There are two points on the horizontal line passing through the foot of a tower in the same side of the tower. The angle of depression of these point from the top of the tower are `45^(@)" and "60^(@)` respectively. Find the distance between the two point if the height if the tower is 150 metres.

Answer

To solve the problem, we need to find the distance between the two points on the horizontal line from the foot of the tower, given the angles of depression and the height of the tower.

### Step-by-Step Solution:

1. **Understand the Setup**:
- Let the height of the tower be \( h = 150 \) meters.
- Let \( A \) be the point where the angle of depression is \( 45^\circ \).
- Let \( B \) be the point where the angle of depression is \( 60^\circ \).
- Let \( O \) be the foot of the tower.

2. **Draw the Diagram**:
- Draw a vertical line representing the tower from point \( O \) to point \( T \) (the top of the tower).
- Draw horizontal lines from points \( A \) and \( B \) to the foot of the tower \( O \).
- The angles of depression from point \( T \) to points \( A \) and \( B \) are \( 45^\circ \) and \( 60^\circ \) respectively.

3. **Use Trigonometry to Find Distances**:
- For point \( A \) (angle of depression \( 45^\circ \)):
\[ \tan(45^\circ) = \frac{h}{OA} \]
Since \( \tan(45^\circ) = 1 \):
\[ 1 = \frac{150}{OA} \implies OA = 150 \text{ meters} \]

- For point \( B \) (angle of depression \( 60^\circ \)):
\[ \tan(60^\circ) = \frac{h}{OB} \]
Since \( \tan(60^\circ) = \sqrt{3} \):
\[ \sqrt{3} = \frac{150}{OB} \implies OB = \frac{150}{\sqrt{3}} = 50\sqrt{3} \text{ meters} \]

4. **Calculate the Distance Between Points \( A \) and \( B \)**:
- The distance \( AB \) is given by:
\[ AB = OA - OB = 150 - 50\sqrt{3} \]

5. **Final Calculation**:
- To find \( 50\sqrt{3} \):
\[ \sqrt{3} \approx 1.732 \implies 50\sqrt{3} \approx 50 \times 1.732 \approx 86.6 \text{ meters} \]
- Therefore:
\[ AB \approx 150 - 86.6 = 63.4 \text{ meters} \]

### Conclusion:
The distance between the two points \( A \) and \( B \) is approximately \( 63.4 \) meters.