The angle of mathematics problem (the problem comes from the synchronous instruction of classroom optimization in Longmen Bookstore, 20 12, the first volume of mathematics)

Solution: according to the meaning of the question ∠ doe: ∠ BOE: ∠ doc: ∠ coa =1:2: 3: 4.

So the ratio of each portion is120/(1+2+1+2) = 20, so ∠COD= 1*20=20.

∠COD= 1*20=20 because ∠COE=∠EOD+∠COD.

So ∠COE 20+20 = 40∠COE = 40.

Solving the known OE bisection angle AOC

Then angle AOE= angle COE

Because ∠ AOE+∠ ECO+COB = 180.

And because ∠ cob; ; ∠AOE = 2； five

So 2a+5a+5a= 180.

a= 1 5

Then ∠ aod = 15× 8 = 120.

So ∠DOB= 180 minus 120.

=60

Solution: According to the nature of folding, we can get

∠BFE =∠B′FE，∠CFG =∠B′FG，

∴∠EFG= 1/2∠BFC=90。

So the answer is 90.