- #1

TonyC

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When I work the problem, I come up with tan 130 degrees.

Where am I going wrong?

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- Thread starter TonyC
- Start date

- #1

TonyC

- 86

- 0

When I work the problem, I come up with tan 130 degrees.

Where am I going wrong?

- #2

TD

Homework Helper

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[tex]50^\circ = \frac{{5\pi }}{{18}}[/tex]

We know that:

[tex]\cot \left( \alpha \right) = \tan \left( {\frac{\pi }{2} - \alpha } \right)[/tex]

So:

[tex]\cot \left( {\frac{{5\pi }}{{18}}} \right) = \tan \left( {\frac{\pi }{2} - \frac{{5\pi }}{{18}}} \right) = \tan \left( {\frac{{2\pi }}{9}} \right) = \tan \left( {40^\circ } \right)[/tex]

- #3

TonyC

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That makes perfect sense now.....

The light has come on, thank you

The light has come on, thank you

- #4

TD

Homework Helper

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No problem

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