(1) Let A be AD⊥BC, and the extension line of BC is at point D, as shown in figure 1:
In Rt△ADC, AC=4,
∫∠C = 150,
∴∠ACD=30,
∴AD= 1/2AC=2,
CD=AC? cos30 =4×√3/2=2√3,
In Rt△ABD, tanB=AD/BD=2/BD= 1/8,
∴BD= 16,
∴bc=bd-cd= 16-2√3;
(2) Take a little m on the side of BC, so that CM=AC, and then connect AM, as shown in Figure 2:
∫∠ACB = 150,
∴∠AMC=∠MAC= 15,
tan 15 = tan∠AMD = AD/MD = 2/(4+2√3)= 1/(2+√3)≈ 1/(2+ 1.7)≈0.27≈0.3