At one side of a road, there os a house and on the other side there is a tower. House and tower are on the same horizontal plane. The angle of depression of the top and foot of the house, from the top of the tower are `30^(@)" and "45^(@)` respectively. If the height of the house is 10 m, then find the distance between the house and the tower.

Answer

To solve the problem, we will use trigonometric ratios and the information given about the angles of depression from the tower to the house. Let's break down the solution step by step.

### Step 1: Understand the Geometry
We have a tower (E) and a house (C) on opposite sides of a road. The height of the house (CD) is given as 10 m. Let the height of the tower be \( H \) (EC), and the distance between the house and the tower (CB) be \( X \).

### Step 2: Set Up the Angles of Depression
From the top of the tower (E), the angle of depression to the top of the house (C) is \( 30^\circ \), and the angle of depression to the foot of the house (D) is \( 45^\circ \).

### Step 3: Using the Angle of Depression to Form Equations
1. For the angle of depression to the top of the house (30°):
\[ \tan(30^\circ) = \frac{H - 10}{X} \]
We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so:
\[ \frac{1}{\sqrt{3}} = \frac{H - 10}{X} \]
Rearranging gives:
\[ H - 10 = \frac{X}{\sqrt{3}} \quad \text{(1)} \]

2. For the angle of depression to the foot of the house (45°):
\[ \tan(45^\circ) = \frac{H}{X} \]
Since \( \tan(45^\circ) = 1 \):
\[ 1 = \frac{H}{X} \]
This implies:
\[ H = X \quad \text{(2)} \]

### Step 4: Substitute Equation (2) into Equation (1)
Substituting \( H = X \) into Equation (1):
\[ X - 10 = \frac{X}{\sqrt{3}} \]

### Step 5: Solve for X
Multiplying both sides by \( \sqrt{3} \) to eliminate the fraction:
\[ \sqrt{3}(X - 10) = X \]
Expanding gives:
\[ \sqrt{3}X - 10\sqrt{3} = X \]
Rearranging terms:
\[ \sqrt{3}X - X = 10\sqrt{3} \]
Factoring out \( X \):
\[ X(\sqrt{3} - 1) = 10\sqrt{3} \]
Thus,
\[ X = \frac{10\sqrt{3}}{\sqrt{3} - 1} \]

### Step 6: Rationalizing the Denominator
To rationalize the denominator:
\[ X = \frac{10\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{10\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = \frac{10\sqrt{3}(\sqrt{3} + 1)}{2} \]
This simplifies to:
\[ X = 5\sqrt{3}(\sqrt{3} + 1) = 5(3 + \sqrt{3}) = 15 + 5\sqrt{3} \]

### Step 7: Calculate the Numerical Value
Using \( \sqrt{3} \approx 1.732 \):
\[ X \approx 15 + 5 \times 1.732 = 15 + 8.66 = 23.66 \text{ m} \]

### Final Answer
The distance between the house and the tower is approximately **23.66 meters**.