解答:解:如图,作AE⊥BC于点E,DF⊥BC于点F,可得矩形AEFD,那么EF=AD=190(mm),∴BE=AE×cot∠B≈49.014(mm),同理可得CF≈49.014(mm),∴BC=BE+EF+CF=288.028≈288.0(mm).∴它的里口宽BC是288.0mm.