@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
·
8 months ago

Log in to reply

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

## Comments

Sort by:

TopNewestIt should be \(\dfrac14 N s^2 \cot \dfrac \pi N \). – Pi Han Goh · 8 months ago

Log in to reply

\( \tan(\frac{90^{\circ}(n-2)}{n}) = \tan( 90^{\circ} - \frac{180^{\circ}}{n}) = \cot(\frac{180^{\circ}}{n}) \) – Siddhartha Srivastava · 8 months ago

Log in to reply

– Pi Han Goh · 8 months 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 · 8 months agoLog in to reply

– Pi Han Goh · 8 months 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 · 7 months, 3 weeks ago

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

– Pi Han Goh · 7 months, 3 weeks ago

Haha no problem!Log in to reply