Question

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 8m. Find the height of the tree.

Answer 1

Let AC was the original tree. Due to storm, it was broken into two parts. The broken part A'B is making 30° with the ground. 
In ∆A'BC,
\(\frac{BC}{ A'C} = tan 30°\)

\(\frac{Bc} 8 = \frac{1 }{ \sqrt3}\)

\(BC = (\frac{8}{\sqrt3})m\)

\(\frac{A'C}{ A'B} = cos 30°\)

\(\frac{8}{A'B} = \frac{\sqrt3}2\)

\(A'B = (\frac{16}{\sqrt3})m\)

Height of the tree = \(A’B + BC\)
= \((\frac{16}{\sqrt3} + \frac{8}{ \sqrt3})m = \frac{24}{ \sqrt3 }m\)

= \(8\sqrt3m\)

Hence, the height of the tree is \(8\sqrt3m\).

Answer 2

Given:
Angle with the ground: \(30^\circ\)
Distance from the foot of the tree to where the top touches the ground: 8 meters
Let L be the length of the broken part of the tree. Using the cosine function:
\(\cos(30^\circ) = \frac{8}{L}\)

\(\frac{\sqrt{3}}{2} = \frac{8}{L}\)

\(L = \frac{8 \times 2}{\sqrt{3}} = \frac{16}{\sqrt{3}} = \frac{16 \sqrt{3}}{3}\)

The standing height of the tree, h, using the sine function:
\(\sin(30^\circ) = \frac{h}{L}\)

\(\frac{1}{2} = \frac{h}{L}\)

\(h = \frac{L}{2} = \frac{16 \sqrt{3}}{6} = \frac{8 \sqrt{3}}{3}\)

Total height of the tree before it broke:
\(H = h + L = \frac{8 \sqrt{3}}{3} + \frac{16 \sqrt{3}}{3} = \frac{24 \sqrt{3}}{3} = 8 \sqrt{3} \text{meters}\)

So, the answer is \(8 \sqrt{3}m\)