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December 19, 2014

December 19, 2014

Posted by **Cobra7710** on Friday, September 2, 2011 at 8:48pm.

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2

- Algebra II -
**Ms. Sue**, Friday, September 2, 2011 at 8:56pmTimmy, Ryland, Cobra, Timmy -- you must be having an identity crisis tonight.

Please use the same name for all of your posts.

- Algebra II -
**Reiny**, Friday, September 2, 2011 at 9:07pmcosØ = -√3/2

1. find the reference angle

(using the positive result √3/2 , use inverse cos to get and angle of 30°

( Make yourself familiar with the ratio of sides of the 30-60-90 triangle)

2. Where is the angle Ø ?

Since the cosine is negative in II and III

in II : Ø = 180-30 or 150°

or

in III : Ø = 180+30 = 210°

- Algebra II -
**Cobra7710**, Saturday, September 3, 2011 at 4:06amoh so we only want reference angle. i thought values meant sin,cos,and tan. how do you know it only wanted reference angle?

- Algebra II -
**Reiny**, Saturday, September 3, 2011 at 7:29amdidn't it say,

"Find the values of Ø" ,**not**find the values of sinØ etc ?

Finding the reference angle is only the first step, from there you have to decide what to do with that reference angle depending on the position of Ø and the values of cosØ

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