An observer on the top of a cliff, 200 m above the sea-level, observes the angles of depression of the two ships to be `45^(@) and 30^(@)` respectively. Find the distance between the ships, if the ships are
on the opposite sides of the cliff

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a cliff that is 200 m high, and from the top of the cliff, an observer sees two ships at angles of depression of 45° and 30°. We need to find the distance between the two ships, which are on opposite sides of the cliff.

### Step 2: Draw a Diagram
1. Draw a vertical line representing the cliff (PQ) with a height of 200 m.
2. Mark point P at the top of the cliff and point Q at the bottom (sea level).
3. Draw two horizontal lines from point P to represent the lines of sight to the ships (A and B).
4. Mark the angle of depression to ship A as 30° and to ship B as 45°.

### Step 3: Use Trigonometry to Find Distances
We will use the tangent of the angles of depression to find the horizontal distances from the base of the cliff to each ship.

#### For Ship B (Angle of Depression = 45°):
- In triangle PQB:
- Height (PQ) = 200 m
- Angle of depression = 45°

Using the tangent function:
\[ \tan(45°) = \frac{PQ}{BQ} \]
\[ 1 = \frac{200}{BQ} \]
Thus, \( BQ = 200 \, \text{m} \).

#### For Ship A (Angle of Depression = 30°):
- In triangle PAQ:
- Height (PQ) = 200 m
- Angle of depression = 30°

Using the tangent function:
\[ \tan(30°) = \frac{PQ}{AQ} \]
\[ \frac{1}{\sqrt{3}} = \frac{200}{AQ} \]
Cross-multiplying gives:
\[ AQ = 200\sqrt{3} \]

### Step 4: Calculate the Value of AQ
Using the approximate value of \(\sqrt{3} \approx 1.732\):
\[ AQ = 200 \times 1.732 \approx 346.4 \, \text{m} \]

### Step 5: Find the Total Distance Between the Ships
The total distance between the two ships (AB) is the sum of AQ and BQ:
\[ AB = AQ + BQ = 346.4 \, \text{m} + 200 \, \text{m} = 546.4 \, \text{m} \]

### Final Answer
The distance between the two ships is **546.4 meters**.

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