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41. The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electric pole and an angle of elevation of 30° with the bottom of an electric pole and an angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
  A.  10 metres
  B.  13 metres
  C.  15 metres
  D.  12 metres
     
   
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 = 30°, θ2 = 60° Height of tower = 15
Using these values in the shortcut, we get:
Height of tower =
m x tan60° / tan60° − tan30°

15 =
m x √3 / √3 − 1/√3

m = 15 x
(3 − 1)/√3 / √3

m = 15 x
2 / √3 x √3

m = 15 x
2 / 3
= 5 x 2 = 10
Hence, the height of the electric pole is 10 metres.

42. From the top of a building 30 m high, the top and bottom of a tower are observed to have angles of depression 30° and 45° respectively. The height of the tower is :
  A.  20
(
1 −
1 / √3
)
m
  B.  30
(
1 −
1 / √3
)
m
  C.  30
(
1 +
1 / √3
)
m
  D.  10
(
1 −
1 / √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 = 30°, θ2 = 45° Height of tower = 30
Using these values in the shortcut, we get:
Height of tower =
m x tan45° / tan45° − tan30°

30 =
m x 1 / 1 − 1/√3

m = 30
(
1 −
1 / √3
)

Hence, the height of the tower is 30
(
1 −
1 / √3
)
metres.

43. From the foot of a tower the angle of elevation of the top of a column is 60° and from the top of the tower which is 25 m high, the angle of elvation is 30°. The height of the column is:
  A.  32.5 m
  B.  37.5 m
  C.  35.5 m
  D.  39.5 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 = 25, θ1 = 30°, θ2 = 60°
Using these values in the shortcut, we get:
Height of tower =
25 x tan60° / tan60° − tan30°

=
25 x √3 / √3 − 1/√3

=
25 x √3 x √3 / (√3 x √3 − 1)

=
25 x 3 / (3 − 1)

=
25 x 3 / 2
= 37.5
Hence, the height of the column is 37.5 metres.

44. A balloon is connected to a meterorological station by a cable of length 20 m, inclined at 60° to the horizontal. Find the height of balloon from the ground. Assuming that there is no slack in the cable.
  A.  100 √3 m
  B.  110 √3 m
  C.  100 √2 m
  D.  110 √2 m
     
   
View Answer

Suppose B be the balloon and AB be the vertical height.

Let C be the meteorological station and CB be the cable.
Then, BC = 200m and ∠ACB = 60°
Then
AB / BC
= sin60° =
√3 / 2

or
AB / 200
=
√3 / 2

∴ AB = 100√3 m

45. From the top of a cliff 25 m high the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. Find the height of the tower.
  A.  70 m
  B.  50 m
  C.  80 m
  D.  85 m
     
   
View Answer

Suppose AB be the cliff and CD be the tower.
Then AB = 25 m

Draw a dotted line from B to C.
We get, BE ⊥ CD
Let ∠EBD = ∠ACB = θ
Now
DE / BE
= tanθ and
AB / AC
= tanθ
∴
DE / BE
=
AB / AC
, So, DE = AB [Since BE = AC]
∴ CD = CE + DE = AB + AB = 2AB = 50 m

46. In a rectangle, if the angle between a diagonal and a side is 30° and the length of diagonal is 6 cm, the area of the rectangle is:
  A.  9 √3 cm 2
  B.  9 √5 cm 2
  C.  7 √3 cm 2
  D.  5 √2 cm 2
     
   
View Answer


Suppose ABCD be the rectangle in which ∠BAC = 30° and AC = 6 cm
Now
AB / AC
= cos30° =
√3 / 2

⇒
AB / 6
=
√3 / 2

⇒ AB = 3√3 cm
BC / AC
= sin30° =
1 / 2

⇒
BC / 6
=
1 / 2

⇒ BC = 3 cm
∴ Area of the rectangle = AB x BC = 9√3 cm2

47. The length of a string between a kite and a point on the ground is 90 m. The string makes an angle of 60° with the level ground. If there is no slack in the string, the height of the kite is:
  A.  42√3 m
  B.  45√5 m
  C.  47√3 m
  D.  45√3 m
     
   
View Answer

Suppose K be the position of the kite and HK be the string so that

HK = 90 m and ∠AHK = 60°
∴
AK / HK
= sin60° =
√3 / 2

⇒
AK / 90
=
√3 / 2

⇒ AK = 45√3 cm
∴ Height of the kite is 45√3 metres.

48. From the top of a pillar of height 20 m, the angles of elevation and depression of the top and bottom of another pillar are 30° and 45° respectively. The height of the second pillar is:
  A.  
20(√2 + 1) / √3
m
  B.  
20(√3 + 2) / √3
m
  C.  
20(√3 + 1) / √3
m
  D.  
20(√2 + 1) / √2
m
     
   
View Answer

Suppose AB and CD be two pillars in which AB = 20 m.

Let BE ⊥ DC. Then,
∠DBE = 30° and ∠EBC = ∠ACB = 45°
Let DE = x. Clearly, EC = AB = 20 m
AC / AB
= cot45° = 1
⇒
AC / 20
= 1
⇒ AC = 20 m.
∴ BE = AC = 20 m.
Now,
DE / BE
= tan30° =
1 / √3

x / 20
=
1 / √3

⇒ x =
20 / √3
m.
∴ Height of the pillar CD = 20 + x
= 20 +
20 / √3

=
20(√3 + 1) / √3

49. The angles of elevation of the top of a tower from two points distant 30 m and 40 m on either side from the base and in the same straight line with it are complementary. The height of the tower is:
  A.  30.64 m
  B.  34.64 m
  C.  32.64 m
  D.  36.64 m
     
   
View Answer

Suppose AB be the tower and C, D be the points of observation.

Then, AC = 30 m and AD = 40 m.
Let ∠ACB = θ. Then, ∠ADB = 90° − θ
Now, tanθ =
AB / AC
=
AB / 30

tan(90° − θ) =
AB / AD
=
AB / 40

or, cotθ =
AB / 40

∴ tanθ x cotθ =
AB2 / 1200

or, AB = √1200 = 20√3 = ( 20 x 1.7321) = 34.64 m

50. Two posts are k metres apart and the height of one is double that of the other. if from the middle point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary, then the height of the shorter post is:
  A.  
k / 2√2
m
  B.  
k / 3√2
m
  C.  
k / 2√3
m
  D.  
k / 3√3
m
     
   
View Answer

Suppose AB and CD be the two posts such that AB = 2CD.

Let M be the midpoint of CA.
Let ∠CMD = θ and ∠AMB = 90° − θ.
Clearly, CM = MA =
1 / 2
k
Let CD = h. Then, AB = 2h
Now,
AB / AM
= tan(90° − θ) = cotθ
⇒
2h / k/2
= cotθ
⇒ cotθ =
4h / k
............(eq 1)
CD / CM
= tanθ
⇒ tanθ =
h / k/2
=
2h / k
............... (eq 2)
Multiplying (eq 1) and (eq 2), we get
4h / k
x
2h / k
= 1
∴ h2 =
k2 / 8

⇒ h =
k / 2√2
metres.

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