From two points , lying on the same horizontal line , the angles of elevation of the top of the pilar are `theta and phi(thetaltphi)` . If the height of the pillar is h m and the two points lie on the same sides of the pillar , then the the distance between the two points is

Options

A. `h(tantheta-tanphi)` metre
B. `h(cotphi-cottheta)` metre
C. `h(cottheta-cotphi)` metre
D. `h((tanthetatanphi)/(tanphi-tantheta))` metre

Answer

To solve the problem step by step, we will use trigonometric principles to derive the distance between the two points based on the angles of elevation and the height of the pillar.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a pillar of height \( h \) meters.
- There are two points \( A \) and \( B \) on the same horizontal line, with angles of elevation \( \phi \) and \( \theta \) respectively, where \( \phi > \theta \).
- Let \( x \) be the horizontal distance from point \( A \) to the base of the pillar, and \( D \) be the distance between points \( A \) and \( B \).

2. **Setting Up the Triangles**:
- From point \( A \), we can form triangle \( AOC \) where \( O \) is the base of the pillar.
- From point \( B \), we can form triangle \( BOC \).

3. **Using Trigonometric Ratios**:
- For triangle \( AOC \):
\[ \tan(\phi) = \frac{h}{x} \quad \text{(1)} \]
- For triangle \( BOC \):
\[ \tan(\theta) = \frac{h}{D + x} \quad \text{(2)} \]

4. **Rearranging the Equations**:
- From equation (1), we can express \( x \):
\[ x = \frac{h}{\tan(\phi)} \quad \text{(3)} \]
- From equation (2), we can express \( D + x \):
\[ D + x = \frac{h}{\tan(\theta)} \quad \text{(4)} \]

5. **Substituting for \( x \)**:
- Substitute equation (3) into equation (4):
\[ D + \frac{h}{\tan(\phi)} = \frac{h}{\tan(\theta)} \]

6. **Solving for \( D \)**:
- Rearranging gives:
\[ D = \frac{h}{\tan(\theta)} - \frac{h}{\tan(\phi)} \]
- Factor out \( h \):
\[ D = h \left( \frac{1}{\tan(\theta)} - \frac{1}{\tan(\phi)} \right) \]

7. **Using Cotangent**:
- Recall that \( \frac{1}{\tan(x)} = \cot(x) \):
\[ D = h \left( \cot(\theta) - \cot(\phi) \right) \]

### Final Result:
The distance \( D \) between the two points is given by:
\[ D = h \left( \cot(\theta) - \cot(\phi) \right) \]