如图,直线AB,CD相交于点O,OE评分角BOD,OF平分角COE,角AOD比角BOE=7:1,求AOF度
人气:104 ℃ 时间:2020-03-23 06:29:48
解答
OE评分角BOD,∠AOD︰∠BOE=7︰1
∠AOD︰∠BOD=7︰2
∠AOD+∠BOD=180º
∠AOD=180º×7/(7+2)=140º,∠BOD=60º,∠BOE=∠DOE=30º
∠COE=180º-∠DOE=180º-30º=150º
OF平分∠COE
∠COF=∠EOF=½∠COE=½×150º=75º
∠AOC=180º-∠AOD=180º-140º=40º
∴∠AOF=∠AOC+∠COF=40º+75º=115º
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