- #1

- 352

- 2

So can someone post the formula for this?

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- Thread starter DaleSwanson
- Start date

- #1

- 352

- 2

So can someone post the formula for this?

- #2

- 14

- 0

Let a be the adjacent, let b be the opposite, and let c be the hypotenuse and A the area. Then

[tex]

\[

c = \sqrt{\frac{2A}{\cos \theta \sin \theta}}

\]

\[

a= \sqrt{2A \cot \theta}

\]

\[

b = \sqrt{2A \tan \theta}

\]

[/tex]

[tex]

\[

c = \sqrt{\frac{2A}{\cos \theta \sin \theta}}

\]

\[

a= \sqrt{2A \cot \theta}

\]

\[

b = \sqrt{2A \tan \theta}

\]

[/tex]

Last edited:

- #3

- 352

- 2

Perfect, thanks.

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