Hello Huzaifa,
We can split these velocities into their x- and y-components. Let the East and West directions be along the x-axis and the North and South directions be along the y-axis. Let the North direction be on the positive y-direction and the East direction be on the positive x-axis.
Airplane velocity components:
Since we know the direction of the airplane is to the South East, the angle from the North (bearing) = 135°
This lies in the 4th quadrant of the unit circle:
Vy = 600 sin(360-45°) = -600×1/√2 = -424 km/hr
Vx = 600 cos(360-45°) = 600×1/√2 = 424 km/hr
Wind velocity components
Bearing = 45° which lies in the first quadrant of the unity circle diagram
Vy = 60 sin(45°) = 60×1/√2 = 42.4 km/hr
Vx = 60 cos(45°) = 60×1/√2 = 42.4 km/hr
We combine the x-components from the airplane + wind to get the new Vx:
Vx = 424 + 42.4 = 466.4 km/hr
We combine the y-components from the airplane + wind to get the new Vy:
Vy = -424 + 42.4 = -381.6 km/hr
Now to find the resultant magnitude (Vr) of the new Vx and Vy we use pythagoras theorem:
Vr = √(Vx2 +Vy2)
Vr = √[(466.4)2 +(-381.6)2] = 602.6 km/hr
The angle between this resultant velocity (Vr) is found by using:
Arctan(-Vy/Vx) = arctan(-381.6/466.4) = -39.3°
You need to show both the magnitude and the angle with the x-axis for a complete answer.
Feel free to ask follow-up questions. I hope this helps.
