An observer standing on a 300 m high tower observes two boats in the same direction, their angle of depression are `60^(@) and 30^(@)` respectively. The distance between the boats is

Options

A. 173.2 m
B. 346.4 m
C. 25 m
D. 72 m

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have an observer on a 300 m high tower observing two boats. The angles of depression to the boats are 60° and 30°. We need to find the distance between the two boats.

### Step 2: Draw the Diagram
1. Draw a vertical line representing the tower (300 m).
2. Mark the top of the tower as point A and the base as point E.
3. Draw horizontal lines from point A to the boats (B1 and B2) at angles of depression of 60° and 30°, respectively.

### Step 3: Identify the Right Triangles
- Triangle AEB1 (for the boat at 60°)
- Triangle AEB2 (for the boat at 30°)

### Step 4: Use Trigonometry to Find Distances
1. For triangle AEB1:
- Angle of depression = 60°
- Using the tangent function:
\[ \tan(60°) = \frac{\text{height}}{\text{distance to B1}} = \frac{300}{y} \]
- We know \(\tan(60°) = \sqrt{3}\):
\[ \sqrt{3} = \frac{300}{y} \implies y = \frac{300}{\sqrt{3}} = 100\sqrt{3} \text{ m} \]

2. For triangle AEB2:
- Angle of depression = 30°
- Using the tangent function:
\[ \tan(30°) = \frac{\text{height}}{\text{distance to B2}} = \frac{300}{x + y} \]
- We know \(\tan(30°) = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{300}{x + 100\sqrt{3}} \implies x + 100\sqrt{3} = 300\sqrt{3} \]
- Rearranging gives:
\[ x = 300\sqrt{3} - 100\sqrt{3} = 200\sqrt{3} \text{ m} \]

### Step 5: Calculate the Distance Between the Boats
The distance between the two boats (B1 and B2) is \(x\):
\[ x = 200\sqrt{3} \text{ m} \]

### Step 6: Approximate the Value
Using \(\sqrt{3} \approx 1.732\):
\[ x \approx 200 \times 1.732 = 346.4 \text{ m} \]

### Final Answer
The distance between the boats is approximately **346.4 m**.

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