A ladder rests against a vertical wall at inclination `alpha` to the horizontal. Its foot is pulled away from the wall through a distance p so that it's upper end slides q down the wall and then ladder make an angle `beta` to the horizontal then p/q =

Options

A. `( cos beta - cos alpha)/(sin alpha - sin beta)`.
B. `( cos alpha - sin alpha)/(sin alpha - sin beta)`.
C. `( cos beta - cos alpha)/(sin beta - cos beta)`.
D. None

Explanation

To solve the problem step by step, we will analyze the situation involving the ladder, the wall, and the angles involved.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let the ladder be represented by line segment \( AB \), where \( A \) is the top of the ladder resting against the wall and \( B \) is the foot of the ladder on the ground.
- Initially, the ladder makes an angle \( \alpha \) with the horizontal ground.

2. **Initial Positions**:
- Let the height of the ladder at point \( A \) from the ground be \( h \).
- The foot of the ladder \( B \) is at a distance \( d \) from the wall.

3. **Initial Length of the Ladder**:
- The length of the ladder \( L \) can be expressed using trigonometric functions:
\[ L = \frac{h}{\sin(\alpha)} = \frac{d}{\cos(\alpha)} \]

4. **After Pulling the Ladder**:
- The foot of the ladder is pulled away from the wall by a distance \( p \), making the new position of \( B \) at \( d + p \).
- The upper end \( A \) slides down the wall by a distance \( q \), making the new height \( h - q \).

5. **New Length of the Ladder**:
- The length of the ladder remains the same, so we can express it again:
\[ L = \frac{h - q}{\sin(\beta)} = \frac{d + p}{\cos(\beta)} \]

6. **Setting Up the Equations**:
- From the initial position:
\[ L = \frac{h}{\sin(\alpha)} \quad (1) \]
- From the new position:
\[ L = \frac{h - q}{\sin(\beta)} \quad (2) \]
- Equating equations (1) and (2):
\[ \frac{h}{\sin(\alpha)} = \frac{h - q}{\sin(\beta)} \]

7. **Rearranging the Equation**:
- Cross-multiplying gives:
\[ h \sin(\beta) = (h - q) \sin(\alpha) \]
- Expanding and rearranging:
\[ h \sin(\beta) = h \sin(\alpha) - q \sin(\alpha) \]
\[ h (\sin(\beta) - \sin(\alpha)) = -q \sin(\alpha) \]
- Therefore:
\[ h = \frac{-q \sin(\alpha)}{\sin(\beta) - \sin(\alpha)} \quad (3) \]

8. **Using the New Length Equation**:
- From the new position, we also have:
\[ L = \frac{d + p}{\cos(\beta)} \quad (4) \]
- Equating (1) and (4):
\[ \frac{h}{\sin(\alpha)} = \frac{d + p}{\cos(\beta)} \]

9. **Rearranging Again**:
- Cross-multiplying gives:
\[ h \cos(\beta) = (d + p) \sin(\alpha) \]
- Rearranging:
\[ h = \frac{(d + p) \sin(\alpha)}{\cos(\beta)} \quad (5) \]

10. **Equating (3) and (5)**:
- Setting the two expressions for \( h \) equal:
\[ \frac{-q \sin(\alpha)}{\sin(\beta) - \sin(\alpha)} = \frac{(d + p) \sin(\alpha)}{\cos(\beta)} \]

11. **Simplifying**:
- Cancel \( \sin(\alpha) \) from both sides (assuming \( \sin(\alpha) \neq 0 \)):
\[ \frac{-q}{\sin(\beta) - \sin(\alpha)} = \frac{d + p}{\cos(\beta)} \]
- Rearranging gives:
\[ p = \frac{(d + q)(\cos(\beta))}{\sin(\beta) - \sin(\alpha)} \]

12. **Final Ratio**:
- We can express \( \frac{p}{q} \):
\[ \frac{p}{q} = \frac{\cos(\beta) - \cos(\alpha)}{\sin(\alpha) - \sin(\beta)} \]

### Final Result:
\[ \frac{p}{q} = \frac{\cos(\beta) - \cos(\alpha)}{\sin(\alpha) - \sin(\beta)} \]