The formula for interior angles of a polygon is: 180 x (number of sides - 2) <- http://www.math-prof.com/Geom/Geom_Ch_30.asp If the interior (INT) angles of a triangle a, b, c is equal to 180⁰ then the exterior (EXT) angles equal to 900⁰. a_full_angle = 360⁰ b_full_angle = 360⁰ c_full_angle = 360⁰ a_full_angle + b_full_angle + c_full_angle = 1080⁰ full_angle_total = 1080⁰ a_INT = 40⁰ b_INT = 60⁰ c_INT = 80⁰ a_INT + b_INT + c_INT = 180⁰ triangle_INT_total = 180⁰ a_EXT = 320⁰ b_EXT = 300⁰ c_EXT = 280⁰ a_EXT + b_EXT + c_EXT = 900⁰ triangle_EXT_total = 900⁰ a_full_angle = 360⁰ = a_INT + a_EXT b_full_angle = 360⁰ = b_INT + b_EXT c_full_angle = 360⁰ = c_INT + c_EXT full_angle_total = 1080⁰ = triangle_INT_total + triangle_EXT_total If the interior angles of a Quadrilateral (four sided polygon) equal 360⁰ then the exterior angles equal: 360⁰ x 4 = 1440⁰ 1440⁰ – 360⁰ = 1080⁰ If the interior angles of a Pentagon (five sided polygon) equal 540⁰ then the exterior angles equal: 360⁰ x 5 = 1800⁰ 1800⁰ – 540⁰ = 1260⁰ If the interior angles of a Hexagon (six sided polygon) equal 720⁰ then the exterior angles equal: 360⁰ x 6 = 2160⁰ 2160⁰ – 720⁰ = 1440⁰ If the interior angles of a Heptagon (seven sided polygon) equal 900⁰ then the exterior angles equal: 360⁰ x 7 = 2520⁰ 2520⁰ – 900⁰ = 1620⁰ If the interior angles of an Octagon (eight sided polygon) equal 1080⁰ then the exterior angles equal: 360⁰ x 8 = 2880⁰ 2880⁰ – 1080⁰ = 1800⁰ Etc... Is the formula for the exterior angles of a polygon? n = number of sides of the polygon (360 x n) – (180 x (n - 2)) Is there a better way to do this?