#### Answer

1- $ \triangle AEB$ is an equilateral so each angle is equal to $ 60^{\circ} $
And all side of equilateral are congruent and they are equal to the sides of the square ABCD.
2- in a triangle BEC is an isosceles triangle ( EB and CB are congruent ) so $ \angle CEB = \angle ECB= 2x+30= 90, x= 75 $
And the same procedure for triangle DEC.
3- in a triangle DEC, where the goal is to find $\angle DEC $ , we find that $ \angle EDC=\angle ECD= 90-75=15 $
4- last step that since $ \triangle DEC = 15+ 15 + \angle DEC = 150 $
We conclude that $ \angle DEC= 150^{\circ} $.

#### Work Step by Step

m$\angle$A+m$\angle$D=180
m$\angle$D=180-58
m$\angle$D=122$^{\circ}$
m$\angle$B+m$\angle$C=180
m$\angle$B+125=180
m$\angle$B=180-125
m$\angle$B=55$^{\circ}$