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To solve this problem, let’s analyze each part.

### Part 1: How far from home is she?

1. **First, we break down her path:**
   - She walks **3 miles west**.
   - Then she walks **4 miles southwest**.

2. **Coordinate Setup and Vector Addition:**
   - Let’s assume she starts at the origin \((0, 0)\).
   - Walking 3 miles west means moving in the negative \(x\)-direction, so her position is now \((-3, 0)\).
   - Walking 4 miles southwest implies a movement at a \(45^\circ\) angle south of west. The southwest direction splits her 4-mile walk equally between the \(x\)- and \(y\)-axes.
   
3. **Calculating the Displacement in Southwest Direction:**
   - The displacement components for 4 miles southwest (at a \(45^\circ\) angle) are:
     \[
     x = -4 \cos(45^\circ) = -4 \times \frac{\sqrt{2}}{2} = -2\sqrt{2} \approx -2.83
     \]
     \[
     y = -4 \sin(45^\circ) = -4 \times \frac{\sqrt{2}}{2} = -2\sqrt{2} \approx -2.83
     \]

4. **Finding the Final Coordinates:**
   - Her position after both movements is:
     \[
     x = -3 + (-2.83) = -5.83
     \]
     \[
     y = 0 + (-2.83) = -2.83
     \]

5. **Distance from Home:**
   - The straight-line distance from the origin \((0, 0)\) to \((-5.83, -2.83)\) is:
     \[
     d = \sqrt{(-5.83)^2 + (-2.83)^2}
     \]
     \[
     d = \sqrt{34 + 8} \approx \sqrt{42} \approx 6.48 \text{ miles}
     \]

   So, she is approximately **6.48 miles** from home.

---

### Part 2: In what direction must she walk to head directly home?

1. **Direction Calculation:**
   - To find the direction angle \(\theta\), measured as “degrees North of East,” we calculate the angle of the vector \((-5.83, -2.83)\) relative to the positive \(x\)-axis (east).
   - The tangent of the angle is:
     \[
     \tan(\theta) = \frac{|y|}{|x|} = \frac{2.83}{5.83}
     \]
     \[
     \theta = \arctan\left(\frac{2.83}{5.83}\right) \approx 25.5^\circ
     \]

   Therefore, to head directly home, she should walk **25.5 degrees North of East**.

---

### Summary

- **Distance from home:** approximately **6.48 miles**
- **Direction to walk home:** approximately **25.5 degrees North of East**

Would you like a deeper explanation on any of these steps?

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Here are 5 related questions that expand on this problem:

1. What if she walked 5 miles north instead of southwest; how far would she be from home then?
2. How would the answer change if the angle of the southwest walk was different, say, 30 degrees instead of 45?
3. If she took a different path with multiple turns, how would we calculate her total displacement?
4. How does using trigonometric functions help in finding displacement and direction in navigation problems?
5. What if she walked an additional distance at a different angle; how would we calculate her new position?

**Tip:** Always draw a diagram for navigation problems—it helps visualize distances and angles better!

A person leaves home and walks 3 miles west, then 4 miles southwest. How far from home is she? In what direction must she walk to head directly home?

Learn how to calculate the straight-line distance and direction from the starting point for a two-part displacement problem involving movements west and southwest. This solution includes trigonometric concepts, vector components, and the Pythagorean Theorem, ideal for students in Grades 9-11.

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