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kittiwitti1

  • one year ago

A two-person tent is to be made so that the height at the center is a = 6 feet (see the figure below). If the sides of the tent are to meet the ground at an angle 60°, and the tent is to be b = 8 feet in length, how many square feet of material will be needed to make the tent? (Assume that the tent has a floor and is closed at both ends, and give your answer in exact form.) Picture reference: http://prntscr.com/8a6azy I did it two ways and got different answers, not sure which one I did wrong...

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  1. Jhannybean
    • one year ago
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    |dw:1440836698456:dw|

  2. kittiwitti1
    • one year ago
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    I tried doing it on my own and got this:\[\frac{360}{\sqrt{3}}ft^{2}\]

  3. kittiwitti1
    • one year ago
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    I feel like that's wrong somehow...

  4. Jhannybean
    • one year ago
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    Alright, let's check.

  5. Jhannybean
    • one year ago
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    You split the area into 3 separate sections, yes? The triangle faces and the rectangular sides?

  6. kittiwitti1
    • one year ago
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    Yes.

  7. Jhannybean
    • one year ago
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    And then added the base?

  8. kittiwitti1
    • one year ago
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    The base is the same as the sides in area, right? :o

  9. Jhannybean
    • one year ago
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    |dw:1440837218920:dw|

  10. kittiwitti1
    • one year ago
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    These are the formulas I got. Triangle:\[h=6,b=\frac{6}{\sqrt{3}},hyp=\frac{12}{\sqrt{3}}\] Square:\[w=\frac{12}{3},l=8\]

  11. Jhannybean
    • one year ago
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    we need to find x first, and then double it to find the area of the front facing triangluar portion.

  12. kittiwitti1
    • one year ago
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    I think if I did anything wrong I probably messed up there

  13. Jhannybean
    • one year ago
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    I havent worked it out yet so one minute! :P

  14. kittiwitti1
    • one year ago
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    |dw:1440837311425:dw| Okay lol

  15. kittiwitti1
    • one year ago
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    |dw:1440837440575:dw| Do these look right to you?

  16. kittiwitti1
    • one year ago
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    I mean correctly done* lol that sounded rude. ^^: I found another proof online with different digits for A and B though, and got a different answer: http://mathhelpboards.com/questions-other-sites-52/aju051000s-questions-yahoo-answers-involving-trigonometry-6263.html

  17. Jhannybean
    • one year ago
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    Yeah. the \(\dfrac{12}{\sqrt{3}}\) looks correct, and that would be the side of the triangular prism... so \(A_2 = \dfrac{12}{\sqrt{3}} \cdot 8 \)

  18. kittiwitti1
    • one year ago
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    Yes

  19. kittiwitti1
    • one year ago
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    I got it like this:\[2\times\frac{1}{2}\times(6\times\frac{12}{\sqrt{3}})+3(8\times\frac{12}{\sqrt{3}})\]

  20. kittiwitti1
    • one year ago
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    Before the plus: Triangles. After the plus: Rectangles

  21. kittiwitti1
    • one year ago
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    I thought the link's method was too complicated and tried to do it myself...but the answers don't match. Not sure which one's wrong

  22. kittiwitti1
    • one year ago
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    I mean I got an answer of \[\frac{360}{\sqrt{3}}\] from doing it myself but it doesn't equal the answer I got from the method in the URL I posted:\[\ne\frac{240}{\sqrt{3}}\]

  23. Jhannybean
    • one year ago
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    Arg... latex!

  24. kittiwitti1
    • one year ago
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    LOL Brb I have a mosquito in my house. -_-

  25. Jhannybean
    • one year ago
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    |dw:1440838289662:dw| \[SA = \color{red}{A_1} +\color{blue}{A_2} +\color{green}{A_3}\]\[SA = \color{red}{2\left(\frac{ 12}{\sqrt{3}} \cdot 8\right)} +\color{blue}{2\left(\frac{1}{2} \cdot \frac{12}{\sqrt{3}} \cdot 6\right)} + \color{green}{\left(\frac{12}{\sqrt{3}} \cdot 8\right)} \]

  26. kittiwitti1
    • one year ago
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    I used 3 instead of 2 for red because I just stuck all the rectangular areas together.

  27. Jhannybean
    • one year ago
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    yeah that works too.

  28. Jhannybean
    • one year ago
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    \[SA = \color{red}{2\left(\frac{ 12}{\sqrt{3}} \cdot 8\right)} +\color{blue}{2\left(\frac{1}{2} \cdot \frac{12}{\sqrt{3}} \cdot 6\right)} + \color{green}{\left(\frac{12}{\sqrt{3}} \cdot 8\right)}\]\[SA = 3\left(\frac{12}{\sqrt{3}} \cdot 8 \right) +2\left(\frac{1}{2}\cdot \frac{12}{\sqrt{3}} \cdot 6\right)\]

  29. kittiwitti1
    • one year ago
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    \[3\left(8\times\frac{12}{\sqrt{3}}\right)+2\left(\frac{1}{2}\times6\times\frac{12}{\sqrt{3}}\right)=\frac{72}{\sqrt{3}}+3\left(\frac{96}{\sqrt{3}}\right)\]So far so good?

  30. kittiwitti1
    • one year ago
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    I mean is it good so far* (what is wrong with my grammar)

  31. Jhannybean
    • one year ago
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    \[SA = \frac{288}{\sqrt{3}} +\frac{72}{\sqrt{3}}\]

  32. Jhannybean
    • one year ago
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    Then simplify it some more, and you get \[SA= \frac{360\sqrt{3}}{3} = 120\sqrt{3}\]

  33. kittiwitti1
    • one year ago
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    Yes. I did get that... But what's up with the other one giving me\[\frac{240}{\sqrt{3}}\]

  34. Jhannybean
    • one year ago
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    Just saw the link you posted earlier, the height being different would result in a different area for the triangle..hence different answer? that's why this guy is getting his \(SA = \dfrac{244}{\sqrt{3}}\).

  35. kittiwitti1
    • one year ago
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    a = 6ft b = 8ft θ = 60° \[\tan60°=\frac{a}{\frac{w}{2}}=\frac{2a}{w}\]\[w=2a\cot60°=\frac{8}{\sqrt{3}}ft\]\[\sin60°=\frac{a}{s}\]\[s=a\csc60°=\frac{8}{\sqrt{3}}ft\]

  36. kittiwitti1
    • one year ago
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    I did redo it with my own numerical values from my problem, but I still got a different answer.

  37. Jhannybean
    • one year ago
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    OK let me work it out first, one min.

  38. Jhannybean
    • one year ago
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    So they started ff by solving for half the base first, given the height of the triangle. |dw:1440839311853:dw|

  39. Jhannybean
    • one year ago
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    Sorry if im a ltitle slow in my responses right now, Kind of getting drowsy.

  40. kittiwitti1
    • one year ago
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    Okay, I'll type out the rest in case you guys can spot an error\[A_{T}=2\times\frac{1}{2}\times\frac{8}{\sqrt{3}}ft\times6ft=\frac{48}{\sqrt{3}}ft^{2}\]\[A_{R}=3\times\frac{8}{\sqrt{3}}\times8ft=64\sqrt{3}ft^{2}\] \[A=A_{T}+A_{R}=64\sqrt{3}ft^{2}+\frac{48}{\sqrt{3}}ft^{2}=\]\[64\sqrt{3}+\frac{48}{\sqrt{3}}\left(\frac{\sqrt{3}}{\sqrt{3}}\right)=\frac{48\sqrt{3}}{3}+64\sqrt{3}\]

  41. kittiwitti1
    • one year ago
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    BLEH that took too long...\[3\left(\frac{48\sqrt{3}}{3}\right)+3(64\sqrt{3})=48\sqrt{3}+192\sqrt{3}=240\sqrt{3}\]Not sure what I did wrong... And sorry it wasn't this -\[\frac{240}{3}\]

  42. kittiwitti1
    • one year ago
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    *It wasn't 240/root 3

  43. Jhannybean
    • one year ago
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    I think however the guy solved it is a little too complicated and the way we solved it earlier is just easier. Don't you think?

  44. kittiwitti1
    • one year ago
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    So then for the next equation (which just has different values) I should do the same? http://prntscr.com/8a6nx4

  45. kittiwitti1
    • one year ago
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    Yeah I think it was too long-winded and complex as well xD

  46. Jhannybean
    • one year ago
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    |dw:1440839957574:dw| \[\tan(60) = \frac{6}{\dfrac{w}{2}} = \frac{12}{w} \iff w = 12\cot(60^\circ)\] etc...

  47. Jhannybean
    • one year ago
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    Okay, now you just solve it with \(a=5\) instead of \(a=6\)

  48. kittiwitti1
    • one year ago
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    Yeah, so I don't need to go the long-winded method... and just change the values in the equation right? :o

  49. Jhannybean
    • one year ago
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    |dw:1440840212338:dw| and yes. I'll start you off here by helping you find the area of the triangular face.

  50. kittiwitti1
    • one year ago
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    It's okay, you're tired. I can handle this one :) I was afraid I had to do the long method but this one I can do. ^^

  51. kittiwitti1
    • one year ago
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    But thank you :D

  52. Jhannybean
    • one year ago
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    Oh here's an easier way, if it helps at all. split the base of the triangle in half, and label it x. Then you have 2 x's. label the entire base y. So first you'll solve for x, and when you DOUBLE it, you'll find y :) Then you just use the y-value in the rest of your calculations.

  53. kittiwitti1
    • one year ago
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    What o-o

  54. Jhannybean
    • one year ago
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    So what I would do is... |dw:1440840514569:dw|

  55. kittiwitti1
    • one year ago
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    I think I did that subconsciously xD

  56. kittiwitti1
    • one year ago
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    I got 6/sqrt 3 then x2 = 12/sqrt3

  57. Jhannybean
    • one year ago
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    \[\tan(60) = \frac{5}{x} \iff x = \frac{5}{\tan(60)}\]\[y=2\left(\frac{5}{\tan(60)}\right)\]

  58. Jhannybean
    • one year ago
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    YEah, then you're good. no need for a long fancy explanation like that guy did.

  59. kittiwitti1
    • one year ago
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    What is the weird arrow you put there lol Is there a latex code?

  60. Jhannybean
    • one year ago
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    That means its related. I could write either or interchangeably.

  61. Jhannybean
    • one year ago
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    `\[\iff\]`

  62. kittiwitti1
    • one year ago
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    Oh. Is there a latex code x_x?|dw:1440840784961:dw|

  63. kittiwitti1
    • one year ago
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    Alright so far? (got the code thanks)

  64. Jhannybean
    • one year ago
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    Yep, you gt it.

  65. kittiwitti1
    • one year ago
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    Thank you! :D I can handle it from here. Sorry to make you stay up to help me ^^;

  66. Jhannybean
    • one year ago
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    just remember that the triangle is equilateral, therefore the base will be the same as the hypotenuse.

  67. Jhannybean
    • one year ago
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    No its fine :)

  68. Jhannybean
    • one year ago
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    Good luck~

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