Question

BC is a tower, B is its base. A is a point on a horizontal line passing through B, the angle of elevation of C from A is 60°. From another point D on AB, the angle of elevation is found to be 30°, then BD =

• A. 2AB
• B. \(\frac{1}{2}\)AB
• C. 3AB
• D. \(\frac{1}{3}\)AB

Answer 1

The Correct Option is C

Let the height of the tower \( BC = h \), and the horizontal distance \( AB = x \).
From triangle \( \triangle ABC \),
\[ \tan(60^\circ) = \frac{h}{x} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow h = \sqrt{3}x \]
Now consider point \( D \) on \( AB \) such that \( BD = y \). So, point \( D \) is at distance \( x - y \) from the base \( B \).
In triangle \( \triangle DBC \),
\[ \tan(30^\circ) = \frac{h}{y} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{y} \Rightarrow y = \sqrt{3}h \]
Substitute \( h = \sqrt{3}x \) into the above equation:
\[ y = \sqrt{3}(\sqrt{3}x) = 3x \]
Hence, \( BD = 3AB \).

The correct option is (C): \(3AB\)