A vertical tower stands on level ground. From a point P, the angle of elevation to the top is 30°. After moving 100 m closer to the tower along the line joining P to the foot of the tower, the angle of elevation becomes 60°. The height of the tower is:
\(50\sqrt3\) meters
Let the tower's height be h. Using the standard result for this configuration: \(h = \dfrac{d}{\cot30^\circ-\cot60^\circ}\), where d=100 m is the distance moved closer.
\(\cot30^\circ = \sqrt3\) and \(\cot60^\circ = \dfrac{1}{\sqrt3}\), so \(h = \dfrac{100}{\sqrt3-\tfrac{1}{\sqrt3}} = \dfrac{100}{\tfrac{2}{\sqrt3}} = \dfrac{100\sqrt3}{2} = 50\sqrt3\).
Hence, the height of the tower is \(50\sqrt3\) metres.
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