The tops of two poles of height 40 m and 25 m are connected by a wire. If the wire makes an angle `30^(@)` with the horizontal, then the length of the wire is

Options

A. 30 m
B. 20 m
C. 15 m
D. 25 m

Answer

To solve the problem, we need to find the length of the wire connecting the tops of two poles of heights 40 m and 25 m, making an angle of 30° with the horizontal.

### Step-by-Step Solution:

1. **Identify the heights of the poles:**
- Height of the first pole (Pole A) = 40 m
- Height of the second pole (Pole B) = 25 m

2. **Calculate the vertical distance between the tops of the poles:**
- Vertical distance (h) = Height of Pole A - Height of Pole B
- \( h = 40 \, \text{m} - 25 \, \text{m} = 15 \, \text{m} \)

3. **Understand the geometry of the situation:**
- The wire makes an angle of 30° with the horizontal.
- We can visualize a right triangle where:
- The vertical side (opposite) is the height difference (15 m).
- The angle with the horizontal is 30°.
- The length of the wire is the hypotenuse (let's denote it as L).

4. **Use the sine function to relate the height and the length of the wire:**
- From trigonometry, we know that:
\[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \]
- Here, \( \theta = 30° \), the opposite side is 15 m, and the hypotenuse is L.
- Therefore:
\[ \sin(30°) = \frac{15}{L} \]

5. **Substitute the value of \( \sin(30°) \):**
- We know that \( \sin(30°) = \frac{1}{2} \).
- So, we can write:
\[ \frac{1}{2} = \frac{15}{L} \]

6. **Solve for L:**
- Rearranging the equation gives:
\[ L = \frac{15}{\frac{1}{2}} = 15 \times 2 = 30 \, \text{m} \]

7. **Conclusion:**
- The length of the wire connecting the tops of the two poles is **30 meters**.