A twon B is 12 km south and 18 km west of a town A. Show that the bearing of B from A is `S 56^(@)20'` W. Also, find the distance of B from A.

Explanation

To solve the problem step by step, we will first visualize the situation and then calculate the required distance and bearing.

### Step 1: Visualize the Position of Towns A and B
- Town A is at a reference point (0, 0).
- Town B is located 12 km south and 18 km west of Town A.

### Step 2: Set Up the Coordinate System
- In a Cartesian coordinate system:
- Moving south corresponds to the negative y-direction.
- Moving west corresponds to the negative x-direction.
- Therefore, the coordinates of Town B can be represented as:
- A (0, 0)
- B (-18, -12)

### Step 3: Calculate the Distance from A to B
- We can use the Pythagorean theorem to find the distance \( d \) between points A and B:
\[ d = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(-18 - 0)^2 + (-12 - 0)^2} = \sqrt{(-18)^2 + (-12)^2} = \sqrt{324 + 144} = \sqrt{468} \]
- Now, we simplify \( \sqrt{468} \):
\[ \sqrt{468} = \sqrt{36 \times 13} = 6\sqrt{13} \]
- Approximating \( \sqrt{13} \approx 3.60555 \):
\[ d \approx 6 \times 3.60555 \approx 21.63 \text{ km} \]

### Step 4: Calculate the Bearing of B from A
- The bearing is measured clockwise from the north direction.
- We need to find the angle \( \theta \) using the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{12}{18} = \frac{2}{3} \]
- To find \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{2}{3}\right) \]
- Using a calculator, we find:
\[ \theta \approx 56.31^\circ \]
- Since Town B is located south and west of Town A, the bearing is:
\[ \text{Bearing} = S 56^\circ 20' W \]

### Final Results
- The distance from A to B is approximately \( 21.63 \) km.
- The bearing of B from A is \( S 56^\circ 20' W \).