Two hikers leave their camp (P) at the same time. Hiker A walks due east for 9 km and Hiker B walks due south for 9.5 km.
a) How far apart are the two hikers at this point, correct to 2 decimal places?
b) What is the bearing of Hiker A from Hiker B, correct to the nearest degree?
c) What is the bearing of Hiker B from Hiker A, correct to the nearest degree?
d) Suppose Hiker B now heads off on a true bearing of 150 ° for 11 km. How far east would he be from his starting point?
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### (a) Distance Between the Two Hikers

The hikers' movements create a right triangle, where:
- Hiker A is at \((9, 0)\).
- Hiker B is at \((0, -9.5)\).

The distance \(d\) between them can be calculated using the Pythagorean theorem:

\[
d = \sqrt{(9 - 0)^2 + (0 - (-9.5))^2}
\]
\[
d = \sqrt{9^2 + 9.5^2}
\]
\[
d = \sqrt{81 + 90.25}
\]
\[
d = \sqrt{171.25}
\]
\[
d \approx 13.09 \text{ km}
\]

So, the distance between the two hikers is approximately **13.09 km**.

### (b) Bearing of Hiker A from Hiker B

The bearing from Hiker B to Hiker A can be found using the inverse tangent function:

\[
\theta = \tan^{-1}\left(\frac{9}{9.5}\right)
\]
\[
\theta \approx \tan^{-1}(0.947) \approx 43.6^\circ
\]

Since the bearing is measured clockwise from the north, the bearing of Hiker A from Hiker B is:

\[
360^\circ - 43.6^\circ \approx 316^\circ
\]

Therefore, the bearing of Hiker A from Hiker B is approximately **316°**.

### (c) Bearing of Hiker B from Hiker A

Similarly, the bearing from Hiker A to Hiker B can be found using the inverse tangent function:

\[
\theta = \tan^{-1}\left(\frac{9.5}{9}\right)
\]
\[
\theta \approx \tan^{-1}(1.056) \approx 46.4^\circ
\]

Since the bearing is measured clockwise from the north, the bearing of Hiker B from Hiker A is:

\[
180^\circ + 46.4^\circ = 226^\circ
\]

Therefore, the bearing of Hiker B from Hiker A is approximately **226°**.

### (d) Eastward Distance from the Starting Point for Hiker B

Hiker B now walks 11 km on a true bearing of \(150^\circ\). We can break this distance into its eastward (\(x\)) and southward (\(y\)) components:

\[
x = 11 \times \sin(150^\circ)
\]
\[
x = 11 \times 0.5 = 5.5 \text{ km}
\]

So, Hiker B would be **5.5 km east** of his starting point.

### Summary

1. **Distance between the hikers**: 13.09 km.
2. **Bearing of Hiker A from Hiker B**: 316°.
3. **Bearing of Hiker B from Hiker A**: 226°.
4. **Eastward distance for Hiker B**: 5.5 km.

---

Would you like more details on any part of the solution or have additional questions?

Here are some related questions:

1. How would the distance change if both hikers traveled 12 km instead?
2. If both hikers change their directions by 45°, what would be their new coordinates?
3. How would the bearing calculations change if the angles were measured counterclockwise?
4. If Hiker A walked another 5 km east, what would be the new distance between them?
5. How would the solution change if the Earth’s curvature is considered?

**Tip**: For non-right triangles, you can use the law of cosines to find the distance between two points.

Two hikers leave their camp (P) at the same time. Hiker A walks due east for 9 km and Hiker B walks due south for 9.5 km. a) How far apart are the two hikers at this point, correct to 2 decimal places? b) What is the bearing of Hiker A from Hiker B, correct to the nearest degree? c) What is the bearing of Hiker B from Hiker A, correct to the nearest degree? d) Suppose Hiker B now heads off on a true bearing of 150° for 11 km. How far east would he be from his starting point?

Solve a geometric problem involving two hikers walking in different directions. Learn how to calculate their distance using the Pythagorean theorem, find bearings using inverse tangent, and determine eastward displacement using trigonometry. Suitable for students in Grades 9-12.

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