A ladder rests against a vertical wall at an angle `alpha` to the horizontal. If the foot is pulled away through a distance 2m, then it slides a distance 5 m down the wall, finally making an angle `beta ` with the horizontal. The value of `tan((alpha+beta)/(2))` is equal to

Options

A. `(2)/(5)`
B. `(5)/(2)`
C. 10
D. None of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Geometry
We have a ladder of length \( L \) resting against a vertical wall at an angle \( \alpha \) to the horizontal. When the foot of the ladder is pulled away by 2 m, the top of the ladder slides down by 5 m, making a new angle \( \beta \) with the horizontal.

### Step 2: Set Up the Initial and Final Positions
Let:
- \( A \) be the point where the ladder touches the wall.
- \( B \) be the point where the ladder touches the ground.
- \( P \) be the point where the foot of the ladder was initially.
- \( Q \) be the point where the foot of the ladder is after being pulled away.

From the problem:
- The distance \( AP \) (initial height of the ladder on the wall) can be expressed as \( L \sin \alpha \).
- The distance \( AQ \) (final height of the ladder on the wall) can be expressed as \( L \sin \beta \).
- The horizontal distance \( BP \) is \( L \cos \alpha \).
- The horizontal distance \( BQ \) is \( L \cos \beta \).

### Step 3: Write the Equations
From the problem, we know:
1. \( AQ - AP = 5 \) (the height decreased by 5 m)
2. \( BQ - BP = 2 \) (the foot was pulled away by 2 m)

This gives us:
\[ L \sin \beta - L \sin \alpha = -5 \quad \text{(1)} \]
\[ L \cos \beta - L \cos \alpha = 2 \quad \text{(2)} \]

### Step 4: Simplify the Equations
From equations (1) and (2), we can factor out \( L \):
\[ L (\sin \beta - \sin \alpha) = -5 \quad \text{(3)} \]
\[ L (\cos \beta - \cos \alpha) = 2 \quad \text{(4)} \]

### Step 5: Divide the Equations
Dividing equation (3) by equation (4):
\[ \frac{\sin \beta - \sin \alpha}{\cos \beta - \cos \alpha} = \frac{-5}{2} \]

### Step 6: Use Trigonometric Identities
Using the identity for the difference of sines and cosines:
\[ \frac{\sin \beta - \sin \alpha}{\cos \beta - \cos \alpha} = \frac{2 \cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\beta - \alpha}{2}\right)}{-2 \sin\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\beta - \alpha}{2}\right)} \]
This simplifies to:
\[ -\cot\left(\frac{\alpha + \beta}{2}\right) = \frac{5}{2} \]

### Step 7: Find \( \tan\left(\frac{\alpha + \beta}{2}\right) \)
Taking the reciprocal gives:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = -\frac{2}{5} \]

### Conclusion
Thus, the value of \( \tan\left(\frac{\alpha + \beta}{2}\right) \) is:
\[ \frac{2}{5} \]