Two points P and Q are at the distance of x and y (where `ygtx)` respectivelyf from the base of a building and on a straight line.If the angles of elevation of the top of the building from points P and Q are complementary ,then what is the height of the building ?

Options

A. xy
B. `sqrt(((y)/(x)))`
C. `sqrt(((x)/(y)))`
D. `sqrt((xy))`

Answer

To solve the problem step by step, we can follow the reasoning laid out in the video transcript.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a building with height \( h \).
- Point \( P \) is at a distance \( x \) from the base of the building.
- Point \( Q \) is at a distance \( y \) from the base of the building (where \( y > x \)).
- The angles of elevation from points \( P \) and \( Q \) to the top of the building are complementary, meaning they add up to \( 90^\circ \).

2. **Setting Up the Angles**:
- Let the angle of elevation from point \( P \) be \( \theta \).
- Therefore, the angle of elevation from point \( Q \) will be \( 90^\circ - \theta \).

3. **Using Trigonometric Ratios**:
- From point \( P \):
\[ \tan(\theta) = \frac{h}{x} \]
This implies:
\[ h = x \tan(\theta) \]

- From point \( Q \):
\[ \tan(90^\circ - \theta) = \cot(\theta) = \frac{h}{y} \]
This implies:
\[ h = y \cot(\theta) \]

4. **Equating the Two Expressions for Height**:
- Since both expressions equal \( h \), we can set them equal to each other:
\[ x \tan(\theta) = y \cot(\theta) \]

5. **Using the Identity for Cotangent**:
- Recall that \( \cot(\theta) = \frac{1}{\tan(\theta)} \). Substituting this into the equation gives:
\[ x \tan(\theta) = \frac{y}{\tan(\theta)} \]

6. **Multiplying Through by \( \tan(\theta) \)**:
- To eliminate the fraction, multiply both sides by \( \tan(\theta) \):
\[ x \tan^2(\theta) = y \]

7. **Rearranging the Equation**:
- We can rearrange this to find \( \tan^2(\theta) \):
\[ \tan^2(\theta) = \frac{y}{x} \]

8. **Finding the Height**:
- Now substituting \( \tan^2(\theta) \) back into the expression for height:
\[ h = x \tan(\theta) = x \sqrt{\tan^2(\theta)} = x \sqrt{\frac{y}{x}} = \sqrt{xy} \]

### Final Answer:
The height of the building \( h \) is given by:
\[ h = \sqrt{xy} \]