Let's break down the problem shown in the image.

Given:

• Airspeed: The airplane has an airspeed of 590 km/h at a bearing of 42° north of east.
• Wind velocity: The wind velocity is 60 km/h in the direction of 28° north of west.

To find:

• The resultant ground speed of the airplane (i.e., the actual speed and direction relative to the ground).

Solution process:

1. Convert vectors into components:

   • Airspeed (590 km/h): This is at 42° north of east. We can resolve this into:

     • $v_{x1} = 590 \times \cos(42°)$ (eastward component)
     • $v_{y1} = 590 \times \sin(42°)$ (northward component)

   • Wind velocity (60 km/h): This is at 28° north of west, or 180° - 28° = 152° measured counterclockwise from east. Resolving this gives:

     • $v_{x2} = 60 \times \cos(152°)$ (westward component)
     • $v_{y2} = 60 \times \sin(152°)$ (northward component)

2. Sum components:

   • The total east-west component will be $v_{x1} + v_{x2}$.
   • The total north-south component will be $v_{y1} + v_{y2}$.

3. Calculate resultant velocity:

   • The magnitude of the resultant velocity (ground speed) can be found using: $v_{res} = \sqrt{(v_{x1} + v_{x2})^2 + (v_{y1} + v_{y2})^2}$

Let me calculate this.The actual ground speed of the aircraft is approximately 572.3 km/h.

Would you like further details or have any other questions?

Here are five related questions to explore:

1. How would the resultant velocity change if the wind speed increased?
2. What would happen if the wind direction changed to directly west?
3. How would you calculate the angle of the resultant velocity relative to east?
4. How does the magnitude of the airspeed vector affect the resultant velocity?
5. How would this problem change if the airplane were flying directly north?

Tip: When resolving vectors into components, always use consistent angle reference points, like measuring angles from the positive x-axis (east).