1. **State the problem:** A sandpile forms a mound where the slant side RT is 20 meters and makes a 40° angle with the ground at point R. We need to find the volume of the sandpile. 2. **Identify the shape and formula:** The sandpile can be approximated as a cone. The volume of a cone is given by: $$V = \frac{1}{3} \pi r^2 h$$ where $r$ is the radius of the base and $h$ is the height. 3. **Find the height $h$ of the cone:** The slant side RT is the hypotenuse of a right triangle with angle 40° at R. The height $h$ is the side opposite the 40° angle: $$h = 20 \sin 40^\circ$$ Calculate: $$h = 20 \times 0.6428 = 12.856 \text{ meters}$$ 4. **Find the radius $r$ of the base:** The radius is the adjacent side to the 40° angle: $$r = 20 \cos 40^\circ$$ Calculate: $$r = 20 \times 0.7660 = 15.32 \text{ meters}$$ 5. **Calculate the volume:** $$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (15.32)^2 (12.856)$$ Calculate $r^2$: $$r^2 = (15.32)^2 = 234.7$$ Then: $$V = \frac{1}{3} \pi \times 234.7 \times 12.856$$ $$V = \frac{1}{3} \times 3.1416 \times 234.7 \times 12.856$$ $$V = \frac{1}{3} \times 9481.5 = 3160.5 \text{ cubic meters}$$ 6. **Final answer:** The volume of the sandpile is approximately **3161 cubic meters** to the nearest cubic meter.