Two fire spotter towers, Tower A (western tower) and Tower B (eastern tower), are 115115 miles apart on an east-west line. From Tower A, a fire is spotted at a bearing of N, 42, to the power o , EN42 
o
 E . From Tower B, the same fire is spotted from a bearing of N, 15, to the power o , WN15 
o
 W . Find the distance from Tower A to the fire. Round your answer to the nearest tenth of a mile.

This problem involves determining the distance from Tower A to the fire using the Law of Sines. The scenario involves two towers, A and B, located along an east-west line, and each tower spots a fire at specific bearings.

### Step 1: Interpret the bearings
- **From Tower A**: The fire is spotted at a bearing of N 42° E. This means the angle between the north-south line and the line to the fire is 42° eastward.
- **From Tower B**: The fire is spotted at a bearing of N 15° W. This means the angle between the north-south line and the line to the fire is 15° westward.

Since the towers are 115 miles apart on an east-west line, we need to create a triangle where:
- The distance between the towers is the base of the triangle (115 miles),
- The angles of observation create two angles inside the triangle.

### Step 2: Calculate the interior angles of the triangle
The bearings from both towers give us two angles:
- **Angle at Tower A (∠A)**: Since the fire is located 42° east of north, the interior angle at Tower A with respect to the east-west line is 90° - 42° = 48°.
- **Angle at Tower B (∠B)**: Since the fire is located 15° west of north, the interior angle at Tower B with respect to the east-west line is 90° - 15° = 75°.

### Step 3: Find the third angle of the triangle
The sum of the angles in any triangle is 180°. Let ∠C be the angle opposite the 115-mile side between Tower A and Tower B.

\[
∠C = 180° - 48° - 75° = 57°
\]

### Step 4: Apply the Law of Sines
The Law of Sines states:

\[
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
\]

Here:
- \(a\) is the distance from Tower A to the fire (what we're solving for),
- \(b\) is the distance from Tower B to the fire,
- \(c = 115\) miles is the distance between the two towers,
- ∠A = 48°, ∠B = 75°, and ∠C = 57°.

Now, using the Law of Sines:

\[
\frac{115}{\sin 57°} = \frac{a}{\sin 75°}
\]

### Step 5: Solve for \(a\)
First, find the sine values:
- \(\sin 57° \approx 0.8387\)
- \(\sin 75° \approx 0.9659\)

Now substitute the values:

\[
\frac{115}{0.8387} = \frac{a}{0.9659}
\]

\[
a = \frac{115 \times 0.9659}{0.8387}
\]

\[
a \approx 132.5 \, \text{miles}
\]

### Final Answer:
The distance from Tower A to the fire is approximately **132.5 miles**.

Would you like more details or have any questions?

Here are 5 related questions:
1. How would the solution change if the distance between the towers was different?
2. Can we apply the Law of Cosines instead of the Law of Sines in this problem?
3. How would the problem change if the bearings were reversed?
4. What if the bearings were given in radians instead of degrees?
5. How can this problem be extended to three towers instead of two?

**Tip:** When working with bearings and angles in trigonometry, always ensure you understand the relationship between the direction and the north-south line, as small misinterpretations can lead to significant errors in calculations.

Two fire spotter towers, Tower A (western tower) and Tower B (eastern tower), are 115 miles apart on an east-west line. From Tower A, a fire is spotted at a bearing of N 42° E. From Tower B, the same fire is spotted from a bearing of N 15° W. Find the distance from Tower A to the fire. Round your answer to the nearest tenth of a mile.

This problem involves calculating the distance from Tower A to a fire using trigonometry. The towers are 115 miles apart, and bearings from both towers are given. The Law of Sines and triangle geometry are applied to find the solution, suitable for students in Grades 10-12.

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