Question

The wind is blowing N 35.0 degrees W at 1.60 * 10^2 mph. A plane has an engine speed of 3.20 * 10^2 mph. Where should the pilot point the plane in order to fly straight W?

Answer 1

Answer: 24.2° SouthWest

Explanation:

First step: DRAW A PICTURE of the vectors from head to tail (see image)

I created a perpendicular from the resultant vector to the vertex of the given vectors so I could use Pythagorean Theorem to find the length of the perpendicular. Then I used that value to find the angle of the plane.

Perpendicular (x):

cos 35° = adjacent/hypotenuse

cos 35° = x/160

→ x = 160 cos 35°

Angle (θ):

sin θ = opposite/hypotenuse

sin θ = x/320

sin θ = 160 cos 35°/320

θ = arcsin (160 cos 35°/320)

θ = 24.2°

Direction is down (south) and left (west)