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vera_ewing

  • one year ago

Math question

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  1. vera_ewing
    • one year ago
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  2. Michele_Laino
    • one year ago
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    hint: \[\sin \theta = \pm \sqrt {1 - {{\left( {\cos \theta } \right)}^2}} \]

  3. Michele_Laino
    • one year ago
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    hint: \[\large \begin{gathered} \sin \theta = \pm \sqrt {1 - {{\left( {\cos \theta } \right)}^2}} = \pm \sqrt {1 - \frac{{16}}{{49}}} = \pm \sqrt {\frac{{33}}{{49}}} = ...? \hfill \\ \hfill \\ \tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{{\left( { \pm \sqrt {\frac{{33}}{{49}}} } \right)}}{{ - \frac{4}{7}}} = ...? \hfill \\ \end{gathered} \]

  4. vera_ewing
    • one year ago
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    I'm not sure...

  5. Michele_Laino
    • one year ago
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    hint: \[ \pm \sqrt {\frac{{33}}{{49}}} = \pm \frac{{\sqrt {33} }}{7}\]

  6. vera_ewing
    • one year ago
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    @Michele_Laino Okay, so sin0 = (sqrt)33/7 ?

  7. anonymous
    • one year ago
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    not to butt in but this is somewhat easier if you draw a triangle

  8. anonymous
    • one year ago
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    |dw:1436190352602:dw|

  9. anonymous
    • one year ago
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    there is picture of an angle whose cosine is \(\frac{4}{7}\) find the "opposite" side via pythagoras then you can take any trig ratio you like

  10. Michele_Laino
    • one year ago
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    since we have 2 square roots, then: \[\sin \theta = \pm \frac{{\sqrt {33} }}{7}\]

  11. vera_ewing
    • one year ago
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    How do we find tanθ Michele?

  12. Michele_Laino
    • one year ago
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    it is simple, you have to compute this ratio: \[\large \tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{{\left( { \pm \frac{{\sqrt {33} }}{7}} \right)}}{{ - \frac{4}{7}}} = \left( { \pm \frac{{\sqrt {33} }}{7}} \right) \times \left( { - \frac{7}{4}} \right) = ...?\]

  13. vera_ewing
    • one year ago
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    -7(sqrt)33/28 @Michele_Laino ?

  14. anonymous
    • one year ago
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    |dw:1436190696248:dw|

  15. anonymous
    • one year ago
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    memorize the trig ratios, look at the triangle and you can find any trig ratio you like

  16. anonymous
    • one year ago
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    for example \[\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}=\pm\frac{\sqrt{33}}{7}\]

  17. anonymous
    • one year ago
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    \[\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=?\]

  18. anonymous
    • one year ago
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    there is really very very little work to solve these just find the missing side of the triangle and use the trig ratios you already know

  19. vera_ewing
    • one year ago
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    tanθ = (sqrt)33/4 right?

  20. anonymous
    • one year ago
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    yes of course

  21. vera_ewing
    • one year ago
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    Thank you.

  22. anonymous
    • one year ago
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    yw

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