问题补充:
直线AB与CD相交于点O,若∠AOD=81°,∠EOC:∠AOE=1:2,求∠BOD和∠BOE的度数.
答案:
∵∠BOD=180°-∠AOD
=180°-81°
=99°∵∠EOC:∠AOE=1:2=k
∴∠EOC=k ∠AOE=2k
∵∠AOC=∠DOB=99°(对顶角相等)
∴∠DOB=∠EOC+∠AOE
=k+2k=3k=99°∴k=33°
∴∠EOC=k=33°
∵∠COB=∠AOD=81°(对顶角相等)
∴∠BOE=∠EOC+∠COB
=33°+81°
=114°
时间:2019-10-08 12:56:11
直线AB与CD相交于点O,若∠AOD=81°,∠EOC:∠AOE=1:2,求∠BOD和∠BOE的度数.
∵∠BOD=180°-∠AOD
=180°-81°
=99°∵∠EOC:∠AOE=1:2=k
∴∠EOC=k ∠AOE=2k
∵∠AOC=∠DOB=99°(对顶角相等)
∴∠DOB=∠EOC+∠AOE
=k+2k=3k=99°∴k=33°
∴∠EOC=k=33°
∵∠COB=∠AOD=81°(对顶角相等)
∴∠BOE=∠EOC+∠COB
=33°+81°
=114°