Question

The angle of elevation of the top of a building from the foot of a tower is \( 30^\circ \), and the angle of elevation of the top of the tower from the foot of the building is \( 60^\circ \). If the tower is 50 m high, find the height of the building.

Hint:

Use right-angle trigonometry for height and distance problems involving angles of elevation and depression.

Answer 1

Let the height of the building be \( h \) meters and the distance between the building and the tower be \( x \) meters.
Using the right triangle formed by the tower:
\[ \tan 60^\circ = \frac{50}{x}. \] \[ \sqrt{3} = \frac{50}{x}. \] \[ x = \frac{50}{\sqrt{3}} = \frac{50\sqrt{3}}{3}. \] Now, using the right triangle formed by the building:
\[ \tan 30^\circ = \frac{h}{x}. \] \[ \frac{1}{\sqrt{3}} = \frac{h}{\frac{50\sqrt{3}}{3}}. \] \[ h = \frac{50\sqrt{3}}{3} \times \frac{1}{\sqrt{3}}. \] \[ h = \frac{50}{3} = 16.67 \text{ m}. \] Thus, the height of the building is approximately 16.67 meters.