#### Question

A 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.

#### Solution

Let AC be the original height of the tree. Suppose BD be the broken part of the tree which is rested at D from the base of the tree.

Here, CD = 20 m and ∠BDC = 60º.

In right ∆BCD,

\[\tan60^\circ = \frac{BC}{CD}\]

\[ \Rightarrow \sqrt{3} = \frac{BC}{20}\]

\[ \Rightarrow BC = 20\sqrt{3} m . . . . . \left( 1 \right)\]

Also,

\[\cos60^\circ = \frac{CD}{BD}\]

\[ \Rightarrow \frac{1}{2} = \frac{20}{BD}\]

\[ \Rightarrow BD = 40 m . . . . . \left( 2 \right)\]

∴ Height of the tree = AB + BC = BD + BC =

\[\left( 40 + 20\sqrt{3} \right) m\] [Using (1) and (2)]

Thus, the height of the tree is \[\left( 40 + 20\sqrt{3} \right) m\]