From an aeroplane above a straight road the angles of depression of two positions at a distance 20 m apart on the road are observed to be `30^(@) and 45^(@).` The height of the aeroplane above the ground is

Options

A. `10 sqrt3 m`
B. `10 ( sqrt3 -1) m`
C. `10 ( sqrt3 +1 ) m `
D. `20 m`

Answer

To solve the problem step by step, we will use the concept of angles of depression and the properties of right triangles.

### Step 1: Understand the Problem
We have an aeroplane at a height \( h \) above the ground. The angles of depression to two points on the road are \( 30^\circ \) and \( 45^\circ \). The distance between these two points on the road is \( 20 \) m.

### Step 2: Set Up the Diagram
1. Let \( A \) be the position of the aeroplane.
2. Let \( B \) and \( C \) be the two points on the road, where \( BC = 20 \) m.
3. The angle of depression to point \( B \) is \( 30^\circ \) and to point \( C \) is \( 45^\circ \).

### Step 3: Define the Variables
- Let \( AD = h \) (the height of the aeroplane).
- Let \( BD = x \) (the horizontal distance from point \( B \) to the point directly below the aeroplane).
- Therefore, \( CD = BD + BC = x + 20 \).

### Step 4: Use the Tangent Function
For triangle \( ABD \) (where angle \( ADB = 30^\circ \)):
\[ \tan(30^\circ) = \frac{AD}{BD} \implies \frac{1}{\sqrt{3}} = \frac{h}{x} \implies h = \frac{x}{\sqrt{3}} \tag{1} \]

For triangle \( ACD \) (where angle \( ADC = 45^\circ \)):
\[ \tan(45^\circ) = \frac{AD}{CD} \implies 1 = \frac{h}{x + 20} \implies h = x + 20 \tag{2} \]

### Step 5: Set the Equations Equal
From equations (1) and (2):
\[ \frac{x}{\sqrt{3}} = x + 20 \]

### Step 6: Solve for \( x \)
Multiply both sides by \( \sqrt{3} \):
\[ x = \sqrt{3}(x + 20) \]
\[ x = \sqrt{3}x + 20\sqrt{3} \]
Rearranging gives:
\[ x - \sqrt{3}x = 20\sqrt{3} \]
\[ x(1 - \sqrt{3}) = 20\sqrt{3} \]
\[ x = \frac{20\sqrt{3}}{1 - \sqrt{3}} \tag{3} \]

### Step 7: Substitute \( x \) back to find \( h \)
Using equation (1):
\[ h = \frac{x}{\sqrt{3}} = \frac{20\sqrt{3}}{(1 - \sqrt{3})\sqrt{3}} = \frac{20\sqrt{3}}{3 - \sqrt{3}} \]

### Step 8: Rationalize the Denominator
Multiply the numerator and denominator by \( 3 + \sqrt{3} \):
\[ h = \frac{20\sqrt{3}(3 + \sqrt{3})}{(3 - \sqrt{3})(3 + \sqrt{3})} = \frac{20\sqrt{3}(3 + \sqrt{3})}{9 - 3} = \frac{20\sqrt{3}(3 + \sqrt{3})}{6} \]
\[ h = \frac{10\sqrt{3}(3 + \sqrt{3})}{3} \]

### Step 9: Calculate \( h \)
Calculate the numerical value of \( h \):
\[ h = 10\sqrt{3} - 10 \approx 10(1.732) - 10 \approx 17.32 - 10 = 7.32 \text{ m} \]

### Final Answer
The height of the aeroplane above the ground is approximately \( 7.32 \) meters.

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