A snake observes an eagle perching on the top of a pole 20 m high. Its elevation from snake s eye is `45^@` before it hies off horizontally straight away from the snake and after one second the elevation of the eagle reduces to `30^@`.The speed of the eagle is

Options

A. 14.64m//s
B. 17.71m//s
C. 12m//s
D. None of these

Explanation

To solve the problem step by step, we will break down the information given and apply trigonometric principles to find the speed of the eagle.

### Step-by-Step Solution:

1. **Understand the Situation**:
- The eagle is perched on top of a pole that is 20 meters high.
- The snake observes the eagle at an elevation angle of \(45^\circ\) initially.
- After the eagle flies off horizontally for 1 second, the elevation angle observed by the snake reduces to \(30^\circ\).

2. **Set Up the Diagram**:
- Let point \(B\) be the position of the eagle when the angle of elevation is \(45^\circ\).
- Let point \(M\) be the position of the eagle after it has flown off horizontally for 1 second.
- Let point \(O\) be the position of the snake's eye level, which is at the base of the pole.

3. **Calculate the Distance \(OD\)**:
- In triangle \(BOD\) (where \(D\) is the top of the pole):
\[ \tan(45^\circ) = \frac{BD}{OD} \]
Since \(\tan(45^\circ) = 1\):
\[ 1 = \frac{20}{OD} \implies OD = 20 \text{ meters} \]

4. **Calculate the Distance \(ON\)**:
- In triangle \(MON\) (where \(N\) is the new position of the eagle):
\[ \tan(30^\circ) = \frac{MN}{ON} \]
Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{MN}{ON} \]
- Here, \(MN = BD = 20\) meters (the height of the eagle remains the same):
\[ \frac{1}{\sqrt{3}} = \frac{20}{ON} \implies ON = 20\sqrt{3} \text{ meters} \]

5. **Calculate the Distance \(DN\)**:
- We know that \(ON = OD + DN\):
\[ DN = ON - OD = 20\sqrt{3} - 20 \]
- Simplifying this gives:
\[ DN = 20(\sqrt{3} - 1) \text{ meters} \]

6. **Calculate the Speed of the Eagle**:
- The eagle flies horizontally for 1 second, so its speed \(v\) is given by:
\[ v = \frac{DN}{\text{time}} = \frac{20(\sqrt{3} - 1)}{1} = 20(\sqrt{3} - 1) \text{ m/s} \]
- Approximating \(\sqrt{3} \approx 1.732\):
\[ v \approx 20(1.732 - 1) = 20 \times 0.732 = 14.64 \text{ m/s} \]

### Final Answer:
The speed of the eagle is approximately **14.64 m/s**.