答案:2.90° 点拨:如答图,延长AB交DC于点F,
因为∠1:∠2:∠3 = 9:2:1,所以设∠1 = 9x,∠2 = 2x,∠3 = x. 由∠1 + ∠2 + ∠3 = 180°,得9x + 2x + x = 180°,解得x = 15°,故∠1 = 9×15° = 135°,∠2 = 2×15° = 30°,∠3 = 15°,所以∠DCB = ∠E = ∠3 = 15°,∠2 = ∠EAB = ∠D = 30°,所以∠EAC = 60°,∠DCA = 30°,所以∠4 = ∠EAC + ∠DCA = 60° + 30° = 90°.