Question

A storm broke a tree and the treetop rested $20m$ from the base of the tree, making an angle of ${60}^{\circ }$ with the horizontal. Find the height of the tree.

Answer 1

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, $CD=20m$ and $\angle BDC={60}^{\circ }$
In right angles triangle $BCD$,
$\tan {60}^{\circ }=\frac{BC}{CD}$

$\Rightarrow \sqrt{3}=\frac{BC}{20m}$

$\Rightarrow BC=20\sqrt{3}m$ .....(1)

Also, In right angles triangle $BCD$,
$\cos {60}^{\circ }=\frac{BD}{CD}$

$\Rightarrow \frac{1}{2}=\frac{20m}{BD}$

$\Rightarrow BD=40m$ ---(2)

Therefore, Height of the tree $=BD+BC=40+20\sqrt{3}m$
Thus, the height of the tree is $40+20\sqrt{3}m$