{1}^n180\] \[\begin{align}\sum\limits_{1}^{n}\text{interior angle} &= \sum\limits_{1}^n180 - \sum\limits_{1}^n\text{exterior angle} \\~\\ &=180n - 360\\~\\ &=180(n-2) \end{align}\]

  20. anonymous
    • one year ago
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    Well that just dumbed everything down exponentially, haha. Yeah, that's very basic, concise, makes sense :) Awesome, thanks :3

  21. ganeshie8
    • one year ago
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    XD that signed angles thingy is a bit confusing though, it always trips me and makes me doubt the exterior angle sum thm...

  22. anonymous
    • one year ago
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    Actually yeah, what exactly did you mean when you said "signed" angles?

  23. ganeshie8
    • one year ago
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    for convex polygon, it is easy to see that the external angles must add up to 360 for one complete revolution : |dw:1436932101810:dw|

  24. ganeshie8
    • one year ago
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    for definiteness, lets agree that we always measure angle with respect to that blue extended line, |dw:1436932485018:dw| if the polygon is convex, notice that all the angles will be measured counter clockwise. so we don't have to worry about "sense" here

  25. ganeshie8
    • one year ago
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    But if we allow the polygons to be concave, some angles will be measured counterclockwise and some clockwise |dw:1436932611913:dw| we need to distinguish between these two

  26. anonymous
    • one year ago
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    Oh, duh. Signed as in positive or negative with respect to the direction of rotation. I have no idea why I was thinking of signed as in labels or something else. Brain fart, lol.

  27. ganeshie8
    • one year ago
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    Yes haha thats it I guess. For "concave polygons", most of the usual theorems work if we replace "angle" by "signed angle"

  28. anonymous
    • one year ago
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    Gotcha. This course so far has basically been a bunch of random proofs and then homework that is not aided at all by the proofs done in class or in the textbook, haha. In the end, though, I think being forced to investigate it so much will help me out (gotta find some positive). Thanks again :)

  29. ganeshie8
    • one year ago
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    Are you doing axiomatic geometry ?

  30. ganeshie8
    • one year ago
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    I remember ikram doing this course sometime ago and it was very interesting

  31. anonymous
    • one year ago
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    It doesnt really have a specific name. The course itself is called "College Geometry". I guess right now we're just dealing with Euclidean stuff, but I know we will be touching on spherical and hyperbolic geometry at some point as well. I don't believe it ever becomes anything beyond calc 1 integration. Although I think if we get far enough (doubt it), there is a chapter that requires line integrals.

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