An Exciting Journey of an Airplane from City A to City B

How far does the plane fly?

What is the total distance covered by an airplane when it flies in a straight line from city A to city B, with city B located 90 kilometers north and 180 kilometers west of city A?

Answer:

The airplane flies approximately 201.25 kilometers from city A to city B.

When an airplane embarks on a journey from city A to city B, it covers a total distance of approximately 201.25 kilometers. This distance is calculated based on the coordinates of both cities, with city B being situated 90 kilometers north and 180 kilometers west of city A.

To determine the total distance covered by the airplane, we utilize the Pythagorean theorem, a fundamental principle in mathematics that relates to the sides of a right-angled triangle. In this scenario, the northward and westward movements of the airplane can be represented as the two perpendicular sides of a right triangle, with the direct flight path serving as the hypotenuse.

By applying the Pythagorean theorem, which states that the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides, we can calculate the flight distance as follows:

Flight distance = √((90 km)^2 + (180 km)^2)

By substituting the given values into the formula, we obtain:

Flight distance = √(8100 km^2 + 32400 km^2) = √40500 km^2 ≈ 201.25 kilometers

Therefore, the airplane covers an approximate distance of 201.25 kilometers on its journey from city A to city B, combining both the northward and westward components of the flight path.

It is fascinating to observe how mathematical concepts such as the Pythagorean theorem can be applied to real-world scenarios, shedding light on the intricate calculations involved in the field of aviation and navigation. The journey of an airplane from one city to another encompasses not only geographical distances but also mathematical principles that define the trajectories and paths of flight.