What is the angle that \vec{C} makes with the positive x axis?

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Discussion Overview

The discussion revolves around determining the angle that the vector \(\vec{C}\) makes with the positive x-axis, given its length and an initial angle. Participants explore the relationship between the vector's components and its orientation in a coordinate system.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant attempts to calculate the components of the vector \(\vec{C}\) using the equations for sine and cosine but finds their results incorrect.
  • Another participant questions the validity of the angles derived, suggesting that the angle should be measured starting from the positive x-axis and going around to the y-axis.
  • There is a discussion about the correct interpretation of the angle, with one participant claiming to have calculated it as 600 degrees, which raises confusion among others.
  • Participants discuss the method of measuring angles in a counterclockwise direction from the x-axis to the y-axis and then to the vector \(\vec{C}\).
  • One participant expresses uncertainty about how they arrived at 600 degrees and seeks clarification on the correct approach.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct angle or the method to calculate it. There are multiple competing interpretations of how to measure the angle from the x-axis to the vector \(\vec{C}\).

Contextual Notes

There are unresolved issues regarding the correct application of trigonometric functions and the interpretation of angles in different quadrants. The discussion includes potential misunderstandings about the diagram referenced by participants.

Paymemoney
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Homework Statement


Find the components of the vector [tex]\vec{C}[/tex] with length c = 1.00 and angle [tex]\phi = 30.0^{\circ}[/tex] as shown.

NV_ct.jpg


Homework Equations


[tex]opp=hypcos(\theta)[/tex]
[tex]adj=hypcos(\theta)[/tex]


The Attempt at a Solution



tried to use the above equations to answer the question, however when i looked in the answers it was incorrect.
Answer i got was x=0.866 and y=0.5

When i realized that it was between the vector and the y axis, i found that the angle didn't make contact with the x-axis, so my question is What is the angle that [tex]\vec{C}[/tex] makes with the positive x axis?

P.S
 
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Hi Paymemoney! :smile:
Paymemoney said:

Homework Equations


[tex]opp=hypcos(\theta)[/tex]
[tex]adj=hypcos(\theta)[/tex]

erm :redface: … they can't both be right, can they? :biggrin:
When i realized that it was between the vector and the y axis, i found that the angle didn't make contact with the x-axis, so my question is What is the angle that [tex]\vec{C}[/tex] makes with the positive x axis?

Start at the positive x-axis, go round to the y-axis, then go round some more … that's your angle! :wink:

But the main thing you've done wrong is that the dotted line marked "x" on your diagram isn't x, is it? :smile:
 
tiny-tim said:
Hi Paymemoney! :smile:


erm :redface: … they can't both be right, can they? :biggrin:

oops, meant to be [tex]opp=hypsin(\theta)[/tex]

tiny-tim said:
Start at the positive x-axis, go round to the y-axis, then go round some more … that's your angle! :wink:

But the main thing you've done wrong is that the dotted line marked "x" on your diagram isn't x, is it? :smile:

ok so i got 600 degrees, what is the next step i should take to solve [tex]\vec{C}[/tex]
 
Hi Paymemoney! :smile:

(have a degree: º and a theta: θ and a phi: φ :wink:)
Paymemoney said:
ok so i got 600 degrees …

hold it!

however did you get 600º ? :confused:
 
well i went around the the axis 2 times o_o
 
Paymemoney said:
well i went around the the axis 2 times o_o

Still not 600º. :confused:

Anyway, just go from the x-axis to the y-axis anticlockwise, and then carry on to C.
 
ok this is what i have done:

vectorquess.JPG


tell me if this is correct.
 
Yes, that's exactly correct. :smile:

The "first quadrant" (top-right) is 0º to 90º, the "second quadrant" (top-left) is 90º to 180º, and so on. :wink:
 

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