Solution

Given $$\angle {110^{o}}.$$
But we know that sum of adjacent angles of a parallelogram is $$180^o$$
$$\angle A+ \angle B=180^o$$
$$\angle B=180^O-110^O=70^O$$
Also $$\angle B +\angle C=180^o$$ [Since $$\angle B$$ and $$\angle C$$ are adjacent angles]
$$70^o+\angle C=180^o$$
$$\angle C=180^o-70^o=110^o$$
Now, $$\angle C+\angle D=180^o$$ [Since $$\angle C$$ and $$\angle D$$ are adjacent angles]
$$110^o + \angle D=180^o$$
$$\angle D=180^o-110^o=70^o$$