Angles in Polygons

Triangulate to count interior angles; exteriors always total 360^ .

The idea

At each vertex of a polygon, extend one side past the corner; the angle between that extension and the next side is the exterior angle at that vertex, and it completes the interior angle to $180^{\circ}$.

Theorem (Angle sums of a convex polygon).

In a convex polygon with $n$ sides, the interior angles add to $(n-2)\cdot 180^{\circ}$, and the exterior angles, one at each vertex, add to $360^{\circ}$.

The figure shows a pentagon, $n = 5$, with the diagonals from one vertex drawn in. Only the number of sides matters, not the polygon's shape. If the polygon is also regular — all sides equal and all angles equal — each sum splits evenly: each of the $n$ interior angles is $\frac{(n-2)\cdot 180^{\circ}}{n}$, and each exterior angle is $\frac{360^{\circ}}{n}$.

The exterior total does not depend on $n$ at all: walk once around the polygon, and the exterior angles are the turns you make at the corners, which together amount to one full revolution.

Ways to work on it

• Walkthrough. Derive (n-2)· 180^ and meet the exterior-angle rule.
• Proof. See why — triangulate for (n-2)180^ , then prove exteriors total 360^ .
• Practice. Interior-angle sum and each angle of a random regular polygon.
• Hardest. Given an interior angle, solve for the number of sides.

Not sure where to start? Take the ten-question placement test.