Resultant vector of an isosceles triangle?

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Homework Statement


what is the resultant vector of an isosceles triangle?


Homework Equations


R^2=a^2+b^2-4abcos(theta)

The Attempt at a Solution


The books answer R=2acos(theta/2)
Using the formula above, and knowing that a=b in a isosceles triangle I am getting:
R=sqrt[2a^2+2a^2cos(theta)]
 

Answers and Replies

  • #2
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It depends a little.

isoscelesvectors.png

If the two vectors are as shown in the 1st diagram, where the sum is CA + AB in the directions shown, the sum is the vector CB.
By the cosine rule its length R would be
R2 = a2 + a2 - 2a.a cos θ
R2 = 2a2 - 2a2 cos θ

If, on the other hand, you mean the two vectors AB + AC as in the lower diagram with directions shown, the sum is vector AC in the diagram on the right.
It looks like this is what the book's answer refers to.
 
  • #3
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Thanks for the response. Now I have one more question about that. In the attachment, the book talks about (1-cos(theta))=2sin^2(theta/2) gives the magnitude of R as R=2Asin(theta/2). I don't see how they jump from the first equation to the second. Can anyone explain?
 

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  • #4
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Thanks for the response. Now I have one more question about that. In the attachment, the book talks about (1-cos(theta))=2sin^2(theta/2) gives the magnitude of R as R=2Asin(theta/2). I don't see how they jump from the first equation to the second. Can anyone explain?

Yes, using the cos rule on the triangle gives
D2 = 2A2 - 2A2 cos θ
D2 = 2A2 (1- cos θ)
Using the identity for (1 - cos θ) gives
D2 = 2A2 (2 sin2 (θ/2))
D2 = 4A2 sin2 (θ/2)

D =
 

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