# Resultant vector of an isosceles triangle?

atypical

## 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

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

Stonebridge
It depends a little.

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.

atypical
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?

#### Attachments

• Picture 5.png
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Stonebridge
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 =