two fire towers A and B are 30 kilometers apart. the bearing from A to B is N 65 degrees E. A fire is spotted by a ranger in each tower, and its bearing from A and B are N 28 degrees E and N 16.5 degrees W respectively. Find the distance of the fire from each tower.

This problem can be solved using trigonometry, specifically the Law of Sines. We can model this situation as a triangle with the fire and the two towers as the vertices.

### Step 1: Diagram and labeling
- Let \( A \) and \( B \) represent the two fire towers.
- The distance between \( A \) and \( B \) is 30 km.
- The fire is located at point \( F \), and we need to find the distances \( AF \) and \( BF \).

**Bearings given:**
- The bearing from \( A \) to \( B \) is \( N 65^\circ E \), meaning the angle between north and the line joining \( A \) and \( B \) is \( 65^\circ \) clockwise from the north.
- The bearing from \( A \) to the fire is \( N 28^\circ E \), which makes an angle of \( 28^\circ \) clockwise from the north.
- The bearing from \( B \) to the fire is \( N 16.5^\circ W \), which makes an angle of \( 16.5^\circ \) counterclockwise from the north.

### Step 2: Angles in the triangle
To apply the Law of Sines, we need to calculate all the angles of the triangle formed by points \( A \), \( B \), and \( F \).

- The interior angle at \( A \) is the difference between the bearings at \( A \). Since the bearing from \( A \) to \( B \) is \( 65^\circ \), and from \( A \) to the fire is \( 28^\circ \), the angle \( \angle A \) in the triangle is:
  \[
  \angle A = 65^\circ - 28^\circ = 37^\circ
  \]

- The interior angle at \( B \) is the sum of the bearing from \( B \) to the fire (since it is counterclockwise) and the bearing from \( A \) to \( B \). So, the angle \( \angle B \) in the triangle is:
  \[
  \angle B = 65^\circ + 16.5^\circ = 81.5^\circ
  \]

- The third angle \( \angle F \) is found by subtracting the sum of \( \angle A \) and \( \angle B \) from \( 180^\circ \):
  \[
  \angle F = 180^\circ - (37^\circ + 81.5^\circ) = 61.5^\circ
  \]

### Step 3: Applying the Law of Sines
The Law of Sines states:
\[
\frac{AF}{\sin(\angle B)} = \frac{BF}{\sin(\angle A)} = \frac{AB}{\sin(\angle F)}
\]
Where:
- \( AB = 30 \, \text{km} \)
- \( \angle A = 37^\circ \)
- \( \angle B = 81.5^\circ \)
- \( \angle F = 61.5^\circ \)

Now, we can solve for \( AF \) and \( BF \):
\[
\frac{AF}{\sin(81.5^\circ)} = \frac{30}{\sin(61.5^\circ)}
\]
Solving for \( AF \):
\[
AF = \frac{30 \cdot \sin(81.5^\circ)}{\sin(61.5^\circ)}
\]
Next, solve for \( BF \):
\[
BF = \frac{30 \cdot \sin(37^\circ)}{\sin(61.5^\circ)}
\]

Let me calculate these values for you.The distances of the fire from the two towers are:

- The distance from Tower \( A \) to the fire (\( AF \)) is approximately \( 33.76 \, \text{km} \).
- The distance from Tower \( B \) to the fire (\( BF \)) is approximately \( 20.54 \, \text{km} \).

Would you like further details on the calculation process, or have any other questions?

Here are 5 related questions:
1. How does the Law of Sines apply to solving triangles with bearings?
2. What is the significance of bearing angles in navigation problems?
3. Can the Law of Cosines also be used to solve this problem? If so, how?
4. How can one use triangulation to locate a point on a map?
5. How would the solution change if the distance between towers \( A \) and \( B \) was different?

**Tip:** When solving problems involving bearings, it’s helpful to draw a diagram to visualize the angles clearly.

Two fire towers A and B are 30 kilometers apart. The bearing from A to B is N 65 degrees E. A fire is spotted by a ranger in each tower, and its bearing from A and B are N 28 degrees E and N 16.5 degrees W respectively. Find the distance of the fire from each tower.

Solve a problem involving two fire towers A and B, 30 km apart, where bearings to a fire are given as N 28° E and N 16.5° W. Learn how to apply the Law of Sines to find the distances from each tower to the fire. This problem is suitable for students in Grades 10-12 studying trigonometry.

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