- #1

francisg3

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1. x=(2r sinθ)/3θ where x=r/2

2. r/2=(2r sinθ)/3θ

3. r=(4r sinθ)/3θ

4. 1=(4 sinθ)/3θ

5. 1/4= sinθ/3θ

6. 3θ/4= sin^-1(θ)

I don't know where to go from there. All I need is an expression in the form of θ= ...

Thanks!

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- Thread starter francisg3
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- #1

francisg3

- 32

- 0

1. x=(2r sinθ)/3θ where x=r/2

2. r/2=(2r sinθ)/3θ

3. r=(4r sinθ)/3θ

4. 1=(4 sinθ)/3θ

5. 1/4= sinθ/3θ

6. 3θ/4= sin^-1(θ)

I don't know where to go from there. All I need is an expression in the form of θ= ...

Thanks!

- #2

tiny-tim

Science Advisor

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(try using the X

1. x=(2r sinθ)/3θ where x=r/2

2. r/2=(2r sinθ)/3θ

3. r=(4r sinθ)/3θ

4. 1=(4 sinθ)/3θ

5. 1/4= sinθ/3θ

6. 3θ/4= sin^-1(θ)

I don't know where to go from there. All I need is an expression in the form of θ= ...

Thanks!

i'm confused …

6. has sin

were they

anyway, when you have A = sin

you can take sin of both sides, giving you sinA = B

- #3

HallsofIvy

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