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21. A ladder leaning against a vertical wall makes an angle of 45° with the ground. The foot of the ladder is 3 m from the wall. Find the length of the ladder.
  A.  2√3 metres
  B.  3√2 metres
  C.  2√2 metres
  D.  3√3 metres
     
   
View Answer

AB = 3tan45° = 3m

∴ AC = √(32 + 32) = 3√2
∴ the length of the ladder is 3√2 metres.

22. The ratio of the length of a rod and its shadow is 1: √3. The angle of elevation of the sun is:
  A.  30°
  B.  25°
  C.  35°
  D.  40°
     
   
View Answer


Let AB be the rod and AC be its shadow.
Let ∠ACB = θ
Let AB = z
Then, AC = √(3)z
∴ tanθ =
AB / AC
=
z / √3 x z
=
1 / √3

∴ tanθ = tan30°
Hence, θ = 30°

23. When the sun is 30° above the horizontal, the length of shadow cast by a building 50 m high is:
  A.  50√2 m
  B.  55√3 m
  C.  55√2 m
  D.  50√3 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 = ?
θ° = 30°
Height = 50
Using these values in the shortcut, we get:
50 = z x tan30°
z =
50 / tan30°

z =
50 / 1/√3

z = 50√3
∴ the height is 50√3.

24. A straight tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an anagle of 30° with the ground. The distance from the foot of tht tree to the point where the top touches the ground is 10 metres. The height of the tree is:
  A.  10√2 m
  B.  15√3 m
  C.  10√3 m
  D.  15√5 m
     
   
View Answer

AC = 10 x tan30° =
10 / √3
m

AB =
√[
102 +
(
10 / √3
)
2
]
=
20 / √3

∴ Height of the tree = AB(AD) + AC =
20 / √3
+
10 / √3
= 10√3
Hence, the height of the tree is 100√3 metres.

25. 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 160m, find the height of the first tower.
  A.  105 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.

26. A person of height 2 m wants to get a fruit which is on a pole of height
10 / 3
m. If he stands at a distance of
4 / √3
m from the foot of the pole, then the angle at which he should throw the stone; so that it hits the fruit is:
  A.  60°
  B.  45°
  C.  30°
  D.  90°
     
   
View Answer


2m =
10 / 3
−
4 / √3
tanθ
or, tanθ x
4 / √3
=
4 / 3

or, tanθ =
4 / 3
x
√3 / 4
=
1 / √3

tanθ = tan30°
Hence, θ = 30°

27. The distance between two multi-storeyed buildings is 60m. The angle of depression of the top of the first building as seen from the top of the second building which is 150 m high is 30°. The height of the first building is:
  A.  111.36 m
  B.  105.36 m
  C.  125.36 m
  D.  115.36 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= 150, z = 60, θ = 30°
Using these values in the shortcut, we get:
Required height = (150 − 60 x tan30°)
= 150 − 60 x
1 / √3

= 150 − 60 x
1 x √3 / √3 x √3

= 150 − 60 x
√3 / 3

= 150 − 20√3
= 150 − 20 x 1.732
= 150 − 34.64 = 115.36
Hence, the height of first building is 115.36 metres.

28. The heights of two poles are 80m and 62.5m. If the line joining their tops makes an angle of 45° with the horizontal, then the distance between the poles is:
  A.  13.5 m
  B.  17.5 m
  C.  19.5 m
  D.  21.5 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 = 80, z = ?, θ = 45°, Height = 62.5
Using these values in the shortcut, we get:
62.5 = (80 − z x tan45°)
62.5 = 80 − z
z = 80 − 62.5 = 17.5
Hence, the distance between the poles is 17.5 metres.

29. Two poles of equal heights stand on either sides of a roadway which is 120 m wide. At a point on the roadway between the poles, the elevations of the tops of the pole are 60° and 30°. Find the heights of the poles and the position of the point.
  A.  Height: 30√3 m; Position of the point P from B: 30 m
  B.  Height: 30√2 m; Position of the point P from B: 30 m
  C.  Height: 30√3 m; Position of the point P from B: 20 m
  D.  Height: 30√2 m; Position of the point P from B: 20 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 and the
(ii) position of the point P from B(see the figure) =
ztanθ2 / tanθ1 + tanθ2
units.


Here, θ1 = 60°, z = 120, θ2 = 30°
Using these values in the shortcut, we get:
Height of the pole =
120 x √3 x 1/√3 / √3 + 1/√3
= 30√3
Hence, the height of the pole is 30√3 metres.
(ii) Position of the point P from B =
120 x 1/√3 / √3 + 1/√3
= 30
Hence, the position of the point P from B is 30 metres.

30. Two poles of equal heights are standing opposite to each other on either side 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.  10 m
  B.  11 m
  C.  13 m
  D.  15 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.
(iii) position of the point P from D(see the figure) =
ztanθ1 / 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 / 1/√3 + √3

=
15√3 / 2
= 7.5 x 1732 = 12.975 ≈13
Hence, the height of the pole is 13 metres.

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