Polygon ABCDEF is a regular hexagon. What is the measure in degrees of angle ABE?
Using the degree angle formula, we have the number of degrees in one angle is:
( 6 - 2 ) $\cdot$ 180 / 6 =120
Angle ABE is a bisector of angle ABC, so 120 / 2 = $\boxed{60^\circ}$
Let A = (0, 2)
Let B = (sqrt 3 , 1)
Let C = (sqrt 3 - 1)
Let D = ( 0 , - 2)
Let E = (-sqrt 3 , -1)
AB = sqrt [ 3 + 1] = 2
BE = sqrt [ (2sqrt 3)^2 + 2^2] = sqrt [16] = 4
AE = sqrt [ 3 + 3^2] = sqrt [12]
By the Law of Cosines
AE^2 = AB^2 + BE^2 - 2 ( AB * BE) cos ABE
12 = 4 + 16 - 2( 2 * 4) cos ABE
-8 / ( -16) = cos ABE =1/2
arccos 1/2 = 60° = ABE