The angles of depression of the top and bottom of 12,m tall building from the top of a multistoried building are `30^(@)` and `45^(@)` respectively. Find the height of the multi storeyed building and the distance between the two buildings.

Answer

To solve the problem step by step, we will use the concepts of trigonometry, specifically the tangent function, which relates angles to the ratios of the sides of right triangles.

### Step 1: Define the problem
Let:
- \( AB \) be the height of the multi-storied building.
- \( CD \) be the height of the 12 m tall building.
- \( x \) be the horizontal distance between the two buildings.
- The angle of depression to the top of the building \( CD \) is \( 30^\circ \).
- The angle of depression to the bottom of the building \( CD \) is \( 45^\circ \).

### Step 2: Set up the right triangles
From the top of the multi-storied building \( A \):
- The angle of depression to the top of building \( CD \) (point \( C \)) is \( 30^\circ \).
- The angle of depression to the bottom of building \( CD \) (point \( D \)) is \( 45^\circ \).

### Step 3: Use the tangent function for the angle of depression
For the triangle \( AEC \) (where \( E \) is the point directly below \( A \) on the ground):
- The height from \( A \) to \( C \) (which is \( AB - 12 \)) is the opposite side, and \( x \) is the adjacent side.
- Using the tangent function:
\[ \tan(30^\circ) = \frac{AB - 12}{x} \]
Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we can write:
\[ \frac{1}{\sqrt{3}} = \frac{AB - 12}{x} \quad \text{(Equation 1)} \]

For the triangle \( ADF \):
- The height from \( A \) to \( D \) is \( AB \) (the full height of the multi-storied building) and the horizontal distance is still \( x \).
- Using the tangent function:
\[ \tan(45^\circ) = \frac{AB}{x} \]
Since \( \tan(45^\circ) = 1 \), we can write:
\[ 1 = \frac{AB}{x} \quad \text{(Equation 2)} \]
This implies:
\[ x = AB \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 2 into Equation 1
Substituting \( x = AB \) into Equation 1:
\[ \frac{1}{\sqrt{3}} = \frac{AB - 12}{AB} \]
Cross-multiplying gives:
\[ AB = (AB - 12) \sqrt{3} \]
Expanding and rearranging:
\[ AB \sqrt{3} - AB = -12 \]
\[ AB (\sqrt{3} - 1) = 12 \]
Thus,
\[ AB = \frac{12}{\sqrt{3} - 1} \]

### Step 5: Rationalize the denominator
To simplify \( AB \):
\[ AB = \frac{12(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{12(\sqrt{3} + 1)}{3 - 1} = \frac{12(\sqrt{3} + 1)}{2} = 6(\sqrt{3} + 1) \]
Using \( \sqrt{3} \approx 1.732 \):
\[ AB \approx 6(1.732 + 1) = 6(2.732) \approx 16.392 \text{ m} \]

### Step 6: Find the height of the multi-storied building
Thus, the height of the multi-storied building is:
\[ AB \approx 16.39 \text{ m} \]

### Step 7: Find the distance between the two buildings
Using Equation 2:
\[ x = AB \approx 16.39 \text{ m} \]

### Step 8: Final calculations
The total height of the multi-storied building is:
\[ AB = 12 + (AB - 12) \approx 12 + 16.39 \approx 28.39 \text{ m} \]

### Summary of Results
- Height of the multi-storied building: \( 28.39 \text{ m} \)
- Distance between the two buildings: \( 16.39 \text{ m} \)