As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are `30^@`and `45^@`. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Options

A. `75(sqrt3-1)`m
B. `85(sqrt3-1)`m
C. `75(sqrt3+1)`m
D. None

Explanation

To solve the problem, we will follow these steps:

### Step 1: Draw the Diagram
We start by drawing a diagram of the situation described in the problem. We have a lighthouse (AB) that is 75 meters high. The two ships are located at points C and D on the sea level (BC). The angles of depression to the ships from the top of the lighthouse are 30° and 45° respectively.

### Step 2: Identify the Right Triangles
From the top of the lighthouse (point A), we can form two right triangles:
1. Triangle ABD for the ship at point D (angle of depression = 45°).
2. Triangle ABC for the ship at point C (angle of depression = 30°).

### Step 3: Calculate Distance BD (Ship D)
For triangle ABD:
- The height of the lighthouse (AB) = 75 m.
- Angle of depression to ship D = 45°.

Using the tangent function:
\[ \tan(45°) = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{BD} \]
\[ 1 = \frac{75}{BD} \]
Thus,
\[ BD = 75 \text{ m}. \]

### Step 4: Calculate Distance BC (Ship C)
For triangle ABC:
- The height of the lighthouse (AB) = 75 m.
- Angle of depression to ship C = 30°.

Using the tangent function:
\[ \tan(30°) = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{BC} \]
\[ \frac{1}{\sqrt{3}} = \frac{75}{BC} \]
Cross-multiplying gives:
\[ BC = 75\sqrt{3} \text{ m}. \]

### Step 5: Calculate Distance CD (Distance Between Ships)
Since the ships are on the same side of the lighthouse, the total distance from the lighthouse to ship C (BC) is the sum of the distances BD and CD:
\[ BC = BD + CD. \]
Let CD be represented as \(x\):
\[ 75\sqrt{3} = 75 + x. \]
Rearranging gives:
\[ x = 75\sqrt{3} - 75. \]

### Final Answer
Thus, the distance between the two ships (CD) is:
\[ CD = 75(\sqrt{3} - 1) \text{ m}. \]