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Important Terms used in Heights and Distances

 

Trigonometry is used in calculating relation between Heights and Distances along with other attributes likes angle of elevation etc.

Horizontal ray: -
A ray parallel to the surface of the earth emerging from the eye of the observer is known as horizontal ray.

Ray of vision: -
The ray from the eye of the observer towards the object under observation is known as the ray of vision or ray of sight.

Angle of Elevation: -  
If the object under observation is above the horizontal ray passing through the point of observation, the measure of the angle formed by the horizontal ray and the ray of vision is known as angle of elevation. 

 Angle of depression: -  

If the object under observation is below the horizontal ray passing through the point of observation, the measure of the angle formed by the horizontal ray and the ray of vision is known as angle of depression.

 

 

Solved Examples

Example

From a cliff 150m above the shore line the angle of depression of a ship is 19 0 30 '.  Find the distance from the ship to a point on the shore directly below the observer.

Solution

Let OB be the cliff of height 150 m, A be the position of the ship the angle of depression of the ship is 19 0 30’,   ABC = 19 0 30’.
 ABC = 19 0 30' http://www.tutorvista.com/js/jsMath/wysiwyg_asciimath/mimetex/mimetex.cgi?%5Cdisplaystyle%5CRightarrow OAB = ABC = 19 0 30’

 

 

In Right angled triangle Δ AOB,

we want to find angle A = adjacent side , OB = opposite side = 150m

cot (19 0 30')   =  = 

OA = 150 cot (19 0 30') = 150 tan (90 -    19 0 30')

OA = 150 * tan ( 70 0 30')   = 150 ( 2.8239) = 423.59 m

 

 

Example

An airplane at an altitude of 500 m observes the angle of depression of opposite point on the two banks of a river to be 32 0 28’ and 40 0 12’ respectively.  Find in meters the width of the river.


 

Solution

A is the position of the airplane, DC is the river.
From figure, angle DAP = 32 0 18’, angle ADB = 32 0 18’ and angle CAQ = 40 0 12’, angle ACB = 40 0 12’ (alternate interior angles) 
In right angled ΔDBA:                               In right angled ΔABC:
Cot (32 0 18 ‘) = DB/AB                            Cot (40 0 12’) = BC / AB
DB = AB cot (32 0 18’)                              BC = 500 cot (40 0 12’)
DB = 500 tan (900 – 32 0 18’)                   BC = 500 tan (90 0 – 40 0 12’) 
= 500 tan (57 0 42 ‘)                                = 500 tan (49 0 48’)
= 500 (1.5818)                                         = 500(1.1833)
DB = 790.9                                               BC = 591.65
 
 Width of the river DC = DB +BC = 790.9 +591.65 = 1382.55m

 

 

Example

There is a small island in the middle of a 100 m wide river and a tall tree stands on the island. P and Q are points directly opposite to each other on two banks and in line with the tree. If the angles of elevation of the top of the tree from P and Q are respectively 30° and 45°, find the height of the tree.

Solution

Step 1: 
Let OA be the tree. Height of the tree is 'h' meter.
 

 
Step 2: 
Consider ΔPOA and ΔQOA.
tan 30°=   and tan 45° =   
 
Step 3:
Substitute the values

  • OP = √3h and OQ = h
  • OP + OQ = √3h + h
  • PQ = (√3 + 1)h
  • 100 = (√3 + 1)h
  • h =  m

Step 4: 
Rationalize the denominator and find h
            h =  m
               = 50(1.732 - 1) = 36.6 m
Step 5: 
Write the solution
Height of the tree = 36.6 m





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