A vertical pole stands on the level ground. From a point on the ground, 25m away from the foot of the pole , the angle of elevation of its top is found to be `60^(@)`. Find the height of the pole.

Options

A. `25sqrt(3)`
B. `50sqrt(3)`
C. `55sqrt(3)`
D. `65sqrt(3)`

Explanation

To find the height of the vertical pole, we can use the concept of trigonometry, specifically the tangent function, which relates the angle of elevation to the opposite side (height of the pole) and the adjacent side (distance from the pole).

### Step-by-Step Solution:

1. **Identify the Triangle**:
- Let the height of the pole be \( h \) meters.
- The distance from the point on the ground to the foot of the pole is 25 meters.
- The angle of elevation from this point to the top of the pole is \( 60^\circ \).

2. **Use the Tangent Function**:
- In the right triangle formed, we can use the tangent of the angle of elevation:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
- Here, \( \theta = 60^\circ \), the opposite side is the height of the pole \( h \), and the adjacent side is the distance from the pole, which is 25 meters.
- Therefore, we can write:
\[ \tan(60^\circ) = \frac{h}{25} \]

3. **Substitute the Value of Tangent**:
- We know that \( \tan(60^\circ) = \sqrt{3} \) or approximately \( 1.732 \).
- Substituting this value into the equation gives:
\[ \sqrt{3} = \frac{h}{25} \]

4. **Solve for \( h \)**:
- To find \( h \), multiply both sides by 25:
\[ h = 25 \cdot \sqrt{3} \]
- Now, substituting the approximate value of \( \sqrt{3} \):
\[ h = 25 \cdot 1.732 \]
- Calculating this gives:
\[ h \approx 43.3 \text{ meters} \]

5. **Conclusion**:
- The height of the pole is approximately \( 43.3 \) meters.