Question

The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45° respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (use \(\sqrt{ 3}\) = 1.73)

Answer 1

Step 1: Understanding the problem:
We are given the following information:
- The height of the first building (height of the building with the top and bottom observed) is \( 8 \, \text{m} \).
- The angles of depression of the top and bottom of the building from the top of the multi-storeyed building are \( 30^\circ \) and \( 45^\circ \) respectively.
- We need to find:
1. The height of the multi-storeyed building.
2. The distance between the two buildings.
Let's assume the following:
- The height of the multi-storeyed building is \( h \, \text{m} \).
- The distance between the two buildings is \( d \, \text{m} \).
The problem involves two right-angled triangles formed by the line of sight from the top of the multi-storeyed building to the top and bottom of the 8 m building. The distance between the two buildings is along the horizontal, which is the base of the right-angled triangles.

Step 2: Set up the triangles:
For both the top and bottom of the 8 m building, we can use the tangent function of the respective angles of depression.
1. For the top of the 8 m building, the angle of depression is \( 30^\circ \). The height difference between the two buildings is \( h - 8 \, \text{m} \) and the distance between them is \( d \). From the tangent of \( 30^\circ \):
\[ \tan 30^\circ = \frac{h - 8}{d} \] Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), we get:
\[ \frac{1}{\sqrt{3}} = \frac{h - 8}{d} \] Thus, we have the first equation:
\[ d = \sqrt{3}(h - 8) \quad \text{(Equation 1)} \] 2. For the bottom of the 8 m building, the angle of depression is \( 45^\circ \). The height difference between the two buildings is \( h \) and the distance between them is \( d \). From the tangent of \( 45^\circ \):
\[ \tan 45^\circ = \frac{h}{d} \] Since \( \tan 45^\circ = 1 \), we get:
\[ 1 = \frac{h}{d} \] Thus, we have the second equation:
\[ d = h \quad \text{(Equation 2)} \] Step 3: Solve the system of equations:
From Equation 2, we know that \( d = h \). Substitute this into Equation 1:
\[ h = \sqrt{3}(h - 8) \] Now, solve for \( h \):
\[ h = \sqrt{3}h - 8\sqrt{3} \] Rearrange the equation:
\[ h - \sqrt{3}h = -8\sqrt{3} \] Factor out \( h \):
\[ h(1 - \sqrt{3}) = -8\sqrt{3} \] Thus, the height of the multi-storeyed building is:
\[ h = \frac{-8\sqrt{3}}{1 - \sqrt{3}} \] Now, we simplify this using \( \sqrt{3} = 1.73 \). Calculate the denominator:
\[ 1 - \sqrt{3} = 1 - 1.73 = -0.73 \] Substitute this into the equation:
\[ h = \frac{-8 \times 1.73}{-0.73} = \frac{-13.84}{-0.73} \approx 19 \] So, the height of the multi-storeyed building is approximately \( 19 \, \text{m} \).

Step 4: Find the distance between the two buildings:
Now, substitute \( h = 19 \, \text{m} \) into Equation 2:
\[ d = h = 19 \, \text{m} \] Conclusion:
The height of the multi-storeyed building is approximately \( 19 \, \text{m} \) and the distance between the two buildings is approximately \( 19 \, \text{m} \).