Home ≫ Arithmetic Aptitute ≫ Height and Distance ≫ Page 4
31. An aeroplane when 3000 m high passes vertically above another at an instant when the angles of elevation at the same ovserving point are 60° and 45° respectively. How many metres lower is one than the other?
  A.  1368 metres
  B.  1168 metres
  C.  1468 metres
  D.  1268 metres
     
   
View Answer

Shortcut:
Consider the following figure,
In the above figure, θ1 and θ2 are given. AC is given. To find AB we have following formula.
AC
(
1 −
tanθ2 / tanθ1
)



Here, θ1 = 60°, AC = 3000, θ2 = 45°
Using these values in the shortcut, we get:
Height of the pole = 3000
[
1 −
1 / √3
]

= 3000
[
1 −
1 / √3
]
≈ 1268
Hence, the height of the pole is 1268 metres.

32. A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 7m. At a point on the plane, the angle of elevation of the bottom of the flagstaff is 30° and that of the top of the flagstaff is 45°. Find the height of the tower.
  A.  
5 / √3 − 1
m
  B.  
7 / √2 − 1
m
  C.  
7 / √3 − 1
m
  D.  
5 / √2 − 1
m
     
   
View Answer

Shortcut:
Consier the following figure,
In the above figure, θ1 and θ2 are given. AC is given.
To find AB we have following formula. AC
(
1 −
tanθ2 / tanθ1
)


Here, Total height(AC) = Height of tower + flagstaff
θ1 = 30°, AC = ?, θ2 = 45°Height = 7
Using these values in the shortcut, we get:
Height of flagstaff = AC
(
1 −
tan30° / tan45°
)

7 = AC
(
1 −
1/√3 / tan45°
)

7 = AC(1 − √3)
AC =
7 / 1 − 1/√3

AC =
7√3 / √3 − 1

AC =
7√3 / √3 − 1

As we know:
AC (Total Height) = AB (Height of tower) + BC (Height of flagstaff)
We have to find the height of tower.
∴ BC = AC − AB
BC =
7√3 / √3 − 1
− 7
=
7 / √3 − 1
− 7
Hence, the height of the tower is
7 / √3 − 1
metres.

33. The length of a string between a kite and a point on the ground is 102 m. If the string makes and angle α with the level ground such that θ = 15/8, how high is the kite?
  A.  80 m
  B.  90 m
  C.  85 m
  D.  95 m
     
   
View Answer

Shortcut:
Consider the following figure,

In the above 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 = 102
Using these values in the shortcut, we get:
Required Height =
[
15 / √[(15)2 + (8)2]
x 102
]

=
[
15 / √(225 + 64)
x 102
]


=
[
15 / 17
x 102
]
= 15 x 6 = 90
Hence, the height of the kite is 90 metres.

34. The angles of depression of two ships from the top of a lighthouse are 45° and 30°. if the ships are 120 m apart, find the height of the lighthouse.
  A.  44 m
  B.  40 m
  C.  42 m
  D.  46 m
     
   
View Answer

Shortcut:
The angles of depression of two ships from the top of a lighthouse are θ1 and θ2. If the ships are 'm' metres apart, then the,
(i) Height of the lighthouse =
mtanθ1tanθ2 / tanθ1 + tanθ2
metres,
(ii) Distance of ship at P from the foot of the lighthouse =
mtanθ1 / tanθ1 + tanθ2
metres
(iii) Distance of ship at Q from the foot of lighthouse =
mtanθ2 / tanθ1 + tanθ2
metres


Here, m = 120, θ1 = 30, θ2 = 45
Using these values in the shortcut, we get:
Required Height =
120 x tan30° x tan45° / tan30° + tan45°

=
120 x (1/√3) x 1 / (1/√3) + 1

=
120 x 1 x 1 / 1 + √3
≈44
Hence, the height of the lighthouse is 44 metres.

35. From the top of a cliff 100√3 m high the angles of depression of two boats which are due south of observer are 60° and 30°. Find the distance between the two boats.
  A.  350 m
  B.  420 m
  C.  450 m
  D.  400 m
     
   
View Answer

Shortcut:
The angles of depression of two ships from the top of a lighthouse are θ1 and θ2. If the ships are 'm' metres apart, then the,
(i) Height of the lighthouse =
mtanθ1tanθ2 / tanθ1 + tanθ2
metres,
(ii) Distance of ship at P from the foot of the lighthouse =
mtanθ1 / tanθ1 + tanθ2
metres
(iii) Distance of ship at Q from the foot of lighthouse =
mtanθ2 / tanθ1 + tanθ2
metres

Here, m = ?, θ1 = 60°, θ2 = 30°, Height of lighthouse = 100√3
Using these values in the shortcut, we get:
Height of lighthouse =
m x tan60° x tan30° / tan60° + tan30°

100√3 =
m x √3 x (1/√3) / √3 + (1/√3)

m = 100√3
√3 + 1/√3 / 1

= 100√3
(√3)2 + 1 / √3

= 100 x (3 + 1) = 100 x 4 = 400
Hence, the distance between boats is 400 metres.

36. A landmark on a river bank is observed from two points A and B on the opposite bank of the river. The lines of sight make equal angles 45° with the bank of the river. If AB = 1 km, then the width of the river is:
  A.  
1 / 3
km
  B.  
1 / 2
km
  C.  
3 / 2
km
  D.  
2 / 3
km
     
   
View Answer

Shortcut:
The angles of depression of two ships from the top of a lighthouse are θ1 and θ2. If the ships are 'm' metres apart, then the,
(i) Height of the lighthouse =
mtanθ1tanθ2 / tanθ1 + tanθ2
metres,
(ii) Distance of ship at P from the foot of the lighthouse =
mtanθ1 / tanθ1 + tanθ2
metres
(iii) Distance of ship at Q from the foot of lighthouse =
mtanθ2 / tanθ1 + tanθ2
metres

Here, m = 1, θ1 = 45°, θ2 = 45°
Using these values in the shortcut, we get:
Width of the river =
1 x tan45° x tan45° / tan45° + tan45°

=
1 x 1 x 1 / 1 + 1

=
1 / 2

Hence, the width of the river is
1 / 2
Km.

37. The angles of elevation of the top of a tower 40 m high from two points on the level ground on its opposite sides are 45° and 60°. The distance between the two points in nearest metres is:
  A.  60 m
  B.  65 m
  C.  63 m
  D.  67 m
     
   
View Answer

Shortcut:
The angles of depression of two ships from the top of a lighthouse are θ1 and θ2. If the ships are 'm' metres apart, then the,
(i) Height of the lighthouse =
mtanθ1tanθ2 / tanθ1 + tanθ2
metres,
(ii) Distance of ship at P from the foot of the lighthouse =
mtanθ1 / tanθ1 + tanθ2
metres
(iii) Distance of ship at Q from the foot of lighthouse =
mtanθ2 / tanθ1 + tanθ2
metres

Here, m = ?, θ1 = 45°, θ2 = 60° Height of tower = 40
Using these values in the shortcut, we get:
Height of tower =
m x tan45° x tan60° / tan45° + tan60°

40 =
m x tan45° x tan60° / tan45° + tan60°

40 =
m x 1 x √3 / 1 + √3

m = 40 x
1 + √3 / √3

m = 40 x
1 + 1.732 / 1.732

m = 40 x
2.732 / 1.732
≈63
Hence, the distance between two points is 63 metres.

38. Two boats approach a lighthouse in mid-sea from opposite directions. The angles of elevation of the top of the lighthouse from the two boats are 30° and 45° respectively. if the distance between the two boats is 100 m, the height of the lighthouse is:
  A.  36.6 m
  B.  32.6 m
  C.  34.6 m
  D.  38.6 m
     
   
View Answer

Shortcut:
The angles of depression of two ships from the top of a lighthouse are θ1 and θ2. If the ships are 'm' metres apart, then the,
(i) Height of the lighthouse =
mtanθ1tanθ2 / tanθ1 + tanθ2
metres,
(ii) Distance of ship at P from the foot of the lighthouse =
mtanθ1 / tanθ1 + tanθ2
metres
(iii) Distance of ship at Q from the foot of lighthouse =
mtanθ2 / tanθ1 + tanθ2
metres

Here, m = 100, θ1 = 30°, θ2 = 45°
Using these values in the shortcut, we get:
Height of tower =
100 x tan30° x tan45° / tan30° + tan45°

=
100 x tan30° x tan45° / tan30° + tan45°

=
100 x 1/√3 x 1 / 1/√3 + 1

=
100 x 1 x 1 / 1 + √3

=
100 / 1 + 1.732

=
100 / 2.732
= 36.6
Hence, the height of the house is 36.6 metres.

39. From the top and bottom of a building of height 120 metres, the angles of elevation of the top of a towers are 30° and 45° respectively. Find the height of the tower.
  A.  254 m
  B.  284 m
  C.  294 m
  D.  274 m
     
   
View Answer

Shortcut:
From the top and bottom of a building of height 'm' units, the angles of elevation of the top of a tower are θ1 and θ2 respectively, then the,
(i) Height of the tower =
mtanθ2 / tanθ2 − tanθ1
units,
(ii) Distance between the building and the tower =
m / tanθ2 − tanθ1
units
(iii) RM(See the below figure) =
mtanθ1 / tanθ2 − tanθ1
units
[Note: If height of the tower is 'H' units, then the distance between the building and the tower is
H / tanθ2
units.]

Here, m = 120, θ1 = 30°, θ2 = 45°
Using these values in the shortcut, we get:
Height of tower =
120 x tan45° / tan45° − tan30°

=
120 x 1 / 1 − tan30°

=
120 / 1 − 1/√3

=
120 / √3 − 1

=
120 / 1.732 − 1

=
120 / 0.732
≈284
Hence, the height of the tower is 284 metres.

40. A tower is 30 m high. An observer from the top of the tower makes an angle of depression of 60° at the base of a building and angle of depression of 45° at the top of the building, what ist he height of the building? Also, find the distance between building and tower.
  A.  10.6 m;
30 / √3
m
  B.  12.6 m;
20 / √2
m
  C.  12.6 m;
30 / √3
m
  D.  15.6 m;
25 / √2
m
     
   
View Answer

Shortcut:
From the top and bottom of a building of height 'm' units, the angles of elevation of the top of a tower are θ1 and θ2 respectively, then the,
(i) Height of the tower =
mtanθ2 / tanθ2 − tanθ1
units,
(ii) Distance between the building and the tower =
m / tanθ2 − tanθ1
units
(iii) RM(See the below figure) =
mtanθ1 / tanθ2 − tanθ1
units
[Note: If height of the tower is 'H' units, then the distance between the building and the tower is
H / tanθ2
units.]

Here, m = ?, θ1 = 45°, θ2 = 60° Height of tower = 30
Using these values in the shortcut, we get:
Height of tower =
m x tan60° / tan60° − tan45°

30 =
m x √3 / √3 − 1

m = 30 x
√3 − 1 / √3

m = 30 x
1.732 − 1 / 1.732

m = 30 x
0.732 / 1.732
≈12.6
Hence, the height of the building is 12.6 metres.
Distance between the building and the tower =
30 / tan60°

=
30 / √3

Hence, the distance between building and tower is
30 / √3
metres.

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