How Do You Return to the Oak Tree After Finding the Treasure?

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Homework Help Overview

The problem involves navigating a series of movements starting from an oak tree to find a treasure, followed by determining the return path to the starting point. The subject area includes vector addition and trigonometry in a two-dimensional plane.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the resultant vector from their movements but questions the discrepancy between their result and the book's answer. Some participants suggest revisiting the angle used in the calculations, specifically regarding the third angle's measurement.

Discussion Status

Participants are actively discussing the angle calculations and the interpretation of directional angles. There is a focus on clarifying the difference between angles measured north of east versus east of north. Some guidance has been offered regarding the correct interpretation of these angles, but no consensus has been reached on the final calculations.

Contextual Notes

The original poster expresses confusion over the angle calculations and seeks to understand the reasoning behind the suggested corrections. There is an emphasis on ensuring clarity in the definitions and measurements used in the problem setup.

eeriana
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Homework Statement


While following a treasure map, you start at an old oak tree. You first walk 825 m directly south, then turn and walk 1.25 km at 30degrees west of north, then 1.00 km 40.0 degrees north of east where you find a treasure. To return to the oak tree, in what direction would you walk and how far?


Homework Equations





The Attempt at a Solution


I thought that I did everything right, but can't seem to get the answer that they have in the back of the book. Any help would be appreciated.

R=A+B+C.
A = 270 degrees B= 120 degrees C = 50 degrees

Ax = Acos\Theta= (825m)(cos 270) = 0
Ay = Asin \Theta= (825)(sin270) =-825m
Bx= (1250m)(cos 120) = -625m
By= (1250) (sin 120) = 1082.53m
Cx=(1000)(cos50) = 642.78m
Cy=(1000)(cos 50) = 766m

Rx= 17.78m Ry = 1023.53 R= \sqrt{}(17.78^2)+(1023.53^2) = 1023.68m

arctan (1023.53/17.78) = 89 degrees

The answer I come up with is 1023.7 m at 1 degree west of south.

The answer in the book is 911 m at 8.9 degrees w of s.

Thanks for the help!
 
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Your third angle should be 40 not 50... is use cos40 not cos50 etc...
 
learningphysics said:
Your third angle should be 40 not 50... is use cos40 not cos50 etc...

Forgive me, but could you explain why. Perhaps I am not understanding the angle calculation, but I thought it was counterclockwise from the + x-axis towards the + y-axis and I thought that with it being 40 degrees north of east, it would be 90-40. I am sorry if this sounds like a stupid question, but I am trying to understand where I am going wrong so that I can understand what to do the next time...

Thank you :)

Amy
 
eeriana said:
Forgive me, but could you explain why. Perhaps I am not understanding the angle calculation, but I thought it was counterclockwise from the + x-axis towards the + y-axis and I thought that with it being 40 degrees north of east, it would be 90-40. I am sorry if this sounds like a stupid question, but I am trying to understand where I am going wrong so that I can understand what to do the next time...

Thank you :)

Amy

If you go counterclockwise from the +x axis towards the y-axis... that's the same as going north of east... The angle east of north is 50... the angle north of east is 40.

going north from east... is the same as going counterclockwise towards the y-axis... because you are going up from the positive x-axis...
 
Last edited:
learningphysics said:
If you go counterclockwise from the +x axis towards the y-axis... that's the same as going north of east... The angle east of north is 50... the angle north of east is 40.

going north from east... is the same as going counterclockwise towards the y-axis... because you are going up from the positive x-axis...

I think I see what you're saying.

Thank you for your help...

Amy
 

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