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Examples III

Example

The angles of depression of two ships from the top of a light house are 45° and 30° towards east. If the ships are 200 m apart, find the height of the light house.

Solution

Let AB be the height of the light house be h metres and let C and D be the positions of the two ships.

Given: CD = 200 m.

In the 
ΔABC, tan 45° = 
 


  1 = 
  BC = h
In the 
ΔABD,
 tan 30° = 

   
   
   h = h + 200
   (-1)h = 200
   h =  = 273.22 m
Hence, the required height of the light house is 273.22 m.


 

Example

From the top of a building 60 m high, the angles of depression of the top and the bottom of a vertical lamp post are observed to be 30° and 60° respectively. Find, 
(a) The horizontal distance between the lamp post and the building.
(b) The difference between the heights of the building and the lamp post.

Solution

(a) Let BD be the height of the building = 60 m. Let AB be (x m) the distance between the lamp post and the building.

In the ΔABD,

tan 60° = 

   = 
 x =  = 34.64 m  .......(1)

(b) Let the height of the lamp post be h m.

In the 
ΔECD,
tan 30° = 

          
           x =  (60-h)
            = (60-h)       [x =  from (1)]
           20 = 60 - h
           h = 40.

The difference between the heights = 60 - 40  = 20 m.


 

Example

An aeroplane at an altitude of 200 m, observes the angles of depression of the opposite points on the two banks of a river to be 45° and 60°. Find the width of the river.

Solution

Let BD = 200 m be the height of the aeroplane and AC width of the river.
In the 
ΔABD, tan 60° = 

 
 AB =  m
In 
ΔDBC,
tan 45° = 

 1 = 
 BC = 200 m
The width of the river
= AC
= AB + BC

 + 200 = 315.47 m.


 

Example

From the top of a hill, the angles of a depression of the two consecutive kilometre stones due last, are found to be 30° and 45°. Find the height of the hill.   

Solution

Let AB = h km be the height of the hill. 
In the

ΔABC,
tan 45° = 

         1 =  
         h = BC
In the 
ΔABD,
tan 30° = 

      
    h = h + 1
     ( - 1)h = 1
 h =  = 1.37 km


 

Example

The angle of elevation of a tower at a point is 45°. After going 40 m towards the foot of the tower the angle of elevation becomes 60°. Find the height of the tower. 

Solution

Let AB (h) be the height of the tower, D and C be the positions of the person.
Angles of elevation are 45° and 60°.

In the ΔABC,
tan 60° = 

  
 BC = 
In the 
ΔABD,
tan 45° = 

 1 = 
⇒  + 40 = h
 h = 40
  h = 

      = 94.65 m

The height of the tower is 94.65 m.

 

 

Example

The horizontal distance between two trees of different heights is 60 m. The angle of depression of the first tree when seen from the top of second tree is 45°. If the height of the second tree is 80 m find the height of the first tree.

 
Solution

Let AB (h) be the height of the first tree in metres and CE = 80 m be the height  of the second tree such that the distance between the trees = BC = 60 m.
 In the 
ΔADE,

tan 45°= 

  1 = 
  60 = 80 - h
 h = 20 m
Hence, the height of the first tree is 20 m.


 

Example

The angle of elevation of the top of the tower, from two points P and Q at distances a and b respectively from the base and in the same straight line with it, are complementary. Prove that the height of the tower is  .

Solution

Let AB (h) be the height of the tower in metres.

Let the angle of elevation from P and Q be a  and 90°-α respectively.

BP = a,    BQ = b
In the 
ΔABP,

  tan α  
 tan α = 
 h = a tan α
In the ΔABQ
 tan (90°-
α) = 
  cot α = 
  h = b cot α
Multiplying (i) and (ii),
h= (a tan 
α) (b cot α) = ab
 h = 


 

Example

The angles of elevation of a cliff from a fixed point A is θ. After going up a distance of k metres towards the top of the cliff at an angle of φ it is found that the angle of elevation is α. Show that the height of the cliff, in metres is  k .

 
Solution

Let PQ (h) be the height of the cliff in h metres.

AB = k
In the 
ΔACB,  = cos φ
 AC = AB cos φ  
        = k cos φ
                              (i)
   = sin φ                                           
  BC = AB sin φ = k sin φ            (ii)
  PR = k sin φ
In the 
ΔAPQ,  = tan θ
   = tan θ
  AP = h cot θ                              (iii)
In the 
ΔBRQ,   = tan α         
  = tan α     (since PR = BC)
       (h - PR) cot α = AP - AC
  (h- k sin φ )cot α = h cot θ - k cos φ                            [From (i), (ii) and (iii)] 
     h[cot α - cot θ ] = k sin φ cot α - k cos α
     h[cot θ-cot α] = k [cot φ - sin θ cos α]

h = k .


 

Example

A round balloon of radius 'a' subtends an angle θ at the eye of the observer while the angle of elevation of its centre is φ . Prove that the height of the centre of the balloon, is a sin φ cosec()

AOB = θ                                                                                    
 BOC =                                     

Solution

Let the height of the centre of the balloon be
CD = h

COD = φ                                  
BC = AC = a
In the 
ΔOBC,

  OC = a cosec .......(1)
In  
ΔODC,
sin φ = 

 CD = OC sin φ

Substitute for OC from (i)

 h = a cosec sin φ

 





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