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BloomLocke367

  • one year ago

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  1. anonymous
    • one year ago
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    what

  2. anonymous
    • one year ago
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    whats the question

  3. BloomLocke367
    • one year ago
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    |dw:1433197250937:dw|

  4. BloomLocke367
    • one year ago
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    @Nnesha

  5. anonymous
    • one year ago
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    hold on

  6. BloomLocke367
    • one year ago
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    @perl

  7. anonymous
    • one year ago
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    okay this link will help https://www.mathsisfun.com/algebra/trig-area-triangle-without-right-angle.html

  8. BloomLocke367
    • one year ago
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    @Nnesha

  9. Nnesha
    • one year ago
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    http://prntscr.com/7c1xmi you need to find 3rd side of triangle it's not a right triangle so law of cosine ?familiar with this method ?

  10. BloomLocke367
    • one year ago
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    kind of, I sort of know what it is, but I don't really know how to use it

  11. BloomLocke367
    • one year ago
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    I have to eat dinner, be back in a sec

  12. Nnesha
    • one year ago
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    alright lol

  13. Nnesha
    • one year ago
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    have fun!

  14. BloomLocke367
    • one year ago
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    I'm back, I wasn't that hungry

  15. BloomLocke367
    • one year ago
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    @Nnesha

  16. Nnesha
    • one year ago
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    ahh alright so did you eat candy ? :P

  17. BloomLocke367
    • one year ago
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    no, I had like a few bites.. I think it's my nerves that have me not hungry.. I'm very stressed.

  18. Nnesha
    • one year ago
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    \[\huge\rm c^2 = a^2 +b^2 -2abcos C\] |dw:1433199653612:dw| plug in value solve for c

  19. BloomLocke367
    • one year ago
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    wait, don't I need to use law of sines for this?

  20. Nnesha
    • one year ago
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    mhm nope bec given is two sides and one angle

  21. BloomLocke367
    • one year ago
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    yes, you use cosine if you have 3 sides, and law of sines if you have 2 sides and an angle.. I'm almost positive

  22. Nnesha
    • one year ago
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    http://mathforum.org/library/drmath/view/69918.html you can use law of cosines to find 3rd side

  23. BloomLocke367
    • one year ago
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    but... https://www.mathsisfun.com/algebra/trig-sine-law.html ...

  24. Nnesha
    • one year ago
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    https://www.mathsisfun.com/algebra/trig-cosine-law.html

  25. BloomLocke367
    • one year ago
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    oh, okay.. sorry for doubting you

  26. Nnesha
    • one year ago
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    nah it's fine! i like it :P

  27. Nnesha
    • one year ago
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    okay solve for c

  28. BloomLocke367
    • one year ago
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    c^2=64+324-2(8)(18)cos(25), so far, right?

  29. BloomLocke367
    • one year ago
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    c^2=388-288cos(25) c^2=100cos(25)

  30. BloomLocke367
    • one year ago
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    right?

  31. Nnesha
    • one year ago
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    hold on >:P

  32. BloomLocke367
    • one year ago
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    okay

  33. Nnesha
    • one year ago
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    yep right solve for c

  34. BloomLocke367
    • one year ago
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    I got c^2=90.63 so c=9.52

  35. BloomLocke367
    • one year ago
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    @Nnesha

  36. BloomLocke367
    • one year ago
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    @Nnesha

  37. Nnesha
    • one year ago
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    hmm i got 11.269

  38. BloomLocke367
    • one year ago
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    how? (sorry, I had to go to bed) @Nnesha

  39. BloomLocke367
    • one year ago
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    @pooja195

  40. geerky42
    • one year ago
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    Why do we need to find value of c? |dw:1433254516249:dw| Obviously, \(\large A_{\text{triangle}} = \dfrac12bh\) We already have base \(\overline{\text{QR}} = 8\) So we need to find h: Using fact that \(\sin\theta = \dfrac{\text{Opp}}{\text{Hyp}}\), we have \(\sin Q = \dfrac{h}{\overline{\text{PQ}}} = \dfrac{h}{18}\) Yield \(18\sin(25^\text o) = h\) So we have \(A_{\text{triangle}} = \dfrac12~\overline{\text{QR}}~h = \dfrac{1}{2}(8)(18\sin(25^\text o)~) = \cdots~?\)

  41. BloomLocke367
    • one year ago
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    I got 30.43

  42. geerky42
    • one year ago
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    Me too. So that's your area.

  43. geerky42
    • one year ago
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    Do you understand what I wrote?

  44. BloomLocke367
    • one year ago
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    oh, okay. kinda, I don't understand how you got the height, but the rest I get.

  45. geerky42
    • one year ago
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    well, we can set up height of triangle here:

  46. geerky42
    • one year ago
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    sorry hold on...

  47. BloomLocke367
    • one year ago
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    okay. Can you help with more?

  48. geerky42
    • one year ago
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    |dw:1433254921641:dw| Then we can treat it as "new triangle" |dw:1433254928278:dw| From here, we have hypotenuse and angle Q and height. So using sine, we can solve for h. Then we use that height to solve for area of original triangle.

  49. geerky42
    • one year ago
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    I can try.

  50. BloomLocke367
    • one year ago
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    ohhhhhhh okay. and it's the same height for both triangles?

  51. geerky42
    • one year ago
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    Same angle and hypotenuse, so yeah same height,

  52. BloomLocke367
    • one year ago
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    alright. I have some more questions, if that's okay. I'll post a new question and tag you

  53. geerky42
    • one year ago
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    sure

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