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Some Applications of
Trigonometry
Objective Section
Fill in the Blanks
Q
...
In Fig
...


[CBSE OD, Set 1, 2020]

Ans
...
2
...
m
...
6

Explanation : In DABC,

AC = 10 m 
(length of ladder)

AB = 8 m (height of window)

Fig
...
The distance of the foot

Now
AC2 = AB2 + BC2

(Pythagoras theorem)

⇒
(10)2 = (8)2 + BC2

⇒
100 – 64 = BC2

⇒
BC2 = 36

⇒
BC = 6 m

Very Short Answer Type Questions
Q
...
T
he ratio of the length of a vertical rod
and the length of its shadow is 1 : 3
...
Let the angle of elevation be q
...


Q
...
If a tower 30 m high, casts a shadow
10 3 m long on the ground, then what
is the angle of elevation of the sun? 



We know that,

Perpendicular
Base
AB
⇒ tan q =
BC




tan q =


[CBSE OD, Term 2, Set 1, 2017]
Ans
...



tan θ =



⇒

 AB = 2
...


Q
...
In Fig
...
If AD = 2
...
(use 3 = 1
...
  3
...
What is the angle of elevation
of the sun?

[CBSE Delhi, Term 2, Set 1, 2017]
Ans
...
1

AB
= 3
BC
1



[CBSE Delhi, Term 2, Set 1, 2016]

Ans
...
54 m
...
54) m = 3
...

Q
...
A ladder, leaning against a wall, makes
an angle of 60° with the horizontal
...
5 m away from the
wall, find the length of the ladder
...
46
=
DC
2

sin 60° =

DC =

3
...
73

∴ The length of the ladder is 4 m
...
  6
...
Find the sun’s altitude
...
Let AB be the ladder leaning against a
wall AC
...
5 m



Then, cos 60° =



⇒

BC
AB

1 2
...
Given AB is the tower and BC is its
­shadow
...

Perpendicular
 [
...
  7
...


[CBSE Delhi, Term 2, Set 1, 2015]
Ans
...




60°

30°
M

MC =

y
tan 60°

y

⇒



x tan 30°
=
y tan 60°

C

⇒



x 1/ 3 1
=
=
y
3
3

M is the mid-point of BC i
...
, BM = MC


\



[CBSE Delhi, Set 1, 2020]

(3 marks each)
=

1
...
4
Solution : Given : AC = 1
...
(ii)

x : y = 1 : 3
...
1
...
4
...
5 m long and CD = 3 m, find (i) tan q
(ii) sec q + cosec q


...
(i) and (ii), we get


A

B



y
x
=
tan 30° tan 60°

D

x

In D ABM, we have
AB
= tan 30°
BM

x

⇒
BM =
tan 30° 
In D CDM, we have
DC
= tan 60°
MC

y
⇒
= tan 60°
MC



1
3

Ans
...


Q
...
On a straight line passing through the
foot of a tower, two points C and D are at
distances of 4 m and 16 m from the foot
respectively
...

[CBSE OD, Term 2, Set 1, 2017]

Ans
...


From ∆ ABD,



⇒



Distance covered in 2 min
...
, speed)
= 150 −


In Δ ABC,
AB
= tan (90° − θ)
BC



In Δ ABD,


h
= cot θ
...
(ii)
16


Multiply eq
...


or



AB
BC =
tan 60°

BC =



=



Speed =

150( 3 − 1)
× 60
3 ×2
4500( 3 − 1)
3

= 4500 − 1500 3
= 4500 − 2598 = 1902 m/hr




Hence, the speed of boat is 1902 m/hr
...
  4
...
If height of the tower
is 50 m, find the height of the hill
...
Let AB be hill and DC be tower
...
  3
...
The angle of depression of the
boat changes from 60° to 45° in 2 minutes
...


[CBSE Delhi, Term 2, Set 1, 2017]
AB
Ans
...

[
...
(i)

1
DC
= tan 30° =
3
BC
3
 BC =
DC = 50 3


...
 5
...
Find the distance of
the hill from the ship and the height of the
hill
...
Let AB and CD be the tower and high
building, respectively
...
(i)



⇒ 1 =



⇒

A

Ans
...


D

45°

E

h

50 m

C

Then,



tan 30° =

⇒




tan 60° =



AB
CB
h

⇒
3 =
CB
h

⇒
CB =

...
(i)

x
y

x

...


Q
...
The angles of depression of the top and
bottom of a 50 m high building from the
top of a tower are 45° and 60° respectively
...
(use 3 = 1
...
(i) and (ii), we get
h

h − 50 =
3



⇒
h −

h
= 50
3

( 3 − 1)
⇒
h = 50
3


⇒

h=

3 +1
50 3
50 3
=
×
3 −1
3 −1
3 +1



⇒


h=

50 3 ( 3 + 1)
3 −1

⇒


h=

150 + 50 3
2



⇒



h = 75 + 25 3


⇒



h = 75 + 25 (1
...
25 m

Hence, the height of the tower is 118
...
25 m
...
 7
...
After
flight of 15 seconds, the angle of elevation
change to 30°
...


[CBSE OD, Term 2, Set 1, 2015]
Ans
...


...
 8
...
If the tower is 30 m high, find the
height of the building
...
Let AB be the tower and CD be a building
of height 30 m and x m respectively
...

A
D

Let BC be the height at which the ­aeroplane
flying
...


Let AB = x m, BD = y m
So,
AD = (x + y) m
In D ABC,
BC

tan 60° =
AB
⇒

3 =

1500 3
x

30°
B






30
= 1 ⇒ y = 30
y



And, in DBDC

In D EAD


ED
AD



Long Answer Type Questions
Q
...
From a point on the ground, the angles
of elevation of the bottom and the top
of a tower fixed at the top of a 20 m high

45°

30
= tan 45°
y




tan 30° =

y

C

Then, in DABC

[
...

x = 30 ×

(4 marks each)

building are 45° and 60° respectively
...



[CBSE OD, Set 1, 2020]

Ans
...




tan 60° =



⇒



⇒

DC
BC

50
BC
50
BC =
3
3 =

And, in a right angled triangle ABC


\ AB = 20 m, ∠APB = 45° and ∠CAB = 60°
Now, in right triangle APB,
AB

tan 45° =
PB


⇒



⇒

20
1 =
PB
PB = 20



⇒

tan 60° =

BC
PB

3 =
20



⇒



Now, AC = BC – AB = 20 3 – 20

BC = 20 3

= 20 ( 3 - 1)
Hence, the height of the tower is

20 ( 3 - 1) m
...


Q
...


The angle of elevation of the top of a
building from the foot of a tower is 30° and
the angle of elevation of the top of a tower
from the foot of the building is 60°
...

[CBSE OD, Set 2, 2020]

Ans
...



\ CD = 50 m ∠ACB = 30° and ∠DBC = 60°
...
67 m
...
67 m
...

Q
...
A vertical tower stands on a horizontal
plane and is surmounted by a vertical
flag-staff of height 6 m
...
Find the height of the tower
...
73) [CBSE Delhi, Set 1, 2020]
Ans
...


and height of the flag-staff AD = 6 m
...
(i)



⇒

1 =

AD + AB
BC



⇒

1 =

(6 + h )
BC


⇒
BC = (6 + h)m 
From (i) and (ii), we get
h 3 = (6 + h)




Ans
...


AB = 7 m

Also, let
DC = h m
...
(ii)

h 3 - h = 6
h( 3 - 1) = 6



h =

6
3 -1

6( 3 + 1)
6
3 +1
=
×
=
2
2
3 -1
3 + 1 ( 3 ) - (1)


6 ( 3 + 1) 6 ( 3 + 1)
=
=
3 -1
2


= 3 ( 3 + 1)


= 3 (1
...
73)

= 8
...
19 m is the height of the tower
...

Q
...
From the top of a 7 m high building the
angle of elevation of the top of a tower is
60° and the angle of depression of its foot
is 45°
...


[CBSE Delhi, Term 2, Set 2, 2017]

[CBSE Delhi, Set 3, 2020]

Now, in DAEC, we have
EC
AE
7

⇒
1 =
AE

⇒
AE = 7
In DADE, we have
DE

tan 60° =
AE
h-7


⇒
3 =
7





⇒
⇒



⇒

tan 45° =

7 3 =h–7

h = 7 3 + 7
h = 7( 3 + 1)


\ The height of the tower is 7( 3 + 1)
meters
...


Q
...
The shadow of a tower standing on a level ground is found to be 40 m longer when the
Sun’s altitude is 30° than when it was 60°
...
(Given

3 = 1
...
 6
...
Deepak
standing on the roof of a 50 m high
building, finds the angle of elevation
of the same bird to be 45°
...
Find the distance of the bird from
Deepak
...
Let Amit be at C point and bird is at A
point, such that ∠ACB = 30°, AB is the
height of bird from point B on ground
and Deepak is at D point, DE is the building of height 50 m
...


Q
...
Two poles of equal heights are standing
opposite to each other on either side of
the road which is 80 m wide
...
Find the
heights of the poles and the distance of
the point P from the poles
...
  8
...
If the
length of the flag-staff is 5 m, find the
height of the tower
...
732)

[CBSE OD, Set 3, 2019]
Ans
...

So, BC = 5 m

Ans
...


Let P be a point on the ground such that
∠APB = 30° and ∠APC = 45°, BC = 5 m
...
AP = x)
x



Let AB = h m and PA = x metres



From right ΔPAB, we have
x PA
cot 30° = =
h AB

x
3 =
h


h= 3x


...
(i)

From right ΔPAC, we have
PA
x
=
cot 45°  =
AC h + 5

 x = h + 5
...
(ii)

Equating the values of h from equation (i)
and (ii) we get
80 − x
⇒
x 3 =
3
⇒
3x = 80 − x
⇒
4x = 80
⇒
x = 20 m
On putting x = 20 in equation (i), we get




80 − x
h=
3 


⇒

h = 3 × 20 = 20 3

Thus, height of poles is 20 3 m and point
P is at a distance of 20 m from left pole
and (80 − 20) i
...
, 60 m from right pole
...
732 − 1
3 −1

5
0
...
83 m
Hence, the height of tower is 6
...


Q
...
A man in a boat rowing away from a
light house 100 m high takes 2 minutes
to change the angle of elevation of the
top of the light house from 60° to 30°
...
[Use 3 = 1
...
Let AB be the light house, C and D be the
two positions of the boat, such that,

In ΔABD




⇒




or



⇒


Now, In ΔABC



tan 60° =






AB
BC

3 =

100
y

y=

100
3

AB
BD

1
100
=
3 x+y
x + y = 100 3
y = 100 3 − x
...
(i)



x = 100 3 −

100
3

1 
= 100  3 −

3

2
200
=
= 100 ×
3
3


=
115
...


...
48 m = 2 min
115·48
∴ Speed of boat =
= 57
...
74 m/min

Q
...
As observed from the top of a 100 m high light house from the sea-level, the angles of
depression of two ships are 30° and 45°
...
[Use 3 = 1
...


Topper’s Answers

 Let AB be the light house and two ships
be at C and D
...
Find the
width of the river
...
732]
[CBSE OD, Term 2, Set 1, 2017]

Ans
...

A
60°

45°
300 m

C

BC
= cot 45°
AB
x
=1
100
x = 100



⇒



⇒



Similarly, in ∆ABD,




⇒



⇒

x

B

y

45°

D

In right ΔABC,

In ∆ABC,


60°




...
(ii)

Distance between two ships = y − x
= 100 3 − 100
[from equation (i) and (ii)]
= 100 ( 3 − 1)
= 100 (1
...
732)

= 73
...
 11
...
Flying at this
height, the angles of depression from
the aeroplane of two points on both
banks of a river in opposite directions



BC
= cot 60°
AB

1
x
300
3
⇒
=
⇒x=
×
3
3
3
300



= 100 3 m



= 100 × 1
...
2 m

In right ΔABD,



⇒



BD
= cot 45°
AB

y
= 1 ⇒ y = 300 m
300
Width of river = x + y



= 173
...
2 m

Q
...
A man observes a car from the top of
a tower, which is moving towards the
tower with a uniform speed
...


[CBSE OD, Term 2, Set 2, 2017]
Ans
...


A
30°

45°
100 m

C

In Δ ABC,

In Δ ABC,





AB
= tan 45°
BC
h
= 1 ⇒ h = x
...
(ii)
x+y
3
3

⇒



From eq
...
732



= 16
...
39 minutes
...
  13
...
Find the distance between
the cars
...
732]

[CBSE OD, Term 2, Set 3, 2017]
Ans
...
732
= 173
...
2 + 100

= 273
...
 14
...
Find the height of the cloud
from the surface of water
...


In Δ CMP,
tan 30° =





⇒



CM
PM

1
h
=
or PM =
PM
3

3 h 
...
(ii)
3
From equation (i) and (ii)
3 =



3h=



h + 120
3



⇒



⇒ 2h = 120



⇒



Height of cloud from surface of water

h

h = 60 m
= h + 60



= 60 + 60



= 120 m
...
  15
...
The angles of depression
of these points from the top of the
tower are 60° and 45° respectively
...


[CBSE Delhi, Term 2, Set 2, 2017]
Ans
...
Let AB be tower of height h
...
  16
...
If the
observer moves 20 m towards the base
of the tower, the angle of elevation of
the top increases by 15°, find the height
of the tower
...
  17
...
From a point Y, 40 m
vertically above X, the angle of elevation
of the top Q of tower is 45°
...
(Use 3 = 1
...
We have, PQ as a vertical tower

Then, in Δ CBD,
In ΔYZQ,
tan 45° =



QZ
YZ





⇒

QZ
=1
YZ





⇒

QZ = YZ
...

(
...
73)



= 54
...
6 m

And PQ = (54
...
6 m
...
 18
...
Find the distance travelled by the
ship during the period of observation
...
73)

[CBSE OD, Term 2, Set 2, 2016]
Ans
...


tan 60° =

tan 30° =


...
(ii)

100 3
3

[from equation (i) & (ii)]



 3 −1
= 100 3 

 3 



=



200 × 1
...
33 m

Q
...
From a point on the ground, the angle of
elevation of the top of a tower is observed
to be 60°
...
Find the height of the tower and
its horizontal distance from the point of
observation
...
We have, PQ as a vertical tower
...
(i)

tan 60° =
3 =

QP
XP
QZ + 40
XP

YZ 3 = QZ + 40
(
...

Let the distance travel by bird from B to C is
x m
...
732



∴

QZ = 20



Hence, speed of flying of the bird =

∴ Height of tower = (40 + 20) m = 60 m



and Horizontal distance = QZ 3


= 20 3 m

Q
...
A bird is sitting on the top of a 80 m
high tree
...

The bird flies away horizontally in such
a way that it remained at a constant
height from the ground
...
Find the speed of
flying of the bird
...
732)

[CBSE Delhi, Term 2, Set 1, 2016]
Ans
...



...
56 m
58
...


...
28 m/s

Q
...
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
60° and 30° respectively
...


[CBSE Delhi, Term 2, Set 2, 2016]
Ans
...
(i)
4

 Let C be the position of the cloud and D
be its reflection in the lake
...
(ii)
9
Multiplying equation (i) and (ii), we get

⇒



⇒



⇒



⇒



And in D ABC,



h
36

⇒ 1 =



\



Note: In this question, it has not been
specified whether two points from tower are taken in same or opposite side we
have taken these points on the same side
of tower
...
(i)
3

BC
AB
H
1
=
3 AB
AB = 3H
...
AB = H + 40



cot 60° =

h h
tan 60°
...
  22
...
The angle of depression of the
reflection of the cloud in the lake, at A is
60°
...


[CBSE OD, Term 2, Set 1, 2015]


Again, in D ABC

(AC)2 = (AB)2 + (BC)2

Ans
...




Let PQ be the surface of the lake, A is the
point vertically above P such at AP = 20m
...
  23
...
If the length
of the flag staff is 5 m, find the height of
the tower
...
Let CB be the tower of x m and AC be the
flag staff of 5 m
...
(i) and (ii),

5m
C

P



⇒

x

60°
30°

B

x

tan 30° = PB
PB =

x
= 3 x
...
(ii)
3

tan 60° =



3x =

x+5
3



⇒



⇒ 3x − x = 5



⇒ 2x = 5



⇒
x = 5/2 = 2
...


...
5 m

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