A lower of height 15 m stands vertically on the ground. From a point on the ground the angle of elevation of the top of the tower is found to be `30^(@).` What is the distance of the point from the foot of the tower?

Options

A. `15 sqrt3 m `
B. `10 sqrt3 `
C. `5 sqrt3`
D. `30`

Answer

To solve the problem, we can 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 foot of the tower.

### Step-by-Step Solution:

1. **Identify the given values:**
- Height of the tower (h) = 15 m
- Angle of elevation (θ) = 30°

2. **Draw a right triangle:**
- Let the foot of the tower be point A, the top of the tower be point B, and the point on the ground from where the angle is measured be point C.
- In triangle ABC, AB is the height of the tower (15 m), and AC is the distance from the point C to the foot of the tower A, which we need to find.

3. **Use the tangent function:**
- The tangent of an angle in a right triangle is defined as the ratio of the opposite side to the adjacent side.
- Here, we have:
\[ \tan(θ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{AC} \]
- Substituting the known values:
\[ \tan(30°) = \frac{15}{AC} \]

4. **Find the value of tan(30°):**
- We know that:
\[ \tan(30°) = \frac{1}{\sqrt{3}} \]

5. **Set up the equation:**
- Now substituting this value into the equation:
\[ \frac{1}{\sqrt{3}} = \frac{15}{AC} \]

6. **Cross-multiply to solve for AC:**
- Cross-multiplying gives:
\[ AC = 15 \sqrt{3} \]

7. **Calculate the distance:**
- The distance from the point on the ground to the foot of the tower is:
\[ AC = 15 \sqrt{3} \approx 25.98 \text{ m} \]

### Final Answer:
The distance of the point from the foot of the tower is \( 15 \sqrt{3} \) meters or approximately 25.98 meters.

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