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October 22, 2014

Posted by **Jo** on Tuesday, December 8, 2009 at 5:25am.

h t t p : / / w w w . j i s k h a . c o m / d i s p l a y . c g i ? i d = 1 2 6 0 2 4 6 6 3 4

- PreCalc -
**Jo**, Tuesday, December 8, 2009 at 5:47amsorry

- Jo -
**Reiny**, Tuesday, December 8, 2009 at 8:23amTo solve 200 = 300sin(5pi/11)(t-4) + 500

we get -2/3 = sin(5pi/11)(t-4)

So I know that (5pi/11)(t-4) is in either III or IV, since the sine is negative in those quadrants

To find the "angle in standard position" I ignore the negative and take the inverse sine

That got me .72972 (you must have had that)

For the angle in III

(5pi/11)(t-4) = pi + .72972

solving this gave me 6.7 (which you had)

For the angle in IV

(5pi/11)(t-4) = 2pi - .72972

solving this gave the the 7.9

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