Asked in: July 2025 board paperA storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60° with the horizontal. Find the height of the tree.

Answer: Height of the tree \(= 40 + 20\sqrt3 \approx 74.6\) m

[Figure: Tree AB stands at B. It breaks at C; the broken part CD touches the ground at D, with BD = 20 m and \(\angle CDB = 60^\circ\); \(\angle CBD = 90^\circ\).]

In right-angled ΔCBD:

\(\tan 60^\circ = \dfrac{CB}{BD}\) \(\therefore \sqrt3 = \dfrac{CB}{20}\) \(\therefore CB = 20\sqrt3\) m

\(\cos 60^\circ = \dfrac{BD}{CD}\) \(\therefore \dfrac{1}{2} = \dfrac{20}{CD}\) \(\therefore CD = 40\) m

Height of tree = CB + CD (the broken part CD was part of the tree)

\(= 20\sqrt3 + 40 = 20 \times 1.73 + 40 = 34.6 + 40 = 74.6\) m.