Two ships are sailing in the sea on either side of a light -house . The angle of depression of the two ships are `45^(@)` each . If the height of the light-house is 300 metres, then the distance between the ships is

Options

A. 600 m
B. ` 600//sqrt(3) m `
C. ` 300 sqrt(3) m `
D. 300 m

Answer

To solve the problem step by step, we will use the properties of right triangles and trigonometric ratios.

### Step 1: Understand the Problem
We have a lighthouse of height 300 meters, and two ships are sailing on either side of the lighthouse. The angle of depression from the top of the lighthouse to each ship is 45 degrees. We need to find the distance between the two ships.

### Step 2: Draw the Diagram
Draw a vertical line representing the lighthouse (AB) with height 300 meters. Mark point A as the top of the lighthouse and point B as the base. Mark points P and Q as the positions of the two ships. The angles of depression from point A to points P and Q are both 45 degrees.

### Step 3: Use Trigonometric Ratios
In right triangle ABP:
- The height of the lighthouse (AB) is the opposite side to the angle of depression.
- The distance from the base of the lighthouse to ship P (PB) is the adjacent side.

Using the tangent function:
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \]
For angle A (which is 45 degrees):
\[ \tan(45^\circ) = \frac{PB}{AB} \]
Since \(\tan(45^\circ) = 1\):
\[ 1 = \frac{PB}{300} \]
This implies:
\[ PB = 300 \text{ meters} \]

### Step 4: Calculate Distance for Ship Q
Now, consider triangle ABQ, which is similar to triangle ABP:
Using the same logic:
\[ \tan(45^\circ) = \frac{BQ}{AB} \]
Thus:
\[ 1 = \frac{BQ}{300} \]
This implies:
\[ BQ = 300 \text{ meters} \]

### Step 5: Find the Total Distance Between the Ships
The total distance between the two ships (PQ) is the sum of the distances PB and BQ:
\[ PQ = PB + BQ = 300 + 300 = 600 \text{ meters} \]

### Final Answer
The distance between the ships is **600 meters**.

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