(1)解:∠DOE=∠AOC-∠BOE=70°-30°=40°(2)解:∠EOF=∠EOD+∠DOF=(70°-35°)+(180°-70°)÷2=90°故 OE⊥OF(3)解:2(70-n)+70°=180°解得 n=15°满意老师的回答敬请采纳,谢谢