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Could someone tell me if this formula is ok? \(N\) is the number of sides and \(s\) is the length of the side

Note by Andrés Dextre 1 year, 10 months ago

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It should be \(\dfrac14 N s^2 \cot \dfrac \pi N \).

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They're both the same

\( \tan(\frac{90^{\circ}(n-2)}{n}) = \tan( 90^{\circ} - \frac{180^{\circ}}{n}) = \cot(\frac{180^{\circ}}{n}) \)

OP wrote \(A = \dfrac19 ns^2 \tan \left( \dfrac{\text{gorn}-2}n \right ) \).

@Pi Han Goh – I'll let OP clarify, but from what I see, the "9" is more likely a "4", "go" is "90" and the "r" is a bracket.

So what I thought he wrote is \( A = \frac{1}{4} ns^2 \tan( \frac{90(n-2)}{n} ) \)

@Siddhartha Srivastava – If you think he wrote is true, then his expression is indeed correct. Mine is just simpler/standard.

@Siddhartha Srivastava – Thats what I wrote. Sorry about the handwritring😁

@Andrés Dextre – Haha no problem!

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italics`**bold**`

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boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

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## Comments

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TopNewestIt should be \(\dfrac14 N s^2 \cot \dfrac \pi N \).

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They're both the same

\( \tan(\frac{90^{\circ}(n-2)}{n}) = \tan( 90^{\circ} - \frac{180^{\circ}}{n}) = \cot(\frac{180^{\circ}}{n}) \)

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OP wrote \(A = \dfrac19 ns^2 \tan \left( \dfrac{\text{gorn}-2}n \right ) \).

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So what

I thoughthe wrote is \( A = \frac{1}{4} ns^2 \tan( \frac{90(n-2)}{n} ) \)Log in to reply

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