We are given the following information:
- The height of the lighthouse is 100 meters.
- The angles of elevation from the two ships to the top of the lighthouse are 30° and 45° respectively.

Let:
- \( h = 100 \, \text{m} \) (height of the lighthouse),
- \( d_1 \) be the distance from the first ship (with 30° angle of elevation) to the lighthouse,
- \( d_2 \) be the distance from the second ship (with 45° angle of elevation) to the lighthouse,
- We need to find the total distance between the two ships, i.e., \( d_1 + d_2 \).

### Step 1: Use trigonometric relations
For the first ship, using the tangent function:
\[
\tan(30^\circ) = \frac{h}{d_1} = \frac{100}{d_1}
\]
\[
d_1 = \frac{100}{\tan(30^\circ)} = \frac{100}{\frac{1}{\sqrt{3}}} = 100\sqrt{3} \approx 173.2 \, \text{m}
\]

For the second ship, using the tangent function:
\[
\tan(45^\circ) = \frac{h}{d_2} = \frac{100}{d_2}
\]
\[
d_2 = \frac{100}{\tan(45^\circ)} = \frac{100}{1} = 100 \, \text{m}
\]

### Step 2: Find the total distance between the two ships
The total distance between the two ships is:
\[
d_1 + d_2 = 100\sqrt{3} + 100 \approx 173.2 + 100 = 273.2 \, \text{m}
\]

Thus, the distance between the two ships is approximately 273 meters. 

Let, BD be the lighthouse and A and C be the positions of the ships.

Then, BD=100 m, ∠BAD=30∘ , ∠BCD=45∘

In △ABD, we have

tan30∘=BDBA [∵tanθ=opposite sideAdjacent sides]

⇒1√3=100BA

⇒BA=100√3

In △CBD, we have

tan45∘=BDBC

⇒1=100BC

⇒BC=100 m

Distance between the two ships =AC=BA+BC
=100√3+100

=100(√3+1)

=100(1.73+1)

=100×2.73=273 m