Usage of cotg; Area of polygon

Hi Forum

Here is a question

http://upload.wikimedia.org/wikipedi...Parameters.png

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

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

If we use only tg we'd get

$\displaystyle na^2tan(\pi/n)$

This $\displaystyle \pi/n$ angle regards the angle of the side $\displaystyle s$, 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.

Re: Usage of cotg; Area of polygon

Quote:

Originally Posted by

**Zellator** Hi Forum

Here is a question

http://upload.wikimedia.org/wikipedi...Parameters.png
$\displaystyle A=[ns^2 cotg (\pi/n)]/4$

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

If we use only tg we'd get

$\displaystyle na^2tan(\pi/n)$

This $\displaystyle \pi/n$ angle regards the

angle of the side $\displaystyle s$, correct?

No that is not correct.

$\displaystyle \Theta = \frac{{2\pi }}{n}$ is the *central angle* of a **regular polygon**.

Thus we have $\displaystyle \cot (\Theta ) = \frac{{2a}}{s}$.

Note that $\displaystyle a$ is the length of the altitude and $\displaystyle s$ is the length of the base.

Re: Usage of cotg; Area of polygon

Quote:

Originally Posted by

**Plato** $\displaystyle \Theta = \frac{{2\pi }}{n}$ is the *central angle* of a **regular polygon**.

Thus we have $\displaystyle \cot (\Theta ) = \frac{{2a}}{s}$.

Note that $\displaystyle a$ is the length of the altitude and $\displaystyle s$ 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 $\displaystyle a$ comes from.

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

Ok this was very confusing before your explanation.

I hope I got it right!

Thanks again Plato!