1. **State the problem:** We have a triangle with sides 5.4 cm and 4.4 cm, and an angle of 40° opposite the 5.4 cm side. We need to find the angle $x$ opposite the 4.4 cm side. 2. **Formula used:** We use the Law of Sines, which states: $$\frac{a}{\sin A} = \frac{b}{\sin B}$$ where $a$ and $b$ are sides opposite angles $A$ and $B$ respectively. 3. **Assign values:** Let side $a = 5.4$ cm, angle $A = 40^\circ$, side $b = 4.4$ cm, and angle $B = x$. 4. **Apply Law of Sines:** $$\frac{5.4}{\sin 40^\circ} = \frac{4.4}{\sin x}$$ 5. **Solve for $\sin x$:** $$\sin x = \frac{4.4 \times \sin 40^\circ}{5.4}$$ 6. **Calculate $\sin 40^\circ$:** $$\sin 40^\circ \approx 0.6428$$ 7. **Substitute and simplify:** $$\sin x = \frac{4.4 \times 0.6428}{5.4} = \frac{2.8283}{5.4}$$ 8. **Simplify fraction:** $$\sin x = \cancel{\frac{2.8283}{5.4}} = 0.5238$$ 9. **Find angle $x$:** $$x = \sin^{-1}(0.5238) \approx 31.6^\circ$$ 10. **Answer:** The angle $x$ is approximately $31.6^\circ$.