From an aeroplane above a straight road the angle 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. `10sqrt(3)`m
B. `10(sqrt(3)-1)`m
C. `10(sqrt(3)+1)m`
D. 20m

Explanation

To solve the problem step by step, we will use the concepts of trigonometry, particularly the tangent function, which relates the height of the aeroplane to the distances on the ground.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have an aeroplane flying at a height \( h \) above the ground.
- There are two points \( C \) and \( D \) on the ground, which are 20 meters apart.
- The angle of depression from the aeroplane to point \( C \) is \( 30^\circ \) and to point \( D \) is \( 45^\circ \).

2. **Setting Up the Diagram**:
- Let \( A \) be the position of the aeroplane.
- Let \( B \) be the vertical line from \( A \) to the ground (height \( h \)).
- Let \( D \) be the point directly below the aeroplane on the ground.
- Let \( C \) be the point on the ground where the angle of depression is \( 30^\circ \).

3. **Using Triangle \( ABD \)**:
- In triangle \( ABD \), where \( \angle ADB = 45^\circ \):
\[ \tan(45^\circ) = \frac{h}{BD} \]
- Since \( \tan(45^\circ) = 1 \):
\[ 1 = \frac{h}{BD} \implies h = BD \]
- Let \( BD = x \), so \( h = x \).

4. **Using Triangle \( ABC \)**:
- In triangle \( ABC \), where \( \angle ACB = 30^\circ \):
\[ \tan(30^\circ) = \frac{h}{BC} \]
- Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \):
\[ \frac{1}{\sqrt{3}} = \frac{h}{BC} \]
- Here, \( BC = BD + CD = x + 20 \):
\[ \frac{1}{\sqrt{3}} = \frac{h}{x + 20} \]
- Substituting \( h = x \):
\[ \frac{1}{\sqrt{3}} = \frac{x}{x + 20} \]

5. **Cross-Multiplying**:
\[ x + 20 = \sqrt{3}x \]
\[ 20 = \sqrt{3}x - x \]
\[ 20 = x(\sqrt{3} - 1) \]
\[ x = \frac{20}{\sqrt{3} - 1} \]

6. **Rationalizing the Denominator**:
\[ x = \frac{20(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{20(\sqrt{3} + 1)}{3 - 1} = \frac{20(\sqrt{3} + 1)}{2} = 10(\sqrt{3} + 1) \]

7. **Finding the Height \( h \)**:
\[ h = x = 10(\sqrt{3} + 1) \]

### Final Answer:
The height of the aeroplane above the ground is:
\[ h = 10(\sqrt{3} + 1) \text{ meters} \]