Tunnel entry flow: locomotive enters the tunnel head+train tail enters the tunnel head+train enters the tunnel completely+locomotive exits the tunnel tail+train tail exits the tunnel tail.

So the train itself is used to enter the tunnel: (30-20)/2=5s.

It took the train 20+5 = 25 seconds to complete the tunnel.

The tunnel is 500m long, so the train length is 5/25*500= 100m.

Second,

1)OB = OC-BC = AC/2-BC =(a b+BC)/2-BC = 1.5cm

2)∠BOD =∠BOC+∠COD =∠BOC+∠AOC/2 =∠BOC+(∠AOB-∠BOC)/2 = 55

3)EF = EB+BF = AB/2+BC/2 =(a b+BC)/2 = AC/2 = 14

Adaptation: Figure 2, ∠AOB=78, OE ∠AOC bisector, OF ∠BOC bisector, and find ∠EOF.

Solution: ∠ EOF = ∠ EOC+∠ COF = ∠ BOC/2+∠ COA/2 = (∠ BOC+∠ COA)/2 = ∠ AOB/2 = 39.