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, we will break it down step by step.

### Step 1: Understand the Problem
We have a vertical tower (AB) that is 50 feet high. The foot of the tower (point B) is at the same level as the midpoint of a vertical flag pole (point E). The angles of depression from the top of the tower (point A) to the top (point C) and bottom (point D) of the pole are given as 15° and 45° respectively.

### Step 2: Draw the Diagram
1. Draw a vertical line representing the tower AB, where AB = 50 ft.
2. Draw the flag pole vertically, with point E being the midpoint of the pole.
3. Mark points C (top of the pole) and D (bottom of the pole).
4. Connect point A (top of the tower) to points C and D, forming angles of depression of 15° and 45° respectively.

### Step 3: Set Up the Problem
Let:
- The height of the pole (CD) be \( h \).
- The distance from point B (foot of the tower) to point D (bottom of the pole) be \( x \).
- The distance from point B to point C (top of the pole) will then be \( h + x \).

### Step 4: Use Trigonometric Ratios
From the angle of depression:
1. For angle 45° (from A to D):
\[ \tan(45°) = \frac{AB}{BD} = \frac{50}{x} \]
Since \( \tan(45°) = 1 \):
\[ 1 = \frac{50}{x} \implies x = 50 \text{ ft} \]

2. For angle 15° (from A to C):
\[ \tan(15°) = \frac{AB}{BC} = \frac{50}{h + x} \]
We know \( x = 50 \):
\[ \tan(15°) = \frac{50}{h + 50} \]

### Step 5: Solve for \( h \)
Using the value of \( \tan(15°) \):
\[ \tan(15°) = 2 - \sqrt{3} \quad \text{(approx. 0.2679)} \]
Setting up the equation:
\[ 2 - \sqrt{3} = \frac{50}{h + 50} \]
Cross-multiplying gives:
\[ (2 - \sqrt{3})(h + 50) = 50 \]
Expanding:
\[ (2 - \sqrt{3})h + 100 - 50\sqrt{3} = 50 \]
Rearranging:
\[ (2 - \sqrt{3})h = 50 + 50\sqrt{3} - 100 \]
\[ (2 - \sqrt{3})h = 50\sqrt{3} - 50 \]
\[ h = \frac{50\sqrt{3} - 50}{2 - \sqrt{3}} \]

### Step 6: Simplify \( h \)
To simplify \( h \):
Multiply numerator and denominator by the conjugate of the denominator:
\[ h = \frac{(50\sqrt{3} - 50)(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} \]
Calculating the denominator:
\[ (2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1 \]
Thus:
\[ h = (50\sqrt{3} - 50)(2 + \sqrt{3}) = 100\sqrt{3} - 100 + 150 - 50\sqrt{3} \]
Combining like terms:
\[ h = 50\sqrt{3} + 50 \]

### Step 7: Find the Length of the Pole
Since \( E \) is the midpoint of the pole:
\[ \text{Length of the pole} = 2h = 2(50\sqrt{3} + 50) = 100\sqrt{3} + 100 \]

### Final Answer
The length of the pole is \( 100\sqrt{3} + 100 \) feet.