Question

The angle of elevation of the top of a tower standing on a horizontal plane from a point A is a. After walking a distance d towards the foot of the tower, the angle of elevation is found to be B. Then the height of the tower is

• A. \(\frac{d}{tan\,a-tan\,ẞ}\)
• B. \(d(cot\,a-cot\,B)\)
• C. \(\frac{d}{cot\,a-cot\,ẞ}\)
• D. \(d(tan\,a-tan\,ẞ)\)

Answer 1

The Correct Option is C

Let the height of the tower be \( h \), and let the initial distance from point \( A \) to the foot of the tower be \( x \). After walking a distance \( d \) towards the tower, the new distance to the foot of the tower becomes \( x - d \).

Step 1: Use the tangent function for both angles of elevation.

From point \( A \), the angle of elevation is \( \alpha \), so:

\[ \tan\alpha = \frac{h}{x} \quad \implies \quad x = \frac{h}{\tan\alpha}. \]

After walking a distance \( d \), the angle of elevation becomes \( \beta \), so:

\[ \tan\beta = \frac{h}{x - d} \quad \implies \quad x - d = \frac{h}{\tan\beta}. \]

Step 2: Substitute \( x = \frac{h}{\tan\alpha} \) into \( x - d = \frac{h}{\tan\beta} \).

\[ \frac{h}{\tan\alpha} - d = \frac{h}{\tan\beta}. \]

Rearrange to isolate \( d \):

\[ d = \frac{h}{\tan\alpha} - \frac{h}{\tan\beta}. \]

Factor out \( h \):

\[ d = h \left( \frac{1}{\tan\alpha} - \frac{1}{\tan\beta} \right). \]

Step 3: Simplify using cotangent identities.

Recall that \( \cot\theta = \frac{1}{\tan\theta} \). Substituting this:

\[ d = h (\cot\alpha - \cot\beta). \]

Solve for \( h \):

\[ h = \frac{d}{\cot\alpha - \cot\beta}. \]

Final Answer: The height of the tower is \( \mathbf{\frac{d}{\cot\alpha - \cot\beta}} \), which corresponds to option \( \mathbf{(3)} \).