Resultant vector of an isosceles triangle?

AI Thread Summary
The discussion centers on calculating the resultant vector of an isosceles triangle using the cosine rule. The initial formula provided is R^2 = a^2 + b^2 - 4abcos(theta), which simplifies to R = 2acos(theta/2) for isosceles triangles where a = b. The conversation explores two different vector configurations, leading to different resultant vectors, and confirms that the book's answer refers to one specific configuration. Additionally, there is a query regarding the transition from the equation (1-cos(theta)) = 2sin^2(theta/2) to the resultant magnitude R = 2Asin(theta/2), which is clarified through the application of trigonometric identities. The discussion effectively addresses the complexities of vector addition in the context of geometry.
atypical
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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)]
 
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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.
 
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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atypical said:
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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