ABCD is a parallelogram in which ∠A = 110°.

It is given that $A B C D$ is a parallelogram in which $\angle A$ is equal to $110^{\circ}$.

Sum of the adjacent angles of a parallelogram is $180^{\circ}$.

$\therefore \angle A+\angle B=180^{\circ}$

$\Rightarrow 110^{\circ}+\angle B=180^{\circ}$

$\Rightarrow \angle B=\left(180^{\circ}-110^{\circ}\right)$

$\Rightarrow \angle B=70^{\circ}$

$\therefore \angle B=70^{\circ}$

Also, $\angle B+\angle C=180^{\circ}$

$\Rightarrow 70^{\circ}+\angle C=180^{\circ}$

$\Rightarrow \angle C=\left(180^{\circ}-70^{\circ}\right)$

$\Rightarrow \angle C=110^{\circ}$

$\therefore \angle C=110^{\circ}$

Further, $\angle C+\angle D=180^{\circ}$

$\Rightarrow 110^{\circ}+\angle D=180^{\circ}$

$\Rightarrow \angle D=\left(180^{\circ}-110^{\circ}\right)$

$\Rightarrow \angle D=70^{\circ}$

$\therefore \angle D=70^{\circ}$