A vertical tower 50ft high stands on a sloping groud. The foot of the tower is at the same level as the middle point of a vertical flag pole. From the top of the tower the angle of depression of the top and the bottom of the pole are `15^(@)` and `45^(@)` respectively. Find the length of the pole.

Answer

To solve the problem step by step, we will use trigonometric relationships based on the angles of depression and the heights involved.

### Step 1: Understand the Setup
We have a vertical tower (OC) that is 50 ft high. The foot of the tower is at the same level as the midpoint (M) of a vertical flagpole (AB). The angles of depression from the top of the tower to the top (A) and bottom (B) of the pole are 15° and 45°, respectively.

### Step 2: Define Variables
Let:
- The height of the flagpole (AB) be denoted as \( h \).
- The distance from the foot of the tower (O) to the base of the flagpole (B) be denoted as \( x \).
- The height from the ground to the midpoint of the flagpole (M) is \( h/2 \).

### Step 3: Set Up the First Triangle (Top of the Pole)
From the top of the tower (C), the angle of depression to the top of the pole (A) is 15°. Therefore, in triangle OAC:
\[ \tan(15°) = \frac{50 - \frac{h}{2}}{x} \]
This can be rearranged to:
\[ 50 - \frac{h}{2} = x \cdot \tan(15°) \tag{1} \]

### Step 4: Set Up the Second Triangle (Bottom of the Pole)
From the top of the tower (C), the angle of depression to the bottom of the pole (B) is 45°. Therefore, in triangle OBC:
\[ \tan(45°) = \frac{50 + \frac{h}{2}}{x} \]
Since \( \tan(45°) = 1 \), we have:
\[ 50 + \frac{h}{2} = x \tag{2} \]

### Step 5: Solve the Equations
Now we have two equations:
1. \( 50 - \frac{h}{2} = x \cdot \tan(15°) \)
2. \( 50 + \frac{h}{2} = x \)

From equation (2), we can express \( x \):
\[ x = 50 + \frac{h}{2} \tag{3} \]

Substituting equation (3) into equation (1):
\[ 50 - \frac{h}{2} = (50 + \frac{h}{2}) \cdot \tan(15°) \]

### Step 6: Expand and Rearrange
Expanding the right-hand side:
\[ 50 - \frac{h}{2} = 50 \cdot \tan(15°) + \frac{h}{2} \cdot \tan(15°) \]
Rearranging gives:
\[ 50 - 50 \cdot \tan(15°) = \frac{h}{2} + \frac{h}{2} \cdot \tan(15°) \]
Factoring out \( \frac{h}{2} \):
\[ 50 - 50 \cdot \tan(15°) = \frac{h}{2} (1 + \tan(15°)) \]

### Step 7: Solve for \( h \)
Now, solving for \( h \):
\[ h = \frac{2(50 - 50 \cdot \tan(15°))}{1 + \tan(15°)} \]

### Step 8: Calculate \( h \)
Using the value of \( \tan(15°) \):
\[ \tan(15°) \approx 0.2679 \]
Substituting this value:
\[ h = \frac{2(50 - 50 \cdot 0.2679)}{1 + 0.2679} = \frac{2(50 - 13.395)}{1.2679} = \frac{2 \cdot 36.605}{1.2679} \approx \frac{73.21}{1.2679} \approx 57.76 \text{ ft} \]

### Step 9: Find the Length of the Pole
Since the length of the pole (AB) is \( h \), we have:
\[ \text{Length of the pole} = h \approx 57.76 \text{ ft} \]

### Final Answer
The length of the flagpole is approximately **57.76 ft**.