Solution: As shown in the figure, the intersection point D is DH⊥AB in H, and dg∨CB intersects AB in G. 。
∫DC∨AB,
Quadrilateral DCBG is a parallelogram.
∴DC=GB,GD=BC= 1 1.
The difference between the two routes is ad+DG-ag.
At Rt△DGH,
DH=DG? sin37 ≈ 1 1×0.60=6.60,
GH=DG? cos37 ≈ 1 1×0.80≈8.80。
In Rt△ADH,
AD=? 2DH≈ 1.4 1×6.60≈9.3 1。
AH=DH≈6.60。
∴ad+dg-ag=(9.3 1+ 1 1)-(6.60+8.80)≈4.9(km).
That is to say, the distance from A to B is 4.9 kilometers less than before.