A hiker walks 4.5 km in one direction, then makes a 45˚ turn to the right and walks another 6.4 km. What is the magnitude of her displacement? (Hint: this angle is not inside your Vector triangle, you will need to find the angle used for your equation)

To determine the magnitude of the hiker's displacement, we can break the problem down into the following steps:

### Step 1: Understanding the Problem

- The hiker walks 4.5 km in one direction.
- Then, she makes a 45° turn to the right and walks another 6.4 km.

The magnitude of the hiker's displacement is the straight-line distance from her starting point to her final position.

### Step 2: Recognizing the Triangle Formed

- The two legs of the triangle are 4.5 km and 6.4 km.
- The angle between these two legs is not 45° because the angle inside the triangle is the difference between 180° and 45°. Therefore, the angle inside the triangle is:

\[
\text{Angle inside triangle} = 180° - 45° = 135°
\]

### Step 3: Applying the Law of Cosines

To find the magnitude of her displacement (let's call it \( d \)), we can use the Law of Cosines:

\[
d^2 = a^2 + b^2 - 2ab \cos(C)
\]

Where:
- \( a = 4.5 \) km
- \( b = 6.4 \) km
- \( C = 135^\circ \) (the angle inside the triangle)

Substitute the values:

\[
d^2 = 4.5^2 + 6.4^2 - 2(4.5)(6.4)\cos(135^\circ)
\]

First, calculate the individual components:

\[
d^2 = 20.25 + 40.96 - 2(4.5)(6.4)\cos(135^\circ)
\]

\[
d^2 = 61.21 - 57.6 \cdot \cos(135^\circ)
\]

The cosine of \( 135^\circ \) is \( \cos(135^\circ) = -\frac{\sqrt{2}}{2} \approx -0.7071 \):

\[
d^2 = 61.21 + 57.6 \times 0.7071
\]

\[
d^2 = 61.21 + 40.75 = 101.96
\]

Finally, take the square root to find \( d \):

\[
d \approx \sqrt{101.96} \approx 10.1 \text{ km}
\]

### Final Answer:
The magnitude of the hiker's displacement is approximately **10.1 km**.

Would you like any further details or have any questions?

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Here are 5 related questions to expand your understanding:

1. How would the displacement change if the hiker turned 60° instead of 45°?
2. What if the hiker turned left instead of right? How would that affect the solution?
3. How does the Law of Cosines help in finding the displacement when the angle between vectors is obtuse?
4. If the hiker walked 8 km in the second leg instead of 6.4 km, what would be her new displacement?
5. Can we use the Pythagorean Theorem instead of the Law of Cosines for any specific angle? If yes, which angle?

**Tip:** When solving problems involving angles and vectors, it's crucial to correctly identify the angles inside the triangle, especially when the turn is described relative to the original direction.

Learn how to calculate the magnitude of a hiker's displacement after walking 4.5 km in one direction, making a 45° turn to the right, and walking another 6.4 km. Use the Law of Cosines to find the displacement, involving vectors and trigonometry. Suitable for Grades 9-12.

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