Correct option is (C) 160°
\(\because\) Sum of all angles in a quadrilateral is \(360^\circ.\)
\(\therefore\) In quadrilateral ABCD,
\(\angle A+\angle B\) \(+\angle C+\angle D\) \(=360^\circ\)
\(\Rightarrow\) \(200^\circ\) \(+\angle C+\angle D\) \(=360^\circ\) \((\because\angle A+\angle B=200^\circ)\)
\(\Rightarrow\) \(\angle C+\angle D\) \(=360^\circ-200^\circ=160^\circ\)