An equilateral triangle shares a common side with a square as shown. What is the number of degrees in \(m\angle CDB\)?
Note that angle DCB = 90 + 60 = 150°
And since DC = BC, then triangle DCB is isosceles with angle CDB = angle CBD
So angle CDB = (180 - 150) / 2 = 30 / 2 = 15°