**Correct option is (C) 160°**

\(\because\) Sum of all angles in a quadrilateral is \(360^\circ.\)

\(\therefore\) In quadrilateral ABCD,

\(\angle A+\angle B\) \(+\angle C+\angle D\) \(=360^\circ\)

\(\Rightarrow\) \(200^\circ\) \(+\angle C+\angle D\) \(=360^\circ\) \((\because\angle A+\angle B=200^\circ)\)

\(\Rightarrow\) \(\angle C+\angle D\) \(=360^\circ-200^\circ=160^\circ\)