STUDENTIALS LEARNING HUB
Khushboo fashion gallery near gate no 3 veer savarkar nagar Bareilly. 8077959645
Test / Exam Name: Maths Standard: 10TH Subject: MATHEMATICS
Student Name: Section: Roll No.:
Questions: 9 Time: 60 Mins Marks: 40
Q1. An aeroplane when flying at a height, of 3125m from the ground passes vertically below another plane at 5 Mark
an instant when the angles of elevation of the two planes from the same point on the ground are 30o and
60o respectively. Find the distance between the two planes at that instant.
Ans: Let CD = x , AB = y
3125 1
∴ = tan 30o =
y 3
⇒ y = 3125 3m
ST
x + 3125
= tan 60o = 3
y
U
x + 3125
= 3 ⇒ x = 3(3125) − 3125
D
3125 3
EN
= 2(3125) = 6250m.
Q2. If the angle of elevation of a cloud from a point h metres above a lake is a and the angle of depression of 5 Mark
TI
its reflection in the lake be b, prove that the distance of the cloud from the point of observation is,
A
2h sec α
.
LS
tan β − tan α
Ans: Let C′ be the image of cloud C. We have ∠CAB = α and ∠BAC′ = β.
LE
Again let BC = x and AC be the distance of cloud from point of observation.
We have to prove that,
A
2h sec α
AC =
R
(tan β − tan α)
N
The corresponding figure is as follows,
IN
G
H
U
B
We use trigonometric ratios.
In △ABC
BC
⇒ tan α =
AB
x
⇒ tan α =
AB
Again in △ABC′
BC′
⇒ tan β =
AB
x + 2h
⇒ tan β =
AB
Now,
x + 2h x
⇒ tan β − tan α = −
AB AB
2h
⇒ tan β − tan α =
AB
2h
⇒ AB =
tan β − tan α
Again in △ABC
AB
⇒ cos α =
AC
, AB
⇒ AC =
cos α
2h sec α
⇒ AC =
(tan β − tan α)
2h sec α
Hence distance of cloud from points of observation is .
(tan β − tan α)
Q3. Prove that (4 − 4 Mark
5 2)
Ans: Let x = 4 − 5 2 be a rational number.
x = 4 − 5 2
2
⇒ x2 = (4 − 5 2)
2
⇒ x2 = (4)2 + (2 3) − 2(4)(5 2)
⇒ x2 = 16 + 50 − 40 2
⇒ x2 − 66 = −40 2
66 − x2
⇒ = 2
ST
40
Since x is a rational number, x2 is also a rational number.
U
⇒ 66 − x2 is a rational number
D
66 − x2
EN
⇒ is a rational number
40
⇒ 2 is a rational number
TI
But 2 is an irrational number, which is a contradiction.
A
Hence, our assumption is wrong.
LS
Thus, (4 − 5 2) is an irrational number.
Q4. A ladder rests against a wall at an angle α to the horizontal. Its foot is pulled away from the wall through 5 Mark
LE
a distance a, so that it slides a distance b down the wall making an angle β with the horizontal. Show
a cos α − cos β
A
that, = .
b sin β − sin α
R
Ans: Let PQ be the ladder such that its top Q is on the wall OQ and bottom P is on the ground. The ladder is pulled
N
away from the wall through a distance a, so that its top Q slides and takes position Q'. So PQ = P'Q'
IN
∠OPQ = α and ∠OP′ Q′ = β. Let PQ = h
G
We have to prove that,
a cos α − cos β
H
=
b sin β − sin α
U
We have the corresponding figure as follows,
B
We use trigonometric ratios.
In △POQ
OQ
⇒ sin α =
PQ
b + y
⇒ sin α =
h
And,
OP
⇒ cos α =
PQ
x
⇒ cos α =
h
Again in △P′ OQ′
OQ′
⇒ sin β = ′ ′
PQ
y
⇒ sin β =
h
And,
′
Khushboo fashion gallery near gate no 3 veer savarkar nagar Bareilly. 8077959645
Test / Exam Name: Maths Standard: 10TH Subject: MATHEMATICS
Student Name: Section: Roll No.:
Questions: 9 Time: 60 Mins Marks: 40
Q1. An aeroplane when flying at a height, of 3125m from the ground passes vertically below another plane at 5 Mark
an instant when the angles of elevation of the two planes from the same point on the ground are 30o and
60o respectively. Find the distance between the two planes at that instant.
Ans: Let CD = x , AB = y
3125 1
∴ = tan 30o =
y 3
⇒ y = 3125 3m
ST
x + 3125
= tan 60o = 3
y
U
x + 3125
= 3 ⇒ x = 3(3125) − 3125
D
3125 3
EN
= 2(3125) = 6250m.
Q2. If the angle of elevation of a cloud from a point h metres above a lake is a and the angle of depression of 5 Mark
TI
its reflection in the lake be b, prove that the distance of the cloud from the point of observation is,
A
2h sec α
.
LS
tan β − tan α
Ans: Let C′ be the image of cloud C. We have ∠CAB = α and ∠BAC′ = β.
LE
Again let BC = x and AC be the distance of cloud from point of observation.
We have to prove that,
A
2h sec α
AC =
R
(tan β − tan α)
N
The corresponding figure is as follows,
IN
G
H
U
B
We use trigonometric ratios.
In △ABC
BC
⇒ tan α =
AB
x
⇒ tan α =
AB
Again in △ABC′
BC′
⇒ tan β =
AB
x + 2h
⇒ tan β =
AB
Now,
x + 2h x
⇒ tan β − tan α = −
AB AB
2h
⇒ tan β − tan α =
AB
2h
⇒ AB =
tan β − tan α
Again in △ABC
AB
⇒ cos α =
AC
, AB
⇒ AC =
cos α
2h sec α
⇒ AC =
(tan β − tan α)
2h sec α
Hence distance of cloud from points of observation is .
(tan β − tan α)
Q3. Prove that (4 − 4 Mark
5 2)
Ans: Let x = 4 − 5 2 be a rational number.
x = 4 − 5 2
2
⇒ x2 = (4 − 5 2)
2
⇒ x2 = (4)2 + (2 3) − 2(4)(5 2)
⇒ x2 = 16 + 50 − 40 2
⇒ x2 − 66 = −40 2
66 − x2
⇒ = 2
ST
40
Since x is a rational number, x2 is also a rational number.
U
⇒ 66 − x2 is a rational number
D
66 − x2
EN
⇒ is a rational number
40
⇒ 2 is a rational number
TI
But 2 is an irrational number, which is a contradiction.
A
Hence, our assumption is wrong.
LS
Thus, (4 − 5 2) is an irrational number.
Q4. A ladder rests against a wall at an angle α to the horizontal. Its foot is pulled away from the wall through 5 Mark
LE
a distance a, so that it slides a distance b down the wall making an angle β with the horizontal. Show
a cos α − cos β
A
that, = .
b sin β − sin α
R
Ans: Let PQ be the ladder such that its top Q is on the wall OQ and bottom P is on the ground. The ladder is pulled
N
away from the wall through a distance a, so that its top Q slides and takes position Q'. So PQ = P'Q'
IN
∠OPQ = α and ∠OP′ Q′ = β. Let PQ = h
G
We have to prove that,
a cos α − cos β
H
=
b sin β − sin α
U
We have the corresponding figure as follows,
B
We use trigonometric ratios.
In △POQ
OQ
⇒ sin α =
PQ
b + y
⇒ sin α =
h
And,
OP
⇒ cos α =
PQ
x
⇒ cos α =
h
Again in △P′ OQ′
OQ′
⇒ sin β = ′ ′
PQ
y
⇒ sin β =
h
And,
′
