A tower subtends an angle of `60^(@)` at a point on the same level as the foot of the tower and at a second point just 10 meters above the first point the angle of depression of the foot of the tower is `15^(@)`. The height of the tower is (in meters)

Options

A. `(10)/(sqrt3)(2-sqrt3)`
B. `10sqrt3(2-sqrt3)`
C. `(10)/(sqrt3)(2+sqrt3)`
D. `10sqrt3(2+sqrt3)`

Answer

To solve the problem step by step, we will use trigonometric relationships to find the height of the tower.

### Step 1: Understand the Geometry
Let:
- \( AB \) be the height of the tower.
- \( C \) be the point on the ground where the angle of elevation to the top of the tower is \( 60^\circ \).
- \( D \) be the point 10 meters above point \( C \) where the angle of depression to the foot of the tower \( A \) is \( 15^\circ \).

### Step 2: Set Up the Right Triangle for the First Point
From point \( C \):
- The angle of elevation to the top of the tower is \( 60^\circ \).
- Let \( BD \) be the horizontal distance from point \( C \) to the base of the tower \( A \), which we will denote as \( x \).

Using the tangent function:
\[ \tan(60^\circ) = \frac{AB}{BC} = \frac{H}{x} \]
Where \( H \) is the height of the tower.

Since \( \tan(60^\circ) = \sqrt{3} \):
\[ \sqrt{3} = \frac{H}{x} \implies H = x \sqrt{3} \quad \text{(Equation 1)} \]

### Step 3: Set Up the Right Triangle for the Second Point
From point \( D \):
- The angle of depression to the foot of the tower \( A \) is \( 15^\circ \).
- The height from point \( D \) to the ground is \( 10 \) meters.

Using the tangent function again:
\[ \tan(15^\circ) = \frac{AB - 10}{BD} = \frac{H - 10}{x} \]

Since \( \tan(15^\circ) = 2 - \sqrt{3} \):
\[ 2 - \sqrt{3} = \frac{H - 10}{x} \implies H - 10 = x(2 - \sqrt{3}) \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
Now we have two equations:
1. \( H = x \sqrt{3} \)
2. \( H - 10 = x(2 - \sqrt{3}) \)

Substituting Equation 1 into Equation 2:
\[ x \sqrt{3} - 10 = x(2 - \sqrt{3}) \]
Rearranging gives:
\[ x \sqrt{3} - x(2 - \sqrt{3}) = 10 \]
Factoring out \( x \):
\[ x(\sqrt{3} - (2 - \sqrt{3})) = 10 \]
\[ x(2\sqrt{3} - 2) = 10 \]
\[ x = \frac{10}{2(\sqrt{3} - 1)} = \frac{5}{\sqrt{3} - 1} \]

### Step 5: Substitute \( x \) Back to Find \( H \)
Now substitute \( x \) back into Equation 1:
\[ H = x \sqrt{3} = \frac{5 \sqrt{3}}{\sqrt{3} - 1} \]

### Step 6: Rationalize the Denominator
To rationalize:
\[ H = \frac{5 \sqrt{3} (\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{5 \sqrt{3} (\sqrt{3} + 1)}{3 - 1} = \frac{5 \sqrt{3} (\sqrt{3} + 1)}{2} \]
\[ = \frac{5(3 + \sqrt{3})}{2} = \frac{15 + 5\sqrt{3}}{2} \]

### Conclusion
The height of the tower \( H \) is:
\[ H = 15 + 5\sqrt{3} \text{ meters} \]