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Ammarah

  • 3 years ago

how do u know when to use sin and cos?

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  1. Ammarah
    • 3 years ago
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    @mathstudent55

  2. IsTim
    • 3 years ago
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    sin for opposite over hypotenus cos for adjacent over hypotenus

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

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

  5. Ammarah
    • 3 years ago
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    |dw:1358230193689:dw|

  6. Ammarah
    • 3 years ago
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    These three please

  7. Ammarah
    • 3 years ago
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    i dont get them

  8. IsTim
    • 3 years ago
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    use that. sorry. i ahve to go soon.

  9. Ammarah
    • 3 years ago
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    it doesnt help please solve these for me

  10. Callisto
    • 3 years ago
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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?

  11. Ammarah
    • 3 years ago
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    ? no

  12. Callisto
    • 3 years ago
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    what is the opposite side to the reference angle theta?

  13. Ammarah
    • 3 years ago
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    x

  14. Callisto
    • 3 years ago
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    and what is the hypotenuse?

  15. Ammarah
    • 3 years ago
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    2 btw what is theta?

  16. Callisto
    • 3 years ago
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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?

  17. opiesche
    • 3 years ago
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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 :)

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

  19. Ammarah
    • 3 years ago
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    what do i do for why then?

  20. Ammarah
    • 3 years ago
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    |dw:1358231828749:dw|

  21. opiesche
    • 3 years ago
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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.

  22. Ammarah
    • 3 years ago
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    i dont get it....how to i solve for y?

  23. Ammarah
    • 3 years ago
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    and does x=9.5?

  24. opiesche
    • 3 years ago
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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.

  25. Ammarah
    • 3 years ago
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    |dw:1358232234127:dw|

  26. opiesche
    • 3 years ago
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    Correct.

  27. Ammarah
    • 3 years ago
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    Ok how about the next triangle...

  28. Ammarah
    • 3 years ago
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    |dw:1358232364105:dw|

  29. Ammarah
    • 3 years ago
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    |dw:1358232378783:dw|

  30. Ammarah
    • 3 years ago
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    so in the left triangle were solving for a.

  31. Ammarah
    • 3 years ago
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    how would u solve it? please solve it for me.......

  32. opiesche
    • 3 years ago
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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, \[\cos(48) = \frac{ 10 }{ a } \]

  33. Ammarah
    • 3 years ago
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    how would i solve be now...i got 14.9 for a

  34. opiesche
    • 3 years ago
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    That's correct. Now solve the last figure?

  35. Ammarah
    • 3 years ago
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    b=11.1?

  36. opiesche
    • 3 years ago
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    Yes!

  37. opiesche
    • 3 years ago
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    You got it. Can you do the last triangle?

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