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.

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 \( 10 \mathrm{~m} \).

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,

$\mathrm{AD}=10 \mathrm{~m}, \angle \mathrm{CDA}=30^{\circ}$

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,

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

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

$\Rightarrow \tan 30^{\circ}=\frac{x}{10}$

$\Rightarrow \frac{1}{\sqrt3}=\frac{x}{10}$

$\Rightarrow x=\frac{10}{\sqrt3} \mathrm{~m}$........(i)

Similarly,

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

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

$\Rightarrow \cos 30^{\circ}=\frac{10}{y}$

$\Rightarrow \frac{\sqrt3}{2}=\frac{10}{y}$

$\Rightarrow y=\frac{10(2)}{\sqrt3}=\frac{20}{\sqrt3} \mathrm{~m}$

Therefore,

$x+y=\frac{10}{\sqrt3}+\frac{20}{\sqrt3}$

$=\frac{10+20}{\sqrt3}$

$=\frac{30}{\sqrt3}$

$=\frac{30\sqrt3}{\sqrt3 \times \sqrt3}$

$=\frac{30\sqrt3}{3}$

$=10\sqrt3=10(1.73)=17.3 \mathrm{~m}$

Therefore, the height of the tree is $17.3 \mathrm{~m}$ .

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 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.
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 tower, from the point on the ground and at a distance of 30 m from its foot, is 30o. Find the height of tower.
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 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.
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?
A statue \( 1.6 \mathrm{~m} \) tall stands on the top of pedestal. From a point on the ground, the angle of elevation of the top of the statue is \( 60^{\circ} \) and from the same point the angle of elevation of the top of the pedestal is \( 45^{\circ} \). Find the height of the pedestal.
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.
A flag-staff stands on the top of 5 m high tower. From a point on the ground, the angle of elevation of the top of the flag-staff is \( 60^{\circ} \) and from the same point, the angle of elevation of the top of the tower is \( 45^{\circ} \). Find the height of the flag-staff.
A person observed the angle of elevation of the top of a tower as \( 30^{\circ} \). He walked \( 50 \mathrm{~m} \) towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as \( 60^{\circ} \). Find the height of the tower.
A circus artist is climbing a \( 20 \mathrm{~m} \) long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is \( 30^{\circ} \).
Kickstart Your Career
Get certified by completing the course

Get Started