Home ≫ Arithmetic Aptitute ≫ Height and Distance ≫ Page 2
11. Two persons standing on the same side of a tower measure the angles of elevation of the top of the ttower as 30° and 45°. If the height of the tower is 30m, the distance between the two persons is approximately:
  A.  18 metres
  B.  20 metres
  C.  22 metres
  D.  24 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.

z = ?
θ1 = 30°
θ2 = 45°
Height = 30
Using these values in the shortcut, we get:
30 =
z x (1/√3) x 1 / 1 − (1/√3)

z =
30 x [1 − (1/√3)] / 1/√3

z =
30 x [(√3 − 1)/√3] / (1/√3)

z = 30(√3 − 1) = ≈ 22
Hence, the distance between two persons is 22 metres.

12. If from the top of a cliff 100 m high, the angles of depressions of two ships out at sea are 60° and 30°, then the distance between the ships is approximately.
  A.  112.3 m
  B.  117.3 m
  C.  119.3 m
  D.  115.3 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.

z = ?
θ1 = 30°
θ2 = 60°
Height = 100
Using these values in the shortcut, we get:
100 =
z x (1/√3) x √3 / √3 − 1/√3

z =
100 x (√3 − 1/√3) / 1

z =
100 x (3 − 1) / √3

z =
100 x 2 / √3

z =
200 / √3

z =
200√3 / √3 x √3

z =
200√3 / 3
= 115.3
Hence, the distance between the ships is 115.3 metres.

13. The angles of depression two ships from the top of the light house are 45° and 30° towards east. If the ships are 100 m apart, then the height of the light house is:
  A.  50(√3 + 1) m
  B.  45(√3 + 1) m
  C.  50(√5 + 1) m
  D.  45(√5 + 1) 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.

z = 100
θ1 = 30°
θ2 = 45°
Height = ?
Using these values in the shortcut, we get:
H =
100 x 1/√3 x 1 / 1 − 1/√3

H =
100 x 1/√3 x 1 / (√3 − 1)/√3

H =
100 / (√3 − 1)

H =
100(√3 + 1) / (√3 − 1)(√3 + 1)

H =
100(√3 + 1) / 3 − 1

H =
100(√3 + 1) / 2
= 50(√3 + 1)
Hence, the height of the light house is 50(√3 + 1) metres.

14. The shadow of a tower standing on a level plance is found to be 60 m longer when the sun's alttitude is 30°, than when it is 45°. The height of the tower is:
  A.  78.96 m
  B.  81.96 m
  C.  84.96 m
  D.  81.16 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.

z = 60
θ1 = 30°
θ2 = 45°
Height = ?
Using these values in the shortcut, we get:
H =
60 x 1/√3 x 1 / 1 − 1/√3

H =
60 x 1/√3 x 1 / (√3 − 1)/√3

H =
60 / (√3 − 1)

H =
60(√3 + 1) / (√3 − 1)(√3 + 1)

H =
60(√3 + 1) / 3 − 1

H =
60(√3 + 1) / 2
= 30(√3 + 1) = 30 x 2.732 = 81.96
Hence, the height of the light house is 81.96 metres.

15. A boat being rowed away from a cliff 150 m high. At the top of the cliff the angle of depression of the boat changes from 60° to 45° in 2 minutes. The speed of the boat is:
  A.  1.5 km/hr
  B.  1.2 km/hr
  C.  2.9 km/hr
  D.  1.9 km/hr
     
   
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.

z = ?
θ1 = 45°
θ2 = 60°
Height = 150
Using these values in the shortcut, we get:
150 =
z x 1 x √3 / √3 − 1

z =
150(√3 − 1) / √3

z = 150
(
1 −
1 / √3
)
= 63.4
∴ the distance covered in 2 minutes is 63.4 metres.
∴ speed of the boat =
63.4 x 60 / 2 x 1000
= 1.9 km/hr.

16. 100√3 m from the foot of a cliff on level ground, the angle of elevation of the top of a cliff is 30°. Find the height of this cliff.
  A.  100 m
  B.  110 m
  C.  105 m
  D.  115 m
     
   
View Answer

A small boy is standing at some distance from a flagpost. When the sees the flag the angle of elevation formed is θ°. If the height of the flagpost is 'Z' units, then the distance of the child from the flagpost is
Z / tanθ°
units.

Distance = 100√3
θ° = 30°
Using these values in the shortcut, we get:
Distance =
Z / tan30°

100√3 =
Z / 1/√3

100√3 =
Z x √3 / 1

100√3 = Z x 100√3
Z =
100√3 / √3
= 100
∴ the height is 100 metres.

17. The angles of elevation of top and bottom of a flag kept on a flagpost from 30 metres distance are 45° and 30° respectively. What is the height of the flag?
  A.  11.68 m
  B.  12.18 m
  C.  12.68 m
  D.  15.68 m
     
   
View Answer

The angles of elevation of top and bottom of a flag kept on a flagpost from 'z' units distance are θ1° and θ2° respectively. Then
(i) the height of the flag is given by [z(θ1 − θ2)] units
(ii) the height of the flagpost is given by
z x θ2 / (θ1 − θ2)
units

z = 30, θ1 = 45°, θ2 = 30°
Using these values in the shortcut, we get:
Required Height = [30 x (tan45° − tan30°)]
=
[
30 x
(
1 −
1 / √3
)]

= 30 x
√3 − 1 / √3

= 30 x
(√3 − 1) x √3 / √3 x √3

= 30 x
(√3 − 1) x √3 / 3

= 10 x (√3 − 1) x √3
= 10 x (3 − √3)
= 30 − 10√3)
= 30 − (10 x 1.732)
= 30 − 17.3 = 12.68
∴ the height is 12.68 metres.

18. An observer standing 72 m away from a building notices that the angles of elevation of the top and the bottom of a flagstaff on the building are respectively 60° and 45°. The height of the flagstaff is:
  A.  50.7 m
  B.  52.7 m
  C.  53.7 m
  D.  52.2 m
     
   
View Answer

The angles of elevation of top and bottom of a flag kept on a flagpost from 'z' units distance are θ1° and θ2° respectively. Then
(i) the height of the flag is given by [z(θ1 − θ2)] units
(ii) the height of the flagpost is given by
z x θ2 / (θ1 − θ2)
units

z = 72
θ1 = 60°
θ2 = 45°
Using these values in the shortcut, we get:
Required Height = [72 x (tan60° − tan45°)]
= [72 x (√3 − 1)]
= [72 x (1.732 − 1)]
= 72(0.732) = 52.7
∴ the height of flagstaff is 52.7 metres.

19. 300 m from the foot of a cliff on level ground, the angle of elevation of the top of a cliff is 30°. Find the height of this cliff.
  A.  171.20 m
  B.  175.20 m
  C.  173.80 m
  D.  173.20 m
     
   
View Answer

'z' units of distance from the foot of a cliff on level ground, the angle of elevation of the top of a cliff is θ°, then the height of the cliff is (z tanθ°) units.

z = 300
θ° = 30°
Using these values in the shortcut, we get:
Required Height = (300 x tan30°)
= 300 x
1 / √3

= 300 x
1 x √3 / √3 x √3

= 300 x
√3 / 3

= 100 x √3
= 100 x 1.732 = 173.20
∴ the height of cliff is 173.20 metres.

20. If a vertical pole 6 m high has a shadow of length 2√3 m, find the angle of elevation of the sun.
  A.  50°
  B.  55°
  C.  60°
  D.  65°
     
   
View Answer

'z' units of distance from the foot of a cliff on level ground, the angle of elevation of the top of a cliff is θ°, then the height of the cliff is (z tanθ°) units.
z = 2√3
θ° = ?
Height = 6
Using these values in the shortcut, we get:
6 = (2√3 x tanθ)
tanθ =
6 / 2√3

tanθ =
3 / √3

tanθ =
3 x √3 / √3 x √3

tanθ =
3√3 / 3

tanθ = √3
tanθ = tan60°
θ = 60°
∴ the angle of elevation of the sun is 60°.

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