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

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°