jmc

Since $\overline{AB}\parallel \overline{CD}$, we have $\angle A+\angle D=180^{\circ }$. Since $\angle A=2\angle D$, we have $2\angle D+\angle D=180^{\circ }$, so $3\angle D=180^{\circ }$, which means $\angle D=60^{\circ }$. Therefore, $\angle A=2\angle D=\boxed{120^{\circ }}$.