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Astrophysics

  • one year ago

@Miss.SweetiePie

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  1. Astrophysics
    • one year ago
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    Hey so I'll go over the question we did previously, this one: Triangle FGH, FH = 7 ft, FG = 12 ft, F = 70 degrees, I forgot the other one so if you need help on any other just post here but I'll do this one right now and you can go over it.

  2. Astrophysics
    • one year ago
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    |dw:1434005425178:dw| so angle G is what we need to find, so here we'll have to use cosine rule twice.

  3. Astrophysics
    • one year ago
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    \[a^2=b^2+c^2-2bccos(A)\] so this is the cosine rule, so the way we can apply this is if we have two sides and angle between them, so in our case we'll find f first. |dw:1434005605935:dw| \[f^2=g^2+h^2-2gh(\cos(f)) \implies f^2 = 7^2+12^2-2(7)(12)\cos(70)\] \[f^2 = 135.54 \implies f = \sqrt{135.54} = 11.64\] so side f is 11.64 ft |dw:1434005723427:dw|

  4. Astrophysics
    • one year ago
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    Now since we have side f, we can use cosine rule again |dw:1434005756197:dw| notice we don't have the angle as that's what we are looking for, but we do have all sides now. So we will rearrange the formula for g, \[g^2 = f^2+h^2 - 2fh \cos(G)\] so this is the formula we will use and we need to find angle G so doing some algebra (ask if you're not sure) we get \[G = \cos^{-1}\left( \frac{ -g^2+f^2+h^2 }{ 2fh } \right) \implies \cos^{-1} \left( \frac{ -(7)^2+(11.64)^2+(12)^2 }{ 2(11.64)(12) } \right)\] plugging all that in we get our final answer to be..drum roll \[34.4~~ degrees\]

  5. Astrophysics
    • one year ago
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    Alright so go over this and let me know if everything makes sense, I tried to do it in a simple way, and if you have another question just ask :), enjoy!

  6. anonymous
    • one year ago
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    Why would G be negative though?

  7. Astrophysics
    • one year ago
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    G is not negative if you mean length g, it's negative because we are solving for angle G, so we rearrange the formula, so when we divide by -2fh it becomes negative, hope that helps!

  8. Astrophysics
    • one year ago
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    \[G = \cos^{-1} \left( \frac{ f^2+h^2-g^2 }{ 2fh } \right)\]

  9. Astrophysics
    • one year ago
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    You can put it as above if that helps

  10. anonymous
    • one year ago
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    Oooh I see what you're saying now! If only it was this simple the first time...Probably because its not all over the place

  11. anonymous
    • one year ago
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    I understand it like that

  12. anonymous
    • one year ago
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    Hold on let me see if I can find the other one that I missed

  13. Astrophysics
    • one year ago
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    Sure :)

  14. anonymous
    • one year ago
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    |dw:1434041568925:dw|

  15. Astrophysics
    • one year ago
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    I think you should show me your attempt on this one first, as I haven't seen this before and if I did all the work that won't help you :)

  16. anonymous
    • one year ago
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    I dont know how I got the answer I picked but I thinnk I figured out how to do it right\[23\times7\sin (90)\]

  17. anonymous
    • one year ago
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    \[so 161\sin (90)\]

  18. anonymous
    • one year ago
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    \[which = 143.93346284\]

  19. Astrophysics
    • one year ago
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    Is that the triangle they gave you? Or you made it

  20. anonymous
    • one year ago
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    thats the one they gave me

  21. anonymous
    • one year ago
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    \[then I divided that number \to get 71.96\]

  22. anonymous
    • one year ago
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    but the correct answer was 72.93

  23. Astrophysics
    • one year ago
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    |dw:1434078245596:dw| ok let me make it more clear, so this is the triangle right

  24. Astrophysics
    • one year ago
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    B got cut off haha, but we know it's there :P

  25. anonymous
    • one year ago
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    lol the 23 is a lil over to the right but yes your right

  26. Astrophysics
    • one year ago
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    Well the AB = 23 so it's ok, alright let me see

  27. anonymous
    • one year ago
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    okay but when you find out what I did wrong tell me so I can try and fix it

  28. Astrophysics
    • one year ago
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    Yes, no problem, I'll work this out right now and then I'll post it :)

  29. anonymous
    • one year ago
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    Okay and I will follow along

  30. Astrophysics
    • one year ago
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    You could just do \[\cos^{-1} (\frac{ 7 }{ 23 }) = 72\]

  31. Astrophysics
    • one year ago
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    |dw:1434079210532:dw|

  32. Astrophysics
    • one year ago
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    OR actually, we can use cosine law

  33. Astrophysics
    • one year ago
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    Yeah I think they might want you to do that, try it out and see if you get it.

  34. Astrophysics
    • one year ago
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    As you can see, there is more than one way to approach such a problem :P

  35. anonymous
    • one year ago
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    sorry mmy computer is messing up...im about to try it

  36. anonymous
    • one year ago
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    so do exactly what I did just use cos^-1?

  37. Astrophysics
    • one year ago
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    Well, I don't really understand what you did haha.

  38. anonymous
    • one year ago
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    i multiplied the sides then times that by sin then divided by 2

  39. anonymous
    • one year ago
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    lol you only put the haha so it wouldnt sound harsh...didnt you???

  40. anonymous
    • one year ago
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    I might not reply til tomorrow cause I have to watch my nephew but dont go ahead in the question I want to understand it completely! (if you dont mind)

  41. Astrophysics
    • one year ago
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    You can also use sine law

  42. Astrophysics
    • one year ago
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    All of them will give you the same answer, and no problem :)

  43. Astrophysics
    • one year ago
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    sine law might be the quickest

  44. Astrophysics
    • one year ago
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    Because you can solve for the corner angle and then add it to 90, and subtract it from 180

  45. Astrophysics
    • one year ago
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    MANY WAYS :P

  46. anonymous
    • one year ago
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    so if i were to do the sin law it would be \[7\times23\sin 90/2? where am I going wrong??? Should it be sin inverse?

  47. Astrophysics
    • one year ago
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    \[\frac{ a }{ \sin A } = \frac{ b }{ \sin B } = \frac{ c }{ \sin C }\] well It's very confusing as I can not see your latex I think you should do this on paper, and try to figure it out, you can ignore one of the ratio and solve for the corner angle as you have the side.

  48. Astrophysics
    • one year ago
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    It's best if you label the triangle first, you should get used to doing that

  49. anonymous
    • one year ago
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    since I only know one degree would the other sin be Sin(x)? I'm sorry I'm slow I dont catch on easy thats why its important for me to understand how to do it

  50. Astrophysics
    • one year ago
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    No, lets ignore that one for now. First label the triangle, otherwise you will screw up the calculations and it will not make sense. |dw:1434080478635:dw|

  51. Astrophysics
    • one year ago
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    \[\frac{ b }{ \sin B } = \frac{ c }{ \sin C } \implies \frac{ 7 }{ \sin \theta } = \frac{ 23 }{ \sin (90) }\]|dw:1434080587003:dw|

  52. Astrophysics
    • one year ago
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    notice we ignore a/sinA so we're left with the ratios above now we can solve for theta

  53. anonymous
    • one year ago
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    okay and the reason the sin under th 7 is theta because its unknown

  54. Astrophysics
    • one year ago
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    Yes, that's the corner angle as I labelled in the triangle

  55. Astrophysics
    • one year ago
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    That's also what we're solving for

  56. Astrophysics
    • one year ago
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    The first way we did it was actually the quickest and fastest, but you should know all :P

  57. anonymous
    • one year ago
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    okay im with you so far... so anytime we dont know a sin we use sin theta?

  58. Astrophysics
    • one year ago
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    No, it's just a variable I labelled, you can put y, your name, what ever :P

  59. Astrophysics
    • one year ago
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    sin(what ever)

  60. anonymous
    • one year ago
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    oh im sorry I feel like im not getting anywhere :'(

  61. anonymous
    • one year ago
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    i'm sorry if I wasted time

  62. Astrophysics
    • one year ago
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    It's ok, I think you need to just try this on paper, it's sort of hard to explain these kinds of problems on the computer as it's hard to know how much you know, but if you use khan academy/ youtube that may be more beneficial than me because I can only do so much on here. There are many links as well: https://www.khanacademy.org/ https://www.mathsisfun.com/algebra/trig-cosine-law.html https://www.mathsisfun.com/algebra/trig-sine-law.html

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