When and how do I use sin and cos regarding 2-D collisions?

AI Thread Summary
The discussion focuses on understanding when to use sine and cosine in the context of two-dimensional collisions. The user seeks clarification on why sine is applied to the East component and cosine to the North component in one scenario, while the roles switch in another. It is emphasized that the application of these trigonometric functions depends on how angles are measured relative to the defined directions. The user ultimately finds that the labeling of components as East and North is linked to the angle's measurement in relation to these directions. The conversation concludes with the user expressing gratitude for the clarification received.
JohnMC
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I think what I'm doing is overkill because all I want to know is when to use sin and cos in the components regarding (isolated) two-dimensional collisions. I'm just showing more content because it might help for context. Also, there are more steps in order to find the final velocity which is why you can see some steps cut off, but I didn't bother posting more pictures of it because I thought it was irrelevant from then on as I am only looking for why sin and cos are used.

Homework Statement


On QUESTION #1 III., why is sin used in the East component and cos used in the North component, when in QUESTION #2 II., a different but ultimately similar question, the sin and cos are switched? Please feel free to click through the images in order for context. Also, how do I know if I should label the components East and North? Why not other compass directions? I hope I'm making some sense :P

QUESTION #1
I. http://bayimg.com/image/oajpoaaca.jpg
II. http://bayimg.com/image/aajajaacb.jpg
III. http://bayimg.com/image/bajadaacb.jpg
IV. http://bayimg.com/image/cajaaaacb.jpg
V. http://bayimg.com/image/cajaoaacb.jpg


QUESTION #2
I. http://bayimg.com/image/cajcpaacb.jpg
II. http://bayimg.com/image/dajcmaacb.jpg
III. http://bayimg.com/image/eajcjaacb.jpg

Homework Equations


N/A


The Attempt at a Solution


Looked at sine laws, properties of triangles, and so far I have found nothing I can apply it to. Please forgive me if I'm asking such a simple question.
 
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it is because of how the angle is measured.

sin =opp/hyp

cos=adj/hyp

in 3, the east is opposite to the angle, so the east component is hyp*sin, similarly the north is adjacent to the angle. Understand?
 
I would like to say that I understand what you mean, but sadly I don't. Although I think I found out a way how and when to use sin and cos. Wordings such as [12.0 E of N] use East for sin and North for cos; while [12.0 N or E] use East for cos and North for sin. Was this what you meant by "because of the how angle is measured"? My apologies for failing to understand
 
JohnMC said:
I would like to say that I understand what you mean, but sadly I don't. Although I think I found out a way how and when to use sin and cos. Wordings such as [12.0 E of N] use East for sin and North for cos; while [12.0 N or E] use East for cos and North for sin. Was this what you meant by "because of the how angle is measured"? My apologies for failing to understand

See this diagram:

http://img16.imageshack.us/img16/3596/diagrm.jpg

Lets call DAC=α (alpha) and CAB=β

sin=opp/hyp and cos=adj/hyp.

Consider triangle ADC,

cos \alpha =\frac{adjacent}{hypotenuse}= \frac{AD}{AC}

sin\alpha = \frac{opposite}{hypotenuse}= \frac{DC}{AC}

Consider triangle ABC,

sin \beta = \frac{opposite}{hypotenuse}=\frac{BC}{AC}

cos \beta=\frac{adjacent}{hypotenuse}= \frac{AB}{AC}

Do you understand what I meant by how the angle is measured now?
 
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Thanks so much! I already knew the properties of sin and cos as shown in your second post, but it was just the thought of using East and North as a property of a triangle in 2-D collisions and matching it with sin and cos that confused me. I read your first post again and I finally figured it out. Again, thank you :)
 
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