1. **State the problem:** We are given a triangle with sides 11.4 m and 6.0 m, and an angle of 31° opposite the 11.4 m side. We need to find the angle $\alpha$ opposite the 6.0 m side to the nearest degree. 2. **Identify the formula:** Use the Law of Sines, which states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ where $a$, $b$, $c$ are sides opposite angles $A$, $B$, $C$ respectively. 3. **Assign known values:** Let side $a = 6.0$ m opposite angle $\alpha$, side $b = 11.4$ m opposite angle $31^\circ$. 4. **Apply Law of Sines:** $$\frac{6.0}{\sin \alpha} = \frac{11.4}{\sin 31^\circ}$$ 5. **Solve for $\sin \alpha$:** $$\sin \alpha = \frac{6.0 \times \sin 31^\circ}{11.4}$$ Calculate $\sin 31^\circ \approx 0.5150$: $$\sin \alpha = \frac{6.0 \times 0.5150}{11.4} = \frac{3.09}{11.4} \approx 0.2711$$ 6. **Find angle $\alpha$:** $$\alpha = \sin^{-1}(0.2711) \approx 15.7^\circ$$ 7. **Check for ambiguous case:** Since $31^\circ + 15.7^\circ = 46.7^\circ < 180^\circ$, the triangle is valid. 8. **Find the third angle:** $$180^\circ - 31^\circ - 15.7^\circ = 133.3^\circ$$ 9. **Compare with options:** The indicated angle measure closest to $\alpha$ is approximately $16^\circ$, but the problem likely asks for the largest angle or a specific angle. Since none of the options match $16^\circ$, and the largest angle is about $133^\circ$, none of the options 98°, 100°, 102° fit. So answer is d. cannot. Final answer: d. cannot