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anonymous

  • one year ago

Really need help, Will FAN and MEDAL A forest ranger at an observation point A sights a fire in the direction N27degrees10'E. another ranger at an obseraion point B, 6.0 miles due east of A, sights the same fire at N52degrees40'W. Approx. the distance from each other the observation points to the fire. I drew a picture but I want to see someone else interpretation of the information

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  1. anonymous
    • one year ago
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    @Michele_Laino if you're not busy can you help me again. I don't remember if I asked you this one

  2. Michele_Laino
    • one year ago
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    N27degrees10'E means that I'm heading towards North, right?

  3. anonymous
    • one year ago
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    Yes, you're in the North and heading East

  4. Michele_Laino
    • one year ago
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    ok!

  5. Michele_Laino
    • one year ago
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    like this? |dw:1432759303858:dw|

  6. anonymous
    • one year ago
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    Yes I have that part

  7. Michele_Laino
    • one year ago
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    so the complete drawing is: |dw:1432759443568:dw|

  8. Michele_Laino
    • one year ago
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    sorry: |dw:1432759585590:dw|

  9. Michele_Laino
    • one year ago
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    is my drawing right?

  10. anonymous
    • one year ago
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    I think so

  11. Michele_Laino
    • one year ago
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    your exercise asks for distances AC and BC, such that: |dw:1432759724446:dw|

  12. Michele_Laino
    • one year ago
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    am I right?

  13. Michele_Laino
    • one year ago
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    |dw:1432759935107:dw|

  14. anonymous
    • one year ago
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    yeah, i think so

  15. Michele_Laino
    • one year ago
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    Now, we can apply the law of sines, so we can write: \[\Large \frac{6}{{\sin \left( {89^\circ 50'} \right)}} = \frac{{AC}}{{\sin \left( {37^\circ 20'} \right)}}\]

  16. Michele_Laino
    • one year ago
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    so: \[\Large AC = \frac{{6 \times \sin \left( {37^\circ 20'} \right)}}{{\sin \left( {89^\circ 50'} \right)}} = ...?\]

  17. Michele_Laino
    • one year ago
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    what is AC?

  18. anonymous
    • one year ago
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    =3.6

  19. Michele_Laino
    • one year ago
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    hint: we can write this: 37° 20'= 37.33 degrees 89°50'=89.83 degrees

  20. Michele_Laino
    • one year ago
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    I got AC=3.638 miles

  21. Michele_Laino
    • one year ago
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    I think your answer is correct!

  22. Michele_Laino
    • one year ago
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    Now, similarly, using the law of sines again, we can write: \[\Large \frac{6}{{\sin \left( {89^\circ 50'} \right)}} = \frac{{BC}}{{\sin \left( {62^\circ 50'} \right)}}\]

  23. Michele_Laino
    • one year ago
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    so we get: \[\Large BC = \frac{{6 \times \sin \left( {62^\circ 50'} \right)}}{{\sin \left( {89^\circ 50'} \right)}} = ...?\] So, what is BC?

  24. Michele_Laino
    • one year ago
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    hint: 62° 50'=62.83 degrees 89° 50'=89.83 degrees

  25. anonymous
    • one year ago
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    5.3

  26. Michele_Laino
    • one year ago
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    that's right! I got 5.338 miles

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