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 find the distance between the two points on the horizontal line passing through the foot of the tower, we will use trigonometric ratios and the given angles of depression.

### Step-by-Step Solution

1. **Identify the Problem Setup**:
- Let the height of the tower be \( AB = 150 \) meters.
- Let point \( C \) be where the angle of depression is \( 60^\circ \) and point \( D \) where the angle of depression is \( 45^\circ \).
- We need to find the distance \( CD \).

2. **Understanding Angles of Depression**:
- The angle of depression from point \( A \) (top of the tower) to point \( D \) is \( 45^\circ \).
- The angle of depression from point \( A \) to point \( C \) is \( 60^\circ \).
- By alternate interior angles, the angles at points \( B \) (foot of the tower) are also \( 45^\circ \) and \( 60^\circ \) respectively.

3. **Finding Distance \( BD \)**:
- In triangle \( ABD \):
\[ \tan(45^\circ) = \frac{AB}{BD} \]
\[ \tan(45^\circ) = 1 \Rightarrow 1 = \frac{150}{BD} \]
\[ BD = 150 \text{ meters} \]

4. **Finding Distance \( BC \)**:
- In triangle \( ABC \):
\[ \tan(60^\circ) = \frac{AB}{BC} \]
\[ \tan(60^\circ) = \sqrt{3} \Rightarrow \sqrt{3} = \frac{150}{BC} \]
\[ BC = \frac{150}{\sqrt{3}} = 150 \times \frac{\sqrt{3}}{3} = 50\sqrt{3} \text{ meters} \]

5. **Calculating \( BC \) in Decimal**:
- Approximating \( \sqrt{3} \approx 1.732 \):
\[ BC \approx 50 \times 1.732 = 86.6 \text{ meters} \]

6. **Finding Distance \( CD \)**:
- The distance between points \( C \) and \( D \) is:
\[ CD = BD - BC \]
\[ CD = 150 - 86.6 = 63.4 \text{ meters} \]

### Final Answer:
The distance between the two points \( C \) and \( D \) is \( 63.4 \) meters.