A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 10 degrees to the right of her original course, and flies 2 h in the new direction. If she maintains a constant speed of 645 mi/h, how far is she from her starting position?  Round to two decimal places, if necessary.

This problem can be solved using the Law of Cosines, as it describes a triangle where two sides and the included angle are known.

### Step 1: Determine the distances traveled
- In the first leg of the journey, the pilot flies for 1 hour 30 minutes, which is equivalent to \( 1.5 \) hours. Given the speed of 645 miles per hour:
  \[
  d_1 = 645 \times 1.5 = 967.5 \text{ miles}
  \]
  
- In the second leg of the journey, the pilot flies for 2 hours at the same speed:
  \[
  d_2 = 645 \times 2 = 1290 \text{ miles}
  \]

### Step 2: Apply the Law of Cosines
The angle between the two legs of the journey is 10°. Using the Law of Cosines to find the distance \( D \) between the pilot’s starting position and her final position:
\[
D^2 = d_1^2 + d_2^2 - 2 \cdot d_1 \cdot d_2 \cdot \cos(\theta)
\]
where:
- \( d_1 = 967.5 \text{ miles} \)
- \( d_2 = 1290 \text{ miles} \)
- \( \theta = 10^\circ \)

Substituting the known values:
\[
D^2 = 967.5^2 + 1290^2 - 2 \cdot 967.5 \cdot 1290 \cdot \cos(10^\circ)
\]

### Step 3: Calculate the result
First, find the squares of \( d_1 \) and \( d_2 \), and the cosine of 10° (in radians):
\[
967.5^2 = 936067.5, \quad 1290^2 = 1664100, \quad \cos(10^\circ) \approx 0.9848
\]

Now substitute:
\[
D^2 = 936067.5 + 1664100 - 2 \cdot 967.5 \cdot 1290 \cdot 0.9848
\]
\[
D^2 = 936067.5 + 1664100 - 2468495.85
\]
\[
D^2 \approx 130671.65
\]

Taking the square root:
\[
D \approx \sqrt{130671.65} \approx 361.53 \text{ miles}
\]

### Final Answer:
The pilot is approximately **361.53 miles** from her starting position.

Would you like more details or have any questions about this process?

Here are five related questions you might find helpful:
1. How would the result change if the course correction was larger than 10°?
2. What would happen to the distance if the pilot flew slower?
3. How does the Law of Cosines relate to the Pythagorean theorem?
4. Can you solve a similar problem where the speed changes during each leg of the journey?
5. How can you visualize this type of problem using a graph or map?

**Tip**: In problems involving non-right triangles, the Law of Cosines is your go-to formula when you know two sides and the angle between them.

A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 10 degrees to the right of her original course, and flies 2 h in the new direction. If she maintains a constant speed of 645 mi/h, how far is she from her starting position? Round to two decimal places, if necessary.

Solve a real-world problem involving a pilot flying two legs of a journey, with a course correction of 10 degrees. Learn how to apply the Law of Cosines to calculate the final distance of 361.53 miles from the starting position. Ideal for students in Grades 10-12.

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