@Pi Han Goh
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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 \( A = \frac{1}{4} ns^2 \tan( \frac{90(n-2)}{n} ) \)
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Siddhartha Srivastava
·
9 months, 1 week ago

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@Siddhartha Srivastava
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If you think he wrote is true, then his expression is indeed correct. Mine is just simpler/standard.
–
Pi Han Goh
·
9 months, 1 week ago

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TopNewestIt should be \(\dfrac14 N s^2 \cot \dfrac \pi N \). – Pi Han Goh · 9 months, 1 week ago

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\( \tan(\frac{90^{\circ}(n-2)}{n}) = \tan( 90^{\circ} - \frac{180^{\circ}}{n}) = \cot(\frac{180^{\circ}}{n}) \) – Siddhartha Srivastava · 9 months, 1 week ago

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– Pi Han Goh · 9 months, 1 week ago

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

So what

I thoughthe wrote is \( A = \frac{1}{4} ns^2 \tan( \frac{90(n-2)}{n} ) \) – Siddhartha Srivastava · 9 months, 1 week agoLog in to reply

– Pi Han Goh · 9 months, 1 week ago

If you think he wrote is true, then his expression is indeed correct. Mine is just simpler/standard.Log in to reply

– Andrés Dextre · 9 months ago

Thats what I wrote. Sorry about the handwritring😁Log in to reply

– Pi Han Goh · 9 months ago

Haha no problem!Log in to reply