When looked at a plane road from an aeroplane, anles of depression of 2 consecutive direction pillars are `45^@` and `60^@` and the given pillars shows distance in km. Then find the height of the aeroplane.
समतलीय सड़क के ऊपर हवाई जहाज से देखे जाने पर एक पार्श्व के दो क्रमागत किमी. - निर्देशक प्रस्तर के अवनमन कोण क्रमशः `45^@` तथा `60^@` है। तो हवाई जहाज की ऊँचाई निकालिए।

Options

A. A. 2.366 m
B. B. 2.5 m
C. C. 2.8 m
D. D. 2.1 m

Explanation

To find the height of the aeroplane based on the angles of depression to two consecutive direction pillars, we can follow these steps:

### Step 1: Understand the Problem
We have two angles of depression from the aeroplane to two consecutive pillars on the ground. The angles are 45° and 60°. We need to find the height of the aeroplane.

### Step 2: Draw the Diagram
- Let the height of the aeroplane be \( h \).
- Let the distance from the base of the aeroplane to the first pillar (with angle of depression 60°) be \( x \).
- The distance to the second pillar (with angle of depression 45°) will be \( x + 1 \) (since they are consecutive).

### Step 3: Use Trigonometry for the First Pillar
For the first pillar (angle of depression = 60°):
- In triangle formed by the height \( h \) and distance \( x \):
\[ \tan(60°) = \frac{h}{x} \]
- We know that \( \tan(60°) = \sqrt{3} \):
\[ \sqrt{3} = \frac{h}{x} \implies h = x \sqrt{3} \quad \text{(Equation 1)} \]

### Step 4: Use Trigonometry for the Second Pillar
For the second pillar (angle of depression = 45°):
- In triangle formed by the height \( h \) and distance \( x + 1 \):
\[ \tan(45°) = \frac{h}{x + 1} \]
- We know that \( \tan(45°) = 1 \):
\[ 1 = \frac{h}{x + 1} \implies h = x + 1 \quad \text{(Equation 2)} \]

### Step 5: Solve the Equations
Now we have two equations:
1. \( h = x \sqrt{3} \)
2. \( h = x + 1 \)

Setting them equal to each other:
\[ x \sqrt{3} = x + 1 \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ x \sqrt{3} - x = 1 \]
\[ x(\sqrt{3} - 1) = 1 \]
\[ x = \frac{1}{\sqrt{3} - 1} \]

### Step 7: Rationalizing the Denominator
To simplify \( x \):
\[ x = \frac{1(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{\sqrt{3} + 1}{3 - 1} = \frac{\sqrt{3} + 1}{2} \]

### Step 8: Substitute \( x \) Back to Find \( h \)
Now substituting \( x \) back into either equation for \( h \):
Using \( h = x + 1 \):
\[ h = \frac{\sqrt{3} + 1}{2} + 1 = \frac{\sqrt{3} + 1 + 2}{2} = \frac{\sqrt{3} + 3}{2} \]

### Step 9: Calculate the Height
Now we can calculate \( h \):
Using \( \sqrt{3} \approx 1.732 \):
\[ h \approx \frac{1.732 + 3}{2} = \frac{4.732}{2} \approx 2.366 \text{ meters} \]

### Conclusion
The height of the aeroplane is approximately **2.366 meters**.

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