Final answer:
To compensate for the crosswind blowing from the southwest at 60 km/hr, the pilot should steer approximately 5.39° west of north to fly directly north, based on vector analysis and applying trigonometry to the wind's components relative to the plane's airspeed of 450 km/hr.
Step-by-step explanation:
The question involves determining the angle the airplane needs to point to compensate for the crosswind and still fly due north. This challenge can be addressed by using vector analysis, a part of trigonometry and physics.
Firstly, we need to consider the vector representing the wind and the vector of the plane's airspeed. The wind is blowing from the southwest, which means it has a northeastward component that needs to be countered by the plane. To fly straight north, the plane must aim slightly west of north to negate the eastward push of the wind. Let's assign the wind vector a magnitude of 60 km/hr. The plane's airspeed vector has a magnitude of 450 km/hr.
We can represent the wind vector as having both a northward (let's call it 'y') and an eastward ('x') component. In this case, since the wind is from the southwest, the 'x' and 'y' components would be equal, as the wind direction is 45 degrees from south and west, initially aligning with northwest. The components of the wind vector are:
x = y = 60 km/hr * (sin(45)) = 60 km/hr * (0.7071) ≈ 42.43 km/hr
Given that the plane needs to counteract this eastward component to fly due north, we can calculate the angle using trigonometry. The opposite side is the wind's westward component, and the adjacent side is the total airspeed. Using the arctangent function:
θ = arctan(42.43 / 450)
Computing this gives an angle of:
θ ≈ arctan(0.0943)
θ ≈ 5.39°
Therefore, the pilot should aim approximately 5.39° west of north to fly due north given the wind conditions