1. **Stating the problem:** We have a bridge across a valley that is 150 m long. The valley walls make angles of 60° and 54° with the bridge. We need to find the depth of the valley, which is the vertical height from the bridge to the valley floor. 2. **Understanding the triangle:** The bridge forms the base of a triangle with the valley walls. The angles at the base are 60° and 54°, and the side opposite these angles is the height (depth) we want to find. 3. **Using the Law of Sines:** The sum of angles in a triangle is 180°, so the angle opposite the bridge is: $$180^\circ - 60^\circ - 54^\circ = 66^\circ$$ 4. **Labeling the triangle:** Let the triangle be ABC with AB = 150 m (bridge), angle A = 60°, angle B = 54°, and angle C = 66°. 5. **Applying the Law of Sines:** $$\frac{AB}{\sin C} = \frac{AC}{\sin B} = \frac{BC}{\sin A}$$ We know AB = 150 m and angle C = 66°, so: $$\frac{150}{\sin 66^\circ} = \frac{BC}{\sin 60^\circ}$$ 6. **Solving for BC:** $$BC = \frac{150 \times \sin 60^\circ}{\sin 66^\circ}$$ Calculate the sines: $$\sin 60^\circ \approx 0.8660, \quad \sin 66^\circ \approx 0.9135$$ So: $$BC = \frac{150 \times 0.8660}{0.9135} \approx 142.2 \text{ m}$$ 7. **Finding the depth (height):** The depth is the vertical height from the valley floor to the bridge, which corresponds to the side opposite angle B (54°). Using Law of Sines again: $$\frac{AB}{\sin C} = \frac{AC}{\sin B}$$ We want AC, which is the height: $$AC = \frac{150 \times \sin 54^\circ}{\sin 66^\circ}$$ Calculate sines: $$\sin 54^\circ \approx 0.8090$$ So: $$AC = \frac{150 \times 0.8090}{0.9135} \approx 132.8 \text{ m}$$ 8. **Conclusion:** The depth of the valley is approximately 133 meters to the nearest meter.