Question

O是直线AB上一点,∠COD是直角,OE平分∠AOD,OF平分∠BOC；

(1)如图①，当∠BOC=30°时，求∠EOF的度数；
（2）将∠COD绕O点旋转，如图②，求∠EOF的度数
图①

图②

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Answer 1

1）因为∠COD是直角,OE平分∠AOD,OF平分∠BOC,∠BOC=30°
所以，∠AOD=180°-∠COD-∠BOC=60°
∠EOF=∠EOD+∠COF+∠COD
=1/2∠AOD+1/2∠BOC+90°
=135°

2）因为∠COD是直角,OE平分∠AOD,OF平分∠BOC
所以，∠EOF=∠EOD+∠COD-∠COF
=1/2∠AOD-1/2∠BOC+90°
=1/2(180°-∠BOD)-1/2∠BOC+90°
=180°-1/2∠COD

=135°

Answer 2

解:∠AOC=2∠DOE;
理由:∵∠COD直角OE平∠BOC
∴∠COE=∠BOE=90°-∠DOE
即∠AOC=180°-∠BOC=180°-2∠COE=180°-2(90°-∠DOE)
∴:∠AOC=2∠DOE;

Answer 3

(1) ∠EOF=150° （2)∠EOF=150°