1. **State the problem:** Kenneth is looking at a whale from a pier 30 feet above the water, with his eye level 3 feet above the pier, making a total height of $30 + 3 = 33$ feet above the water. The angle of depression to the whale is $20^\circ$. We need to find the horizontal distance from Kenneth's binoculars to the whale. 2. **Identify the right triangle:** The vertical leg is the height from Kenneth's eyes to the water, which is 33 feet. The angle of depression is $20^\circ$, which equals the angle between the horizontal line at Kenneth's eye level and the line of sight to the whale. 3. **Use trigonometry:** The angle of depression corresponds to the angle between the horizontal and the line of sight. The horizontal distance to the whale is the adjacent side, and the vertical height is the opposite side of the angle. 4. **Formula:** Using the tangent function, $$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$$ where $\theta = 20^\circ$, opposite = 33 ft, and adjacent = distance to whale $d$. 5. **Set up the equation:** $$\tan(20^\circ) = \frac{33}{d}$$ 6. **Solve for $d$:** $$d = \frac{33}{\tan(20^\circ)}$$ 7. **Calculate $\tan(20^\circ)$:** $$\tan(20^\circ) \approx 0.36397$$ 8. **Calculate $d$:** $$d = \frac{33}{0.36397} \approx 90.7$$ 9. **Answer:** The whale is approximately **90.7 feet** from Kenneth's binoculars.