Two ships are sailing in the sea on the tow sides of a light house . The angles of elevation of the top of the light house as observed from the two ships are `30^(@)and45^(@)` respectively . If the light house is 100 m high , the distance be - tween the two ships is :(take `sqrt(3)=1.73)`

Options

A. a)173 metre
B. b)200 metre
C. c)273 metre
D. d)300 metre

Answer

To solve the problem, we need to find the distance between the two ships based on the angles of elevation to the top of the lighthouse and its height.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Height of the lighthouse (AC) = 100 m
- Angle of elevation from Ship A (AB) = 30°
- Angle of elevation from Ship B (AD) = 45°

2. **Use Trigonometric Ratios:**
- For Ship A (angle 30°):
\[ \tan(30°) = \frac{\text{opposite}}{\text{adjacent}} = \frac{AC}{CB} \]
\[ \tan(30°) = \frac{100}{CB} \]
Since \(\tan(30°) = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{100}{CB} \]
Rearranging gives:
\[ CB = 100 \sqrt{3} \]

3. **Calculate CB:**
- Substitute \(\sqrt{3} = 1.73\):
\[ CB = 100 \times 1.73 = 173 \text{ m} \]

4. **For Ship B (angle 45°):**
\[ \tan(45°) = \frac{AC}{CD} \]
\[ \tan(45°) = \frac{100}{CD} \]
Since \(\tan(45°) = 1\):
\[ 1 = \frac{100}{CD} \]
Rearranging gives:
\[ CD = 100 \text{ m} \]

5. **Calculate the Total Distance Between the Two Ships:**
- The total distance (AB + AD) is:
\[ AB + AD = CB + CD = 173 + 100 = 273 \text{ m} \]

### Final Answer:
The distance between the two ships is **273 meters**.

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