Answer to Question #147500 in Calculus for Kornay

Question

A plane leaves an airport traveling at 400 mph in the direction N 45° E. A wind is blowing at 40 mph in the direction N 45° W. What is the ground speed of the plane?

Answer 1

The figure below represents a right angled triangle. Using pythagoras theorem we have that

$GS^2 = 400^2 + 40^2$ where GS represents the ground speed, this implies that

$GS = \sqrt{400^2 + 40^2}$ , therefore

GS = 401.995mph

Using sine rule

$\frac{A}{sinA} = \frac{B}{sinB}$

= $\frac{401.995}{sin90} = \frac{40}{sinx}$ ,therefore $x=5.9105^0$

this implies the direction of the ground speed is $50.9105^0$

Therefore the ground speed is 401.995mph at $N50.91^0E$