From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a $20\ m$ high building are $45^o$ and $60^o$ respectively. Find the height of the tower.

Given:

From a point on the ground the angles of elevation of the bottom and top of a transmission tower fixed at the top of \( 20 \mathrm{~m} \) high building are \( 45^{\circ} \) and \( 60^{\circ} \) respectively.

To do:

We have to find the height of the transmission tower.

Solution:

Let $AB$ be the high building and $BC$ be the length of the transmission tower.

Let point $D$ be the point of observation.

From the figure,

$\mathrm{AB}=20 \mathrm{~m}, \angle \mathrm{BDA}=45^{\circ}, \angle \mathrm{CDA}=60^{\circ}$

Let the height of the transmission tower be $\mathrm{BC}=h \mathrm{~m}$ and the distance between the point of observation and the foot of the building be $\mathrm{AD}=x \mathrm{~m}$.

This implies,

$\mathrm{AC}=20+h \mathrm{~m}$

We know that,

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

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

$\Rightarrow \tan 45^{\circ}=\frac{20}{x}$

$\Rightarrow 1(x)=20$

$\Rightarrow x=20 \mathrm{~m}$

Similarly,

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

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

$\Rightarrow \tan 60^{\circ}=\frac{20+h}{x}$

$\Rightarrow \sqrt3=\frac{20+h}{20}$

$\Rightarrow 20+h=20\sqrt3 \mathrm{~m}$

$\Rightarrow h=20(\sqrt3-1) \mathrm{~m}$

Therefore, the height of the transmission tower is $20(\sqrt3-1) \mathrm{~m}$ .

Related Articles From a point on the ground the angles of elevation of the bottom and top of a transmission tower fixed at the top of \( 20 \mathrm{~m} \) high building are \( 45^{\circ} \) and \( 60^{\circ} \) respectively. Find the height of the transimission tower.
The angles of elevation and depression of the top and the bottom of a tower from the top of a building, $60\ m$ high, are $30^{o}$ and $60^{o}$ respectively. Find the difference between the heights of the building and the tower and the distance between them.
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.
From the top of a $7\ m$ high building, the angle of elevation of the top of a cable tower is $60^o$ and the angle of depression of its foot is $45^o$. Determine the height of the tower.
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.
From the top of a building \( 15 \mathrm{~m} \) high the angle of elevation of the top of a tower is found to be \( 30^{\circ} \). From the bottom of the same building, the angle of elevation of the top of the tower is found to be \( 60^{\circ} \). Find the height of the tower and the distance between the tower and building.
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?
The angle of elevation of top of tower from certain point is $30^o$. if the observer moves $20\ m$ towards the tower, the angle of elevation of the top increases by $15^o$. Find the height of the tower.
The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is $60^{o}$. From a point Y, $40\ m$ vertically above X, the angle of elevation of the top Q of tower is $45^{o}$. Find the height of the tower PQ and the distance PX. $( Use\ \sqrt{3} \ =\ 1.73)$
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.
The angles of elevation of the top of a rock from the top and foot of a \( 100 \mathrm{~m} \) high tower are respectively \( 30^{\circ} \) and \( 45^{\circ} \). Find the height of the rock.
From the top of a \( 50 \mathrm{~m} \) high tower, the angles of depression of the top and bottom of a pole are observed to be \( 45^{\circ} \) and \( 60^{\circ} \) respectively. Find the height of the pole..
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $60^o$. From another point $20\ m$ away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is $30^o$ (see the given figure). Find the height of the tower and the width of the canal."
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.
Kickstart Your Career
Get certified by completing the course

Get Started