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 the 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 \( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \) separately for both elevation and depression angles. Equating the horizontal distances gives the required relation.

Answer 1

Step 1: Understanding the figure.
Let \( AB \) be the building and \( CD \) be the tower, where \( CD = 40 \, \text{m} \). Let the horizontal distance between the building and the tower be \( x \, \text{m} \). Let the height of the building be \( h \, \text{m} \).
Step 2: From the top of the building, angle of elevation to top of tower is \( 60^\circ \).
Hence, in triangle \( EBC \): \[ \tan 60^\circ = \frac{CD - h}{x} \] \[ \sqrt{3} = \frac{40 - h}{x} \Rightarrow x = \frac{40 - h}{\sqrt{3}} \quad \text{(i)} \]
Step 3: From the top of the building, angle of depression to foot of tower is \( 45^\circ \).
In triangle \( EBD \): \[ \tan 45^\circ = \frac{h}{x} \] \[ 1 = \frac{h}{x} \Rightarrow x = h \quad \text{(ii)} \]
Step 4: Equate (i) and (ii).
\[ h = \frac{40 - h}{\sqrt{3}} \] \[ h\sqrt{3} = 40 - h \] \[ h(\sqrt{3} + 1) = 40 \] \[ h = \frac{40}{\sqrt{3} + 1} \] Rationalize the denominator: \[ h = 40 \times \frac{\sqrt{3} - 1}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = 40 \times \frac{\sqrt{3} - 1}{2} \] \[ h = 20(\sqrt{3} - 1) \] Step 5: Final Answer.
\[ \boxed{\text{Height of the building } = 20(\sqrt{3} - 1) \, \text{m}} \]