A 20 m high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower, as seen from the foot of the pole, is `60^(@)` and the angle of elevation of the top of the pole as seen from the foot of the tower is `30^(@)` . Find :
the horizontal distance between the pole and the tower.

Answer

To solve the problem step by step, we will use trigonometric ratios, particularly the tangent function, which relates the angles of elevation to the opposite and adjacent sides of the right triangles formed by the pole, tower, and the ground.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a vertical pole (CD) of height 20 m.
- We have a vertical tower (AB) with an unknown height (let's call it 'h').
- The angle of elevation to the top of the tower (A) from the foot of the pole (C) is 60°.
- The angle of elevation to the top of the pole (D) from the foot of the tower (B) is 30°.

2. **Setting Up the Diagram**:
- Let the horizontal distance between the pole and the tower be 'd'.
- From point C (foot of the pole), the angle of elevation to point A (top of the tower) is 60°.
- From point B (foot of the tower), the angle of elevation to point D (top of the pole) is 30°.

3. **Using Triangle CBD**:
- In triangle CBD, we know:
- Height of the pole (CD) = 20 m
- Angle CBD = 30°
- Using the tangent function:
\[ \tan(30°) = \frac{CD}{BD} \]
- Substituting the known values:
\[ \tan(30°) = \frac{20}{d} \]
- We know that \(\tan(30°) = \frac{1}{\sqrt{3}}\), so:
\[ \frac{1}{\sqrt{3}} = \frac{20}{d} \]
- Cross-multiplying gives:
\[ d = 20 \sqrt{3} \]

4. **Calculating d**:
- Now, we calculate \(d\):
\[ d \approx 20 \times 1.732 \approx 34.64 \text{ m} \]

5. **Using Triangle ADB**:
- Now, we can also verify using triangle ADB:
- In triangle ADB, we have:
- Height of the tower (AB) = h (unknown)
- Angle ADB = 60°
- Using the tangent function:
\[ \tan(60°) = \frac{h}{d} \]
- We know that \(\tan(60°) = \sqrt{3}\), so:
\[ \sqrt{3} = \frac{h}{d} \]
- Substituting \(d = 20 \sqrt{3}\):
\[ h = d \cdot \sqrt{3} = (20 \sqrt{3}) \cdot \sqrt{3} = 20 \cdot 3 = 60 \text{ m} \]

6. **Final Result**:
- The horizontal distance between the pole and the tower is approximately **34.64 meters**.