https://file.aiquickdraw.com/mathbot/file/1728556477918gqo5zt5c.jpg

Let's break down the problem step by step.

### Information from the problem:
1. **Bearing of X from P** is \( 023^\circ \).
2. **Distance from X to Y** is 95 km (X is due north of Y).
3. **Bearing of Y from P** is \( 113^\circ \).

### Let's solve each part:

#### i) **Draw a diagram showing the above information:**

- **Point P** is the origin. 
- From P, draw a line at a **bearing of \(023^\circ\)**. This line represents the direction of point **X** from **P**.
- **Y** is located 95 km due south of **X** (since X is due north of Y). 
- The bearing of **Y** from **P** is \(113^\circ\). This means from P, another line is drawn at \(113^\circ\), which points to Y.

(If you'd like, I can sketch the diagram to better visualize the relationships between the points.)

#### ii) **Find the size of \( \angle XPY \):**

To find \( \angle XPY \), subtract the bearing of X from the bearing of Y:

\[
\angle XPY = 113^\circ - 23^\circ = 90^\circ
\]

Thus, \( \angle XPY = 90^\circ \). This indicates that triangle \( \Delta XPY \) is a right-angled triangle.

#### iii) **Find the distance between X and P, to the nearest kilometre:**

Now, since \( \Delta XPY \) is a right-angled triangle, we can use trigonometry to find the distance \( PX \).

Given:
- \( XY = 95 \) km (as X is 95 km north of Y),
- \( \angle XPY = 90^\circ \).

We can calculate \( PX \) using the sine or cosine rule, but here it's simpler to use Pythagoras' theorem because it's a right triangle.

\[
PX = XY \cdot \cos(90^\circ) = 95 \, \text{km}
\]

Thus, the distance between **X** and **P** is **95 km**.

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Do you want any further details or have questions? Below are some relative questions:

1. How can bearings be applied in navigational problems?
2. What is the significance of bearings in geometrical constructions?
3. Can we calculate distances if the bearing angles differ from 90°?
4. What trigonometric functions are commonly used in solving bearing problems?
5. How can Pythagoras’ theorem be extended to 3D navigational problems?

**Tip:** When solving bearing problems, always ensure that bearings are measured clockwise from the north to avoid confusion.

The bearing of X from P is 023°. X is 95 km due North of Y, which is on a bearing of 113° from P. Find the size of ∠XPY and the distance between X and P.

This problem involves calculating bearings between points X, P, and Y, and using Pythagoras' theorem to find the distance between X and P. Learn how to apply bearings and right-angle trigonometry to solve navigation problems.

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