NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

NCERT
SOLUTIONS
CLASS - 10th

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Class : 10th
Subject : Maths
Chapter : 9
Chapter Name : Some Applications of Trigonometry

Exercise 9.1

Q1 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 .
∘

It can be observed from the figure that AB is the pole. In △ABC,

AB ∘
= sin 30
AC

AB 1
=
20 2

20
AB = = 10
2

Therefore, the height of the pole is 10m .

Page : 203 , Block Name : Exercise 9.1

Q2 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.

Let AC was the original tree. Due to storm, it was broken into parts. The broken part
A B is making 30 with the ground.
′ ∘

′
In ΔA BC

BC ∘
′
= tan 30
A C

BC 1
=
8 √3

8
BC = ( )m
√3
′
A C ∘
′
= cos 30
A B

8 √3

′
=
A B 2

′ 16
A B = ( )m
√3

′
Height of tree = A B + BC

16 8 24
= ( + )m = m
√3 √3 √3

= 8√3m

Hence, the height of the tree is 8√3m.

Page : 203 , Block Name : Exercise 9.1

Q3 A contractor plans to install two slides for the children to play in a park. For the children
below the age of 5 years, she 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, where as for the elder children she wants to have a
∘

steep side 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?

Answer. It can be observed that AC and PR are the slides for younger and elder children
respectively.

In △ABC,

AB ∘
= sin 30
AC

1.5 1
=
AC 2

AC = 3m

In ΔPQR
,

PQ
= sin 60
PR

3 √3
=
PR 2

6
PR = = 2√3m
√3

Therefore, the length of these slides are 3 m and 2√3m.

Page : 203 , Block Name : Exercise 9.1

Q4 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.
∘

Let AB be the tower and the angle of elevation from point C(on ground) is 30 .
∘

In △ABC ,
AB ∘
= tan 30
BC

AB 1
=
30 √3

30
AB = = 10√3m
√3

Therefore, the height of the tower is 10√3m.

Q5 A kite is ying 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.

Let K be the kite and the string is tied to point P on the ground.
In ΔKLP,

KL ∘
= sin 60
KP

60 √3
=
KP 2

120
KP = = 40√3m
√3

Hence, the length of the string is 40√3m .

Q6 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.

Let the boy was standing at point S initially. He walked towards the building and reached at
point T.
It can be observed that
PR = PQ − RQ

57
= (30 − 1.5)m = 28.5m = m
2

In ΔPAR,

PR ∘
= tan 30
AR

57 1
=
ARR √3

57
AR = ( √3) m
2

In △PRB,
PR ∘
= tan 60
BR

57
= √3
2BR

57 19√3
BR = = ( )m
2
2√3

ST = AB

57√3 19√3
= AR − BR = ( − )m
2 2

38√3
= ( ) m = 19√3m
2

Hence, he walked 19√3 m towards the building.

Q7 From a point on the ground, the angles of elevation of the bottom and the top of a
transmission tower xed at the top of a 20 m high building are 45 and 60 respectively. Find
∘ ∘

the height of the tower.

Let BC be the building, AB be the transmission tower, and D be the point on the ground from
where the elevation angles are to be measured.
In △BCD,
BC ∘
= tan 45
CD

20
= 1
CD

CD = 20m

In ΔACD

AC ∘
= tan 60
CD

AB+BC
= √3
CD

AB+20
= √3
20

AB = (20√3 − 20)m

= 20(√3 − 1)m

Therefore, the height of the transmission tower is 20(√3 − 1)m.

Q8 A statue, 1.6 m tall, stands on a top of pedestal, from a point on the ground, the
angle of elevation of the top of 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.
∘

Let AB be the statue, BC be the pedestal, and D be the point on the ground from where the
elevation angles are to be measured.

In ΔBCD,

BC ∘
= tan 45
CD

BC
= 1
CD

BC = CD

In △ACD,

AB+BC ∘
= tan 60
CD

AB+BC
= √3
BC

1.6 + BC = BC√3

BC(√3 − 1) = 1.6

(1.6)(√3 + 1)
BC =
(√3 − 1)(√3 + 1)

1.6(√3 + 1)
=
2 2
(√3) − (1)

1.6(√3 + 1)
= = 0.8(√3 + 1)
2

Therefore, the height of the pedestal is 0.8 (√3 + 1)m.

Q9 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, nd the height of the building.
∘

Let AB be the building and CD be the tower.
In ΔCDB1

CD ∘
= tan 60
BD

50
= √3
BD

50
BD =
√3

In △ABD,

AB ∘
= tan 30
BD

50 1 50 2
AB = × = = 16
√3 √3 3 3

Therefore, the height of the building is 16 2

3
m .

Q10 Two poles of equal heights are standing opposite each other an 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 poles and the
∘ ∘

distance of the point from the poles.

Let AB and CD be the poles and O is the point from where the elevation angles are measured.
In △ABO,

AB ∘
= tan 60
BO

AB
= √3
BO

AB
BO =
√3

In ΔCDO ,

CD ∘
= tan 30
DO

CD 1
=
80−BO √3

CD√3 = 80 − BO

AB
CD√3 = 80 −
√3

AB
CD√3 + = 80
√3

Since the poles are of equal heights,
CD = AB

1
CD [√3 + ] = 80
√3

3+1
CD ( ) = 80
√3

CD = 20√3m

AB CD 20√3
BO = = = ( ) m = 20m
√3 √3 √3

DO = BD − BO = (80 − 20)m = 60m

Therefore, the height of poles is 20√3 m and the point is 20 m and 60 m far from these poles.

Q11 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.

Let AB be a building and CD be a cable tower.

In △ABD,

AB ∘
= tan 45
BD

7
= 1
BD

BD = 7m

In ΔACE,

AC = BD = 7m

CE
∘
= tan 60
AE

CE
= √3
7

CE = 7√3m

CD = CE + ED = (7√3 + 7)m

Therefore, the height of the cable tower is 7(√3 + 1)m.
AB ∘
= tan 60
BC

AB
= √3
BC

AB
BC =
√3

In △ABD
AB ∘
= tan 30
BD

AB 1
=
BC+CD √3

AB 1
=
AB
+20 √3
√3

AB√3 1
=
AB+20√3 √3

3AB = AB + 20√3

2AB = 20√3

AB = 10√3m

AB 10√3
BC = = ( ) m = 10m
√3 √3

Therefore, the height of the tower is 10√3 m and the width of the canal is 10 m.

Q12 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.
∘ ∘

Let AB be a building and CD be a cable tower.
In ΔABD,
AB ∘
= tan 45
BD

7
= 1
BD

BD = 7m

In ΔACE,
AE = BD = 7 m
CE ∘
= tan 60
AE

CE
= √3
7

CE = 7√3m

CD = CE + ED = (7√3 + 7)m

= 7(√3 + 1)m

Therefore, the height of the cable tower is 7(√3 + 1)m

Q13 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, nd the distance between the two ships.

Let AB be the lighthouse and the two ships be at point C and D respectively.

AB ∘
= tan 45
BC

In triangle ABC , 75
= 1
BC

BC = 75m

In △ABD ,

AB ∘
= tan 30
BD

75 1
=
BC+CD √3

75 1
=
75+CD √3

75√3 = 75 + CD

75(√3 − 1)m = CD

Therefore, the distance between the ships is 75(√3 − 1) m.

Q14 A 1.2 m 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 the angle of elevation reduces to 30
∘ ∘

Let the initial position A of balloon change to B after some time and CD be the girl.In △ ACE,
AE ∘
= tan 60
CE

AF−EF ∘
= tan 60
CE

88.2−1.2
= √3
CE

87
= √3
CE

g7
CE = = 29√3m
√3

In ΔBCG,

BG ∘
= tan 30
CG

88.2−1.2 1
=
CG √3

Distance travelled by balloon = EG = CG − CE

= (87√3 − 29√3)m

= 58√3m

Page : 205 , Block Name : Exercise 9.1

Q15 A straight highway leads to the foot of a tower. A man standing at the top of the
tower observes a car as 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.

Let AB be the tower.

Initial position of the car is C, which changes to D after six seconds.

In △ADB ,

AB ∘
= tan 60
DB

AB
= √3
DB

AB
DB =
√3

In △ABC

AB ∘
= tan 30
BC

AB 1
=
BD+DC √3

AB√3 = BD + DC

AB
AB√3 = + DC
√3

AB 1
DC = AB√3 − = AB (√3 − )
√3 √3

2AB
=
√3

2AB
Time taken by the car to travel distance DC (i.e., ) = 6 seconds
√3

AB 6 AB
Time taken by the car to travel distance DB ( i.e., ) = ×
2AB
√3 √3
√3

6
= = 3 seconds
2

Page : 205 , Block Name : Exercise 9.1

Q16 The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 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 m .

Let AQ be the tower and R, S are the points 4 m, 9 m away from the base of the tower
respectively.
The angles are complementary. Therefore, if one angle is θ, the other will be 90 − θ.
In △AQR,

AQ
= tan θ
QR

AQ
= tan θ
4

In ΔAQS,
AQ
= tan(90 − θ)
SQ

AQ
= cot θ
9

On multiplying equations ( i) and (ii), we obtain

AQ AQ
( )( ) = (tan θ) ⋅ (cot θ)
4 9

2
AQ
= 1
36

2
AQ = 36

AQ = √36 = ±6

However, height cannot be negative.

Therefore, the height of the tower is 6 m.

Page : 205 , Block Name : Exercise 9.1