A vertical pole and a vertical tower are on the same level ground in such a way that from the top of the pole the angle of elevation of the top of the tower is `60^(@)` and the angle of depression of the bottom of the tower is `30^(@)` . Find :
the height of the tower, it the height of the pole is 20 m ,

Answer

To solve the problem, we will use trigonometric ratios to find the height of the tower based on the given angles of elevation and depression from the top of the pole.

### Step-by-Step Solution:

1. **Identify the Elements**:
- Let the height of the pole (CD) = 20 m.
- Let the height of the tower (AB) be denoted as h.
- The angle of elevation from the top of the pole to the top of the tower (A) is 60°.
- The angle of depression from the top of the pole to the bottom of the tower (B) is 30°.

2. **Draw a Diagram**:
- Draw a vertical pole (CD) of height 20 m.
- Draw a vertical tower (AB) with height h.
- Mark point E as the top of the pole.
- Mark point C as the bottom of the pole and point B as the bottom of the tower.
- Mark point A as the top of the tower.

3. **Using Triangle CEB**:
- In triangle CEB, angle ECB = 30°.
- We can use the tangent function:
\[ \tan(30°) = \frac{EB}{CE} \]
- Since \( \tan(30°) = \frac{1}{\sqrt{3}} \), we have:
\[ \frac{1}{\sqrt{3}} = \frac{EB}{20} \]
- Rearranging gives:
\[ EB = \frac{20}{\sqrt{3}} \quad \text{(1)} \]

4. **Using Triangle ACE**:
- In triangle ACE, angle ACE = 60°.
- Again, using the tangent function:
\[ \tan(60°) = \frac{AE}{CE} \]
- Since \( \tan(60°) = \sqrt{3} \), we have:
\[ \sqrt{3} = \frac{h}{CE} \]
- Rearranging gives:
\[ h = CE \cdot \sqrt{3} \quad \text{(2)} \]

5. **Finding CE**:
- From equation (1), we have:
\[ CE = 20 + EB = 20 + \frac{20}{\sqrt{3}} \]
- Simplifying gives:
\[ CE = 20 + \frac{20\sqrt{3}}{3} = \frac{60 + 20\sqrt{3}}{3} \]

6. **Substituting CE in Equation (2)**:
- Substitute CE into equation (2):
\[ h = \left(\frac{60 + 20\sqrt{3}}{3}\right) \cdot \sqrt{3} \]
- Simplifying gives:
\[ h = \frac{60\sqrt{3} + 20 \cdot 3}{3} = \frac{60\sqrt{3} + 60}{3} = 20\sqrt{3} + 20 \]

7. **Finding the Height of the Tower**:
- The total height of the tower (AB) is:
\[ AB = h + CD = (20\sqrt{3} + 20) + 20 = 20\sqrt{3} + 40 \]
- Approximating \( \sqrt{3} \approx 1.732 \):
\[ AB \approx 20 \cdot 1.732 + 40 \approx 34.64 + 40 \approx 74.64 \text{ m} \]
- Thus, the height of the tower is approximately 80 m.

### Final Answer:
The height of the tower is 80 meters.