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

Let the height of the multi-storeyed building be h, and the distance between the two buildings be d.

1. Using the angle of depression of 45°:

\[ \tan(45^\circ) = \frac{8}{d_1} \]

Since \(\tan(45^\circ) = 1\), we have:

\[ d_1 = 8 \, \text{m} \]

This is the horizontal distance from the base of the multi-storeyed building to the base of the 8 m tall building.

2. Using the angle of depression of 30°:

\[ \tan(30^\circ) = \frac{h - 8}{d_1 + 14} \]

Substituting \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\) and \(d_1 = 8\):

\[ \frac{1}{\sqrt{3}} = \frac{h - 8}{8 + 14} \]

Simplifying:

\[ \frac{1}{\sqrt{3}} = \frac{h - 8}{22} \]

\[ h - 8 = \frac{22}{\sqrt{3}} \approx 12.67 \]

Therefore:

\[ h = 8 + 12.67 = 20.67 \, \text{m} \]

Thus, the height of the multi-storeyed building is approximately 20.67 m, and the distance between the two buildings is 14 m.