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anonymous

  • one year ago

Max observes the zoo and the library from a helicopter flying at a height of 400 times square root of 3 feet above the ground, as shown below: Here are the choices: A) 800 Feet B) 600 Feet C) 200 Feet D) 400 Feet The picture for this problem will be shown below:

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  1. anonymous
    • one year ago
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  2. ganeshie8
    • one year ago
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    what are you trying to find here ?

  3. anonymous
    • one year ago
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    Oh, sorry, I forgot to put that in haha. What is the distance between the zoo and the library?

  4. anonymous
    • one year ago
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    Hello? @ganeshie8

  5. anonymous
    • one year ago
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    |dw:1441031827999:dw|

  6. anonymous
    • one year ago
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    \[ZL=GL-GZ\] Use \[\triangle GHZ\] to find GZ and \[\triangle GHL\] to find GL

  7. anonymous
    • one year ago
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    the tangent function will be your help here

  8. anonymous
    • one year ago
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    Okay, I'm understanding so far. How would I write that formula out though? Well, not just tell me but, explain how I'd work it out.

  9. anonymous
    • one year ago
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    In \[\triangle GHZ\] \[\tan(60)=\frac{GH}{GZ}\] see the point now?

  10. anonymous
    • one year ago
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    Okay, so basically there are 2 problems here? But yes, I'm understanding so far since those are the opposite and adjacent sides.

  11. anonymous
    • one year ago
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    In a sense, yes, you will have to use 2 separate triangles to find 2 values

  12. anonymous
    • one year ago
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    for you reference, I shall provide the values of tan30 and tan60 \[\tan(30)=\frac{1}{\sqrt{3}}\] \[\tan(60)=\sqrt{3}\]

  13. anonymous
    • one year ago
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    Why is tan(30) = 1/3? and tan(60) = 3?

  14. anonymous
    • one year ago
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    Well, it's a function, functions take in variables (angles in this case) and return some value depending on the value of the variable and it's 1/sqrt(3) and sqrt(3) For example consider \[x^2+7x\] What you are asking is something like why will x=1 give us the answer as 8?

  15. anonymous
    • one year ago
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    One moment please. :)

  16. anonymous
    • one year ago
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    Sure

  17. anonymous
    • one year ago
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    Okay, I'm back. I'm still a bit confused. I understand functions need variables. But are we using sqrt because of the 400sqrt{3} ??

  18. anonymous
    • one year ago
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    Ok let's take my same example \[x^2+7x\] When x=1, we get answer as 8 Similarly consider \[\tan(x)\] when \[x=60\] degrees we get the answer as \[\sqrt{3}\] It's one of those things you need to remember, I can't help much if you ask "why" is this so?

  19. anonymous
    • one year ago
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    Infact you are given the height of the tower as PRECISELY as 400sqrt{3} so that your calculations would be easier

  20. anonymous
    • one year ago
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    sorry height of helicopter

  21. anonymous
    • one year ago
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    Ok, now I understand what you're trying to stay, just for the sake of curiosity, let me tell you that tan(x) can represented as an infinite series of x, \[\tan(x)=x+\frac{2x^3}{3!}+\frac{16x^5}{5!}+\frac{272x^7}{7!}+....\] This is mclaurin series for tan(x), basically mclaurin series are a way of representing functions as an infinite polynomial where x is in radians let's convert 60 to radians \[60 \times \frac{\pi}{180}=\frac{\pi}{3}\] Now when you put x=pi/3, for each term you add, your value will get closer and closer to \[\sqrt{3}\] Now don't ask where I pulled that series out from, it's totally out of bounds fornow, I did this for the sake of visualization, and don't go around wasting time putting pi/3 in all those terms, you will get the answer as sqrt{3} the ! means \[3!=3 \times 2 \times 1\] \[5!=5 \times 4 \times 3 \times 2 \times 1\] It's a way of writing product from that number to 1

  22. anonymous
    • one year ago
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    I'm sorry if I can't find an easier way to tell you, but for now just remember that \[\tan(60)=\sqrt{3}\] and \[\tan(30)=\frac{1}{\sqrt{3}}\]

  23. anonymous
    • one year ago
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    It's okay, I don't want you to try and help me understand for too long. I'll ask my teacher about it as well and look over my notes again, but let's just continue.

  24. anonymous
    • one year ago
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    You don't need to worry about asking my help for long, If I was busy I wouldn't come to the website in the first place, right?

  25. anonymous
    • one year ago
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    Let's go back to triangle GHZ

  26. anonymous
    • one year ago
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    \[\tan(60)=\frac{400\sqrt{3}}{GZ}\] From this we can find GZ \[\sqrt{3}=\frac{400\sqrt{3}}{GZ}\] Dividing by root 3 and multiplying by GZ throughout we get \[\frac{GZ}{\sqrt{3}} \times \sqrt{3}=\frac{GZ}{\sqrt{3}} \times \frac{400\sqrt{3}}{GZ}\] root 3 cancels on left side, GZ and root 3 cancel on right side we get \[GZ=400\] Similarly find GL using triangle GHL and tan 30

  27. anonymous
    • one year ago
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    Are you understanding?

  28. anonymous
    • one year ago
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    Yes, I'm understanding. You're finding the distances right?

  29. anonymous
    • one year ago
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    Yep, and we're gonna subtract them for ZL, we're indirectly finding this distance ZL

  30. anonymous
    • one year ago
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    Can you find GL for me now?

  31. anonymous
    • one year ago
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    Oh okay, so you're putting GH and GZ together to find ZL so we know the distance from the two?

  32. anonymous
    • one year ago
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    \[GL-GZ=ZL\] Look at the figure, if you remove the length GZ from GL, you're left with ZL

  33. anonymous
    • one year ago
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    Right, like you said before we'd subtract the two for ZL.

  34. anonymous
    • one year ago
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    I'll try figuring out GL now.

  35. anonymous
    • one year ago
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    \[\tan (30) = \frac{ 400\sqrt{3} }{ GL }\] \[\sqrt{3}= \frac{ 400\sqrt{3} }{ GL}\] Is this right so far? Because I remember you saying that tan(30) = 1/sqrt3 ??

  36. anonymous
    • one year ago
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    \[\tan(30)=\frac{1}{\sqrt{3}}\] But you've written \[\sqrt{3}\] Slight error there

  37. anonymous
    • one year ago
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    \[\frac{1}{\sqrt{3}}=\frac{400\sqrt{3}}{GL}\]

  38. anonymous
    • one year ago
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    How would you write the next step then if I'd divide by root?

  39. anonymous
    • one year ago
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    Multiply by sqrt{3} throughout

  40. anonymous
    • one year ago
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    Oh okay, just a second. I'll continue working out the problem.

  41. anonymous
    • one year ago
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    \[\frac{ GL }{ \sqrt{3} } \times \sqrt{3}= \frac{ GL }{ \sqrt{3}} \times \frac{ 400\sqrt{3} }{ GL }\] Root 3 cancles on the left side, GL and root 3 cancel on right side. So far is this good?

  42. anonymous
    • one year ago
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    Your right side doesn't look good, we multiplied by root 3 throughout, so how are you getting a root 3 on the denominator on the right side?

  43. anonymous
    • one year ago
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    Wouldn't you do that for both?

  44. anonymous
    • one year ago
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    \[\frac{1}{\sqrt{3}}=\frac{400\sqrt{3}}{GL}\] If you multiply by GL and root 3 throughout, \[\frac{1}{\sqrt{3}}\times GL \times \sqrt{3}=\frac{400\sqrt{3}}{GL} \times GL \times \sqrt{3}\]

  45. anonymous
    • one year ago
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    So it changes because \[\tan \left( 30 \right) = \frac{ 1 }{ \sqrt{3} } \] ?

  46. anonymous
    • one year ago
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    What changes?

  47. anonymous
    • one year ago
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    The equation since for the other one it was \[\tan \left( 60 \right)= \sqrt{3}\] Wouldn't the answer and equation become different?

  48. anonymous
    • one year ago
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    Of course, both the equations from triangle 1 and triangle 2 are different

  49. anonymous
    • one year ago
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    Okay.. I'm lost now. I'm reviewing the module where this should've been taught to me, I'm not finding it anywhere which is why I'm very confused since my teacher never broke this down for me in the first place. So, the equations would be different, then how come GL's equation isn't the same as GZ, I know they're supposed to be different but in what way?

  50. anonymous
    • one year ago
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    Because the angles are different, what we are doing is just applying trigonometry to both the triangles separately, it does not necessarily mean that both will give same equations \[\sqrt{3}=\frac{400\sqrt{3}}{GZ}\] and \[\frac{1}{\sqrt{3}}=\frac{400\sqrt{3}}{GL}\] Think of it as 2 different triangles |dw:1441037237930:dw|

  51. anonymous
    • one year ago
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    Okay, and couldn't you simplify \[400\sqrt{3}\] ?

  52. anonymous
    • one year ago
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    You cannot further simplify \[400\sqrt{3}\] However you can approximate(that is not required here) \[400\sqrt{3}\approx400 \times 1.732\]

  53. anonymous
    • one year ago
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    Okay, so this is what we'd have: \[\frac{ 1 }{ \sqrt{3}} \times GL \times \sqrt{3} = \frac{ 400\sqrt{3} }{ GL } \times GL \times \sqrt{3} \] Right? What I'm confused about it how we'd get our answer from there?

  54. anonymous
    • one year ago
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    For the first problem for GZ, you never told me how you broke down the equation and got your final answer, so that's what I'm stuck on mainly.

  55. anonymous
    • one year ago
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    ok I'll start with GZ \[\sqrt{3}=\frac{400\sqrt{3}}{GZ}\] Let's take it one at a time Divide by root 3 we get \[\frac{\sqrt{3}}{\sqrt{3}}=\frac{400\sqrt{3}}{GZ \sqrt{3}}\] This becomes \[1=\frac{400}{GZ}\] Now we multiply by GZ \[GZ=\frac{400}{GZ} \times GZ\] \[GZ=400\]

  56. anonymous
    • one year ago
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    Oh, okay, so GZ would stay 400 since GZ isn't a number in the first place?

  57. anonymous
    • one year ago
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    It's like your x, I've used GZ you can also do like \[\sqrt{3}=\frac{400\sqrt{3}}{x}\] It's an unknown value, but the rules of maths still apply, a value known or unknown to us, so if we multiply \[\frac{1}{GZ}\] by \[GZ\] they cancel out and leave us with 1 just like any other number

  58. anonymous
    • one year ago
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    \[\frac{1}{5} \times 5=1\] \[\frac{1}{GZ} \times GZ=1\]

  59. anonymous
    • one year ago
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    It's a variable

  60. anonymous
    • one year ago
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    Oh, stupid me. I understand that now. So for GL, how would that problem get broken down? The equation is different so it wouldn't be the exact steps that happened for GZ.

  61. anonymous
    • one year ago
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    Hm let's have a look \[\frac{1}{\sqrt{3}}=\frac{400\sqrt{3}}{GL}\] Let's multiply by root 3 \[\frac{\sqrt{3}}{\sqrt{3}}=\frac{400\sqrt{3}\sqrt{3}}{GL}\] Now we know that \[\sqrt{a} \times \sqrt{b}=\sqrt{ab}\] So on right we have \[\frac{\sqrt{3}}{\sqrt{3}}=\frac{400\sqrt{3 \times 3}}{GL}\] Left side cancels out because same quantity \[1=\frac{400\sqrt{9}}{GL}\] But we know that square root of 9 is 3 \[1=\frac{400 \times 3}{GL}\] \[1=\frac{1200}{GL}\] \[1 \times GL=\frac{1200}{GL} \times GL\]\[GL=1200\]

  62. anonymous
    • one year ago
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    technically it's not a variable, it's just some unknown number, a variable is a quantity that changes, but as you saw GZ has only one value and that is 400

  63. anonymous
    • one year ago
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    Oh okay! I'm noticing now the whole problem. It seems quite easy now that I see it broken down. So then you'd subtract the two and get 800?

  64. anonymous
    • one year ago
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    Yep

  65. anonymous
    • one year ago
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    Okay, I might leave this up for a bit to take notes off of but I'll be sure to close it when I'm finished. Once again, thank you for helping me throughout this problem for such a long time! I REALLY appreciate it. :)

  66. anonymous
    • one year ago
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    @ScienceAndMath Try this question: |dw:1441039105610:dw| In the triangle ABC above, Find the value of x for which sin(x) is equal to cos(x)

  67. anonymous
    • one year ago
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    Wouldn't you need to use a Trigonometry Function, but there are two: sin and cos..?

  68. anonymous
    • one year ago
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    Yep, try to find sine and cosine and put them equal to each other, see where that can get you

  69. anonymous
    • one year ago
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    \[\sin = \frac{ Opposite }{ Hypotenuse }= \cos = \frac{ Adjacent }{ Hypotenuse }\] \[\sin(x) = \frac{ a }{ c } = \cos(x) = \frac{ b }{ c }\]

  70. anonymous
    • one year ago
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    Good! Now we want to find x for which sin and cosine are equal, so in a sense what we are saying is given: sin(x)=cos(x) find x So if we equate them, we get \[\frac{a}{c}=\frac{b}{c}\] Anything you feel you can cancel out in this equation?

  71. anonymous
    • one year ago
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    We can cancel out the denominators since they are the same? Which leaves you with a = b.

  72. anonymous
    • one year ago
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    Good!! Now if in that triangle (look at the figure above), a and b are equal, does that tell you anything about the triangle?

  73. anonymous
    • one year ago
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    Congruent sides?

  74. anonymous
    • one year ago
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    Yep, what is a triangle with 2 equal sides known as?

  75. anonymous
    • one year ago
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    Isosceles triangle. It'd have two equal sides and two equal angles. But an isosceles triangle is shaped differently..

  76. anonymous
    • one year ago
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    Yep, the angles opposite to the sides are equal, |dw:1441040422979:dw|

  77. anonymous
    • one year ago
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    Wouldn't the property be: \[m<A + m<B + m<C = 180\] ??

  78. anonymous
    • one year ago
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    Yep, sum of all 3 angles in a triangle is always 180 degrees

  79. anonymous
    • one year ago
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    so we get \[x+x+90=180\] The two angles are equal so the 2nd angle is infact also x

  80. anonymous
    • one year ago
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    Oh okay, so would we add or subtract from the problem?

  81. anonymous
    • one year ago
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    Let's club the x's together first \[2x+90=180\] Does that makes sense?

  82. anonymous
    • one year ago
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    Oh, yes it does lol. So I think afterwards you'd subtract subtract 90 from 180 which gives you 90 and then you take 2x and divide that into 90 which is = 45

  83. anonymous
    • one year ago
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    Yes!! So what does this tells us?? \[\sin(45)=\cos(45)?\] Is this true?Yep!! \[\sin(45)=\cos(45)=\frac{1}{\sqrt{2}}\] I'll make a table of common trigonometric values for you in degrees, make sure to remember it well |dw:1441041112353:dw|

  84. anonymous
    • one year ago
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    Okay, I wrote down the chart. I think I'm understanding it more now. I remember in the last problem we used the tan function and the sqrt3 and the 1/sqrt3 since the degrees were 30 and 60. So for Sin and Cos, since they're 45 degrees we'd use 1/sqrt2?

  85. anonymous
    • one year ago
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    Yep, If you are ever asked sine of 45 degrees, you'd say it's 1/sqrt(2)

  86. anonymous
    • one year ago
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    You can also verify tangent for each value, we know that \[\tan(x)=\frac{\sin(x)}{\cos(x)}\] Let's look at 60 degrees \[\tan(60)=\frac{\sin(60)}{\cos(60)}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\sqrt{3}}{2}\times 2=\sqrt{3}\]

  87. anonymous
    • one year ago
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    Actually, tan 90 would be "not defined" but the closer and closer you get to 90 degrees, like 89.9999 the closer and closer tan gets to infinity

  88. anonymous
    • one year ago
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    Oh okay, I noticed that thank you haha. Could you possibly help me on a new problem?

  89. anonymous
    • one year ago
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    Sure

  90. anonymous
    • one year ago
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    Oh, I actually think this problem is similar to the last one lol.

  91. anonymous
    • one year ago
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    Two ropes, AD and BD, are tied to a peg on the ground at point D. The other ends of the ropes are tied to points A and B on a flagpole, as shown below: Angle ADC measures 60° and angle BDC measures 30°. What is the distance between the points A and B on the flagpole? Choice are: A) 20 Feet B) 15 Feet C) 10 Feet D) 25 Feet

  92. anonymous
    • one year ago
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    Not much difference between this and last question, you'd again use tangent function, it's just a vertical figure instead of a horizontal figure now

  93. anonymous
    • one year ago
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    find AC and BC like we did and subtract BC from AC

  94. anonymous
    • one year ago
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    If you get stuck, don't hesitate to ask, even if I'm not here, people will be here to help, I'll stay up for an hour or two more

  95. anonymous
    • one year ago
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    Okay, thank you. I'll see what I can work out now. :)

  96. anonymous
    • one year ago
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    \[\tan \left( 30 \right)= \frac{ 5\sqrt{3} }{ BC }\] \[\tan \left( 60 \right)= \frac{ 5\sqrt{3} }{ AC}\] This right so far?

  97. anonymous
    • one year ago
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    nope, doesn't look good, check your opposite and adjacent

  98. anonymous
    • one year ago
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    Okay. On the opposite it's just the flagpole? Or do you mean the degrees starting from D and going out towards the flagpole the other way?

  99. anonymous
    • one year ago
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    |dw:1441043497380:dw|

  100. anonymous
    • one year ago
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    Right, so \[\tan \left( 30 \right)= \frac{ BC }{ 5\sqrt{3} } \] and \[\tan \left( 60 \right)= \frac{ AC }{ 5\sqrt{3} }\]

  101. anonymous
    • one year ago
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    |dw:1441043990553:dw|

  102. anonymous
    • one year ago
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    Okay, then in the chart I wrote down that you showed me, wouldn't I use sqrt 3 for 60 and 1/sqrt3 for 30?

  103. anonymous
    • one year ago
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    It's okay, thank you for your help! Bye now :)

  104. anonymous
    • one year ago
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    yep

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