1. **State the problem:** Susan is at the top of a 2.1 m high wall at the edge of an 8 m wide beach. She measures the angle of depression to a swimmer in the sea as 6°. We need to find how far out to sea the swimmer is from the base of the wall, rounded to the nearest metre. 2. **Identify the right triangle and angle:** The angle of depression from the top of the wall to the swimmer is 6°, which means the angle between the horizontal line at the top of the wall and the line of sight to the swimmer is 6° downward. 3. **Set up the triangle:** The vertical height of the wall is 2.1 m. The horizontal distance from the base of the wall to the swimmer is what we want to find, call it $x$. 4. **Use the tangent function:** In a right triangle, $$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$$ Here, the opposite side is the height of the wall (2.1 m), and the adjacent side is the horizontal distance $x$. 5. **Write the equation:** $$\tan(6^\circ) = \frac{2.1}{x}$$ 6. **Solve for $x$:** $$x = \frac{2.1}{\tan(6^\circ)}$$ 7. **Calculate $\tan(6^\circ)$:** $$\tan(6^\circ) \approx 0.1051$$ 8. **Calculate $x$:** $$x = \frac{2.1}{0.1051} \approx 19.98$$ 9. **Round to the nearest metre:** The swimmer is approximately 20 metres out to sea from the base of the wall. 10. **Note:** The 8 m width of the beach is not needed to find the horizontal distance from the wall base to the swimmer because the angle of depression is measured from the top of the wall directly to the swimmer. **Final answer:** The swimmer is about **20 metres** out to sea.