an airplane needs to head due north, but there is a wind blowing from the southwest at 40 km/hr. the plane…

Explanation:

Step1: Analyze Wind and Plane Vectors

The wind is from southwest, so its direction is northeast (since wind blowing from southwest moves towards northeast) with components. The wind vector has magnitude 40 km/hr. Its east component is (40\cos(45^\circ)) and north component is (40\sin(45^\circ)) (since southwest to northeast is 45° from north and east). The plane's airspeed vector (relative to air) has magnitude 600 km/hr, let the angle west of north be (\theta). So the plane's velocity relative to ground should have no east component (to go due north). So plane's west component (which is east component negative) should cancel wind's east component. Plane's east component: (600\sin(\theta)) (wait, no: if angle is west of north, then the east - west component is (600\sin(\theta)) west, and north component is (600\cos(\theta)). Wind's east component is (40\cos(45^\circ)) (since from southwest, direction is 45° from north towards east, so east component is (40\cos(45^\circ)), north component (40\sin(45^\circ))). To have net east - west component zero: (600\sin(\theta)=40\cos(45^\circ)) (because plane's west component equals wind's east component).

Step2: Solve for (\theta)

We know that (\cos(45^\circ)=\sin(45^\circ)=\frac{\sqrt{2}}{2}\approx0.7071) So (600\sin(\theta)=40\times0.7071) (600\sin(\theta)=28.284) (\sin(\theta)=\frac{28.284}{600}\approx0.04714) Then (\theta=\arcsin(0.04714)\approx2.70^\circ) (using calculator: (\arcsin(0.04714)) is approximately 2.7 degrees)

Answer:

(\approx2.7) (or more precisely, using (\sin(\theta)=\frac{40\cos(45^\circ)}{600}=\frac{40\times\frac{\sqrt{2}}{2}}{600}=\frac{20\sqrt{2}}{600}=\frac{\sqrt{2}}{30}\approx\frac{1.4142}{30}\approx0.04714), (\theta=\arcsin(0.04714)\approx2.7^\circ))