This test mainly examines the application of triangles in real life.
Solution: Let BC = x m (x > 2) and AC = y m, then AB = y- 1.
In △ ABC, (y-1) 2 = y 2+x 2-2xycs60 is obtained by cosine theorem.
So y = (x > 2)..................8 points.
Method 1: y = (x-2)++4 ≥ 4+2.
If and only if x-2 =, that is, x -2= 2+, the minimum value of y is 4+2.
Method 2: y' = =.
X = 2+ from y' = 0, because when 2 < x < 2+, y' < 0；; When x > 2+ and y' > 0,
So when x = 2+, the minimum value of y is 4+2.
Answer: The shortest length of AC is 4+2m, and the length of BC is 2+ m ................. 15.