From a light house the angle of depression of two ships on opposite sides of the light house are observed to be `30^@` and `45^(@)`. If the height of the light house be 100 metres, then the distance between the ships of the line joining them passes through the foot of the light house is `100(sqrt(3)-1)`.

Answer

To solve the problem, we need to find the distance between two ships observed from a lighthouse, given the angles of depression and the height of the lighthouse.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a lighthouse of height \( AB = 100 \) meters.
- The angle of depression to the first ship (point C) is \( 30^\circ \).
- The angle of depression to the second ship (point D) is \( 45^\circ \).

2. **Drawing the Diagram:**
- Draw a vertical line representing the lighthouse (AB).
- Mark point A at the top of the lighthouse and point B at the base.
- Mark points C and D on the ground, representing the positions of the ships.
- Draw horizontal lines from A to C and A to D, representing the angles of depression.

3. **Finding Distance to Ship C:**
- In triangle \( ABC \):
- The angle of depression \( \angle ACB = 30^\circ \).
- By the properties of angles, \( \angle ABC = 30^\circ \).
- Using the tangent function:
\[ \tan(30^\circ) = \frac{AB}{BC} = \frac{100}{X} \]
- We know \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \):
\[ \frac{1}{\sqrt{3}} = \frac{100}{X} \]
- Rearranging gives:
\[ X = 100\sqrt{3} \text{ meters} \]

4. **Finding Distance to Ship D:**
- In triangle \( ABD \):
- The angle of depression \( \angle ADB = 45^\circ \).
- Thus, \( \angle ABD = 45^\circ \).
- Using the tangent function:
\[ \tan(45^\circ) = \frac{AB}{BD} = \frac{100}{Y} \]
- We know \( \tan(45^\circ) = 1 \):
\[ 1 = \frac{100}{Y} \]
- Rearranging gives:
\[ Y = 100 \text{ meters} \]

5. **Calculating the Total Distance Between the Ships:**
- The total distance \( CD \) between the two ships is given by:
\[ CD = BC + BD = X + Y \]
- Substituting the values we found:
\[ CD = 100\sqrt{3} + 100 \]
- Factoring out 100:
\[ CD = 100(\sqrt{3} + 1) \text{ meters} \]

### Conclusion:
The distance between the two ships is \( 100(\sqrt{3} + 1) \) meters.