A boat is moving away from an observation tower .It makes an angle of depression of `30^(@)` with an observer 's eye when at a distance of 50 metre from the tower . After 8 seconds ,the angle of depression becomes `60^(@)` .By as - suming that it is running in still water , the approximate speed of the boat is :

Options

A. 33 km /hr
B. 42 km/hr
C. 45 km/hr
D. 50 km/hr

Answer

To solve the problem step by step, we will use trigonometric principles to find the height of the tower and the distance covered by the boat.

### Step 1: Understand the scenario
We have an observation tower (point A) and a boat (point C) that is initially 50 meters away from the tower. The angle of depression from the observer's eye (point B) to the boat is 30 degrees.

### Step 2: Calculate the height of the tower
Using the first triangle (triangle ABC) where:
- Angle C = 30 degrees
- Distance BC = 50 meters

Using the tangent function:
\[ \tan(30^\circ) = \frac{AB}{BC} \]
Where AB is the height of the tower.

Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{AB}{50} \]
Cross-multiplying gives:
\[ AB = 50 \cdot \frac{1}{\sqrt{3}} = \frac{50}{\sqrt{3}} \approx 28.87 \text{ meters} \]

### Step 3: Analyze the second position of the boat
After 8 seconds, the angle of depression changes to 60 degrees. Let’s denote the new position of the boat as point D.

### Step 4: Calculate the distance from the tower to the new position of the boat
In triangle ABD, where:
- Angle D = 60 degrees

Using the tangent function again:
\[ \tan(60^\circ) = \frac{AB}{BD} \]
Where BD is the distance from the tower to the new position of the boat.

Since \(\tan(60^\circ) = \sqrt{3}\):
\[ \sqrt{3} = \frac{AB}{BD} \]
Substituting the value of AB:
\[ \sqrt{3} = \frac{50\sqrt{3}}{BD} \]
Cross-multiplying gives:
\[ BD = 50 \text{ meters} \]

### Step 5: Calculate the distance traveled by the boat
The initial distance from the tower to the boat was 50 meters (BC), and now the distance to the new position (BD) is also 50 meters. Therefore, the distance traveled by the boat (CD) is:
\[ CD = BD - BC = 50 - 50 = 0 \text{ meters} \]

### Step 6: Calculate the speed of the boat
Since the boat has traveled a distance of 0 meters in 8 seconds, the speed is:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{0}{8} = 0 \text{ m/s} \]

### Step 7: Convert speed to km/h
To convert from m/s to km/h, we multiply by \( \frac{18}{5} \):
\[ \text{Speed in km/h} = 0 \cdot \frac{18}{5} = 0 \text{ km/h} \]

### Conclusion
The approximate speed of the boat is **0 km/h**.