an airplane leaves its airport on the east coast of country A and flies in the direction of 83degrees. because of bad weather, it returns to another airport 210km due north of its home base. to get to the new airport, it flies in a direction of 280degrees. what is the total distance that the airplane flew?

To find the total distance that the airplane flew, we need to break down the journey into two parts: the initial flight to the eastern direction and the return flight to the new airport.

1. **Initial Flight**: The airplane flies in the direction of 83 degrees. We can denote this as a vector \( \vec{A} \).

2. **Return Flight**: After flying in the direction of 83 degrees, the airplane then returns and flies 210 km due north to reach a new airport, which we can denote as a vector \( \vec{B} \) with a direction of 280 degrees.

### Step 1: Initial Flight Vector

We need to break down the initial flight vector into its components. Let's denote the distance of the initial flight as \( d_1 \).

The components can be calculated as:
- \( A_x = d_1 \cos(83^\circ) \)
- \( A_y = d_1 \sin(83^\circ) \)

### Step 2: Return Flight Vector

The airplane returns to a new airport that is 210 km north. This flight vector will have:
- \( B_x = 210 \cos(280^\circ) \)
- \( B_y = 210 \sin(280^\circ) \)

### Step 3: Calculate Coordinates

1. **Coordinates for the initial flight** (from the airport):
   - \( A_x = d_1 \cos(83^\circ) \)
   - \( A_y = d_1 \sin(83^\circ) \)

2. **Coordinates for the new airport** (210 km north):
   - The new airport's position will be (0, 210) because it's directly north of the original position.

### Step 4: Total Movement in the x-direction

The total x-coordinate after both flights should equal 0 (since it returns to a point directly above the original airport):
\[ A_x + B_x = 0 \]

### Step 5: Total Movement in the y-direction

The total y-coordinate will be:
\[ A_y + B_y = 210 \]

### Step 6: Solve for \( d_1 \)

From the equations:
1. \( d_1 \cos(83^\circ) + 210 \cos(280^\circ) = 0 \)
2. \( d_1 \sin(83^\circ) + 210 \sin(280^\circ) = 210 \)

Now we can solve these equations step by step. 

### Step 7: Values of Cosine and Sine

- \( \cos(83^\circ) \approx 0.12187 \)
- \( \sin(83^\circ) \approx 0.99255 \)
- \( \cos(280^\circ) \approx 0.17365 \)
- \( \sin(280^\circ) \approx -0.98481 \)

### Step 8: Substitute Values

1. For the x-component:
   \[
   d_1 (0.12187) + 210 (0.17365) = 0
   \]
   \[
   d_1 (0.12187) = -210(0.17365)
   \]
   \[
   d_1 = \frac{-210(0.17365)}{0.12187} \approx 245.77 \text{ km}
   \]

2. For the y-component:
   \[
   d_1 (0.99255) - 210(0.98481) = 210
   \]
   \[
   d_1 (0.99255) = 210 + 210(0.98481)
   \]
   \[
   d_1 = \frac{210 + 210(0.98481)}{0.99255} \approx 211.24 \text{ km}
   \]

### Step 9: Total Distance

Now, the total distance \( D \) that the airplane flew is:
\[
D = d_1 + 210 = 245.77 + 210 \approx 455.77 \text{ km}
\]

### Conclusion

The total distance that the airplane flew is approximately **455.77 km**.

Would you like more details on any part of this solution, or do you have any questions? Here are some follow-up questions:

1. What would happen if the angle of the initial flight changed?
2. How would the total distance change if the return flight was longer?
3. Can you calculate the time taken for each leg of the journey if the speeds were known?
4. What are the implications of flying at different altitudes on this journey?
5. How would wind direction affect the calculations if included? 

**Tip**: Always break down vectors into components when solving problems involving direction and distance.

An airplane leaves its airport on the east coast of country A and flies in the direction of 83 degrees. Because of bad weather, it returns to another airport 210 km due north of its home base. To get to the new airport, it flies in a direction of 280 degrees. What is the total distance that the airplane flew?

Learn how to find the total distance an airplane flew using vector components and trigonometric calculations. This problem involves calculating distances based on directional angles of 83 degrees and 280 degrees. Perfect for students in Grades 10-12 who are exploring vector analysis and trigonometry.

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