From the top of a light -house at a height 20 metres above sea -level,the angle of depression of a ship is `30^(@)`.The distance of the ship from the foot of the light house is

Options

A. 20 m
B. `20sqrt(3)m`
C. 30 m
D. `30sqrt(3)m`

Answer

To solve the problem step by step, we will use trigonometric ratios.

### Step 1: Understand the problem
We have a lighthouse that is 20 meters tall, and we need to find the distance of a ship from the foot of the lighthouse when the angle of depression from the top of the lighthouse to the ship is 30 degrees.

### Step 2: Draw a diagram
Visualize the situation:
- Draw a vertical line representing the lighthouse (20 meters high).
- Mark the top of the lighthouse as point A and the foot of the lighthouse as point B.
- Draw a horizontal line from point A to the ship (point C).
- The angle of depression from point A to point C is 30 degrees.

### Step 3: Identify the right triangle
From point A (top of the lighthouse), draw a line down to the horizontal line at point C (the position of the ship). This forms a right triangle ABC:
- AB = height of the lighthouse = 20 meters (perpendicular)
- BC = distance from the foot of the lighthouse to the ship (base)
- Angle A = 30 degrees (angle of depression)

### Step 4: Use trigonometric ratios
In triangle ABC, we can use the tangent function, which is defined as:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
Here, the angle of depression (30 degrees) corresponds to angle A in triangle ABC. Therefore:
\[ \tan(30^\circ) = \frac{AB}{BC} \]
Substituting the known values:
\[ \tan(30^\circ) = \frac{20}{BC} \]

### Step 5: Find the value of \(\tan(30^\circ)\)
We know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
So we can rewrite the equation:
\[ \frac{1}{\sqrt{3}} = \frac{20}{BC} \]

### Step 6: Solve for BC
Cross-multiplying gives:
\[ BC = 20 \cdot \sqrt{3} \]

### Conclusion
The distance of the ship from the foot of the lighthouse is:
\[ BC = 20\sqrt{3} \text{ meters} \]