1. **State the problem:** We have triangle ABC with sides AB = 9.7 m, BC = 12.3 m, and angle ABC = 115°. We need to find angle ACB, labeled as $x$, correct to 3 significant figures. 2. **Identify known values:** - Side AB = 9.7 m - Side BC = 12.3 m - Angle ABC = 115° 3. **Use the Law of Cosines to find side AC:** The Law of Cosines states: $$c^2 = a^2 + b^2 - 2ab\cos(C)$$ Here, side AC is opposite angle B (115°), so: $$AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(115^\circ)$$ Calculate: $$AC^2 = 9.7^2 + 12.3^2 - 2 \times 9.7 \times 12.3 \times \cos(115^\circ)$$ 4. **Calculate each term:** $$9.7^2 = 94.09$$ $$12.3^2 = 151.29$$ $$\cos(115^\circ) = \cos(180^\circ - 65^\circ) = -\cos(65^\circ) \approx -0.4226$$ 5. **Substitute values:** $$AC^2 = 94.09 + 151.29 - 2 \times 9.7 \times 12.3 \times (-0.4226)$$ $$AC^2 = 245.38 + 2 \times 9.7 \times 12.3 \times 0.4226$$ Calculate the product: $$2 \times 9.7 \times 12.3 = 238.62$$ $$238.62 \times 0.4226 \approx 100.85$$ So: $$AC^2 = 245.38 + 100.85 = 346.23$$ 6. **Find AC:** $$AC = \sqrt{346.23} \approx 18.60 \text{ m}$$ 7. **Use the Law of Sines to find angle $x$ (angle ACB):** Law of Sines: $$\frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c}$$ Here, angle B = 115°, side AC = 18.60 m opposite angle B, side AB = 9.7 m opposite angle C ($x$). So: $$\frac{\sin(115^\circ)}{18.60} = \frac{\sin(x)}{9.7}$$ 8. **Solve for $\sin(x)$:** $$\sin(x) = \frac{9.7 \times \sin(115^\circ)}{18.60}$$ Calculate $\sin(115^\circ)$: $$\sin(115^\circ) = \sin(180^\circ - 65^\circ) = \sin(65^\circ) \approx 0.9063$$ So: $$\sin(x) = \frac{9.7 \times 0.9063}{18.60} = \frac{8.79}{18.60} \approx 0.4726$$ 9. **Find angle $x$:** $$x = \sin^{-1}(0.4726) \approx 28.2^\circ$$ 10. **Answer:** The value of $x$ correct to 3 significant figures is: $$\boxed{28.2^\circ}$$