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rational

  • one year ago

show that if three angles in a convex polygon are equal to \(60^{\circ}\) then it must be an equilateral triangle

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  1. ikram002p
    • one year ago
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    |dw:1439061574644:dw|

  2. ikram002p
    • one year ago
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    i'll try other form |dw:1439061670041:dw|

  3. ikram002p
    • one year ago
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    assume x is 60 |dw:1439061819528:dw|

  4. ikram002p
    • one year ago
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    (if it works for quadrilateral then it works for any polygon ) geometric logic so we need to show by contradiction that we cant construct a quadrilateral with 3 60 angle

  5. rational
    • one year ago
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    yes that looks legit

  6. ikram002p
    • one year ago
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    that would make the other angle 180 seems more like not concave nor convex

  7. ikram002p
    • one year ago
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    |dw:1439062235117:dw|

  8. ikram002p
    • one year ago
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    i feel sleepy i might write it more need in morning,gn

  9. anonymous
    • one year ago
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    Sum of all the exterior angles of a polygon is 360 degrees. The exterior angle corresponding to 60 degrees is 120 degrees. Since three angles are 60 degrees, the sum of those exterior angles is 120*3 = 360 degrees. Thus there can be no other non-zero exterior angle corresponding to a vertex. So, these are the only 3 vertices, and the polygon is an equilateral triangle.

  10. ikram002p
    • one year ago
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    more neat**

  11. rational
    • one year ago
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    Nice :) More generally : Any convex polygon with \(n\) angles equal to \(\dfrac{360}{n}\) must be a regular \(\text{n-gon}\)

  12. ikram002p
    • one year ago
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    hehehe ok that was direct

  13. rational
    • one year ago
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    * Any convex polygon with \(n\) angles equal to \(180-\dfrac{360}{n}\) must be a regular \(\text{n-gon}\)

  14. rational
    • one year ago
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    for example, in a square we have \(4\) angles equal to \(180-\frac{360}{4} = 90\)

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