**Height
and Distance**

Trigonometric
ratios are used to solve the problems related to **height and distance**. This
is one of the important application of trigonometry. It helps to find the
**heights** of the objects like tree, building, tower etc. and **distance** between the
objects.

While
finding such height and distances, we have to solve the right angled triangle.
For example, in the figure given below, AB is the height of a tree, BC is the
distance between the tree and the observer at C.

Here, if θ
and BC are known, we can calculate the height of tree AB, or if θ and AB are
known, we can calculate the distance between the tree and the observer by using
trigonometric ratio.

**Angle
of elevation and depression**

In the figure
given below, an observer is looking at the top of a tower. Here, OA is called
the **line of sight** and OB is the **horizontal line**. ∠BOA
is the angle between the upward line of sight and the horizontal line. Here, ∠BOA is called the **angle of elevation**.

Thus, the
angle made by the upward line of sight with horizontal line is known as **angle
of elevation**.

And, in the
figure given below, ∠BOA is the angle between the
downward line of sight and the horizontal line. ∠BOA
here, is called the **angle of depression**.

Thus, the angle
made by downward line of sight with horizontal line is known as **angle of
depression**.

In the
figure given below, the horizontal lines AC and BO are parallel to each other.
The angle of depression CAO and the angle of elevation BAO are alternate angles
between parallel lines.

∴
∠CAO = ∠BAO,
i.e. Angle of depression = Angle of elevation.

The
instrument which is used to measure the angle of elevation or the angle of
depression is called **Theodolite**.

**Worked
Out Examples **

*Example 1: A
man observes the top of a tower 40 m high and finds the elevation of 30°. Find
the distance between the man and the foot of the tower.*

*Solution:*

*Let AB be the height
of the tower and BC be the distance between the foot of the tower and the man.*

*Here, AB = 40 m*

*Angle of elevation, **∠**BCA
= 30°*

*In right angled
triangle ABC, *

*tan30°
= p/b = AB/BC = 40/BC*

*or, 1/√3 = 40/BC*

*or, BC = 40√3 = 40 × 1.732 = 69.28 m*

*So, the required
distance is 69.28 m.*

*Example 2:
An electric pole 20 m high is supported by a wire fixing its one end on the ground
at some distance from the pole. If the wire joining the top of the pole is
inclined to the ground at an angle of 30°, find the length of the wire.*

*Solution:*

*Let AB be the height
of the pole and AC be the length of wire.*

*Here, AB = 20m.*

*Angle of elevation, **∠**BCA
= 30°.*

*In right angled
triangle ABC, *

*sin30°
= p/b = AB/AC = 20/AC*

*or, ½ = 20/AC*

*or, AC = 40.*

*So, the required
length of the wire is 40m.*

*Example 3:
The circumference of a circular pond is 176m and a pillar is fixed at the
centre of the pond. If a person finds the angle of elevation of 60° of the top
of the pillar from any point on the bank of the pond, find the height of the
pillar above the water pillar.*

*Solution:*

*Let AB be the height
of the pillar above the water level and
BC be the radius of the pond.*

*Here, the
circumference of the pond = 176m*

*i.e. 2**π**r = 176*

*or, 2 × 22/7 × BC = 176*

*or, 44/7 × BC = 176*

*or, BC = 176 × 7/44*

*or, BC = 28m*

*The angle of
elevation, **∠**BCA = 60°*

*Now, in right angled
triangle ABC, *

* tan60° = p/b = AB/BC = AB/28*

*or, √3 = AB/28*

*or, AB = 28√3 = 28 × 1.372 = 48.5*

*So, the required
height of the pillar is 48.5m.*

*
*

*Example 4: A
man 1.5m tall standing at a distance of 50m from a tree observes the angle of
elevation of the top of the tree is 45°. Find the height of the tree.*

*Solution:*

*Let CE be the height
of the tree, AB be the height of the man and BC be the distance between the
foot of the tree and the man.*

*Here, AB = 1.5m, BC =
50m, Angle of elevation DAE = 45°*

*In rectangle ABCD,*

*DC = AB = 1.5m and AD
= BC = 50m*

*In right angled
triangle ADE,*

* tan45° = p/b = ED/AD = ED/50*

*or, 1 = ED/50*

*or, ED = 50m*

*Now, CE = ED + DC = 50
+ 1.5 = 51.5m*

*So, the required
height of the tree is 51.5m.*

*Example 5: A
boy, 1.2m tall, is flying a kite. When the length of the string of the kite is
180m, it makes an angle of 30° with the horizontal line. At what height is the
kite from the ground?*

*Solution:*

*Let CK be the height
of the kite from the ground, AB be the height of the boy and AK be the length
of the string.*

*Here, AB = 1.2m, AK =
180m, Angle of elevation DAK = 30°*

*In rectangle ABCD, AB
= DC = 1.2m.*

*In right angled
triangle ADK,*

* sin30° = p/h = KD/AK = KD/180*

*or, ½ = KD/180*

*or, 2KD = 180*

*or, KD = 180/2 = 90m*

*Now, CK = KD + DC = 90
+ 1.2 = 91.2m*

*So, the kite is at a
height of 91.2m from the ground.*

*Example 6: A
man is 1.6m tall and the length of his shadow in the sun is 1.6√3 m. Find the
altitude of the sun.*

*Solution:
*

*Let AB be the height
of the man and BC be the length of his shadow. Let θ be the altitude of the
sun.*

*Here, AB = 1.6 m, BC =
1.6√3 m.*

*In right angled
triangle ABC,*

* tanθ = p/b = AB/BC*

*or, tanθ = 1.6/1.6√3*

*or, tanθ = 1/√3 = tan30°*

*∴**
θ = 30°*

*So, the required
altitude of the sun is 30°.*

*Example 7: A
man √3 m tall is 72 m away from a tower 25√3 m high. Find the angle of elevation
of the top of the tower from his eyes.*

*Solution:*

*Let AB be the height
of the man, CD be the height of the tower and BD be the distance between the
man and the tower.*

*Here, BD = AE = 72 m*

* AB = √3 m*

* CD = 25√3 m*

* CE = CD – ED = 25√3 – AB = 25√3 - √3
= 24√3 m.*

*Now, in right angled
triangle AEC,*

* tanA = p/b = CE/AE*

*or, tanA = 24√3/72 = √3/3*

*or, tanA = 1/√3*

*or, tanA = tan30°*

*∴**
A = 30°*

*So, the required angle
of elevation is 30°.*

*Example 8:
The top of a tree which is broken by the wind makes an angle of 60° with the
ground at a distance 3√3 m from the foot of the tree. Find the height of the
tree before it was broken.*

*Solution:*

*Let AC be the height
of the tree before it was broken, BD be the broken part of the tree and CD be
the distance between the foot of the tree and the point on the ground at which
the top of the tree touched.*

*Here, CD = 3√3 m,
Angle of elevation, **∠**CDB = 60°.*

*In right angled
triangle BCD,*

* tan60° = p/b = BC/CD = BC/3√3*

*or, √3 = BC/3√3*

*or, BC = 9 m*

*Also, cos60° = b/h =
CD/BD*

*or, ½ = 3√3/BD*

*or, BD = 6√3 = 6 × 1.732 = 10.39 m.*

*Then, the height of
the tree before broken = AC*

* =
AB + BC*

* =
BD + 9m*

* =
10.39m + 9m*

* =
19.39m*

*So, the required
height of the tree was 19.39 m.*

*Example 9:
Two vertical poles are fixed 60 m apart. The angle of depression of the top of
the first pole from the top of the second pole which is 150 m high is 30°. Find
the height of the first pole.*

*Solution:*

*Let AB be the height
of the first (shorter) pole, CD be the height of the second (taller) pole and
BD be the distance between them.*

*Here, BD = AE = 60 m*

* CD = 150 m*

*The angle of
depression = angle of elevation, **∠**CAE = 30°.*

*In right angled
triangle CEA,*

* tan30° = p/b = CE/AE = CE/60*

*or, 1/√3 = CE/60*

*or, CE = 60/√3*

*or, CE = 34.64 m*

*Again, in rectangle
ABDE,*

*AB = DE = CD – CE =
150 – 34.64 = 115.36 m*

*So, the required
height of the first pole is 115.36 m.*

*Example 10:
Two men are on the opposite side of a tower of 30 m high. They observed the
angle of elevation of the top of the tower and found to be 30° and 60°
respectively. Find the distance between them.*

*Solution:*

*Let AB be the height
of the tower and CD be the distance between the two men.*

*Here, AB = 30m.*

*Angle of elevations, **∠**BCA
=30° and **∠**BDA = 60°.*

*In right angled
triangle ABC,*

* tan30° = p/b = AB/BC*

*or, 1/√3 = 30/BC*

*or, BC = 30√3 = 30 × 1.732 = 52.05 m*

*Again, in right angled
triangle ABD,*

* tan60° = p/b = AB/BD*

*or, √3 = 30/BD*

*or, BD = 30/√3 = 30/1.732 = 17.32m*

*Now, CD = BC + BD =
52.05 + 17.32 = 69.37 m*

*So, the required
distance between the men is 69.37m.*

* *

*You can comment your questions
regarding the problems on height and distance here.*

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