Question

From the top of a building, the angle of elevation of the top of a tower is \( 60^\circ \). From the top of the building, the angle of depression of foot of the tower is \( 45^\circ \). If the height of the tower is 40 metres, then prove that the height of the building is \( 20(\sqrt{3} - 1) \) metres.

Hint:

Use the properties of right-angled triangles and the tangent function to solve for unknown distances and heights when dealing with angles of elevation and depression.

Answer 1

Let the height of the building be \( h_1 \) and the height of the tower be \( h_2 = 40 \, \text{m} \). Let the distance between the building and the tower be \( d \). Triangle 1: Angle of elevation In the first triangle, the angle of elevation from the top of the building to the top of the tower is \( 60^\circ \). Using the tangent function, we can write: \[ \tan(60^\circ) = \frac{h_2 - h_1}{d} = \frac{40 - h_1}{d}. \] Since \( \tan(60^\circ) = \sqrt{3} \), we have: \[ \sqrt{3} = \frac{40 - h_1}{d} \quad \Rightarrow \quad d = \frac{40 - h_1}{\sqrt{3}}. \] Triangle 2: Angle of depression In the second triangle, the angle of depression to the foot of the tower is \( 45^\circ \). Using the tangent function, we can write: \[ \tan(45^\circ) = \frac{h_1}{d} = \frac{h_1}{d}. \] Since \( \tan(45^\circ) = 1 \), we have: \[ 1 = \frac{h_1}{d} \quad \Rightarrow \quad d = h_1. \] Solving for \( h_1 \) Substitute \( d = h_1 \) into the equation \( d = \frac{40 - h_1}{\sqrt{3}} \): \[ h_1 = \frac{40 - h_1}{\sqrt{3}}. \] Now, solve for \( h_1 \): \[ h_1 \sqrt{3} = 40 - h_1, \] \[ h_1 \sqrt{3} + h_1 = 40, \] \[ h_1(\sqrt{3} + 1) = 40, \] \[ h_1 = \frac{40}{\sqrt{3} + 1}. \] To simplify, multiply both the numerator and denominator by \( \sqrt{3} - 1 \): \[ h_1 = \frac{40(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{40(\sqrt{3} - 1)}{3 - 1} = \frac{40(\sqrt{3} - 1)}{2}. \] Thus: \[ h_1 = 20(\sqrt{3} - 1) \, \text{metres}. \]
Conclusion: The height of the building is \( 20(\sqrt{3} - 1) \, \text{metres}. \)