To solve this problem, we need to determine the actual heading of the airplane (its direction relative to the ground) so that it can counteract the wind and reach its destination directly. The key steps involve vector addition of the airplane's velocity relative to the air and the wind velocity to get the airplane's velocity relative to the ground.

Step-by-Step Solution

1. Define the Given Information in Vector Form:

   • Airplane's cruising speed: $546 \, \text{mph}$ (relative to the air)
   • Wind speed and direction: $86 \, \text{mph}$ at $25^\circ$ north of east
   • Destination direction: $845 \, \text{mi}$ at $44^\circ$ west of north

2. Convert Directions to Standard Cartesian Coordinates:

   • For the wind vector:
     The wind direction is $25^\circ$ north of east. In standard Cartesian coordinates, east is $0^\circ$, and angles are measured counterclockwise.
     The wind vector components ($W_x, W_y$) are: $W_x = 86 \cos(25^\circ), \quad W_y = 86 \sin(25^\circ)$

   • For the desired travel direction (Schletertown to Bennettsville):
     The destination direction is $44^\circ$ west of north, which is equivalent to $90^\circ + 44^\circ = 134^\circ$ in standard Cartesian coordinates. The direction vector components ($D_x, D_y$) for the destination are: $D_x = -\cos(44^\circ), \quad D_y = \sin(44^\circ)$

3. Set Up the Equations for Ground Velocity Vector:

   • The airplane’s velocity vector relative to the ground ($\mathbf{V_g}$) is the sum of the airplane’s velocity vector relative to the air ($\mathbf{V_a}$) and the wind velocity vector ($\mathbf{W}$).
     $\mathbf{V_g} = \mathbf{V_a} + \mathbf{W}$

   • The components of the airplane's velocity vector relative to the air ($\mathbf{V_a}$) are defined by its airspeed and heading angle ($\theta$), which we need to find. We can write: [ V_{a_x} =