How to find the angle between the resultant and the X axis?

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  • #2
cepheid
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I don't understand. Doesn't the diagram show you exactly how to do it? :confused:
 
  • #3
Integral
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Use algebra to isolate sin( [itex] \theta [/itex]) then apply arcsin.
 
  • #4
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I don't understand. Doesn't the diagram show you exactly how to do it? :confused:
The answer is wrong.
 
  • #5
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Use algebra to isolate sin( [itex] \theta [/itex]) then apply arcsin.
Can you please elaborate?
 
  • #6
cepheid
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The answer is wrong.
No, 10.26 degrees is correct.

Can you please elaborate?
Solve for sin(theta) from the equation that you get from the sine law. Once you know sin(theta), then you know theta, because you can apply the inverse sine (a.k.a. arcsine) on your calculator just as Integral mentioned.
 
  • #7
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No, 10.26 degrees is correct.

The computer keeps say it's wrong.

Solve for sin(theta) from the equation that you get from the sine law. Once you know sin(theta), then you know theta, because you can apply the inverse sine (a.k.a. arcsine) on your calculator just as Integral mentioned.
Yeah and i keep getting 10.16.
 
  • #8
cepheid
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Yeah and i keep getting 10.16.
Can you post your calculation steps? Maybe we can see where you went wrong.
 
  • #9
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Can you post your calculation steps? Maybe we can see where you went wrong.
15.02/sin(154)
= 34.2632

6.1/34.2632
= .1780

Sin^- .1780 = 10.26

10.26 + 25 = 35.26 as my final answer
 
  • #10
754
1
deaninator is correct; the answer is wrong.

Break the vectors down to their x- and y-components and solve it that way. You'll see that the resultant vector makes an angle of 40.75[itex]^\circ[/tex] with the x axis.

The problem stems from the fact that the wrong angle is being found in the attachment. The angle you need to find is the one below the resultant vector, not the one above.

Using the sine law, you would find this angle by:

[tex]\frac{9.3}{sin \theta} = \frac{15.02}{sin 154^\circ}[/tex]
 
  • #11
64
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deaninator is correct; the answer is wrong.

Break the vectors down to their x- and y-components and solve it that way. You'll see that the resultant vector makes an angle of 40.75[itex]^\circ[/tex] with the x axis.

The problem stems from the fact that the wrong angle is being found in the attachment. The angle you need to find is the one below the resultant vector, not the one above.

Using the sine law, you would find this angle by:

[tex]\frac{9.3}{sin \theta} = \frac{15.02}{sin 154^\circ}[/tex]
then you add 25 degrees right?
 
  • #13
64
0
deaninator is correct; the answer is wrong.

Break the vectors down to their x- and y-components and solve it that way. You'll see that the resultant vector makes an angle of 40.75[itex]^\circ[/tex] with the x axis.

The problem stems from the fact that the wrong angle is being found in the attachment. The angle you need to find is the one below the resultant vector, not the one above.

Using the sine law, you would find this angle by:

[tex]\frac{9.3}{sin \theta} = \frac{15.02}{sin 154^\circ}[/tex]
Hey do you know a site where I can find where I learn how to break down vectors to their x- and y- components?
 
  • #15
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cool website... i should find more of such kinds..
 
  • #16
754
1
then you add 25 degrees right?
Yes.

25 degrees is the angle between the positive x-axis and the vector on the bottom of the parallelogram (the 6.1 magnitude vector).

[itex]\theta[/tex] is the angle between that vector and the resultant vector.

The sum of the 2 angles is the angle between the resultant and the x-axis.
 
  • #18
D H
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dThe problem stems from the fact that the wrong angle is being found in the attachment.
Alternatively, they found the right angle but added it to the wrong angle. They should have calculated 51 degrees less 10.26 degrees = 40.74 degrees. (The reason they didn't get 40.75 degrees is because that 10.26 degrees should be 10.25 degrees.)
 
  • #19
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Alternatively, they found the right angle but added it to the wrong angle. They should have calculated 51 degrees less 10.26 degrees = 40.74 degrees. (The reason they didn't get 40.75 degrees is because that 10.26 degrees should be 10.25 degrees.)
Can you tell me how you got to to 40.74 please?
 
  • #20
754
1
Referring back to the diagrams in the attachment, you found the angle [itex]\theta[/tex] to be 10.26[itex]^\circ[/tex]

That is the angle between the 9.3 magnitude vector (from the first diagram) and the resultant vector R. (Note that the angle between the 6.1 magnitude vector and the resultant vector is not the same.)

Since the 9.3 magnitude vector makes an angle of 51[itex]^\circ[/tex] with the x-axis and the resultant vector is between that vector and the x-axis, then the resultant vector is 10.26[itex]^\circ[/tex] closer to the x-axis. Therefore the resultant vector makes an angle of 51[itex]^\circ[/tex] - 10.26[itex]^\circ[/tex] = 40.74[itex]^\circ[/tex] with the x-axis.
 

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