Question

O为直线AB上一点,过O点作射线OC,OD平分角AOC,OE在角BOC内,且角BOE＝2角COE,若角DOE＝72度,求

Answer 1

设∠COE=θ,则∠BOE=2θ

设∠AOC=2φ,则∠DOE=72°

∴∠BOE+∠COE+∠AOC=3θ+2φ=θ+2（θ+φ）=180°

θ+144=180°==>θ=36°

∴∠BOE=72°