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# Question Bank: Some Applications of trigonometry Notes | Study Mathematics (Maths) Class 10 - Class 10

## Class 10: Question Bank: Some Applications of trigonometry Notes | Study Mathematics (Maths) Class 10 - Class 10

The document Question Bank: Some Applications of trigonometry Notes | Study Mathematics (Maths) Class 10 - Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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``` Page 1

SOME APPLICATION OF TRIGONOMETRY

HEIGHT AND DISTANCES
1) A ladder 5 m long, leaning against a vertical wall make an angle of
?
60 with the ground.
a) How high on the wall does the ladder reach?
b) How far is the foot of the ladder from the wall?
c) What angle does the ladder make with the wall?
Solution:
a) The height that the ladder reach is PQ

5
60 sin
PQ
?
?

325 . 4
2
5 73 . 1
5
2
3
5 60 sin ?
?
? ? ? ? ?
?
PQ m
b) The distance of the foot of the ladder from the wall is RQ

5
60 cos
RQ
?
?

5 60 cos ? ?
?
RQ = 5 . 2 5
2
1
? ? m
c) The angle that the ladder makes with the wall is angle P
865 . 0
5
325 . 4
5
cos ? ? ? ?
PQ
P

?
30 ) 865 . 0 ( cos
1
? ? ?
?
P (approx.)

2) 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 building increases from
?
30 to
?
60 as he walks towards the building. Find the distance he
walks towards the building.
Solution:
We are given the height of building (AF) = 30m.
The height of the boy (BD) = 1.5m
Therefore, AC = AF-BD = 30 – 1.5 =28.5m
We want to find distance he walked towards the building which is BG
AGC In ? ,
We have,
?
60 cot ?
AC
GC

?     AC GC ? ?
?
60 cot
?     5 . 28
3
1
? ? GC =
3
3
3
5 . 28
?
?      m GC 3 5 . 9 ?

Page 2

SOME APPLICATION OF TRIGONOMETRY

HEIGHT AND DISTANCES
1) A ladder 5 m long, leaning against a vertical wall make an angle of
?
60 with the ground.
a) How high on the wall does the ladder reach?
b) How far is the foot of the ladder from the wall?
c) What angle does the ladder make with the wall?
Solution:
a) The height that the ladder reach is PQ

5
60 sin
PQ
?
?

325 . 4
2
5 73 . 1
5
2
3
5 60 sin ?
?
? ? ? ? ?
?
PQ m
b) The distance of the foot of the ladder from the wall is RQ

5
60 cos
RQ
?
?

5 60 cos ? ?
?
RQ = 5 . 2 5
2
1
? ? m
c) The angle that the ladder makes with the wall is angle P
865 . 0
5
325 . 4
5
cos ? ? ? ?
PQ
P

?
30 ) 865 . 0 ( cos
1
? ? ?
?
P (approx.)

2) 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 building increases from
?
30 to
?
60 as he walks towards the building. Find the distance he
walks towards the building.
Solution:
We are given the height of building (AF) = 30m.
The height of the boy (BD) = 1.5m
Therefore, AC = AF-BD = 30 – 1.5 =28.5m
We want to find distance he walked towards the building which is BG
AGC In ? ,
We have,
?
60 cot ?
AC
GC

?     AC GC ? ?
?
60 cot
?     5 . 28
3
1
? ? GC =
3
3
3
5 . 28
?
?      m GC 3 5 . 9 ?

SOME APPLICATION OF TRIGONOMETRY

ABC In ? ,
We have,
?
30 cot ?
AC
BC

?     AC BC ? ?
?
30 cot
?     5 . 28 3 ? ? BC
?      m BC 3 5 . 28 ?

BG = BC – GC
= 3 5 . 9 3 5 . 28 ?
= m 3 19
Therefore, he walks m 3 19 towards the tower.

3) The angle of elevation of the top of a tower from two points distant p and q from the base and in the same
straight line with it are complementary. Prove that the height of tower is pq
Solution:
Let CD be the tower of height h.
Let AC be the base of the tower.
Let A and B be two points on the same line through the base such that CA = p, and CB = q.
Because the angles at A and B are complementary,
So let ? ? ?DAC
And therefore, ? ? ? ?
?
90 DBC
ACD In ? ,
p
h
AC
CD
? ? ? tan …………….(1)
BCD In ? ,
q
h
CB
CD
? ? ? ) 90 tan( ?
?
or
q
h
? ? cot ………………(2)
Multiplying (1) and (2) we have,
p
h
q
h
? ? ? ? cot tan
?
pq
h
2
tan
1
. tan ?
?
?
?
pq
h
2
1 ?

Page 3

SOME APPLICATION OF TRIGONOMETRY

HEIGHT AND DISTANCES
1) A ladder 5 m long, leaning against a vertical wall make an angle of
?
60 with the ground.
a) How high on the wall does the ladder reach?
b) How far is the foot of the ladder from the wall?
c) What angle does the ladder make with the wall?
Solution:
a) The height that the ladder reach is PQ

5
60 sin
PQ
?
?

325 . 4
2
5 73 . 1
5
2
3
5 60 sin ?
?
? ? ? ? ?
?
PQ m
b) The distance of the foot of the ladder from the wall is RQ

5
60 cos
RQ
?
?

5 60 cos ? ?
?
RQ = 5 . 2 5
2
1
? ? m
c) The angle that the ladder makes with the wall is angle P
865 . 0
5
325 . 4
5
cos ? ? ? ?
PQ
P

?
30 ) 865 . 0 ( cos
1
? ? ?
?
P (approx.)

2) 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 building increases from
?
30 to
?
60 as he walks towards the building. Find the distance he
walks towards the building.
Solution:
We are given the height of building (AF) = 30m.
The height of the boy (BD) = 1.5m
Therefore, AC = AF-BD = 30 – 1.5 =28.5m
We want to find distance he walked towards the building which is BG
AGC In ? ,
We have,
?
60 cot ?
AC
GC

?     AC GC ? ?
?
60 cot
?     5 . 28
3
1
? ? GC =
3
3
3
5 . 28
?
?      m GC 3 5 . 9 ?

SOME APPLICATION OF TRIGONOMETRY

ABC In ? ,
We have,
?
30 cot ?
AC
BC

?     AC BC ? ?
?
30 cot
?     5 . 28 3 ? ? BC
?      m BC 3 5 . 28 ?

BG = BC – GC
= 3 5 . 9 3 5 . 28 ?
= m 3 19
Therefore, he walks m 3 19 towards the tower.

3) The angle of elevation of the top of a tower from two points distant p and q from the base and in the same
straight line with it are complementary. Prove that the height of tower is pq
Solution:
Let CD be the tower of height h.
Let AC be the base of the tower.
Let A and B be two points on the same line through the base such that CA = p, and CB = q.
Because the angles at A and B are complementary,
So let ? ? ?DAC
And therefore, ? ? ? ?
?
90 DBC
ACD In ? ,
p
h
AC
CD
? ? ? tan …………….(1)
BCD In ? ,
q
h
CB
CD
? ? ? ) 90 tan( ?
?
or
q
h
? ? cot ………………(2)
Multiplying (1) and (2) we have,
p
h
q
h
? ? ? ? cot tan
?
pq
h
2
tan
1
. tan ?
?
?
?
pq
h
2
1 ?

SOME APPLICATION OF TRIGONOMETRY

pq h ? ?
2

pq h ? ?

4)  A Jet airway is flying horizontally at a height of 5000 m. above the ground is observed at an elevation of
?
60 ,
And after 30 seconds, the elevation is observed to be
?
30 . Find the speed of the jet airway in km/hr.
Solution:
Initially the jet airway is at X and after 30 seconds its position is at Z.
Therefore, XZ is the distance covered in 30 seconds.
Also
?
60 ? ?XPY ,

?
30 ? ?ZPQ
XY = ZQ = 5000m
In right XYP ?
?
60 cot ?
XY
PY
?
3
5000
? PY ………..( 1)
In right ZQP ?
?
30 cot ?
ZQ
PQ
? 3 5000 ? PQ …………….( 2)
From (1) and (2) we get,
YQ = PQ – PY =
3
5000
3 5000 ? =
?
?
?
?
?
?
?
?
?
3
1
3 5000
YQ =
?
?
?
?
?
?
?
?
3
2
5000 = m
3
17320
3
732 . 1 2 5000
3
3 2 5000
?
? ?
?
? ?

Therefore, distance covered in 30 sec is m
3
17320

Therefore, Speed = s m /
30 3
17320
?
= hr km/ 60 60
1000 30 3
17320
? ?
? ?

Therefore Speed = hr km/ 8 . 692
Therefore speed of  jet airway is 692.8km/hr.

Page 4

SOME APPLICATION OF TRIGONOMETRY

HEIGHT AND DISTANCES
1) A ladder 5 m long, leaning against a vertical wall make an angle of
?
60 with the ground.
a) How high on the wall does the ladder reach?
b) How far is the foot of the ladder from the wall?
c) What angle does the ladder make with the wall?
Solution:
a) The height that the ladder reach is PQ

5
60 sin
PQ
?
?

325 . 4
2
5 73 . 1
5
2
3
5 60 sin ?
?
? ? ? ? ?
?
PQ m
b) The distance of the foot of the ladder from the wall is RQ

5
60 cos
RQ
?
?

5 60 cos ? ?
?
RQ = 5 . 2 5
2
1
? ? m
c) The angle that the ladder makes with the wall is angle P
865 . 0
5
325 . 4
5
cos ? ? ? ?
PQ
P

?
30 ) 865 . 0 ( cos
1
? ? ?
?
P (approx.)

2) 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 building increases from
?
30 to
?
60 as he walks towards the building. Find the distance he
walks towards the building.
Solution:
We are given the height of building (AF) = 30m.
The height of the boy (BD) = 1.5m
Therefore, AC = AF-BD = 30 – 1.5 =28.5m
We want to find distance he walked towards the building which is BG
AGC In ? ,
We have,
?
60 cot ?
AC
GC

?     AC GC ? ?
?
60 cot
?     5 . 28
3
1
? ? GC =
3
3
3
5 . 28
?
?      m GC 3 5 . 9 ?

SOME APPLICATION OF TRIGONOMETRY

ABC In ? ,
We have,
?
30 cot ?
AC
BC

?     AC BC ? ?
?
30 cot
?     5 . 28 3 ? ? BC
?      m BC 3 5 . 28 ?

BG = BC – GC
= 3 5 . 9 3 5 . 28 ?
= m 3 19
Therefore, he walks m 3 19 towards the tower.

3) The angle of elevation of the top of a tower from two points distant p and q from the base and in the same
straight line with it are complementary. Prove that the height of tower is pq
Solution:
Let CD be the tower of height h.
Let AC be the base of the tower.
Let A and B be two points on the same line through the base such that CA = p, and CB = q.
Because the angles at A and B are complementary,
So let ? ? ?DAC
And therefore, ? ? ? ?
?
90 DBC
ACD In ? ,
p
h
AC
CD
? ? ? tan …………….(1)
BCD In ? ,
q
h
CB
CD
? ? ? ) 90 tan( ?
?
or
q
h
? ? cot ………………(2)
Multiplying (1) and (2) we have,
p
h
q
h
? ? ? ? cot tan
?
pq
h
2
tan
1
. tan ?
?
?
?
pq
h
2
1 ?

SOME APPLICATION OF TRIGONOMETRY

pq h ? ?
2

pq h ? ?

4)  A Jet airway is flying horizontally at a height of 5000 m. above the ground is observed at an elevation of
?
60 ,
And after 30 seconds, the elevation is observed to be
?
30 . Find the speed of the jet airway in km/hr.
Solution:
Initially the jet airway is at X and after 30 seconds its position is at Z.
Therefore, XZ is the distance covered in 30 seconds.
Also
?
60 ? ?XPY ,

?
30 ? ?ZPQ
XY = ZQ = 5000m
In right XYP ?
?
60 cot ?
XY
PY
?
3
5000
? PY ………..( 1)
In right ZQP ?
?
30 cot ?
ZQ
PQ
? 3 5000 ? PQ …………….( 2)
From (1) and (2) we get,
YQ = PQ – PY =
3
5000
3 5000 ? =
?
?
?
?
?
?
?
?
?
3
1
3 5000
YQ =
?
?
?
?
?
?
?
?
3
2
5000 = m
3
17320
3
732 . 1 2 5000
3
3 2 5000
?
? ?
?
? ?

Therefore, distance covered in 30 sec is m
3
17320

Therefore, Speed = s m /
30 3
17320
?
= hr km/ 60 60
1000 30 3
17320
? ?
? ?

Therefore Speed = hr km/ 8 . 692
Therefore speed of  jet airway is 692.8km/hr.

SOME APPLICATION OF TRIGONOMETRY

5) If the distance of a person from a tower is 100m and the angle subtends by the top of tower with the ground is
?
30
What is the height of the tower in metres?

Solution:
AB = distance of the man from the tower = 100m
BC = height of the tower= h
The trigonometric function that uses AB and BC is tan A, where A =
?
30
So,
100
30 tan
h
AB
BC
? ?
?

? m h 74 . 57
732 . 1
100
100
3
1
? ? ? ?
Therefore, height of the tower is 57.74m

6) Two pillars of equal height are on either side of a road, which is 60 m wide. The angle of elevation of the top
of the pillars are
?
60 and
?
30 at a appoint on the road between the pillars. Find the position of the point between
the pillars. Also find the height of each pillar.
Solution:
Let PQ and SR be two pillars of equal height ‘h’ m
Let ‘T’ be a point on the road such that
QT = ‘x’meter.
TR = (60 –x) metre.
?
30 ? ?STR
In triangle PQT,

Page 5

SOME APPLICATION OF TRIGONOMETRY

HEIGHT AND DISTANCES
1) A ladder 5 m long, leaning against a vertical wall make an angle of
?
60 with the ground.
a) How high on the wall does the ladder reach?
b) How far is the foot of the ladder from the wall?
c) What angle does the ladder make with the wall?
Solution:
a) The height that the ladder reach is PQ

5
60 sin
PQ
?
?

325 . 4
2
5 73 . 1
5
2
3
5 60 sin ?
?
? ? ? ? ?
?
PQ m
b) The distance of the foot of the ladder from the wall is RQ

5
60 cos
RQ
?
?

5 60 cos ? ?
?
RQ = 5 . 2 5
2
1
? ? m
c) The angle that the ladder makes with the wall is angle P
865 . 0
5
325 . 4
5
cos ? ? ? ?
PQ
P

?
30 ) 865 . 0 ( cos
1
? ? ?
?
P (approx.)

2) 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 building increases from
?
30 to
?
60 as he walks towards the building. Find the distance he
walks towards the building.
Solution:
We are given the height of building (AF) = 30m.
The height of the boy (BD) = 1.5m
Therefore, AC = AF-BD = 30 – 1.5 =28.5m
We want to find distance he walked towards the building which is BG
AGC In ? ,
We have,
?
60 cot ?
AC
GC

?     AC GC ? ?
?
60 cot
?     5 . 28
3
1
? ? GC =
3
3
3
5 . 28
?
?      m GC 3 5 . 9 ?

SOME APPLICATION OF TRIGONOMETRY

ABC In ? ,
We have,
?
30 cot ?
AC
BC

?     AC BC ? ?
?
30 cot
?     5 . 28 3 ? ? BC
?      m BC 3 5 . 28 ?

BG = BC – GC
= 3 5 . 9 3 5 . 28 ?
= m 3 19
Therefore, he walks m 3 19 towards the tower.

3) The angle of elevation of the top of a tower from two points distant p and q from the base and in the same
straight line with it are complementary. Prove that the height of tower is pq
Solution:
Let CD be the tower of height h.
Let AC be the base of the tower.
Let A and B be two points on the same line through the base such that CA = p, and CB = q.
Because the angles at A and B are complementary,
So let ? ? ?DAC
And therefore, ? ? ? ?
?
90 DBC
ACD In ? ,
p
h
AC
CD
? ? ? tan …………….(1)
BCD In ? ,
q
h
CB
CD
? ? ? ) 90 tan( ?
?
or
q
h
? ? cot ………………(2)
Multiplying (1) and (2) we have,
p
h
q
h
? ? ? ? cot tan
?
pq
h
2
tan
1
. tan ?
?
?
?
pq
h
2
1 ?

SOME APPLICATION OF TRIGONOMETRY

pq h ? ?
2

pq h ? ?

4)  A Jet airway is flying horizontally at a height of 5000 m. above the ground is observed at an elevation of
?
60 ,
And after 30 seconds, the elevation is observed to be
?
30 . Find the speed of the jet airway in km/hr.
Solution:
Initially the jet airway is at X and after 30 seconds its position is at Z.
Therefore, XZ is the distance covered in 30 seconds.
Also
?
60 ? ?XPY ,

?
30 ? ?ZPQ
XY = ZQ = 5000m
In right XYP ?
?
60 cot ?
XY
PY
?
3
5000
? PY ………..( 1)
In right ZQP ?
?
30 cot ?
ZQ
PQ
? 3 5000 ? PQ …………….( 2)
From (1) and (2) we get,
YQ = PQ – PY =
3
5000
3 5000 ? =
?
?
?
?
?
?
?
?
?
3
1
3 5000
YQ =
?
?
?
?
?
?
?
?
3
2
5000 = m
3
17320
3
732 . 1 2 5000
3
3 2 5000
?
? ?
?
? ?

Therefore, distance covered in 30 sec is m
3
17320

Therefore, Speed = s m /
30 3
17320
?
= hr km/ 60 60
1000 30 3
17320
? ?
? ?

Therefore Speed = hr km/ 8 . 692
Therefore speed of  jet airway is 692.8km/hr.

SOME APPLICATION OF TRIGONOMETRY

5) If the distance of a person from a tower is 100m and the angle subtends by the top of tower with the ground is
?
30
What is the height of the tower in metres?

Solution:
AB = distance of the man from the tower = 100m
BC = height of the tower= h
The trigonometric function that uses AB and BC is tan A, where A =
?
30
So,
100
30 tan
h
AB
BC
? ?
?

? m h 74 . 57
732 . 1
100
100
3
1
? ? ? ?
Therefore, height of the tower is 57.74m

6) Two pillars of equal height are on either side of a road, which is 60 m wide. The angle of elevation of the top
of the pillars are
?
60 and
?
30 at a appoint on the road between the pillars. Find the position of the point between
the pillars. Also find the height of each pillar.
Solution:
Let PQ and SR be two pillars of equal height ‘h’ m
Let ‘T’ be a point on the road such that
QT = ‘x’meter.
TR = (60 –x) metre.
?
30 ? ?STR
In triangle PQT,

SOME APPLICATION OF TRIGONOMETRY

?
60 tan ?
QT
PQ

? 3 ?
x
h

? x h 3 ? …………….(1)

In right triangle STR
?
30 tan ?
RT
SR

?
3
1
60
?
? x
h

?
3
60 x
h
?
? …………(2)
Plug in the value of h from equation (1), we get
15
60 4
60 3
3
60
3
?
?
? ?
?
?
x
x
x x
x
x

Plug in the value of x in equation (1)
m h 95 . 25 15 73 . 1 ? ? ?
Therefore, height of each pillar = 25.95 m ; position of the point from a pillar making an angle of
?
60 is 15m.

```
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## Mathematics (Maths) Class 10

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