AD垂直于CD,角A=角C=30度,求角ABC的度数?

连接BD
则∠ABD+∠A+∠ADB=180°,∠C +∠CDB+∠CBD =180°
∴∠ABD+∠A+∠ADB+∠C +∠CDB+∠CBD=360°
∵∠A=∠C =30°,∠ADC =90°
∴∠ABD +∠CBD =360°-90°-30°-30°=210°
∴∠ABC=360°-210°=150°