Two tracking stations are on the equator 158 miles apart. A weather balloon is located on a bearing of N42°E from the western st
ation and on a bearing of N14°W from the eastern station. To the nearest tenth of a mile, how far is the balloon from the western station? *
2 answers:
Answer:
199.38
Step-by-step explanation:
To better understand the problem I leave an associated graph.
We must calculate the distance of AC, we know AB
AB = 158 miles.
Now we will apply the law of sine where:
AB / sin (° ACB) = AC / sin (° ABC)
We know the ° ABC = 14 °, the statement gives us the ° CAB = 90 ° - 38 ° = 52 °
Now assuming that within a triangle all the angles must add up to 180 °, then
° ACB = 180 ° - 52 ° - 14 ° = 114 °
Solving for AC:
AC = (AB * sin (° ABC)) / sin (° ACB)
AC = (158 * sin (14 °)) / sin (114 °)
AC = 199.38
Therefore the distance between the balloon from the western station is 199.38 miles
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Step-by-step explanation:
<u>Point-slope form:</u>
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Answer:
well this is the answer
Step-by-step explanation:
