An aeroplane from point A gives an angle of elevation `60^o` and after 20 second when aeroplane is moving with the speed 432 km/hr and makes an angle of `30^o` . then find the height of the aeroplane in meter

Answer

To solve the problem, we will use trigonometric principles to find the height of the aeroplane.

### Step-by-Step Solution:

1. **Convert the speed of the aeroplane to meters per second:**
The speed of the aeroplane is given as 432 km/hr. To convert this to meters per second:
\[ \text{Speed in m/s} = \frac{432 \times 1000}{3600} = 120 \text{ m/s} \]

2. **Calculate the distance travelled in 20 seconds:**
The distance travelled by the aeroplane in 20 seconds can be calculated as:
\[ \text{Distance} = \text{Speed} \times \text{Time} = 120 \text{ m/s} \times 20 \text{ s} = 2400 \text{ m} \]

3. **Set up the right triangles:**
- Let \( h \) be the height of the aeroplane.
- From point A, the angle of elevation is \( 60^\circ \). Using the tangent function:
\[ \tan(60^\circ) = \frac{h}{d_1} \quad \Rightarrow \quad d_1 = \frac{h}{\sqrt{3}} \quad \text{(since } \tan(60^\circ) = \sqrt{3}\text{)} \]
- After 20 seconds, the angle of elevation is \( 30^\circ \). The distance from the new position to the point directly below the aeroplane is \( d_1 + 2400 \):
\[ \tan(30^\circ) = \frac{h}{d_1 + 2400} \quad \Rightarrow \quad d_1 + 2400 = h \sqrt{3} \quad \text{(since } \tan(30^\circ) = \frac{1}{\sqrt{3}}\text{)} \]

4. **Substituting \( d_1 \) into the second equation:**
Substitute \( d_1 = \frac{h}{\sqrt{3}} \) into the second equation:
\[ \frac{h}{\sqrt{3}} + 2400 = h \sqrt{3} \]

5. **Solve for \( h \):**
Rearranging the equation gives:
\[ 2400 = h \sqrt{3} - \frac{h}{\sqrt{3}} \]
\[ 2400 = h \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) \]
\[ 2400 = h \left( \frac{3 - 1}{\sqrt{3}} \right) = h \left( \frac{2}{\sqrt{3}} \right) \]
\[ h = 2400 \cdot \frac{\sqrt{3}}{2} = 1200\sqrt{3} \]

6. **Calculate the numerical value of \( h \):**
Using \( \sqrt{3} \approx 1.732 \):
\[ h \approx 1200 \times 1.732 \approx 2078.8 \text{ m} \]

### Final Answer:
The height of the aeroplane is approximately **2078.8 meters**.