文档介绍:,△ABC中,∠ABC,∠ACB的角平分线交于点O,若∠BOC=132°,则∠A等于多少度?ABCO12132°做几何题的时候要边读题,边将已知条件在图中标示出∠1=∠ABO=∠ABC∠2=∠ACO=∠,△ABC中,∠ABC,∠ACB的角平分线交于点O,若∠BOC=132°,则∠A等于多少度?ABCO12132°想要求∠A∠A+∠ABC+∠ACB=180°就要求∠ABC+∠ACB∠ABC=2∠1∠ACB=2∠2∠ABC+∠ACB=2(∠1+∠2)就要求∠1+∠2解:∵∠BOC+∠1+∠2=180°且∠BOC=132°∴∠1+∠2=180°-∠BOC=48°∵∠ABC,∠ACB的平分线交与点O∴∠ABC=2∠1,∠ACB=2∠2∴∠ABC+∠ACB=2(∠1+∠2)=2×48°=96°∵∠A+∠ABC+∠ACB=180°∴∠A=180°-(∠ABC+∠ACB)=180°-96°=84°.