∴∠DOC=90，

∴∠DOM=45，

∴∠MOC=45 -30 = 15。

∵∠AOC=60，∠AOB= 150，

∴∠BOC=90，

∴∠NOC=45，

∴∠NOD=45 -30 = 15。

∴∠MOC=∠NOD，

(2) ①: ∵ om equals ∠AOD, on the equal part ∠BOC,

∴∠AOD=2∠AOM,∠BOC=2∠BON.

∴∠aob=∠aod+∠boc-∠cod=2∠aom+2∠bon-30 = 150

∴∠AOM+∠BON=90，

∴∠MON= 150 -90 =60

② Let ∠MOC=∠AOC=x,

Then ∠ DOM = 30-x, then 30-x = 2 x,

X = 10，

∠ DOM = 20，∠ nod = 40，

Then ∠ AOC = 10, ∠ nod = 4 ∠ MOC.

Solution: Is the (1) line on ∠ AOC divided equally? Reason:

Let the reverse extension line of on be OD,

∫OM divided equally ∠BOC,

∴∠MOC=∠MOB，

∵OM⊥ON again,

∴∠MOD=∠MON=90，

∴∠COD=∠BON，

∫∠AOD =∠BON (equal to the vertex angle),

∴∠COD=∠AOD，

∴OD sharing ∠AOC,

Is the straight line on the bisector ∠AOC? ..

(2)∵∠BOC= 120

∴∠AOC=60，

∴∠BON=∠COD=30，

That is, when rotated by 60 degrees, on the bisector ∠AOC,

Judging from the meaning of the question, 6t = 60 or 240,

∴t= 10 or 40;

Hope to adopt