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Question-1

A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Fig.).

 

Solution:



From the figure,
AB – Height of the pole
AC – Length of the rope
In D ABC,
Sin 300 =
=

ž  = AB
ž AB = 10 m.
Height of the pole = 10 m.

Question-2

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Solution:

Let AC be the height of the tree.
If AB is the broken part of the tree that bends once touches the ground then AB = BD = x      ..........(i)
In Δ BCD,
tan 30
°   =
=
BC = m
Cos 30
°  =
=
BD =
from (i) AB = BD =
The height of the tree AC = AB + BC
                                    = +
                                    =
The height of the tree AC =
                                    = 8m.

Question-3

A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Solution:
The plan of two slides for the children to play in a park is given in the diagram

In Δ ABC,
AB – height of the slide
AC – Length of the slide
sin 3
0° =
       =
  AC = 1.5 × 2
          = 3m
In Δ DEF,
DF – Length of the slide
DE – Height of the slide
Sin 6
0° =
      =
   DF =
           = ×
           = = 2
The length of the slide for children below 5 years = 3 m.
The length of the slide for elder children = 2 m.

Question-4

The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Solution:

Let AB be the height of the tower
In Δ ABC,
tan 3
0° =
      =
   AB =
= = 10
The height of the tower = 10 m.

Question-5

A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Solution:

Let AB be the height of the tower and AC be the length of the string.
In Δ ABC
Sin 6
0° =
     =
AC =
    =
    =
    = 40 m
The length of the string = 40 m.

Question-6

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Solution:

Let AB = 1.5 m be the height of the boy and CE = 30m be the height of the building.
If BF = x is the distance moved towards the building.
In Δ AED,
cot 3
0° =
      = …….. (i) Here ABCD is a rectangle with AB = CD
since ED = EC – DC
             = 30 – 1.5
             = 28.5 m
substitute ED = 28.5 in (i)
           =
           Þ AD = 28.5
              AD = 49.36m
In Δ GED,
cot 6
0° =
      =
      GD =
By rationalizing the denominator
 = 16.45m
As AG = BF = x and since AG = AD – GD
    AG = 49.36 – 16.45
         = 32.91 m
The distance moved by the boy towards the building = 32.91 m.

Question-7

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Solution:

Let AB = x be the height of the transmission tower.
If   BC = 20 is the height of the building
In Δ BCD
tan 45o =
         1 =
CD = 20 m
In Δ ACD,
tan 600 =
       =
       =
    20 = x + 20
         x = 20 - 20

 Height of the transmission tower = 20( - 1) m.

Question-8

A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Solution:

Let AB = 1.6 m, be the height of the statue BC = x m be the height of pedestal CD be the distance between the foot of the pedestal to a point on the ground.
In Δ ACD,
  tan 6
0° =
since AC = AB + BC = 1.6 + x
        =
        CD = …………………. (i)
In Δ BCD,
tan 45
° =
         1 =
   CD = x
substitute CD = x in (i)
                 x =
          Þ x = 1.6 + x

      Þ x - x = 1.6
         x(-1) = 1.6
                  x =
Rationalising the denominator
                  x =
                    =
                    = 0.8(= 2.19 m
Hence the height of the pedestal = 2.19 m.

Question-9

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Solution:

Let AB be the height of the building.
If CD = 50 m is the height of the tower,
In Δ BCD,
tan 60
° =
      =
      BC =
In Δ ABC,



 The height of the building = 16.

Question-10

Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Solution:
Let AB and CD be the two poles of equal heights opposite to each other on either side heights of the road which is 80 m wide.


 

Let E be a point on the ground such that BE = x m and ED = (80 – x) m.
If suppose you consider AB = CD = y.
In Δ ABE,
tan 60
° =

      =
    y = x........ (i)
In Δ CDE,
tan 30
° =

       =
     Þ y =  .................(ii)
           
  Þ x =
    Þ 3x = 80 – x
3x + x = 80
      4x = 80
           x =
              = 20 m
         BE = 20 m
Substitute x = BE = 20 in (1)
               y =
            Þ y =
                  = 20 m
             ED = 80 - x
                  = 80 – 20
                  = 60 m
Thus we conclude the poles are at a height of 20m and the point E, is at a distance of 20 m from the pole AB, and at a distance of 60 m from the pole CD.


Question-11

A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig.). Find the height of the tower and the width of the canal.
 

 


Solution:
align="center">
Let AB = h be the height of the tower.
BC = x be the width of the canal
In Δ ABC,
tan 600 =
      =
    h = x m .........(i)
In Δ ABD,
tan 300 =
  =
  = =
    h = ………………. (ii)
from (i) and (ii)
    = x
x +20 = 3x
     2x = 20
          x = 10 m
substitute x = 10 m in (i)
               h = x
                  = (10)
                  = 10
                  = 10(1.732) = 17.32 m

Question-12

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Solution:

Let AB be the building of height 7 m, and CE be the height of the cable tower.
In Δ ABE,
tan 45
° =
        1 =
  Þ BE = 7 m
since ABED is a square AD = BE = 7 m
In Δ ACD,
tan 60
° =
      =
  CD = 7 m
Thus the height of the cable tower = CD + DE
                                                = 7 + 7
                                                = 7(+1) m.

Question-13

As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Solution:

Let AB = 75m be the height of the light house C and D are the position of two strips such that CD = x.
In Δ ABC,
tan 45
° =
        1 =
  BC = 75m
In Δ ABD,
tan 300 =
      = 75 + x = 75
      x = 75 - 75
          x = 75( - 1)

Therefore, the distance between the ships is  m.

Question-14

 A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

 


Solution:
  
The height of the girl = 1.2 m
Height of the balloon above the ground level = 88.2 m
BE is the distance traveled by the balloon during the interval.
In Δ CDE,
tan 30
° =
      =
       CE = 88.2m
In Δ ABC,
tan 60
° =
      =
      BC =
      EB = EC – BC
          = 88.2 -
     
          = 88.2
          =.

             = 102 m

Question-15

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Solution:
                                  
Let t be the time taken by the car to reach the foot of the tower.
In Δ ABD, cot 600 =
                    AB = AD cot 600............. (i)
In Δ ACD,
cot 300 =
      AC = AD cot 300 ...............(ii)
Since AC = AB + BC
        BC = AC – AB
Substituting (i) and (ii)
BC = AD cot 300 – AD cot 600
     = AD(cot 300 – cot 600)
Speed of the car while traveling the distance BC = …… (iii)
since speed =
Speed of the car when it travels from A to B = …….. (iv)
Since the car approaches the foot of the tower with the uniform speed.
Þ  
 =  
t(cot 300 - cot 600) = 6 cot 600
                         t(
√3 - ) = 6 × 
 
                         t  =
                   t =
                      =
                       2t = 6
                            t =
                              = 3 seconds
Therefore the time taken by the car to reach the foot of the tower from the point B is 3 seconds.

Question-16

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Solution:
                                                        
Let AB be the height of the tower. In the figure, C and D are the two points which are at a distance of 4 m and 9 m from the foot of the tower.
In Δ ABC,
tan θ =
        =
AB = 4 tan θ ………..(i)
In Δ ABD,
tan (90 - θ) =
          cot θ = [ tan(90 - θ ) = cot θ ]
         AB = 9 cot θ ………..(ii)
from (i) and (ii)
4 tan θ = 9 cot θ
4 tan θ =
tan2 θ =
tan θ =
substitute tan θ
  in (i)
AB = 4 tan θ
    = 4 × = 6 m

∴ The height of the tower = 6 m.
Hence proved.




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