In the figure shown, $ABCD$ is a square and $\triangle CDE$ is equilateral. What is the degree measure of $\angle CBE$?
Note that EC = DC = BC
So triangle ECB is isosceles with EC = BC
So angle ECB = angle CBE
And angle ECB = 90 + 60 = 150°
So angle CBE = ( 180 - 150) / 2 = 30 / 2 = 15°
+195 
