Question

如图一在三角形ABC中角ABC 角ACB的平分线相交于点O角A=40度,求角BOC度数

Answer 1

∵∠A=40°

∴∠ABC+∠ACB=180°-40°=140°

∴∠BOC=180°-1/2∠ABC-1/2∠ACB=180°-1/2(∠ABC+∠ACB)= 180°-1/2×140°=180°-70°=110°