Tree Breakage Geometry Setup

A tree breaks, and the broken part bends to touch the ground, forming a right-angled triangle with the remaining part of the tree and the ground.

• The angle the top makes with the ground is \(\theta = 30^\circ\).
• The distance from the foot of the tree to where the top touches the ground is the adjacent side, \(\text{adjacent} = 18\) m.
• Let \(h\) be the height of the remaining part of the tree (opposite side).
• Let \(x\) be the length of the broken part (hypotenuse).
• The total height of the tree is \(h + x\).

Calculating Remaining Height (h)

We use the tangent function, which relates the opposite side and the adjacent side to the angle:

\(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\)

Substituting the known values:

\(\tan(30^\circ) = \frac{h}{18}\)

Solving for \(h\):

\(h = 18 \times \tan(30^\circ)\)

Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\):

\(h = 18 \times \frac{1}{\sqrt{3}} = \frac{18}{\sqrt{3}}\)

Rationalizing the denominator:

\(h = \frac{18\sqrt{3}}{3} = 6\sqrt{3} \text{ m}\)

Calculating Broken Part (x)

We use the cosine function, which relates the adjacent side and the hypotenuse to the angle:

\(\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\)

Substituting the known values:

\(\cos(30^\circ) = \frac{18}{x}\)

Solving for \(x\):

\(x = \frac{18}{\cos(30^\circ)}\)

Since \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\):

\(x = \frac{18}{\frac{\sqrt{3}}{2}} = \frac{18 \times 2}{\sqrt{3}} = \frac{36}{\sqrt{3}}\)

Rationalizing the denominator:

\(x = \frac{36\sqrt{3}}{3} = 12\sqrt{3} \text{ m}\)

Total Tree Height Calculation

The total height of the tree is the sum of the remaining height (\(h\)) and the length of the broken part (\(x\)):

\(\text{Total Height} = h + x\)

\(\text{Total Height} = 6\sqrt{3} \text{ m} + 12\sqrt{3} \text{ m}\)

\(\text{Total Height} = (6 + 12)\sqrt{3} \text{ m}\)

\(\text{Total Height} = 18\sqrt{3} \text{ m}\)