Question

如圖，直線AB、CD相交于點(diǎn)O，OEAB，O為垂足，如果EOD=35°，則COB=       ；

Answer 1

p;【答案】解析:
∵OEAB,∠EOD=35°
∴∠AOD=90°+35°=125°
∵∠COB=∠AOD
∴∠COB=125°