$A$ man standing on a level plane observes the elevation of the top of a pole to be $\alpha$. He then walks a distance equal to double the height of the pole and then finds that the elevation is now $2\alpha$. Then $\alpha=$

Let the height of the pole be h and the initial distance from it be x, so tan alpha = h / x. The man walks a distance of 2h towards the pole, making the new distance x - 2h and the new angle 2 alpha, so tan(2 alpha) = h / (x - 2h). Using the double angle identity gives 2 tan alpha / (1 - tan^2 alpha) = h / (x - 2h), and substituting tan alpha = h / x results in 2(h/x) / (1 - h^2/x^2) = h / (x - 2h). Solving the equation 2x / (x^2 - h^2) = 1 / (x - 2h) yields 2x^2 - 4hx = x^2 - h^2, which simplifies to x^2 - 4hx + h^2 = 0 and gives the ratio x/h = 2 + sqrt(3). Because tan alpha = h / x = 1 / (2 + sqrt(3)) = 2 - sqrt(3), the angle alpha is 15 degrees.

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