A tower stands vertically on the ground. From a point on the ground, which is `15 m` away from the foot of the tower, the angle of elevation of the top of the tower is found to be `60^@`. Find the height of the tower.

Options

A. `12sqrt3`
B. `15sqrt3`
C. `13sqrt3`
D. None

Explanation

To find the height of the tower, we can use the properties of right triangles and trigonometric ratios. Here's a step-by-step solution:

### Step 1: Understand the Problem
We have a vertical tower (let's denote the height of the tower as \( AB \)) and a point on the ground (point \( C \)) that is 15 meters away from the base of the tower (point \( A \)). The angle of elevation from point \( C \) to the top of the tower \( B \) is \( 60^\circ \).

### Step 2: Draw a Diagram
Draw a right triangle \( ABC \) where:
- \( A \) is the foot of the tower,
- \( B \) is the top of the tower,
- \( C \) is the point on the ground from where the angle of elevation is measured.

In this triangle:
- \( AC = 15 \) m (the distance from point \( C \) to the foot of the tower),
- \( AB = h \) m (the height of the tower),
- \( \angle ACB = 60^\circ \).

### Step 3: Use the Tangent Function
In right triangle \( ABC \), we can use the tangent function, which relates the opposite side to the adjacent side:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
Here, \( \theta = 60^\circ \), the opposite side is \( AB \) (height of the tower), and the adjacent side is \( AC \) (distance from the tower).

Thus, we have:
\[ \tan(60^\circ) = \frac{AB}{AC} \]

### Step 4: Substitute Known Values
Substituting the known values into the equation:
\[ \tan(60^\circ) = \frac{h}{15} \]
We know that \( \tan(60^\circ) = \sqrt{3} \). Therefore, we can write:
\[ \sqrt{3} = \frac{h}{15} \]

### Step 5: Solve for \( h \)
To find \( h \), multiply both sides by 15:
\[ h = 15 \cdot \sqrt{3} \]

### Step 6: Final Answer
Thus, the height of the tower is:
\[ h = 15\sqrt{3} \text{ meters} \]