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anonymous

  • one year ago

Will Medal! The measure of theta, in degrees, is approximately 48.8 degrees? A. True B. False (Picture Below) ***My Answer: B***

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  1. anonymous
    • one year ago
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  2. UsukiDoll
    • one year ago
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    do have any reasons why you think it's B?

  3. UsukiDoll
    • one year ago
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    did you use the law of Cosines? cos A = \[\cos A = \frac{(b^2+c^2-a^2)}{2bc}\] \[\cos B = \frac{(c^2+a^2-b^2)}{2ca} \]

  4. anonymous
    • one year ago
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    I used the formula: \[c ^{2} = a ^{2} + b ^{2} -2ab cosC\]

  5. zeesbrat3
    • one year ago
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    Couldn't you just use arccos?

  6. UsukiDoll
    • one year ago
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    hmmm but I want cos C to be by itself and it's easier to grab the angle A and B first.

  7. anonymous
    • one year ago
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    Now that I see the formulas you gave me, I think I may be doing the wrong formula.

  8. UsukiDoll
    • one year ago
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    @zeesbrat3 arccos is later into the formula... you have to find angle A and angle B first .. since this is all S S S (side side side)

  9. zeesbrat3
    • one year ago
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    Fair enough

  10. UsukiDoll
    • one year ago
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    c is supposed to be the longest side a.k.a hypotenuse

  11. anonymous
    • one year ago
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    Oh ok. The laws of cosines were getting me confused for a second.

  12. anonymous
    • one year ago
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    So we would have a = 5.3 b = 4 and c = 7?

  13. UsukiDoll
    • one year ago
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    I need letters first... we need to redraw the triangle with A B C and then look across for our a b c |dw:1434936158533:dw| ok let's use those values

  14. UsukiDoll
    • one year ago
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    so that would mean ... a = 5.3 , b=4, and c = 7... now we use Cos A to find Angle A

  15. anonymous
    • one year ago
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    Ok

  16. UsukiDoll
    • one year ago
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    \[\cos A = \frac{(b^2+c^2-a^2)}{2bc} \] you just have to plug in those values.. and simplify

  17. anonymous
    • one year ago
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    Great! Thank you! So we would have cos A = 0.66?

  18. anonymous
    • one year ago
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    Then in degrees it would be 37.8 degrees?

  19. UsukiDoll
    • one year ago
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    this is where you need to use \[\cos^{-1}(0.66) \] and then you use a calculator to find the degrees of it

  20. UsukiDoll
    • one year ago
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    Can I just double check this for a sec? ^_^ so fast

  21. anonymous
    • one year ago
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    Oh ok! Yeah go ahead :)

  22. UsukiDoll
    • one year ago
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    \[\cos A = \frac{(4^2+7^2-5.3^2)}{2(4)(7)}\]

  23. anonymous
    • one year ago
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    Oh nevermind I got 48.7 degrees

  24. zeesbrat3
    • one year ago
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    I got \[\cos \theta = .6835\]

  25. UsukiDoll
    • one year ago
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    \[\cos A = \frac{(16+49-28.09)}{56} \rightarrow \frac{36.91}{56} \rightarrow 0.659 \rightarrow 0.66\]

  26. UsukiDoll
    • one year ago
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    \[\cos^{-1}(0.66) \rightarrow 48.7\] I also got 48.7 degrees

  27. zeesbrat3
    • one year ago
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    My bad, I used 54 as the denominator, not 56... ignore the blonde...

  28. UsukiDoll
    • one year ago
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    ok cool... we got angle A = 48.7 degrees... Now we need angle B

  29. anonymous
    • one year ago
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    Oh whoops for the the rest of the equation haha

  30. UsukiDoll
    • one year ago
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    let's try and solve further.

  31. anonymous
    • one year ago
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    Ok

  32. UsukiDoll
    • one year ago
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    \[\cos B = \frac{(c^2+a^2-b^2)}{2ca} \] c = 7, b = 4, a = 5.3

  33. UsukiDoll
    • one year ago
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    back the equation tool gave me a hard time

  34. anonymous
    • one year ago
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    I got around 35.1 degrees (34.95 not rounded)

  35. UsukiDoll
    • one year ago
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    \[\cos B = \frac{(7^2+5.3^2-4^2)}{2(7)(5.3)}\] \[\cos B = \frac{(49+28.09-16)}{74.2}\] \[\cos B = \frac{61.09}{74.2}\] \[\cos B = 0.84\] \[\cos^{-1}(0.84) =31.95 -> 32\] huh that's odd... how did you get 35.1 degrees?

  36. triciaal
    • one year ago
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    another approach when given only the 3 sides use cosine rule to find largest angle opposite longest side

  37. UsukiDoll
    • one year ago
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    oh wait I got the division wrong \[\frac{61.09}{74.2} = 0.823 \]

  38. UsukiDoll
    • one year ago
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    \[\cos^{-1}(0.823) = 34.58\]

  39. UsukiDoll
    • one year ago
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    anyway .. let's just assume that Angle B = 34.58 Angle A = 48.7 to find Angle C.. we use the fact that angles of a triangle must add up to 180 degrees Angle C = 180-48.7-34.58 I will also take the 35.1 into consideration for Angle B Angle C = 180-48.7-35.1

  40. triciaal
    • one year ago
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    |dw:1434937131238:dw|

  41. triciaal
    • one year ago
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    now use the sin rule

  42. UsukiDoll
    • one year ago
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    Assuming that Angle B is 34.58 Angle C = 180-48.7-34.58 = 96.72 Assuming that Angle B is 35.1 Angle C = 180-48.7-35.1=96.20 But I think ... in the drawing... only one angle is being considered.

  43. triciaal
    • one year ago
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    |dw:1434937491614:dw|

  44. triciaal
    • one year ago
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    With rounding error true

  45. UsukiDoll
    • one year ago
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    I had angle A almost close to 48.8 as well because when I divided \[\frac{36.91}{56} \]

  46. UsukiDoll
    • one year ago
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    48.76 since the number after the decimal point is greater than 5 I can round up and it does become 48.8

  47. UsukiDoll
    • one year ago
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    if rounding isn't required then it wouldn't be 48.8 but that's cutting it close..

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