The angle of elevation of a tower at a point d meter away from its base is `30^(@)`. If the tower is 15 m high, then 3d = _________ m

Answer

To solve the problem step by step, we will use the concept of trigonometry, specifically the tangent function, which relates the angle of elevation to the height of the tower and the distance from the base of the tower.

### Step-by-Step Solution:

1. **Understand the Problem**: We have a tower of height 15 meters and a point D meters away from its base where the angle of elevation to the top of the tower is 30 degrees. We need to find the value of 3d.

2. **Draw a Diagram**: Visualize the scenario by drawing a right triangle where:
- The height of the tower (perpendicular) is 15 meters.
- The distance from the base of the tower to the point (base) is d meters.
- The angle of elevation from the point to the top of the tower is 30 degrees.

3. **Use the Tangent Function**: In the right triangle formed, the tangent of the angle of elevation (30 degrees) is given by the formula:
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \]
Here, the opposite side is the height of the tower (15 m) and the adjacent side is the distance from the base (d m). Therefore:
\[ \tan(30^\circ) = \frac{15}{d} \]

4. **Substitute the Value of Tan(30°)**: We know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
So we can write:
\[ \frac{1}{\sqrt{3}} = \frac{15}{d} \]

5. **Cross Multiply to Solve for d**: Cross multiplying gives us:
\[ d = 15 \cdot \sqrt{3} \]

6. **Calculate 3d**: Now, we need to find 3d:
\[ 3d = 3 \cdot (15 \cdot \sqrt{3}) = 45 \cdot \sqrt{3} \]

7. **Final Answer**: Therefore, the value of 3d is:
\[ 3d = 45\sqrt{3} \text{ meters} \]