直线AB,CD,EF相交于点O,∠AOE=35°,∠BOC=2∠AOC,求∠DOF的度数.

/>∵∠AOB＝∠AOC+∠BOC,∠BOC＝2∠AOC∴∠AOB＝∠AOC+2∠AOC＝3∠AOC∵直线AB∴∠AOB＝180∴3∠AOC＝180∴∠AOC＝60∵∠BOD与∠AOC为对顶角∴∠BOD＝60∵∠AOE与∠BOF为对顶角,∠AOE＝35∴∠BOF＝∠AOE＝35∴∠DOF＝∠BOD+∠BOF＝60+35＝95°