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 step by step, we will denote the height of the lighthouse as \( h \) (or \( X \) as mentioned in the transcript). We will also denote the distances from the base of the lighthouse to the ships as follows:

- Let \( S_1 \) be the ship on one side of the lighthouse, and let the distance from the base of the lighthouse to \( S_1 \) be \( y \).
- Let \( S_2 \) be the ship on the other side of the lighthouse, and let the distance from the base of the lighthouse to \( S_2 \) be \( z \).

Given:
- The distance between the two ships \( S_1 \) and \( S_2 \) is \( y + z = 10(\sqrt{3} + 1) \).
- The angle of elevation from \( S_1 \) (where the angle is \( 60^\circ \)) and from \( S_2 \) (where the angle is \( 45^\circ \)).

### Step 1: Set up the equations using trigonometric ratios

1. For ship \( S_2 \) (angle of elevation \( 45^\circ \)):
\[ \tan(45^\circ) = \frac{h}{z} \]
Since \( \tan(45^\circ) = 1 \):
\[ h = z \quad \text{(Equation 1)} \]

2. For ship \( S_1 \) (angle of elevation \( 60^\circ \)):
\[ \tan(60^\circ) = \frac{h}{y} \]
Since \( \tan(60^\circ) = \sqrt{3} \):
\[ \sqrt{3} = \frac{h}{y} \implies h = y\sqrt{3} \quad \text{(Equation 2)} \]

### Step 2: Substitute and solve for \( h \)

From Equation 1, we have:
\[ z = h \]
From Equation 2, we have:
\[ y = \frac{h}{\sqrt{3}} \]

### Step 3: Substitute \( y \) and \( z \) into the distance equation

Substituting \( y \) and \( z \) into the distance equation:
\[ y + z = 10(\sqrt{3} + 1) \]
Substituting the values of \( y \) and \( z \):
\[ \frac{h}{\sqrt{3}} + h = 10(\sqrt{3} + 1) \]

### Step 4: Combine terms

Combine the terms on the left:
\[ h\left(\frac{1}{\sqrt{3}} + 1\right) = 10(\sqrt{3} + 1) \]

### Step 5: Solve for \( h \)

To simplify:
\[ h\left(\frac{1 + \sqrt{3}}{\sqrt{3}}\right) = 10(\sqrt{3} + 1) \]
Now, multiply both sides by \( \sqrt{3} \):
\[ h(1 + \sqrt{3}) = 10\sqrt{3}(\sqrt{3} + 1) \]
Calculating the right side:
\[ 10\sqrt{3}(\sqrt{3} + 1) = 10(3 + \sqrt{3}) = 30 + 10\sqrt{3} \]
Now, divide both sides by \( 1 + \sqrt{3} \):
\[ h = \frac{30 + 10\sqrt{3}}{1 + \sqrt{3}} \]

### Step 6: Rationalize the denominator

Multiply the numerator and denominator by the conjugate of the denominator:
\[ h = \frac{(30 + 10\sqrt{3})(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} \]
Calculating the denominator:
\[ (1 + \sqrt{3})(1 - \sqrt{3}) = 1 - 3 = -2 \]
Calculating the numerator:
\[ (30 + 10\sqrt{3})(1 - \sqrt{3}) = 30 - 30\sqrt{3} + 10\sqrt{3} - 30 = -20\sqrt{3} \]
Thus:
\[ h = \frac{-20\sqrt{3}}{-2} = 10\sqrt{3} \]

### Final Answer:
The height of the lighthouse is \( 10\sqrt{3} \) meters.