anonymous
  • anonymous
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
Mathematics
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anonymous
  • anonymous
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
Mathematics
jamiebookeater
  • jamiebookeater
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
@Michele_Laino if you're not busy can you help me again. I don't remember if I asked you this one
Michele_Laino
  • Michele_Laino
N27degrees10'E means that I'm heading towards North, right?
anonymous
  • anonymous
Yes, you're in the North and heading East

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Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • Michele_Laino
like this? |dw:1432759303858:dw|
anonymous
  • anonymous
Yes I have that part
Michele_Laino
  • Michele_Laino
so the complete drawing is: |dw:1432759443568:dw|
Michele_Laino
  • Michele_Laino
sorry: |dw:1432759585590:dw|
Michele_Laino
  • Michele_Laino
is my drawing right?
anonymous
  • anonymous
I think so
Michele_Laino
  • Michele_Laino
your exercise asks for distances AC and BC, such that: |dw:1432759724446:dw|
Michele_Laino
  • Michele_Laino
am I right?
Michele_Laino
  • Michele_Laino
|dw:1432759935107:dw|
anonymous
  • anonymous
yeah, i think so
Michele_Laino
  • Michele_Laino
Now, we can apply the law of sines, so we can write: \[\Large \frac{6}{{\sin \left( {89^\circ 50'} \right)}} = \frac{{AC}}{{\sin \left( {37^\circ 20'} \right)}}\]
Michele_Laino
  • Michele_Laino
so: \[\Large AC = \frac{{6 \times \sin \left( {37^\circ 20'} \right)}}{{\sin \left( {89^\circ 50'} \right)}} = ...?\]
Michele_Laino
  • Michele_Laino
what is AC?
anonymous
  • anonymous
=3.6
Michele_Laino
  • Michele_Laino
hint: we can write this: 37° 20'= 37.33 degrees 89°50'=89.83 degrees
Michele_Laino
  • Michele_Laino
I got AC=3.638 miles
Michele_Laino
  • Michele_Laino
I think your answer is correct!
Michele_Laino
  • Michele_Laino
Now, similarly, using the law of sines again, we can write: \[\Large \frac{6}{{\sin \left( {89^\circ 50'} \right)}} = \frac{{BC}}{{\sin \left( {62^\circ 50'} \right)}}\]
Michele_Laino
  • Michele_Laino
so we get: \[\Large BC = \frac{{6 \times \sin \left( {62^\circ 50'} \right)}}{{\sin \left( {89^\circ 50'} \right)}} = ...?\] So, what is BC?
Michele_Laino
  • Michele_Laino
hint: 62° 50'=62.83 degrees 89° 50'=89.83 degrees
anonymous
  • anonymous
5.3
Michele_Laino
  • Michele_Laino
that's right! I got 5.338 miles

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