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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Step-by-step explanation:
Data:
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Answer:
125 ft²
Step-by-step explanation:
We know that that surface area of the triangular prism that will be covered with insulation is equal to the area of the two triangular faces, plus the two lateral rectangular faces.
So you get:
SA = 2 [(15)(15)] + 2 [(30)(15)] = 1,125 ft²