Suppose that a wind is blowing in the direction S45°E at a speed of 50 km/h. A pilot is steering a plane in the direction N60°E at an airspeed (speed in still air) of 100 km/h.

Question

The true course, or track, of the plane is the direction of the resultant of the velocity vectors of the plane and the wind. The ground speed of the plane is the magnitude of the resultant. Find the true course and the ground speed of the plane. (Round your answers to one decimal place.)

Answer 1

The wind is represented by a vector pointing to the lower right from the origin. This vector is of length 50, representing a speed of 50 kilometers per hour. The vector points from the origin (0, 0) to the point (50/√2, -50/√2).

How did I discover this point (50/√2, -50/√2)? The direction S45ºE is 45 degrees to the east of south. On a map, south is straight down, and east is to the right. S45ºE is therefore half way between down and right. This direction is called "to the lower right". The vector of length 50 forms the hypotenuse of a 45-45-90 right triangle. (Draw this triangle by drawing a straight line from the point (50/√2, -50/√2) up to the X axis.) The ratio of the sides of a 45-45-90 right triangle is always 1:1:√2. (The first two numbers, 1 and 1, are the same because the triangle is isoceles.) Since the hypotenuse of the triangle is of length 50, each of the two legs of the triangle must be of length 50/√2. (We can check this using the Pythagorean theorem: (50/√2)² + (50/√2)² = 50².)

The path that the plane would take through still air is represented by a vector pointing to the upper right from the origin. This vector is of length 100, representing a speed of 100 kilometers per hour. The vector points from the origin (0, 0) to the point (50, 50√3).

How did I discover this point (50, 50√3)? The direction N60ºE is 60 degrees to the east of north. On a map, north is straight up, and east is to the right. N60ºE is therefore somewhere between up and right; we call this direction "to the upper right". More precisely, this direction is closer to right than to north because 60º is greater than 45°. I'd like to write this direction N60ºE in terms of a smaller angle. So N60ºE is exactly the same direction as E30°N (30 degrees to the north of east.) The vector of length 100 forms the hypotenuse of a 30-60-90 right triangle. (Draw this triangle by drawing a straight line from the point (50, 50√3) down to the X axis.) The ratio of the sides of a 30-60-90 right triangle is always 1:2:√3. (The middle number, 2, is the biggest, so that one must be the length of the hypotenuse.) Since our hypotenuse is of length 100, the two legs must be of length 50 and 50√3. (We can check this using the Pythagorean theorem: (50√3)² + 50² = 100².) The shorter leg must be the vertical one because it's opposite the smaller base angle (30 degrees, at the origin). The longer leg must be the horizontal one because it's opposite the bigger base angle (60 degrees, at the point (50, 50√3). So the shorter leg (vertical, the y coöridinate) is of length 50 and the longer leg (horizontal, the x coöridinate) is of length 50√3.

To sum up the story so far, we have one vector (the wind) pointing from (0, 0) to (50/√2, -50/√2), and another vector (the course through still air) pointing from (0, 0) to (50√3, 50). Their total effect on the airplane is the sum of these two vectors. To get the sum of these two vectors, you add together the two x values (50/√2 and 50√3) and you add together the two y values (-50/√2 and 50). The sum is the vector

(50/√2 + 50√3, -50/√2 + 50)

Use your calculator to compute the values 50/√2 + 50√3 and -50/√2 + 50. Then use the Pythagorean theorem to compute the length of this vector (50/√2 + 50√3, -50/√2 + 50). This length is the ground speed of the plane; remember it's in "kilometers per hour".

The "true course" is the direction in which this vector is pointing. It should be approximately to the upper right (in other words, a few degrees to the north of east) because 50/√2 + 50√3 is positive and -50/√2 + 50 is also positive. The exact angle at which this vector is pointing has a tangent of

"rise"/"run" = (-50/√2 + 50)/(50/√2 + 50√3)

In other words, this angle is arctan (-50/√2 + 50)/(50/√2 + 50√3). I hope you have a calculator that can do arctan. The true course will be E (the arctan)º N. Convert the arctan from radians to degrees and round to one decimal place. Please message me if you need greater detail. Good luck.