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anonymous

  • one year ago

Trig and triangles. Is this possible? One sec, I am typing it up.

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  1. anonymous
    • one year ago
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    Ok, it stats that \( \large a\geq b\space and \space a>h =b * sin A \) there is one triangle Well I have a situation where a is greater than b and a is greater than h so there should be one triangle Well, using Sine Law, I get that there can be two angles for sine A, which should mean there are two triangles correct?

  2. anonymous
    • one year ago
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    o-o some one would have to be a pro to figure that out...

  3. anonymous
    • one year ago
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    It is trig Pre Cal

  4. anonymous
    • one year ago
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    what grade you in?

  5. anonymous
    • one year ago
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    College

  6. abb0t
    • one year ago
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    Yes. Are there choices? I can give you the answer.

  7. anonymous
    • one year ago
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    oh wow

  8. anonymous
    • one year ago
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    No choices. Let me see if I can make one up close to it.

  9. anonymous
    • one year ago
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    a =40, b =10 , B=5 degrees a and b are sides. Ok, If we do \( \large \frac{sin5}{10}=\frac{sin A}{40} = 20.40 \) Now the law states\( \large a\geq b\space and \space a>h =b * sin A \) But clearly we have another angle we can use for A. We can use 159.6, which should mean we have two triangles correct?

  10. anonymous
    • one year ago
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    h = 3.48

  11. anonymous
    • one year ago
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    @Astrophysics @math1234

  12. anonymous
    • one year ago
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    @dan815

  13. anonymous
    • one year ago
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    @Loser66

  14. mathmath333
    • one year ago
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    what is \(h\)

  15. anonymous
    • one year ago
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    h is the height

  16. anonymous
    • one year ago
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    This has to do with Sine Law

  17. anonymous
    • one year ago
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    It has to do with Sine Law and figuring out if there is one or two triangles

  18. mathmath333
    • one year ago
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    height drawn from vertex A to its opposite base?

  19. anonymous
    • one year ago
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    The rule stats that \( \huge \large a\geq b\space and \space a>h =b * sin A \) is one triange but Sine Law says different because I can use another angle for A.

  20. anonymous
    • one year ago
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    I am drawing a picture

  21. Loser66
    • one year ago
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    |dw:1435185125370:dw|

  22. Loser66
    • one year ago
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    \(a\geq b\) and \(h = b sinA\) since BH is the height, we have sin A = h/c hence \(sinA = \dfrac{h}{c}=\dfrac{bsinA}{c}\), hence b =c

  23. anonymous
    • one year ago
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    |dw:1435185224437:dw|

  24. Loser66
    • one year ago
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    |dw:1435185335223:dw|

  25. Loser66
    • one year ago
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    But I am not sure what you want me to do. hehehe....

  26. anonymous
    • one year ago
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    a =40, b =10 , B=5 degrees Well, the rule states \( \large a\geq b\space and \space a>h =b * sin A \) a is greater than b and a is greater than h so we should have one triangle. Ok, but now Sine Law goes \(\large A=\frac{sin5}{10}=\frac{sin A}{40} = 20.40 \) h = 10* sin(20.40) =3.42 So a is greater than h but we can use another angle since 180-20.40 = 159.6 and B only = 5 so 159.6+ 5 = 169.6 so C can equal 10.4 if we use angle 159.6. But this shows we have two triangles using sine law correct but the other rule must not hold true?

  27. anonymous
    • one year ago
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    15 owlbucks to help me understand. To me it seems the rule(\( \large (a\geq b\space and \space a>h =b * sin A ) \) = one triangle) does not hold true.

  28. anonymous
    • one year ago
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    In my studies this is call the Ambiguous Case of (SSA) Side Side Angle

  29. phi
    • one year ago
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    I think they are using the rule to differentiate between these two triangles: |dw:1435187302571:dw|

  30. phi
    • one year ago
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    just using the law of sines allows both the obtuse triangle (with short a) and acute triangle with long a for the obtuse triangle, b sin theta is bigger than side a

  31. anonymous
    • one year ago
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    This is what they stat When two sides and an angle are given, the resulting information may result in one triangle, two triangles or no triangle at all, depending on the relationship between a, h, b. No Triangle = a<h=b sinA One Right Triangle = a = h =b sinA Two Triangles = b>a>h=b sinA One Triangle a >=b and a> h=b sinA

  32. anonymous
    • one year ago
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    With the triangle I have up there, these rules do not hold true.

  33. anonymous
    • one year ago
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    Am I right? Those rules I have do not apply correct?

  34. anonymous
    • one year ago
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    Now they apply to one of the examples they give with the rules, which is a=7, b=5 A=70

  35. phi
    • one year ago
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    I think we have to be clear what side is a and b. Your example a =40, b =10 , B=5 degrees has only one triangle for a solution, so the "rule" should work, as long as we are careful about the definitions

  36. phi
    • one year ago
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    when they say "2 sides and an angle" do they mean the "included angle" ?

  37. anonymous
    • one year ago
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    Yes, that is how I take it.

  38. phi
    • one year ago
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    for this to work One Right Triangle = a = h =b sinA it means side b is the hypotenuse and angle A is opposite side a |dw:1435188557398:dw|

  39. anonymous
    • one year ago
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    What gets me is that if sinA = 20.40 that means there is another Angle for A correct? Because they 180-20.40 = A =159.6 also? But this could be where I am guffing and thinking there could be two triangles. I might need to draw it out with angle 159.6

  40. phi
    • one year ago
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    Looking at your rules (trying to make sense of them) No Triangle = a<h=b sinA means |dw:1435188767810:dw|

  41. anonymous
    • one year ago
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    Correct

  42. phi
    • one year ago
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    so the first two make sense. But note the angle is *not* the included angle between a and b

  43. phi
    • one year ago
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    One Triangle a >=b and a> h=b sinA I think your example that you entered up above fits this one.

  44. anonymous
    • one year ago
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    The rules make sense but it is when I start using Sine Law and when I got A= 20.40 and then there can be another angle for A, which is 180-20.40 = A = 159.6

  45. phi
    • one year ago
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    and this one Two Triangles = b>a>h=b sinA means |dw:1435189157259:dw|

  46. anonymous
    • one year ago
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    A could = 159.6 because b=5 so 180-20.40 = 159.6+5 = 164.6 and C could = 180-164.6 = 15.4

  47. anonymous
    • one year ago
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    But I have not drawn out the triangle to see what it looks like

  48. phi
    • one year ago
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    maybe this is the problem the rule uses 2 adjacent sides a and b, and the angle *opposite side a* (so which angle we know defines side a, and b by default is the the other given side)

  49. phi
    • one year ago
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    Thus in your example: a =40, b =10 , B=5 degrees we then label it this way: The angle is always A: angle A= 5 side a= 10, side b= 40 now it looks like we use this rule Two Triangles = b>a>h=b sinA and you are correct, we do get two triangles

  50. anonymous
    • one year ago
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    Ok I am perplexed. On the problem I posted do the rules hold true? I am given the problem as B=5 not A = 5. but it appear this triangle has two and not 1.

  51. anonymous
    • one year ago
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    I have to find A before the rules work.

  52. phi
    • one year ago
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    The rules require you to label the sides and angle in a very specific way. There is no choice: the angle must be labeled A, and the side opposite angle A is side a and the remaining side becomes b. Now you can use the rule.

  53. phi
    • one year ago
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    If you are given 2 sides and an angle that is *not included between them* label the given angle A (no matter what the problem calls it) label its opposite side a label the remaining side b now you can use the rule.

  54. anonymous
    • one year ago
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    Since they are using A but I have B the rule for me b>=a and b> h =sinB

  55. phi
    • one year ago
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    Here is trying to make 2 triangles with a=40 and b=10

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  56. phi
    • one year ago
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    Here is the other way round, b=40, a=10 now we can make two triangles (see the red and blue lines)

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  57. anonymous
    • one year ago
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    Yes, we can create two triangles but I am wondering about the rules.

  58. phi
    • one year ago
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    What is a problem you want to analyze using the rules?

  59. anonymous
    • one year ago
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    |dw:1435191754375:dw| To me it seems the rule is being applied like b >= a and b> h=a sinB

  60. phi
    • one year ago
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    ok. In that picture, to use the rule you must rename it (as we have already discussed) then when you apply the rule, you will find two triangles are possible.

  61. phi
    • one year ago
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    i.e A = 5, B is unknown , side a=10 , side b= 40

  62. anonymous
    • one year ago
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    It is holding true how I have it right?? b= 10 a = 40 and B =5 No Triangle = b<h=a sinB One Right Triangle = b = h =a sinB Two Triangles = a>b>h=a sinB One Triangle b >=a and b> h=a sinB

  63. anonymous
    • one year ago
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    Sine Law showed me there was two triangle but I was setting up the rule above wrong because that is how they explain it but actually it can be set up like I just did correct?

  64. phi
    • one year ago
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    yes, you can relabel in the rules. But the idea is to be consistent.

  65. anonymous
    • one year ago
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    Yes, I see what I was doing wrong. It is how I was setting up the rules I just posted. I knew there was two triangles thanks to Sine Law but I was confused on the rules because the way the explained it and made it seem like that I had to use b and SineA but I can use a sinB long as it is the same pattern

  66. anonymous
    • one year ago
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    Ty for you time!!!

  67. phi
    • one year ago
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    yw

  68. anonymous
    • one year ago
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    I sent you 20 owlbucks. Thank once again.

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