An airplane flying at a speed of 350 mi/hr flies from a point A in the direction 140° for 30 minutes and then flies in the direction 230° for 45 minutes. Approximate, to the nearest mile, the distance from the airplane to A.

To solve this problem, we'll break down the airplane's flight into two legs and then use vector addition to find the resulting distance from point A.

### Step 1: Break down the flight into two legs
1. **First leg:**
   - Speed = 350 mi/hr
   - Time = 30 minutes = 0.5 hours
   - Direction = 140°
   
   Distance covered in the first leg:
   \[
   d_1 = 350 \times 0.5 = 175 \text{ miles}
   \]

2. **Second leg:**
   - Speed = 350 mi/hr
   - Time = 45 minutes = 0.75 hours
   - Direction = 230°
   
   Distance covered in the second leg:
   \[
   d_2 = 350 \times 0.75 = 262.5 \text{ miles}
   \]

### Step 2: Break down each leg into its components
We'll use trigonometry to break down these distances into their x (east-west) and y (north-south) components.

1. **First leg (140°):**
   \[
   x_1 = 175 \times \cos(140°)
   \]
   \[
   y_1 = 175 \times \sin(140°)
   \]

2. **Second leg (230°):**
   \[
   x_2 = 262.5 \times \cos(230°)
   \]
   \[
   y_2 = 262.5 \times \sin(230°)
   \]

### Step 3: Sum the components to find the resultant vector
The total x and y components will give us the position of the airplane relative to point A.

\[
x_{\text{total}} = x_1 + x_2
\]
\[
y_{\text{total}} = y_1 + y_2
\]

### Step 4: Calculate the distance from A to the airplane
Finally, use the Pythagorean theorem to find the distance from point A to the airplane.

\[
\text{Distance} = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2}
\]

Let's calculate these values.

### Step 1: Calculate the components
For the first leg:
\[
x_1 = 175 \times \cos(140°) \approx 175 \times (-0.7660) = -134.05 \text{ miles}
\]
\[
y_1 = 175 \times \sin(140°) \approx 175 \times 0.6428 = 112.49 \text{ miles}
\]

For the second leg:
\[
x_2 = 262.5 \times \cos(230°) \approx 262.5 \times (-0.6428) = -168.77 \text{ miles}
\]
\[
y_2 = 262.5 \times \sin(230°) \approx 262.5 \times (-0.7660) = -201.08 \text{ miles}
\]

### Step 2: Sum the components
\[
x_{\text{total}} = -134.05 + (-168.77) = -302.82 \text{ miles}
\]
\[
y_{\text{total}} = 112.49 + (-201.08) = -88.59 \text{ miles}
\]

### Step 3: Calculate the distance
\[
\text{Distance} = \sqrt{(-302.82)^2 + (-88.59)^2} \approx \sqrt{91700 + 7850} \approx \sqrt{99550} \approx 315.47 \text{ miles}
\]

### Final Answer
The distance from the airplane to point A is approximately **315 miles**.

Do you want details or have any questions?

---

1. How do you interpret angles when calculating components in navigation problems?
2. Why is the cosine function used for calculating the x-component of displacement?
3. How would the problem change if the plane flew for different durations at each leg?
4. What is the importance of breaking vectors into components in physics?
5. How can this method be applied to a three-dimensional flight path?

**Tip:** Always ensure your calculator is in the correct mode (degrees or radians) when working with trigonometric functions.

Learn how to calculate the distance of an airplane from point A using trigonometry and vector addition. This problem involves breaking down flight legs, calculating vector components, and applying the Pythagorean theorem. Suitable for high school students studying trigonometry and vector mathematics.

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