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Math Help - Usage of cotg; Area of polygon

  1. #1
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    Thumbs up Usage of cotg; Area of polygon

    Hi Forum
    Here is a question

    A=[ns^2 cotg (\pi/n)]/4

    I am confused about the usage of cotg in this formula. How was this formula calculated?

    If we use only tg we'd get
    na^2tan(\pi/n)

    This \pi/n angle regards the angle of the side s, correct?

    I'm no pro in co-trigonometric functions but,
    Will this, in any way, relate to the fundamental propriety of cotg=A/O?
    Trigonometric functions on non right angles are rather strange.

    Thanks.
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  2. #2
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    Re: Usage of cotg; Area of polygon

    Quote Originally Posted by Zellator View Post
    Hi Forum
    Here is a question

    A=[ns^2 cotg (\pi/n)]/4
    I am confused about the usage of cotg in this formula. How was this formula calculated?

    If we use only tg we'd get
    na^2tan(\pi/n)

    This \pi/n angle regards the angle of the side s, correct?
    No that is not correct.
    \Theta  = \frac{{2\pi }}{n} is the central angle of a regular polygon.
    Thus we have \cot (\Theta ) = \frac{{2a}}{s}.
    Note that a is the length of the altitude and s is the length of the base.
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  3. #3
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    Thumbs up Re: Usage of cotg; Area of polygon

    Quote Originally Posted by Plato View Post
    \Theta  = \frac{{2\pi }}{n} is the central angle of a regular polygon.
    Thus we have \cot (\Theta ) = \frac{{2a}}{s}.
    Note that a is the length of the altitude and s is the length of the base.
    Hi Plato
    I guess I didn't know how to put the central angle part in proper terms. Thanks for the explanation.
    Though I'm kind of confused of where does the 2 in a comes from.

    Maybe we have a two there because s is actually \frac{{s}}{2}?

    Ok this was very confusing before your explanation.
    I hope I got it right!
    Thanks again Plato!
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