Question

已知,△ABC中,∠ACB=90°,AC=BC=PC,∠PAC=30°,求∠PBA的度数

Answer 1

∵CA = CP

∴∠CAP = ∠CPA= 30°

∴∠ACP =120°

∵∠ACB = 90°

∴∠BCP = 30°

∵CB = CP

∴∠CBP = ∠CPB = (180°-30°)/2 = 75°

∵∠CPA = 30°

∴∠APB = 75°-30°= 45°

∴在△APB中

∵∠ACB = 90°,AC=BC

∴∠CAB = 45°

∵∠BAP = 30°

∴∠PAB = 45°-30° = 15°

∴∠ABP = 180°-∠PAB - ∠APB = 180°-15°-45° = 120°