Calculation of Math Angle of Grade One in Junior High School

Analysis: (1) as shown in figure (1), according to the meaning of the question ∠MON=∠MOD+∠NOC-∠COD.

= 1/2(∠AOD+∠BOC)-∠COD

= 1/2(∠AOB+∠COD)-∠COD

= 60. That is ∠ mon = 60.

(2) From the meaning of the question, at the beginning, ∠ BOD = 105, ∠COD rotates counterclockwise around the O point at the speed of X per second 10 second.

Then ∠ BOD =105-10x; ∠AOC = 15+ 10x；

So ∠ BOC = 135- 10x, ∠ AOD = 45+ 10x, and the value of x can be obtained by listing the proportional relationship according to the meaning of the question.

Solution: (1) From the meaning of the question, it can be known that ∠ AOB = 150, ∠ COD = 30, om and on share ∠AOD and ∠BOC equally.

∠MON=∠MOD+∠NOC-∠COD

= 1/2(∠AOD+∠BOC)-∠COD

= 1/2(∠AOB+∠COD)-∠COD

=60 ,

You can get ∠ mon = 60.

(2) From the meaning of the question, ∠ BOD =105-10x; ∠AOC = 15+ 10x；

So ∠ BOC = 135- 10x, ∠ AOD = 45+ 10x,

Because ∠ aom: ∠ bon = 7: 1 1,

OM and ON share ∠AOD and ∠ Bank of China,

So ∠ AOD: ∠ BOC = 7: 1 1

Namely (45+10x): (135-10x) = 7:11;