A vertically straight tree, $ 15 \mathrm{~m} $ high, is broken by the wind in such a way that its top just touches the ground and makes an angle of $ 60^{\circ} $ with the ground. At what height from the ground did the tree break?

Given:

A vertically straight tree, \( 15 \mathrm{~m} \) high, is broken by the wind in such a way that its top just touches the ground and makes an angle of \( 60^{\circ} \) with the ground.

To do:

We have to find the height of the tree from the ground where the tree broke.

Solution:  

Let $AB$ be the original height of the tree and $DB$ be the height of the tree from the ground where the tree broke.

Let point $C$ be the point where the top of the broken tree touches the ground.

From the figure,

$\mathrm{AB}=15 \mathrm{~m}, \angle \mathrm{DCB}=60^{\circ}$

Let the height of the tree from the ground where the tree broke be $\mathrm{DB}=x \mathrm{~m}$ and the height of the broken part be $\mathrm{DC}=15-x \mathrm{~m}$.

We know that,

$\sin \theta=\frac{\text { Opposite }}{\text { Hypotenuse }}$

$=\frac{\text { DB }}{DC}$

$\Rightarrow \sin 60^{\circ}=\frac{x}{15-x}$

$\Rightarrow \frac{\sqrt3}{2}=\frac{x}{15-x}$

$\Rightarrow (15-x)\sqrt3=2(x) \mathrm{~m}$

$\Rightarrow 2x+1.732x=15(1.732) \mathrm{~m}$

$\Rightarrow 3.732x=25.98 \mathrm{~m}$

$\Rightarrow x=\frac{25.98}{3.732} \mathrm{~m}$

$\Rightarrow x=6.9 \mathrm{~m}$

Therefore, the tree broke at $6.9 \mathrm{~m}$  from the ground.     

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Updated on: 10-Oct-2022

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