The bearing from A to C is S 51​° E. The bearing from A to B is N 88​° E. The bearing from B to C is S 39​° W. A plane flying at 250 mph takes 3.6 hr to go from A to B. Find the distance from A to C.


To solve this problem, we can use the law of sines and some trigonometric identities based on the given bearings. Here's a step-by-step approach:

### Step 1: Analyze the Bearings
- The bearing from A to C is \( S 51^\circ E \), which means the angle measured clockwise from the south is \( 51^\circ \).
- The bearing from A to B is \( N 88^\circ E \), which means the angle measured clockwise from the north is \( 88^\circ \).
- The bearing from B to C is \( S 39^\circ W \), which means the angle measured clockwise from the south is \( 39^\circ \).

### Step 2: Calculate the Angle \( \angle BAC \)
To find the angle \( \angle BAC \), we consider the difference in direction between the bearings from A to C and from A to B:
- Bearing \( N 88^\circ E \) from A to B corresponds to an angle of \( 88^\circ \) clockwise from the north.
- Bearing \( S 51^\circ E \) from A to C corresponds to an angle of \( 180^\circ - 51^\circ = 129^\circ \) clockwise from the north.

So, the angle \( \angle BAC = 129^\circ - 88^\circ = 41^\circ \).

### Step 3: Calculate the Angle \( \angle ABC \)
Now, calculate the angle \( \angle ABC \) between the directions from A to B and B to C:
- Bearing \( S 39^\circ W \) from B to C corresponds to an angle of \( 180^\circ + 39^\circ = 219^\circ \) clockwise from the north.
- Bearing \( N 88^\circ E \) from A to B corresponds to an angle of \( 88^\circ \) clockwise from the north.

So, the angle \( \angle ABC = 219^\circ - 88^\circ = 131^\circ \).

### Step 4: Calculate the Remaining Angle \( \angle BCA \)
Since the sum of the angles in a triangle is \( 180^\circ \), the remaining angle \( \angle BCA \) is:
\[
\angle BCA = 180^\circ - \angle BAC - \angle ABC = 180^\circ - 41^\circ - 131^\circ = 8^\circ
\]

### Step 5: Calculate the Distance from A to B
The plane travels from A to B at 250 mph for 3.6 hours. Therefore, the distance from A to B is:
\[
AB = 250 \text{ mph} \times 3.6 \text{ hours} = 900 \text{ miles}
\]

### Step 6: Use the Law of Sines to Find AC
Using the Law of Sines:
\[
\frac{AC}{\sin(\angle ABC)} = \frac{AB}{\sin(\angle BCA)}
\]
Substitute the known values:
\[
\frac{AC}{\sin(131^\circ)} = \frac{900}{\sin(8^\circ)}
\]
\[
AC = \frac{900 \times \sin(131^\circ)}{\sin(8^\circ)}
\]
Now, calculate the values:
\[
AC \approx \frac{900 \times 0.7547}{0.1392} \approx \frac{679.23}{0.1392} \approx 4879.25 \text{ miles}
\]

Thus, the distance from A to C is approximately **4879.25 miles**.

Would you like more details, or do you have any other questions?

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Here are 5 related questions:
1. How would the calculation change if the plane's speed varied during the flight?
2. Can the law of sines be used to find the other distances in the triangle? How?
3. How would you determine the distance if the bearings were given in a different format?
4. What is the significance of bearings in navigation, and how are they typically measured?
5. What trigonometric laws could be used if the problem involved an obtuse triangle?

**Tip:** Always double-check the bearings and angles when working with navigation problems, as small errors can lead to significant inaccuracies.

Learn how to solve a navigation problem involving bearings and distances using trigonometric principles. This problem covers the application of the Law of Sines and trigonometric identities to determine distances between points based on given bearings. Suitable for advanced high school students and anyone interested in navigation mathematics.

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