The angle of elevation of the top of a building from the foot of a tower is `30^(@)` . The angle of elevation of the top of the tower from the foot of the building is `60^(@)` . If the tower is 60 m high, find the height of the building.

Answer

To solve the problem, we will use trigonometric ratios. Let's break it down step by step.

### Step 1: Understand the Problem
We have a tower and a building. The height of the tower (let's denote it as \( T \)) is given as 60 m. We need to find the height of the building (let's denote it as \( H \)). The angles of elevation are given as follows:
- From the foot of the tower to the top of the building, the angle of elevation is \( 30^\circ \).
- From the foot of the building to the top of the tower, the angle of elevation is \( 60^\circ \).

### Step 2: Draw a Diagram
Draw a right triangle for both the tower and the building. Label the points:
- Let \( A \) be the foot of the tower.
- Let \( B \) be the top of the tower.
- Let \( C \) be the foot of the building.
- Let \( D \) be the top of the building.

### Step 3: Set Up the Triangles
1. In triangle \( ABC \) (where \( A \) is the foot of the tower, \( B \) is the top of the tower, and \( C \) is the foot of the building):
- The height of the tower \( AB = 60 \) m.
- The angle of elevation \( \angle CAB = 30^\circ \).
- The base \( AC \) is unknown.

2. In triangle \( BDC \) (where \( B \) is the top of the tower, \( D \) is the top of the building, and \( C \) is the foot of the building):
- The height of the tower \( BC = 60 \) m.
- The angle of elevation \( \angle DBC = 60^\circ \).
- The base \( DC \) is also unknown.

### Step 4: Use Trigonometric Ratios
Using the tangent function, we can set up equations for both triangles.

1. For triangle \( ABC \):
\[ \tan(30^\circ) = \frac{AB}{AC} = \frac{60}{AC} \]
We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \):
\[ \frac{1}{\sqrt{3}} = \frac{60}{AC} \]
Cross-multiplying gives:
\[ AC = 60 \sqrt{3} \]

2. For triangle \( BDC \):
\[ \tan(60^\circ) = \frac{BD}{DC} = \frac{H}{DC} \]
We know that \( \tan(60^\circ) = \sqrt{3} \):
\[ \sqrt{3} = \frac{H}{DC} \]
Rearranging gives:
\[ DC = \frac{H}{\sqrt{3}} \]

### Step 5: Relate the Two Bases
Since both triangles share the same base \( AC + DC \):
\[ AC + DC = 60 \sqrt{3} + \frac{H}{\sqrt{3}} \]

### Step 6: Set Up the Equation
Since \( AC + DC \) is equal to the distance between the foot of the tower and the foot of the building, we can set up the equation:
\[ 60 \sqrt{3} + \frac{H}{\sqrt{3}} = 60 \]

### Step 7: Solve for \( H \)
To solve for \( H \), we multiply through by \( \sqrt{3} \) to eliminate the fraction:
\[ 60 \cdot 3 + H = 60 \sqrt{3} \]
\[ 180 + H = 60 \sqrt{3} \]
Subtracting 180 from both sides gives:
\[ H = 60 \sqrt{3} - 180 \]

### Step 8: Calculate \( H \)
Using the approximate value of \( \sqrt{3} \approx 1.732 \):
\[ H \approx 60 \cdot 1.732 - 180 \approx 103.92 - 180 \approx -76.08 \]
This indicates a miscalculation in the setup. Let's correct it.

### Correct Calculation
Using the correct relationship:
1. From triangle \( ABC \):
\[ AC = 60 \sqrt{3} \]
2. From triangle \( BDC \):
\[ DC = \frac{H}{\sqrt{3}} \]
3. The total height from the foot of the tower to the top of the building is:
\[ H + 60 = 60 \sqrt{3} \]
Rearranging gives:
\[ H = 60 \sqrt{3} - 60 \]
\[ H = 60(\sqrt{3} - 1) \]

### Final Calculation
Using \( \sqrt{3} \approx 1.732 \):
\[ H \approx 60(1.732 - 1) = 60(0.732) \approx 43.92 \]

### Conclusion
The height of the building is approximately \( 43.92 \) meters.