Question

From the top of a building 60 metres high, the angles of depression of the top and bottom of a tower are observed to be \( 30^\circ \) and \( 60^\circ \) respectively. Find the height of the tower.

Hint:

When using angles of depression, use the tangent function to relate the height and distance of objects.

Answer 1

Let the height of the tower be \( h \) metres. Let the distance between the building and the tower be \( x \) metres. From the problem, we have two right-angled triangles: one formed by the top of the building and the top of the tower, and another formed by the top of the building and the bottom of the tower. Triangle 1: Angle of depression of the top of the tower The angle of depression to the top of the tower is \( 30^\circ \). From the tangent function: \[ \tan(30^\circ) = \frac{60 - h}{x}. \] Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have: \[ \frac{1}{\sqrt{3}} = \frac{60 - h}{x} \quad \Rightarrow \quad x = \sqrt{3}(60 - h). \] Triangle 2: Angle of depression of the bottom of the tower The angle of depression to the bottom of the tower is \( 60^\circ \). From the tangent function: \[ \tan(60^\circ) = \frac{60}{x}. \] Since \( \tan(60^\circ) = \sqrt{3} \), we have: \[ \sqrt{3} = \frac{60}{x} \quad \Rightarrow \quad x = \frac{60}{\sqrt{3}} = 20\sqrt{3}. \] Set up the equation for \( h \) Now substitute \( x = 20\sqrt{3} \) into the equation \( x = \sqrt{3}(60 - h) \): \[ 20\sqrt{3} = \sqrt{3}(60 - h). \] Cancel \( \sqrt{3} \) from both sides: \[ 20 = 60 - h \quad \Rightarrow \quad h = 40. \]
Conclusion: The height of the tower is \( 40 \) metres.