如图所示,在正六边形ABCDEF中,AC与BD交于点O,求角AOB
问题描述:
如图所示,在正六边形ABCDEF中,AC与BD交于点O,求角AOB
答
∵正六边形
∴∠ABC=120°
且AB=BC,∴△ABC等腰,即∠BAC=∠BCA=30° ,同理可得∠OBC=30°
∴∠ABO=∠ABC-∠OBC=120°-30°=90°
因此在△AOB中,∠AOB=180°-∠OAB-∠ABO=180°-30°-90°=60°