# A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle $30^o$ 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.

Given:

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of \( 30^{\circ} \) with the ground.

The distance between the foot of the tree to the point where the top touches the ground is \( 8 \mathrm{~m} \).

To do:

We have to find the height of the tree.

Solution:

Let $AB$ be the original height of the tree and $CD$ be the broken part of the tree that touches the ground at $D$.

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

From the figure,

$\mathrm{AD}=8 \mathrm{~m}, \angle \mathrm{CDA}=30^{\circ}$

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

We know that,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { CA }}{DA}$

$\Rightarrow \tan 30^{\circ}=\frac{x}{8}$

$\Rightarrow \frac{1}{\sqrt3}=\frac{x}{8}$

$\Rightarrow x=\frac{8}{\sqrt3} \mathrm{~m}$........(i)

Similarly,

$\cos \theta=\frac{\text { Adjacent }}{\text { Hypotenuse }}$

$=\frac{\text { DA }}{CD}$

$\Rightarrow \cos 30^{\circ}=\frac{8}{y}$

$\Rightarrow \frac{\sqrt3}{2}=\frac{8}{y}$

$\Rightarrow y=\frac{8(2)}{\sqrt3}=\frac{16}{\sqrt3} \mathrm{~m}$

Therefore,

$x+y=\frac{8}{\sqrt3}+\frac{16}{\sqrt3}$

$=\frac{8+16}{\sqrt3}$

$=\frac{24}{\sqrt3}$

$=\frac{24\sqrt3}{\sqrt3 \times \sqrt3}$

$=\frac{24\sqrt3}{3}$

$=8\sqrt3 \mathrm{~m}$

Therefore, the height of the tree is $8\sqrt3 \mathrm{~m}$.

Related Articles

- A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of \( 30^{\circ} \) with the ground. The distance between the foot of the tree to the point where the top touches the ground is \( 8 \mathrm{~m} \). Find the height of the tree.
- A tree breaks due to the storm and the broken part bends so that the top of the tree touches the ground making an angle of \( 30^{\circ} \) with the ground. The distance from the foot of the tree to the point where the top touches the ground is 10 metres. Find the height of the tree.
- A tree is broken at a height of $5\ m$ from the ground and its top touches the ground at a distance of $12\ m$ from the base of the tree. Find the original height of the tree.
- A \( 8 \mathrm{~m} \) high bamboo tree standing erect on the ground breaks at the height of \( 3 \mathrm{~m} \) from the ground. Its broken part remains attached to the main part. Find distance between the top of the tree and the base of the tree on the ground.
- 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?
- The horizontal distance between two trees of different heights is \( 60 \mathrm{~m} \). The angle of depression of the top of the first tree when seen from the top of the second tree is \( 45^{\circ} \). If the height of the second tree is \( 80 \mathrm{~m} \), find the height of the first tree.
- The angle of elevation of the top of a tower from a point on the ground, which is $30\ m$ away from the foot of the tower is $30^o$. Find the height of the tower.
- The angle of elevation of the top of tower, from the point on the ground and at a distance of 30 m from its foot, is 30o. Find the height of tower.
- A statue, $1.6\ m$ tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is $60^o$ and from the same point the angle of elevation of the top of the pedestal is $45^o$. Find the height of the pedestal.
- From the top of a 7 m high building, the angle of the elevation of the top of a tower is $60^{o}$ and the angle of the depression of the foot of the tower is $30^{o}$. Find the height of the tower.
- A tower stands vertically on the ground. From a point on the ground, \( 20 \mathrm{~m} \) away from the foot of the tower, the angle of elevation of the top of the tower is \( 60^{\circ} \). What is the height of the tower?
- The angle of elevation of the top of a building from the foot of a tower is $30^o$ and the angle of elevation of the top of the tower from the foot of the building is $60^o$. If the tower is $50\ m$ high, find the height of the building.
- The angle of elevation of the top of a tower from a point \( A \) on the ground is \( 30^{\circ} \). On moving a distance of 20 metres towards the foot of the tower to a point \( B \) the angle of elevation increases to \( 60^{\circ} \). Find the height of the tower and the distance of the tower from the point \( A \).
- The angle of elevation of the top of a tower $30\ m$ high from the foot of another tower in the same plane is $60^o$ and the angle of elevation of the top of the second tower from the foot of the first tower is $30^o$. then find the distance between the two towers.
- The angle of the elevation of the top of vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 30°. Find the height of the tower.

##### Kickstart Your Career

Get certified by completing the course

Get Started