Question

如图，直线AB.CD相较于点O，OD平分角AOF,OE垂直CD于O，角AOF=50°求角COB,角EOB,角BOF的度数

和下面的式子没关系

Answer 1

• ∠AOF=50°，OD平分∠AOF，∴∠AOD=50°÷2=25°，∵∠COB与∠AOD是对顶角，∴∠COB=25°，∠DOF=25°

• ∵OE⊥CD∴∠EOC=90°

• 又∵∠EOB=∠EOC+∠COB=90°+25°=115°

• ∵∠BOF=平角COD-∠COB-∠DOF=180°-25°-25°=130°

Answer 2

1、角BOF=180°-角AOF
∵角AOF=50°
∴角BOF=180-50°=130°
2、∵OD平分角AOF，角AOF=50°
∴角AOD=角DOF=1/2角AOF=1/2*50°=25°
又∵直线AB.CD相较于点O
所以角COB=角AOD=25°
3、∵OE垂直CD于O
∴角EOC=90°
所以角EOB=角EOC+角COB=90度+25度=115度

Answer 3

角COB 因为角AOF=50,OD平分角AOF 所以角AOD=50x1/2=25 因为角COB与角AOD是对顶角 所以角COB=角DOF=25 角EOB 因为OE垂直CD 所以角EOC=90 所以角EOB=角EOC 角COB=115 角BOF 所以角BOF=角COD-角COB-角DOF =180-25-25 =130

Answer 4

• ∠AOF=50°，OD平分∠AOF，∴∠AOD=50°÷2=25°，∵∠COB与∠AOD是对顶角，∴∠COB=25°，∠DOF=25°

• ∵OE⊥CD∴∠EOC=90°

• 又∵∠EOB=∠EOC+∠COB=90°+25°=115°

• ∵∠BOF=平角COD-∠COB-∠DOF=180°-25°-25°=130