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Examples II

Example

On the same side of a tower two objects are observed from the top of the tower. Their angles of depression are 45° and 60°. Find the distance between the objects, if the height of the tower is 300 m.

Solution

Let DC be the height of the tower and A, B be the positions of the two objects.  
Let the distance between the objects be x m.
In the 
Δ BCD, tan 60°= 
           =  

          BC =  m  (1)
In the 
Δ ACD,
       tan 45° = 

            1 = 

   x + BC = 300
  x +  = 300
     x = 300 -  = 126.79 m
Hence, the distance between the two objects is 126.79 m.

  
 
Example

Two vertical poles are fixed 60 m apart. The angle of elevation of the top of the first as seen from the top of the second, which is 160 m high, is 30°. Find the height of the first pole.

Solution

Let AP and CQ be the two vertical poles. Let the height of the first pole AP be h m.

Given: CQ = 160 m , PQ = 60 m.  

In the ΔABC, tan 30°= 
              
    (160-h) = 60
                h = 160 - 
 Height of the first pole = 125.36 m.

 
Example

A vertical stick 10 cm long, casts a shadow 6 cm long on the ground. Under similar conditions, a tower casts a shadow 10 m long. Find the height of the tower.

Solution

Let CB = 10 cm be the height of the stick and EF = h m be the height of the tower.
Given:

DE = 10 m, AB = 6 cm

In

Δ ABC,

tan 
θ 
In the 
Δ DEF,

tan 
θ = 
           h =  = 16.67 m
The required height of the tower is 16.67 m.
 
Example

The pilot of a helicopter, at an altitude of 1200 m finds that two ships are sailing toward it in the same direction. The angles of depression of the ships as observed from the helicopter are 60° and 45° respectively. Find the distance between the ships.

Solution

Let CD be the height at which the helicopter is flying CD = 1200 m.

Let A and B the position of the two ships. 

In the ΔBCD, tan 60° = 
 
 BC =  m     ...    (1)
In the 
ΔACD,
          tan 45° = 

                1 = 
         x + BC = 1200

Substituting for BC from (1)
    x +  = 1200
               x = 1200 -  = 507.16 m
Hence, the distance between the ships is 507.16 m.

 
Example

A tree 12 m high is broken by the wind in such a way that its top touches the ground and makes an angle of 60° with the ground. At what height from the bottom, the tree is broken by the wind?

Solution

Let ACB be the total length of the tree broken AC + CB = 12 m. 
Let BC (h) be the height from the bottom of the tree, broken by the wind.
CD = CA = 12 - h

In the triangle ABC,

sin 60° = 
 
 2h = 12-h
 h =  = 5.57 m
Hence, the tree is broken at a height of 5.57 from the bottom. 

 

 

Example

A surveyor wants to find out the height of a hill. He observes that the angle of elevation of the top of the hill at points A and B, 300 m apart, lying on a straight line, passing through the base of the top on the same side of the hill, are 30° and 45° respectively. What is the height of the hill?

Solution

Let the height of the hill be h m.
Given:

AB = 300 m 

In the ΔBCD,
       tan 45° = 

            1 = 
          BC = h
In the 
ΔACD,

tan 30°= 
        
      300 + BC =  h
    300+h=  h
h =  = 409.84 m
Hence, the height of the hill is 409.84 m.

 

 

Example

The string of a kite is 250 m long and it makes an angle of 60° with the horizontal. Find the height of the kite  above the ground (Assuming the string to be taut).

Solution

Let BC be the height of the kite in h m.
AC = 250 m is the length of the taut string. 

In the ΔACB,

sin 60° =

  
 h =  x 250 = 216.5 m
The required height of the kite from the ground is 216.5 m.

 





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