Question

如图直线AB,CD,EF交于一点O,GO⊥EF且∠GOB=30°,∠AOC等于40°,求∠COE?

Answer 1

解：由图可知：角AOC+角COE+角EOG+角GOB=180度（平角）
所以，40+角COE+90+30=180
求得：角COE=20度

Answer 2

∠AOF = ∠EOB = ∠EOG + ∠BOG = 90° + 30° = 120°;
∠COE = ∠EOF - ∠AOF - ∠COA = 180° - 120° - 40° = 20°.

Answer 3

20° 。,∠AOB=180°，180°-30°-40°-90°=20°