Question

已知,如图K一27一13,点P是△ABC的两外角平分线的交点,∠A=m,求∠BPC的度数(用m表示).

Answer 1

∠BPC=180°-1/2[(180°-∠ABC)+(180°-∠ACB)]

=1/2(∠ABC+∠ACB)

∠ABC+∠ACB=180°-∠A=180°-m

∴∠BPC=1/2(180°-m)=90°-1/2m