1) A boat leaves a dock al point A and travels for a distance of 15 km to point B on a bearing of $$ 135^{\circ} $$. The boat then changes comse and travels for a distance of 8 km to a point C on a bearing of $$ 060^{\circ} $$. a) Illustrate the above infomation in a cleals labelled diagram. The diagtam should show the (i) north direction (ii) bearing $$ 135^{\circ} $$ and $$ 060^{\circ} $$ (iii) distances 8 km and 15 km b) Calculate (i) The distance $$ A C $$ (ii) $$ \angle B C A $$ (iii) The bearing of $$ A $$ lrom $$ C $$.

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UpStudy AI · Accepted Answer

Sure, let's break down the problem step by step.

## Part a) Illustration

While I can't draw a diagram directly here, I can describe how to construct it clearly:

1. **Draw the North Direction:**
   - Start by drawing a vertical line pointing upwards labeled "North."

2. **Plot Point A:**
   - Mark a point labeled **A** at the intersection of your coordinate system.

3. **Plot Point B:**
   - From point **A**, draw a line at a **bearing of $$135^{\circ}$$**. 
   - **Bearing $$135^{\circ}$$** means:
     - Start from the north direction and rotate **135 degrees clockwise**.
     - This points towards the **southeast** direction.
   - The distance from **A to B** is **15 km**. Mark point **B** at the end of this line and label the distance.

4. **Plot Point C:**
   - From point **B**, draw a line at a **bearing of $$060^{\circ}$$**.
   - **Bearing $$060^{\circ}$$** means:
     - Start from the north direction and rotate **60 degrees clockwise**.
     - This points towards the **northeast** direction.
   - The distance from **B to C** is **8 km**. Mark point **C** at the end of this line and label the distance.

5. **Label Directions and Bearings:**
   - Clearly indicate the north direction.
   - Mark the angles $$135^{\circ}$$ and $$060^{\circ}$$ from the north.
   - Ensure all distances (15 km and 8 km) are labeled.

**Visual Representation:**

```
N
|
|
|\
| \ 60°
|  \  
|   \  
A----B----C
 \135°
```

- **A** is the starting point.
- **AB** is 15 km at a $$135^{\circ}$$ bearing.
- **BC** is 8 km at a $$060^{\circ}$$ bearing.
- **N** indicates the north direction.

*Note: For an accurate diagram, it's recommended to use graph paper or drawing software to scale the bearings and distances appropriately.*

## Part b) Calculations

### Given:
- **AB = 15 km** at $$135^{\circ}$$ bearing.
- **BC = 8 km** at $$060^{\circ}$$ bearing.

We'll use coordinate geometry to solve the problem.

#### Step 1: Assign Coordinates
- Let’s place point **A** at the origin $$(0, 0)$$.
- **North** is the positive y-direction, and **East** is the positive x-direction.

#### Step 2: Calculate Coordinates of Point B
- Bearing $$135^{\circ}$$ corresponds to:
  - **East component (x):** $$15 \times \sin(135^{\circ}) = 15 \times \frac{\sqrt{2}}{2} \approx 10.606$$ km East
  - **North component (y):** $$15 \times \cos(135^{\circ}) = 15 \times (-\frac{\sqrt{2}}{2}) \approx -10.606$$ km South
- **Coordinates of B:** $$(10.606, -10.606)$$ km

#### Step 3: Calculate Coordinates of Point C
- Bearing $$060^{\circ}$$ corresponds to:
  - **East component (x):** $$8 \times \sin(60^{\circ}) = 8 \times \frac{\sqrt{3}}{2} \approx 6.928$$ km East
  - **North component (y):** $$8 \times \cos(60^{\circ}) = 8 \times 0.5 = 4$$ km North
- **Coordinates of C relative to B:** $$(6.928, 4)$$ km
- **Absolute Coordinates of C:** 
  - $$x = 10.606 + 6.928 \approx 17.534$$ km East
  - $$y = -10.606 + 4 = -6.606$$ km South
  - **C:** $$(17.534, -6.606)$$ km

### b)(i) Calculate the Distance $$AC$$

Using the distance formula:
$$
AC = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2} = \sqrt{(17.534 - 0)^2 + (-6.606 - 0)^2}
$$
$$
AC = \sqrt{17.534^2 + 6.606^2} = \sqrt{307.3 + 43.69} = \sqrt{350.99} \approx 18.73 \text{ km}
$$

**Answer:** $$AC \approx 18.73$$ km

### b)(ii) Calculate $$\angle BCA$$

We have triangle $$ABC$$ with sides:
- $$AB = 15$$ km
- $$BC = 8$$ km
- $$AC \approx 18.73$$ km

Using the **Law of Cosines** to find $$\angle BCA$$ (angle at C):

$$
\cos(\angle BCA) = \frac{AC^2 + BC^2 - AB^2}{2 \times AC \times BC}
$$
$$
\cos(\angle BCA) = \frac{18.73^2 + 8^2 - 15^2}{2 \times 18.73 \times 8}
$$
$$
\cos(\angle BCA) = \frac{350.99 + 64 - 225}{299.68} = \frac{189.99}{299.68} \approx 0.6335
$$
$$
\angle BCA \approx \cos^{-1}(0.6335) \approx 50^\circ
$$

**Answer:** $$\angle BCA \approx 50^\circ$$

### b)(iii) Determine the Bearing of $$A$$ from $$C$$

To find the **bearing from C to A**, follow these steps:

1. **Determine the Vector from C to A:**
   - **East component:** $$0 - 17.534 = -17.534$$ km (West)
   - **North component:** $$0 - (-6.606) = 6.606$$ km (North)

2. **Calculate the Bearing Angle:**
   - **Tan of the angle:** $$\tan(\theta) = \frac{\text{West component}}{\text{North component}} = \frac{17.534}{6.606} \approx 2.655$$
   - **Angle:** $$\theta = \tan^{-1}(2.655) \approx 69^\circ$$

3. **Interpret the Bearing:**
   - Since the direction is **West** of **North**, the bearing is written as **N69°W**.
   - Alternatively, bearings can be expressed in degrees clockwise from North:
     $$
     \text{Bearing} = 360^\circ - 69^\circ = 291^\circ
     $$

**Answer:** The bearing of $$A$$ from $$C$$ is **N69°W** or **291°**.

---

**Summary of Answers:**

b)
1. The distance $$AC \approx 18.73$$ km.
2. $$\angle BCA \approx 50^\circ$$.
3. The bearing of $$A$$ from $$C$$ is **N69°W** or **291°**.
**Part a) Illustration:** - Draw a vertical line labeled "North." - Plot point **A** at the origin. - From **A**, draw a line at $$135^{\circ}$$ bearing (southeast) for 15 km to point **B**. - From **B**, draw a line at $$060^{\circ}$$ bearing (northeast) for 8 km to point **C**. - Label all directions and distances clearly. **Part b) Calculations:** 1. **Distance $$AC \approx 18.73$$ km. 2. **Angle $$BCA \approx 50^\circ$$. 3. **Bearing of $$A$$ from $$C$$ is N69°W or 291°.

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