Processing math: 100%
MCQExams
0:0:2
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 2 - MCQExams.com
CBSE
Class 10 Maths
Some Applications Of Trigonometry
Quiz 2
At a point 15 metres away from the base of a 15 metres high house, the angle of elevation of the top is
Report Question
0%
45
0
0%
30
0
0%
60
0
0%
90
0
Explanation
tan
θ
=
15
15
=
1
θ
=
45
0
The angle of elevation of a cloud from a point h m above the level of water in a lake is
α
and the angle of depression of its reflection in the lake is
β
.Then the height of the cloud above the water level is
h
s
i
n
(
β
−
α
)
s
i
n
(
β
+
α
)
Report Question
0%
True
0%
False
Explanation
From Figure
Point
A
is the cloud
Point
D
is view point
F
D
is parallel to
B
E
and
C
D
transversal
So
∠
F
D
C
=
∠
D
C
E
From Reflection property
∠
B
C
A
=
∠
D
C
E
=
β
From
Δ
A
B
C
sin
β
=
A
B
A
C
⋯
(
1
)
From
Δ
C
D
E
sin
β
=
D
E
D
C
⋯
(
2
)
From
(
1
)
&
(
2
)
⟹
A
B
A
C
=
D
E
D
C
⟹
A
B
=
D
E
D
C
×
A
C
⋯
(
3
)
From
Δ
A
C
D
By Exterior angle is sum of two interior opposite angles
2
β
=
α
+
β
+
∠
D
A
C
∠
D
A
C
=
β
−
α
From
sin
rule
⟹
A
C
sin
(
α
+
β
)
=
D
C
sin
(
β
−
α
)
⟹
A
C
D
C
=
sin
(
β
−
α
)
sin
(
α
+
β
)
⋯
(
4
)
Put it in
(
3
)
⟹
A
C
=
D
E
sin
(
β
−
α
)
sin
(
α
+
β
)
From Question
D
E
=
h
⟹
A
C
=
h
sin
(
β
−
α
)
sin
(
α
+
β
)
The angles of elevation of the top of a tower from two points on the ground at distances a metres and b metres from the base of the tower and in the same straight line are complementary. The height of the tower is
√
a
b
metres.
Report Question
0%
True
0%
False
Explanation
Let AB be the tower and let C and D be the two points of the observer. Then,
AC = a metres and AD = b metres
Let
∠
A
C
B
=
θ
. Then,
∠
A
D
B
=
(
90
o
−
θ
)
.
Let AB = h metres
In right
△
C
A
B
, we have
tan
θ
=
A
B
A
D
tan
θ
=
h
a
h
=
a
tan
θ
......(1)
In right
△
D
A
B
, we have,
tan
(
90
o
−
θ
)
=
A
B
A
D
cot
θ
=
h
b
h
=
b
cot
θ
.......(2)
From 1 and 2, we get,
h
2
=
a
b
h
=
√
a
b
m
e
t
r
e
s
A
B
is vertical pole with
B
at the ground level and
A
at the top. A man find that the angle of elevation of the point
A
from a certain point
C
on the ground is
60
o
. He moves away from the pole along the line
B
C
to a point
D
such that
C
D
=
7
m
. From
D
the angle of elevation of the point
A
is
45
o
. Then the height of the pole is:
Report Question
0%
7
√
3
2
1
√
3
−
1
m
0%
7
√
3
2
(
√
3
+
1
)
m
0%
7
√
3
2
(
√
3
−
1
)
m
0%
7
√
3
2
1
√
3
+
1
m
Explanation
Then in triangle
A
B
C
t
a
n
60
∘
=
A
B
B
C
x
=
h
√
3
----- (i)
in
Δ
A
B
D
t
a
n
45
∘
=
A
B
B
D
1
=
h
x
+
7
h
=
x
+
7
h
=
h
√
3
+
7
from (i)
h
=
7
√
3
√
3
−
1
h
=
7
√
3
2
(
√
3
+
1
)
Therefore the height of the pole is
7
√
3
2
(
√
3
+
1
)
m
A bridge above the river makes an angle of
45
o
with the bank of river. If length of bridge above the river is
150
m
then breadth of river will be
Report Question
0%
75
m
0%
50
√
2
m
0%
150
m
0%
75
√
2
m
Explanation
Let
A
C
be the bridge and
A
B
be the width of river.
In
Δ
A
B
C
A
B
A
C
=
sin
45
∘
A
B
150
=
1
√
2
A
B
=
150
√
2
2
=
75
√
2
therefore width of the river is
75
√
2
m
.
A flagstaff stands on the middle of a square tower. A man on the ground, opposite to the middle of one face and distant from it
100
m, just see the flag ; on his receding another
100
m, the tangents of the elevation of the top of the tower and the top of the flagstaff are found to be
1
2
and
5
9
. Find the height of the flagstaff, the ground being horizontal
Report Question
0%
20
0%
25
0%
30
0%
35
Explanation
h
2
b
=
h
1
100
h
1
200
=
1
2
⇒
h
1
=
100
h
1
+
h
2
200
+
b
=
5
9
100
+
b
200
+
b
=
5
9
b
=
100
4
⇒
b
=
h
2
=
25
As you ride the Ferris wheel, your distance from the ground varies sinusoidally with time. An equation to model the motion is y=20cos(
π
4
(
t
−
3
)
)
+
23
. Predict your height above the ground at a time of 1 seconds.
Report Question
0%
20.86
f
t
0%
23
f
t
0%
8.14
f
t
0%
18.96
Explanation
y
=
20
c
o
s
(
π
4
(
t
−
3
)
)
+
23
At
t
=
1
second
Height above the ground,
y
=
20
c
o
s
(
π
4
(
1
−
3
)
)
+
23
y
=
20
c
o
s
(
−
2
π
4
)
+
23
⟹
y
=
23
f
t
. [ Since ,
c
o
s
(
−
π
4
)
=
0
]
The shadow of a tower, when the angle of elevation of the sun is
45
o
, is found to be 10 metres longer than when the angle of elevation is
60
o
. Find the height of the tower.
Report Question
0%
15
+
5
√
3
m
0%
12
+
5
√
3
m
0%
15
+
√
3
m
0%
12
+
√
3
m
Explanation
Let AB be the tower and let AC and AD be its shadows when the angles of elevation of the sum are
60
o
and
45
o
, respectively. Then,
∠
A
C
B
=
60
o
,
∠
A
D
B
=
45
o
,
C
D
=
10
m
Let
A
B
=
h
m
e
t
r
e
s
and
A
C
=
x
m
e
t
r
e
s
In right
△
CAB, we have,
tan
60
o
=
A
B
A
C
√
3
=
h
x
x
=
h
√
3
......(1)
In right
△
DAB, we have,
tan
45
o
=
A
B
A
D
1
=
h
10
+
x
x
=
h
−
10
.....(2)
From 1 and 2, we get,
h
√
3
=
h
−
10
h
=
√
3
h
−
10
√
3
√
3
h
−
h
=
10
√
3
h
(
√
3
−
1
)
=
10
√
3
h
=
5
√
3
(
√
3
+
1
)
h
=
15
+
5
√
3
m
The shadow of a
6
m
high tower is
15
m
and at the same point of time length of shadow of a tree is
25
m
. What is the height of the tree?
Report Question
0%
21
m
0%
10
m
0%
35
m
0%
n
o
n
e
o
f
t
h
e
s
e
Explanation
H
i
g
h
t
o
f
t
h
e
t
o
w
e
r
L
e
n
g
t
h
o
f
s
h
a
d
o
w
o
f
t
h
e
t
o
w
e
r
=
H
i
g
h
t
o
f
t
h
e
t
r
e
e
L
e
n
g
t
h
o
f
s
h
a
d
o
w
o
f
t
h
e
t
r
e
e
⟹
6
15
=
H
25
⟹
H
=
25
×
6
15
∴
H
=
10
m
OPtion B is correct.
If a ladder
13
m
is placed against a wait such that its roots at a distance from the wall, then the height of the top of the ladder from the ground :
Report Question
0%
10
m
0%
11
m
0%
12
m
0%
None of these
Explanation
If a ladder is
13
m long is placed at a distance of
12
m from a wall, then the weight of the wall is:
A
.
T
.
Q
,
h
=
√
(
13
)
2
−
(
12
)
2
=
√
169
−
144
=
√
25
=
5
m
A tower stands vertically on the ground. From a point on the ground which is
30
m
away from the foot of a tower, the angle of elevation of the top of the tower is found to be
45
o
. Find the height of tower.
Report Question
0%
15
0%
40
0%
30
0%
20
Explanation
REF.Image.
30
m
away /
45
∘
inclination.
Let the height of tower be h m
so
A
B
=
h
m
Distance of the point from
foot of tower =
30
m
C
B
=
30
m
Angle of elevation
=
45
∘
∠
A
C
B
=
45
∘
Since tower is vertical to ground
∠
A
B
C
=
90
∘
Now
t
a
n
C
=
s
i
d
e
o
p
p
t
o
a
n
g
l
e
C
s
i
d
e
a
d
j
a
c
e
n
t
t
o
a
n
g
l
e
C
tan
C
=
A
B
C
B
tan
45
=
A
B
C
B
1
=
H
30
H
=
30
Hence height of tower =
30
m
Answer = option
C
=
30
m
A man observes the angle of elevation of a balloon to be
30
0
at a point
A
. He then walks towards the balloon and at a certain place
B
, he finds the angle of elevation to be
60
0
. He further walks in the direction of the balloon and finds it to be directly over him at a height of
1
2
k
m
, then the distance
A
B
is:
Report Question
0%
1
√
2
k
m
0%
1
√
3
k
m
0%
1
√
4
k
m
0%
1
√
5
k
m
Explanation
h
=
500
m
tan
60
0
=
h
b
⇒
b
=
500
√
3
tan
30
0
=
h
d
+
b
⇒
d
+
500
√
3
=
500
√
3
⇒
d
=
1000
√
3
m
⇒
d
=
1
√
3
k
m
The horizontal distance between two towers is
60
m
and angular depression of the top of the first as seen from the second, which is
150
m
in height, is
30
0
. The height of the first tower is
Report Question
0%
(
150
+
20
√
3
)
m
0%
(
150
+
15
√
3
)
m
0%
(
150
−
20
√
5
)
m
0%
(
150
−
20
√
3
)
m
Explanation
Let AB and CD be the two towers
Height of the second tower, AB =150m and
Distance between two towers BD =60m
Let's assume the height of the first tower is
h
m
.
We know, In right angled
△
,
tan
(
Θ
)
=
O
p
p
o
s
i
t
e
A
d
j
a
c
e
n
t
tan
30
0
=
A
E
E
C
tan
30
0
=
150
−
h
60
.....[AE=AB-ED] and [EC=BD=60]
⇒
1
√
3
=
150
−
h
60
⇒
h
√
3
=
150
√
3
−
60
⇒
h
=
150
−
20
√
3
The height of the first tower is
150
−
20
√
3
m.
A kite is flying with the string inclined at
30
o
to the horizon. The height of the kite above the ground, when the string is
15
m
long is
Report Question
0%
15
m
0%
30
m
0%
15
2
m
0%
15
3
m
Explanation
sin
30
0
=
A
B
A
C
=
h
15
⇒
1
2
=
h
15
⇒
h
=
15
2
m
From the top of the tree, a man observes the angle of depression of a point which is at a distance of
40
m
from the foot is
75
0
. The height of the tree is
:
Report Question
0%
40
√
3
m
0%
40
(
2
+
√
3
)
m
0%
21
√
3
m
0%
3
√
21
m
Explanation
Let say AC be the tree and the Observer is at A who is observing point B.
Given: BC = 40m
∠
B
=
75
o
= angle of depression {alternate angles}
In
△
A
B
C
,
tan
75
0
=
h
40
⇒
tan
(
45
o
+
30
o
)
=
h
40
⇒
tan
45
o
+
tan
30
o
1
−
tan
45
o
tan
30
o
=
h
40
∵
tan
A
+
tan
B
1
−
tan
A
tan
B
=
t
a
n
(
A
+
B
)
⇒
1
+
1
√
3
1
−
1
×
1
√
3
=
h
40
⇒
(
√
3
+
1
)
(
√
3
−
1
)
=
h
40
⇒
(
3
+
1
+
2
√
3
2
)
=
h
40
{rationalizing by multiplying and dividing by
√
3
+
1
}
⇒
(
2
+
√
3
)
=
h
40
⇒
h
=
40
(
2
+
√
3
)
m
From the top of a hill
h
meters high, the angle of depression of the top and the bottom of a pillar are
α
,
β
respectively. Then the height(in meters) of the pillar is
Report Question
0%
h
(
tan
β
−
tan
α
)
tan
β
0%
h
(
tan
α
−
tan
β
)
tan
α
0%
h
(
tan
β
+
tan
α
)
tan
β
0%
h
(
tan
β
+
tan
α
)
tan
α
Explanation
tan
α
=
h
−
d
a
tan
β
=
h
a
tan
α
tan
β
=
h
−
d
h
⇒
tan
α
tan
β
=
1
−
d
h
⇒
d
h
=
tan
β
−
tan
α
tan
β
⇒
d
=
h
(
tan
β
−
tan
α
tan
β
)
From the top of a building
h
metres, the angle of depression of an object on the ground is
α
, the distance of the object from the foot of the building is
Report Question
0%
h
cot
α
0%
h
tan
α
0%
h
cos
α
0%
h
sin
α
Explanation
tan
α
=
h
d
⇒
d
=
h
tan
α
⇒
d
=
h
cot
α
The angle of elevation of an object from a point
P
on the level ground is
α
. Moving
d
meters on the ground towards the object, the angle of elevation is found to be
β
, then the height (in meters) of the object is
Report Question
0%
d
tan
α
0%
d
cot
β
0%
d
cot
α
+
cot
β
0%
d
cot
α
−
cot
β
Explanation
Let the height of the object
A
B
be
h
metres.
Given,
∠
A
C
B
=
α
,
∠
A
D
B
=
β
,
C
D
=
d
metres
Let
D
B
=
x
metres.
Then, in right-angled
△
A
B
D
,
cot
β
=
x
h
⇒
x
=
h
cot
β
.
.
.
(
i
)
In right-angled
△
A
C
B
,
cot
α
=
x
+
d
h
⇒
x
+
d
=
h
cot
α
.
.
.
(
i
i
)
Let us subtract the equations:
(
i
i
)
–
(
i
)
⇒
d
=
h
cot
α
–
h
cot
β
⇒
d
=
h
(
cot
α
−
cot
β
)
⇒
h
=
d
cot
α
−
cot
β
The flag staff of height
10
metres is placed on the top of a tower of height
30
metres. At the top of a tower of height
40
metres, the flag staff and the tower subtend equal angles then the distance between the two towers (in metres) is
Report Question
0%
40
√
2
0%
10
√
2
0%
20
√
2
0%
30
√
2
Explanation
tan
2
θ
=
40
d
...(i)
⇒
2
tan
θ
1
−
tan
2
θ
=
40
d
tan
θ
=
10
d
...(ii)
Substituting (ii) in (i), we get
2
10
d
1
−
(
10
)
2
d
2
=
40
d
⇒
d
d
2
−
10
2
=
2
d
⇒
d
2
=
2
d
2
−
2
×
10
2
⇒
d
2
=
2
×
10
2
⇒
d
=
10
√
2
m
Straight pole(AB) subtends a right angle at a point
D
of another pole at a distance of
30
meters from
A
, the top of
A
being
60
0
above the horizontal line joining the point
B
to the pole
A
. The length of the pole
A
is, in meters
Report Question
0%
20
√
3
0%
40
√
3
0%
60
√
3
0%
40
√
3
Explanation
tan
60
=
h
2
30
h
2
=
30
√
3
tan
30
0
=
h
1
30
h
1
=
10
√
3
h
1
+
h
2
= length of the pole =
40
√
3
The upper part of a tree broken over by the wind makes an angle of
60
0
with the ground and touches the ground at a distance of 50 metres from the foot. The height of the tree in metres is
Report Question
0%
124.2
0%
186.6
0%
243.2
0%
164.2
Explanation
We will consider that the tree is broken at A.
So the total heigh of the tree will be
A
C
+
A
B
Let
A
C
=
L
1
and
A
B
=
L
2
⇒
Height of the tree
=
L
1
+
L
2
In the
△
A
B
C
,
∠
C
=
90
∘
,
sin
60
∘
=
A
C
A
B
L
2
sin
60
0
=
L
1
.....(1)
tan
60
0
=
L
1
50
⇒
L
1
=
50
√
3
m
Substituting the value of
L
1
in (1)
L
2
×
√
3
2
=
50
√
3
⇒
L
2
=
100
m
⇒
h
=
L
1
+
L
2
⇒
h
=
100
+
50
√
3
=
186.6
m
∴
Height of the tree =
=
186.6
m
The angle of elevation of the top of a flagstaff when observed from a point at a distance
60
meters from its foot is
30
0
. The height of the flagstaff (in meters) is:
Report Question
0%
20
√
3
0%
10
√
3
0%
60
√
3
0%
30
√
3
Explanation
Let
A
B
=
h
be the height of the flagstaff and
C
be the point at a distance of
60
m
from the foot.
In
△
A
B
C
,
tan
30
0
=
A
B
B
C
=
h
60
⇒
1
√
3
=
h
60
⇒
h
=
60
√
3
⇒
h
=
20
√
3
m
A tower subtends an angle
α
at a point
A
on the same level as the foot of the tower
B
is a point vertically above
A
and
A
B
=
h
metres. The angle of depression of the foot of the tower from
B
is
β
. The height of the tower is
Report Question
0%
h
tan
α
cot
β
0%
h
tan
α
tan
β
0%
h
cot
α
cot
β
0%
h
cot
α
tan
β
Explanation
tan
α
=
h
1
d
tan
β
=
h
d
tan
α
tan
β
=
h
1
h
h
1
=
h
tan
α
cot
β
In a prison wall there is a window of
1
metre height,
24
metres from the ground. An observer at a height of
10
m
from ground, standing at a distance from the wall finds the angle of elevation of the top of the window and the top of the wall to be
45
0
and
60
0
respectively. The height of the wall above the window is
Report Question
0%
15
√
3
0%
15
(
1
−
1
√
3
)
0%
15
(
√
3
−
1
)
0%
14
(
√
3
−
1
)
Explanation
We can see that ,
C
E
=
D
F
=
10
Therefore,
B
C
=
B
E
−
C
E
=
24
−
10
=
14
in
△
B
C
D
tan
45
o
=
14
b
⇒
b
=
14
Now, In
△
A
C
D
tan
60
o
=
a
+
14
14
⇒
√
3
=
a
+
14
14
⇒
a
=
14
(
√
3
−
1
)
Therefore, Answer is
14
(
√
3
−
1
)
The horizontal distance between two towers is
30
meters. From the foot of the first tower the angle of elevation of the top of the second tower is
60
o
. From the top of the second tower the angle of depression of the top of the first is
30
o
. The height of the small tower is:
Report Question
0%
20
(
√
3
+
1
)
mts
0%
20
(
√
3
−
1
)
mts
0%
20
√
3
mts
0%
20
mts
Explanation
tan
60
o
=
h
2
30
...{1}
tan
30
o
=
h
2
−
h
1
30
...{2}
tan
30
o
=
h
2
30
−
h
1
30
From {1} and {2}
⇒
h
1
30
=
tan
60
o
−
tan
30
o
⇒
h
1
30
=
3
−
1
√
3
⇒
h
1
=
30
×
2
√
3
⇒
h
1
=
20
√
3
m
If from the top of a tower of
60
metre high, the angles of depression of the top and floor of a house are
α
and
β
respetivley and if the height of the house is
60
sin
(
β
−
α
)
x
, then
x
=
Report Question
0%
sin
α
sin
β
0%
cos
α
cos
β
0%
sin
α
cos
β
0%
cos
α
sin
β
Explanation
tan
α
=
60
−
h
d
tan
β
=
60
d
tan
α
tan
β
=
60
−
h
60
tan
α
tan
β
=
1
−
h
60
h
60
=
tan
β
−
tan
α
tan
β
h
60
=
sin
β
cos
α
−
sin
α
cos
β
cos
α
sin
β
h
60
=
sin
(
β
−
α
)
sin
β
cos
α
h
=
60
sin
(
β
−
α
)
cos
α
sin
β
∴
x
=
cos
α
sin
β
On the level ground the angle of elevation of the top of a tower is
30
0
On moving 20 metres nearer tower, the angle of elevation is found to be
60
0
The height of the towerin metres is
Report Question
0%
10
√
3
0%
8
√
3
0%
6
√
3
0%
5
√
3
Explanation
tan
60
=
h
d
d
=
h
√
3
tan
30
=
h
20
+
d
1
√
3
=
h
20
+
h
√
3
20
√
3
+
h
=
3
h
h
=
10
√
3
lf the shadow of a tower is
√
3
times of its height, the altitude of the sun is
Report Question
0%
15
0
0%
30
0
0%
45
0
0%
60
0
Explanation
Let the height of tower be
h
.
tan
θ
=
h
√
3
h
⇒
tan
θ
=
1
√
3
⇒
θ
=
30
o
Two poles of heights
6
m and
11
m stand vertically on a plane ground. If the distance between their feet is
12
m, find the distance between their tips.
Report Question
0%
13
m
0%
15
m
0%
14
m
0%
none of the above
Explanation
From the given figure, the distance between the tips of the poles forms a right angled triangle with
5
m and
12
m
as its hypotenuse.
Pythagoras theorem for right triangle,
H
y
p
2
=
(
S
i
d
e
1
)
2
+
(
S
i
d
e
2
)
2
So, required distance
=
√
5
2
+
12
2
=
√
25
+
144
=
√
169
=
13
m
The angle of elevation of the top of a hill when observed from a certain point on the horizontal plane through its base is
30
0
. After walking 120 meters towards it on level ground the elevation is found to be
60
0
. Find the height of the hill(in meters).
Report Question
0%
120
0%
60
√
3
0%
120
√
3
0%
60
Explanation
From the triangle
A
B
C
, let
Height of the hill
(
B
C
)
=
h
A
D
=
120
m
and
D
B
=
d
In the
△
D
B
C
,
∠
B
=
90
∘
tan
60
∘
=
B
C
D
B
⇒
√
3
=
h
d
........(1)
In the
△
A
B
C
,
∠
B
=
90
∘
tan
30
∘
=
B
C
A
B
⇒
1
√
3
=
h
d
+
120
......(2)
Dividing (1) by (2)
h
d
h
d
+
120
=
√
3
1
√
3
⇒
d
+
120
d
=
3
⇒
3
d
=
d
+
120
⇒
d
=
60
Substituting the value of
d
in (1), we get
√
3
=
h
60
⇒
h
=
60
√
3
∴
Height of the hill
=
60
√
3
m
0:0:2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
1
Not Answered
29
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 10 Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page