A person stands at a point A due south of a tower and observes his elevation is `60^(@)`. He then walks westwards towards B where the elevation is `45^(@)`. At a point C on AB produced he finds it to be `30^(@)`, then AB=2BC.

Answer

To solve the problem step by step, we will analyze the situation involving the tower and the angles of elevation from points A, B, and C.

### Step 1: Understand the Setup
- Let the height of the tower be \( H \).
- Point A is due south of the tower.
- The angle of elevation from point A to the top of the tower is \( 60^\circ \).
- The angle of elevation from point B (which is west of A) to the top of the tower is \( 45^\circ \).
- The angle of elevation from point C (further west from B) to the top of the tower is \( 30^\circ \).
- We know that \( AB = 2BC \).

### Step 2: Set Up the Relationships
1. **From Point A**:
- Using the angle of elevation \( 60^\circ \):
\[ \tan(60^\circ) = \frac{H}{d_A} \]
where \( d_A \) is the distance from A to the base of the tower.
\[ \sqrt{3} = \frac{H}{d_A} \implies d_A = \frac{H}{\sqrt{3}} \]

2. **From Point B**:
- Using the angle of elevation \( 45^\circ \):
\[ \tan(45^\circ) = \frac{H}{d_B} \]
where \( d_B \) is the distance from B to the base of the tower.
\[ 1 = \frac{H}{d_B} \implies d_B = H \]

3. **From Point C**:
- Using the angle of elevation \( 30^\circ \):
\[ \tan(30^\circ) = \frac{H}{d_C} \]
where \( d_C \) is the distance from C to the base of the tower.
\[ \frac{1}{\sqrt{3}} = \frac{H}{d_C} \implies d_C = H \sqrt{3} \]

### Step 3: Relate the Distances
- The distances between the points are related as follows:
- \( AB = d_B - d_A \)
- \( BC = d_C - d_B \)

### Step 4: Express AB and BC
1. **Calculate \( AB \)**:
\[ AB = d_B - d_A = H - \frac{H}{\sqrt{3}} = H \left(1 - \frac{1}{\sqrt{3}}\right) = H \left(\frac{\sqrt{3} - 1}{\sqrt{3}}\right) \]

2. **Calculate \( BC \)**:
\[ BC = d_C - d_B = H \sqrt{3} - H = H(\sqrt{3} - 1) \]

### Step 5: Use the Given Condition
According to the problem, \( AB = 2BC \):
\[ H \left(\frac{\sqrt{3} - 1}{\sqrt{3}}\right) = 2 \cdot H(\sqrt{3} - 1) \]

### Step 6: Simplify the Equation
1. Cancel \( H \) (assuming \( H \neq 0 \)):
\[ \frac{\sqrt{3} - 1}{\sqrt{3}} = 2(\sqrt{3} - 1) \]
2. Cross-multiply:
\[ \sqrt{3} - 1 = 2\sqrt{3}(\sqrt{3} - 1) \]
3. Simplify:
\[ \sqrt{3} - 1 = 2(3 - \sqrt{3}) \implies \sqrt{3} - 1 = 6 - 2\sqrt{3} \]
4. Rearranging gives:
\[ 3\sqrt{3} = 7 \implies \sqrt{3} = \frac{7}{3} \]

### Conclusion
The relationships derived from the angles of elevation and the distances between points A, B, and C lead us to the conclusion that the conditions given in the problem hold true.