Home ≫ Arithmetic Aptitute ≫ Height and Distance ≫ Page 6
51. The banks of a river are parallel. A swimmer starts from a point on one of the banks and swims in a straight line inclined to the bank at 45° and reqches the opposite bank at a point 20 m from the point opposite to the starting point. The breadth of the river is:
  A.  10.10 metres
  B.  10.14 metres
  C.  14.14 metres
  D.  12.12 metres
     
   
View Answer

Suppose A be the starting point and B, the end point of the swimmer. Then,

AB = 20 m and ∠BAC = 45°
Now,
BC / AB
= sin45° =
1 / √2

⇒ BC =
20 x √2 / 2
= 14.14

52. The angle of elevation of an aeroplane from a point on the ground is 45°. After 15 seconds's flight, the elevation changes to 30°. If the aeroplane is flying at a height of 3000 m, the speed of the plane is:
  A.  525 km/hr
  B.  515 km/hr
  C.  537 km/hr
  D.  527 km/hr
     
   
View Answer

Let A and B be the two positions of the plane and let O be the point of observation and OD be the horizontal. Draw AC⊥OD and BD⊥OD.

Then, ∠DOB = 30° and ∠DOA = 45° and AC = BD = 3000 m.
∴
OD / DB
= cot30° = √3 ⇒ OD = (3000 x √3) m
OC / AC
= cot45° = 1
⇒ OC = 3000 m
Distance covered in 15 sec = AB = CD = OD − OC = (3000√3 − 3000) = 2196 metres.
∴ Speed of the plane =
2196 / 15
x
1 / 1000
x 60 x 60 = 527 km/hr.

53. A man on a cliff observes a fishing trawler at an angle of depression of 30° which is approaching the shore to the point immediately beneath the observer with a uniform speed. 6 minutes later, the angle of depression of the trawler is found to be 60°. The time taken by the trawler to reach the shore is:
  A.  3 minutes
  B.  2 minutes
  C.  4 minutes
  D.  5 minutes
     
   
View Answer

Let AB be the cliff and C and D be the two positions of the fishing trawler.

Then, ∠ACB = 30° and ∠ADB = 60°
Let AB = h
Now,
AD / AB
= cot60° =
1 / √3

⇒ AD =
h / √3

and
AC / AB
= cot30° = √3
⇒ AC = √(3)h
CD = AC − AD = √(3)h −
h / √3
=
2h / √3

Let u m/min be the uniform speed of the trawler.
Distance covered in 6 min = 6 u metres.
∴ CD = 6u
⇒
2h / √3
= 6u
⇒ h = 3√(3)u
Now, AD =
h / √3
=
3√3u / √3
= 3u
Time taken by trawler to reach A =
Distance / Speed
=
3u / u
= 3
Hence, trawler will take 3 minutes to reach the shore.

54. A flagstaff of height
1 / 5
of the height of a tower is mounted on the top of the tower. If the angle of elevation of the top of the flagstaff as seen from the ground is 45° and the angle of elevation of the top of the tower as seen from the same place is θ, then the value of tanθ is:
  A.  
2 / 3
  B.  
5 / 6
  C.  
1 / 6
  D.  
7 / 6
     
   
View Answer

Let AB be the tower and BC the flagstaff.

Let AB = h
Then, BC =
1 / 5
h.
Let O be the observer,
Then, ∠AOC = 45°
∠AOB = θ
Now,
OA / AC
= cot45° = 1 ⇒
OA / h + (1/5)h
= 1
⇒ OA =
6 / 5
h
tanθ
AB / OA
=
h / (6/5)h
=
5 / 6

Hence, the value of tanθ is
5 / 6
.

55. If the angle of elevation of a cloud from a point 200 m above a lake is 30° and the angle of depression of its reflection in the lake is 60°, then the height of the cloud above the lake is:
  A.  300 m
  B.  340 m
  C.  380 m
  D.  400 m
     
   
View Answer

Let C be the cloud and C' be its reflection in the lake.

Let CS = C'S' = x
Now,
BC / AB
= tan30° =
1 / √3

⇒ x − 200 =
AB / √3

Also,
BC' / AB
= tan60° = √3
⇒ x + 200 = (AB)√3
∴ √3(x − 200) =
x + 200 / √3
.
or, x = 400
∴ CS = 400
Hence, the height of the cloud above the lake is 400 metres.

56. The angle of elevation of a lampost changes from 30° to 60° when a man walks towards it. If the height of the lampost is 10√3 metres, find the distance travelled by man.
  A.  20 metres
  B.  10 metres
  C.  15 metres
  D.  25 metres
     
   
View Answer

A man wishes to find the height of a flagspost which stands on a horizontal plane, at a point on this plane he finds the angle of elevation of the top of the flagspost to be θ1. On walking 'z' units towards the tower he finds the corresponding angle of elevation to be θ2. Then the height (H) of the flagpost is given by
[
ztanθ1tanθ2 / tanθ2 − tanθ1
]
units and the value of DB(below given) is given by
ztanθ1 / tanθ2 − tanθ1
units.

[Note: The angle of elevation of a lampost changes from θ1 to θ2, when a man walks towards it. If the height of the lampost is H metres, then the distance travelled by man is given by
[
H(tanθ2 − tanθ1) / tanθ1 x tanθ2
]
metres. 2. If the time for which man walks towards lampost is given as 't' see then speed of the man can be calculated by the formula given below.
Speed of the man =
[
H / t
x
(tanθ2 − tanθ1) / tanθ1 x tanθ2
]

H = 10√3
θ1 = 30°
θ2 = 60°
Using these values in the shortcut, we get:
=
[
10√3 x (√3 − 1/√3) / √3 x 1/√3
]
= 20 metres

57. The angle of elevation of a lamppost changes from 30° to 60° when a man walks 20 m towards it. What is the height of the lampost?
  A.  12.32 m
  B.  15.32 m
  C.  17.32 m
  D.  19.32 m
     
   
View Answer

A man wishes to find the height of a flagspost which stands on a horizontal plane, at a point on this plane he finds the angle of elevation of the top of the flagspost to be θ1. On walking 'z' units towards the tower he finds the corresponding angle of elevation to be θ2. Then the height (H) of the flagpost is given by
[
ztanθ1tanθ2 / tanθ2 − tanθ1
]
units and the value of DB(below given) is given by
ztanθ1 / tanθ2 − tanθ1
units.

Note: The angle of elevation of a lampost changes from θ1 to θ2, when a man walks towards it. If the height of the lampost is H metres, then the distance travelled by man is given by
[
H(tanθ2 − tanθ1) / tanθ1 x tanθ2
]
metres.
2. If the time for which man walks towards lampost is given as 't' see then speed of the man can be calculated by the formula given below.
Speed of the man =
[
H / t
x
(tanθ2 − tanθ1) / tanθ1 x tanθ2
]

z = 20
θ1 = 30°
θ2 = 60°
Using these values in the shortcut, we get:
=
[
20 x tan60° x tan30° / tan60° − tan30°
]

=
[
20 x √3 x 1/√3 / √3 − 1/√3
]

=
[
20 x √3 / √3 x √3 − 1
]

=
[
20√3 / 3 − 1
]

=
[
20√3 / 2
]
= 10 x 1.732 = 17.32

58. The horizontal distance between two towers is 50 √3 m. The angle of depression of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 160 m, find the height of the first tower.
  A.  100 m
  B.  110 m
  C.  115 m
  D.  120 m
     
   
View Answer

Shortcut:
The horizontal distance between two towers is 'z' units. The angle of depression of the first tower when seen from the top of the second tower is θ°.
(i) If the height of the second tower is 'h1' units then the height of the first tower is given by (h1 − ztanθ) units.
(ii) If the height of the first tower is given as 'h2' units then the height of the second tower is given by (h2 + ztanθ).

Here, h1= 160, z = 50√3, θ = 30°
Using these values in the shortcut, we get:
Required height = (160 − 50√3 x tan30°)
= 160 − 50√3 x
1 / √3

=160 − 50 = 110
Hence, the height of first tower is 110 metres.

59. Two poles of equal heights are standing opposite to each other on either sie of a road, which is 30 m wide. From a point between them on the road, the angles of elevation of the tops are 30° and 60°. The height of each pole is:
  A.  11 m
  B.  15 m
  C.  17 m
  D.  13 m
     
   
View Answer

Shortcut:
Two poles of equal heights stand on either sides of a roadway which is 'z' units wide. At a point of the roadway between the poles, the elevations of the tops of the pole are θ1° and θ2°, then the
(i) the heights of poles =
ztanθ1tanθ2 / tanθ1 + tanθ2
units
(ii) position of the point P from B(see the figure) =
ztanθ2 / tanθ1 + tanθ2
units.

Here, θ1 = 30°, z = 30, θ2 = 60°
Using these values in the shortcut, we get:
Height of the pole =
30 x 1/√3 x √3 / √3 + 1/√3

=
15√3 / 2
= 7.5 x 1.732 = 12.975 m ≈ 13 m
Hence, the height of the pole is 13 metres.

60. The length of a string between a kite and a point on the ground is 85 m. If the string makes an angle θ with the level ground such tant tanθ = 15/8, how high is the kite, when there is no slack in the string?
  A.  70 m
  B.  80 m
  C.  75 m
  D.  85 m
     
   
View Answer

Shortcut:
Consider the following figure,

In this figure, AC = m;
tanθ =
a / b
then,
(i) AB =
[
a / √(a2 + b2)
x m
]
and
(ii) BC =
[
b / √(a2 + b2)
x m
]


Here, a = 15, b = 8, m = 85
Using these values in the shortcut, we get:
Required Height =
[
15 / √[(15)2 + (8)2]
x 85
]

=
[
15 / √(225 + 64)
x 85
]

=
[
15 / 17
x 85
]
= 15 x 5 = 75
Hence, the height of the kite is 75 metres.

Copyright © 2020-2022. All rights reserved. Designed, Developed and content provided by Anjula Graphics & Web Desigining .