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 pole, if the height of the tower is 75 m

Answer

To solve the problem, we will use trigonometric ratios in right triangles formed by the pole and the tower.

### Step-by-Step Solution:

1. **Identify the given information:**
- Height of the tower (AB) = 75 m
- Angle of elevation from the top of the pole (C) to the top of the tower (A) = 60°
- Angle of depression from the top of the pole (C) to the bottom of the tower (B) = 30°
- Let the height of the pole (CD) = h m.

2. **Draw a diagram:**
- Draw a vertical tower (AB) and a vertical pole (CD) on the same level ground.
- Mark points A (top of the tower), B (bottom of the tower), C (top of the pole), and D (bottom of the pole).
- Draw horizontal lines from C to B and C to A.

3. **Use the angle of elevation (60°):**
- In triangle AEC, where E is the foot of the pole:
- \( \tan(60°) = \frac{AE}{CE} \)
- Since \( \tan(60°) = \sqrt{3} \), we have:
\[ \sqrt{3} = \frac{AE}{CE} \]
- Therefore, \( AE = CE \cdot \sqrt{3} \) (1)

4. **Use the angle of depression (30°):**
- In triangle BEC:
- \( \tan(30°) = \frac{BE}{CE} \)
- Since \( \tan(30°) = \frac{1}{\sqrt{3}} \), we have:
\[ \frac{1}{\sqrt{3}} = \frac{BE}{CE} \]
- Therefore, \( BE = CE \cdot \frac{1}{\sqrt{3}} \) (2)

5. **Relate the heights:**
- The total height of the tower (AB) can be expressed as:
\[ AB = AE + BE \]
- Substituting equations (1) and (2) into this equation:
\[ 75 = CE \cdot \sqrt{3} + CE \cdot \frac{1}{\sqrt{3}} \]
- Factor out \( CE \):
\[ 75 = CE \left( \sqrt{3} + \frac{1}{\sqrt{3}} \right) \]
- Simplifying the expression inside the parentheses:
\[ \sqrt{3} + \frac{1}{\sqrt{3}} = \frac{3 + 1}{\sqrt{3}} = \frac{4}{\sqrt{3}} \]
- Therefore:
\[ 75 = CE \cdot \frac{4}{\sqrt{3}} \]
- Solving for \( CE \):
\[ CE = \frac{75 \cdot \sqrt{3}}{4} \]

6. **Find the height of the pole (h):**
- From the triangle BEC, we have:
\[ h = CE - BE \]
- Substitute \( BE \) from equation (2):
\[ BE = CE \cdot \frac{1}{\sqrt{3}} = \frac{75 \cdot \sqrt{3}}{4} \cdot \frac{1}{\sqrt{3}} = \frac{75}{4} \]
- Thus:
\[ h = CE - BE = \frac{75 \cdot \sqrt{3}}{4} - \frac{75}{4} \]
- Factor out \( \frac{75}{4} \):
\[ h = \frac{75}{4} \left( \sqrt{3} - 1 \right) \]

7. **Calculate the height of the pole:**
- Using \( \sqrt{3} \approx 1.732 \):
\[ h \approx \frac{75}{4} \cdot (1.732 - 1) \approx \frac{75}{4} \cdot 0.732 \approx 13.65 \text{ m} \]

### Final Answer:
The height of the pole is approximately **13.65 m**.