The distance between two pillars of length 16 metres and 9 metres is x metres . If two angles of elevation of their respective top from the bottom of the other are compementary to each other , then the value of x (in metres ) is

Options

A. 15
B. 16
C. 12
D. 9

Answer

To solve the problem, we need to find the distance \( x \) between two pillars of heights 16 meters and 9 meters, given that the angles of elevation from the top of one pillar to the top of the other are complementary.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two pillars: one is 16 meters tall and the other is 9 meters tall. The distance between the bases of these two pillars is \( x \) meters. The angles of elevation from the top of one pillar to the top of the other are complementary.

2. **Setting Up the Angles**:
Let \( \theta \) be the angle of elevation from the top of the 9-meter pillar to the top of the 16-meter pillar. Therefore, the angle of elevation from the top of the 16-meter pillar to the top of the 9-meter pillar will be \( 90^\circ - \theta \).

3. **Using the Tangent Function**:
The tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side. For the angle \( \theta \):
\[ \tan(\theta) = \frac{16}{x} \]
For the angle \( 90^\circ - \theta \):
\[ \tan(90^\circ - \theta) = \cot(\theta) = \frac{9}{x} \]

4. **Relating the Tangents**:
We know that:
\[ \tan(90^\circ - \theta) = \frac{1}{\tan(\theta)} \]
Therefore, we can write:
\[ \frac{9}{x} = \frac{1}{\tan(\theta)} \]

5. **Substituting for \( \tan(\theta) \)**:
From the first equation, we have:
\[ \tan(\theta) = \frac{16}{x} \]
Substituting this into the equation gives:
\[ \frac{9}{x} = \frac{x}{16} \]

6. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 9 \cdot 16 = x^2 \]
Simplifying this:
\[ x^2 = 144 \]

7. **Taking the Square Root**:
Taking the square root of both sides, we find:
\[ x = 12 \]

### Final Answer:
The value of \( x \) is \( 12 \) meters.

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