A man in a boat rowing away from a lighthouse 150 m high, takes2 minutes to change the angle of elevation of the top of the lighthouse from `60^(@)` to `45^(@)` . Find the speed of the boat .

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the problem and draw a diagram
We have a lighthouse that is 150 m high. A man in a boat rows away from the lighthouse, changing the angle of elevation of the top of the lighthouse from 60° to 45° in 2 minutes.

### Step 2: Set up the scenario
Let:
- Point A be the top of the lighthouse.
- Point B be the base of the lighthouse.
- Point C be the position of the boat when the angle of elevation is 60°.
- Point D be the position of the boat when the angle of elevation is 45°.

### Step 3: Use trigonometry to find distances
1. **For angle C (60°)**:
- We know that \( \tan(60°) = \frac{AB}{BC} \)
- Here, \( AB = 150 \) m (height of the lighthouse) and \( \tan(60°) = \sqrt{3} \).
- Therefore, \( \sqrt{3} = \frac{150}{BC} \).
- Rearranging gives us \( BC = \frac{150}{\sqrt{3}} \).

2. **Rationalizing \( BC \)**:
- \( BC = \frac{150 \sqrt{3}}{3} = 50\sqrt{3} \) m.

3. **For angle D (45°)**:
- We know that \( \tan(45°) = \frac{AB}{BD} \).
- Thus, \( 1 = \frac{150}{BD} \).
- Rearranging gives us \( BD = 150 \) m.

### Step 4: Find distance CD
- The distance \( CD \) is given by \( CD = BD - BC \).
- Substituting the values we found:
\[ CD = 150 - 50\sqrt{3} \]

### Step 5: Calculate the distance
- Factoring out 50:
\[ CD = 50(3 - \sqrt{3}) \text{ m} \]

### Step 6: Find the speed of the boat
- Speed is defined as distance divided by time.
- The time taken is 2 minutes, which is \( 2 \times 60 = 120 \) seconds.
- Therefore, the speed \( v \) is:
\[ v = \frac{CD}{\text{time}} = \frac{50(3 - \sqrt{3})}{120} \]

### Step 7: Simplify the speed
- Simplifying gives:
\[ v = \frac{5(3 - \sqrt{3})}{12} \text{ m/s} \]

### Step 8: Numerical approximation
- Using \( \sqrt{3} \approx 1.732 \):
\[ 3 - \sqrt{3} \approx 3 - 1.732 = 1.268 \]
- Thus:
\[ v \approx \frac{5 \times 1.268}{12} \approx \frac{6.34}{12} \approx 0.528 \text{ m/s} \]

### Step 9: Round off the speed
- Rounding to two decimal places, the speed of the boat is approximately:
\[ v \approx 0.53 \text{ m/s} \]

### Final Answer
The speed of the boat is **0.53 m/s**.