解:(1)∵BD为△ABC的高 ∴∠BDC=90° ∴∠1=90°-∠BCA 同理∠2=90°-∠ABC ∵∠ABC+AC=180°-50°=130° ∴∠BOC=180°- (∠1+∠2) =180°-(90°-∠ABC+90°-∠ACB) =180°-180°+∠ABC+∠ACB=130° (2)∵BM、 CN分别为△ABC的角平分线 ∴∠1= ∠ABC ∠2= ∠ACB ∵∠A=50° ∴∠ABC+∠ACB=180°-50°=130° ∴∠BPC=180°-(∠1+∠2) =180°-( ∠ABC+ ∠ACB) = 180°- (∠ABC+∠ACB) =180°- ×130° =115° |