1. **State the problem:** We need to find the horizontal distance from the base of a 110-foot tower to an airplane on the runway, given the angle of depression from the top of the tower is 15°. 2. **Identify the right triangle and relevant sides:** The tower height is the vertical leg ($110$ ft), the distance from the base to the airplane is the horizontal leg (unknown, call it $x$), and the angle of depression is $15^\circ$. 3. **Relate angle of depression to angle in triangle:** The angle of depression from the top corresponds to the angle between the horizontal leg and the hypotenuse inside the triangle, so the angle at the top of the tower inside the triangle is $15^\circ$. 4. **Use trigonometric ratio:** The tangent of the angle relates the opposite side (tower height) to the adjacent side (distance $x$): $$\tan(15^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{110}{x}$$ 5. **Solve for $x$:** $$x = \frac{110}{\tan(15^\circ)}$$ 6. **Calculate $\tan(15^\circ)$:** $$\tan(15^\circ) \approx 0.2679$$ 7. **Substitute and compute:** $$x = \frac{110}{0.2679} \approx 410.7$$ 8. **Round to nearest tenth:** $$x \approx 410.7 \text{ feet}$$ **Final answer:** The airplane is approximately 410.7 feet from the base of the tower.