The upper part of a tree broken by the wind m | vedclass

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Question

(C) Let the tree be $AC$ before breaking,where $C$ is the root and $A$ is the top. Let it break at point $D$ such that the top $A$ touches the ground

Vedclass · Accepted Answer

$17.32$

Vedclass · Answer

(C) Let the tree be $AC$ before breaking,where $C$ is the root and $A$ is the top. Let it break at point $D$ such that the top $A$ touches the ground at point $B$. In $	riangle BCD$,$\angle DBC = 30^\circ$ and $BC = 10 \, m$. Using trigonometry: $	an(30^\circ) = \frac{CD}{BC} \implies \frac{1}{\sqrt{3}} = \frac{CD}{10} \implies CD = \frac{10}{\sqrt{3}} \, m$. $\cos(30^\circ) = \frac{BC}{BD} \implies \frac{\sqrt{3}}{2} = \frac{10}{BD} \implies BD = \frac{20}{\sqrt{3}} \, m$. The original height of the tree is $CD + AD = CD + BD$ (since $AD = BD$). Height $= \frac{10}{\sqrt{3}} + \frac{20}{\sqrt{3}} = \frac{30}{\sqrt{3}} = 10\sqrt{3} \, m$. Using $\sqrt{3} \approx 1.732$,the height is $10 	imes 1.732 = 17.32 \, m$.

The upper part of a tree broken by the wind makes an angle of $30^\circ$ with the ground. The distance from the root to the point where the top of the tree touches the ground is $10 \, m$. What was the original height of the tree in meters?

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