$A$ tower of height $h$' standing at the centre of a square with sides of length $a$' makes the same angle $\alpha$ at each of the four corners then $a^{2}=$
- $ h^{2}\cot^{2}\alpha$
- $ h^{2}\tan^{2}\alpha$
- $ 2h^{2}\cot^{2}\alpha$
- $ 2h^{2}\tan^{2}\alpha$
Let the tower be at the center (0,0,0) with height h. The corners of the square are at (+-a/2, +-a/2, 0). The distance from the center to a corner is sqrt((a/2)^2 + (a/2)^2) = sqrt(a^2/4 + a^2/4) = a/sqrt(2). The angle alpha is formed by the tower height h and the distance from the base to the corner. So tan(alpha) = h / (a/sqrt(2)) = h * sqrt(2) / a. Thus, a = h * sqrt(2) / tan(alpha) = h * sqrt(2) * cot(alpha). Squaring both sides: a^2 = 2 * h^2 * cot^2(alpha).
A square with side length a has a diagonal length of a times the square root of 2, meaning the distance from the center to any corner is half of that, or a divided by the square root of 2. This distance, the height h, and the line of sight to the top of the tower form a right triangle where the angle of elevation is alpha. Using the trigonometric ratio for tangent, we have tan alpha equals h divided by (a divided by the square root of 2), which rearranges to a equals h times the square root of 2 times cot alpha. Squaring both sides of this equation yields a squared equals 2 times h squared times cot squared alpha.