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The formula for the inner angles of a polygon with 99 vertices?

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JT

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May 1, 2013, 5:55:12 PM5/1/13
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((99/2)-1)/99)*360=176,3636363636364

Isn't this formula easier then the one you use?

1treePetrifiedForestLane

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May 1, 2013, 5:56:43 PM5/1/13
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tdhis is how the terminology arose,
There is no where, thereinsville.

JT

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May 1, 2013, 6:06:49 PM5/1/13
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On 1 Maj, 23:56, 1treePetrifiedForestLane <Space...@hotmail.com>
wrote:
> tdhis is how the terminology arose,
> There is no where, thereinsville.

Or it could also be so simple that the angle of the 99 side polygon is
48,5/99 subturns.

>
>
>
>
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> > ((99/2)-1)/99)*360=176,3636363636364
>
> > Isn't this formula easier then the one you use?

Or why not just write

JT

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May 1, 2013, 6:18:46 PM5/1/13
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So the formula for a the inner angle of a regular polygon as expressed
in subturns, (maybe revolutions is better?) would be ((n/2)-1)/n

JT

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May 1, 2013, 6:19:38 PM5/1/13
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On 2 Maj, 00:06, JT <jonas.thornv...@gmail.com> wrote:
48,5/99 revolutions sounds better doesn't it.

William Elliot

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May 1, 2013, 11:21:34 PM5/1/13
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Wally W.

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May 1, 2013, 11:43:33 PM5/1/13
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On Wed, 1 May 2013 20:21:34 -0700, William Elliot wrote:

>

Syntax error. Please place empty strings in quotes.


Robin Chapman

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May 2, 2013, 4:43:06 AM5/2/13
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On 01/05/2013 22:55, JT wrote:
> ((99/2)-1)/99)*360=176,3636363636364

wrong ...

JT

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May 2, 2013, 9:03:18 AM5/2/13
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The answer or the calculation?

JT

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May 2, 2013, 9:13:56 AM5/2/13
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On 2 Maj, 10:43, Robin Chapman <R.J.Chap...@ex.ac.uk> wrote:
You do realise that in a uniform polygon with 99 vertices the sum of
inner angles are 99/2-1=48,5 revolutions, and you have to divide the
revolutions into the number of vertices.
To get the individual inner angle of each vertice?

JT

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May 2, 2013, 9:24:34 AM5/2/13
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Ok just learned the singular form is vertex, will see how long it stick
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