Question

如图,在三角形ABC中,角ABC,角acb的平分线相交于点o,你能找出角a与角

Answer 1

解:∠ABC+∠ACB=180°-∠A.
OB,OC分别平分∠ABC,∠ACB,则:
∠OBC+∠OCB=(1/2)(∠ABC+∠ACB)=90°-(1/2)∠A.
故∠BOC=180°-(∠OBC+∠OCB)=180°-[90°-(1/2)∠A]=90°+(1/2)∠A.