60º Solution: Since ABCD is a cyclic quadrilateral $\angle\text{B}+\angle\text{D}=180^\circ$ $60^\circ+\angle\text{D}=180^\circ$ $\angle\text{D}=120^\circ$ Now since AD is parallel to BC $\angle\text{C}+\angle\text{D}=180^\circ$ $\angle\text{C}+120^\circ=180^\circ$ $\angle\text{C}=60^\circ$
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