draw a neat diagram for each question. 

a) martin walked from home (H) to the local store (S) on a bearing of 035 degrees. he then wlaked on a bearing of 125 degrees to the park ( P) which is 450m due east of his home.

i) show that angle HSP = 90 degrees

ii ) find the distance between martin's home and the store (1dp)

Let's break this problem into parts, then I'll explain the steps to create the diagrams for each part of the question and solve the problem.

### Step 1: Understanding the Problem

- **Home (H)** is the starting point.
- **Store (S)** is at a bearing of \( 035^\circ \) from \( H \).
- **Park (P)** is at a bearing of \( 125^\circ \) from \( S \) and is 450 meters due east of \( H \).

### Step 2: Diagram Explanation

We'll create two diagrams:
1. **Diagram 1** showing the positions of \( H \), \( S \), and \( P \) based on the given bearings.
2. **Diagram 2** focusing on the right triangle formed to show \( \angle HSP \) and calculate the distance.

### Step 3: Solve the Problem

#### **Part (i): Show that \( \angle HSP = 90^\circ \)**

Given:
- The bearing of \( S \) from \( H \) is \( 035^\circ \), so the angle \( \angle SHP = 35^\circ \) from the north.
- The bearing of \( P \) from \( S \) is \( 125^\circ \), which means that from the north at \( S \), the direction to \( P \) is \( 125^\circ - 035^\circ = 90^\circ \).

This indicates that \( \angle HSP = 90^\circ \), as the angle between the two bearings is a right angle.

#### **Part (ii): Find the Distance Between Martin's Home and the Store**

To find the distance \( HS \), we can use trigonometry because we have a right triangle:

1. \( HP = 450 \) meters (since \( P \) is due east of \( H \)).
2. The angle \( \angle HSP = 90^\circ \).

Using the sine rule:

\[
\sin(35^\circ) = \frac{HS}{450}
\]

Solving for \( HS \):

\[
HS = 450 \times \sin(35^\circ) \approx 450 \times 0.5736 = 258.1 \, \text{meters (to 1 decimal place)}
\]

### Diagrams (to be visualized):
1. **Diagram 1**:
   - \( H \) at the origin.
   - \( S \) at a \( 035^\circ \) angle from \( H \) (north is \( 0^\circ \)).
   - \( P \) at a \( 125^\circ \) angle from \( S \).
   - \( P \) is directly east of \( H \) (450m).

2. **Diagram 2**:
   - Triangle \( HSP \) showing the right angle at \( S \).
   - Label the sides with \( HS = 258.1m \) and \( HP = 450m \).

These diagrams will demonstrate the relationship between the points and confirm that \( \angle HSP = 90^\circ \) as well as provide the required distance.

Would you like to see a visual representation of these diagrams?

Learn how to solve a problem involving bearings and distances using trigonometry and the sine rule. This problem demonstrates the calculation of angles and distances in a right triangle scenario. Suitable for students in Grades 9-12.

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