Geometry is fun

[BdS]

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?