解:∵∠EOD=28°46′,OD平分∠COE,
∴∠COD=∠EOD=28°46′,
∵∠AOB=40°,
∴∠COB=180°﹣∠AOB﹣∠EOD﹣∠COD,
=180°﹣40°﹣28°46′﹣28°46′,
=82°28′.
故答案为:82 °28′.
A.∠COD=∠AOB
B.∠AOD=∠AOB
C.∠BOD=∠AOD
D.∠BOC=∠AOD
如图,O是直线AB上一点,OC为任一条射线,OD平分∠BOC,OE平分∠AOC.
(1)指出图中∠AOD的补角,∠BOE的补角;
(2)若∠BOC=68°,求∠COD和∠EOC的度数;
(3)∠COD与∠EOC具有怎样的数量关系?
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