Jetliner Displacement: Magnitude & Direction in Northwest - 66*, 27*

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A jetliner takes off from LA, reaching an altitude of 5 km and a position 25 km west and 10 km north of the airport after 5 minutes. The magnitude of its displacement is calculated to be 27 km, with a direction specified as 66 degrees west of north and 27 degrees above the horizontal. The discussion emphasizes the need to construct separate triangles for calculating angles, clarifying that the hypotenuse should not be assumed initially. Participants confirm the angles derived from their calculations, noting adjustments in understanding the components involved. The conversation ultimately reinforces the importance of correctly applying trigonometric principles to determine displacement in three-dimensional space.
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Homework Statement



A jetliner takes off from LA. after 5 min, the jet has an altitude of 5 km above the level of the airport and is 25 km west and 10 km north of the airport. What are the magnitude and the diretion of its displacement during this time interval? Specify the direction by stating two angles: the angle of the displacements direction west of north and the angle of the displacement's direction above the horizontal.

Homework Equations



for West of North:
construct a triangle with adjacent side 10 km and hypotenuse 25 km
then use cox=10/25 to find the angle

For above the horizontal:
construct a triangle with opposite side 5 km and hypotenuse 10
then use tanx=5/10 to find angle

For magnitude:

use the magnitude formula for: [-25 km, 10 km, 5 km]


The Attempt at a Solution



Is horizontal always the Z component/up?

west of north angle: 66*
above horizontal angle: 27*
direction of displacement: northwest
magnitude: 27 km
 
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possum30540 said:
for West of North:
construct a triangle with adjacent side 10 km and hypotenuse 25 km
then use cox=10/25 to find the angle

For above the horizontal:
construct a triangle with opposite side 5 km and hypotenuse 10
then use tanx=5/10 to find angle

It's been awhile since I've done any of this so somebody correct me if otherwise, but for west of north and for above the horizontal, you should not be using any distance as the hypotenuse.

You will need two different triangles for each situation which you did, but the hypotenuse is not known until you calculate it but you don't need it.
 
Okay . .I understand now that I look at it again. Thank you.

for north of west:
construct triangle with opposite 25 and adjacent 10

for above horizontal:
construct triangle with opposite 5 and adjacent 10
 
I believe so yes, that is what I had written down I think. I erased it but I think the angles I had were something like 68.198* and 26.565*
 
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