If from the top of a tower, 60 meters high, the angles of depression of the top an floor of a house are `3^(@) and 60^(@)` respectively, then the height (in meters) of the house is

Answer

To solve the problem, we need to determine the height of the house given the height of the tower and the angles of depression to the top and the floor of the house.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a tower of height 60 meters.
- The angle of depression to the top of the house is \(30^\circ\) (corrected from the transcript).
- The angle of depression to the floor of the house is \(60^\circ\).

2. **Setting Up the Diagram:**
- Let \(A\) be the top of the tower, \(B\) be the top of the house, and \(C\) be the floor of the house.
- The height of the house is denoted as \(H\).
- The horizontal distance from the base of the tower to the house is \(x\).

3. **Using the Angle of Depression to the Top of the House:**
- From point \(A\) (top of the tower), the angle of depression to point \(B\) (top of the house) is \(30^\circ\).
- In triangle \(ABM\) (where \(M\) is the point vertically below \(A\) on the ground), we can use the tangent function:
\[ \tan(30^\circ) = \frac{60 - H}{x} \]
- We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), so:
\[ \frac{1}{\sqrt{3}} = \frac{60 - H}{x} \quad \text{(1)} \]

4. **Using the Angle of Depression to the Floor of the House:**
- From point \(A\), the angle of depression to point \(C\) (floor of the house) is \(60^\circ\).
- In triangle \(ACM\), we can use the tangent function again:
\[ \tan(60^\circ) = \frac{60}{x} \]
- We know that \(\tan(60^\circ) = \sqrt{3}\), so:
\[ \sqrt{3} = \frac{60}{x} \quad \text{(2)} \]

5. **Solving for \(x\):**
- From equation (2), we can express \(x\):
\[ x = \frac{60}{\sqrt{3}} = 20\sqrt{3} \]

6. **Substituting \(x\) back into equation (1):**
- Substitute \(x\) into equation (1):
\[ \frac{1}{\sqrt{3}} = \frac{60 - H}{20\sqrt{3}} \]
- Cross-multiplying gives:
\[ 20\sqrt{3} = \sqrt{3}(60 - H) \]
- Dividing both sides by \(\sqrt{3}\):
\[ 20 = 60 - H \]

7. **Solving for \(H\):**
- Rearranging gives:
\[ H = 60 - 20 = 40 \]

### Final Answer:
The height of the house is \(40\) meters.