Two towers of the same height stand on either side of a road 60 m wide. At a point on the road between the towers, the elevations of the towers are `60^(@)` and `30^(@)`. Find the height of the towers and the position of the point.

Answer

To solve the problem, we need to find the height of the towers and the position of the point on the road where the angles of elevation are given. Let's denote the height of the towers as \( h \) and the distance from the first tower to the point as \( x \). The distance from the second tower to the point will then be \( 60 - x \) since the total width of the road is 60 m.

### Step-by-Step Solution:

1. **Identify the Angles and Set Up the Triangles**:
- The angle of elevation from the point to the first tower is \( 60^\circ \).
- The angle of elevation from the point to the second tower is \( 30^\circ \).

2. **Use the Tangent Function**:
- For the first tower (with angle \( 60^\circ \)):
\[ \tan(60^\circ) = \frac{h}{x} \]
Since \( \tan(60^\circ) = \sqrt{3} \), we can write:
\[ \sqrt{3} = \frac{h}{x} \quad \Rightarrow \quad h = x \sqrt{3} \quad \text{(Equation 1)} \]

- For the second tower (with angle \( 30^\circ \)):
\[ \tan(30^\circ) = \frac{h}{60 - x} \]
Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we can write:
\[ \frac{1}{\sqrt{3}} = \frac{h}{60 - x} \quad \Rightarrow \quad h = \frac{(60 - x)}{\sqrt{3}} \quad \text{(Equation 2)} \]

3. **Equate the Two Expressions for Height**:
- From Equation 1 and Equation 2, we have:
\[ x \sqrt{3} = \frac{(60 - x)}{\sqrt{3}} \]

4. **Cross-Multiply to Solve for \( x \)**:
- Multiplying both sides by \( \sqrt{3} \):
\[ 3x = 60 - x \]
- Rearranging gives:
\[ 3x + x = 60 \quad \Rightarrow \quad 4x = 60 \quad \Rightarrow \quad x = 15 \]

5. **Substitute \( x \) Back to Find \( h \)**:
- Substitute \( x = 15 \) into Equation 1:
\[ h = 15 \sqrt{3} \]

### Final Answers:
- The height of the towers is \( 15 \sqrt{3} \) meters.
- The position of the point is 15 meters from the first tower.