Question

如图,直线AB与CD相交于点O,∠AOE=∠BOE=∠COF=∠DOF=90°

如果∠AOC=1/5∠EOF，求
∠AOC

Answer 1

∵∠AOC+∠COE=∠AOE=90°
∴∠COE=90°-∠AOC
∴∠EOF=∠COF+∠COE=90°+(90°-∠AOC)=180°-∠AOC
而∠AOC=1/5∠EOF
∴∠AOC=1/5×(180°-∠AOC)
解得∠AOC=30°

Answer 2

因为∠AOE=∠BOE=∠COF=∠DOF=90°
∠AOC=1/5∠EOF
∠EOF=90+90-∠AOC
5∠AOC=180--∠AOC
∠AOC=30度