We can solve this problem by using the Law of Sines and applying it to the triangle formed by the two tracking stations and the weather balloon.

Step 1: Understand the situation

We are given:

• Two tracking stations located 134 miles apart on the equator.
• The bearing from the western station to the weather balloon is $N 38^\circ E$.
• The bearing from the eastern station to the weather balloon is $N 19^\circ E$.

Step 2: Calculate the angle between the two bearings

The total angle between the two bearings will be the difference between $38^\circ$ and $19^\circ$, since they are measured from the north. This means: $\text{Angle between bearings} = 38^\circ - 19^\circ = 19^\circ$ This is the angle at the western station.

Step 3: Apply the Law of Sines

Now we have a triangle where:

• The distance between the stations (base) is 134 miles.
• The angle between the two bearings (at the western station) is $19^\circ$.
• We need to find the distance from the western station to the balloon.

Let the distance from the western station to the balloon be $x$, and the distance from the eastern station to the balloon be $y$.

From the Law of Sines, we have: $\frac{x}{\sin(19^\circ)} = \frac{134}{\sin(180^\circ - 19^\circ)}$ Since $\sin(180^\circ - 19^\circ) = \sin(19^\circ)$, this simplifies to: $\frac{x}{\sin(19^\circ)} = \frac{134}{\sin(161^\circ)}$

Now, we can solve for $x$: $x = \frac{134 \cdot \sin(19^\circ)}{\sin(161^\circ)}$

Using a calculator to find the sine values: $\sin(19^\circ) \approx 0.32557, \quad \sin(161^\circ) \approx 0.30902$

Thus: $x = \frac{134 \cdot 0.32557}{0.30902} \approx \frac{43.579}{0.30902} \approx 141.5 \, \text{miles}$

Final Answer:

The weather balloon is approximately 142 miles from the western station, rounding to the nearest mile.

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Would you like a detailed breakdown of any step or further questions on this problem?

Here are some related questions you might find interesting:

1. How can the Law of Sines be applied in non-right triangles?
2. What other methods could be used to solve this problem, such as using vectors or coordinate geometry?
3. How would the angle change if the balloon's bearings were different (e.g., N 45°E and N 30°E)?
4. How would this problem change if the stations were not on the equator, but at different latitudes?
5. What other real-world applications involve using bearings and distances to find unknown positions?

Tip: When solving for distances and angles in non-right triangles, the Law of Sines is a powerful tool when you know either two angles and one side or two sides and one angle.