From a point P on a level ground the angle of elevation to the top of the tower is `30^(@)` .If the tower is 100 metre high , the distance of point P from the foot of the tower is (Take `sqrt(3)=1.73)`

Options

A. 149 metre
B. 156 metre
C. 173 metre
D. 188 metre

Answer

To solve the problem, we will use trigonometric ratios. We are given the height of the tower and the angle of elevation from point P to the top of the tower.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Height of the tower (AB) = 100 meters
- Angle of elevation (∠APB) = 30 degrees

2. **Draw a Right Triangle:**
- Let P be the point on the ground.
- Let A be the foot of the tower (the base).
- Let B be the top of the tower.
- The triangle formed is triangle APB, where:
- AB is the height of the tower (perpendicular).
- AP is the distance from point P to the foot of the tower (base).
- PB is the line of sight from point P to the top of the tower.

3. **Use the Tangent Function:**
- In triangle APB, we can use the tangent of the angle of elevation:
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \]
- Here, \(\theta = 30^\circ\), the opposite side is AB (height of the tower = 100 m), and the adjacent side is AP (the distance we want to find).
- Therefore:
\[ \tan(30^\circ) = \frac{AB}{AP} \]
- We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\).

4. **Set Up the Equation:**
- Plugging in the values:
\[ \frac{1}{\sqrt{3}} = \frac{100}{AP} \]

5. **Cross-Multiply to Solve for AP:**
- Cross-multiplying gives:
\[ AP = 100 \cdot \sqrt{3} \]

6. **Substitute the Value of \(\sqrt{3}\):**
- Given that \(\sqrt{3} \approx 1.73\):
\[ AP = 100 \cdot 1.73 = 173 \text{ meters} \]

7. **Conclusion:**
- The distance of point P from the foot of the tower is **173 meters**.