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anonymous

  • one year ago

In triangle FGH, FH=7feet FG=12feet and measurement F=70 degrees. Find measurement G round to the nearest tenth

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  1. anonymous
    • one year ago
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    @wolf1728 @Astrophysics

  2. wolf1728
    • one year ago
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    Hello !!! I got "bumped" and I entered this topic!!!

  3. anonymous
    • one year ago
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    |dw:1433834150459:dw|

  4. anonymous
    • one year ago
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    yay and did I do that correctly?

  5. wolf1728
    • one year ago
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    I guess you did (and good diagram)

  6. anonymous
    • one year ago
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    Thats what I meant *did i do the diagram correctly*

  7. wolf1728
    • one year ago
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    Yes you did!!

  8. wolf1728
    • one year ago
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    Where's Astro?

  9. anonymous
    • one year ago
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    @Astrophysics

  10. anonymous
    • one year ago
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    not sure

  11. anonymous
    • one year ago
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    can you help me get it started and i do it and you tell me if im doing it wrong or not

  12. anonymous
    • one year ago
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    you guys there?

  13. wolf1728
    • one year ago
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    Two sides and included angle, we use Law of Cosines

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  14. Astrophysics
    • one year ago
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    This should not be a right triangle

  15. Astrophysics
    • one year ago
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    Otherwise we could just say it's 20 degrees ;)

  16. wolf1728
    • one year ago
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    b=7 c =12 Angle A=70 degrees a^2 = (b^2 + c^2) -(2*b*c * cos(A))

  17. Astrophysics
    • one year ago
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    You can use the sine law

  18. anonymous
    • one year ago
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    So one side should be longer...i was thinking the same! but 20 isnt a answer choice!

  19. Astrophysics
    • one year ago
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    That would be a whole lot easier

  20. anonymous
    • one year ago
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    that is what wolf included correct

  21. Astrophysics
    • one year ago
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    Not quite, he is using cosine law

  22. Astrophysics
    • one year ago
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    But, I will let wolf continue :)

  23. anonymous
    • one year ago
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    does it matter which one is used or?

  24. wolf1728
    • one year ago
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    a^2 = 7^2 + 12^2 -(2*7*12*cos(70) a^2 = 49 + 144 - (108*.34202) a^2 = 193 -36.93816 a^2 = 156.06184 a = sq root(156.06184) a = 12.49

  25. anonymous
    • one year ago
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    I dont think thats correct its not a choice I have maybe we have to use Cos*b*

  26. anonymous
    • one year ago
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    hold on im looking

  27. wolf1728
    • one year ago
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    How about 11.642 ?

  28. anonymous
    • one year ago
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    how did you get that

  29. anonymous
    • one year ago
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    its one of the answers but I am trying to learn

  30. Astrophysics
    • one year ago
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    Well lets redraw the triangle and put everything in place, and continue from there |dw:1433835265958:dw|

  31. Astrophysics
    • one year ago
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    Oh I guess wolf already did it xD

  32. anonymous
    • one year ago
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    Ok I was just a little off on the graph

  33. wolf1728
    • one year ago
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    a^2 = 49 + 144 - (168*.34202) a^2 = 193 - (57.45936) a^2=135.54064 a = 11.6421922334 ARRGGGHHHH !!!!!!!!!!! The other time I made an arithmetic mistake.

  34. Astrophysics
    • one year ago
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    I try not to make right triangles, but always end up doing so haha, and good job wolf I believe in you! :P

  35. wolf1728
    • one year ago
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    was 11.642 an answer?

  36. anonymous
    • one year ago
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    11.6 yes

  37. wolf1728
    • one year ago
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    all right !!!!!!!!!!!!!!!!

  38. Astrophysics
    • one year ago
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    Alright cool! :)

  39. Astrophysics
    • one year ago
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    Did you understand everything, sorry about the first question it was a bit messy, I'll go look at it again later...

  40. anonymous
    • one year ago
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    wait dont go anywhere I am gonna show you the work for a problem and I just would like for you to check

  41. anonymous
    • one year ago
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    slowly but surely I know a lot more than I did before I came here

  42. wolf1728
    • one year ago
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    all right !!

  43. anonymous
    • one year ago
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    Okay so Im finding the area of a triangle and it looks like

  44. anonymous
    • one year ago
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    |dw:1433835840125:dw|

  45. anonymous
    • one year ago
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    so I did 12^2-7^2=b and got 144-49=95

  46. wolf1728
    • one year ago
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    There is a side angle side formula for triangle area Area = (1/2) * side 1 * sine (included angle) * side 2

  47. anonymous
    • one year ago
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    then i did sin*40*95 and got 61.6 and divided it by 2 and got 30.53

  48. wolf1728
    • one year ago
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    Using side angle side formula I get 26.997

  49. anonymous
    • one year ago
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    that was my second choice, when my calculator was on rads i got a different answer than I did when it was on degrees I just didn't know which I should have been on

  50. wolf1728
    • one year ago
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    You were using the formula?

  51. anonymous
    • one year ago
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    I had seen another question similar to this and I was following along but with my question instead

  52. wolf1728
    • one year ago
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    You can find a side angle side triangle area calculator here http://www.1728.org/triang.htm (It's my own website)

  53. anonymous
    • one year ago
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    Really???!!! thats so cool! I'm gonna check it out! and If you dont mind checking 2 more questions with me

  54. wolf1728
    • one year ago
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    Wow - two more?? Well I'll try - go ahead

  55. anonymous
    • one year ago
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    im sorry I just want to understand it and the calculator isnt working....ill add more to your testimonial

  56. wolf1728
    • one year ago
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    okay - by the way my site has got about 250 pages I'd say half the pages are all kinds of calculators

  57. anonymous
    • one year ago
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    \[\sin ^{2} \Theta \over 1- \cos \theta \]

  58. anonymous
    • one year ago
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    have to simplify it

  59. anonymous
    • one year ago
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    I am going to keep that site itll help me a lot

  60. wolf1728
    • one year ago
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    Hmmmm that's probably a trigonometric identity. (I'm glad you like my site).

  61. anonymous
    • one year ago
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    can you help?

  62. anonymous
    • one year ago
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    I have answer choices if its easier

  63. wolf1728
    • one year ago
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    I was just doing a web search for trig identities. Maybe I can solve it myself.

  64. anonymous
    • one year ago
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    I belive you can youve been doing great i feel im in comfortable hands

  65. wolf1728
    • one year ago
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    sine(angle) = y/hypotenuse cosine = x/hyp

  66. anonymous
    • one year ago
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    theres no triangle its just says to simplify the trigonomic equation

  67. anonymous
    • one year ago
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    im looking up trig identies

  68. wolf1728
    • one year ago
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    Okay and I made a quick graphic

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  69. anonymous
    • one year ago
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    \[ http://www.math.com/tables/trig/identities.htm would this help\]

  70. wolf1728
    • one year ago
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    okay let me see that page in the meantime, I started with this sine^2 = (y^2 / hyp^2) 1 - cos = (1 -x/hyp)

  71. anonymous
    • one year ago
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    does it help?

  72. wolf1728
    • one year ago
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    Yes I think I found something

  73. anonymous
    • one year ago
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    where at

  74. anonymous
    • one year ago
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    I might have found something too

  75. wolf1728
    • one year ago
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    Almost at the top of the page they state: sin^2(x) + cos^2(x) = 1 and you have to simplify sin^2(Θ) / (1−cos(θ))

  76. anonymous
    • one year ago
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    like this sin^2 (a) * (1 + cos(a)) / (1 - cos^2 (a)) = 1 + cos(a

  77. wolf1728
    • one year ago
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    sin^2(x) + cos^2(x) = 1 Therefore, sin^2(x) = 1 -cos^2(x) SO, we can say sin^2(Θ) / (1−cos(θ)) equals (1 -cos^2(x)) / (1 - cos(x))

  78. wolf1728
    • one year ago
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    Hmmm now to simplify that

  79. anonymous
    • one year ago
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    which seems tricky

  80. anonymous
    • one year ago
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    the -cos 2 and the cos would leave one -cos right?

  81. wolf1728
    • one year ago
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    I don't know - that would be considered "cancelling terms" which should NOT be done. I got: (1 -cos^2(x)) / (1 - cos(x)) = (1 - (cos(x) * cos(x))) / (1 -cos(x))

  82. anonymous
    • one year ago
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    theres a discrepancy here somewhere

  83. anonymous
    • one year ago
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    Ill keep looking at it and I will message you if there was something wrong with the question...youve helped me a lot I dont want you to feel like im over doing it

  84. anonymous
    • one year ago
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    whats your name btw?

  85. wolf1728
    • one year ago
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    I'm thinking of trying one more thing - actuallly calculating values of sine and cosine My name is Robert - but wolf is okay too

  86. anonymous
    • one year ago
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    Mucho Gusto! and how are you going to do that? im summer and sorry my computer froze up

  87. wolf1728
    • one year ago
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    Okay I looked up sine 35 degrees 0.57358 squaring that we get 0.3289940164 cosine 35 degrees = 0.81915 1 minus that = 0.18085 sine 45 degrees 0.70711 squaring that 0.5000045521 cosine 45 degrees 0.70711 1 minus that = 0.29289 Well, that doesn't show much does it? Maybe it's time to quit.

  88. anonymous
    • one year ago
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    lol maybe our brains are fried!

  89. wolf1728
    • one year ago
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    Yes, we've been doing this stuff for about 2 hours?

  90. anonymous
    • one year ago
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    roughly yes lol...are you tired?

  91. wolf1728
    • one year ago
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    Yes I am tired - think I'll call it a night.

  92. wolf1728
    • one year ago
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    whoops - just got "bumped" from openstudy

  93. anonymous
    • one year ago
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    Alright goodnight and ill let my mom know you helped me im sure shell do a better testimonial

  94. anonymous
    • one year ago
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    what does that mean

  95. wolf1728
    • one year ago
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    that's okay - and thanks. Might as well get going. It was good meeting you SweetiePie

  96. wolf1728
    • one year ago
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    "bumped" means getting tossed off openstudy

  97. anonymous
    • one year ago
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    oh im sorry i hope it wasnt my fault

  98. wolf1728
    • one year ago
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    no - sometimes it just happens

  99. anonymous
    • one year ago
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    sweetie pies daughter

  100. anonymous
    • one year ago
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    welll im gonna let you go i really appreciate all the help night

  101. wolf1728
    • one year ago
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    Thank you and I was glad to help - see you later

  102. anonymous
    • one year ago
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    :)

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is replying to Can someone tell me what button the professor is hitting...

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