# Usage of cotg; Area of polygon

• Jun 14th 2011, 04:11 AM
Zellator
Usage of cotg; Area of polygon
Hi Forum
Here is a question
$\displaystyle A=[ns^2 cotg (\pi/n)]/4$

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.
• Jun 14th 2011, 06:02 AM
Plato
Re: Usage of cotg; Area of polygon
Quote:

Originally Posted by Zellator
Hi Forum
Here is a question
$\displaystyle A=[ns^2 cotg (\pi/n)]/4$

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.
• Jun 14th 2011, 09:09 AM
Zellator
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!