(1)由已知得∠BOC=180°-∠AOC=150°, 又∠COD是直角,OE平分∠BOC, ∴∠DOE=∠COD-∠BOC=90°-×150°=15°;
(2)由(1)∴∠DOE=∠COD-∠BOC=90°, ∴∠DOE=90°-(180°-∠AOC), ∴∠DOE=∠AOC=α;
(3)∠AOC=2∠DOE; 理由:∵∠COD是直角,OE平分∠BOC, ∴∠COE=∠BOE=90°-∠DOE, 则得∠AOC=180°-∠BOC=180°-2∠COE=180°-2(90°-∠DOE), 所以得:∠AOC=2∠DOE; ②4∠DOE-5∠AOF=180° 理由:设∠DOE=x,∠AOF=y, 左边=∠AOC-4∠AOF=2∠DOE-4∠AOF=2x-4y, 右边=2∠BOE+∠AOF=2(90-x)+y=180-2x+y, 所以,2x-4y=180-2x+y即4x-5y=180, 所以,4∠DOE-5∠AOF=180°. |