Describing all points 4000 miles from the north pole I'd like to describe all of the points on the Earth's surface that are exactly 4000 miles from the North Pole. I know that this will eventually give me an equation for a circle; I want to find that equation assuming that the Earth's radius is 3960 miles and that the center of the Earth is at $(0,0)$. I started by drawing some triangles: from the center of the Earth to the North Pole is one side, from the center to the surface is another, and from the North Pole to the surface is the third. The lengths of these sides are 3960, 3960, and 4000 respectively. 
Using the law of cosines should let me find the size of each angle, but I'm not getting angles that add up to 180. I get approximately $60.9908^\circ$ for the angle opposite the 4000 side and about $59.6653^\circ$ for the other two angles, which adds up to $180.321^\circ$. I didn't do any rounding; is this just an issue with my calculator making some internal rounding errors? Usually copy-pasting the long decimals keeps everything working well through the end but this is a pretty significant error (at least in my mind). 
When I try to find the radius (which I think should be equal to the altitude running from the surface to the side connecting the center and the North Pole), I get different values depending on which angles I use, which I don't really like. Is there any way to resolve this? Am I thinking about it in the wrong way?
Edit: It turns out that instead of getting 60.9908$^\circ$ I should have been getting 60.669$^\circ$, which resolves all of my problems. I hope the problem proves of some interest to anyone else!
 A: When I do the law of cosines on a $3960-3960-4000$ triangle, I get the angle opposite $4000$ to be $\arccos 1-\frac{4000^2}{2\cdot 3960^2}\approx \arccos 0.48984798 \approx 60.669410^\circ$ and the angle opposite $3960$ to be $\arccos \frac {4000}{2 \cdot 3960} \approx \arccos 0.5050505051 \approx 59.665295^\circ$.  The three angles add to $180^\circ$ within rounding error.  You didn't show how you calculated the angles, so I can't guess where the error is.
A: Unfortunately, this method requires some familiarity with parametric functions.  However, it does give you a more precise answer by considering the distance along a great circle instead of the straight-line distance.
Motivated by spherical coordinates, we know that the earth can be described by the following parametric function:
$$f(u, v) = \left[ \begin{array}{ccc}
3960\cos(u)\sin(v) \\
3960\sin(u)\sin(v)\\
3960\cos(v)\end{array} \right]$$
Where $3960$ is the radius of the earth and $u$ and $v$ are angles such that $0 \leq u \leq 2\pi$ and $0 \leq v \leq \pi$.  In particular, think of $v$ as an angle measured with respect to the south pole that encodes the latitude of your point.  That is, if you are at the north pole, then $v = 0$.  If you are at the equator, then $v = \frac{\pi}{2}$, and if you are at the south pole, then $v = \pi$.  
Now, we know that the circumference of the earth is given by $c = 2\pi \cdot 3960$ miles, and so half the circumference is $\pi \cdot 3960$ miles.  If we travel $4000$ miles from the north pole, then we have traveled $\frac{4000}{\pi \cdot 3960} \approx 0.3125$ the total distance from the north to the south pole.  Hence, at the latitude circle, $v =0.3125\pi$ radians.
We can now plug this value of $v$ back into the original function above to get:
$$f(u) = \left[ \begin{array}{ccc}
3960\sin(0.3125\pi)\cos(u) \\
3960\sin(0.3125\pi)\sin(v)\\
3960\cos(0.3125\pi)\end{array} \right]$$
Now, notice that the $z$ coordinate is just a constant.  It is simply giving you the distance from the center of the earth to the center of the latitude circle (i.e. its 'height').  If you're used to parametric functions, then you'll quickly recognize the $x$ and $y$ coordinates as describing a circle of radius $3960\sin(0.3125\pi) \approx 3292.61966$ miles, which is what you were looking for!
