Question

An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50 mi/h) is blowing toward the south. (a) If the airspeed of the plane (its speed in still air) is 320.0 km/h (about 200 mi/h), in which direction should the pilot head?

Answer 1

Answer:

14.48° North of West.

Explanation:

The conditions mentioned in the question has been drawn in the figure below where,

• The vector CA represents the velocity of airplane,
• θ is the angle from west pointing the direction of airplane in which it is required to travel,
• Vector AB is the direction of wind, and,
• Vector CB is the resultant direction.

As ΔABC is a right angle triangle, for calculating the direction of airplane so that it can travel to west, we need to calculate the direction of vector CA which can be calculated with the help of sine theta.

So,

$Sin$θ$= \frac{80}{320} = \frac{1}{4}$

Thus the value of θ = 14.48°

So, the pilot should head towards 14.48° North of West.