As observed from the top of a light house the angle of depression of two ships in opposite direction are respectively `60^(@) and 45^(@)`. If distance between the two ships in `200((sqrt(3)+1)/sqrt(3))` meter then find the height of the light house.

Answer

To solve the problem, we need to find the height of the lighthouse based on the angles of depression to two ships and the distance between them. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have a lighthouse from which two ships are observed at angles of depression of 60° and 45°. The distance between the two ships is given as \(200 \cdot \frac{\sqrt{3}+1}{\sqrt{3}}\) meters.

### Step 2: Set Up the Diagram
Draw a diagram with the lighthouse at point A, the first ship at point B, and the second ship at point C. The angles of depression to ships B and C are 60° and 45°, respectively. The horizontal line from the top of the lighthouse to the ships represents the line of sight.

### Step 3: Identify Right Triangles
From the top of the lighthouse (point A), draw vertical lines down to the horizontal line at the level of the ships. This creates two right triangles:
- Triangle ABD for the ship at B (angle of depression = 45°)
- Triangle ACD for the ship at C (angle of depression = 60°)

### Step 4: Use Trigonometric Ratios
For triangle ABD:
- The angle of depression is 45°, so the angle at point B is also 45°.
- Using the tangent function:
\[ \tan(45°) = \frac{h}{BD} \implies 1 = \frac{h}{BD} \implies h = BD \]

For triangle ACD:
- The angle of depression is 60°, so the angle at point C is also 60°.
- Using the tangent function:
\[ \tan(60°) = \frac{h}{CD} \implies \sqrt{3} = \frac{h}{CD} \implies h = \sqrt{3} \cdot CD \]

### Step 5: Express CD in Terms of BD
Let \(BD = x\). The distance between the two ships is given as:
\[ BC = BD + CD = x + CD = 200 \cdot \frac{\sqrt{3}+1}{\sqrt{3}} \]
Thus,
\[ CD = 200 \cdot \frac{\sqrt{3}+1}{\sqrt{3}} - x \]

### Step 6: Substitute CD in the Height Equation
Substituting \(CD\) into the height equation from triangle ACD:
\[ h = \sqrt{3} \cdot \left(200 \cdot \frac{\sqrt{3}+1}{\sqrt{3}} - x\right) \]

### Step 7: Set the Two Height Equations Equal
Now we have two expressions for \(h\):
1. \(h = x\)
2. \(h = \sqrt{3} \cdot \left(200 \cdot \frac{\sqrt{3}+1}{\sqrt{3}} - x\right)\)

Setting them equal:
\[ x = \sqrt{3} \cdot \left(200 \cdot \frac{\sqrt{3}+1}{\sqrt{3}} - x\right) \]

### Step 8: Solve for x
Distributing \(\sqrt{3}\):
\[ x = 200(\sqrt{3}+1) - \sqrt{3}x \]
Combining like terms:
\[ x + \sqrt{3}x = 200(\sqrt{3}+1) \]
\[ x(1 + \sqrt{3}) = 200(\sqrt{3}+1) \]
\[ x = \frac{200(\sqrt{3}+1)}{1+\sqrt{3}} \]

### Step 9: Find the Height
Now substituting back to find \(h\):
\[ h = x = 200 \text{ meters} \]

### Final Answer
The height of the lighthouse is **200 meters**.