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Ammarah

how do u know when to use sin and cos?

  • one year ago
  • one year ago

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  1. Ammarah
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    @mathstudent55

    • one year ago
  2. IsTim
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    sin for opposite over hypotenus cos for adjacent over hypotenus

    • one year ago
  3. Ammarah
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    I know but like in a problem...i dont get which one i should use

    • one year ago
  4. Ammarah
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    Use a sine or cosine ration to find the value of each variable. Round decimals to the nearest tenth.

    • one year ago
  5. Ammarah
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    |dw:1358230193689:dw|

    • one year ago
  6. Ammarah
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    These three please

    • one year ago
  7. Ammarah
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    i dont get them

    • one year ago
  8. IsTim
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    use that. sorry. i ahve to go soon.

    • one year ago
  9. Ammarah
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    it doesnt help please solve these for me

    • one year ago
  10. Callisto
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    |dw:1358230939592:dw| Refer to the reference angle. Sin => opposite side (to the reference angle) / hypotenuse In this diagram, can you determine what sine theta is?

    • one year ago
  11. Ammarah
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    ? no

    • one year ago
  12. Callisto
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    what is the opposite side to the reference angle theta?

    • one year ago
  13. Ammarah
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    x

    • one year ago
  14. Callisto
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    and what is the hypotenuse?

    • one year ago
  15. Ammarah
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    2 btw what is theta?

    • one year ago
  16. Callisto
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    Sorry :| that was supposed to be z :| If you know x and z, you can find theta. Because sin (theta) = x/z Got it?

    • one year ago
  17. opiesche
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    Ammara, so in an equation, Callisto is saying is that in the figure above, \[\sin(\theta) = \frac{ x }{ z }\] So, in your figure, the first problem: you know the hypotenuse (18) and the angle opposite x (32 degrees). So you know that, \[\sin(32) = \frac{ x }{ 18 }\] Then solve for x :)

    • one year ago
  18. opiesche
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    And that's the sine of 32 degrees of course, not radians.

    • one year ago
  19. Ammarah
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    what do i do for why then?

    • one year ago
  20. Ammarah
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    |dw:1358231828749:dw|

    • one year ago
  21. opiesche
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    Well, you know that the angle opposite x is 32 degrees, and that the angle opposite the hypotentuse is 90 degrees. You also know the total sum of all interior angles of a triangle - so from that you can figure out the remaining angle you don't know, and do the same thing.

    • one year ago
  22. Ammarah
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    i dont get it....how to i solve for y?

    • one year ago
  23. Ammarah
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    and does x=9.5?

    • one year ago
  24. opiesche
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    Or alternatively, use the cosine which is the adjacent angle over the hypotenuse: \[\cos(\theta) = \frac{ y }{ z }\] Substitute for the adjacent angle (32 degrees again) and for the hypotenuse (18), and solve for y the same way you solved for x above.

    • one year ago
  25. Ammarah
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    |dw:1358232234127:dw|

    • one year ago
  26. opiesche
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    Correct.

    • one year ago
  27. Ammarah
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    Ok how about the next triangle...

    • one year ago
  28. Ammarah
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    |dw:1358232364105:dw|

    • one year ago
  29. Ammarah
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    |dw:1358232378783:dw|

    • one year ago
  30. Ammarah
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    so in the left triangle were solving for a.

    • one year ago
  31. Ammarah
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    how would u solve it? please solve it for me.......

    • one year ago
  32. opiesche
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    Correct. And you know that the cosine of the angle (48 degrees) is the same as 10 divided by a - careful here, because a is the hypotenuse: \[\cos(\Theta) = \frac{ 10 }{ a }\] So, \[\cos(48) = \frac{ 10 }{ a } \]

    • one year ago
  33. Ammarah
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    how would i solve be now...i got 14.9 for a

    • one year ago
  34. opiesche
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    That's correct. Now solve the last figure?

    • one year ago
  35. Ammarah
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    b=11.1?

    • one year ago
  36. opiesche
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    Yes!

    • one year ago
  37. opiesche
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    You got it. Can you do the last triangle?

    • one year ago
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