Answer :
Answer:
The pilot has to fly the plane at an angle 83° north of west.
Explanation:
The airspeed of an airplane is 600 km/hr ( north direction)
In figure, [tex]V_{PA}=600\text{ km/h}[/tex]
A wind blowing northeast at 90 km/h
In figure, [tex]V_{A}=90\text{ km/h}[/tex]
Let actual speed of airplane and direction of plane,
Speed = x km/h
Direction = Ф (north of west, refer figure)
Now, we divide each velocity in horizontal and vertical component.
- For wind, [tex]V_{A}=90\text{ km/h}[/tex]
Horizontal component, [tex]V_{\text{Horizontal}}=90\cos45^\circ\text{ km/h}[/tex]
Vertical component, [tex]V_{\text{Vertical}}=90\sin45^\circ\text{ km/h}[/tex]
- For airplane,
Horizontal component, [tex]V_{\text{Horizontal}}=x\cos\theta\text{ km/h}[/tex]
Vertical component, [tex]V_{\text{Vertical}}=x\sin\theta\text{ km/h}[/tex]
If the airplane needs to head due north.
Horizontal component must be cancel out.
Therefore, [tex]x\cos\theta=90\cos45^\circ----------(1)[/tex]
Speed of airplane is 600 km/h due to north.
So, Sum of vertical component must be equal to 600
Therefore, [tex]x\sin\theta+90\sin45^\circ=600-----------(2)[/tex]
Using eq(1) and eq(2) we get,
[tex]\tan\theta=\dfrac{600\sqrt{2}-90}{90}[/tex]
[tex]\theta=83.23^\circ[/tex]
[tex]\theta\approx 83^\circ[/tex]
Hence, The pilot has to fly the plane at an angle 83° north of west.
