1. **Problem statement:** We have triangle MNR with angle $M = 120^\circ$, side $MN = 9$, side $NR = 16$, and we need to find angle $x = \angle R$. 2. **Formula used:** Use the Law of Cosines to find side $MR$ first: $$MR^2 = MN^2 + NR^2 - 2 \cdot MN \cdot NR \cdot \cos(M)$$ 3. **Calculate $MR^2$:** $$MR^2 = 9^2 + 16^2 - 2 \cdot 9 \cdot 16 \cdot \cos(120^\circ)$$ Recall $\cos(120^\circ) = -\frac{1}{2}$, so $$MR^2 = 81 + 256 - 2 \cdot 9 \cdot 16 \cdot \left(-\frac{1}{2}\right)$$ $$MR^2 = 337 + 144 = 481$$ 4. **Find $MR$:** $$MR = \sqrt{481}$$ 5. **Use Law of Cosines again to find angle $x = \angle R$:** $$\cos(x) = \frac{NR^2 + MR^2 - MN^2}{2 \cdot NR \cdot MR}$$ Substitute values: $$\cos(x) = \frac{16^2 + (\sqrt{481})^2 - 9^2}{2 \cdot 16 \cdot \sqrt{481}} = \frac{256 + 481 - 81}{2 \cdot 16 \cdot \sqrt{481}} = \frac{656}{32 \sqrt{481}}$$ 6. **Simplify the fraction:** $$\cos(x) = \frac{656}{32 \sqrt{481}} = \frac{41}{2 \sqrt{481}}$$ 7. **Calculate $x$:** $$x = \cos^{-1}\left(\frac{41}{2 \sqrt{481}}\right) \approx 53.1^\circ$$ **Final answer:** $$x \approx 53.1^\circ$$