Resolving vector components using trigonometry

In summary, the question is asking to find the components of a vector with length 1.00 and angle 20.0° with respect to the x-axis. The x-component is found by using cos20° and the y-component is found by using sin20°. It is important to check the angle mode on a calculator when encountering unexpected results.
  • #1
NewtonIsaac
5
0

Homework Statement


Find the components of the vector A with length a = 1.00 and angle=20.0° with respect to the x-axis as shown.

Enter the x component followed by the y component, separated by a comma.


Homework Equations


none


The Attempt at a Solution



To find the X-component I did the following:

cos20°=x/1
x=cos20°
x=.408082062

The Y-component was found in a similar fashion:

sin20°=y/1
y=sin20°
y=.912945251

My answer was entered as x,y substituting the relevant values in for x and y. However, the online system I'm using keeps telling me I'm wrong. Can anyone point me in the right direction?
 
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  • #2
How can the sine of the angle be greater than the cosine? The angle is less than 45 degrees.
 
  • #3
I apologize, sometimes I can be mathematically challenged. I don't understand what you are getting at. Here is a picture of my attempt at a solution, in case that helps.
 

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  • #4
Here is the actual figure itself.



NV_ct.jpg
 
  • #5
Newton said:
I apologize, sometimes I can be mathematically challenged. I don't understand what you are getting at. Here is a picture of my attempt at a solution, in case that helps.
Your method is correct, it is just that your numbers are wrong. [itex] \cos(45^o) = \sin(45^o) [/itex] and since cos approaches 45 degrees from above and sin from below, you would expect that the cosine of an angle less then 45 degrees be greater than the sine of that angle, as Muphrid stated.
 
  • #6
CAF123 said:
Your method is correct, it is just that your numbers are wrong. [itex] \cos(45^o) = \sin(45^o) [/itex] and since cos approaches 45 degrees from above and sin from below, you would expect that the cosine of an angle less then 45 degrees be greater than the sine of that angle, as Muphrid stated.


Thank you. It's at times like these I feel incredibly stupid. I had my calculator set to radians instead of degrees. Sorry for wasting everyone's time, including my own. Problem solved.
 
  • #7
What CAF123 said. It should be apparent just looking at the diagram that it makes no sense for the cosine to be smaller than the sine. The vector is much more in the x-direction than it is in the y-direction.

The reason I point this out is as a general check when doing numerical calculations. It should make you question whether you typed in the numbers to your calculator correctly or if it was in the correct mode. Your angles are in degrees. I suspect you left your calculator in radians.

Edit: good, you figured it out! Leaving the calculator in the wrong angle mode is common, which is why having an intuitive check on what makes sense is really, really useful. That's the only reason I tried to point you in that direction.
 
  • #8
Muphrid said:
What CAF123 said. It should be apparent just looking at the diagram that it makes no sense for the cosine to be smaller than the sine. The vector is much more in the x-direction than it is in the y-direction.

The reason I point this out is as a general check when doing numerical calculations. It should make you question whether you typed in the numbers to your calculator correctly or if it was in the correct mode. Your angles are in degrees. I suspect you left your calculator in radians.

Edit: good, you figured it out! Leaving the calculator in the wrong angle mode is common, which is why having an intuitive check on what makes sense is really, really useful. That's the only reason I tried to point you in that direction.

Yeah, I was actually not aware of how to check that intuitively. More often than not I have found that math does not always come intuitively to me. But now I understand, thank you for your help.
 

1. How do I resolve vector components using trigonometry?

To resolve vector components using trigonometry, you will need to use the trigonometric functions sine, cosine, and tangent. First, draw a diagram of the vector and label the angle and side lengths. Then, use the appropriate trigonometric function to find the component of the vector in the desired direction.

2. What is the difference between magnitude and direction of a vector?

The magnitude of a vector refers to its size or length, while the direction refers to the angle that the vector makes with a reference axis. Both magnitude and direction are necessary to fully describe a vector.

3. Can I resolve vector components using trigonometry in three dimensions?

Yes, you can resolve vector components using trigonometry in three dimensions. In this case, you will need to use the three-dimensional versions of the trigonometric functions (sine, cosine, and tangent) and consider the vector in terms of its components in each dimension.

4. What is the purpose of resolving vector components using trigonometry?

Resolving vector components using trigonometry allows you to break down a vector into its vertical and horizontal components, which can be useful in solving problems involving motion or forces in different directions. It also allows you to determine the magnitude and direction of a vector more easily.

5. Are there any limitations to resolving vector components using trigonometry?

One limitation of resolving vector components using trigonometry is that it assumes the vector is in a two-dimensional plane. In reality, many vectors exist in three-dimensional space, which may require a different approach. Additionally, trigonometry may not be the most efficient method for resolving vector components in certain situations, such as when dealing with complex systems or non-linear motion.

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