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Falco276 Group Title

There are many ropes keeping a hot air balloon from floating away before a balloon race. One of these ropes is fixed to the ground at a 45° angle. Another is fixed to the ground at a 30° angle.

  • one year ago
  • one year ago

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  1. Falco276 Group Title
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    If the hot air balloon is 21 feet off the ground, what is the distance between the ground directly underneath the balloon and the second rope rounded to the nearest hundredth of a foot? Type the numeric answer only in the box below.

    • one year ago
  2. Falco276 Group Title
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    • one year ago
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  3. Falco276 Group Title
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    @GCR92

    • one year ago
  4. GCR92 Group Title
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    ok let me see

    • one year ago
  5. Falco276 Group Title
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    I think I know. The vertical angle measurement is equal to the horizontal measurement of the right angle.

    • one year ago
  6. Falco276 Group Title
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    21 feet by 21 feet. I just have to find out the measurement to rope 2

    • one year ago
  7. GCR92 Group Title
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    they are asking you the distance between rope 2 and underneath the balloon right?

    • one year ago
  8. Falco276 Group Title
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    yes, exactly.

    • one year ago
  9. GCR92 Group Title
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    just to make sure i understood well, i remind you i'm not english!! hehehe

    • one year ago
  10. GCR92 Group Title
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    ok

    • one year ago
  11. GCR92 Group Title
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    lemme get a paper so i can explain better

    • one year ago
  12. Falco276 Group Title
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    |dw:1358712594988:dw|

    • one year ago
  13. GCR92 Group Title
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    the distance between rope 2 and 1 is x right? as the drawing says

    • one year ago
  14. Falco276 Group Title
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    yes, but I found the half of X which is 45.

    • one year ago
  15. Falco276 Group Title
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    look at the drawing, am I right?

    • one year ago
  16. Falco276 Group Title
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    I only have to fing the measurement that is the question mark.

    • one year ago
  17. Falco276 Group Title
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    *find

    • one year ago
  18. GCR92 Group Title
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    you have 2 angles if you notice

    • one year ago
  19. GCR92 Group Title
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    even though they dont give you a number, you can see that the angle underneath the balloon is 90, a right triangle

    • one year ago
  20. Falco276 Group Title
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    i noticed that.

    • one year ago
  21. GCR92 Group Title
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    apply pitagoras

    • one year ago
  22. GCR92 Group Title
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    ah well you need to find the sides first, nevermind

    • one year ago
  23. Falco276 Group Title
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    sides?

    • one year ago
  24. GCR92 Group Title
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    you gonna solve the problem like this

    • one year ago
  25. GCR92 Group Title
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    x: distance from rope 2 to underneath the balloon - distance from rope 1 to underneath the balloon

    • one year ago
  26. Falco276 Group Title
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    distance from rope 1 to underneath the balloon = 45.

    • one year ago
  27. Falco276 Group Title
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    Im trying to figure out rope 2

    • one year ago
  28. GCR92 Group Title
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    how do you know its 45? 45 is the given angle

    • one year ago
  29. Falco276 Group Title
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    no, not from the angle. On how I got 45 was that the vertical measurement leading to the balloon has the same measurement that leads to rope 1.

    • one year ago
  30. Falco276 Group Title
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    NO NO NO NO!!!!!

    • one year ago
  31. GCR92 Group Title
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    Im solving it, wait

    • one year ago
  32. Falco276 Group Title
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    I have it wrong!

    • one year ago
  33. GCR92 Group Title
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    it has a lot to do with sinus and cosinus

    • one year ago
  34. Falco276 Group Title
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    its 21!!!!!! Its 21!!!!

    • one year ago
  35. GCR92 Group Title
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    well first you need to remember that sinus = op side / h

    • one year ago
  36. GCR92 Group Title
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    we are measuring from rope 1 to balloon now, so sin 30 right?

    • one year ago
  37. GCR92 Group Title
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    i tell you that sin 30 = 1/2

    • one year ago
  38. GCR92 Group Title
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    so you gonna put 1/2 = 21ft / h right?

    • one year ago
  39. GCR92 Group Title
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    solve for h and then you have the hypotenuse

    • one year ago
  40. GCR92 Group Title
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    what do you think you can do next

    • one year ago
  41. GCR92 Group Title
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    there are many ways to solve this problem

    • one year ago
  42. Falco276 Group Title
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    um, okay.

    • one year ago
  43. GCR92 Group Title
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    also remember that all the angles of a triangle = 180, if this doesn't happen, it's not a triangle

    • one year ago
  44. GCR92 Group Title
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    so if you have a 90º angle, a 30º angle...what's the other angle?

    • one year ago
  45. Falco276 Group Title
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    wait, so were trying to find a measurement form rope 1 to the balloon, am I right?

    • one year ago
  46. GCR92 Group Title
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    yep

    • one year ago
  47. GCR92 Group Title
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    i got it already, just waiting to see how you proceed

    • one year ago
  48. GCR92 Group Title
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    that problem is too difficult for you, i mean, they should teach you trigonommetry first, there's no way you can do that without using trigonommetry

    • one year ago
  49. Falco276 Group Title
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    I never took trigonometery before.

    • one year ago
  50. GCR92 Group Title
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    I know you haven't heard of the trigonometric functions

    • one year ago
  51. GCR92 Group Title
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    you need to do it 2 steps

    • one year ago
  52. Falco276 Group Title
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    no, never. I just take basic geometry.

    • one year ago
  53. GCR92 Group Title
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    ok look

    • one year ago
  54. GCR92 Group Title
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    |dw:1358713814089:dw|

    • one year ago
  55. GCR92 Group Title
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    do you remember what was the sinus and what was the cosinus?

    • one year ago
  56. Falco276 Group Title
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    (sigh) I don't remember. (honest, i don't want to be mean to you.)

    • one year ago
  57. GCR92 Group Title
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    hahaha, dont worry about it, i'll explain again, but you should take note of it because you are going to need it in EVERY problem

    • one year ago
  58. Falco276 Group Title
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    okay, I'll take your note.

    • one year ago
  59. GCR92 Group Title
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    |dw:1358714056382:dw|

    • one year ago
  60. GCR92 Group Title
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    Now you need to MEMORIZE this scheme, mathematics are also a subject to study!

    • one year ago
  61. GCR92 Group Title
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    now ill draw you a triangle and you'll tell me which is the oppsite and which is the adjacent side ok?

    • one year ago
  62. Falco276 Group Title
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    ok.

    • one year ago
  63. GCR92 Group Title
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    |dw:1358714498016:dw|

    • one year ago
  64. Falco276 Group Title
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    I think the opposite side is left from angle A the adj side is right from Angle A.

    • one year ago
  65. GCR92 Group Title
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    nope

    • one year ago
  66. GCR92 Group Title
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    just read its name

    • one year ago
  67. GCR92 Group Title
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    opposite = infront of the angle

    • one year ago
  68. GCR92 Group Title
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    adjacent = sidewards the angle

    • one year ago
  69. GCR92 Group Title
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    alongside of it

    • one year ago
  70. Falco276 Group Title
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    fine! Isn't the adjecent side the- How do I Explain it! Isnt the adjecent side the sidewards of the angle?

    • one year ago
  71. GCR92 Group Title
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    draw it

    • one year ago
  72. Falco276 Group Title
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    |dw:1358715165594:dw|

    • one year ago
  73. GCR92 Group Title
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    no, the other one, that's the hypotenuse

    • one year ago
  74. Falco276 Group Title
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    |dw:1358715296994:dw|

    • one year ago
  75. Falco276 Group Title
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    @hba

    • one year ago
  76. GCR92 Group Title
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    yep

    • one year ago
  77. hba Group Title
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    Yes @Falco276

    • one year ago
  78. Falco276 Group Title
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    @hba please help me!

    • one year ago
  79. hba Group Title
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    @GCR92 is helping you,Just follow him :)

    • one year ago
  80. GCR92 Group Title
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    I have to explain the whole trigonometry before you are able to solve that problem...

    • one year ago
  81. Falco276 Group Title
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    (sigh) ok.

    • one year ago
  82. GCR92 Group Title
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    I explain as simple as it is, but not simpler

    • one year ago
  83. Falco276 Group Title
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    ok

    • one year ago
  84. GCR92 Group Title
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    ok i put the turbo now

    • one year ago
  85. GCR92 Group Title
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    we are going to solve 2 triangles

    • one year ago
  86. GCR92 Group Title
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    |dw:1358715724015:dw|

    • one year ago
  87. Falco276 Group Title
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    wait. I think I have it.

    • one year ago
  88. GCR92 Group Title
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    we are going to use the sinus to solve this one, why sinus? because the problem has given us the opposite side, wich is 21 ft

    • one year ago
  89. Falco276 Group Title
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    For the 45° triangle

    • one year ago
  90. GCR92 Group Title
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    and sinus = op side / h

    • one year ago
  91. GCR92 Group Title
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    and we can make a relationship

    • one year ago
  92. GCR92 Group Title
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    an equation

    • one year ago
  93. Falco276 Group Title
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    tan Ө = opposite / adjacent tan Ө = height of the balloon off the ground / x tan 45° = 15 / x x = 15 / tan 45° x = 15 / 1 x = 15

    • one year ago
  94. Falco276 Group Title
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    is this correct?

    • one year ago
  95. GCR92 Group Title
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    where did you get 15?

    • one year ago
  96. Falco276 Group Title
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    oh, no i meant 21 (facepalm)

    • one year ago
  97. GCR92 Group Title
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    You could've done it if you had the value of the x side

    • one year ago
  98. GCR92 Group Title
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    but that's what we are looking for

    • one year ago
  99. GCR92 Group Title
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    to use the tangent you need to have the value of 2 sides

    • one year ago
  100. GCR92 Group Title
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    I'd like to continue now

    • one year ago
  101. Falco276 Group Title
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    |dw:1358716138389:dw|

    • one year ago
  102. GCR92 Group Title
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    why?

    • one year ago
  103. Falco276 Group Title
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    |dw:1358716271703:dw|

    • one year ago
  104. Falco276 Group Title
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    |dw:1358716355941:dw|

    • one year ago
  105. Falco276 Group Title
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    now we take the other triangle.

    • one year ago
  106. GCR92 Group Title
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    the adjacent side is not 21, 21 is the only data they give us, you can't invent it

    • one year ago
  107. GCR92 Group Title
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    listen i dont want to be mean either but i want to help you and i want to finish, if you think you don't need to do what i tell you to solve it, go ahead

    • one year ago
  108. GCR92 Group Title
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    sin 30 = 1/2, we do that in order to calculate H

    • one year ago
  109. GCR92 Group Title
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    1/2 = 21 / h

    • one year ago
  110. GCR92 Group Title
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    h= 42

    • one year ago
  111. GCR92 Group Title
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    now we take the cosinus in order to calculate X

    • one year ago
  112. GCR92 Group Title
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    cos 30 = sqrt 3 / 2

    • one year ago
  113. GCR92 Group Title
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    sqrt 3 / 2 = 42 / x

    • one year ago
  114. GCR92 Group Title
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    x = 36.373

    • one year ago
  115. Falco276 Group Title
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    |dw:1358716402610:dw|

    • one year ago
  116. Falco276 Group Title
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    |dw:1358717381175:dw|

    • one year ago
  117. Falco276 Group Title
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    |dw:1358717481086:dw|

    • one year ago
  118. GCR92 Group Title
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    you can never write tan and let it alone

    • one year ago
  119. GCR92 Group Title
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    it is always tan OF SOMETHING

    • one year ago
  120. GCR92 Group Title
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    tan of x, tan of alfa, tan of 20

    • one year ago
  121. Falco276 Group Title
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    so the answer is 21 sq. root 3

    • one year ago
  122. GCR92 Group Title
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    ill solve it

    • one year ago
  123. GCR92 Group Title
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    2 triangles

    • one year ago
  124. GCR92 Group Title
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    take the 30º triangle and then, sin 30 = 1/2

    • one year ago
  125. GCR92 Group Title
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    1/2 = 21 / h, h = 42

    • one year ago
  126. GCR92 Group Title
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    now we have the h of the first triangle

    • one year ago
  127. GCR92 Group Title
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    now we can calculate the distance from the rope to underneath the balloon, that distance will be given by the cosinus

    • one year ago
  128. GCR92 Group Title
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    cos 30 = sqrt 3 / 2

    • one year ago
  129. GCR92 Group Title
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    sqrt 3 /2 = x/42

    • one year ago
  130. GCR92 Group Title
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    x = 36,373

    • one year ago
  131. GCR92 Group Title
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    now we go to the second triangle and apply exactly the same steps but with 45º

    • one year ago
  132. Falco276 Group Title
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    maybe my way of solving this problem was wrong. sorry.

    • one year ago
  133. Falco276 Group Title
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    i tried my best.

    • one year ago
  134. GCR92 Group Title
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    I know but mathematics doesn't magically happen, you need to study a lot( and suffer lol) dont worry

    • one year ago
  135. GCR92 Group Title
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    if you do the same with the second triangle

    • one year ago
  136. GCR92 Group Title
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    sin 45 = sqrt 2 / 2

    • one year ago
  137. GCR92 Group Title
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    sqrt 2 / 2 = 21 / h

    • one year ago
  138. GCR92 Group Title
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    h = 29.698

    • one year ago
  139. GCR92 Group Title
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    now we calculate the cosinus

    • one year ago
  140. GCR92 Group Title
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    cos 45 = sqrt 2 / 2

    • one year ago
  141. GCR92 Group Title
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    sqrt 2 / 2 = x / 28,698

    • one year ago
  142. GCR92 Group Title
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    x = 20,292

    • one year ago
  143. GCR92 Group Title
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    distance from underneath the ballon to the second rope = 36, 373 + 20,292 = 56,665 feet

    • one year ago
  144. GCR92 Group Title
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    and it's solved

    • one year ago
  145. Falco276 Group Title
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    wait, but earlier you said that x is 36- never mind :P

    • one year ago
  146. GCR92 Group Title
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    that's the distance between underneath the balloon and rope 2

    • one year ago
  147. Falco276 Group Title
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    ah, i see.

    • one year ago
  148. GCR92 Group Title
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    and 20,292 ft is the distance between underneath the balloon and rope 1

    • one year ago
  149. GCR92 Group Title
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    sorry

    • one year ago
  150. Falco276 Group Title
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    got it.

    • one year ago
  151. GCR92 Group Title
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    36ft is the distance between rope 2 and 1

    • one year ago
  152. GCR92 Group Title
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    then you calculate from rope 1 to underneath the ballon and it gives you 20 ft

    • one year ago
  153. GCR92 Group Title
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    you add it and you obtain the final result

    • one year ago
  154. Falco276 Group Title
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    okay.

    • one year ago
  155. GCR92 Group Title
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    what i did is totally useless, i mean you can't learn in 20 min, i always believed that a teacher is need in order to learn something, anyway, wish you luck with this!

    • one year ago
  156. GCR92 Group Title
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    needed*

    • one year ago
  157. GCR92 Group Title
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    you did well! at least you had ideas! you only need to learn the theory well

    • one year ago
  158. GCR92 Group Title
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    i also think your teachers are crazy though

    • one year ago
  159. Falco276 Group Title
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    XD Thanks Gonzalo! I promise to keep the theory at heart. Yes, I finally understood the answer and also the work. I'll tell you if need help on anything else.

    • one year ago
  160. GCR92 Group Title
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    You are welcome, sure, see you! :D

    • one year ago
  161. Falco276 Group Title
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    thanks! (thumbs up)

    • one year ago
  162. GCR92 Group Title
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    I also can be wrong, i mean if your teachers yells at you it is my fault, just don't take that in account, it's been ages since i dont take trigonometry :p

    • one year ago
  163. Falco276 Group Title
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    oh, i understand, dont worry! my teachers are really nice.

    • one year ago
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