Some

Applications
of Trigonometry 9
NCERT SOLUTIONS

What's inside
– In-Chapter Q's (solved)
– Textbook Exercise Q's (solved)
 Exercise – 9.1
1. A circus artist is climbing a 20 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° (See Figure Below). 

Sol. : In the given figure, AB is the height of the pole and the length of the rope is 20 m, which is
tied at the top of the pole.

In right angled DABC,

sin 30° = AB
AC

1
2 = AB
20

AB = 20 =
2 10 m
Hence, the height of the pole is 10 m.
2. 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° 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.
Sol. : Let AC be the height of the tree, which is broken from B due to storm and the broken part BC
touches the ground at point D making an angle of 30° with the ground.

m
Given : AD = 8 m
∠BDA = 30°

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 In ∆ABD,
tan 30° = AB
AD
⇒ 1
= AB
8
3
⇒ AB = 8
3
Again, in ∆ABD,
cos 30° = AD
BD

⇒ 2
3
= 8
BD 
⇒ BD = 8×2
= 16
m
3 3
Q AC = AB + BC
\ AC = 8 + BD[∵ BC = BD]
3
⇒ AC = 8
+ 16
3 3
= 24
=8 3 m
3
Hence, the height of the tree is 8 3 m.
3. A contractor plans to install two slides for the children to play in a park. For the children
below the age of 5 years, he prefers to have a slide whose top is at a height of 1.5 m,
and is inclined at an angle of 30° to the ground, whereas for elder children, he wants to
have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What
should be the length of the slide in each case?
Sol. : (i) For children below the age of 5 years :
Let the slide be AC.
The height of the slide AB = 1.5 m and the slide AC
makes an angle of 30° with the ground.
A

1.5 m

30°
C B
In right angle DABC,
sin 30° = AB
AC

⇒ 1
2 = 1.5
AC
\ AC = 1.5 × 2 = 3 m
(ii) For children above the age of 5 years :
Let the slide be AC. The height of the slide AB = 3 m and the slide AC makes an angle of 60°
with the ground.
In right angled DABC,

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 A

3m

60°
C B

sin 60° = AB
AC

⇒ 2
3
= 3
AC

\ AC = 3× 2
3
= 2 3 m
4. 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°. Find the height of the tower.
Sol. : Let there is a tower AB of height h metres, which is at a distance of 30 metres from a point C
on the ground.
The angle of elevation of the top of the tower from point C is 30°.
A

hm

30°
C B
30 m
Then, in right angle DABC,
tan 30° = BC AB

⇒ 1
= h
30
3
⇒ h= 30
= 30
×
3
3 3 3

\ h = 303 3 = 10 3 m
5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is
temporarily tied to a point on the ground. The inclination of the string with the ground
is 60°. Find the length of the string, assuming that there is no slack in the string.
Sol. : Let the string AC be of length x m and the kite is flying at a height BC = 60 m from the ground
and the string is temporarily tied at point A. The string of the kite makes an angle of 60° with
the ground.

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 C

x 60 m

60°
A B
In right angled DABC,

sin 60° = BC
AC

⇒ 2
3
= 60
x
⇒ x= 60 × 2
= 120
3 3

\ x= 120
×
3
3 3

= 120 3
3 = 40 3m

Hence, the length of string be 40 3 m.

6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of
elevation from his eyes to the top of the building increases from 30° to 60° as he walks
towards the building. Find the distance he walked towards the building.
Sol. : Given : Height of building, AB = 30 m
and height of the boy CD = 1.5 m
Let the boy’s eye be C and the angle of elevation of the top of the building from the eye are
30° and 60°. Let he walked x metres towards the building.
B

28.5 m

30° 60°
C x y F
1.5 m E 1.5 m
D A
So, AB = 30 m
BF = AB – AF = 30 – 1.5
= 28.5 m
CE = x and EF = y
In right angled ∆BCF,
tan 30° = BF
CF
⇒ 1
= 28.5
x+ y
3

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 ⇒ x + y = 28.5 3 ...(i)
In right angled ∆BEF,
tan 60° = BF EF
⇒ 3 = 28.5
y
⇒ y= 28.5
...(ii)
3
Put the value of y in equation (i),
x + 28.5 = 28.5 3
3
⇒ x = 28.5 3 – 28.5
3

= 28.5 3 × 3 – 28.5
3
28.5 ]3 – 1g
=
3

= 28.5 × 2
× 3
3 3

= 28.5 × 2 3
3 = 9.5 × 2 3

⇒ x = 19 3 m
Hence, the boy walked 19 3 metres towards the building.
7. 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° and 60° respectively.
Find the height of the tower.
Sol. : Let the height of the transmission tower AD = h m. The height of the building BD = 20 m
and the tower fixed at the top of the building. From a point C on the ground, the angles of
elevation of the bottom D and top A of the tower are 45° and 60° respectively.
A

20 m

60°
45°
C B

Now, in right angled ∆DBC,

tan 45° = BD
BC

⇒ 1= 20
BC
\ BC = 20 m ...(i)

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 In right angled DABC,
tan 60° = BC =
AB AD + BD
BC
⇒ 3 = h + 20
20 [From eqn. (i) BC = 20]
⇒ 20 3 = h + 20
\ h = 20 3 – 20
= 20 ^ 3 – 1h m
Hence, the height of transmission tower is 20 ^ 3 – 1h m.
8. 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° and from the same point the angle of
elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Sol. : Let the pedestal is AB whose height is h metres. A statue, BD = 1.6 m tall stands on the top
of the pedestal.
Let a point C on the ground from where the angle of elevation of the top of the statue is 60°
and the angle of elevation of the top of the pedestal is 45°.
D

1.6 m

60°
45°
C x A

Let, AC = x
In right angled ∆ABC,
tan 45° = AC AB

⇒ 1= h
x
\ x = h...(i)
In right angled ∆DAC,
tan 60° = DA AC =
AB + BD
AC
⇒ 3 = h + 1.6
x
⇒ 3 x = h + 1.6
Substituting the value of x from eqn. (i), we get
3 h = h + 1.6

⇒ 3h – h = 1.6
⇒ h^ 3 – 1h = 1.6

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 ⇒ h= 1.6
× 3 +1
3 –1 3 +1
1.6 ^ 3 + 1h
= 3–1
1.6 ^
= 2 3 + 1h

\ h = 0.8 ^ 3 + 1h
Hence, the height of pedestal = 0.8 ^ 3 + 1h m
9. The angle of elevation of the top of a building from the foot of the tower is 30° and the
angle of elevation of the top of the tower from the foot of the building is 60°. If the tower
is 50 m high, find the height of the building.
Sol. : Let the building is CD, whose height is h metres.
The tower AB = 50 m high. The angle of elevation of the top of the building from the foot of
the tower is ∠CBD = 30° and the angle of elevation of the top of the tower from the foot of
building is 60°.
A

50 m

h
60° 30°
C xm B

Let, BC = x m
In right angled ∆ABD,
tan 60° = BC AB

⇒ 3 = 50
x
⇒ x= 50
...(i)
3
In right angled ∆BCD,
tan 30° = CD BC
⇒ 1
= h
x
3
⇒ x= h 3
Substituting the value of x from eqn. (i), we get
50
=h 3
3
⇒ 50 = h 3 × 3 = 3h
\ h = 50
3 = 16 3 m
2

Hence, the height of the building is 2

16 3 m.

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 10. Two poles of equal heights are standing opposite each other on either side of the road,
which is 80 m wide. From a point between them on the road, the angles of elevation of
the top of the poles are 60° and 30°, respectively. Find the height of the poles and the
distances of the point from the poles.
Sol. : Let AB and CD be two poles of equal length placed on either side of a road 80 m wide.
Let the angles of elevation of the top of the poles from a point E on the road be 60° and 30°.

i.e., ∠AEB = 60°, ∠CED = 30°

Let, AB = CD = h m,
and BE = x m
then DE = (80 – x) m
In right angled ∆ABE,
tan 60° = BE AB

⇒ 3 = h
x
⇒ h= 3 x...(i)
In right angled ∆CDE,
tan 30° = CD DE
⇒ 1
= h
80 – x
3
⇒ 80 – x = 3 h...(ii)
On putting the value of h from eqn. (i) into eqn. (ii), we get
80 – x = 3 ^ 3 xh
⇒ 80 – x = 3x
⇒ 80 = 3x + x = 4x
\ x = 80
4 = 20 m
On putting the value of x in eqn. (i),
h = 3 x = 3 × 20
= 20 3 m
and DE = 80 – x = 80 – 20 = 60 m
Hence, the height of poles is 20 3 m. The distance between the poles are 20 m and 60 m re-
spectively.
11. 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°. 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°. Find the height of the tower
and the width of the canal.

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 A

60°
30°
D
20m C B

Let, the height of the tower AB be h m and x m be the width of the canal BC.
Sol. :
In right angled ∆ABC,
tan 60° = BC AB

⇒ 3 = h
x
⇒ h= 3x ...(i)
In right angled ∆ABD,
tan 30° = BD AB
= AB
BC + CD
⇒ 1
= h
x + 20
3
⇒ x + 20 = h 3 ...(ii)
Putting the value of h in eqn. (ii),
x + 20 = x 3 × 3
⇒ x + 20 = 3x
⇒ 20 = 3x – x
⇒ 2x = 20 ⇒ x = 20 2 =10 m
On putting the value of x in eqn. (i),
h = 3 x = 3 × 10
\ h = 10 3 m
Hence, the height of the tower is 10 3 m and the width of the canal is 10 m.
12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is
60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Sol. : Given that the angle of elevation of the top of only tower AB from a 7m high building CD is
60° and the angle of depression of the foot is 45°.
B

hm

60°
C E
45°

7m 7m

45°
D A
xm
i.e., ∠BCE = 60°

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 and ∠ACE = 45°
\ CE || AD
then ∠ACE = ∠CAD
= 45° [Alternate internal angle]
Let, the distance between the tower and the building be AD = x metres.
In right angle ∆CDA,
tan 45° = CD AD
⇒ 1 = 7x
⇒ x=7m
In right angle ∆BEC,
tan 60° = CE BE

⇒ 3 = x [Q AD = CE]
h

⇒ h= 3x
⇒ h = 3 × 7
\ h= 7 3 m
Hence, height of tower AB = AE + BE
= 7 + h
= 7 + 7 3
= 7 ^1 + 3 h m
13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of
depression of two ships are 30° and 45°. If one ship is exactly behind the other on the
same side of the lighthouse, find the distance between the two ships.
Sol. : The height of the lighthouse from sea level AC is CD = 75 m.
Let A and B be the positions of two ships on the sea level. Let the distance between two ships A
and B be AB = 5 m.
The angles of depression of the two ships from the top D of the lighthouse are ∠ODA = 30° and
∠ODB = 45°.
Here, ∠DAC = ∠ODA = 30°
and ∠CBD = ∠ODB = 45°
Let, BC = x
D
O
30°
45°

75 m

30° 45°
A C
y B x

In right angle ∆DBC,

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 tan 45° = CD
BC
⇒ 1= 75
x
\ x = 75 m
In right angle ∆ACD,
tan 30° = CD AC =
CD
AB + BC
1
= 75
y+x
3
x + y = 75 3
⇒ 75 + y = 75 3
⇒ y = 75 3 – 75
\ y = 75 ^ 3 – 1h
Hence, the distance between the two ships is 75 ^ 3 – 1h m.
14. 1.2m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2
m from the ground. The angle of elevation of the balloon from the eyes of the girl at any
instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance
travelled by the balloon during the interval.

Sol. : Let AB be the horizontal line on which a girl is standing whose length AD = 1.2 m.
Let, FH = EB = 88.2 m is the height of the balloon from the horizontal line AB. The angle of
elevation of the balloon from the girl’s eye D are 60° and 30°.

E
F

88.2 m
60°
30°
D C
G
1.2 m 1.2 m

A H B
x y

i.e., ∠FDC = 60°

and ∠EDC = 30°
\ CE = FG

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 = 88.2 – 1.2 = 87 m
Let, AH = DG = x m
and HB = GC = y m
In right angle ∆FGD,
tan 60° = FG
DG
⇒ 3 = 87
x
\ x= 87
...(i)
3
In right angle ∆ECD,
tan 30° = EC
DC = EC
DG + GC

⇒ 1
= 87
x+ y
3

⇒ x + y = 87 3

⇒ 87
+ y = 87 3
3

⇒ y = 87 3 – 87
3

= 87 d 3–
1
3
n

= 87 d 3 –31 n = 87 × 2
3

= 87 × 2
× 3
3 3

= 87 × 2 × 3
3

\ y = 29 × 2 × 3
= 58 3 m
Hence, the distance covered by the balloon = 58 3 m.
15. A straight highway leads to the foot of a tower. A man standing at the top of the tower
observes a car at an angle of depression of 30°, which is approaching the foot of the
tower with a uniform speed. Six seconds later, the angle of depression of the car is found
to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Sol. : Let AB be a tower whose height is h metres. A boy standing at the top of a tower observes
a car at an angle of depression of 30°, which is approaching the height of the tower with a
uniform speed. After 6 seconds the angle of depression of the car becomes 60°.

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 A
O
30°
60°

h m

30° 60°
C D B
x y

∠OAC = 30°, ∠OAD = 60°

Now, ∠OAC = ∠ACB = 30°
[Alternate interior angle]
∠OAD = ∠ADB = 60°­
[Alternate interior angle]
Let, CD = x and BD = y
In right angle DABD,
tan 60° = AB
BD

⇒ 3 = h
y

⇒ h= 3 y ...(i)
In right angle DABC,
tan 30° = AB AB
BC = BD + DC

⇒ 1
= h
y+x
3

⇒ y+x= h 3 ...(ii)

Putting the value of h from. eqn. (i) into eqn. (ii) we get,
y+x= 3y× 3

⇒ y + x = 3y
⇒ x = 3y – y
x
x = 2y ⇒ y = 2
Q CD = Time taken to cover a distance of x m = 6 sec.
\ y = Time taken to cover x
2 m distance = 6
2 = 3 sec.
Hence, the car will take 3 seconds to reach the tower.

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