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The problem provided in the image involves a plane flying at a speed of 330 miles per hour on a bearing of \( N65^\circ E \), and a ground speed of 370 miles per hour. The direction of the ground speed vector is \( 30^\circ \). We are asked to find both the speed of the wind and its direction angle.

### Step-by-step Solution:
We can approach this using vector addition, where the velocity of the plane relative to the ground is the sum of the velocity of the plane relative to the air and the wind speed vector.

1. **Velocity of the plane relative to the air (true airspeed):**
   \[
   V_p = 330 \, \text{miles per hour} \quad \text{at a bearing of } N65^\circ E.
   \]
   This means the angle measured from the north is \( 65^\circ \).

2. **Ground speed (resultant velocity):**
   \[
   V_g = 370 \, \text{miles per hour} \quad \text{at an angle of } 30^\circ \text{ from the north}.
   \]

3. **Wind speed vector:**
   The wind vector \( V_w \) will be determined by solving the vector equation:
   \[
   V_g = V_p + V_w
   \]
   This can be solved using trigonometry and vector components.

### Vector Components
- Break the velocities into northward and eastward components.
   - Plane’s velocity components:
     \[
     V_{pN} = 330 \cos(65^\circ)
     \]
     \[
     V_{pE} = 330 \sin(65^\circ)
     \]
   - Ground speed components:
     \[
     V_{gN} = 370 \cos(30^\circ)
     \]
     \[
     V_{gE} = 370 \sin(30^\circ)
     \]

4. **Find the wind’s components:**
   The wind's components are found by subtracting the plane’s velocity components from the ground speed components:
   \[
   V_{wN} = V_{gN} - V_{pN}
   \]
   \[
   V_{wE} = V_{gE} - V_{pE}
   \]

5. **Calculate wind speed and direction:**
   The magnitude of the wind speed \( V_w \) is:
   \[
   V_w = \sqrt{V_{wN}^2 + V_{wE}^2}
   \]
   The direction of the wind can be found using the arctangent function:
   \[
   \theta_w = \tan^{-1} \left( \frac{V_{wE}}{V_{wN}} \right)
   \]

Let me now calculate the exact values using these steps.The speed of the wind is approximately **213.9 miles per hour**, and its direction angle is about **-32.2 degrees**. The negative angle indicates that the wind is blowing to the west of north.

Would you like further details or have any questions about this process?

Here are some related questions to explore:
1. What is the significance of the bearing angle in navigation problems?
2. How does wind speed affect an aircraft's ground speed?
3. Can we solve similar problems using graphical vector addition?
4. What would happen if the wind speed was larger than the airspeed of the plane?
5. How would we handle a scenario where the wind direction is exactly opposite the plane's direction?

**Tip:** When solving vector problems involving angles, always ensure that you measure the angles consistently from a common reference point (like north).

A plane is flying at a speed of 330 miles per hour on a bearing of N65°E. Its ground speed is 370 miles per hour and its true course, given by the direction angle of the ground speed vector, is 30°. Find the speed, in miles per hour, and the direction angle, in degrees, of the wind.

Solve a vector problem involving an airplane flying at 330 mph with a ground speed of 370 mph. Learn how to find the wind speed and direction using vector addition and trigonometry.

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