A light plane is headed due south with a speed relative to still air of 165 km/hkm/h . After 1.00 hh, the pilot notices that they have covered only 135 kmkm and their direction is not south but southeast 45.0

 v_air = 118.10 km / h , θ = 36 º  (36 north east)

Explanation:

In this exercise we can look for air velocity, the easiest method is to decompose the speed in a reference system, we used trigonometry

        d = 135 km

        θ= 45.0 south-east

This Angle measured from the positive side of the x-axis is

         θ = 360-45

         θ = 315º

         sin 315 = y / d

        cos 315 = x / d

        y = d sin315

        x = d cos 315

        y = 135 sin 315 = -95.46 km

        x = 135 cos 315 = 95.46 km

Airplane distance after 1 hour

        D₁ = v / t =

        D₁ = 165/1 = 165 km

        = -165 j ^

         D₁ₓ = 0 i^

Total distance traveled

       D₁ₓ = x - x_air

       D_{1y} = y - y_air

       x_air = x -D₁ₓ

       y_air = y -D_{1y}

       x_air = 95.46 - 0

       x_air = 95.46 km

       y_air = -95.46 - (-165)

       y_air = 69.54 km

As the air has a constant speed we can use the speed formula

        v = d / t

        vₓ = x / t

        vₓ = 95.46 km / h

        = 69.54 km / h

To give the module let's use Pythagoras

        v_air = √ (vₓ² + v_{y}²)

        v_air = √ (95.46² + 69.54²)

        v_air = 118.10 km / h

For the direction let's use trigonometry

       tan θ = v_{y} / vₓ

       θ = tan⁻¹ v_{y} / vₓ

       θ = tan⁻¹ 69.54 / 95.46

       θ = 36 º

In cardinal notation is 36 north east