**Answer:**

**Explanation:**

From the given information:

The coordinate axis is situated in the east and north direction.

So, the north will be the y-axis and the east will be the x-axis

Similarly, the velocity of the plane in regard to the air in the coordinate system will be

where:

= velocity of the plane in regard to the air

v = velocity

θ = angle of inclination of the plane with respect to the horizontal

replacing v = 180 km/ and θ = 20° in above equation, then:

The velocity of the airplane in the coordinate system as:

where;

= velocity of the airplane

= velocity

∅ = angle of inclination with regard to the base axis;

Then; replacing = 150 km/h and ∅ = 30°

Therefore, the velocity of the plane in the system is :

--- (1)

( 150 cos 30° - 180 cos 20°)i + ( 150 sin 30° - 180sin 20°)j

(-39.24 km/h)i + (13.44 km/h) j

The magnitude is:

= 41.48 km/h

The airplane is moving at an angle of the inverse tangent to the abscissa and ordinate.

The angle of motion is:

tan θ = 39.24/13.44

tan θ = 2.9

θ =

θ = 70.97°

The angle of motion is 70.97° from west of north with a velocity of 41.48 km/h.