已知AC⊥BD, ∠ADB=3° ∠BDC=21° ∠ABD=27° 求∠ACB

设 AB=1 则AC=2 BC=√3 BD=DC=√6/2 A作AE⊥DB 交DB延线于E , ∠EBA=∠BAE=45° ,即 EB=EA=√2/2 , ED=(√6+√2)/2 所 tan∠ADB =EA/ED = (√2/2 )/(√6+√2)/2 =(√3 -1)/2 ≈0.366025403