∴∠AOB=80 +20 = 100。

∫OM is the bisector of ∞∠AOB.

∴∠AOM= 100 ÷2=50。

And ∵ON is the bisector of ∠AOC, ∠ AOC = 80.

∴∠AON=80 ÷2=40。

∴∠mon=∠aom-∠aon=50-. 40 = 10。

(2)∫∠AOC = 80，∠BOC=x。

∴∠AOB=80 +x

And ∵OM is the angular bisector of ∠AOB.

∴∠AOM=(80 +x )÷2=40 +x/2

And ∵ON is the bisector of ∠AOC, ∠ AOC = 80.

∴∠AON=80 ÷2=40

∴∠mon=∠aom-∠aon=(40+x/2)-40 = x/2

(3)∫∠AOC = m，∠BOC=x。

∴∠AOB=m +x

And ∵OM is the angular bisector of ∠AOB.

∴∠AOM=(m +x )÷2=(m+x)/2

And ∵ON is the bisector of ∠AOC, ∠ AOC = m.

∴∠AON=m/2

∴∠mon=∠aom-∠aon=(m+x)/2—m/2 = x/2

(4) The degree ∠ mon is related to ∠BOC.

Degree ∠MON is ∠BOC's 1/2.

Using letters to represent numbers can clearly show the relationship between two numbers.