Question

已知在三角形ABC中角A=62度,BOCO分别是角ABC,角ACB的平分线,且BOCO相交于点O,求角

Answer 1

∠OBC=1／2∠ABC,∠OCB=1／2∠ACB,

∠ABC+∠ACB+∠A=180º,

∠OBC+∠0CB=1/2 (∠ABC+∠ACB)=1／2﹙180º-62º)=59º

∠OBC+∠0CB+∠BOC=180º

∠BOC=180º-59º=121º.