1. **State the problem:** We are given triangle $\triangle KNO$ with $m\angle NKO = 30^\circ$ and angles $\angle NKO \cong \angle KNM \cong \angle MNT$. We need to find $m\angle NMT$ and $m\angle NMK$. 2. **Analyze the given information:** - $m\angle NKO = 30^\circ$ - $\angle NKO \cong \angle KNM \cong \angle MNT$ means these three angles are equal. 3. **Assign variables:** Let each of these congruent angles be $x$. So, $$x = m\angle NKO = m\angle KNM = m\angle MNT = 30^\circ$$ 4. **Find $m\angle NMT$:** Since $M$ and $T$ lie on segment $KO$, and $\angle MNT$ is $30^\circ$, consider triangle $\triangle NMT$. In $\triangle NMT$, the angles are $m\angle NMT$, $m\angle NMK$, and $m\angle MNT$ (which is $30^\circ$). 5. **Find $m\angle NMK$:** Similarly, in triangle $\triangle NMK$, the angles are $m\angle NMK$, $m\angle KNM$ (which is $30^\circ$), and $m\angle NKM$. 6. **Use angle sum property in triangles:** - In $\triangle NMK$: $$m\angle NMK + m\angle KNM + m\angle NKM = 180^\circ$$ $$m\angle NMK + 30^\circ + m\angle NKM = 180^\circ$$ - In $\triangle NMT$: $$m\angle NMT + m\angle MNT + m\angle NTM = 180^\circ$$ $$m\angle NMT + 30^\circ + m\angle NTM = 180^\circ$$ 7. **Note that $m\angle NKM$ and $m\angle NTM$ are supplementary because points $K, M, T, O$ lie on the same line segment $KO$ with $M$ and $T$ between $K$ and $O$. Therefore, $$m\angle NKM + m\angle NTM = 180^\circ$$ 8. **Combine equations:** From step 6, $$m\angle NMK + 30^\circ + m\angle NKM = 180^\circ$$ $$m\angle NMT + 30^\circ + m\angle NTM = 180^\circ$$ Add these two equations: $$m\angle NMK + m\angle NMT + 30^\circ + m\angle NKM + 30^\circ + m\angle NTM = 360^\circ$$ Simplify: $$m\angle NMK + m\angle NMT + m\angle NKM + m\angle NTM + 60^\circ = 360^\circ$$ Using step 7, $$m\angle NKM + m\angle NTM = 180^\circ$$ Substitute: $$m\angle NMK + m\angle NMT + 180^\circ + 60^\circ = 360^\circ$$ Simplify: $$m\angle NMK + m\angle NMT + 240^\circ = 360^\circ$$ Subtract 240°: $$m\angle NMK + m\angle NMT = 120^\circ$$ 9. **Assuming symmetry and equal angles $m\angle NMK = m\angle NMT$ (since $\angle KNM$ and $\angle MNT$ are equal and the figure is symmetric), then: $$2 \times m\angle NMK = 120^\circ$$ Divide both sides by 2: $$m\angle NMK = \cancel{\frac{2}{2}} \times \frac{120^\circ}{\cancel{2}} = 60^\circ$$ Therefore, $$m\angle NMK = 60^\circ$$ And $$m\angle NMT = 60^\circ$$ **Final answers:** $$m\angle NMT = 60^\circ$$ $$m\angle NMK = 60^\circ$$