If from the top of a tower, 60 meters high, the angles of depression of the top an floor of a house are `30^(@) and 60^(@)` respectively, then the height (in meters) of the house is

Answer

To solve the problem step by step, we will use trigonometric ratios and the properties of angles of depression.

### Step-by-Step Solution:

1. **Understand the Problem:**
- We have a tower that is 60 meters high.
- The angles of depression to the top and the floor of a house are 30° and 60°, respectively.
- We need to find the height of the house.

2. **Draw a Diagram:**
- Draw a vertical line representing the tower (60 meters).
- Mark the top of the tower as point A and the bottom as point B.
- Draw a horizontal line from point A to represent the line of sight to the top of the house (point C) and another line to the bottom of the house (point D).
- The angle of depression to point C (top of the house) is 30°, and to point D (floor of the house) is 60°.

3. **Identify Angles:**
- The angle of depression from A to C is 30°, which means the angle at point A (between the horizontal line and the line AC) is also 30°.
- The angle of depression from A to D is 60°, which means the angle at point A (between the horizontal line and the line AD) is also 60°.

4. **Set Up the Right Triangles:**
- For triangle ABC (top of the house):
- Let the horizontal distance from the tower to the house be \( x \).
- The height from point B to point C is \( 60 - H \) (where \( H \) is the height of the house).
- Using the tangent function:
\[ \tan(30°) = \frac{60 - H}{x} \]
- We know that \( \tan(30°) = \frac{1}{\sqrt{3}} \), so:
\[ \frac{1}{\sqrt{3}} = \frac{60 - H}{x} \quad \text{(1)} \]

- For triangle ABD (floor of the house):
- The height from point B to point D is 60 meters.
- Using the tangent function:
\[ \tan(60°) = \frac{60}{x} \]
- We know that \( \tan(60°) = \sqrt{3} \), so:
\[ \sqrt{3} = \frac{60}{x} \quad \text{(2)} \]

5. **Solve for x:**
- From equation (2):
\[ x = \frac{60}{\sqrt{3}} = 20\sqrt{3} \]

6. **Substitute x into Equation (1):**
- Substitute \( x \) back into equation (1):
\[ \frac{1}{\sqrt{3}} = \frac{60 - H}{20\sqrt{3}} \]
- Cross-multiply:
\[ 20\sqrt{3} = \sqrt{3}(60 - H) \]
- Simplifying gives:
\[ 20 = 60 - H \]
- Rearranging gives:
\[ H = 60 - 20 = 40 \]

7. **Conclusion:**
- The height of the house is \( H = 40 \) meters.

### Final Answer:
The height of the house is **40 meters**.