Question

已知三角形ABC内接于⊙O,∠ABC=25°,∠ACB=75°,过A点作⊙O的切线交BC的延长线于点P,则∠APB多少度

Answer 1

∠0AB=∠0BA

∠0BC=∠0CB

∠0AC=∠OCA

∠BAC=180°-∠ABC-∠ACB=80°

∠0AB+∠OAC=80

∠0BC+∠0BA=25

∠0CA+∠0CB=75

解得∠0AB=15

∠BAP=∠0AB+90=105

∠APB=180-105-25=50