Question

已知,如图,直线ab,cd相交于点o,oe⊥ab,of平分∠boc,∠doe=72度,求∠bof和∠aof的度数

Answer 1

∵OE⊥AB,即∠BOE=90°

∠DOE=72°

∴∠BOD=90°-∠DOE=90°-72°=18°

∴∠AOC=∠BOD=180°

∴∠BOC=180°-∠AOC=180°-18°=162°

∵OF平分∠BOC

∴∠BOF=∠COF=1/2∠BOC=1/2×162°=81°

∴∠AOF=∠AOC+∠COF=18°+81°=99°