Calculating the Magnitude of Vector B with Trig | Homework Help

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The discussion centers on calculating the magnitude of vector B given that the resultant of vectors A and B is 20 units, A has a magnitude of 8 units, and the angle between them is 40 degrees. Multiple participants calculated the magnitude of B and consistently arrived at 13.20 units. They confirmed their calculations using trigonometric methods and the cosine law, suggesting that the discrepancy lies in the provided answer choices. The consensus is that the calculated value is likely correct, indicating a possible error in the answer options. The thread highlights the importance of verifying calculations in vector problems.
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Homework Statement



The resultant of vectors A and B has a magnitude of 20 units. A has a magnitude of 8 units, and the angle between A and B is 40 degrees. Calculate the magnitude of B.

Homework Equations



Trig, resolving vector components.

The Attempt at a Solution



I worked out the equations and got |B| = 13.20 units. But, this isn't any of my choices. Am I right, or am I missing something here?
 
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Your result is correct.

ehild
 
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Ampere said:

Homework Statement



The resultant of vectors A and B has a magnitude of 20 units. A has a magnitude of 8 units, and the angle between A and B is 40 degrees. Calculate the magnitude of B.

Homework Equations



Trig, resolving vector components.

The Attempt at a Solution



I worked out the equations and got |B| = 13.20 units. But, this isn't any of my choices. Am I right, or am I missing something here?

That's what I got. But maybe we made the same mistake? Mind showing you you got that?
 
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Sure. Align A along the +x axis, with B at 40 degrees above that.

Then:

Ax = 8
Ay = 0
Bx = Bcos(40)
By = Bsin(40)

So Rx = 8+Bcos(40)
Ry = Bsin(40)

The magnitude of the resultant R would be sqrt(Rx^2 + Ry^2), which is equal to 20, so

20 = sqrt((8+Bcos(40))^2 + (Bsin(40))^2). Solving for B yields 13.20.

For the record, my choices were

12.6
16.2
14.8
18.4

But I don't see how you could get any of them.
 
Ampere said:
Sure. Align A along the +x axis, with B at 40 degrees above that.

Then:

Ax = 8
Ay = 0
Bx = Bcos(40)
By = Bsin(40)

So Rx = 8+Bcos(40)
Ry = Bsin(40)

The magnitude of the resultant R would be sqrt(Rx^2 + Ry^2), which is equal to 20, so

20 = sqrt((8+Bcos(40))^2 + (Bsin(40))^2). Solving for B yields 13.20.

For the record, my choices were

12.6
16.2
14.8
18.4

But I don't see how you could get any of them.

Same thing I did. If ehild got the same result as us I have little doubt that it's correct. Probably just a mistake in the choices.
 
Awesome, thanks.
 
Dick said:
Same thing I did. If ehild got the same result as us I have little doubt that it's correct. Probably just a mistake in the choices.

I used Cosine Law, with 140°angle between vectors A and B, and got the same result.

ehild
 
Dick said:
Same thing I did. If ehild got the same result as us I have little doubt that it's correct. Probably just a mistake in the choices.

I used Cosine Law, and got the same result.

ehild
 
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