A pilot is flying from Daytona Beach to Prescott. The two cities are 1874 miles away from one another, and the angle from Daytona Beach to Prescott is W28N. At takeoff, there is a 12 mph wind blowing N37E that blows constantly throughout the entire 4-hour flight. At what angle should the pilot leave Daytona Beach to fly in a straight line and land in Prescott?

This problem involves vector addition and trigonometry. We need to calculate the correct angle that the pilot should fly to compensate for the wind blowing during the flight.

### Step 1: Break down the vectors

1. **The vector of the plane's desired path**:  
   The desired direction of flight from Daytona Beach to Prescott is \( W28^\circ N \), which means 28 degrees north of west or \( 360^\circ - 28^\circ = 332^\circ \) from the positive x-axis in standard position (measured counterclockwise from east).

2. **The wind vector**:  
   The wind is blowing at 12 mph in the direction \( N37^\circ E \), which is 37 degrees east of north or \( 90^\circ - 37^\circ = 53^\circ \) from the positive x-axis.

### Step 2: Set up the vector components

For both vectors, we will decompose them into their \(x\) (east-west) and \(y\) (north-south) components.

1. **Plane's desired velocity vector**:  
   The speed of the plane in still air is not given, but we know it needs to counteract the wind to fly in a straight path. Let's assume the plane's airspeed is \( v_p \).

   - \( V_p = (v_p \cos 332^\circ, v_p \sin 332^\circ) \)
   - \( V_p = (v_p \cos 28^\circ, -v_p \sin 28^\circ) \)

2. **Wind vector**:  
   The wind speed is 12 mph, and the direction is 53° from the x-axis.

   - \( V_w = (12 \cos 53^\circ, 12 \sin 53^\circ) \)

   Calculating the components:
   \[
   V_w = (12 \times \cos 53^\circ, 12 \times \sin 53^\circ) = (7.21, 9.58)
   \]

### Step 3: Calculate the required angle

The pilot's path plus the wind vector must result in the desired direction. This means the combined velocity vector of the plane and wind must equal the desired velocity vector toward Prescott. Let the angle the pilot should fly be \( \theta \).

- The new velocity vector is the sum of the plane's heading and the wind:
  \[
  (v_p \cos \theta, v_p \sin \theta) + (7.21, 9.58) = (v_p \cos 28^\circ, -v_p \sin 28^\circ)
  \]

To solve for \( \theta \), we use vector addition and trigonometric identities to balance both the x- and y-components. I'll now calculate the angle numerically.

The pilot should fly at an angle of approximately \( 122.23^\circ \) counterclockwise from the east, or \( W32.23^\circ S \) (32.23 degrees south of west), to compensate for the wind and land in Prescott.

Would you like further details or explanations?

Here are some follow-up questions:
1. How can vector addition help in other navigational problems involving wind or currents?
2. What is the significance of breaking vectors into their x- and y-components?
3. How does trigonometry apply to solving real-world navigation problems?
4. What would happen if the wind speed changed during the flight?
5. How does airspeed differ from ground speed in aviation calculations?

**Tip**: In vector problems, always ensure your angles and directions are consistent with the coordinate system you are using.

Solve a navigation problem where a pilot must adjust their course to account for wind interference. This problem involves vector addition and trigonometry, using key concepts such as angle calculation and wind velocity compensation. Ideal for advanced high school or college students.

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