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Falco276

  • 2 years ago

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.

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  1. Falco276
    • 2 years ago
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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.

  2. Falco276
    • 2 years ago
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  3. Falco276
    • 2 years ago
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    @GCR92

  4. GCR92
    • 2 years ago
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    ok let me see

  5. Falco276
    • 2 years ago
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    I think I know. The vertical angle measurement is equal to the horizontal measurement of the right angle.

  6. Falco276
    • 2 years ago
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    21 feet by 21 feet. I just have to find out the measurement to rope 2

  7. GCR92
    • 2 years ago
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    they are asking you the distance between rope 2 and underneath the balloon right?

  8. Falco276
    • 2 years ago
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    yes, exactly.

  9. GCR92
    • 2 years ago
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    just to make sure i understood well, i remind you i'm not english!! hehehe

  10. GCR92
    • 2 years ago
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    ok

  11. GCR92
    • 2 years ago
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    lemme get a paper so i can explain better

  12. Falco276
    • 2 years ago
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    |dw:1358712594988:dw|

  13. GCR92
    • 2 years ago
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    the distance between rope 2 and 1 is x right? as the drawing says

  14. Falco276
    • 2 years ago
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    yes, but I found the half of X which is 45.

  15. Falco276
    • 2 years ago
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    look at the drawing, am I right?

  16. Falco276
    • 2 years ago
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    I only have to fing the measurement that is the question mark.

  17. Falco276
    • 2 years ago
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    *find

  18. GCR92
    • 2 years ago
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    you have 2 angles if you notice

  19. GCR92
    • 2 years ago
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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

  20. Falco276
    • 2 years ago
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    i noticed that.

  21. GCR92
    • 2 years ago
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    apply pitagoras

  22. GCR92
    • 2 years ago
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    ah well you need to find the sides first, nevermind

  23. Falco276
    • 2 years ago
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    sides?

  24. GCR92
    • 2 years ago
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    you gonna solve the problem like this

  25. GCR92
    • 2 years ago
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    x: distance from rope 2 to underneath the balloon - distance from rope 1 to underneath the balloon

  26. Falco276
    • 2 years ago
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    distance from rope 1 to underneath the balloon = 45.

  27. Falco276
    • 2 years ago
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    Im trying to figure out rope 2

  28. GCR92
    • 2 years ago
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    how do you know its 45? 45 is the given angle

  29. Falco276
    • 2 years ago
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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.

  30. Falco276
    • 2 years ago
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    NO NO NO NO!!!!!

  31. GCR92
    • 2 years ago
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    Im solving it, wait

  32. Falco276
    • 2 years ago
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    I have it wrong!

  33. GCR92
    • 2 years ago
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    it has a lot to do with sinus and cosinus

  34. Falco276
    • 2 years ago
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    its 21!!!!!! Its 21!!!!

  35. GCR92
    • 2 years ago
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    well first you need to remember that sinus = op side / h

  36. GCR92
    • 2 years ago
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    we are measuring from rope 1 to balloon now, so sin 30 right?

  37. GCR92
    • 2 years ago
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    i tell you that sin 30 = 1/2

  38. GCR92
    • 2 years ago
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    so you gonna put 1/2 = 21ft / h right?

  39. GCR92
    • 2 years ago
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    solve for h and then you have the hypotenuse

  40. GCR92
    • 2 years ago
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    what do you think you can do next

  41. GCR92
    • 2 years ago
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    there are many ways to solve this problem

  42. Falco276
    • 2 years ago
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    um, okay.

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

  44. GCR92
    • 2 years ago
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    so if you have a 90º angle, a 30º angle...what's the other angle?

  45. Falco276
    • 2 years ago
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    wait, so were trying to find a measurement form rope 1 to the balloon, am I right?

  46. GCR92
    • 2 years ago
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    yep

  47. GCR92
    • 2 years ago
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    i got it already, just waiting to see how you proceed

  48. GCR92
    • 2 years ago
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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

  49. Falco276
    • 2 years ago
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    I never took trigonometery before.

  50. GCR92
    • 2 years ago
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    I know you haven't heard of the trigonometric functions

  51. GCR92
    • 2 years ago
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    you need to do it 2 steps

  52. Falco276
    • 2 years ago
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    no, never. I just take basic geometry.

  53. GCR92
    • 2 years ago
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    ok look

  54. GCR92
    • 2 years ago
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    |dw:1358713814089:dw|

  55. GCR92
    • 2 years ago
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    do you remember what was the sinus and what was the cosinus?

  56. Falco276
    • 2 years ago
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    (sigh) I don't remember. (honest, i don't want to be mean to you.)

  57. GCR92
    • 2 years ago
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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

  58. Falco276
    • 2 years ago
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    okay, I'll take your note.

  59. GCR92
    • 2 years ago
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    |dw:1358714056382:dw|

  60. GCR92
    • 2 years ago
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    Now you need to MEMORIZE this scheme, mathematics are also a subject to study!

  61. GCR92
    • 2 years ago
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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?

  62. Falco276
    • 2 years ago
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    ok.

  63. GCR92
    • 2 years ago
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    |dw:1358714498016:dw|

  64. Falco276
    • 2 years ago
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    I think the opposite side is left from angle A the adj side is right from Angle A.

  65. GCR92
    • 2 years ago
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    nope

  66. GCR92
    • 2 years ago
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    just read its name

  67. GCR92
    • 2 years ago
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    opposite = infront of the angle

  68. GCR92
    • 2 years ago
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    adjacent = sidewards the angle

  69. GCR92
    • 2 years ago
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    alongside of it

  70. Falco276
    • 2 years ago
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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?

  71. GCR92
    • 2 years ago
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    draw it

  72. Falco276
    • 2 years ago
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    |dw:1358715165594:dw|

  73. GCR92
    • 2 years ago
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    no, the other one, that's the hypotenuse

  74. Falco276
    • 2 years ago
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    |dw:1358715296994:dw|

  75. Falco276
    • 2 years ago
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    @hba

  76. GCR92
    • 2 years ago
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    yep

  77. hba
    • 2 years ago
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    Yes @Falco276

  78. Falco276
    • 2 years ago
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    @hba please help me!

  79. hba
    • 2 years ago
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    @GCR92 is helping you,Just follow him :)

  80. GCR92
    • 2 years ago
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    I have to explain the whole trigonometry before you are able to solve that problem...

  81. Falco276
    • 2 years ago
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    (sigh) ok.

  82. GCR92
    • 2 years ago
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    I explain as simple as it is, but not simpler

  83. Falco276
    • 2 years ago
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    ok

  84. GCR92
    • 2 years ago
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    ok i put the turbo now

  85. GCR92
    • 2 years ago
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    we are going to solve 2 triangles

  86. GCR92
    • 2 years ago
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    |dw:1358715724015:dw|

  87. Falco276
    • 2 years ago
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    wait. I think I have it.

  88. GCR92
    • 2 years ago
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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

  89. Falco276
    • 2 years ago
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    For the 45° triangle

  90. GCR92
    • 2 years ago
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    and sinus = op side / h

  91. GCR92
    • 2 years ago
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    and we can make a relationship

  92. GCR92
    • 2 years ago
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    an equation

  93. Falco276
    • 2 years ago
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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

  94. Falco276
    • 2 years ago
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    is this correct?

  95. GCR92
    • 2 years ago
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    where did you get 15?

  96. Falco276
    • 2 years ago
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    oh, no i meant 21 (facepalm)

  97. GCR92
    • 2 years ago
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    You could've done it if you had the value of the x side

  98. GCR92
    • 2 years ago
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    but that's what we are looking for

  99. GCR92
    • 2 years ago
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    to use the tangent you need to have the value of 2 sides

  100. GCR92
    • 2 years ago
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    I'd like to continue now

  101. Falco276
    • 2 years ago
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    |dw:1358716138389:dw|

  102. GCR92
    • 2 years ago
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    why?

  103. Falco276
    • 2 years ago
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    |dw:1358716271703:dw|

  104. Falco276
    • 2 years ago
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    |dw:1358716355941:dw|

  105. Falco276
    • 2 years ago
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    now we take the other triangle.

  106. GCR92
    • 2 years ago
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    the adjacent side is not 21, 21 is the only data they give us, you can't invent it

  107. GCR92
    • 2 years ago
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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

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

  109. GCR92
    • 2 years ago
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    1/2 = 21 / h

  110. GCR92
    • 2 years ago
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    h= 42

  111. GCR92
    • 2 years ago
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    now we take the cosinus in order to calculate X

  112. GCR92
    • 2 years ago
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    cos 30 = sqrt 3 / 2

  113. GCR92
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    sqrt 3 / 2 = 42 / x

  114. GCR92
    • 2 years ago
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    x = 36.373

  115. Falco276
    • 2 years ago
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    |dw:1358716402610:dw|

  116. Falco276
    • 2 years ago
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    |dw:1358717381175:dw|

  117. Falco276
    • 2 years ago
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    |dw:1358717481086:dw|

  118. GCR92
    • 2 years ago
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    you can never write tan and let it alone

  119. GCR92
    • 2 years ago
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    it is always tan OF SOMETHING

  120. GCR92
    • 2 years ago
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    tan of x, tan of alfa, tan of 20

  121. Falco276
    • 2 years ago
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    so the answer is 21 sq. root 3

  122. GCR92
    • 2 years ago
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    ill solve it

  123. GCR92
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    2 triangles

  124. GCR92
    • 2 years ago
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    take the 30º triangle and then, sin 30 = 1/2

  125. GCR92
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    1/2 = 21 / h, h = 42

  126. GCR92
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    now we have the h of the first triangle

  127. GCR92
    • 2 years ago
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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

  128. GCR92
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    cos 30 = sqrt 3 / 2

  129. GCR92
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    sqrt 3 /2 = x/42

  130. GCR92
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    x = 36,373

  131. GCR92
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    now we go to the second triangle and apply exactly the same steps but with 45º

  132. Falco276
    • 2 years ago
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    maybe my way of solving this problem was wrong. sorry.

  133. Falco276
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    i tried my best.

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

  135. GCR92
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    if you do the same with the second triangle

  136. GCR92
    • 2 years ago
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    sin 45 = sqrt 2 / 2

  137. GCR92
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    sqrt 2 / 2 = 21 / h

  138. GCR92
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    h = 29.698

  139. GCR92
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    now we calculate the cosinus

  140. GCR92
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    cos 45 = sqrt 2 / 2

  141. GCR92
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    sqrt 2 / 2 = x / 28,698

  142. GCR92
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    x = 20,292

  143. GCR92
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    distance from underneath the ballon to the second rope = 36, 373 + 20,292 = 56,665 feet

  144. GCR92
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    and it's solved

  145. Falco276
    • 2 years ago
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    wait, but earlier you said that x is 36- never mind :P

  146. GCR92
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    that's the distance between underneath the balloon and rope 2

  147. Falco276
    • 2 years ago
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    ah, i see.

  148. GCR92
    • 2 years ago
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    and 20,292 ft is the distance between underneath the balloon and rope 1

  149. GCR92
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    sorry

  150. Falco276
    • 2 years ago
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    got it.

  151. GCR92
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    36ft is the distance between rope 2 and 1

  152. GCR92
    • 2 years ago
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    then you calculate from rope 1 to underneath the ballon and it gives you 20 ft

  153. GCR92
    • 2 years ago
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    you add it and you obtain the final result

  154. Falco276
    • 2 years ago
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    okay.

  155. GCR92
    • 2 years ago
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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!

  156. GCR92
    • 2 years ago
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    needed*

  157. GCR92
    • 2 years ago
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    you did well! at least you had ideas! you only need to learn the theory well

  158. GCR92
    • 2 years ago
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    i also think your teachers are crazy though

  159. Falco276
    • 2 years ago
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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.

  160. GCR92
    • 2 years ago
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    You are welcome, sure, see you! :D

  161. Falco276
    • 2 years ago
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    thanks! (thumbs up)

  162. GCR92
    • 2 years ago
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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

  163. Falco276
    • 2 years ago
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    oh, i understand, dont worry! my teachers are really nice.

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