A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

From figure,

Let AB be the left over tree and BC be the broken part of tree.

C is the top of tree and A is the foot of the tree.

In △ABC,

tan 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Side adjacent to angle 30°}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{AC} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{8} \\[1em] \Rightarrow AB = \dfrac{8}{\sqrt{3}} \text{ m}.$

In right angle triangle ABC,

By pythagoras theorem, we get :

⇒ BC² = AB² + AC²

$\Rightarrow BC^2 = \Big(\dfrac{8}{\sqrt{3}}\Big)^2 + 8^2 \\[1em] \Rightarrow BC^2 = \dfrac{64}{3} + 64 \\[1em] \Rightarrow BC^2 = \dfrac{64 + 192}{3} \\[1em] \Rightarrow BC^2 = \dfrac{256}{3} \\[1em] \Rightarrow BC = \sqrt{\dfrac{256}{3}} \\[1em] \Rightarrow BC = \dfrac{16}{\sqrt{3}} \text{ m}.$

Height of tree = AB + BC

$= \dfrac{8}{\sqrt{3}} + \dfrac{16}{\sqrt{3}} \\[1em] = \dfrac{24}{\sqrt{3}} \\[1em] = \dfrac{24}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] = \dfrac{24\sqrt{3}}{3} \\[1em] = 8\sqrt{3} \text{ m}.$

Hence, height of tree = $8\sqrt{3}$ m.