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?
1 answer:
Answer: 24.2° SouthWest
<u>Step-by-step explanation:</u>
First step: DRAW A PICTURE of the vectors from head to tail <em>(see image)</em>
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.
<u>Perpendicular (x):</u>
cos 35° = adjacent/hypotenuse
cos 35° = x/160
→ x = 160 cos 35°
<u>Angle (θ):</u>
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)
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