Geometry: Degrees in more than 4 corners A triangle has  180 degrees and a rectangle 360 but a figure with 5 or more? is it still 360?
 A: To complement Arturo's answer, it is true that the sum of the interior angles in a $n$-gon is $(n-2)\pi$, even if the polygon is not convex. Here's a sketch of a proof: First prove that every polygon, even non-convex, has a diagonal totally contained inside the polygon. Now cut the polygon into two parts along that diagonal. By induction, if the two parts have $a$ and $b$ sides, the sums of their interior angles is $(a-2)\pi$ and $(b-2)\pi$. So, the sum of the interior angles in the original polygon is $(a-2+b-2)\pi = (a+b-4)\pi = (n+2-4)\pi = (n-2)\pi$, completing the induction. That $a+b=n+2$ comes from the fact that the diagonal is counted twice in the parts but not at all in the original polygon.
A: Here is another way of deriving the answer $(n-2)\pi$. The sum of (signed) external angles is $2\pi$, as you make one net turn if you go around the polygon (the external angle is by how much you turn if you go around the polygon and pass from one edge to the following edge -for non-convex polygon the angle may be negative). The (internal) angle between two edges is $\pi$ minus the external angle, hence the sum of all (internal) angles is $n\pi - 2\pi$.
edit: here is a picture of the external angles.

A: This is rather poorly phrased, but I think you are asking about the sum of the interior angles in a $n$-gon.
If the figure has $n$ sides, then the interior angles add up to $(n-2)\times 180^{\circ}$. For $n=3$ (triangle) you get $180$, for $n=4$ (quadrilateral) you get $360$; for a pentagon $(n=5)$ you get $540$, etc.
Added. This can be proven by induction. To avoid technical issues, we restrict to convex polygons. Thanks to lhf for pointing this out. 
In the case of $n=3$, the proof goes back to Euclid.
Assume the result holds for a figure with $n$ sides, and consider a figure with $n+1$ sides. Number the vertices $1,2,\ldots,n,n+1$. Now draw a line between corners $n-1$ and $n+1$ (a diagonal); the original figure is cut into two different figures, a figure with $n$ sides and a triangle. The figure with $n$ sides has interior angles adding up to $(n-2)\times 180^{\circ}$ by the induction hypothesis; the triangle has interior angles adding up to $180^{\circ}$; so the original figure has interior angles adding up to
$$(n-2)\times 180^{\circ} + 180^{\circ} = (n-1)\times 180^{\circ} = \Bigl( (n+1)-2\Bigr)\times 180^{\circ},$$
as desired. 
