A boat is being rowed away from a cliff 150 meter high. At the top of the cliff the angle of depression of the boat changes from `60^(@)` to `45^(@)` in 2 minutes. Find the speed of the boat.

Answer

To solve the problem, we need to find the speed of the boat as it moves away from the cliff. We will use the concept of angles of depression and some trigonometry.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a cliff of height \( h = 150 \) meters.
- The angle of depression changes from \( 60^\circ \) to \( 45^\circ \) in 2 minutes.

2. **Setting Up the Diagram**:
- Let \( A \) be the top of the cliff and \( B \) be the position of the boat when the angle of depression is \( 60^\circ \).
- Let \( C \) be the position of the boat when the angle of depression is \( 45^\circ \).
- The horizontal distance from the base of the cliff to the boat at position \( B \) is \( x_1 \) and at position \( C \) is \( x_2 \).

3. **Using Trigonometry**:
- For angle \( 60^\circ \):
\[ \tan(60^\circ) = \frac{h}{x_1} \implies \sqrt{3} = \frac{150}{x_1} \implies x_1 = \frac{150}{\sqrt{3}} = 50\sqrt{3} \text{ meters} \]

- For angle \( 45^\circ \):
\[ \tan(45^\circ) = \frac{h}{x_2} \implies 1 = \frac{150}{x_2} \implies x_2 = 150 \text{ meters} \]

4. **Calculating the Distance Travelled by the Boat**:
- The distance travelled by the boat in 2 minutes is:
\[ \text{Distance} = x_2 - x_1 = 150 - 50\sqrt{3} \]

5. **Calculating the Speed of the Boat**:
- The speed of the boat is given by:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{150 - 50\sqrt{3}}{2} \text{ meters/minute} \]

6. **Final Calculation**:
- To find a numerical value, we can approximate \( \sqrt{3} \approx 1.732 \):
\[ 50\sqrt{3} \approx 50 \times 1.732 = 86.6 \]
\[ \text{Distance} \approx 150 - 86.6 = 63.4 \text{ meters} \]
\[ \text{Speed} \approx \frac{63.4}{2} = 31.7 \text{ meters/minute} \]

### Final Answer:
The speed of the boat is approximately \( 31.7 \) meters per minute.