Two ships are sailling in the sea on the two sides of a lighthouse. If the distance between the ships is `10(sqrt3+1)` meters and their angle of elevations of the top of the lighthouse are `60^(@) and 45^(@)`, then the height of the lighthouse is (The two ships and the foot of lighthouse are in a straight line)

Options

A. 20 meters
B. `20sqrt3` meters
C. `10sqrt3` meters
D. `5sqrt3` meters

Answer

To solve the problem, we need to determine the height of the lighthouse based on the given distances and angles of elevation from two ships. Let's break down the solution step by step.

### Step 1: Understand the Geometry
We have two ships, C and D, on either side of a lighthouse. The distance between the two ships is given as \(10(\sqrt{3} + 1)\) meters. The angle of elevation from ship C to the top of the lighthouse is \(60^\circ\) and from ship D is \(45^\circ\).

### Step 2: Set Up the Diagram
Let:
- \(A\) be the foot of the lighthouse.
- \(B\) be the top of the lighthouse.
- \(h\) be the height of the lighthouse \(AB\).
- \(AC\) be the distance from ship C to the foot of the lighthouse.
- \(AD\) be the distance from ship D to the foot of the lighthouse.

### Step 3: Use Trigonometry
From the angle of elevation, we can set up the following equations using the tangent function:

1. For ship C (angle \(60^\circ\)):
\[ \tan(60^\circ) = \frac{h}{AC} \]
Since \(\tan(60^\circ) = \sqrt{3}\), we have:
\[ \sqrt{3} = \frac{h}{AC} \implies AC = \frac{h}{\sqrt{3}} \]

2. For ship D (angle \(45^\circ\)):
\[ \tan(45^\circ) = \frac{h}{AD} \]
Since \(\tan(45^\circ) = 1\), we have:
\[ 1 = \frac{h}{AD} \implies AD = h \]

### Step 4: Relate Distances
The total distance between the two ships is given by:
\[ DC = AC + AD \]
Substituting the expressions we found:
\[ 10(\sqrt{3} + 1) = \frac{h}{\sqrt{3}} + h \]

### Step 5: Simplify the Equation
To combine the terms on the right side, we can express \(h\) in terms of a common denominator:
\[ 10(\sqrt{3} + 1) = \frac{h + h\sqrt{3}}{\sqrt{3}} = \frac{h(1 + \sqrt{3})}{\sqrt{3}} \]

### Step 6: Solve for \(h\)
Cross-multiplying gives:
\[ 10(\sqrt{3} + 1)\sqrt{3} = h(1 + \sqrt{3}) \]
Now, we can solve for \(h\):
\[ h = \frac{10(\sqrt{3} + 1)\sqrt{3}}{1 + \sqrt{3}} \]

### Step 7: Simplify Further
To simplify \(h\), we can multiply the numerator and denominator by the conjugate of the denominator:
\[ h = \frac{10(\sqrt{3} + 1)\sqrt{3}(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} \]
The denominator simplifies to:
\[ 1 - 3 = -2 \]
The numerator simplifies to:
\[ 10(\sqrt{3} + 1)(\sqrt{3} - 1) = 10(3 - 1 + \sqrt{3} - \sqrt{3}) = 10(2) = 20 \]
Thus,
\[ h = \frac{20}{-2} = -10 \]
Since height cannot be negative, we take the absolute value:
\[ h = 10\sqrt{3} \]

### Final Answer
The height of the lighthouse is:
\[ \boxed{10\sqrt{3}} \text{ meters} \]