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.