Question

速度解题2222222222222 (10 9:46:32)

Answer 1

∠ABC+∠ACB+∠BAC=180°

∠ABC=66° ∠ACB=54°

∠BAC=180°-∠ABC-∠ACB=180°-66-54=60°

BE垂直AC与E,所以△ABE为直角三角形 则 ∠AEB=90°

∠BAE=∠BAC=60°

∠ABE=180°-∠AEB-∠BAE=180°-90°-60°=30°

同理可证∠ACF=30°