1. **Problem Statement:** You have a drag strip 404 m long and bleachers 3 m deep set 20 m away from the track. (a) Find the position in the bleachers where you must turn your head the farthest from straight ahead to watch the race, and find the greatest angle to the nearest hundredth. 2. **Setup and Variables:** Let the horizontal axis represent the length of the track from 0 m (start) to 404 m (finish). The bleachers are located 20 m away vertically from the track, and have depth 3 m. Define $x$ as the horizontal position along the bleachers from the front seat (closest to the track) to the back seat (3 m deep). The vertical distance from the track to the front of the bleachers is 20 m, so the vertical distance from the track to a seat at depth $x$ is $20 + x$ meters. 3. **Angle Calculation:** The angle you turn your head to watch the race at position $x$ is the angle between the line of sight to the start and the line of sight to the finish. The angle to the start is $\theta_1 = \arctan\left(\frac{20+x}{0}\right)$ but since the start is at horizontal 0, the angle to the start from the seat at horizontal position $s$ along the track is $\theta_1 = \arctan\left(\frac{20+x}{s}\right)$ where $s$ is the horizontal distance from the seat to the start. Similarly, the angle to the finish is $\theta_2 = \arctan\left(\frac{20+x}{404 - s}\right)$. The total angle turned is $\theta = \theta_2 - \theta_1$. 4. **Finding the position that maximizes the angle:** Since the bleachers are 3 m deep, $x$ ranges from 0 to 3. The horizontal position along the bleachers is fixed at 0 (front edge) since the bleachers are beside the track, so the horizontal distance to the track is fixed. The angle is maximized at the front seat ($x=0$) because increasing $x$ increases the vertical distance, which reduces the angle. 5. **Calculate the greatest angle at the front seat ($x=0$):** The angle to the start is $\theta_1 = \arctan\left(\frac{20}{0}\right)$ which tends to $90^\circ$ but since the start is directly horizontal, the angle is $\arctan(\infty) = 90^\circ$. The angle to the finish is $\theta_2 = \arctan\left(\frac{20}{404}\right)$. Calculate $\theta_2$: $$\theta_2 = \arctan\left(\frac{20}{404}\right) = \arctan(0.0495) \approx 2.84^\circ$$ The total angle turned is: $$\theta = 90^\circ - 2.84^\circ = 87.16^\circ$$ 6. **Answer for (a):** The position requiring the greatest head turn is the front seat ($x=0$), and the greatest angle is approximately $87.16^\circ$. --- 7. **Part (b) - Algebraic model for distance $d$ visible by turning head angle $\theta$ from front middle seat:** The front middle seat is at depth $x=1.5$ m (middle of 3 m depth). Let $d$ be the horizontal distance along the track visible by turning head angle $\theta$ from the straight forward position. Using right triangle trigonometry: $$d = (20 + 1.5) \tan(\theta) = 21.5 \tan(\theta)$$ 8. **Domain for $\theta$:** The angle $\theta$ must be such that the visible distance $d$ is between 0 and the length of the track 404 m. So, $$0 \leq d \leq 404$$ $$0 \leq 21.5 \tan(\theta) \leq 404$$ Divide by 21.5: $$0 \leq \tan(\theta) \leq \frac{404}{21.5} \approx 18.79$$ Since $\tan(\theta)$ is positive, $\theta$ is between 0 and $\arctan(18.79)$: $$0 \leq \theta \leq \arctan(18.79) \approx 86.95^\circ$$ --- **Final answers:** (a) The front seat requires the greatest head turn, approximately $87.16^\circ$. (b) The distance visible is modeled by: $$d = 21.5 \tan(\theta)$$ with domain: $$0 \leq \theta \leq 86.95^\circ$$