In the figure BCDE is a square and ABC is equilateral then ∠ADC is :

15^o

According to the question,

∠BCD = \({90}^\circ\)

∠ACB = \({60}^\circ\)

So, ∠ACD = (\({90\; + \;60}^\circ\)) = \({150}^\circ\)

AB = BC = AC (as ABC is an equilateral triangle)

BC = CD = DE = EB (as it is a square)

Hence, AC = CD = BC

So, \(\Delta \)ACD is an isosceles triangle where AC = CD and ∠CAD = ∠ADC.

Now, ∠ADC = (\({180\; - \;150}^\circ\))/2 = \({15}^\circ\)

Therefore, ∠ADC is \({15}^\circ\).