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MaryJSmith

  • 2 years ago

What is the length of the portion BC of the cloth?

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  1. MaryJSmith
    • 2 years ago
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  2. MaryJSmith
    • 2 years ago
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    8cos35 degrees 8sin 35 degrees sin35 degrees/8 cos35 degrees/8

  3. saifoo.khan
    • 2 years ago
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    use sin ratio.

  4. MaryJSmith
    • 2 years ago
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    idk what that is..

  5. saifoo.khan
    • 2 years ago
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    \[\sin \theta = \frac{opposite}{hypotenuse}\]

  6. MaryJSmith
    • 2 years ago
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    idk what all that means..

  7. saifoo.khan
    • 2 years ago
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    |dw:1333128731997:dw|Dont sleep in class. ;)

  8. MaryJSmith
    • 2 years ago
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    I'm home schooled.. I can't sleep in class , haha . I just don't understand this stuff & I have HORRIBLE memory :) Ha !

  9. saifoo.khan
    • 2 years ago
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    lol. Ok. this is easy i can teach you. :)

  10. MaryJSmith
    • 2 years ago
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    So what is it exactly that I'm supposed to do here ??

  11. saifoo.khan
    • 2 years ago
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    There are three basic ratios. Sin ratio, cos ratio, tan ratio. you have to remember these.. Not them down somewhere.

  12. MaryJSmith
    • 2 years ago
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    okay..

  13. saifoo.khan
    • 2 years ago
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    |dw:1333128936512:dw|

  14. saifoo.khan
    • 2 years ago
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    |dw:1333129034555:dw|

  15. saifoo.khan
    • 2 years ago
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    Now these are the three sides. now it depends upon the question which two ratio you have to use.. you have to judge it by yourself.

  16. saifoo.khan
    • 2 years ago
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    Now in your question.. you are given hypotenuse and you have to find the opposite side.. so u will be using sin ratio.

  17. MaryJSmith
    • 2 years ago
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    i still don't get it though . like , what #'s do i plug in & where ?? it's so confusing to me!!

  18. inkyvoyd
    • 2 years ago
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    Ok, so the basic idea is that if you have a right angle, and You know the second angle, you know the ratio of the two sides. Now all of the trig functions essentially take in an angle, and out put a ratio.

  19. MaryJSmith
    • 2 years ago
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    i'm still confused :(

  20. inkyvoyd
    • 2 years ago
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    Do you want to talk it over in chat?

  21. MaryJSmith
    • 2 years ago
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    No.. too overwhelming . Just mssg me .

  22. inkyvoyd
    • 2 years ago
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    Can we do something like tiny chat? Messaging is too slow. If you have a skype, you could chat with me there (or create an anonymous one)

  23. MaryJSmith
    • 2 years ago
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    i cant , im also takin a test & i can only be on here & my test (im bein supervised)

  24. inkyvoyd
    • 2 years ago
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    Wait, you are allowed to go here and on your test?

  25. MaryJSmith
    • 2 years ago
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    Yes

  26. inkyvoyd
    • 2 years ago
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    Alright, Trigonometry is like an extension of triangle congruence.

  27. inkyvoyd
    • 2 years ago
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    *triangle similarity

  28. inkyvoyd
    • 2 years ago
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    For triangle similarity, we can prove that two triangles are similar given certain info (SAS, AA, SSS, etc). But if they are similar, then how can we figure out the side lengths they have?

  29. MaryJSmith
    • 2 years ago
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    idk..

  30. inkyvoyd
    • 2 years ago
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    The answer is that we use trigonometry. The basic idea of trig is that in a right triangle where we know all the degree measures, we can determine the ratio between the sides.

  31. inkyvoyd
    • 2 years ago
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    Tell me if you are following what I'm saying, ok

  32. MaryJSmith
    • 2 years ago
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    I'm not.. haha sorry . I am so confused !! I'm a visual/audio learner..

  33. inkyvoyd
    • 2 years ago
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    ok, who is supervising you? I have file that will help you a lot, but it's for a geometry software called "GeoGebra"

  34. MaryJSmith
    • 2 years ago
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    i have geogebra

  35. inkyvoyd
    • 2 years ago
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    Alright, take a look at this file. Tell me if it helps.

  36. MaryJSmith
    • 2 years ago
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    I'm just gonna guess cos theres like 4 or 5 probs like this & I have 15 mins to finish 20 more problems !!!!!

  37. inkyvoyd
    • 2 years ago
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    sigh. if they are typable, put them into wolfram alpha dot com or something,

  38. Mikellaalbertson
    • one year ago
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    So what's the answer

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