Two tracking stations are on the equator 158 miles apart. A weather balloon is located on a bearing of N42°E from the western station 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? *

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

it would open into a square depending on the type of envelope i’m not sure about the second part