1. **Stating the problem:** Given trapezium KLMN with angles \(\angle K = 120^\circ\) and \(\angle M = 110^\circ\), and \(NL = NM\). Point R lies on the extension of \(NL\) beyond L. We need to find the measure of \(\angle KNR\). 2. **Understanding the figure and given data:** - \(KLMN\) is a trapezium, so \(KL \parallel NM\). - \(NL = NM\) means triangle \(NLM\) is isosceles with \(NL = NM\). - Point R is on the line extended from \(NL\) beyond L. 3. **Key properties and formulas:** - Sum of interior angles in any quadrilateral is \(360^\circ\). - Since \(KL \parallel NM\), consecutive interior angles are supplementary. - In triangle \(NLM\), since \(NL = NM\), angles opposite these sides are equal. 4. **Calculate \(\angle NLM\) and \(\angle NML\):** Since \(\angle M = 110^\circ\) and \(\angle K = 120^\circ\), and \(KL \parallel NM\), angles \(\angle LKN\) and \(\angle NMK\) are supplementary to \(\angle K\) and \(\angle M\) respectively. 5. **Calculate \(\angle LKN\):** \(\angle LKN + \angle NMK = 180^\circ\) because they are consecutive interior angles. 6. **Calculate \(\angle LKN\):** \(\angle LKN = 180^\circ - 110^\circ = 70^\circ\) 7. **Calculate \(\angle NMK\):** \(\angle NMK = 180^\circ - 120^\circ = 60^\circ\) 8. **Sum of angles in trapezium:** \(\angle K + \angle L + \angle M + \angle N = 360^\circ\) 9. **Calculate \(\angle L + \angle N\):** \(120^\circ + \angle L + 110^\circ + \angle N = 360^\circ\) \(\angle L + \angle N = 360^\circ - 230^\circ = 130^\circ\) 10. **In triangle \(NLM\), since \(NL = NM\), \(\angle NLM = \angle NML\). Let each be \(x\). Then:** \(x + x + \angle LNM = 180^\circ\) \(2x + \angle LNM = 180^\circ\) 11. **Calculate \(\angle LNM\):** Since \(\angle N = \angle LNM = 130^\circ - \angle L\) (from step 9), but \(\angle L\) is part of trapezium angles, we need to find \(\angle LNM\) differently. 12. **Note that \(\angle LNM\) is the angle at N in triangle NLM, and since R lies on extension of NL, \(\angle KNR\) is related to \(\angle LKN\) and \(\angle LNM\). 13. **Using the fact that \(NL = NM\), triangle NLM is isosceles with base LM. So \(\angle NLM = \angle NML = x\). Then:** \(2x + \angle LNM = 180^\circ\) 14. **From trapezium, \(\angle L + \angle N = 130^\circ\), and \(\angle L = x\) (since \(\angle L = \angle NLM\)) and \(\angle N = \angle LNM\). So:** \(x + \angle LNM = 130^\circ\) 15. **From step 13, \(2x + \angle LNM = 180^\circ\), subtract step 14:** \((2x + \angle LNM) - (x + \angle LNM) = 180^\circ - 130^\circ\) \(x = 50^\circ\) 16. **Then \(\angle LNM = 130^\circ - x = 130^\circ - 50^\circ = 80^\circ\)** 17. **Now, \(\angle KNR\) is the angle between lines KN and NR. Since R lies on extension of NL beyond L, \(\angle KNR = \angle LKN + \angle LNM = 70^\circ + 80^\circ = 150^\circ\). But this is the external angle at N, so the internal angle \(\angle KNR = 180^\circ - 150^\circ = 30^\circ\).** 18. **Final answer:** \[\boxed{30^\circ}\] **Answer choice:** c. 30