Calculating Magnitude of Vector: Solving for A & Theta

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Homework Help Overview

The discussion revolves around calculating the magnitude of a vector, specifically focusing on the relationship between the vector's components and the angle involved. The original poster seeks assistance in understanding how to derive the magnitude from given components and angles.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the magnitude of a vector and its components, particularly using trigonometric functions. The original poster attempts to clarify their understanding of how to isolate the magnitude "A" from the equation involving the x component "Ax" and the angle "theta". Some participants question the clarity of the original poster's request and explore the implications of using cosine in this context.

Discussion Status

The discussion is ongoing, with participants providing insights into the relationships between vector components and angles. Some guidance has been offered regarding the use of trigonometric functions, but there is no explicit consensus on the best approach to solve for "A" or "theta".

Contextual Notes

There is mention of potential confusion regarding the terms used, such as the distinction between the magnitude and the components of the vector. The original poster indicates that they have specific values for "Ax" and "theta", but the overall context suggests that further clarification may be needed to fully address their questions.

MarXphysics
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1.I only need to know how to get magnitude out of a vector



2. Suchs as 25m = a cos (60 degrees)


3. I tried 25m/cos/60degrees but it don't seem right

please assist I know it isn't a real problem it's just the step that I need help. If you could write it in fomula form AX=A cos (theta) solving for "A". And if you can show me how to solve for "theta" as well please.
 
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magnitude is simply the length of the vector. if you are using cos then it sounds like what you really need is the x or y component.
 
I see that I may have been unclear Ax is the x component. I have Ax and I have ethe angle theta what I do no thave is the magnitude.
 
Ax is written A(sub)x(/sub) except that you use brackets.
 
what do you mean solve for A? that's trivial. A*cos(theta) is no different from A*B.

theta is a little harder. give me a minute.
 
Thanksyou
I have Ax = A cos [tex]\Theta[/tex]
where I know Ax and [tex]\Theta[/tex]
how do I find A
 

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