Question

A tower subtends an angle of \( 30^\circ \) at a point on the same level as the foot of the tower. At a second point \( h \) meters above the first, the depression of the font of the tower is \( 60^\circ \). What is the horizontal distance of the tower from the point?

• A. \( 2h \cot 60^\circ \)
• B. \( 2h \cot 60^\circ \)
• C. \( h \tan 60^\circ \)
• D. \( 2h \tan 60^\circ \)

Hint:

When solving problems with angles of depression or elevation, use trigonometric ratios to relate the heights and distances.

Answer 1

The Correct Option is B

We are given that a tower subtends an angle of \( 30^\circ \) at a point on the same level as the foot of the tower, and at a second point \( h \) meters above the first, the depression of the font of the tower is \( 60^\circ \). We are tasked with finding the horizontal distance of the tower from the point. Let the height of the tower be \( h_1 \) and the horizontal distance from the first point to the foot of the tower be \( x \). Using trigonometry, we can find the relationships: - For the first angle of \( 30^\circ \), we use \( \tan 30^\circ = \frac{h_1}{x} \), which gives \( x = \frac{h_1}{\tan 30^\circ} = h_1 \cot 30^\circ \). - For the second angle of \( 60^\circ \), we use \( \tan 60^\circ = \frac{h_1 - h}{x} \), which gives \( x = \frac{h_1 - h}{\tan 60^\circ} \). Equating the two expressions for \( x \), we find: \[ h_1 \cot 30^\circ = \frac{h_1 - h}{\tan 60^\circ} \] Solving for the horizontal distance, we find that the distance is \( 2h \cot 60^\circ \).