Mary,

I know that the sum of the exterior angles pf a polygon is 360^o. I remember this because of a really neat proof that Walter Whitely sent us a while ago. It just involves walking around the perimeter of the polygon and adding up the angles of each turn at the vertices.

At each vertex the external angle measures 180^o minus the internal angle. Hence for a pentagon

360^o = sum of (external angles)
= sum of (180^o - internal angles)
= 5 × 180^o - sum of (internal angles)

What is the sum of the measures of the internal angles?

Harley