Using Trigonometry vs. Geometry I have a friend who is trying to build a wooden retaining wall around a trampoline.  He wants the wall to be in the shape of a regular polygon with sides between $24"-30"$.  He asked me to figure out how many sides he should make it, and then calculate the angles so that he knows how to cut the wood.  The trampoline has a diameter of $16'$.
So I attacked the problem at first using Geometry.  I found that the area of the trampoline is approximately $201.1$ $ft^2$.  Then I used the regular polygon formula $A=\frac{a\cdot p}{2}$, where $a$ is the apothem and $p$ is the perimeter for all sides from $24"-30"$.  To make a long story short.  I found that a polygon of $24$ sides with $26"$ sides would best do the trick, and another mathematics friend of mine concurred.  Since we have 24 sides, using the internal angle formula, I know that the sum of the interir angles is $A(24)=180(24-2)=3960$ which means each of the $24$ angles should be $165$ degrees.  Halving this gives me cuts of $82.5$ degrees.
However, using Trig, if the long side of the triangle is $96"$ and the short side is $13"$, then $\tan{\theta}=\frac{96}{13}$ which implies that $\theta=82.288$, not $82.5$
Why the discrepancy?  I know these numbers are close, but shouldn't the math agree?  I don't see where my error lies if there is an error in my math.  
 A: EDIT: I had it the wrong way round.
I think it may be due to the fact that you're using the distance to the base of the triangle (the long side) as half the diameter of the trampoline. If you look at this diagram this isn't strictly accurate since there's a little difference.

Which as Blue states the you can work out the circum radius of the polygon with:
$$ \frac{26}{2 \sin \left(\frac{\pi }{24}\right)}= 13 \csc \left(\frac{\pi }{24}\right)\approx 99.5969 $$
Then you still need to account for the difference $d$ in the second diagram, you can find the distance with $d$ accounted for using:
$$\sqrt{r^2-\frac{c^2}{4}} \rightarrow \sqrt{\left(\frac{26}{2 \sin \left(\frac{\pi }{24}\right)}\right)^2-\frac{26^2}{4}}=\sqrt{169 \csc ^2\left(\frac{\pi }{24}\right)-169} \approx 98.7448$$
where $r$ is the circum radius and $c$ is the length of the side of the polygon.
Thus you can then find the angle to be:
$$ \tan ^{-1}\left(\frac{1}{13} \sqrt{169 \csc ^2\left(\frac{\pi }{24}\right)-169}\right) \times \frac{180}{\pi}=82.5^\circ $$
Of course you could just use the inverse cosine: $$\cos ^{-1}\left(\frac{26/2}{\left(13 \csc \left(\frac{\pi }{24}\right)\right)}\right) \times \frac{180}{\pi}=82.5^\circ$$ to skip a step from the above working but I thought that since you were using $\tan$ I would too.


Diagram for comment

