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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
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino if you're not busy can you help me again. I don't remember if I asked you this one

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1N27degrees10'E means that I'm heading towards North, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, you're in the North and heading East

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1like this? dw:1432759303858:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes I have that part

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so the complete drawing is: dw:1432759443568:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1sorry: dw:1432759585590:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1is my drawing right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1your exercise asks for distances AC and BC, such that: dw:1432759724446:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1432759935107:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1Now, 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
 one year ago
Best ResponseYou've already chosen the best response.1so: \[\Large AC = \frac{{6 \times \sin \left( {37^\circ 20'} \right)}}{{\sin \left( {89^\circ 50'} \right)}} = ...?\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1hint: we can write this: 37° 20'= 37.33 degrees 89°50'=89.83 degrees

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I got AC=3.638 miles

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think your answer is correct!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1Now, 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
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.1hint: 62° 50'=62.83 degrees 89° 50'=89.83 degrees

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1that's right! I got 5.338 miles
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