To solve this problem, we need to use trigonometry and the sine and cosine rules. Let's break it down step by step.

1. **Understanding the Bearings and Positions:**
- Town B is on a bearing of 116° from town A.
- Town C is on a bearing of 204° from town B.
- Town C is due south and 23 km from town A.

2. **Visualizing the Problem:**
- Draw a diagram with town A at the origin.
- Draw a line from A to B at an angle of 116° from the north.
- Draw a line from B to C at an angle of 204° from the north.
- Draw a vertical line from A to C, which is 23 km.

3. **Finding the Angles:**
- The bearing of 116° from A to B means the angle between the north direction and line AB is 116°.
- The bearing of 204° from B to C means the angle between the north direction and line BC is 204°.
- Since town C is due south of town A, the angle between the north direction and line AC is 180°.

4. **Calculating the Angle at Town B:**
- The angle between line AB and line AC is 116° - 180° = -64° (which is equivalent to 296° in a positive direction).
- The angle between line BC and line AC is 204° - 180° = 24°.
- Therefore, the angle at town B (angle ABC) is 180° - 24° = 156°.

5. **Using the Sine Rule:**
- We need to find the distance BC.
- In triangle ABC, we know:
- Angle BAC = 116°
- Angle ABC = 156°
- Side AC = 23 km

- Using the sine rule:
$$
\frac{BC}{\sin(116°)} = \frac{23}{\sin(156°)}
$$

6. **Solving for BC:**
- Calculate the sines:
$$
\sin(116°) \approx 0.8829
$$
$$
\sin(156°) \approx 0.4067
$$

- Substitute these values into the sine rule:
$$
\frac{BC}{0.8829} = \frac{23}{0.4067}
$$

- Solve for BC:
$$
BC = \frac{23 \times 0.8829}{0.4067} \approx 49.9 \text{ km}
$$

Therefore, the distance of town B from town C is approximately 49.9 km to 3 significant figures.

Question

To solve this problem, we need to use trigonometry and the sine and cosine rules. Let's break it down step by step.

1. **Understanding the Bearings and Positions:**
   - Town B is on a bearing of 116° from town A.
   - Town C is on a bearing of 204° from town B.
   - Town C is due south and 23 km from town A.

2. **Visualizing the Problem:**
   - Draw a diagram with town A at the origin.
   - Draw a line from A to B at an angle of 116° from the north.
   - Draw a line from B to C at an angle of 204° from the north.
   - Draw a vertical line from A to C, which is 23 km.

3. **Finding the Angles:**
   - The bearing of 116° from A to B means the angle between the north direction and line AB is 116°.
   - The bearing of 204° from B to C means the angle between the north direction and line BC is 204°.
   - Since town C is due south of town A, the angle between the north direction and line AC is 180°.

4. **Calculating the Angle at Town B:**
   - The angle between line AB and line AC is 116° - 180° = -64° (which is equivalent to 296° in a positive direction).
   - The angle between line BC and line AC is 204° - 180° = 24°.
   - Therefore, the angle at town B (angle ABC) is 180° - 24° = 156°.

5. **Using the Sine Rule:**
   - We need to find the distance BC.
   - In triangle ABC, we know:
     - Angle BAC = 116°
     - Angle ABC = 156°
     - Side AC = 23 km

- Using the sine rule:
     $$
     \frac{BC}{\sin(116°)} = \frac{23}{\sin(156°)}
     $$

6. **Solving for BC:**
   - Calculate the sines:
     $$
     \sin(116°) \approx 0.8829
     $$
     $$
     \sin(156°) \approx 0.4067
     $$

- Substitute these values into the sine rule:
     $$
     \frac{BC}{0.8829} = \frac{23}{0.4067}
     $$

- Solve for BC:
     $$
     BC = \frac{23 \times 0.8829}{0.4067} \approx 49.9 \text{ km}
     $$

Therefore, the distance of town B from town C is approximately 49.9 km to 3 significant figures.

Knowee AI · Accepted Answer