From the top of a building, the angle of elevation and depression of top and bottom of a tower are `60^(@) and 30^(@)` respectively. If the height of the building is `5` m, then find the height of the tower.

Options

A. `10sqrt3`m
B. 15 m
C. `15sqrt3`m
D. 20 m

Explanation

To solve the problem, we will use trigonometric ratios, specifically the tangent function, which relates the angles of elevation and depression to the heights and distances involved.

### Step-by-Step Solution:

1. **Understand the Problem:**
- We have a building (AB) of height 5 m.
- From the top of the building (point A), the angle of elevation to the top of the tower (point C) is 60°.
- The angle of depression to the bottom of the tower (point D) is 30°.
- We need to find the height of the tower (CD).

2. **Identify the Triangles:**
- Triangle ABD (for angle of depression) and triangle ACD (for angle of elevation).

3. **Calculate the Distance BD (horizontal distance from the building to the tower):**
- In triangle ABD, we have:
- Height of the building (AB) = 5 m (perpendicular)
- Angle of depression (∠BAD) = 30°
- Using the tangent function:
\[ \tan(30°) = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{5}{BD} \]
- Rearranging gives:
\[ BD = 5\sqrt{3} \text{ m} \]

4. **Calculate the Height of the Tower (CD):**
- In triangle ACD, we have:
- Angle of elevation (∠CAD) = 60°
- The horizontal distance (AD) = BD = 5√3 m
- Using the tangent function again:
\[ \tan(60°) = \frac{CD}{AD} \implies \sqrt{3} = \frac{CD}{5\sqrt{3}} \]
- Rearranging gives:
\[ CD = 5\sqrt{3} \cdot \sqrt{3} = 15 \text{ m} \]

5. **Total Height of the Tower (EC):**
- The total height of the tower (EC) is the sum of the height of the building (AB) and the height of the tower (CD):
\[ EC = AB + CD = 5 + 15 = 20 \text{ m} \]

### Final Answer:
The height of the tower is **20 meters**.