An observer, $ 1.5 \mathrm{~m} $ tall, is $ 28.5 \mathrm{~m} $ away from a tower $ 30 \mathrm{~m} $ high. Determine the angle of elevation of the top of the tower from his eye.
Given:
An observer, \( 1.5 \mathrm{~m} \) tall, is \( 28.5 \mathrm{~m} \) away from a tower \( 30 \mathrm{~m} \) high.
To do:
We have to determine the angle of elevation of the top of the tower from his eye.
Solution:
Let $AB$ be the tower and $CD$ be the observer who is $28.5\ m$ away from the tower.
From the figure,
$AB = 30\ m, CD=1.5\ m, AC=28.5\ m$
This implies,
$AE=CD=1.5\ m, DE=AC=28.5\ m$ and $BE=30-1.5=28.5\ m$
Let \( \theta \) be the angle of elevation of the top of the tower from the eye of the observer.
In $\Delta \mathrm{BDE}$,
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$\Rightarrow \tan \theta=\frac{\mathrm{BE}}{\mathrm{DE}}$
$=\frac{28.5}{28.5}$
$=1$
$=\tan 45^{\circ}$ [Since $\tan 45^{\circ}=1$]
$\Rightarrow \theta=45^{\circ}$
Therefore, the angle of elevation of the top of the tower from his eye is $45^{\circ}$.
Related Articles
- An observer \( 1.5 \) metres tall is \( 20.5 \) metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.
- An observer $1.6$ metres tall is $10.5$ metres away from a tower $22$ metres high.Determine the angle of elevation of the top of the tower from the eye of the observer.
- The angles of elevation of the top of a tower from two points at a distance of \( 4 \mathrm{~m} \) and \( 9 \mathrm{~m} \) from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is \( 6 \mathrm{~m} \).
- From the top of a \( 7 \mathrm{~m} \) high building, the angle of elevation of the top of a cable tower is \( 60^{\circ} \) and the angle of depression of its foot is \( 45^{\circ} . \) Determine 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 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 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 of the building from the foot of the tower is \( 30^{\circ} \) and the angle of the top of the tower from the foot of the building is \( 60^{\circ} \). If the tower is \( 50 \mathrm{~m} \) high, find the height of the building.
- 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 hill at the foot of a tower is \( 60^{\circ} \) and the angle of elevation of the top of the tower from the foot of the hill is \( 30^{\circ} \). If the tower is \( 50 \mathrm{~m} \) high, what is the height of the hill?
- A \( 1.5 \mathrm{~m} \) tall boy is standing at some distance from a \( 30 \mathrm{~m} \) tall building. The angle of elevation from his eyes to the top of the building increases from \( 30^{\circ} \) to \( 60^{\circ} \) as he walks towards the building. Find the distance he walked towards the building.
- 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.
- 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.
- 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 T.V. Tower stands vertically on a bank of a river. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is \( 60^{\circ} \). From a point \( 20 \mathrm{~m} \) away this point on the same bank, the angle of elevation of the top of the tower is \( 30^{\circ} \). Find the height of the tower and the width of the river.
Kickstart Your Career
Get certified by completing the course
Get Started