ABCD is a square field with AB=x. A vertical pole OP of height 2x stands at the centre O of the square field. If \angle APO=\theta, then what is \cot\theta equal to?

Explanation

The distance from a corner A to the center O of a square with side x is half the diagonal, so AO = \frac{x\sqrt{2}}{2} = \frac{x}{\sqrt{2}}. In the right-angled \triangle AOP, the height OP = 2x. The angle \angle APO = \theta, hence \cot\theta = \frac{OP}{AO} = \frac{2x}{x/\sqrt{2}} = 2\sqrt{2}.

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