Question

Each side of a square subtends an angle of \( 60^\circ \) at the top of a tower \( h \) meters high standing in the center of the square. If \( a \) is the length of each side of the square, then

• A. \( 2a^2 = h^2 \)
• B. \( 2h^2 = a^2 \)
• C. \( 3a^2 = 2h^2 \)
• D. \( 2h^2 = 3a^2 \)

Hint:

When solving problems involving angles of elevation and squares, use the symmetry of the shape and apply trigonometric identities such as tangent for right triangles.

Answer 1

The Correct Option is B

Step 1: Understand the setup of the problem.
We are given a square with each side of length \( a \), and a tower of height \( h \) located at the center of the square. The angle between the line of sight from the top of the tower to two opposite corners of the square is \( 60^\circ \).

Step 2: Use geometry to analyze the situation.
The square is symmetric, and the angle of elevation from the center of the square to the top of the tower subtends a \( 60^\circ \) angle at the corner of the square. By dividing the square into two right-angled triangles, we can apply trigonometric ratios.

Step 3: Set up the trigonometric relationship.
Consider one of the right-angled triangles formed by half the diagonal of the square, the height of the tower \( h \), and the distance from the center of the square to the corner. The angle of elevation is \( 60^\circ \), so we can use the tangent function:
\[ \tan(60^\circ) = \frac{h}{\frac{a}{\sqrt{2}}} \] because the distance from the center to the corner of the square is \( \frac{a}{\sqrt{2}} \), which is half the length of the diagonal.

Step 4: Solve for \( h \).
We know that \( \tan(60^\circ) = \sqrt{3} \), so: \[ \sqrt{3} = \frac{h}{\frac{a}{\sqrt{2}}} \quad \implies \quad h = \frac{a \sqrt{2}}{\sqrt{3}} \]

Step 5: Square both sides to simplify.
Squaring both sides to eliminate the square root: \[ h^2 = \frac{a^2 \cdot 2}{3} \] Thus, we get: \[ 2h^2 = \frac{a^2}{3} \] So, the correct relation is \( 2h^2 = a^2 \), corresponding to option (B).