Two houses are collinear with the base of a tower and are at distance 3 m and 12 m (on the same side) from the base of the tower. The angles of elevation from these two houses of the top of the tower are complementary. What is the height of the tower ?

Options

A. 4 m
B. 6 m
C. 7 . 5 m
D. 36 m

Answer

To solve the problem step by step, we will use trigonometric ratios and the properties of complementary angles.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let the height of the tower be \( h \).
- The distance from the first house (point A) to the base of the tower is 3 m.
- The distance from the second house (point B) to the base of the tower is 12 m.
- The angles of elevation from these two houses to the top of the tower are complementary. If we denote the angle of elevation from house A as \( \alpha \), then the angle of elevation from house B will be \( 90^\circ - \alpha \).

2. **Setting Up the Trigonometric Equations**:
- For house A (3 m from the tower):
\[ \tan(\alpha) = \frac{h}{3} \quad \text{(Equation 1)} \]
- For house B (12 m from the tower):
\[ \tan(90^\circ - \alpha) = \frac{h}{12} \]
Since \( \tan(90^\circ - \alpha) = \cot(\alpha) \), we can rewrite this as:
\[ \cot(\alpha) = \frac{h}{12} \quad \text{(Equation 2)} \]

3. **Relating the Two Equations**:
- From Equation 1, we have:
\[ \tan(\alpha) = \frac{h}{3} \implies \cot(\alpha) = \frac{3}{h} \]
- Substitute this into Equation 2:
\[ \frac{3}{h} = \frac{h}{12} \]

4. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ 3 \cdot 12 = h \cdot h \]
\[ 36 = h^2 \]

5. **Finding the Height**:
- Taking the square root of both sides:
\[ h = \sqrt{36} = 6 \text{ m} \]

### Final Answer:
The height of the tower is **6 meters**.