The angle of elevation of the top of a tree from a point on the ground is 30°. After moving 10 m towards the tree, the angle of elevation becomes 60°. Find the height of the tree.
\(5\sqrt{3}\) m
Let the height of the tree be \(h\) m and the distance from the second point (after moving 10 m) to the base of the tree be \(d\) m.
From the second point: \(\tan 60° = \frac{h}{d}\), so \(\sqrt{3} = \frac{h}{d}\), giving \(d = \frac{h}{\sqrt{3}}\).
From the first point (10 m farther): \(\tan 30° = \frac{h}{d + 10}\), so \(\frac{1}{\sqrt{3}} = \frac{h}{d + 10}\), giving \(d + 10 = h\sqrt{3}\).
Substituting \(d = \frac{h}{\sqrt{3}}\): \(\frac{h}{\sqrt{3}} + 10 = h\sqrt{3}\).
Multiply through by \(\sqrt{3}\): \(h + 10\sqrt{3} = 3h\), so \(2h = 10\sqrt{3}\), giving \(h = 5\sqrt{3}\) m.
Hence, the height of the tree is \(5\sqrt{3}\) m.
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