Question

已知,如图2,三角形ABC的两个外角平分线BP,CP交于点P,角A等于48度,求角P的度数

Answer 1

∠CBP=(180-∠ABC)/2,∠BCP=(180-∠ACB)/2

∠P=180-(∠CBP+∠BCP)=180-[(180-∠ABC)/2+(180-∠ACB)/2]

=∠ABC/2+∠ACB/2=(180-∠A)/2=(180-48)/2=66(度）