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anonymous

  • 5 years ago

if a two ships left the same port at the same time but ship A left at N35'E45 degrees at a speed of 25 mph and ship B left at N24'56degrees at a speed of 25 mph, how far apart would they be in 4 hours? i dont understand what formula to use law of sines? law of cosines? help!!

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  1. anonymous
    • 5 years ago
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    Hi, this site keeps crashing on me. Use cosine rule - the ships will have an angle between them equal to the difference of their angles, and they will have traveled 100miles in 4 hours, so you have your side lengths. The distance is the distance between them .

  2. anonymous
    • 5 years ago
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    the second one is actually S24'E56 so so do i find the hypotenuse?

  3. anonymous
    • 5 years ago
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    Cosine rule...

  4. anonymous
    • 5 years ago
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    If you can wait, I can send you a solution later. I can't do anything on the machine I'm on right now.

  5. anonymous
    • 5 years ago
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    yes i can wait! and ship b is 23 mph sorry for all of the mistakes

  6. anonymous
    • 5 years ago
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    it's okay. later.

  7. anonymous
    • 5 years ago
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    Okay, this is going to be an application of the cosine rule. Do you know how to get angles when they're given in this form first?

  8. anonymous
    • 5 years ago
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    yes, you put them in the calculator using 2nd angle

  9. anonymous
    • 5 years ago
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    Yeah, but I mean geometrically. I'm drawing a picture of the situation.

  10. anonymous
    • 5 years ago
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    no and ok awesome

  11. anonymous
    • 5 years ago
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  12. anonymous
    • 5 years ago
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    Now, the angle (when you're given it in this form, i.e. North blah blah) is swept out from the north pole towards where the object is. Here, ship B has the smaller angle (blue) and ship A has the larger. What you're being asked to find is the distance between the ships at the end of 4 hours.

  13. anonymous
    • 5 years ago
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    The distance is the line in pink. Now, since speed is *defined* as \[(speed)=\frac{(distance)}{(time)}\]it means,\[(distance)=(speed) \times (time)\]

  14. anonymous
    • 5 years ago
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    oh, but wouldnt ship b be in the 4th quadrant since its angle is SE?

  15. anonymous
    • 5 years ago
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    oh ok, so then you would get 100 mi and 92 mi which give you the side lengths?

  16. anonymous
    • 5 years ago
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    The distance ship A travels in 4 hours is then,\[d_A=25\frac{mi}{h}\times 4 h=100 miles\]

  17. anonymous
    • 5 years ago
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    Ship B will, in 4 hours, travel,\[d_B=23\frac{mi}{h} \times 4 h = 92 miles\]

  18. anonymous
    • 5 years ago
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    Now you have the two distances - the blue and red lines.

  19. anonymous
    • 5 years ago
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    All you need to do is now find the angle BETWEEN the two ships - this will stay the same since they don't change course. The angle between them is the angle in pink. It's given by the difference between the larger and smaller angles.

  20. anonymous
    • 5 years ago
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    \[\theta = 35^o45' -24^o56'=10^o49'\]

  21. anonymous
    • 5 years ago
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    You fine so far?

  22. anonymous
    • 5 years ago
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    ok that makes sense, but what if the angle was supposed to be in the 4th quadrant since i wrote it wrong

  23. anonymous
    • 5 years ago
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    i mean the 3rd

  24. anonymous
    • 5 years ago
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    I'm not sure what you mean.

  25. anonymous
    • 5 years ago
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    since ship b was S24E56 but i wrote it as N24E56

  26. anonymous
    • 5 years ago
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    like ship b was supposed to be S24E56 not N24E56

  27. anonymous
    • 5 years ago
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    Give me a second.

  28. anonymous
    • 5 years ago
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    ok

  29. anonymous
    • 5 years ago
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  30. anonymous
    • 5 years ago
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    To say S24E56 means "24 degrees, 56 minutes from magnetic South in the Easterly direction." So you move from the 'South' line and sweep upwards towards the East by 24 degrees, 56 minutes.

  31. anonymous
    • 5 years ago
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    The first letter tellls you what pole to start from, then the second letter tells you whether to sweep left or right (left = West, right = East).

  32. anonymous
    • 5 years ago
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    The problem is set up in the same way - you're still going to have to use the cosine rule because you need to find the distance opposite an angle, where that angle is BETWEEN two sides you know the value of.

  33. anonymous
    • 5 years ago
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    so when im finding the angle, do i have to include the 56 minutes part when using the law of cosine?

  34. anonymous
    • 5 years ago
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  35. anonymous
    • 5 years ago
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    Is this what you've got?

  36. anonymous
    • 5 years ago
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    yes, exactly!

  37. anonymous
    • 5 years ago
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    Okay...

  38. anonymous
    • 5 years ago
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    Now we have to find the angle IN BETWEEN from the info. we're given.

  39. anonymous
    • 5 years ago
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    so to do this, do i subtract the two give angles from 180?

  40. anonymous
    • 5 years ago
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  41. anonymous
    • 5 years ago
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    Note, from your picture, that the red angle and the green angle will sum to 90 degrees. The blue angle and purple angle will sum to 90 degrees also.

  42. anonymous
    • 5 years ago
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    We need (green) + (purple), since this is the angle in between. We have, (red) + (green) + (purple) + (blue) = 90 + 90 = 180 degrees. We know that (red) = 35degrees, 45 minutes and that (blue) = 24 degrees, 56 minutes, so

  43. anonymous
    • 5 years ago
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    (green) + (purple) = 180 - 35deg.45min - 24deg.56min. = 119deg.19min.

  44. anonymous
    • 5 years ago
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    This is your angle in between the two lines. We have the distances calculated above, so we have the following now:

  45. anonymous
    • 5 years ago
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  46. anonymous
    • 5 years ago
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    (They're both the same). Okay, so now we can use the cosine formula: \[a^2=100^2+92^2-2(100)(92)\cos(119^o19')\]to get\[a^2 \approx 27473\]which means that the distance is (after taking the square root),\[a \approx 165.75\]miles.

  47. anonymous
    • 5 years ago
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    That's it.

  48. anonymous
    • 5 years ago
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    thank you. i totally understand it now!

  49. anonymous
    • 5 years ago
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    Good :) Good luck with it. If you need anymore help, come back to the site - I might be around.

  50. anonymous
    • 5 years ago
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    ok. thank you!! =)

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