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2233 clothes cos
Answer:

(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