A plane with a heading of 50 degrees north of east has an air speed of 400 mph. If a 35 mph wind is blowing from the north, what are the plane's groundspeed and true course

v = 376.06 mph ,    θ = 41.23º

Explanation:

This exercise can be solved with the addition of vectors, we have two vectors: the speed of the airplane 400 mph in the 45 direction to the north east and the speed of the wind coming from the north, therefore the wind direction is south.

One way to solve these exercises is to use trigonometry to break down the speed of the plane.

     v₁ₓ = v₁ cos 45

      = v₁ sin45

      v₁ₓ = 400 0.707

       v₁ₓ = 282.84 mph

       v_{1y} = 282.84 mph

Wind speed is

        v₂ = 35 mph

Let's add each speed

      vₓ = v₁ₓ

      vₓ = 282.84 mph

      v_{y} = v_{1y} - v₂

      v_{y} = 282.84 - 35

      v_{y} = 247.84 mph

The result can be done in two ways

       v = (282.84 i ^ + 247.84 j ^) mph

And in the form of a module and angle, for which we use the Pythagorean theorem and trigonometry

      v = √ (vₓ²+ v_{y}²)

      v = √ (282.84² + 247.84²)

      v = 376.06 mph

      tan θ = v_{y} / vₓ

      θ = tan⁻¹ (247.84 / 282.84)

      θ = 41.23º

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