1. **Problem statement:** Given a triangle with sides $CB=531$ m, $CA=302$ m, and an angle of $49.3^\circ$ at vertex $B$, find the distance $AB$. 2. **Formula used:** We can use the Law of Cosines to find the unknown side $AB$: $$AB^2 = CB^2 + CA^2 - 2 \times CB \times CA \times \cos(\angle B)$$ 3. **Substitute the known values:** $$AB^2 = 531^2 + 302^2 - 2 \times 531 \times 302 \times \cos(49.3^\circ)$$ 4. **Calculate each term:** $$531^2 = 281961$$ $$302^2 = 91204$$ 5. **Calculate the cosine term:** $$\cos(49.3^\circ) \approx 0.6523$$ 6. **Calculate the product:** $$2 \times 531 \times 302 \times 0.6523 = 209432.5$$ 7. **Put it all together:** $$AB^2 = 281961 + 91204 - 209432.5 = 163732.5$$ 8. **Take the square root to find $AB$:** $$AB = \sqrt{163732.5} \approx 404.6 \text{ m}$$ **Final answer:** The distance $AB$ is approximately $404.6$ meters.