To solve the problem of the airplane needing to head due north while compensating for the wind blowing from the northeast, we'll need to use vector addition to determine the angle at which the pilot needs to fly.
First, we'll represent the airplane's velocity and the wind's velocity as vectors. Since the wind is blowing from the northeast, it has both a southward component and a westward component relative to due north.
Let's define our axes such that the north direction is positive y, and east is positive x. The airplane needs to end up flying due north, which is along the positive y-axis.
We'll denote:
- \( V_p \) = Velocity of the plane with respect to the air (airspeed) = 700 km/hr (directed at an angle we need to find, east of north),
- \( V_w \) = Velocity of the wind = 50 km/hr (from the northeast, so it has equal components to the south and west, at a -45-degree angle to the axes),
- \( V_r \) = Resultant velocity of the plane with respect to the ground (due north).
To cancel out the wind's effect and ensure that the plane ends up flying due north, the eastward (x) component of the plane's velocity must be equal in magnitude and opposite in direction to the eastward component of the wind's velocity.
Given that the wind velocity \( V_w \) is at a -45-degree angle, its components can be found using trigonometry:
\[ V_{wx} = V_w \cos(-45^\circ) \]
\[ V_{wy} = V_w \sin(-45^\circ) \]
Since \( \cos(-45^\circ) = \sin(-45^\circ) = \frac{\sqrt{2}}{2} \), and we're looking for the eastward component, \( V_{wx} \) will be positive and \( V_{wy} \) negative because it's to the south:
\[ V_{wx} = 50 \cdot \frac{\sqrt{2}}{2} \]
\[ V_{wx} = 25\sqrt{2} \]
Similarly, \( V_{wy} = -25\sqrt{2} \) but this component doesn't factor into our calculation for the required angle, since it affects only the north/south direction.
Now, to counteract this eastward wind component, the eastward component of the plane's velocity \( V_{px} \) must be equal to \( -V_{wx} \):
\[ V_{px} = -V_{wx} \]
\[ V_{px} = -25\sqrt{2} \] km/hr
The plane's airspeed is \( V_p = 700 \) km/hr, so using trigonometry with \( V_{px} \) and \( V_p \), we can find the angle \( \theta \) east of north that the plane should head:
\[ \sin(\theta) = \frac{V_{px}}{V_p} \]
\[ \sin(\theta) = \frac{-25\sqrt{2}}{700} \]
\[ \sin(\theta) = \frac{-25\sqrt{2}}{700} \approx -0.05071 \]
Taking the inverse sine (arcsin) to solve for \( \theta \):
\[ \theta = \arcsin(-0.05071) \]
The value of \( \theta \) obtained will be negative, indicating the angle west of north, but we want the angle east of north, so we take the absolute value:
\[ |\theta| \approx \arcsin(0.05071) \]
Using a calculator to find \( \theta \) in degrees:
\[ |\theta| \approx 2.905 \] degrees
Therefore, the pilot will have to fly approximately 2.905 degrees east of north to compensate for the northeast wind and ensure the resultant path is due north.