Question

已知△ABC ,∠ABC、∠ACB的平分线BE、CF交于点O(1)若∠ABC=80°,∠ACB=60° 求∠A、∠BOC

Answer 1

∵∠ABC＝80,∠ACB＝60

∴∠A＝180-∠ABC-∠ACB＝180-80-60＝40

∵BO平分∠ABC,CO平分∠ACB

∴∠ABO＝∠CBO＝∠ABC/2＝40,∠ACO＝∠BCO＝∠ACB/2＝30

∴∠BOC＝180-（∠CBO+∠BCO）＝180-（40+30）＝110°

∵∠ABC＝70,∠ACB＝50

∴∠A＝180-∠ABC-∠ACB＝180-70-50＝60

∵BO平分∠ABC,CO平分∠ACB

∴∠ABO＝∠CBO＝∠ABC/2＝35,∠ACO＝∠BCO＝∠ACB/2＝25

∴∠BOC＝180-（∠CBO+∠BCO）＝180-（35+25）＝120°

3、∠BOC＝90+∠A,对任意三角形都成立

证明：

∵BO平分∠ABC,CO平分∠ACB

∴∠ABO＝∠CBO＝∠ABC/2,∠ACO＝∠BCO＝∠ACB/2

∴∠BOC＝180-（∠CBO+∠BCO）

＝180-（∠ABC+∠ACB）/2

＝180-（180-∠A）/2

＝90+∠A/2