Find $\angle ACB$

Let $\angle {ACB}=2α$ . Draw a line $\overline {FG}$ through $C$ perpendicular to $\overline {AB}$ intersecting $\overline {AB}$ at $F$ and $\overline {DE}$ at $G$ . Then $|\overline {EG}|=|\overline {AF}|=\dfrac{a}{2}, |\overline {GC}|=\dfrac{a}{2}\tan 15\degree, |\overline {CF}|=\dfrac{a}{2}\cot α\implies \dfrac{a}{2}(\tan 15\degree+\cot α)=a\implies \cot α=3\implies α=30\degree\implies \angle {ACB}=2α=\boxed {60\degree}$ .